Research ArticleA New Approach in Pressure Transient AnalysisUsing Numerical Density Derivatives to Improve Diagnosis ofFlow Regimes and Estimation of Reservoir Properties forMultiple Phase Flow

Victor Torkiowei Biu and Shi-Yi Zheng

London South Bank University UK

Correspondence should be addressed to Victor Torkiowei Biu biutlsbuacuk

Received 5 April 2015 Accepted 7 June 2015

Academic Editor Alireza Bahadori

Copyright copy 2015 V T Biu and S-Y Zheng This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

This paper presents the numerical density derivative approach (another phase of numerical welltesting) in which each fluidrsquosdensities around the wellbore are measured and used to generate pressure equivalent for each phase using simplified pressure-density correlation as well as new statistical derivative methods to determine each fluid phasersquos permeabilities and the averageeffective permeability for the system with a new empirical model Also density related radial flow equations for each fluid phaseare derived and semilog specialised plot of density versus Horner time is used to estimate 119896 relative to each phase Results from 2examples of oil and gas condensate reservoirs show that the derivatives of the fluid phase pressure-densities equivalent display thesame wellbore and reservoir fingerprint as the conventional bottom-hole pressure BPR method It also indicates that the averageeffective 119896ave ranges between 43 and 57mD for scenarios (a) to (d) in Example 10 and 404mD for scenarios (a) to (b) in Example 20using the new fluid phase empiricalmodel for119870 estimationThis is within the 119896 value used in the simulationmodel and likewise thatestimated from the conventional BPRmethod Results also discovered that in all six scenarios investigated the heavier fluid such aswater and the weighted average pressure-density equivalent of all fluid gives exact effective 119896 as the conventional BPRmethodThisapproach provides an estimate of the possible fluid phase permeabilities and the of each phase contribution to flow at a givenpoint Hence at several dp1015840 stabilisation points the relative 119896 can be generated

1 Introduction

Several sets of well and reservoir models have been gen-erated with pressure derivatives with different boundaryconditions Likewise several type curves which account fordifferent combinations of wellbore reservoir characteristicsand boundary effects with associated flow regimes for com-putation of well and reservoir parameters have been used tosimplify well test interpretation This demonstrates that thelog-log plot of the pressure derivative is a powerful tool forreservoir model identification in pressure transient analysis

However in practice each current method of transientdata analysis has its own strengths and limitations withno single pressure and production data analysis methodcapable of handling all types of data and reservoir types with

clear reliable results [1] The log derivative and derivativetype curve which have remained reference flow regimersquosdiagnostic tools for over four decades are the only unifiedapproach for welltest interpretation and are applicable in awide range of situations

The derivative method which is the greatest break-through in welltest analysis was first introduced by Tiabin 1976 [2 3] and developed by French mathematicianDominique Bourdet in 1983 [4] It has remained the referencesolution for identifying flow regime boundary response anduse for diagnosing complex reservoir features till date Thisapproach has helped to reduce the uncertainties surroundingthe interpretation of welltest data because key regions ofradial flow and boundary features have been adequatelydiagnosed However due to the nonunique solution of

Hindawi Publishing CorporationJournal of Petroleum EngineeringVolume 2015 Article ID 214084 16 pageshttpdxdoiorg1011552015214084

2 Journal of Petroleum Engineering

the mathematical fluid flow equation mostly in heteroge-neous reservoirmost engineers in the industry are compelledto use analytical model and type curve solutions to matchcomplex model which is oftentimes not realistic Assump-tions made are ignored while pursuing a perfect match andresults obtained from this approach are often misleading[5]

This marked the beginning of numerical well test-ing in the industry by Zheng 2006 [5] although theapproach started from the early 1990s [6ndash10] Zheng mademore advances in 2006 providing more solutions to thenonunique problems mostly in heterogeneous reservoirsthrough numerical welltesting thereby promoting its appli-cation More papers have been published by researchers onthe subject thereby reflecting the advancement of numericalwelltesting and its application in solving various reservoirengineering practical problems

One of the main limitations of the pressure derivativeis that the measured pressure data must be constructedinto derivative data by means of numerical differentiationOftentimes derivative data from real field are very noisyand difficult to interpret resulting in various smoothingtechniques developed by researchers on this subject It is prac-tically believed that smoothing of pressure derivative dataoften alters the characteristics of the data Also it is difficult todistinguish between fluid and reservoir fingerprints in criticalsaturated reservoirs

Another limitation of these derivatives is diagnosing flowregimes in complex reservoir structures such as complexfaulted systems and high permeability streak with interbed-ded shales which is common in deep water turbidite sys-tems channel-levee lobe and channelized deposits Alsoin situations of multiphase flow around the wellbore thederivative data are always noisy and difficult to interpretresulting in the application of deconvolution and varioussmoothing techniques to obtain a perceived representativemodel which often might not be Additionally the ana-lytical solution for transient pressure analysis is limited tosingle phase flow which in real case is never the situationPresently there are few literatures or research on multiphasetransient pressure analysis However the combination ofthe new statistical approach [11] and the density deriva-tive approach serves as a support tool for better inter-pretation and estimation of reservoir properties in theseconditions

The diagnosis of flow which appears as distinctive pat-terns in the pressure derivative curve is a vital point inwelltest interpretations since each flow regime reflects thegeometry of the flow streamlines in the tested formationHence for each flow regime identified a set of well andorreservoir parameters can be estimated using the region ofthe transient data that exhibits the characteristic patternbehaviour [11] In the study the pressure derivative formula-tion from Horne (1995) [12] and the new statistical approachby Biu and Zheng (2015) [11] would be used throughout theanalysis

The mathematical formulation for pressure derivative byHorne (1995) [12] is given as

(120597119901

120597 ln 119905) = 119905 (

120597119901

120597119901)

119894

minus119860

119860 =ln (119905119894119905119894minus119896

) Δ119901119894+119895

ln (119905119894+119895

119905119894) ln (119905

119894+119895119905119894minus119896

)+119861

119861 =ln (119905119894+119895

119905119894minus119896

1199052

119894) Δ119901119894

ln (119905119894+119895

119905119894) ln (119905

119894119905119894minus119896

)

minusln (119905119894+119895

119905119894) Δ119901119894+119896

ln (119905119894119905119894minus119896

) ln (119905119894+119895

119905119894minus119896

)

(1)

Also the mathematical formulation for the new statisticalderivative approach by Biu and Zheng (2015) [11] is given asfollows

Model 1 Consider the following

StatDev (119894) = SQRT (pdd (119894) 119909Δdev (119894) 119909Δ2119875 (119894)) (2)

Model 2 (the exponential function) Consider the following

StatExp (119894)

= SQRT (EXP (SQRT (Δ2119875))) 119909pdd (119894) 119909Δ

2119875 (119894)

(3)

Model 3 Consider the following

StatdDev (119894)

= SQRT (pdd (119894) 119909Δdev (119894) 119909Δ2119875 (119894) 119909Δ

2119875 (119894))

(4)

Model 4 (the time function) Consider the following

StattDev (119894) = STDEV (Δ119905119905 (119894) Δ119905119905 (119894 + 1) StatDev (119894)

StatDev (119894 + 1)) (5)

Equations (1) to (5) are the derivative and statistical modelsused for flow regime diagnosis behaviours and estimation ofwellbore and reservoir properties using the log-log derivativeplot

2 Theoretical Concept ofthe Density Derivatives

The basic concepts involved in the derivation of fluid flowequation include

(i) conservation of mass equation(ii) transport rate equation (eg Darcyrsquos law)(iii) equation of state

Consider flow in a cylindrical coordinate with flow butwith flow in angular and 119911-directions neglected as shown inFigure 1 the equations are given as follows

Mass rate inminusMass rate out = Mass rate storage (6)

Journal of Petroleum Engineering 3

r

r r + Δr

r + Δr

Figure 1 Schematics of basic fluid flows concept

Equation (6) represents the conservation of mass Since thefluid is moving the equation

119902 = minus119896

120583119860

120597119901

120597119903(7)

is applied By conservingmass in an elemental control volumeas shown in Figure 1 and applying transport rate equation thefollowing equation is obtained

minus[2120587119903ℎ119896

120583120588120597119901

120597119903]

119903

= minus[2120587119903119896ℎ

120583120588120597119901

120597119903]

119903+Δ119903

+ 2120587119903Δ119903ℎ120597

120597119905(120588120601)

(8)

Expand the equation using Taylor series

1119903

120597

120597119903[119903119896120588

120583

120597119901

120597119903] =

120597

120597119905[120588120601] (9)

Equations (6) to (9) apply to both liquid and gas Equation(9) is known as the general diffusivity equation and for eachfluid the density or pressure term in (9) can be replaced bythe correct expression in terms of density or pressure

For small or constant compressibility liquid

120588 = 120588119900119890119862[119901minus119901

119900] (10)

Substituting for pressure in the equation the diffusivityequation in terms of density is given as

1205972120588

1205971199032+1119903

120597120588

120597119903=

120601120583 [119888 + 119888119903]

119896

120597120588

120597119905 (11)

1205972120588

1205971199032+1119903

120597120588

120597119903=

120601120583119888119905

119896

120597120588

120597119905 (12)

Equation (12) is known as the density diffusivity equationwhich can also be rewritten in the form of pressure

Over the decades the transient test analysis has appliedthe general diffusivity equation in pressure term to generateseveral nonunique solutions using several pressure-rate data

Invariably as in pressure term the density term alsoimplored

1205972120588

120597119903+1119903

120597120588

120597119903=1119903

120597

120597119903(119903

120597120588

120597119903) = 0 (13)

For inner boundary condition

[119903120597120588

120597119903]

119903119908

=119902120583

2120587119896ℎ= Constant (14)

For outer boundary condition

120588 = 120588119890

at 119903 = 119903119890 (15)

Presently there are permanent downhole gauges (PDG) withdensity measurement tool along with pressure and temper-ature installed during flowing and shut-in testing condi-tions but the data are not interpreted or used for reservoirmonitoring For simplification and application of the densityderivative in existing welltest software the density-pressureequivalent equation is formulated

3 Software Suitability (Pressure Equivalent)

To apply the numerical density approach in existing softwarethe pressure equivalent of the fluid density changes at thewellbore is generated from the relationship below

Using the isothermal compressibility coefficient 119862 interms of density

119862 =1120588

120597120588

120597119901 (16)

31 For Small or Constant Compressibility Fluid Such as Oiland Water Consider the following

minus119862int

119901

119901119900

119889119901 = int

120588

1205880

120597120588

120588 (17)

Integrating

119890119862[1198750minus119875] =

120588

1205880

119862 [1198750 minus119875] = ln [120588

1205880]

(18)

119875 = 1198750 minusln [1205881205880]

119862 (19)

or applying the 119890119909 expansion series

119890119909= 1+119909+

1199092

2+

1199093

3+ sdot sdot sdot +

119909119899

119899 (20)

Because the term 119862[1198750 minus 119875] is very small the 119890119909 term can be

approximated as

119890119909= 1+119909 (21)

Therefore (14) can be rewritten as

120588 = 1205880 [1minus119862 (1198750 minus119875)]

120588

1205880= 1minus119862 [1198750 minus119875]

there4 119875 = 1198750 minus[1205881205880 + 1

119862]

(22)

4 Journal of Petroleum Engineering

For small compressibility fluid such as oil and water either(19) or (22) is used to generate the pressure equivalent fromwell fluid density obtained from reservoir simulation or PDGtool This pressure is then analyzed in any available well testsoftwares

32 For Compressible Fluid in Isothermal Conditions Con-sider the following

119862119892= minus

1V[120597V120597119901

]

119879

(23)

For real gas equation of state

V =119899119877119879119911

119875 (24)

Differentiating with respect to pressure at constant tempera-ture

(120597V120597119901

)

119879

= 119899119877119879[1119901

(120597119911

120597119901)minus

119911

1199012 ] (25)

Substituting (23) into (25)

119862119892=

1119901

minus1119911[119889119911

119889119901] (26)

In terms of density

119862119892=

1119901

minus1120588

[120597120588

120597119901] (27)

This equation is applicable for real gas conditionFor compressible fluid

119862119892=

1119901

minus1120588

[120597120588

120597119901]

1120588

[120597120588

120597119901] =

1119901

minus119862119892

int

120588

1205880

120597120588

120588= int

119901

1199010

[120597119901

119901minus 120597119901119862

119892]

ln [120588

1205880] = ln [

119901

1199010]minus [119901minus1199010] 119862119892

(28)

Applying the power series for ln119901

ln [119901] = [119901 minus 1] minus[119901 minus 1]2

2+ sdot sdot sdot +

[minus1]119899 [119901 minus 1]119899

119899

+ sdot sdot sdot 0 lt 119901 le 2(29)

Limit ln119909 to the 1st term only

[120588

1205880]minus 1 = [

119901

119901119900

]minus 1minus [119901minus1199010] 119862119892

[120588

1205880] = [

119901

119901119900

]minus [119901minus1199010] 119862119892

(30)

[120588

1205880] =

119901 minus 119901120588119900119888119892minus 119901119900

2119888119892

119901119900

119901119900120588

120588119900

minus119901119900

2119888119892= 119901 [1minus119901

119900119888119892]

(31)

119875 =119875119900120588 minus 119875119900

2120588119900119888119892

120588119900(1 minus 119875

119900119888119892) (32)

For compressible fluids such as gas (32) is used to generatethe pressure equivalent from the fluid density obtained fromreservoir simulation or PDG tool at the well and then thepressure is analyzed in any available well test softwares

Also outside the available welltest software the pressurederivative can be generated from the pressure equivalentobtained from the oil gas and water densities at the well byapplying (1) formulated by Horne (1995) [12] or (2) to (4) byBiu and Zheng (2015) [11]

4 Density Weighted Average (DWA)

While (19) or (22) and (32) give the pressure equivalentfor independent fluid phases such as gas oil and waterthe weighted average method is used to obtain the densityequivalent for a two- or three-phase combinationThe equiv-alent pressure derived from the density for all three fluidcomponents such as gas oil and water is given as

119875119894=

120588119892119901119892+ 120588119900119901119900+ 120588119908119901119908

120588119892+ 120588119900+ 120588119908

(33)

Equation (33) comprises all the fluid phases in the system andas such will be comparable to the conventional bottom-holepressure measurement during the derivative analysis

41 Empirical Model Correlation between Fluid Phases119870 Anempirical model integrating the fluid phasersquos permeabilitiesfor a given system is formulated to determine the averagereservoir permeability The mathematical model is given as

119896ave = 4radic1198961199001198961199081198962119892 (34)

where 119896119900= oil phase permeability 119896

119892= gas phase permeabil-

ity and 119896119908= water phase permeability

With the estimation of the phasersquos permeabilities it istherefore possible to estimate the possible relative permeabil-ity to each phase and the percentage contribution to flow byeach phase at one point analysis hence at several points therelative 119896 can be generated

To illustrate applicability of this approach 6 scenarios inconventional oil reservoir and gas condensate reservoir are

Journal of Petroleum Engineering 5

Local grid refinement

Pressure (psia) Pressure (barsa)

PROD1

PROD1

35591 35630 35669 35708 35747 3093430899 30970 31006 31041

Figure 2 Simulation model for gas cap + oil + water reservoir and gas condensate + water showing local grid refinement around the well

Table 1 Reservoir and fluid data for Example 1

Parameters Design valueEclipse model Black oilModel dimension 10 times 5 times 5Length by width ft by ft 500 times 400Thickness ℎ ft 250Permeability 119870119909 by 119870119910 mD 500 by 500Porosity 20Well diameter ft 060Initial water saturation 119878119908

119894 22

Permeability 119870 mD 50Gas oil contact (GOC) ft 8820Oil water contact (OWC) ft 90000Initial pressure 119875

119894 psia 40000

Formation temperature 119879 oF 2000

investigated using numerical model built with a commercialsimulator Local grid refinement (LGR) is imposed aroundthe well to capture sharp change in fluid densities as shownin Figure 2

The simulation software keywords LBPR LDENOLDENW LDENG AND WBHP were outputs to obtain thedensity and pressure change around the well and as far as theperturbation could extend

Example 1 Table 1 presents a summary of the well andreservoir synthetic data used for the buildup and drawdownsimulated scenarios with additional information given belowIt is required to generate the pressure equivalent and deriva-tive for each phase compare their diagnostic signaturesand also determine the phases permeabilities and averagereservoir permeability

Assumption (i) Oil reservoir + gas cap is completed with onewell

(ii) LGR is imposed around the well and far across toaccount for pressure and density changes

Gas oil and water densities around the local grid refine-ment (wellbore) andWBHP were output using the simulatorkeywords The following scenarios were evaluated

(a) Flowing+ buildup sequence well perforated ℎ119901= 30 ft

between oil and water layer Net sand thickness ℎ =250 ft

(b) Flowing+buildup sequence well perforated ℎ119901= 30 ft

inside the oil layer Net sand thickness ℎ = 250 ft(c) Flowing+ buildup sequence well perforated ℎ

119901= 30 ft

between gas and oil layer Net sand thickness ℎ =250 ft

(d) Falloff test flowing + buildup sequence well perfo-rated ℎ

119901= 30 ft inside the oil layer Net sand thickness

ℎ = 250 ftThe schematic of the 4 scenarios is depicted in Figures 3and 4 Figure 5 shows the production shut-in and injectionsequence for the 4 scenarios in Example 1

In scenario (a) the well is completed between the oil andwater layer to mimic multiphase conditions at the wellboreand some distance away from the well and also see the effecton pressure distribution fluid densities changes around thewellbore and estimate fluid phase permeabilities 119896 using thespecialised plot and 119896ave from the empirical model for three-phase conditions

The derivative in Figure 6 shows a good radial flow butdrop in derivative at late time due to constant pressuresupport (likely aquifer support) A continuous drop is seenfrom 10 hrs in the model parameters The derivatives forall output parameters display the same well and reservoirsignatures (3 flow periods early to late time response)but with different 119889119901

1015840 stabilisation The well bottom-holepressure (BPRrarrBHP) response shows good overlay withpressure equivalent of density weighted average (PDENDWArarr PDENA) and pressure equivalent of water density (PDEN-WAT rarr PDENW) while pressure equivalent of gas den-sity (PDENGAS rarr PDENG) and pressure equivalent ofoil density (PDENOIL rarr PDENO) differ completely ThePDENDWA gives a better fingerprint that is less noisy

A permeability value of 508mD is estimated from thebottom-hole pressure BPR where 119896 = 1627119902119861120583119898ℎ and119898 is

6 Journal of Petroleum Engineering

Gas

Oil

Water

Case 1 production test

Net sand thickness h = 250ftwater layerWell perforated hp = 30ft between oil andFlowing + buildup sequence

(a)

Gas

Oil

Water

Case 2 production test Flowing + buildup sequence

Net sand thickness h = 250ftWell perforated hp = 30ft inside the oil layer

(b)

Gas

Oil

Water

Case 3 production test Flowing + buildup sequence

Net sand thickness h = 250ftoil layerWell perforated hp = 30ft between gas and

(c)

Gas

Oil

Water

Case 4 falloff test Flowing + buildup sequence

Net sand thickness h = 250ftwater layerWell perforated hp = 30ft between oil and

(d)

Figure 3 Schematics of well perforation interval and sand thickness for oil + gas cap + water reservoir

obtained from the specialised plot This is an approximate ofthe input value in the simulation model Also the simulatoroutputs PDENDWA and PDENWAT in the simulation givethe same 119896 value while PDENGAS and PDENOIL differgiving 158 and 4037mD respectively At ℎ = 50 f t the best119896 estimate is obtained depicting ℎ = 50 f t as the thicknesscontributing to flowAt ℎ gt 50ft 119896 drops below the 119896 imputedin the model

Using (34) 119896ave = 473mD is obtained which isapproximately close to that of BPR hence a good estimateof the fluid phase permeabilities A summary of the result isshown in Tables 2 and 5

In scenario (b) thewell is completedwithin the oil sectionto capture the pressure and fluid densities changes around thewell The derivative in Figure 7 shows a good radial flow butwith noisy numerical artefact It is likely that the boundaryresponse is masked by numerical artefact Also a continuousdrop is seen after 10 hrs in the derivative

Using the estimated fluid phase permeabilities 119896ave from(34) is 472mD as shown in Table 3 which is the same asscenario (a) This is in line with the uniform 119896 used in thesimulation model Also the BPR gives a permeability value of500mD which is the same for PDENDWA and PDENWATbut differs with PDENGAS and PDENOIL that give 121 and

Journal of Petroleum Engineering 7

Water

Case 1 production test

Gas + condensate

Flowing + buildup sequence

condensate and water layerNet sand thickness h = 45m

Well perforated hp = 90m between gas

(a)

Water

Case 2 production test

Gas + condensate

Net sand thickness h = 45m

Flowing + buildup sequence

condensate layerWell perforated hp = 90m inside gas

(b)

Figure 4 Schematics of well perforation interval and sand thickness for gas condensate + water reservoir

00

5000

0 50 100 150 200 250 300 350

Prod

uctio

nin

ject

ion

rate

bop

d

Duration

Test design sequence

Case dCases a to c

minus25000

minus20000

minus15000

minus10000

minus5000

Figure 5 Production and injection test sequence for scenarios (a)to (d) (buildup and falloff)

Table 2 119870 estimates for new approach versus conventionalapproach for scenario (a)

Parameters Numerical density119896 (mD)

Equivalentℎ (ft)

BHP 508PDENA 508PDENG 158 50PDENO 4037PDENW 508

4844mD respectively At ℎ = 150ft (60 of sand thicknessand 83 of oil thickness) the estimated 119896 is still within range

00

00

01

10

100

0000 0001 0010 0100 1000 10000 100000Log time (h)

BHPPENDAPDENO

PDENGPDENW

508mD

4037mD

158mD

k =1627qB120583

mh

Case a derivatives for conventional and numerical density models

logdp998400

Figure 6 Derivative and 119870 estimation for scenario (a)

(119896 = 489mD)This indicates ℎ = 50ft or 83 of oil thicknesscontributing to flow

To test this approach in the gas column the well wascompleted in between the gas and oil layer which is consid-ered as scenario (c) and scenario (d) with the well completedwithin the oil layer but with water injection after flowingand shut-in sequence In both scenarios the multiphasefluid distribution is triggered at the wellbore in order tocapture the density changes for each phase and calculate fluidphase permeabilities First the well fluid densities equivalent

8 Journal of Petroleum Engineering

Case b derivatives for conventional and numerical density models

00

01

10

100

0000 0001 0010 0100 1000 10000 100000Log time (h)

BHPPENDAPDENO

PDENGPDENW

4842mDk =1627qB120583

mh

120mD

489mD

logdp998400

Figure 7 Derivative and 119870 estimation for scenario (b)

0001

0010

0100

1000

10000

0000 0001 0010 0100 1000 10000 100000Log time (h)

BHPPENDAPDENO

PDENGPDENW

412mD

209mD

Case c derivatives for conventional and numerical density models

4842mD

k =1627qB120583

mh

logdp998400

Figure 8 Derivative and 119870 estimation for scenario (c)

Table 3 119870 estimates for new approach versus conventionalapproach for scenario (b)

Parameters Numerical density119896 (mD)

Equivalentℎ (ft)

BHP 500PDENA 500PDENG 121 50PDENO 4842PDENW 500

pressures and pressures at bottom-hole for flowing andbuildup test are generated

The derivative for both scenarios declines after 3 hoursprobably due to fluid redistribution but is noisy (numericalartefact) in scenario (c) as shown in Figure 8 A good radial

Table 4 119870 estimates for new approach versus conventionalapproach for scenario (d)

Parameters Numerical Density119896 (mD) Equivalent ℎ (ft)

Scenario (d)

BHP 500 100PDENA 500PDENG 175 50PDENO 490 250PDENW 500 100

stabilisation for both the conventional method BPR and thedensity outputs PDENDWA PDENWAT PDENGAS andPDENOIL is observed Likewise as obtained in scenarios (a)and (b) a good 119870ave value of 57 and 52mD from (34) isobtained for scenarios (c) and (d) respectivelyThis is slightlyhigher than 119896 = 50mD imputed in the simulation model

For each of the fluid phase permeabilities estimates 119896value of 50mD is obtained using the conventional BPR forscenario (c) which is in line with the uniform 119896 in thesimulation model This is the same for the PDENDWA andPDENWAT Also at ℎ = 150ft (60 of sand thickness and83 of oil thickness) the estimated 119896 is still within range(119896 = 412mD)This also indicates that ℎ = 150ft or 83 of oilthickness is contributing to flowThe 119896 values for PDENGASandPDENOIL are 209 and 4842mD respectively Summaryof the result is shown in Table 4

However in scenario (d) the permeability value of500mD using the conventional BPR which is the samefor PDENDWA and PDENWAT is only achieved if ℎ =

100ft (40 of sand thickness and 56 of oil thickness)PDENGAS and PDENOIL differ giving 350mD at ℎ =

50ft and 1960mD at ℎ = 250ft respectively This indicatesthe impact of water injection on densities and pressureschanges around the well and consequently its impact on sandthickness contributing to flow The drop in derivative curvein Figure 9 depicts the impact of the injected water

In summary for scenarios (a) to (d) the 119896 value of500mD is achieved if the thickness contributing to flowranges from 50 to 150 ft Generally results indicate that in all4 scenarios investigated the heavier fluid such as water andthe weighted average pressure-density equivalent of all fluidgive exact effective 119896 as the BPR likewise the density outputsPDENDWA and PDENWAT Also results from Table 5 showthat the empirical model from (34) with all fluid phasepermeabilities gives an average effective 119896ave of 47ndash57mDwhich is within that used in the simulationmodel and also thesame as the estimated conventional approach This approachprovides an estimate of the possible fluid phase permeabilitiesand theof each phase contribution to flow hence at severalpoints the relative 119896 can be generated as shown in Table 5

Example 2 To capture the influence of highly compressiblefluid on estimated fluid phase permeabilities an example ongas condensate reservoir (volatile system) was tested Table 6presents a summary of the well and reservoir synthetic dataused for the buildup and drawdown simulated scenarios with

Journal of Petroleum Engineering 9

Table 5 Comparison of 119870 estimates between conventional and numerical density parameters

Scenario Numerical density method Calculated119870 (mD)

Conventional119896 (mD)

Simulationmodel 119870 (mD) Relative 119870

phasecontribution to flowPhases 119870 (mD)

(a)Gas 158 003 30Oil 4037 473 508 500 086 860

Water 508 011 110

(b)Gas 12 003 20Oil 4844 432 500 500 113 890

Water 50 012 90

(c)Gas 209 004 40Oil 4842 570 500 500 086 87

Water 50 009 90

(d)Gas 175 003 30Oil 490 523 500 500 095 880

Water 50 010 90

1000

Case d derivatives for conventional and numerical density models

0010

0100

10000

BHPPENDAPDENO

PDENGPDENW

0100 1000 10000 100000Log time (h)

k =1627qB120583

mh

209mD

500mD

4842mD

logdp998400

Figure 9 Derivative and 119870 estimation for scenario (d)

additional information given below It is required to generatethe pressure equivalent and derivative for each fluid phasecompare their diagnostic signatures and also determine thephases permeabilities and average reservoir permeability

Assumption (i) Condensate reservoir is completed with onewell

(ii) LGR is imposed around the well and far across toaccount for pressure and density changes

Gas condensate andwater densities around the local gridrefinement (wellbore) andWBHPwere output using the sim-ulatorrsquos keywords The following scenarios were evaluated

(a) Flowing + buildup sequence well perforated ℎ119901

=

30ft between gas condensate and water layer Netsand thickness ℎ = 150ft

Table 6 Summary of reservoir simulations data

Parameters Design valueEclipse model Black oilModel dimension 9 times 3 times 3Length by width ft by ft 1312 times 984Thickness ℎ ft 150Permeability 119870119909 by 119870119910 mD 4000 by 3000Porosity 30Well diameter ft 115Initial water saturation 119878119908

119894 60

Permeability 119870 mD 400Gas oil contact (GOC) ft 6890Oil water contact (OWC) ft 6890Initial pressure 119875

119894 psi 4495

Formation temperature 119879 ∘C 120 2000

(b) Flowing + buildup sequence well perforated ℎ119901

=

30ft inside gas condensate layer Net sand thicknessℎ = 150ft

Figure 10 shows the production and shut-in sequence for 2scenarios in Example 2

As in Example 1 the derivative in scenario (a) ofExample 2 also shows good radial flowwith no late time effectas seen in Figure 11 A negative half-slope fingerprint is seen at03 to 30 hrs of buildup and a good stabilisation from 40 hrsin the model parameters

The derivatives for all output parameters display the samewell and reservoir fingerprint (2 flow periods early to middletime response) but with different 1198891199011015840 stabilisation

Permeability value of 3707mD was obtained from theconventionalmethod BPRwhere 119896 = 1637119902119879119898ℎ is the samefor the PDENDWA and PDENWAT and 119898 is obtained fromthe specialised plot This is in line with the uniform 119896 (119896119909 =

400mD and 119896119910 = 300mD) imputed in the simulationmodel

10 Journal of Petroleum Engineering

00

5000000

10000000

15000000

20000000

25000000

0 40 80 120 160

Prod

uctio

n ra

te (s

cfd

)

Duration

Test design sequence

Cases a to b

Figure 10 Production test sequence (buildup) for Example 2

10000

Case a derivatives for conventional and numerical density models

00100

01000

BHPPENDAPDENO

PDENGPDENW

1422mD

3707mD

35747mD

0000 0001 0010 0100 1000 10000

100000

100000Log time (h)

k =1627qB120583

mh

logdp998400

Figure 11 Derivative and 119870 estimation for scenario (a)

However PDENGAS and PDENOIL differ giving 350mDand 35747mD respectively if ℎ = 148ft

Also the derivative in Figure 12 shows good radial flowand a negative half-slope fingerprint is seen at 05 to 20 hrsof buildup and a good stabilisation from 30 hrs in the modelparameters After 3 hrs of shut-in somenoisy data (numericalartefact) is seen in all the fluid phase derivatives but still goodenough to identify 119889119901

1015840 stabilisationFor each of the fluid phase permeabilities 119896 value of

3409mD at ℎ = 148ft is obtained using BPR which is thesame for the PDENDWA and PDENWAT PDENGAS andPDENOIL differ giving 1490 and 35141mD at ℎ = 148ftrespectively

For scenarios (a) and (b) a good 119896ave value of 4050and 4030mD is obtained respectively from the empiricalmodel (34) integrating all fluid phase permeabilities whichis within that used in the simulationmodel and the estimated

0001000

0010000

0100000

1000000

10000000

0000 0001 0010 0100 1000 10000 100000

Case b derivatives for conventional and numerical density models

BHPPENDAPDENO

PDENGPDENW

Log time (h)

k =1627qB120583

mh

1490mD

3410mD

35140mD

logdp998400

Figure 12 Derivative and119870 estimation for scenario (b)

Table 7 Summary of reservoir modelling properties imputed ineclipse model for case study 2

Parameters Numerical density119896 (mD)

Equivalentℎ (ft)

Scenario(a)

BHP 371PDENA 371PDENG 142 148PDENO 3575PDENW 371

Scenario(b)

BHP 341PDENA 341PDENG 149 148PDENO 3514PDENW 341

conventional approach From the 2 scenarios investigated ithas been demonstrated that the heavier fluid such as waterand the weighted average pressure-density equivalent of allfluid give exact effective 119896 as the conventional method BPRA summary of the result is shown in Table 7

This approach estimates the fluid phase permeabilitiesand the of each phase contribution to flow at a given pointhence at several points the relative 119896 can be generated asshown in Table 8

5 Density Related Radial Flow EquationDerivation for Each Fluid Phase

Radial diffusivity equation is given as

1119903

120597120588

120597119903[119903

120597120588

120597119903] =

120601120583119888

119896

120597120588

120597119905 (35)

Journal of Petroleum Engineering 11

Table 8 Comparison of 119870 estimates between conventional and numerical density parameters

Scenario Numerical density method Calculated119870 (mD)

Conventional119896 (mD)

Simulationmodel 119870 (mD) Relative 119870

phasecontribution to flowPhases 119870 (mD)

(a)Gas 1422 004 30Oil 35747 4046 3707 400 089 870

Water 3707 009 90

(b)Gas 149 004 40Oil 35141 4038 3409 400 089 880

Water 3409 008 90

For oil

1119903

120597120588

120597119903[119903

120597120588119900

120597119903] =

120601120583119900119888

119896119900

120597120588119900

120597119905 (36a)

For water phase

1119903

120597120588

120597119903[119903

120597120588119908

120597119903] =

120601120583119908119888

119896119908

120597120588119908

120597119905 (36b)

For gas phase

1119903

120597120588

120597119903[119903

120597120588119892

120597119903] =

120601120583119892119888

119896119892

120597120588119892

120597119905 (36c)

51 From Equation for Slight and Small Compressibility Suchas Water and Oil From (22)

119875 = 119875119900minus

120588120588119900+ 1

119888119900

(37)

Differentiating with respect to 120588

120597119875

120597120588= minus

1119888119900120588119900

120597119875 = minus120597120588

119888119900120588119900

(38)

From Darcy flow equation

119902 = minus2120587119896ℎ120583

119903120597120588

120597119903 (39)

Substitute for 120597119875

119902 =2120587119896ℎ120583

119903

119888119900120588119900

120597120588

120597119903 (40)

To derive the analytical density transient equation for eachphase the following assumptionconditions are applicable

For oil phase from (35)

1119903

120597120588

120597119903[119903

120597120588

120597119903] =

120601120583119888

119896

120597120588

120597119905 (41)

Initial condition is as follows

120588 (119903 119905 = 0) = 120588119894 (42a)

BC at the wellbore is as follows

lim119903rarr 0

2120587119896ℎ120583

119903

119888119900120588119900

120597120588

120597119903= 119876 (42b)

BC at infirmity is as follows

lim119903rarrinfin

120588 (119903 119905) = 120588119894 (42c)

Using the Boltzmann transformation consider the followingAssuming 120578 = 120601120583119888119903

2119896119905

120597120588

120597119903=

120597120588

120597120578

120597120578

120597119903=2120601120583119888119903119896119905

120597120588

120597120578=

1206011205831198881199032

119896119905

2119903

120597120588

120597120578=2120578119903

120597120588

120597120578 (43)

Therefore1119903

120597

120597119903[119903

120597120588

120597119903] =

21205781199032

120597

120597120578[2120578

120597120588

120597120578] =

41205781199032

120597

120597120578[120578

120597120588

120597120578] (44)

Differentiating with respect to 119903 is equal to differentiatingwith respect to 120578 multiply by 2120578119903

From (35) the LHS is resolved as follows

120597120588

120597119905=

120597120588

120597120578

120597120578

120597119905=

120601120583119888119903

1198961199052120597120588

120597120578= minus

120578

119905

120597120588

120597120578 (45)

120601120583119888

119896

120597120588

120597119905=

minus120601120583119888

119896

120578

119905

120597120588

120597119905=

minus1206011205831198881199032

119896119905

120578

1199032120597120588

120597120578=

minus1205782

1199032120597120588

120597120578 (46)

Equating (44) and (46) we have

120597

120597120578[120578

120597120588

120597120578] =

minus120578

4120597120588

120597120578 (47)

This is the simplified ordinary differential equation of 120588 as afunction of 120578

Apply boundary conditions as followsInitial condition is as follows

lim119903rarrinfin

120588 (120578) = 120588119894 (48a)

Substituting (43) into (40)

lim119903rarr 0

2120587119896ℎ120583

119903

119888119900120588119900

120597120588

120597119903= 119876 (48b)

lim119903rarr 0

[120578120597120588

120597120578] =

120601120583119888119900120588119900

4120587119896ℎ (49)

Assume that119898 = 120578(120597120588120597120578)

12 Journal of Petroleum Engineering

From (47)

120597119898

120597120578=

minus119898

4 (50)

Integrating from 120578 = 0 to 120578

997904rArr int

119898(120578)

119898(0)

120597119898

120597120578= minusint

120578

0

120597120578

4

997904rArr ln[119898 (120578)

119898 (0)] =

minus120578

4

997904rArr 119898(120578) = 119898 (0) 119890minus1205784

(51)

Recall boundary condition

lim119903rarr 0

[120578120597120588

120597120578] =

119876120583119888119900120588119900

4120587119896ℎ

119898 (0) =119876120583119888119900120588119900

4120587119896ℎ

119898 (120578) =119876120583119888119900120588119900

4120587119896ℎ119890minus1205784

(52)

Recall that119898 = 120578(120597120588120597120578)Therefore

120597120588 (120578)

120597120578=

120583119876119888119900120588119900

4120587119896ℎ119890minus1205784

120578 (53)

Integrating from 120578 = infin where 120588 = 120588119894

int

120588(120578)

120588119894

120597120588 = minusint

120578

infin

120583119876119888119900120588119900

4120587119896ℎ119890minus1205784

120578120597120578

120588 (120578) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎminusint

infin

120578

119890minus1205784

120578120597120578

(54)

where

120578 =120601120583119888119903

2

119896119905 (55)

Assuming 119906 = 1205784 then 120597120578120578 = 120597119906119906 and 119906 = 12060112058311988811990324119896119905

Hence

120588 = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎminusint

infin

12060112058311988811990324119896119905

119890minus1205784

120578120597120578 (56)

where

minus119864119894 (minus119909) = int

infin

119909

119890minus119906

119906120597119906 (57)

Known as the exponential integral function defined by [13]

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎ119864119894 (minus119909) (58)

where 119909 = 12060112058311988811990324119896119905 and

minus119864119894 (minus119909) = int

infin

119909

119890minus119906

119906120597119906 (59)

At large times 119909 will be small applying Taylorrsquos series andintegrating

minus119864119894 (minus119909) = ln (120574)

ln (120574) = ln (1781) = 05772(60)

where 120574 is known as Eulerrsquos numberFor 14119909 gt 25 consider the followingThen

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎ[ln119909+ ln 120574]

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎln[

1206011205831198881199032120574

4119896119905]

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎln [

22461198961199051206011205831198881199032

]

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎ[ln [

22461198961199051206011205831198881199032

]+ 080907]

(61)

Plotting 120588119908(119905) versus ln(119905) will yield a straight line at longer

time and the slope of the line is given as120597120588119908

120597 ln 119905= 119898oil =

120583119876119888119900120588119900

4120587119896ℎ (62)

Therefore

119896ℎ =120583119876119888119900120588119900

4120587119898oil (63)

52 For Water Phase The radial density equation is given as

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎ[ln [

22461198961199051206011205831198881199032

]+ 080907] (64)

Plotting 120588119908(119905) versus ln(119905) will yield a straight line at longer

time and the slope of the line is given as120597120588119908

120597 ln 119905= 119898water =

120583119876119888119900120588119900

4120587119896ℎ (65)

where

119896ℎ =120583119876119888119900120588119900

4120587119898water (66)

53 For Gas Phase From equation for compressible fluidsuch as gas consider the following

From (32)

119875 =119875119900120588 minus 119875119900

2120588119900119888119892

120588119900(1 minus 119875

119900119888119892)

120597119875

120597120588=

119875119900

120588119900(1 minus 119875

119900119888119892)

120597119875 =119875119900

120588119900(1 minus 119875

119900119888119892)120597120588

(67)

Journal of Petroleum Engineering 13

From the fundamental

119899 =119875119881

119911119877119879 (68)

At standard condition

119875119881

119911119879=

119875sc119881sc119879sc

119875 (5615119902)119911119879

=119875sc119876sc119879sc

119902 =119875sc119879sc

(119911119879

119875)

119876sc5615

119902 =119875sc119879sc

119876sc5615

(119911119879

119875) =

2120587119896ℎ120583

119903120597119901

120597119903

119902 =119875sc119879sc

119876sc5615

(119911119879) =2120587119896ℎ120583

119903119875120597119901

120597119903

(69)

where

120597119875 =119875119900

120588119900(1 minus 119875

119900119888119892)120597120588

120572 =119875119900

120588119900(1 minus 119875

119900119888119892)

120597119875 = 120572120597120588

119875 = 120588119877119911119879

(70)

Substitute into the equation where

119898(120588119894) minus119898 (120588

119908119891) =

2120583

int

120588119894

120588119908119891

120588120597120588

119898 (120588119894) minus119898 (120588

119908119891) =

120588119894

2minus 120588119908119891

2

120583

120588 = radic120588119894

2+ 120588119908119891

2

2

(71)

From the diffusivity equation for gas

1119903

120597120588

120597119903[119903

120597120588119892

120597119903] =

120601120583119892119888

119896119892

120597120588119892

120597119905 (72)

Initial condition is as follows

120588 (119903 119905 = 0) = 120588119894 (73)

BC at the wellbore is as follows

lim119903rarr 0

4120587119896ℎ120572119877120583

(5615119879sc

119875sc) 119903120588

120597120588

120597119903= 119876 (74)

BC at infirmity is as follows

lim119903rarrinfin

120588 (119903 119905) = 120588119894 (75)

where119877119875sc and119879sc are known as temperature pressure andgas constant at standard condition

Therefore integrating from 120578 = infin where 120588 = 120588119894

int

120588(120578)

120588119894

120588120597120588 = minusint

120578

infin

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)119890minus1205784

120578120597120578

1205882(120578) = 120588

119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

minusint

infin

120578

119890minus1205784

120578120597120578

(76)

where

120578 =120601120583119888119903

2

119896119905 (77)

Assume that 119906 = (1205784) (120597120578120578) = 120597119906119906 and 119906 = 12060112058311988811990324119896119905

Hence

1205882= 120588119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)minusint

infin

12060112058311988811990324119896119905

119890minus1205784

120578 (78)

where minus119864119894(minus119909) = intinfin

119909(119890minus119906

119906)120597119906Known as the exponential integral function defined by

[13]

1205882= 120588119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)minus119864119894 (minus119909) (79)

where 119909 = 12060112058311988811990324119896119905 and minus119864119894(minus119909) = int

infin

119909(119890minus119906

119906)120597119906At large times 119909will be small applying Taylorrsquos series and

integrating

minus119864119894 (minus119909) = ln (120574)

ln (120574) = ln (1781) = 05772(80)

where 120574 is known as Eulerrsquos numberFor 14119909 gt 25 consider the followingThen

1205882(119903 119905) = 120588

119894

2+

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) [ln119909+ ln 120574] (81)

1205882(119903 119905) = 120588

119894

2+

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) ln[120601120583119888119903

2120574

4119896119905]

1205882(119903 119905) = 120588

119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) ln [22461198961199051206011205831198881199032

]

119898 (120588119894) minus119898 (120588

119908119891) =

2120583

int

120588119894

120588119908119891

120588 120597120588

119898 (120588119908119891

) = 119898 (120588119894) minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

sdot [ln [22461198961199051206011205831198881199032

]+ 080907]

120597119898 (120588119908)

120597 ln 119905= 119898gas =

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

(82)

14 Journal of Petroleum Engineering

4160041650417004175041800418504190041950420004205042100

560856095609561056105611561156125612

ODENGDEN

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

1205880

(Ibm

ft)

120597120588w120597lnt

= moil =1627120583Qc01205880

kh= minus0009

120597m(120588w)

120597lnt= mgas =

1637120583Q

kh120572R= minus12195

120588g2

(Ibm

ft)

Figure 13 Oil and gas phase density versus log Horner time

6279

6279

6279

6279

6279

6279

6279

6279

01 1 10 100 1000 10000 1000001000000Horner time (h)

Density versus log Horner time

WDEN

120588w

(Ibm

ft)

120597120588w120597lnt

= mwater =1627120583Qc01205880

kh= minus00003

Figure 14 Water phase density versus log Horner time

Plotting 120588119908

2(119905) or 119898(120588

119908) versus ln(119905) will yield a straight line

at longer time and the slope of the line is given as follows

119896ℎ =120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)1

119898gas (83)

Example 3 Data from Example 1(a) was used for this caseand the semilog plot of density for each phase (oil water andgas) is plotted against log of Horner timelowastlowast

119876 used for 119896 calculation is the average rate for allflowing periods

From Figures 13 and 14 the calculated slopes of the radialflow for oil and water phases are given as

120597120588119908

120597 ln 119905= 119898oil =

1627120583119876119888119900120588119900

119896ℎ= minus 0009

120597120588119908

120597 ln 119905= 119898water =

1627120583119876119888119900120588119900

119896ℎ= minus 00003

(84)

The estimated phase permeabilities are

119896oil =1627120583119876119888

119900120588119900

ℎ119898oil= 445mD

119896water =1627120583119876119888

119900120588119900

ℎ119898water= 50mD

(85)

Also for the gas phase consider the following

ODENGDEN

422004225042300423504240042450425004255042600

5613561356145614561556155616

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

1205880

(Ibm

ft)

120597120588w120597lnt

= moil =1627120583Qc01205880

kh= minus0009

120597m(120588w)

120597lnt= mgas =

1637120583Q

kh120572R= minus12502

120588g2

(Ibm

ft)

Figure 15 Oil and gas phase density versus log Horner time

WDEN

6279627962796279627962796280628062806280

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

120588w

(Ibm

ft)

120597120588w120597lnt

= mwater =1627120583Qc01205880

kh= minus00003

Figure 16 Water phase density versus log Horner time

The calculated slope from the semilog plot shown inFigure 13 is given as

120597119898 (120588119908)

120597 ln 119905= 119898gas =

1637120583119876119896ℎ120572119877

= minus 12195 (86)

And estimated gas permeability is

119896gas =1637120583119876

119900

ℎ120572119877119898gas= 170mD (87)

Using the empirical model for all three phases the averageestimated permeability is as follows

119896ave =4radic50 times 445 times 1702

119896ave = 50mD

(88)

Example 4 Data from Example 1(c) was used for this caseand the semilog plot of density for each phase (oil water andgas) is plotted against log of Horner time

From Figures 15 and 16 the calculated slopes of the radialflow for oil and water phases are given as

120597120588119908

120597 ln 119905= 119898oil =

1627120583119876119888119900120588119900

119896ℎ= minus 0009

120597120588119908

120597 ln 119905= 119898water =

1627120583119876119888119900120588119900

119896ℎ= minus 00003

(89)

Journal of Petroleum Engineering 15

The estimated phase permeabilities are

119896oil =1627120583119876119888

119900120588119900

ℎ119898oil= 4352mD

119896water =1627120583119876119888

119900120588119900

ℎ119898water= 49mD

(90)

Also for the gas phase consider the following

The calculated slope from the semilog plot shown inFigure 15 is given as

120597119898 (120588119908)

120597 ln 119905= 119898gas =

1637120583119876119896ℎ120572119877

= minus 12502 (91)

And estimated gas permeability is

119896gas =1637120583119876

119900

ℎ120572119877119898gas= 179mD (92)

Using the empirical model for all three phases the averageestimated permeability is as follows

119896ave =4radic49 times 4352 times 1792

119896ave = 51mD

(93)

6 Conclusion

The following inferences were drawn from the six scenariosreviewed

(i) Results from the three-phase fluid empirical modeldeveloped for average effective 119896ave estimation rangebetween 47 and 57mD for Example 1 and give404mD for Example 2 which is within that used inthe simulation model and also estimated from BPR

(ii) The BPR PDENDWA and PDENWAT give the samestabilisation and the same 119896 estimates making it suit-able for interpretation of pressure transient analysis

(iii) It has been demonstrated in all 6 scenarios inves-tigated that the heavier fluid such as water andthe weighted average pressure-density equivalent ofall fluid give exact effective 119896 as the conventionalmethod

(iv) For oil reservoir system (scenarios (a) to (d)) 119896 =

500mD is achieved if the thickness contributing toflow ranges from 50 to 150 ft while for gas condensatereservoir (scenarios (a) to (b)) 119896 ranges between340 and 371mD which is achieved if the net sandthickness ℎ = 148ft is contributing to flow

(v) This approach also provides an estimate of the pos-sible fluid phase permeabilities and the of eachphase contribution to flow at a given point hence atseveral 1198891199011015840 stabilisation points the relative 119896 can begenerated

(vi) The derivatives for all output parameters display thesame wellbore and reservoir fingerprint as the BPRmethod

(vii) Generally where output parameters (BPR PDE-NOIL PDENGAS PDENWAT and PDENDW)depict different fingerprint the density derivative willserve as support to distinguish between reservoir andnonreservoir response

Nomenclature

119875 Pressure psi119879 Temperature oF119903 Radius ft119896 Permeability mDOslash Porosity fraction120583 Viscosity cp119905 Time hrs119902 Production rate bblday119861 Formation volume factor rbStb119862119905 Total compressibility psiminus1119903119908 Wellbore radius ftΔ119901 Change in pressure psiaℎ Formation thickness ft119860 Drainage area acres119875119908119891 Bottom-hole flowing pressure psi119875119894 Initial pressure psi

119885 Difference between two pointtime series119894 Subscript of an observed variable119888 Subscript of a calculated variableSTEYX SSE of data point119899 Number of data pointsCov Covariance of data point120575 Standard deviation119905119901 Cumulative production time119862119904 Wellbore storage constant119909 Mean of data point

Abbreviations

LBPR Local grid bottom-hole pressureLDENO Local grid oil densityLDENW Local grid water densityLDENG Local grid gas densityWBHP Well bottom-hole pressureBPR Well bottom-hole pressurePDENOIL Pressure equivalent of LDENOPDENGAS Pressure equivalent of LDENGPDENWAT Pressure equivalent of LDENWPDENDWA Pressure equivalent of density weighted

average (LDENO LDENG and LDENW)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Jalali S D Mohaghegh and R Gaskari ldquoIdentifying infilllocations and underperformer wells in mature fields usingmonthly production rate data Carthage Field Cotton Valley

16 Journal of Petroleum Engineering

Formation Texasrdquo in Proceedings of the SPE Eastern RegionalMeeting SPE Paper 104550 Canton Ohio USA October 2006

[2] D Tiab and A Kumar ldquoApplication of PD function to inter-ference analysisrdquo in Proceedings of the 51st Annual Fall Meetingof the Society of Petroleum Engineers of AIME Paper SPE 6053New Orleans La USA 1976

[3] D Tiab and A Kumar ldquoDetection and location of two parallelsealing fault around a wellrdquo in Proceedings of the 51st AnnualFall Meeting of the Society of Petroleum Engineers of AIME SPEPaper 6056 New Orleans La USA 1976

[4] D Bourdet T M Whittle A A Douglas and Y M Pirard ldquoAnew set of type curves simplifies well test analysisrdquo World Oilvol 196 no 6 pp 95ndash106 1983

[5] S Y Zheng ldquoFighting against non-unique solution problemsin heterogeneous reservoirs through numerical well testingrdquo inProceedings of the SPE Asia Pacific Oil and Gas Conference andExhibition SPE 100951 Adelaide Australia September 2006

[6] R N Horne and K O Temeng ldquoRecognition and location ofpinch-out boundaries by pressure transient analysisrdquo SPE 9905Society of Petroleum Engineers 1981

[7] L C Ayestaran R D Nurmi G A K Shehab and W S ElsildquoWelltest design and interpretation improved by integrated welltesting and geological effortsrdquo SPE 17945 1989

[8] G J Massonnat and D Bandiziol ldquoInterdependence betweengeology and well test interpretationrdquo SPE 22740 1991

[9] K F Du and G Stewart ldquoReservoir description from well testinterpretationrdquo in North Sea Oil and Gas ReservoirmdashIII pp339ndash356 Norwegian Institute of Technology (NTH) 1994

[10] S Y Zheng Well testing and characterisation of meanderingfluvial channel reservoirs [PhD thesis] Heriot-Watt University1997

[11] V T Biu and S Y Zheng ldquoA new approach in pressure transientanalysis part I improved diagnosis of flow regimes in oil and gaswellsrdquo in Proceedings of the 3rd EAGEAAPGWorkshop on TightReservoirs in theMiddle East AbuDhabi UnitedArab Emirates2015

[12] R N Horne Modern Welltest Analysis Petroway Palo AltoCalif USA 1995

[13] C S Matthews and D G Russell Pressure Buildup and FlowTests in Wells vol 1 ofMonograph Series SPE Richardson TexUSA 1967

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2 Journal of Petroleum Engineering

the mathematical fluid flow equation mostly in heteroge-neous reservoirmost engineers in the industry are compelledto use analytical model and type curve solutions to matchcomplex model which is oftentimes not realistic Assump-tions made are ignored while pursuing a perfect match andresults obtained from this approach are often misleading[5]

This marked the beginning of numerical well test-ing in the industry by Zheng 2006 [5] although theapproach started from the early 1990s [6ndash10] Zheng mademore advances in 2006 providing more solutions to thenonunique problems mostly in heterogeneous reservoirsthrough numerical welltesting thereby promoting its appli-cation More papers have been published by researchers onthe subject thereby reflecting the advancement of numericalwelltesting and its application in solving various reservoirengineering practical problems

One of the main limitations of the pressure derivativeis that the measured pressure data must be constructedinto derivative data by means of numerical differentiationOftentimes derivative data from real field are very noisyand difficult to interpret resulting in various smoothingtechniques developed by researchers on this subject It is prac-tically believed that smoothing of pressure derivative dataoften alters the characteristics of the data Also it is difficult todistinguish between fluid and reservoir fingerprints in criticalsaturated reservoirs

Another limitation of these derivatives is diagnosing flowregimes in complex reservoir structures such as complexfaulted systems and high permeability streak with interbed-ded shales which is common in deep water turbidite sys-tems channel-levee lobe and channelized deposits Alsoin situations of multiphase flow around the wellbore thederivative data are always noisy and difficult to interpretresulting in the application of deconvolution and varioussmoothing techniques to obtain a perceived representativemodel which often might not be Additionally the ana-lytical solution for transient pressure analysis is limited tosingle phase flow which in real case is never the situationPresently there are few literatures or research on multiphasetransient pressure analysis However the combination ofthe new statistical approach [11] and the density deriva-tive approach serves as a support tool for better inter-pretation and estimation of reservoir properties in theseconditions

The diagnosis of flow which appears as distinctive pat-terns in the pressure derivative curve is a vital point inwelltest interpretations since each flow regime reflects thegeometry of the flow streamlines in the tested formationHence for each flow regime identified a set of well andorreservoir parameters can be estimated using the region ofthe transient data that exhibits the characteristic patternbehaviour [11] In the study the pressure derivative formula-tion from Horne (1995) [12] and the new statistical approachby Biu and Zheng (2015) [11] would be used throughout theanalysis

The mathematical formulation for pressure derivative byHorne (1995) [12] is given as

(120597119901

120597 ln 119905) = 119905 (

120597119901

120597119901)

119894

minus119860

119860 =ln (119905119894119905119894minus119896

) Δ119901119894+119895

ln (119905119894+119895

119905119894) ln (119905

119894+119895119905119894minus119896

)+119861

119861 =ln (119905119894+119895

119905119894minus119896

1199052

119894) Δ119901119894

ln (119905119894+119895

119905119894) ln (119905

119894119905119894minus119896

)

minusln (119905119894+119895

119905119894) Δ119901119894+119896

ln (119905119894119905119894minus119896

) ln (119905119894+119895

119905119894minus119896

)

(1)

Also the mathematical formulation for the new statisticalderivative approach by Biu and Zheng (2015) [11] is given asfollows

Model 1 Consider the following

StatDev (119894) = SQRT (pdd (119894) 119909Δdev (119894) 119909Δ2119875 (119894)) (2)

Model 2 (the exponential function) Consider the following

StatExp (119894)

= SQRT (EXP (SQRT (Δ2119875))) 119909pdd (119894) 119909Δ

2119875 (119894)

(3)

Model 3 Consider the following

StatdDev (119894)

= SQRT (pdd (119894) 119909Δdev (119894) 119909Δ2119875 (119894) 119909Δ

2119875 (119894))

(4)

Model 4 (the time function) Consider the following

StattDev (119894) = STDEV (Δ119905119905 (119894) Δ119905119905 (119894 + 1) StatDev (119894)

StatDev (119894 + 1)) (5)

Equations (1) to (5) are the derivative and statistical modelsused for flow regime diagnosis behaviours and estimation ofwellbore and reservoir properties using the log-log derivativeplot

2 Theoretical Concept ofthe Density Derivatives

The basic concepts involved in the derivation of fluid flowequation include

(i) conservation of mass equation(ii) transport rate equation (eg Darcyrsquos law)(iii) equation of state

Consider flow in a cylindrical coordinate with flow butwith flow in angular and 119911-directions neglected as shown inFigure 1 the equations are given as follows

Mass rate inminusMass rate out = Mass rate storage (6)

Journal of Petroleum Engineering 3

r

r r + Δr

r + Δr

Figure 1 Schematics of basic fluid flows concept

Equation (6) represents the conservation of mass Since thefluid is moving the equation

119902 = minus119896

120583119860

120597119901

120597119903(7)

is applied By conservingmass in an elemental control volumeas shown in Figure 1 and applying transport rate equation thefollowing equation is obtained

minus[2120587119903ℎ119896

120583120588120597119901

120597119903]

119903

= minus[2120587119903119896ℎ

120583120588120597119901

120597119903]

119903+Δ119903

+ 2120587119903Δ119903ℎ120597

120597119905(120588120601)

(8)

Expand the equation using Taylor series

1119903

120597

120597119903[119903119896120588

120583

120597119901

120597119903] =

120597

120597119905[120588120601] (9)

Equations (6) to (9) apply to both liquid and gas Equation(9) is known as the general diffusivity equation and for eachfluid the density or pressure term in (9) can be replaced bythe correct expression in terms of density or pressure

For small or constant compressibility liquid

120588 = 120588119900119890119862[119901minus119901

119900] (10)

Substituting for pressure in the equation the diffusivityequation in terms of density is given as

1205972120588

1205971199032+1119903

120597120588

120597119903=

120601120583 [119888 + 119888119903]

119896

120597120588

120597119905 (11)

1205972120588

1205971199032+1119903

120597120588

120597119903=

120601120583119888119905

119896

120597120588

120597119905 (12)

Equation (12) is known as the density diffusivity equationwhich can also be rewritten in the form of pressure

Over the decades the transient test analysis has appliedthe general diffusivity equation in pressure term to generateseveral nonunique solutions using several pressure-rate data

Invariably as in pressure term the density term alsoimplored

1205972120588

120597119903+1119903

120597120588

120597119903=1119903

120597

120597119903(119903

120597120588

120597119903) = 0 (13)

For inner boundary condition

[119903120597120588

120597119903]

119903119908

=119902120583

2120587119896ℎ= Constant (14)

For outer boundary condition

120588 = 120588119890

at 119903 = 119903119890 (15)

Presently there are permanent downhole gauges (PDG) withdensity measurement tool along with pressure and temper-ature installed during flowing and shut-in testing condi-tions but the data are not interpreted or used for reservoirmonitoring For simplification and application of the densityderivative in existing welltest software the density-pressureequivalent equation is formulated

3 Software Suitability (Pressure Equivalent)

To apply the numerical density approach in existing softwarethe pressure equivalent of the fluid density changes at thewellbore is generated from the relationship below

Using the isothermal compressibility coefficient 119862 interms of density

119862 =1120588

120597120588

120597119901 (16)

31 For Small or Constant Compressibility Fluid Such as Oiland Water Consider the following

minus119862int

119901

119901119900

119889119901 = int

120588

1205880

120597120588

120588 (17)

Integrating

119890119862[1198750minus119875] =

120588

1205880

119862 [1198750 minus119875] = ln [120588

1205880]

(18)

119875 = 1198750 minusln [1205881205880]

119862 (19)

or applying the 119890119909 expansion series

119890119909= 1+119909+

1199092

2+

1199093

3+ sdot sdot sdot +

119909119899

119899 (20)

Because the term 119862[1198750 minus 119875] is very small the 119890119909 term can be

approximated as

119890119909= 1+119909 (21)

Therefore (14) can be rewritten as

120588 = 1205880 [1minus119862 (1198750 minus119875)]

120588

1205880= 1minus119862 [1198750 minus119875]

there4 119875 = 1198750 minus[1205881205880 + 1

119862]

(22)

4 Journal of Petroleum Engineering

For small compressibility fluid such as oil and water either(19) or (22) is used to generate the pressure equivalent fromwell fluid density obtained from reservoir simulation or PDGtool This pressure is then analyzed in any available well testsoftwares

32 For Compressible Fluid in Isothermal Conditions Con-sider the following

119862119892= minus

1V[120597V120597119901

]

119879

(23)

For real gas equation of state

V =119899119877119879119911

119875 (24)

Differentiating with respect to pressure at constant tempera-ture

(120597V120597119901

)

119879

= 119899119877119879[1119901

(120597119911

120597119901)minus

119911

1199012 ] (25)

Substituting (23) into (25)

119862119892=

1119901

minus1119911[119889119911

119889119901] (26)

In terms of density

119862119892=

1119901

minus1120588

[120597120588

120597119901] (27)

This equation is applicable for real gas conditionFor compressible fluid

119862119892=

1119901

minus1120588

[120597120588

120597119901]

1120588

[120597120588

120597119901] =

1119901

minus119862119892

int

120588

1205880

120597120588

120588= int

119901

1199010

[120597119901

119901minus 120597119901119862

119892]

ln [120588

1205880] = ln [

119901

1199010]minus [119901minus1199010] 119862119892

(28)

Applying the power series for ln119901

ln [119901] = [119901 minus 1] minus[119901 minus 1]2

2+ sdot sdot sdot +

[minus1]119899 [119901 minus 1]119899

119899

+ sdot sdot sdot 0 lt 119901 le 2(29)

Limit ln119909 to the 1st term only

[120588

1205880]minus 1 = [

119901

119901119900

]minus 1minus [119901minus1199010] 119862119892

[120588

1205880] = [

119901

119901119900

]minus [119901minus1199010] 119862119892

(30)

[120588

1205880] =

119901 minus 119901120588119900119888119892minus 119901119900

2119888119892

119901119900

119901119900120588

120588119900

minus119901119900

2119888119892= 119901 [1minus119901

119900119888119892]

(31)

119875 =119875119900120588 minus 119875119900

2120588119900119888119892

120588119900(1 minus 119875

119900119888119892) (32)

For compressible fluids such as gas (32) is used to generatethe pressure equivalent from the fluid density obtained fromreservoir simulation or PDG tool at the well and then thepressure is analyzed in any available well test softwares

Also outside the available welltest software the pressurederivative can be generated from the pressure equivalentobtained from the oil gas and water densities at the well byapplying (1) formulated by Horne (1995) [12] or (2) to (4) byBiu and Zheng (2015) [11]

4 Density Weighted Average (DWA)

While (19) or (22) and (32) give the pressure equivalentfor independent fluid phases such as gas oil and waterthe weighted average method is used to obtain the densityequivalent for a two- or three-phase combinationThe equiv-alent pressure derived from the density for all three fluidcomponents such as gas oil and water is given as

119875119894=

120588119892119901119892+ 120588119900119901119900+ 120588119908119901119908

120588119892+ 120588119900+ 120588119908

(33)

Equation (33) comprises all the fluid phases in the system andas such will be comparable to the conventional bottom-holepressure measurement during the derivative analysis

41 Empirical Model Correlation between Fluid Phases119870 Anempirical model integrating the fluid phasersquos permeabilitiesfor a given system is formulated to determine the averagereservoir permeability The mathematical model is given as

119896ave = 4radic1198961199001198961199081198962119892 (34)

where 119896119900= oil phase permeability 119896

119892= gas phase permeabil-

ity and 119896119908= water phase permeability

With the estimation of the phasersquos permeabilities it istherefore possible to estimate the possible relative permeabil-ity to each phase and the percentage contribution to flow byeach phase at one point analysis hence at several points therelative 119896 can be generated

To illustrate applicability of this approach 6 scenarios inconventional oil reservoir and gas condensate reservoir are

Journal of Petroleum Engineering 5

Local grid refinement

Pressure (psia) Pressure (barsa)

PROD1

PROD1

35591 35630 35669 35708 35747 3093430899 30970 31006 31041

Figure 2 Simulation model for gas cap + oil + water reservoir and gas condensate + water showing local grid refinement around the well

Table 1 Reservoir and fluid data for Example 1

Parameters Design valueEclipse model Black oilModel dimension 10 times 5 times 5Length by width ft by ft 500 times 400Thickness ℎ ft 250Permeability 119870119909 by 119870119910 mD 500 by 500Porosity 20Well diameter ft 060Initial water saturation 119878119908

119894 22

Permeability 119870 mD 50Gas oil contact (GOC) ft 8820Oil water contact (OWC) ft 90000Initial pressure 119875

119894 psia 40000

Formation temperature 119879 oF 2000

investigated using numerical model built with a commercialsimulator Local grid refinement (LGR) is imposed aroundthe well to capture sharp change in fluid densities as shownin Figure 2

The simulation software keywords LBPR LDENOLDENW LDENG AND WBHP were outputs to obtain thedensity and pressure change around the well and as far as theperturbation could extend

Example 1 Table 1 presents a summary of the well andreservoir synthetic data used for the buildup and drawdownsimulated scenarios with additional information given belowIt is required to generate the pressure equivalent and deriva-tive for each phase compare their diagnostic signaturesand also determine the phases permeabilities and averagereservoir permeability

Assumption (i) Oil reservoir + gas cap is completed with onewell

(ii) LGR is imposed around the well and far across toaccount for pressure and density changes

Gas oil and water densities around the local grid refine-ment (wellbore) andWBHP were output using the simulatorkeywords The following scenarios were evaluated

(a) Flowing+ buildup sequence well perforated ℎ119901= 30 ft

between oil and water layer Net sand thickness ℎ =250 ft

(b) Flowing+buildup sequence well perforated ℎ119901= 30 ft

inside the oil layer Net sand thickness ℎ = 250 ft(c) Flowing+ buildup sequence well perforated ℎ

119901= 30 ft

between gas and oil layer Net sand thickness ℎ =250 ft

(d) Falloff test flowing + buildup sequence well perfo-rated ℎ

119901= 30 ft inside the oil layer Net sand thickness

ℎ = 250 ftThe schematic of the 4 scenarios is depicted in Figures 3and 4 Figure 5 shows the production shut-in and injectionsequence for the 4 scenarios in Example 1

In scenario (a) the well is completed between the oil andwater layer to mimic multiphase conditions at the wellboreand some distance away from the well and also see the effecton pressure distribution fluid densities changes around thewellbore and estimate fluid phase permeabilities 119896 using thespecialised plot and 119896ave from the empirical model for three-phase conditions

The derivative in Figure 6 shows a good radial flow butdrop in derivative at late time due to constant pressuresupport (likely aquifer support) A continuous drop is seenfrom 10 hrs in the model parameters The derivatives forall output parameters display the same well and reservoirsignatures (3 flow periods early to late time response)but with different 119889119901

1015840 stabilisation The well bottom-holepressure (BPRrarrBHP) response shows good overlay withpressure equivalent of density weighted average (PDENDWArarr PDENA) and pressure equivalent of water density (PDEN-WAT rarr PDENW) while pressure equivalent of gas den-sity (PDENGAS rarr PDENG) and pressure equivalent ofoil density (PDENOIL rarr PDENO) differ completely ThePDENDWA gives a better fingerprint that is less noisy

A permeability value of 508mD is estimated from thebottom-hole pressure BPR where 119896 = 1627119902119861120583119898ℎ and119898 is

6 Journal of Petroleum Engineering

Gas

Oil

Water

Case 1 production test

Net sand thickness h = 250ftwater layerWell perforated hp = 30ft between oil andFlowing + buildup sequence

(a)

Gas

Oil

Water

Case 2 production test Flowing + buildup sequence

Net sand thickness h = 250ftWell perforated hp = 30ft inside the oil layer

(b)

Gas

Oil

Water

Case 3 production test Flowing + buildup sequence

Net sand thickness h = 250ftoil layerWell perforated hp = 30ft between gas and

(c)

Gas

Oil

Water

Case 4 falloff test Flowing + buildup sequence

Net sand thickness h = 250ftwater layerWell perforated hp = 30ft between oil and

(d)

Figure 3 Schematics of well perforation interval and sand thickness for oil + gas cap + water reservoir

obtained from the specialised plot This is an approximate ofthe input value in the simulation model Also the simulatoroutputs PDENDWA and PDENWAT in the simulation givethe same 119896 value while PDENGAS and PDENOIL differgiving 158 and 4037mD respectively At ℎ = 50 f t the best119896 estimate is obtained depicting ℎ = 50 f t as the thicknesscontributing to flowAt ℎ gt 50ft 119896 drops below the 119896 imputedin the model

Using (34) 119896ave = 473mD is obtained which isapproximately close to that of BPR hence a good estimateof the fluid phase permeabilities A summary of the result isshown in Tables 2 and 5

In scenario (b) thewell is completedwithin the oil sectionto capture the pressure and fluid densities changes around thewell The derivative in Figure 7 shows a good radial flow butwith noisy numerical artefact It is likely that the boundaryresponse is masked by numerical artefact Also a continuousdrop is seen after 10 hrs in the derivative

Using the estimated fluid phase permeabilities 119896ave from(34) is 472mD as shown in Table 3 which is the same asscenario (a) This is in line with the uniform 119896 used in thesimulation model Also the BPR gives a permeability value of500mD which is the same for PDENDWA and PDENWATbut differs with PDENGAS and PDENOIL that give 121 and

Journal of Petroleum Engineering 7

Water

Case 1 production test

Gas + condensate

Flowing + buildup sequence

condensate and water layerNet sand thickness h = 45m

Well perforated hp = 90m between gas

(a)

Water

Case 2 production test

Gas + condensate

Net sand thickness h = 45m

Flowing + buildup sequence

condensate layerWell perforated hp = 90m inside gas

(b)

Figure 4 Schematics of well perforation interval and sand thickness for gas condensate + water reservoir

00

5000

0 50 100 150 200 250 300 350

Prod

uctio

nin

ject

ion

rate

bop

d

Duration

Test design sequence

Case dCases a to c

minus25000

minus20000

minus15000

minus10000

minus5000

Figure 5 Production and injection test sequence for scenarios (a)to (d) (buildup and falloff)

Table 2 119870 estimates for new approach versus conventionalapproach for scenario (a)

Parameters Numerical density119896 (mD)

Equivalentℎ (ft)

BHP 508PDENA 508PDENG 158 50PDENO 4037PDENW 508

4844mD respectively At ℎ = 150ft (60 of sand thicknessand 83 of oil thickness) the estimated 119896 is still within range

00

00

01

10

100

0000 0001 0010 0100 1000 10000 100000Log time (h)

BHPPENDAPDENO

PDENGPDENW

508mD

4037mD

158mD

k =1627qB120583

mh

Case a derivatives for conventional and numerical density models

logdp998400

Figure 6 Derivative and 119870 estimation for scenario (a)

(119896 = 489mD)This indicates ℎ = 50ft or 83 of oil thicknesscontributing to flow

To test this approach in the gas column the well wascompleted in between the gas and oil layer which is consid-ered as scenario (c) and scenario (d) with the well completedwithin the oil layer but with water injection after flowingand shut-in sequence In both scenarios the multiphasefluid distribution is triggered at the wellbore in order tocapture the density changes for each phase and calculate fluidphase permeabilities First the well fluid densities equivalent

8 Journal of Petroleum Engineering

Case b derivatives for conventional and numerical density models

00

01

10

100

0000 0001 0010 0100 1000 10000 100000Log time (h)

BHPPENDAPDENO

PDENGPDENW

4842mDk =1627qB120583

mh

120mD

489mD

logdp998400

Figure 7 Derivative and 119870 estimation for scenario (b)

0001

0010

0100

1000

10000

0000 0001 0010 0100 1000 10000 100000Log time (h)

BHPPENDAPDENO

PDENGPDENW

412mD

209mD

Case c derivatives for conventional and numerical density models

4842mD

k =1627qB120583

mh

logdp998400

Figure 8 Derivative and 119870 estimation for scenario (c)

Table 3 119870 estimates for new approach versus conventionalapproach for scenario (b)

Parameters Numerical density119896 (mD)

Equivalentℎ (ft)

BHP 500PDENA 500PDENG 121 50PDENO 4842PDENW 500

pressures and pressures at bottom-hole for flowing andbuildup test are generated

The derivative for both scenarios declines after 3 hoursprobably due to fluid redistribution but is noisy (numericalartefact) in scenario (c) as shown in Figure 8 A good radial

Table 4 119870 estimates for new approach versus conventionalapproach for scenario (d)

Parameters Numerical Density119896 (mD) Equivalent ℎ (ft)

Scenario (d)

BHP 500 100PDENA 500PDENG 175 50PDENO 490 250PDENW 500 100

stabilisation for both the conventional method BPR and thedensity outputs PDENDWA PDENWAT PDENGAS andPDENOIL is observed Likewise as obtained in scenarios (a)and (b) a good 119870ave value of 57 and 52mD from (34) isobtained for scenarios (c) and (d) respectivelyThis is slightlyhigher than 119896 = 50mD imputed in the simulation model

For each of the fluid phase permeabilities estimates 119896value of 50mD is obtained using the conventional BPR forscenario (c) which is in line with the uniform 119896 in thesimulation model This is the same for the PDENDWA andPDENWAT Also at ℎ = 150ft (60 of sand thickness and83 of oil thickness) the estimated 119896 is still within range(119896 = 412mD)This also indicates that ℎ = 150ft or 83 of oilthickness is contributing to flowThe 119896 values for PDENGASandPDENOIL are 209 and 4842mD respectively Summaryof the result is shown in Table 4

However in scenario (d) the permeability value of500mD using the conventional BPR which is the samefor PDENDWA and PDENWAT is only achieved if ℎ =

100ft (40 of sand thickness and 56 of oil thickness)PDENGAS and PDENOIL differ giving 350mD at ℎ =

50ft and 1960mD at ℎ = 250ft respectively This indicatesthe impact of water injection on densities and pressureschanges around the well and consequently its impact on sandthickness contributing to flow The drop in derivative curvein Figure 9 depicts the impact of the injected water

In summary for scenarios (a) to (d) the 119896 value of500mD is achieved if the thickness contributing to flowranges from 50 to 150 ft Generally results indicate that in all4 scenarios investigated the heavier fluid such as water andthe weighted average pressure-density equivalent of all fluidgive exact effective 119896 as the BPR likewise the density outputsPDENDWA and PDENWAT Also results from Table 5 showthat the empirical model from (34) with all fluid phasepermeabilities gives an average effective 119896ave of 47ndash57mDwhich is within that used in the simulationmodel and also thesame as the estimated conventional approach This approachprovides an estimate of the possible fluid phase permeabilitiesand theof each phase contribution to flow hence at severalpoints the relative 119896 can be generated as shown in Table 5

Example 2 To capture the influence of highly compressiblefluid on estimated fluid phase permeabilities an example ongas condensate reservoir (volatile system) was tested Table 6presents a summary of the well and reservoir synthetic dataused for the buildup and drawdown simulated scenarios with

Journal of Petroleum Engineering 9

Table 5 Comparison of 119870 estimates between conventional and numerical density parameters

Scenario Numerical density method Calculated119870 (mD)

Conventional119896 (mD)

Simulationmodel 119870 (mD) Relative 119870

phasecontribution to flowPhases 119870 (mD)

(a)Gas 158 003 30Oil 4037 473 508 500 086 860

Water 508 011 110

(b)Gas 12 003 20Oil 4844 432 500 500 113 890

Water 50 012 90

(c)Gas 209 004 40Oil 4842 570 500 500 086 87

Water 50 009 90

(d)Gas 175 003 30Oil 490 523 500 500 095 880

Water 50 010 90

1000

Case d derivatives for conventional and numerical density models

0010

0100

10000

BHPPENDAPDENO

PDENGPDENW

0100 1000 10000 100000Log time (h)

k =1627qB120583

mh

209mD

500mD

4842mD

logdp998400

Figure 9 Derivative and 119870 estimation for scenario (d)

additional information given below It is required to generatethe pressure equivalent and derivative for each fluid phasecompare their diagnostic signatures and also determine thephases permeabilities and average reservoir permeability

Assumption (i) Condensate reservoir is completed with onewell

(ii) LGR is imposed around the well and far across toaccount for pressure and density changes

Gas condensate andwater densities around the local gridrefinement (wellbore) andWBHPwere output using the sim-ulatorrsquos keywords The following scenarios were evaluated

(a) Flowing + buildup sequence well perforated ℎ119901

=

30ft between gas condensate and water layer Netsand thickness ℎ = 150ft

Table 6 Summary of reservoir simulations data

Parameters Design valueEclipse model Black oilModel dimension 9 times 3 times 3Length by width ft by ft 1312 times 984Thickness ℎ ft 150Permeability 119870119909 by 119870119910 mD 4000 by 3000Porosity 30Well diameter ft 115Initial water saturation 119878119908

119894 60

Permeability 119870 mD 400Gas oil contact (GOC) ft 6890Oil water contact (OWC) ft 6890Initial pressure 119875

119894 psi 4495

Formation temperature 119879 ∘C 120 2000

(b) Flowing + buildup sequence well perforated ℎ119901

=

30ft inside gas condensate layer Net sand thicknessℎ = 150ft

Figure 10 shows the production and shut-in sequence for 2scenarios in Example 2

As in Example 1 the derivative in scenario (a) ofExample 2 also shows good radial flowwith no late time effectas seen in Figure 11 A negative half-slope fingerprint is seen at03 to 30 hrs of buildup and a good stabilisation from 40 hrsin the model parameters

The derivatives for all output parameters display the samewell and reservoir fingerprint (2 flow periods early to middletime response) but with different 1198891199011015840 stabilisation

Permeability value of 3707mD was obtained from theconventionalmethod BPRwhere 119896 = 1637119902119879119898ℎ is the samefor the PDENDWA and PDENWAT and 119898 is obtained fromthe specialised plot This is in line with the uniform 119896 (119896119909 =

400mD and 119896119910 = 300mD) imputed in the simulationmodel

10 Journal of Petroleum Engineering

00

5000000

10000000

15000000

20000000

25000000

0 40 80 120 160

Prod

uctio

n ra

te (s

cfd

)

Duration

Test design sequence

Cases a to b

Figure 10 Production test sequence (buildup) for Example 2

10000

Case a derivatives for conventional and numerical density models

00100

01000

BHPPENDAPDENO

PDENGPDENW

1422mD

3707mD

35747mD

0000 0001 0010 0100 1000 10000

100000

100000Log time (h)

k =1627qB120583

mh

logdp998400

Figure 11 Derivative and 119870 estimation for scenario (a)

However PDENGAS and PDENOIL differ giving 350mDand 35747mD respectively if ℎ = 148ft

Also the derivative in Figure 12 shows good radial flowand a negative half-slope fingerprint is seen at 05 to 20 hrsof buildup and a good stabilisation from 30 hrs in the modelparameters After 3 hrs of shut-in somenoisy data (numericalartefact) is seen in all the fluid phase derivatives but still goodenough to identify 119889119901

1015840 stabilisationFor each of the fluid phase permeabilities 119896 value of

3409mD at ℎ = 148ft is obtained using BPR which is thesame for the PDENDWA and PDENWAT PDENGAS andPDENOIL differ giving 1490 and 35141mD at ℎ = 148ftrespectively

For scenarios (a) and (b) a good 119896ave value of 4050and 4030mD is obtained respectively from the empiricalmodel (34) integrating all fluid phase permeabilities whichis within that used in the simulationmodel and the estimated

0001000

0010000

0100000

1000000

10000000

0000 0001 0010 0100 1000 10000 100000

Case b derivatives for conventional and numerical density models

BHPPENDAPDENO

PDENGPDENW

Log time (h)

k =1627qB120583

mh

1490mD

3410mD

35140mD

logdp998400

Figure 12 Derivative and119870 estimation for scenario (b)

Table 7 Summary of reservoir modelling properties imputed ineclipse model for case study 2

Parameters Numerical density119896 (mD)

Equivalentℎ (ft)

Scenario(a)

BHP 371PDENA 371PDENG 142 148PDENO 3575PDENW 371

Scenario(b)

BHP 341PDENA 341PDENG 149 148PDENO 3514PDENW 341

conventional approach From the 2 scenarios investigated ithas been demonstrated that the heavier fluid such as waterand the weighted average pressure-density equivalent of allfluid give exact effective 119896 as the conventional method BPRA summary of the result is shown in Table 7

This approach estimates the fluid phase permeabilitiesand the of each phase contribution to flow at a given pointhence at several points the relative 119896 can be generated asshown in Table 8

5 Density Related Radial Flow EquationDerivation for Each Fluid Phase

Radial diffusivity equation is given as

1119903

120597120588

120597119903[119903

120597120588

120597119903] =

120601120583119888

119896

120597120588

120597119905 (35)

Journal of Petroleum Engineering 11

Table 8 Comparison of 119870 estimates between conventional and numerical density parameters

Scenario Numerical density method Calculated119870 (mD)

Conventional119896 (mD)

Simulationmodel 119870 (mD) Relative 119870

phasecontribution to flowPhases 119870 (mD)

(a)Gas 1422 004 30Oil 35747 4046 3707 400 089 870

Water 3707 009 90

(b)Gas 149 004 40Oil 35141 4038 3409 400 089 880

Water 3409 008 90

For oil

1119903

120597120588

120597119903[119903

120597120588119900

120597119903] =

120601120583119900119888

119896119900

120597120588119900

120597119905 (36a)

For water phase

1119903

120597120588

120597119903[119903

120597120588119908

120597119903] =

120601120583119908119888

119896119908

120597120588119908

120597119905 (36b)

For gas phase

1119903

120597120588

120597119903[119903

120597120588119892

120597119903] =

120601120583119892119888

119896119892

120597120588119892

120597119905 (36c)

51 From Equation for Slight and Small Compressibility Suchas Water and Oil From (22)

119875 = 119875119900minus

120588120588119900+ 1

119888119900

(37)

Differentiating with respect to 120588

120597119875

120597120588= minus

1119888119900120588119900

120597119875 = minus120597120588

119888119900120588119900

(38)

From Darcy flow equation

119902 = minus2120587119896ℎ120583

119903120597120588

120597119903 (39)

Substitute for 120597119875

119902 =2120587119896ℎ120583

119903

119888119900120588119900

120597120588

120597119903 (40)

To derive the analytical density transient equation for eachphase the following assumptionconditions are applicable

For oil phase from (35)

1119903

120597120588

120597119903[119903

120597120588

120597119903] =

120601120583119888

119896

120597120588

120597119905 (41)

Initial condition is as follows

120588 (119903 119905 = 0) = 120588119894 (42a)

BC at the wellbore is as follows

lim119903rarr 0

2120587119896ℎ120583

119903

119888119900120588119900

120597120588

120597119903= 119876 (42b)

BC at infirmity is as follows

lim119903rarrinfin

120588 (119903 119905) = 120588119894 (42c)

Using the Boltzmann transformation consider the followingAssuming 120578 = 120601120583119888119903

2119896119905

120597120588

120597119903=

120597120588

120597120578

120597120578

120597119903=2120601120583119888119903119896119905

120597120588

120597120578=

1206011205831198881199032

119896119905

2119903

120597120588

120597120578=2120578119903

120597120588

120597120578 (43)

Therefore1119903

120597

120597119903[119903

120597120588

120597119903] =

21205781199032

120597

120597120578[2120578

120597120588

120597120578] =

41205781199032

120597

120597120578[120578

120597120588

120597120578] (44)

Differentiating with respect to 119903 is equal to differentiatingwith respect to 120578 multiply by 2120578119903

From (35) the LHS is resolved as follows

120597120588

120597119905=

120597120588

120597120578

120597120578

120597119905=

120601120583119888119903

1198961199052120597120588

120597120578= minus

120578

119905

120597120588

120597120578 (45)

120601120583119888

119896

120597120588

120597119905=

minus120601120583119888

119896

120578

119905

120597120588

120597119905=

minus1206011205831198881199032

119896119905

120578

1199032120597120588

120597120578=

minus1205782

1199032120597120588

120597120578 (46)

Equating (44) and (46) we have

120597

120597120578[120578

120597120588

120597120578] =

minus120578

4120597120588

120597120578 (47)

This is the simplified ordinary differential equation of 120588 as afunction of 120578

Apply boundary conditions as followsInitial condition is as follows

lim119903rarrinfin

120588 (120578) = 120588119894 (48a)

Substituting (43) into (40)

lim119903rarr 0

2120587119896ℎ120583

119903

119888119900120588119900

120597120588

120597119903= 119876 (48b)

lim119903rarr 0

[120578120597120588

120597120578] =

120601120583119888119900120588119900

4120587119896ℎ (49)

Assume that119898 = 120578(120597120588120597120578)

12 Journal of Petroleum Engineering

From (47)

120597119898

120597120578=

minus119898

4 (50)

Integrating from 120578 = 0 to 120578

997904rArr int

119898(120578)

119898(0)

120597119898

120597120578= minusint

120578

0

120597120578

4

997904rArr ln[119898 (120578)

119898 (0)] =

minus120578

4

997904rArr 119898(120578) = 119898 (0) 119890minus1205784

(51)

Recall boundary condition

lim119903rarr 0

[120578120597120588

120597120578] =

119876120583119888119900120588119900

4120587119896ℎ

119898 (0) =119876120583119888119900120588119900

4120587119896ℎ

119898 (120578) =119876120583119888119900120588119900

4120587119896ℎ119890minus1205784

(52)

Recall that119898 = 120578(120597120588120597120578)Therefore

120597120588 (120578)

120597120578=

120583119876119888119900120588119900

4120587119896ℎ119890minus1205784

120578 (53)

Integrating from 120578 = infin where 120588 = 120588119894

int

120588(120578)

120588119894

120597120588 = minusint

120578

infin

120583119876119888119900120588119900

4120587119896ℎ119890minus1205784

120578120597120578

120588 (120578) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎminusint

infin

120578

119890minus1205784

120578120597120578

(54)

where

120578 =120601120583119888119903

2

119896119905 (55)

Assuming 119906 = 1205784 then 120597120578120578 = 120597119906119906 and 119906 = 12060112058311988811990324119896119905

Hence

120588 = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎminusint

infin

12060112058311988811990324119896119905

119890minus1205784

120578120597120578 (56)

where

minus119864119894 (minus119909) = int

infin

119909

119890minus119906

119906120597119906 (57)

Known as the exponential integral function defined by [13]

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎ119864119894 (minus119909) (58)

where 119909 = 12060112058311988811990324119896119905 and

minus119864119894 (minus119909) = int

infin

119909

119890minus119906

119906120597119906 (59)

At large times 119909 will be small applying Taylorrsquos series andintegrating

minus119864119894 (minus119909) = ln (120574)

ln (120574) = ln (1781) = 05772(60)

where 120574 is known as Eulerrsquos numberFor 14119909 gt 25 consider the followingThen

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎ[ln119909+ ln 120574]

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎln[

1206011205831198881199032120574

4119896119905]

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎln [

22461198961199051206011205831198881199032

]

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎ[ln [

22461198961199051206011205831198881199032

]+ 080907]

(61)

Plotting 120588119908(119905) versus ln(119905) will yield a straight line at longer

time and the slope of the line is given as120597120588119908

120597 ln 119905= 119898oil =

120583119876119888119900120588119900

4120587119896ℎ (62)

Therefore

119896ℎ =120583119876119888119900120588119900

4120587119898oil (63)

52 For Water Phase The radial density equation is given as

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎ[ln [

22461198961199051206011205831198881199032

]+ 080907] (64)

Plotting 120588119908(119905) versus ln(119905) will yield a straight line at longer

time and the slope of the line is given as120597120588119908

120597 ln 119905= 119898water =

120583119876119888119900120588119900

4120587119896ℎ (65)

where

119896ℎ =120583119876119888119900120588119900

4120587119898water (66)

53 For Gas Phase From equation for compressible fluidsuch as gas consider the following

From (32)

119875 =119875119900120588 minus 119875119900

2120588119900119888119892

120588119900(1 minus 119875

119900119888119892)

120597119875

120597120588=

119875119900

120588119900(1 minus 119875

119900119888119892)

120597119875 =119875119900

120588119900(1 minus 119875

119900119888119892)120597120588

(67)

Journal of Petroleum Engineering 13

From the fundamental

119899 =119875119881

119911119877119879 (68)

At standard condition

119875119881

119911119879=

119875sc119881sc119879sc

119875 (5615119902)119911119879

=119875sc119876sc119879sc

119902 =119875sc119879sc

(119911119879

119875)

119876sc5615

119902 =119875sc119879sc

119876sc5615

(119911119879

119875) =

2120587119896ℎ120583

119903120597119901

120597119903

119902 =119875sc119879sc

119876sc5615

(119911119879) =2120587119896ℎ120583

119903119875120597119901

120597119903

(69)

where

120597119875 =119875119900

120588119900(1 minus 119875

119900119888119892)120597120588

120572 =119875119900

120588119900(1 minus 119875

119900119888119892)

120597119875 = 120572120597120588

119875 = 120588119877119911119879

(70)

Substitute into the equation where

119898(120588119894) minus119898 (120588

119908119891) =

2120583

int

120588119894

120588119908119891

120588120597120588

119898 (120588119894) minus119898 (120588

119908119891) =

120588119894

2minus 120588119908119891

2

120583

120588 = radic120588119894

2+ 120588119908119891

2

2

(71)

From the diffusivity equation for gas

1119903

120597120588

120597119903[119903

120597120588119892

120597119903] =

120601120583119892119888

119896119892

120597120588119892

120597119905 (72)

Initial condition is as follows

120588 (119903 119905 = 0) = 120588119894 (73)

BC at the wellbore is as follows

lim119903rarr 0

4120587119896ℎ120572119877120583

(5615119879sc

119875sc) 119903120588

120597120588

120597119903= 119876 (74)

BC at infirmity is as follows

lim119903rarrinfin

120588 (119903 119905) = 120588119894 (75)

where119877119875sc and119879sc are known as temperature pressure andgas constant at standard condition

Therefore integrating from 120578 = infin where 120588 = 120588119894

int

120588(120578)

120588119894

120588120597120588 = minusint

120578

infin

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)119890minus1205784

120578120597120578

1205882(120578) = 120588

119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

minusint

infin

120578

119890minus1205784

120578120597120578

(76)

where

120578 =120601120583119888119903

2

119896119905 (77)

Assume that 119906 = (1205784) (120597120578120578) = 120597119906119906 and 119906 = 12060112058311988811990324119896119905

Hence

1205882= 120588119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)minusint

infin

12060112058311988811990324119896119905

119890minus1205784

120578 (78)

where minus119864119894(minus119909) = intinfin

119909(119890minus119906

119906)120597119906Known as the exponential integral function defined by

[13]

1205882= 120588119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)minus119864119894 (minus119909) (79)

where 119909 = 12060112058311988811990324119896119905 and minus119864119894(minus119909) = int

infin

119909(119890minus119906

119906)120597119906At large times 119909will be small applying Taylorrsquos series and

integrating

minus119864119894 (minus119909) = ln (120574)

ln (120574) = ln (1781) = 05772(80)

where 120574 is known as Eulerrsquos numberFor 14119909 gt 25 consider the followingThen

1205882(119903 119905) = 120588

119894

2+

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) [ln119909+ ln 120574] (81)

1205882(119903 119905) = 120588

119894

2+

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) ln[120601120583119888119903

2120574

4119896119905]

1205882(119903 119905) = 120588

119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) ln [22461198961199051206011205831198881199032

]

119898 (120588119894) minus119898 (120588

119908119891) =

2120583

int

120588119894

120588119908119891

120588 120597120588

119898 (120588119908119891

) = 119898 (120588119894) minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

sdot [ln [22461198961199051206011205831198881199032

]+ 080907]

120597119898 (120588119908)

120597 ln 119905= 119898gas =

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

(82)

14 Journal of Petroleum Engineering

4160041650417004175041800418504190041950420004205042100

560856095609561056105611561156125612

ODENGDEN

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

1205880

(Ibm

ft)

120597120588w120597lnt

= moil =1627120583Qc01205880

kh= minus0009

120597m(120588w)

120597lnt= mgas =

1637120583Q

kh120572R= minus12195

120588g2

(Ibm

ft)

Figure 13 Oil and gas phase density versus log Horner time

6279

6279

6279

6279

6279

6279

6279

6279

01 1 10 100 1000 10000 1000001000000Horner time (h)

Density versus log Horner time

WDEN

120588w

(Ibm

ft)

120597120588w120597lnt

= mwater =1627120583Qc01205880

kh= minus00003

Figure 14 Water phase density versus log Horner time

Plotting 120588119908

2(119905) or 119898(120588

119908) versus ln(119905) will yield a straight line

at longer time and the slope of the line is given as follows

119896ℎ =120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)1

119898gas (83)

Example 3 Data from Example 1(a) was used for this caseand the semilog plot of density for each phase (oil water andgas) is plotted against log of Horner timelowastlowast

119876 used for 119896 calculation is the average rate for allflowing periods

From Figures 13 and 14 the calculated slopes of the radialflow for oil and water phases are given as

120597120588119908

120597 ln 119905= 119898oil =

1627120583119876119888119900120588119900

119896ℎ= minus 0009

120597120588119908

120597 ln 119905= 119898water =

1627120583119876119888119900120588119900

119896ℎ= minus 00003

(84)

The estimated phase permeabilities are

119896oil =1627120583119876119888

119900120588119900

ℎ119898oil= 445mD

119896water =1627120583119876119888

119900120588119900

ℎ119898water= 50mD

(85)

Also for the gas phase consider the following

ODENGDEN

422004225042300423504240042450425004255042600

5613561356145614561556155616

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

1205880

(Ibm

ft)

120597120588w120597lnt

= moil =1627120583Qc01205880

kh= minus0009

120597m(120588w)

120597lnt= mgas =

1637120583Q

kh120572R= minus12502

120588g2

(Ibm

ft)

Figure 15 Oil and gas phase density versus log Horner time

WDEN

6279627962796279627962796280628062806280

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

120588w

(Ibm

ft)

120597120588w120597lnt

= mwater =1627120583Qc01205880

kh= minus00003

Figure 16 Water phase density versus log Horner time

The calculated slope from the semilog plot shown inFigure 13 is given as

120597119898 (120588119908)

120597 ln 119905= 119898gas =

1637120583119876119896ℎ120572119877

= minus 12195 (86)

And estimated gas permeability is

119896gas =1637120583119876

119900

ℎ120572119877119898gas= 170mD (87)

Using the empirical model for all three phases the averageestimated permeability is as follows

119896ave =4radic50 times 445 times 1702

119896ave = 50mD

(88)

Example 4 Data from Example 1(c) was used for this caseand the semilog plot of density for each phase (oil water andgas) is plotted against log of Horner time

From Figures 15 and 16 the calculated slopes of the radialflow for oil and water phases are given as

120597120588119908

120597 ln 119905= 119898oil =

1627120583119876119888119900120588119900

119896ℎ= minus 0009

120597120588119908

120597 ln 119905= 119898water =

1627120583119876119888119900120588119900

119896ℎ= minus 00003

(89)

Journal of Petroleum Engineering 15

The estimated phase permeabilities are

119896oil =1627120583119876119888

119900120588119900

ℎ119898oil= 4352mD

119896water =1627120583119876119888

119900120588119900

ℎ119898water= 49mD

(90)

Also for the gas phase consider the following

The calculated slope from the semilog plot shown inFigure 15 is given as

120597119898 (120588119908)

120597 ln 119905= 119898gas =

1637120583119876119896ℎ120572119877

= minus 12502 (91)

And estimated gas permeability is

119896gas =1637120583119876

119900

ℎ120572119877119898gas= 179mD (92)

Using the empirical model for all three phases the averageestimated permeability is as follows

119896ave =4radic49 times 4352 times 1792

119896ave = 51mD

(93)

6 Conclusion

The following inferences were drawn from the six scenariosreviewed

(i) Results from the three-phase fluid empirical modeldeveloped for average effective 119896ave estimation rangebetween 47 and 57mD for Example 1 and give404mD for Example 2 which is within that used inthe simulation model and also estimated from BPR

(ii) The BPR PDENDWA and PDENWAT give the samestabilisation and the same 119896 estimates making it suit-able for interpretation of pressure transient analysis

(iii) It has been demonstrated in all 6 scenarios inves-tigated that the heavier fluid such as water andthe weighted average pressure-density equivalent ofall fluid give exact effective 119896 as the conventionalmethod

(iv) For oil reservoir system (scenarios (a) to (d)) 119896 =

500mD is achieved if the thickness contributing toflow ranges from 50 to 150 ft while for gas condensatereservoir (scenarios (a) to (b)) 119896 ranges between340 and 371mD which is achieved if the net sandthickness ℎ = 148ft is contributing to flow

(v) This approach also provides an estimate of the pos-sible fluid phase permeabilities and the of eachphase contribution to flow at a given point hence atseveral 1198891199011015840 stabilisation points the relative 119896 can begenerated

(vi) The derivatives for all output parameters display thesame wellbore and reservoir fingerprint as the BPRmethod

(vii) Generally where output parameters (BPR PDE-NOIL PDENGAS PDENWAT and PDENDW)depict different fingerprint the density derivative willserve as support to distinguish between reservoir andnonreservoir response

Nomenclature

119875 Pressure psi119879 Temperature oF119903 Radius ft119896 Permeability mDOslash Porosity fraction120583 Viscosity cp119905 Time hrs119902 Production rate bblday119861 Formation volume factor rbStb119862119905 Total compressibility psiminus1119903119908 Wellbore radius ftΔ119901 Change in pressure psiaℎ Formation thickness ft119860 Drainage area acres119875119908119891 Bottom-hole flowing pressure psi119875119894 Initial pressure psi

119885 Difference between two pointtime series119894 Subscript of an observed variable119888 Subscript of a calculated variableSTEYX SSE of data point119899 Number of data pointsCov Covariance of data point120575 Standard deviation119905119901 Cumulative production time119862119904 Wellbore storage constant119909 Mean of data point

Abbreviations

LBPR Local grid bottom-hole pressureLDENO Local grid oil densityLDENW Local grid water densityLDENG Local grid gas densityWBHP Well bottom-hole pressureBPR Well bottom-hole pressurePDENOIL Pressure equivalent of LDENOPDENGAS Pressure equivalent of LDENGPDENWAT Pressure equivalent of LDENWPDENDWA Pressure equivalent of density weighted

average (LDENO LDENG and LDENW)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Jalali S D Mohaghegh and R Gaskari ldquoIdentifying infilllocations and underperformer wells in mature fields usingmonthly production rate data Carthage Field Cotton Valley

16 Journal of Petroleum Engineering

Formation Texasrdquo in Proceedings of the SPE Eastern RegionalMeeting SPE Paper 104550 Canton Ohio USA October 2006

[2] D Tiab and A Kumar ldquoApplication of PD function to inter-ference analysisrdquo in Proceedings of the 51st Annual Fall Meetingof the Society of Petroleum Engineers of AIME Paper SPE 6053New Orleans La USA 1976

[3] D Tiab and A Kumar ldquoDetection and location of two parallelsealing fault around a wellrdquo in Proceedings of the 51st AnnualFall Meeting of the Society of Petroleum Engineers of AIME SPEPaper 6056 New Orleans La USA 1976

[4] D Bourdet T M Whittle A A Douglas and Y M Pirard ldquoAnew set of type curves simplifies well test analysisrdquo World Oilvol 196 no 6 pp 95ndash106 1983

[5] S Y Zheng ldquoFighting against non-unique solution problemsin heterogeneous reservoirs through numerical well testingrdquo inProceedings of the SPE Asia Pacific Oil and Gas Conference andExhibition SPE 100951 Adelaide Australia September 2006

[6] R N Horne and K O Temeng ldquoRecognition and location ofpinch-out boundaries by pressure transient analysisrdquo SPE 9905Society of Petroleum Engineers 1981

[7] L C Ayestaran R D Nurmi G A K Shehab and W S ElsildquoWelltest design and interpretation improved by integrated welltesting and geological effortsrdquo SPE 17945 1989

[8] G J Massonnat and D Bandiziol ldquoInterdependence betweengeology and well test interpretationrdquo SPE 22740 1991

[9] K F Du and G Stewart ldquoReservoir description from well testinterpretationrdquo in North Sea Oil and Gas ReservoirmdashIII pp339ndash356 Norwegian Institute of Technology (NTH) 1994

[10] S Y Zheng Well testing and characterisation of meanderingfluvial channel reservoirs [PhD thesis] Heriot-Watt University1997

[11] V T Biu and S Y Zheng ldquoA new approach in pressure transientanalysis part I improved diagnosis of flow regimes in oil and gaswellsrdquo in Proceedings of the 3rd EAGEAAPGWorkshop on TightReservoirs in theMiddle East AbuDhabi UnitedArab Emirates2015

[12] R N Horne Modern Welltest Analysis Petroway Palo AltoCalif USA 1995

[13] C S Matthews and D G Russell Pressure Buildup and FlowTests in Wells vol 1 ofMonograph Series SPE Richardson TexUSA 1967

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Journal of Petroleum Engineering 3

r

r r + Δr

r + Δr

Figure 1 Schematics of basic fluid flows concept

Equation (6) represents the conservation of mass Since thefluid is moving the equation

119902 = minus119896

120583119860

120597119901

120597119903(7)

is applied By conservingmass in an elemental control volumeas shown in Figure 1 and applying transport rate equation thefollowing equation is obtained

minus[2120587119903ℎ119896

120583120588120597119901

120597119903]

119903

= minus[2120587119903119896ℎ

120583120588120597119901

120597119903]

119903+Δ119903

+ 2120587119903Δ119903ℎ120597

120597119905(120588120601)

(8)

Expand the equation using Taylor series

1119903

120597

120597119903[119903119896120588

120583

120597119901

120597119903] =

120597

120597119905[120588120601] (9)

Equations (6) to (9) apply to both liquid and gas Equation(9) is known as the general diffusivity equation and for eachfluid the density or pressure term in (9) can be replaced bythe correct expression in terms of density or pressure

For small or constant compressibility liquid

120588 = 120588119900119890119862[119901minus119901

119900] (10)

Substituting for pressure in the equation the diffusivityequation in terms of density is given as

1205972120588

1205971199032+1119903

120597120588

120597119903=

120601120583 [119888 + 119888119903]

119896

120597120588

120597119905 (11)

1205972120588

1205971199032+1119903

120597120588

120597119903=

120601120583119888119905

119896

120597120588

120597119905 (12)

Equation (12) is known as the density diffusivity equationwhich can also be rewritten in the form of pressure

Over the decades the transient test analysis has appliedthe general diffusivity equation in pressure term to generateseveral nonunique solutions using several pressure-rate data

Invariably as in pressure term the density term alsoimplored

1205972120588

120597119903+1119903

120597120588

120597119903=1119903

120597

120597119903(119903

120597120588

120597119903) = 0 (13)

For inner boundary condition

[119903120597120588

120597119903]

119903119908

=119902120583

2120587119896ℎ= Constant (14)

For outer boundary condition

120588 = 120588119890

at 119903 = 119903119890 (15)

Presently there are permanent downhole gauges (PDG) withdensity measurement tool along with pressure and temper-ature installed during flowing and shut-in testing condi-tions but the data are not interpreted or used for reservoirmonitoring For simplification and application of the densityderivative in existing welltest software the density-pressureequivalent equation is formulated

3 Software Suitability (Pressure Equivalent)

To apply the numerical density approach in existing softwarethe pressure equivalent of the fluid density changes at thewellbore is generated from the relationship below

Using the isothermal compressibility coefficient 119862 interms of density

119862 =1120588

120597120588

120597119901 (16)

31 For Small or Constant Compressibility Fluid Such as Oiland Water Consider the following

minus119862int

119901

119901119900

119889119901 = int

120588

1205880

120597120588

120588 (17)

Integrating

119890119862[1198750minus119875] =

120588

1205880

119862 [1198750 minus119875] = ln [120588

1205880]

(18)

119875 = 1198750 minusln [1205881205880]

119862 (19)

or applying the 119890119909 expansion series

119890119909= 1+119909+

1199092

2+

1199093

3+ sdot sdot sdot +

119909119899

119899 (20)

Because the term 119862[1198750 minus 119875] is very small the 119890119909 term can be

approximated as

119890119909= 1+119909 (21)

Therefore (14) can be rewritten as

120588 = 1205880 [1minus119862 (1198750 minus119875)]

120588

1205880= 1minus119862 [1198750 minus119875]

there4 119875 = 1198750 minus[1205881205880 + 1

119862]

(22)

4 Journal of Petroleum Engineering

For small compressibility fluid such as oil and water either(19) or (22) is used to generate the pressure equivalent fromwell fluid density obtained from reservoir simulation or PDGtool This pressure is then analyzed in any available well testsoftwares

32 For Compressible Fluid in Isothermal Conditions Con-sider the following

119862119892= minus

1V[120597V120597119901

]

119879

(23)

For real gas equation of state

V =119899119877119879119911

119875 (24)

Differentiating with respect to pressure at constant tempera-ture

(120597V120597119901

)

119879

= 119899119877119879[1119901

(120597119911

120597119901)minus

119911

1199012 ] (25)

Substituting (23) into (25)

119862119892=

1119901

minus1119911[119889119911

119889119901] (26)

In terms of density

119862119892=

1119901

minus1120588

[120597120588

120597119901] (27)

This equation is applicable for real gas conditionFor compressible fluid

119862119892=

1119901

minus1120588

[120597120588

120597119901]

1120588

[120597120588

120597119901] =

1119901

minus119862119892

int

120588

1205880

120597120588

120588= int

119901

1199010

[120597119901

119901minus 120597119901119862

119892]

ln [120588

1205880] = ln [

119901

1199010]minus [119901minus1199010] 119862119892

(28)

Applying the power series for ln119901

ln [119901] = [119901 minus 1] minus[119901 minus 1]2

2+ sdot sdot sdot +

[minus1]119899 [119901 minus 1]119899

119899

+ sdot sdot sdot 0 lt 119901 le 2(29)

Limit ln119909 to the 1st term only

[120588

1205880]minus 1 = [

119901

119901119900

]minus 1minus [119901minus1199010] 119862119892

[120588

1205880] = [

119901

119901119900

]minus [119901minus1199010] 119862119892

(30)

[120588

1205880] =

119901 minus 119901120588119900119888119892minus 119901119900

2119888119892

119901119900

119901119900120588

120588119900

minus119901119900

2119888119892= 119901 [1minus119901

119900119888119892]

(31)

119875 =119875119900120588 minus 119875119900

2120588119900119888119892

120588119900(1 minus 119875

119900119888119892) (32)

For compressible fluids such as gas (32) is used to generatethe pressure equivalent from the fluid density obtained fromreservoir simulation or PDG tool at the well and then thepressure is analyzed in any available well test softwares

Also outside the available welltest software the pressurederivative can be generated from the pressure equivalentobtained from the oil gas and water densities at the well byapplying (1) formulated by Horne (1995) [12] or (2) to (4) byBiu and Zheng (2015) [11]

4 Density Weighted Average (DWA)

While (19) or (22) and (32) give the pressure equivalentfor independent fluid phases such as gas oil and waterthe weighted average method is used to obtain the densityequivalent for a two- or three-phase combinationThe equiv-alent pressure derived from the density for all three fluidcomponents such as gas oil and water is given as

119875119894=

120588119892119901119892+ 120588119900119901119900+ 120588119908119901119908

120588119892+ 120588119900+ 120588119908

(33)

Equation (33) comprises all the fluid phases in the system andas such will be comparable to the conventional bottom-holepressure measurement during the derivative analysis

41 Empirical Model Correlation between Fluid Phases119870 Anempirical model integrating the fluid phasersquos permeabilitiesfor a given system is formulated to determine the averagereservoir permeability The mathematical model is given as

119896ave = 4radic1198961199001198961199081198962119892 (34)

where 119896119900= oil phase permeability 119896

119892= gas phase permeabil-

ity and 119896119908= water phase permeability

With the estimation of the phasersquos permeabilities it istherefore possible to estimate the possible relative permeabil-ity to each phase and the percentage contribution to flow byeach phase at one point analysis hence at several points therelative 119896 can be generated

To illustrate applicability of this approach 6 scenarios inconventional oil reservoir and gas condensate reservoir are

Journal of Petroleum Engineering 5

Local grid refinement

Pressure (psia) Pressure (barsa)

PROD1

PROD1

35591 35630 35669 35708 35747 3093430899 30970 31006 31041

Figure 2 Simulation model for gas cap + oil + water reservoir and gas condensate + water showing local grid refinement around the well

Table 1 Reservoir and fluid data for Example 1

Parameters Design valueEclipse model Black oilModel dimension 10 times 5 times 5Length by width ft by ft 500 times 400Thickness ℎ ft 250Permeability 119870119909 by 119870119910 mD 500 by 500Porosity 20Well diameter ft 060Initial water saturation 119878119908

119894 22

Permeability 119870 mD 50Gas oil contact (GOC) ft 8820Oil water contact (OWC) ft 90000Initial pressure 119875

119894 psia 40000

Formation temperature 119879 oF 2000

investigated using numerical model built with a commercialsimulator Local grid refinement (LGR) is imposed aroundthe well to capture sharp change in fluid densities as shownin Figure 2

The simulation software keywords LBPR LDENOLDENW LDENG AND WBHP were outputs to obtain thedensity and pressure change around the well and as far as theperturbation could extend

Example 1 Table 1 presents a summary of the well andreservoir synthetic data used for the buildup and drawdownsimulated scenarios with additional information given belowIt is required to generate the pressure equivalent and deriva-tive for each phase compare their diagnostic signaturesand also determine the phases permeabilities and averagereservoir permeability

Assumption (i) Oil reservoir + gas cap is completed with onewell

(ii) LGR is imposed around the well and far across toaccount for pressure and density changes

Gas oil and water densities around the local grid refine-ment (wellbore) andWBHP were output using the simulatorkeywords The following scenarios were evaluated

(a) Flowing+ buildup sequence well perforated ℎ119901= 30 ft

between oil and water layer Net sand thickness ℎ =250 ft

(b) Flowing+buildup sequence well perforated ℎ119901= 30 ft

inside the oil layer Net sand thickness ℎ = 250 ft(c) Flowing+ buildup sequence well perforated ℎ

119901= 30 ft

between gas and oil layer Net sand thickness ℎ =250 ft

(d) Falloff test flowing + buildup sequence well perfo-rated ℎ

119901= 30 ft inside the oil layer Net sand thickness

ℎ = 250 ftThe schematic of the 4 scenarios is depicted in Figures 3and 4 Figure 5 shows the production shut-in and injectionsequence for the 4 scenarios in Example 1

In scenario (a) the well is completed between the oil andwater layer to mimic multiphase conditions at the wellboreand some distance away from the well and also see the effecton pressure distribution fluid densities changes around thewellbore and estimate fluid phase permeabilities 119896 using thespecialised plot and 119896ave from the empirical model for three-phase conditions

The derivative in Figure 6 shows a good radial flow butdrop in derivative at late time due to constant pressuresupport (likely aquifer support) A continuous drop is seenfrom 10 hrs in the model parameters The derivatives forall output parameters display the same well and reservoirsignatures (3 flow periods early to late time response)but with different 119889119901

1015840 stabilisation The well bottom-holepressure (BPRrarrBHP) response shows good overlay withpressure equivalent of density weighted average (PDENDWArarr PDENA) and pressure equivalent of water density (PDEN-WAT rarr PDENW) while pressure equivalent of gas den-sity (PDENGAS rarr PDENG) and pressure equivalent ofoil density (PDENOIL rarr PDENO) differ completely ThePDENDWA gives a better fingerprint that is less noisy

A permeability value of 508mD is estimated from thebottom-hole pressure BPR where 119896 = 1627119902119861120583119898ℎ and119898 is

6 Journal of Petroleum Engineering

Gas

Oil

Water

Case 1 production test

Net sand thickness h = 250ftwater layerWell perforated hp = 30ft between oil andFlowing + buildup sequence

(a)

Gas

Oil

Water

Case 2 production test Flowing + buildup sequence

Net sand thickness h = 250ftWell perforated hp = 30ft inside the oil layer

(b)

Gas

Oil

Water

Case 3 production test Flowing + buildup sequence

Net sand thickness h = 250ftoil layerWell perforated hp = 30ft between gas and

(c)

Gas

Oil

Water

Case 4 falloff test Flowing + buildup sequence

Net sand thickness h = 250ftwater layerWell perforated hp = 30ft between oil and

(d)

Figure 3 Schematics of well perforation interval and sand thickness for oil + gas cap + water reservoir

obtained from the specialised plot This is an approximate ofthe input value in the simulation model Also the simulatoroutputs PDENDWA and PDENWAT in the simulation givethe same 119896 value while PDENGAS and PDENOIL differgiving 158 and 4037mD respectively At ℎ = 50 f t the best119896 estimate is obtained depicting ℎ = 50 f t as the thicknesscontributing to flowAt ℎ gt 50ft 119896 drops below the 119896 imputedin the model

Using (34) 119896ave = 473mD is obtained which isapproximately close to that of BPR hence a good estimateof the fluid phase permeabilities A summary of the result isshown in Tables 2 and 5

In scenario (b) thewell is completedwithin the oil sectionto capture the pressure and fluid densities changes around thewell The derivative in Figure 7 shows a good radial flow butwith noisy numerical artefact It is likely that the boundaryresponse is masked by numerical artefact Also a continuousdrop is seen after 10 hrs in the derivative

Using the estimated fluid phase permeabilities 119896ave from(34) is 472mD as shown in Table 3 which is the same asscenario (a) This is in line with the uniform 119896 used in thesimulation model Also the BPR gives a permeability value of500mD which is the same for PDENDWA and PDENWATbut differs with PDENGAS and PDENOIL that give 121 and

Journal of Petroleum Engineering 7

Water

Case 1 production test

Gas + condensate

Flowing + buildup sequence

condensate and water layerNet sand thickness h = 45m

Well perforated hp = 90m between gas

(a)

Water

Case 2 production test

Gas + condensate

Net sand thickness h = 45m

Flowing + buildup sequence

condensate layerWell perforated hp = 90m inside gas

(b)

Figure 4 Schematics of well perforation interval and sand thickness for gas condensate + water reservoir

00

5000

0 50 100 150 200 250 300 350

Prod

uctio

nin

ject

ion

rate

bop

d

Duration

Test design sequence

Case dCases a to c

minus25000

minus20000

minus15000

minus10000

minus5000

Figure 5 Production and injection test sequence for scenarios (a)to (d) (buildup and falloff)

Table 2 119870 estimates for new approach versus conventionalapproach for scenario (a)

Parameters Numerical density119896 (mD)

Equivalentℎ (ft)

BHP 508PDENA 508PDENG 158 50PDENO 4037PDENW 508

4844mD respectively At ℎ = 150ft (60 of sand thicknessand 83 of oil thickness) the estimated 119896 is still within range

00

00

01

10

100

0000 0001 0010 0100 1000 10000 100000Log time (h)

BHPPENDAPDENO

PDENGPDENW

508mD

4037mD

158mD

k =1627qB120583

mh

Case a derivatives for conventional and numerical density models

logdp998400

Figure 6 Derivative and 119870 estimation for scenario (a)

(119896 = 489mD)This indicates ℎ = 50ft or 83 of oil thicknesscontributing to flow

To test this approach in the gas column the well wascompleted in between the gas and oil layer which is consid-ered as scenario (c) and scenario (d) with the well completedwithin the oil layer but with water injection after flowingand shut-in sequence In both scenarios the multiphasefluid distribution is triggered at the wellbore in order tocapture the density changes for each phase and calculate fluidphase permeabilities First the well fluid densities equivalent

8 Journal of Petroleum Engineering

Case b derivatives for conventional and numerical density models

00

01

10

100

0000 0001 0010 0100 1000 10000 100000Log time (h)

BHPPENDAPDENO

PDENGPDENW

4842mDk =1627qB120583

mh

120mD

489mD

logdp998400

Figure 7 Derivative and 119870 estimation for scenario (b)

0001

0010

0100

1000

10000

0000 0001 0010 0100 1000 10000 100000Log time (h)

BHPPENDAPDENO

PDENGPDENW

412mD

209mD

Case c derivatives for conventional and numerical density models

4842mD

k =1627qB120583

mh

logdp998400

Figure 8 Derivative and 119870 estimation for scenario (c)

Table 3 119870 estimates for new approach versus conventionalapproach for scenario (b)

Parameters Numerical density119896 (mD)

Equivalentℎ (ft)

BHP 500PDENA 500PDENG 121 50PDENO 4842PDENW 500

pressures and pressures at bottom-hole for flowing andbuildup test are generated

The derivative for both scenarios declines after 3 hoursprobably due to fluid redistribution but is noisy (numericalartefact) in scenario (c) as shown in Figure 8 A good radial

Table 4 119870 estimates for new approach versus conventionalapproach for scenario (d)

Parameters Numerical Density119896 (mD) Equivalent ℎ (ft)

Scenario (d)

BHP 500 100PDENA 500PDENG 175 50PDENO 490 250PDENW 500 100

stabilisation for both the conventional method BPR and thedensity outputs PDENDWA PDENWAT PDENGAS andPDENOIL is observed Likewise as obtained in scenarios (a)and (b) a good 119870ave value of 57 and 52mD from (34) isobtained for scenarios (c) and (d) respectivelyThis is slightlyhigher than 119896 = 50mD imputed in the simulation model

For each of the fluid phase permeabilities estimates 119896value of 50mD is obtained using the conventional BPR forscenario (c) which is in line with the uniform 119896 in thesimulation model This is the same for the PDENDWA andPDENWAT Also at ℎ = 150ft (60 of sand thickness and83 of oil thickness) the estimated 119896 is still within range(119896 = 412mD)This also indicates that ℎ = 150ft or 83 of oilthickness is contributing to flowThe 119896 values for PDENGASandPDENOIL are 209 and 4842mD respectively Summaryof the result is shown in Table 4

However in scenario (d) the permeability value of500mD using the conventional BPR which is the samefor PDENDWA and PDENWAT is only achieved if ℎ =

100ft (40 of sand thickness and 56 of oil thickness)PDENGAS and PDENOIL differ giving 350mD at ℎ =

50ft and 1960mD at ℎ = 250ft respectively This indicatesthe impact of water injection on densities and pressureschanges around the well and consequently its impact on sandthickness contributing to flow The drop in derivative curvein Figure 9 depicts the impact of the injected water

In summary for scenarios (a) to (d) the 119896 value of500mD is achieved if the thickness contributing to flowranges from 50 to 150 ft Generally results indicate that in all4 scenarios investigated the heavier fluid such as water andthe weighted average pressure-density equivalent of all fluidgive exact effective 119896 as the BPR likewise the density outputsPDENDWA and PDENWAT Also results from Table 5 showthat the empirical model from (34) with all fluid phasepermeabilities gives an average effective 119896ave of 47ndash57mDwhich is within that used in the simulationmodel and also thesame as the estimated conventional approach This approachprovides an estimate of the possible fluid phase permeabilitiesand theof each phase contribution to flow hence at severalpoints the relative 119896 can be generated as shown in Table 5

Example 2 To capture the influence of highly compressiblefluid on estimated fluid phase permeabilities an example ongas condensate reservoir (volatile system) was tested Table 6presents a summary of the well and reservoir synthetic dataused for the buildup and drawdown simulated scenarios with

Journal of Petroleum Engineering 9

Table 5 Comparison of 119870 estimates between conventional and numerical density parameters

Scenario Numerical density method Calculated119870 (mD)

Conventional119896 (mD)

Simulationmodel 119870 (mD) Relative 119870

phasecontribution to flowPhases 119870 (mD)

(a)Gas 158 003 30Oil 4037 473 508 500 086 860

Water 508 011 110

(b)Gas 12 003 20Oil 4844 432 500 500 113 890

Water 50 012 90

(c)Gas 209 004 40Oil 4842 570 500 500 086 87

Water 50 009 90

(d)Gas 175 003 30Oil 490 523 500 500 095 880

Water 50 010 90

1000

Case d derivatives for conventional and numerical density models

0010

0100

10000

BHPPENDAPDENO

PDENGPDENW

0100 1000 10000 100000Log time (h)

k =1627qB120583

mh

209mD

500mD

4842mD

logdp998400

Figure 9 Derivative and 119870 estimation for scenario (d)

additional information given below It is required to generatethe pressure equivalent and derivative for each fluid phasecompare their diagnostic signatures and also determine thephases permeabilities and average reservoir permeability

Assumption (i) Condensate reservoir is completed with onewell

(ii) LGR is imposed around the well and far across toaccount for pressure and density changes

Gas condensate andwater densities around the local gridrefinement (wellbore) andWBHPwere output using the sim-ulatorrsquos keywords The following scenarios were evaluated

(a) Flowing + buildup sequence well perforated ℎ119901

=

30ft between gas condensate and water layer Netsand thickness ℎ = 150ft

Table 6 Summary of reservoir simulations data

Parameters Design valueEclipse model Black oilModel dimension 9 times 3 times 3Length by width ft by ft 1312 times 984Thickness ℎ ft 150Permeability 119870119909 by 119870119910 mD 4000 by 3000Porosity 30Well diameter ft 115Initial water saturation 119878119908

119894 60

Permeability 119870 mD 400Gas oil contact (GOC) ft 6890Oil water contact (OWC) ft 6890Initial pressure 119875

119894 psi 4495

Formation temperature 119879 ∘C 120 2000

(b) Flowing + buildup sequence well perforated ℎ119901

=

30ft inside gas condensate layer Net sand thicknessℎ = 150ft

Figure 10 shows the production and shut-in sequence for 2scenarios in Example 2

As in Example 1 the derivative in scenario (a) ofExample 2 also shows good radial flowwith no late time effectas seen in Figure 11 A negative half-slope fingerprint is seen at03 to 30 hrs of buildup and a good stabilisation from 40 hrsin the model parameters

The derivatives for all output parameters display the samewell and reservoir fingerprint (2 flow periods early to middletime response) but with different 1198891199011015840 stabilisation

Permeability value of 3707mD was obtained from theconventionalmethod BPRwhere 119896 = 1637119902119879119898ℎ is the samefor the PDENDWA and PDENWAT and 119898 is obtained fromthe specialised plot This is in line with the uniform 119896 (119896119909 =

400mD and 119896119910 = 300mD) imputed in the simulationmodel

10 Journal of Petroleum Engineering

00

5000000

10000000

15000000

20000000

25000000

0 40 80 120 160

Prod

uctio

n ra

te (s

cfd

)

Duration

Test design sequence

Cases a to b

Figure 10 Production test sequence (buildup) for Example 2

10000

Case a derivatives for conventional and numerical density models

00100

01000

BHPPENDAPDENO

PDENGPDENW

1422mD

3707mD

35747mD

0000 0001 0010 0100 1000 10000

100000

100000Log time (h)

k =1627qB120583

mh

logdp998400

Figure 11 Derivative and 119870 estimation for scenario (a)

However PDENGAS and PDENOIL differ giving 350mDand 35747mD respectively if ℎ = 148ft

Also the derivative in Figure 12 shows good radial flowand a negative half-slope fingerprint is seen at 05 to 20 hrsof buildup and a good stabilisation from 30 hrs in the modelparameters After 3 hrs of shut-in somenoisy data (numericalartefact) is seen in all the fluid phase derivatives but still goodenough to identify 119889119901

1015840 stabilisationFor each of the fluid phase permeabilities 119896 value of

3409mD at ℎ = 148ft is obtained using BPR which is thesame for the PDENDWA and PDENWAT PDENGAS andPDENOIL differ giving 1490 and 35141mD at ℎ = 148ftrespectively

For scenarios (a) and (b) a good 119896ave value of 4050and 4030mD is obtained respectively from the empiricalmodel (34) integrating all fluid phase permeabilities whichis within that used in the simulationmodel and the estimated

0001000

0010000

0100000

1000000

10000000

0000 0001 0010 0100 1000 10000 100000

Case b derivatives for conventional and numerical density models

BHPPENDAPDENO

PDENGPDENW

Log time (h)

k =1627qB120583

mh

1490mD

3410mD

35140mD

logdp998400

Figure 12 Derivative and119870 estimation for scenario (b)

Table 7 Summary of reservoir modelling properties imputed ineclipse model for case study 2

Parameters Numerical density119896 (mD)

Equivalentℎ (ft)

Scenario(a)

BHP 371PDENA 371PDENG 142 148PDENO 3575PDENW 371

Scenario(b)

BHP 341PDENA 341PDENG 149 148PDENO 3514PDENW 341

conventional approach From the 2 scenarios investigated ithas been demonstrated that the heavier fluid such as waterand the weighted average pressure-density equivalent of allfluid give exact effective 119896 as the conventional method BPRA summary of the result is shown in Table 7

This approach estimates the fluid phase permeabilitiesand the of each phase contribution to flow at a given pointhence at several points the relative 119896 can be generated asshown in Table 8

5 Density Related Radial Flow EquationDerivation for Each Fluid Phase

Radial diffusivity equation is given as

1119903

120597120588

120597119903[119903

120597120588

120597119903] =

120601120583119888

119896

120597120588

120597119905 (35)

Journal of Petroleum Engineering 11

Table 8 Comparison of 119870 estimates between conventional and numerical density parameters

Scenario Numerical density method Calculated119870 (mD)

Conventional119896 (mD)

Simulationmodel 119870 (mD) Relative 119870

phasecontribution to flowPhases 119870 (mD)

(a)Gas 1422 004 30Oil 35747 4046 3707 400 089 870

Water 3707 009 90

(b)Gas 149 004 40Oil 35141 4038 3409 400 089 880

Water 3409 008 90

For oil

1119903

120597120588

120597119903[119903

120597120588119900

120597119903] =

120601120583119900119888

119896119900

120597120588119900

120597119905 (36a)

For water phase

1119903

120597120588

120597119903[119903

120597120588119908

120597119903] =

120601120583119908119888

119896119908

120597120588119908

120597119905 (36b)

For gas phase

1119903

120597120588

120597119903[119903

120597120588119892

120597119903] =

120601120583119892119888

119896119892

120597120588119892

120597119905 (36c)

51 From Equation for Slight and Small Compressibility Suchas Water and Oil From (22)

119875 = 119875119900minus

120588120588119900+ 1

119888119900

(37)

Differentiating with respect to 120588

120597119875

120597120588= minus

1119888119900120588119900

120597119875 = minus120597120588

119888119900120588119900

(38)

From Darcy flow equation

119902 = minus2120587119896ℎ120583

119903120597120588

120597119903 (39)

Substitute for 120597119875

119902 =2120587119896ℎ120583

119903

119888119900120588119900

120597120588

120597119903 (40)

To derive the analytical density transient equation for eachphase the following assumptionconditions are applicable

For oil phase from (35)

1119903

120597120588

120597119903[119903

120597120588

120597119903] =

120601120583119888

119896

120597120588

120597119905 (41)

Initial condition is as follows

120588 (119903 119905 = 0) = 120588119894 (42a)

BC at the wellbore is as follows

lim119903rarr 0

2120587119896ℎ120583

119903

119888119900120588119900

120597120588

120597119903= 119876 (42b)

BC at infirmity is as follows

lim119903rarrinfin

120588 (119903 119905) = 120588119894 (42c)

Using the Boltzmann transformation consider the followingAssuming 120578 = 120601120583119888119903

2119896119905

120597120588

120597119903=

120597120588

120597120578

120597120578

120597119903=2120601120583119888119903119896119905

120597120588

120597120578=

1206011205831198881199032

119896119905

2119903

120597120588

120597120578=2120578119903

120597120588

120597120578 (43)

Therefore1119903

120597

120597119903[119903

120597120588

120597119903] =

21205781199032

120597

120597120578[2120578

120597120588

120597120578] =

41205781199032

120597

120597120578[120578

120597120588

120597120578] (44)

Differentiating with respect to 119903 is equal to differentiatingwith respect to 120578 multiply by 2120578119903

From (35) the LHS is resolved as follows

120597120588

120597119905=

120597120588

120597120578

120597120578

120597119905=

120601120583119888119903

1198961199052120597120588

120597120578= minus

120578

119905

120597120588

120597120578 (45)

120601120583119888

119896

120597120588

120597119905=

minus120601120583119888

119896

120578

119905

120597120588

120597119905=

minus1206011205831198881199032

119896119905

120578

1199032120597120588

120597120578=

minus1205782

1199032120597120588

120597120578 (46)

Equating (44) and (46) we have

120597

120597120578[120578

120597120588

120597120578] =

minus120578

4120597120588

120597120578 (47)

This is the simplified ordinary differential equation of 120588 as afunction of 120578

Apply boundary conditions as followsInitial condition is as follows

lim119903rarrinfin

120588 (120578) = 120588119894 (48a)

Substituting (43) into (40)

lim119903rarr 0

2120587119896ℎ120583

119903

119888119900120588119900

120597120588

120597119903= 119876 (48b)

lim119903rarr 0

[120578120597120588

120597120578] =

120601120583119888119900120588119900

4120587119896ℎ (49)

Assume that119898 = 120578(120597120588120597120578)

12 Journal of Petroleum Engineering

From (47)

120597119898

120597120578=

minus119898

4 (50)

Integrating from 120578 = 0 to 120578

997904rArr int

119898(120578)

119898(0)

120597119898

120597120578= minusint

120578

0

120597120578

4

997904rArr ln[119898 (120578)

119898 (0)] =

minus120578

4

997904rArr 119898(120578) = 119898 (0) 119890minus1205784

(51)

Recall boundary condition

lim119903rarr 0

[120578120597120588

120597120578] =

119876120583119888119900120588119900

4120587119896ℎ

119898 (0) =119876120583119888119900120588119900

4120587119896ℎ

119898 (120578) =119876120583119888119900120588119900

4120587119896ℎ119890minus1205784

(52)

Recall that119898 = 120578(120597120588120597120578)Therefore

120597120588 (120578)

120597120578=

120583119876119888119900120588119900

4120587119896ℎ119890minus1205784

120578 (53)

Integrating from 120578 = infin where 120588 = 120588119894

int

120588(120578)

120588119894

120597120588 = minusint

120578

infin

120583119876119888119900120588119900

4120587119896ℎ119890minus1205784

120578120597120578

120588 (120578) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎminusint

infin

120578

119890minus1205784

120578120597120578

(54)

where

120578 =120601120583119888119903

2

119896119905 (55)

Assuming 119906 = 1205784 then 120597120578120578 = 120597119906119906 and 119906 = 12060112058311988811990324119896119905

Hence

120588 = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎminusint

infin

12060112058311988811990324119896119905

119890minus1205784

120578120597120578 (56)

where

minus119864119894 (minus119909) = int

infin

119909

119890minus119906

119906120597119906 (57)

Known as the exponential integral function defined by [13]

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎ119864119894 (minus119909) (58)

where 119909 = 12060112058311988811990324119896119905 and

minus119864119894 (minus119909) = int

infin

119909

119890minus119906

119906120597119906 (59)

At large times 119909 will be small applying Taylorrsquos series andintegrating

minus119864119894 (minus119909) = ln (120574)

ln (120574) = ln (1781) = 05772(60)

where 120574 is known as Eulerrsquos numberFor 14119909 gt 25 consider the followingThen

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎ[ln119909+ ln 120574]

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎln[

1206011205831198881199032120574

4119896119905]

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎln [

22461198961199051206011205831198881199032

]

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎ[ln [

22461198961199051206011205831198881199032

]+ 080907]

(61)

Plotting 120588119908(119905) versus ln(119905) will yield a straight line at longer

time and the slope of the line is given as120597120588119908

120597 ln 119905= 119898oil =

120583119876119888119900120588119900

4120587119896ℎ (62)

Therefore

119896ℎ =120583119876119888119900120588119900

4120587119898oil (63)

52 For Water Phase The radial density equation is given as

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎ[ln [

22461198961199051206011205831198881199032

]+ 080907] (64)

Plotting 120588119908(119905) versus ln(119905) will yield a straight line at longer

time and the slope of the line is given as120597120588119908

120597 ln 119905= 119898water =

120583119876119888119900120588119900

4120587119896ℎ (65)

where

119896ℎ =120583119876119888119900120588119900

4120587119898water (66)

53 For Gas Phase From equation for compressible fluidsuch as gas consider the following

From (32)

119875 =119875119900120588 minus 119875119900

2120588119900119888119892

120588119900(1 minus 119875

119900119888119892)

120597119875

120597120588=

119875119900

120588119900(1 minus 119875

119900119888119892)

120597119875 =119875119900

120588119900(1 minus 119875

119900119888119892)120597120588

(67)

Journal of Petroleum Engineering 13

From the fundamental

119899 =119875119881

119911119877119879 (68)

At standard condition

119875119881

119911119879=

119875sc119881sc119879sc

119875 (5615119902)119911119879

=119875sc119876sc119879sc

119902 =119875sc119879sc

(119911119879

119875)

119876sc5615

119902 =119875sc119879sc

119876sc5615

(119911119879

119875) =

2120587119896ℎ120583

119903120597119901

120597119903

119902 =119875sc119879sc

119876sc5615

(119911119879) =2120587119896ℎ120583

119903119875120597119901

120597119903

(69)

where

120597119875 =119875119900

120588119900(1 minus 119875

119900119888119892)120597120588

120572 =119875119900

120588119900(1 minus 119875

119900119888119892)

120597119875 = 120572120597120588

119875 = 120588119877119911119879

(70)

Substitute into the equation where

119898(120588119894) minus119898 (120588

119908119891) =

2120583

int

120588119894

120588119908119891

120588120597120588

119898 (120588119894) minus119898 (120588

119908119891) =

120588119894

2minus 120588119908119891

2

120583

120588 = radic120588119894

2+ 120588119908119891

2

2

(71)

From the diffusivity equation for gas

1119903

120597120588

120597119903[119903

120597120588119892

120597119903] =

120601120583119892119888

119896119892

120597120588119892

120597119905 (72)

Initial condition is as follows

120588 (119903 119905 = 0) = 120588119894 (73)

BC at the wellbore is as follows

lim119903rarr 0

4120587119896ℎ120572119877120583

(5615119879sc

119875sc) 119903120588

120597120588

120597119903= 119876 (74)

BC at infirmity is as follows

lim119903rarrinfin

120588 (119903 119905) = 120588119894 (75)

where119877119875sc and119879sc are known as temperature pressure andgas constant at standard condition

Therefore integrating from 120578 = infin where 120588 = 120588119894

int

120588(120578)

120588119894

120588120597120588 = minusint

120578

infin

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)119890minus1205784

120578120597120578

1205882(120578) = 120588

119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

minusint

infin

120578

119890minus1205784

120578120597120578

(76)

where

120578 =120601120583119888119903

2

119896119905 (77)

Assume that 119906 = (1205784) (120597120578120578) = 120597119906119906 and 119906 = 12060112058311988811990324119896119905

Hence

1205882= 120588119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)minusint

infin

12060112058311988811990324119896119905

119890minus1205784

120578 (78)

where minus119864119894(minus119909) = intinfin

119909(119890minus119906

119906)120597119906Known as the exponential integral function defined by

[13]

1205882= 120588119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)minus119864119894 (minus119909) (79)

where 119909 = 12060112058311988811990324119896119905 and minus119864119894(minus119909) = int

infin

119909(119890minus119906

119906)120597119906At large times 119909will be small applying Taylorrsquos series and

integrating

minus119864119894 (minus119909) = ln (120574)

ln (120574) = ln (1781) = 05772(80)

where 120574 is known as Eulerrsquos numberFor 14119909 gt 25 consider the followingThen

1205882(119903 119905) = 120588

119894

2+

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) [ln119909+ ln 120574] (81)

1205882(119903 119905) = 120588

119894

2+

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) ln[120601120583119888119903

2120574

4119896119905]

1205882(119903 119905) = 120588

119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) ln [22461198961199051206011205831198881199032

]

119898 (120588119894) minus119898 (120588

119908119891) =

2120583

int

120588119894

120588119908119891

120588 120597120588

119898 (120588119908119891

) = 119898 (120588119894) minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

sdot [ln [22461198961199051206011205831198881199032

]+ 080907]

120597119898 (120588119908)

120597 ln 119905= 119898gas =

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

(82)

14 Journal of Petroleum Engineering

4160041650417004175041800418504190041950420004205042100

560856095609561056105611561156125612

ODENGDEN

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

1205880

(Ibm

ft)

120597120588w120597lnt

= moil =1627120583Qc01205880

kh= minus0009

120597m(120588w)

120597lnt= mgas =

1637120583Q

kh120572R= minus12195

120588g2

(Ibm

ft)

Figure 13 Oil and gas phase density versus log Horner time

6279

6279

6279

6279

6279

6279

6279

6279

01 1 10 100 1000 10000 1000001000000Horner time (h)

Density versus log Horner time

WDEN

120588w

(Ibm

ft)

120597120588w120597lnt

= mwater =1627120583Qc01205880

kh= minus00003

Figure 14 Water phase density versus log Horner time

Plotting 120588119908

2(119905) or 119898(120588

119908) versus ln(119905) will yield a straight line

at longer time and the slope of the line is given as follows

119896ℎ =120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)1

119898gas (83)

Example 3 Data from Example 1(a) was used for this caseand the semilog plot of density for each phase (oil water andgas) is plotted against log of Horner timelowastlowast

119876 used for 119896 calculation is the average rate for allflowing periods

From Figures 13 and 14 the calculated slopes of the radialflow for oil and water phases are given as

120597120588119908

120597 ln 119905= 119898oil =

1627120583119876119888119900120588119900

119896ℎ= minus 0009

120597120588119908

120597 ln 119905= 119898water =

1627120583119876119888119900120588119900

119896ℎ= minus 00003

(84)

The estimated phase permeabilities are

119896oil =1627120583119876119888

119900120588119900

ℎ119898oil= 445mD

119896water =1627120583119876119888

119900120588119900

ℎ119898water= 50mD

(85)

Also for the gas phase consider the following

ODENGDEN

422004225042300423504240042450425004255042600

5613561356145614561556155616

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

1205880

(Ibm

ft)

120597120588w120597lnt

= moil =1627120583Qc01205880

kh= minus0009

120597m(120588w)

120597lnt= mgas =

1637120583Q

kh120572R= minus12502

120588g2

(Ibm

ft)

Figure 15 Oil and gas phase density versus log Horner time

WDEN

6279627962796279627962796280628062806280

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

120588w

(Ibm

ft)

120597120588w120597lnt

= mwater =1627120583Qc01205880

kh= minus00003

Figure 16 Water phase density versus log Horner time

The calculated slope from the semilog plot shown inFigure 13 is given as

120597119898 (120588119908)

120597 ln 119905= 119898gas =

1637120583119876119896ℎ120572119877

= minus 12195 (86)

And estimated gas permeability is

119896gas =1637120583119876

119900

ℎ120572119877119898gas= 170mD (87)

Using the empirical model for all three phases the averageestimated permeability is as follows

119896ave =4radic50 times 445 times 1702

119896ave = 50mD

(88)

Example 4 Data from Example 1(c) was used for this caseand the semilog plot of density for each phase (oil water andgas) is plotted against log of Horner time

From Figures 15 and 16 the calculated slopes of the radialflow for oil and water phases are given as

120597120588119908

120597 ln 119905= 119898oil =

1627120583119876119888119900120588119900

119896ℎ= minus 0009

120597120588119908

120597 ln 119905= 119898water =

1627120583119876119888119900120588119900

119896ℎ= minus 00003

(89)

Journal of Petroleum Engineering 15

The estimated phase permeabilities are

119896oil =1627120583119876119888

119900120588119900

ℎ119898oil= 4352mD

119896water =1627120583119876119888

119900120588119900

ℎ119898water= 49mD

(90)

Also for the gas phase consider the following

The calculated slope from the semilog plot shown inFigure 15 is given as

120597119898 (120588119908)

120597 ln 119905= 119898gas =

1637120583119876119896ℎ120572119877

= minus 12502 (91)

And estimated gas permeability is

119896gas =1637120583119876

119900

ℎ120572119877119898gas= 179mD (92)

Using the empirical model for all three phases the averageestimated permeability is as follows

119896ave =4radic49 times 4352 times 1792

119896ave = 51mD

(93)

6 Conclusion

The following inferences were drawn from the six scenariosreviewed

(i) Results from the three-phase fluid empirical modeldeveloped for average effective 119896ave estimation rangebetween 47 and 57mD for Example 1 and give404mD for Example 2 which is within that used inthe simulation model and also estimated from BPR

(ii) The BPR PDENDWA and PDENWAT give the samestabilisation and the same 119896 estimates making it suit-able for interpretation of pressure transient analysis

(iii) It has been demonstrated in all 6 scenarios inves-tigated that the heavier fluid such as water andthe weighted average pressure-density equivalent ofall fluid give exact effective 119896 as the conventionalmethod

(iv) For oil reservoir system (scenarios (a) to (d)) 119896 =

500mD is achieved if the thickness contributing toflow ranges from 50 to 150 ft while for gas condensatereservoir (scenarios (a) to (b)) 119896 ranges between340 and 371mD which is achieved if the net sandthickness ℎ = 148ft is contributing to flow

(v) This approach also provides an estimate of the pos-sible fluid phase permeabilities and the of eachphase contribution to flow at a given point hence atseveral 1198891199011015840 stabilisation points the relative 119896 can begenerated

(vi) The derivatives for all output parameters display thesame wellbore and reservoir fingerprint as the BPRmethod

(vii) Generally where output parameters (BPR PDE-NOIL PDENGAS PDENWAT and PDENDW)depict different fingerprint the density derivative willserve as support to distinguish between reservoir andnonreservoir response

Nomenclature

119875 Pressure psi119879 Temperature oF119903 Radius ft119896 Permeability mDOslash Porosity fraction120583 Viscosity cp119905 Time hrs119902 Production rate bblday119861 Formation volume factor rbStb119862119905 Total compressibility psiminus1119903119908 Wellbore radius ftΔ119901 Change in pressure psiaℎ Formation thickness ft119860 Drainage area acres119875119908119891 Bottom-hole flowing pressure psi119875119894 Initial pressure psi

119885 Difference between two pointtime series119894 Subscript of an observed variable119888 Subscript of a calculated variableSTEYX SSE of data point119899 Number of data pointsCov Covariance of data point120575 Standard deviation119905119901 Cumulative production time119862119904 Wellbore storage constant119909 Mean of data point

Abbreviations

LBPR Local grid bottom-hole pressureLDENO Local grid oil densityLDENW Local grid water densityLDENG Local grid gas densityWBHP Well bottom-hole pressureBPR Well bottom-hole pressurePDENOIL Pressure equivalent of LDENOPDENGAS Pressure equivalent of LDENGPDENWAT Pressure equivalent of LDENWPDENDWA Pressure equivalent of density weighted

average (LDENO LDENG and LDENW)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Jalali S D Mohaghegh and R Gaskari ldquoIdentifying infilllocations and underperformer wells in mature fields usingmonthly production rate data Carthage Field Cotton Valley

16 Journal of Petroleum Engineering

Formation Texasrdquo in Proceedings of the SPE Eastern RegionalMeeting SPE Paper 104550 Canton Ohio USA October 2006

[2] D Tiab and A Kumar ldquoApplication of PD function to inter-ference analysisrdquo in Proceedings of the 51st Annual Fall Meetingof the Society of Petroleum Engineers of AIME Paper SPE 6053New Orleans La USA 1976

[3] D Tiab and A Kumar ldquoDetection and location of two parallelsealing fault around a wellrdquo in Proceedings of the 51st AnnualFall Meeting of the Society of Petroleum Engineers of AIME SPEPaper 6056 New Orleans La USA 1976

[4] D Bourdet T M Whittle A A Douglas and Y M Pirard ldquoAnew set of type curves simplifies well test analysisrdquo World Oilvol 196 no 6 pp 95ndash106 1983

[5] S Y Zheng ldquoFighting against non-unique solution problemsin heterogeneous reservoirs through numerical well testingrdquo inProceedings of the SPE Asia Pacific Oil and Gas Conference andExhibition SPE 100951 Adelaide Australia September 2006

[6] R N Horne and K O Temeng ldquoRecognition and location ofpinch-out boundaries by pressure transient analysisrdquo SPE 9905Society of Petroleum Engineers 1981

[7] L C Ayestaran R D Nurmi G A K Shehab and W S ElsildquoWelltest design and interpretation improved by integrated welltesting and geological effortsrdquo SPE 17945 1989

[8] G J Massonnat and D Bandiziol ldquoInterdependence betweengeology and well test interpretationrdquo SPE 22740 1991

[9] K F Du and G Stewart ldquoReservoir description from well testinterpretationrdquo in North Sea Oil and Gas ReservoirmdashIII pp339ndash356 Norwegian Institute of Technology (NTH) 1994

[10] S Y Zheng Well testing and characterisation of meanderingfluvial channel reservoirs [PhD thesis] Heriot-Watt University1997

[11] V T Biu and S Y Zheng ldquoA new approach in pressure transientanalysis part I improved diagnosis of flow regimes in oil and gaswellsrdquo in Proceedings of the 3rd EAGEAAPGWorkshop on TightReservoirs in theMiddle East AbuDhabi UnitedArab Emirates2015

[12] R N Horne Modern Welltest Analysis Petroway Palo AltoCalif USA 1995

[13] C S Matthews and D G Russell Pressure Buildup and FlowTests in Wells vol 1 ofMonograph Series SPE Richardson TexUSA 1967

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4 Journal of Petroleum Engineering

For small compressibility fluid such as oil and water either(19) or (22) is used to generate the pressure equivalent fromwell fluid density obtained from reservoir simulation or PDGtool This pressure is then analyzed in any available well testsoftwares

32 For Compressible Fluid in Isothermal Conditions Con-sider the following

119862119892= minus

1V[120597V120597119901

]

119879

(23)

For real gas equation of state

V =119899119877119879119911

119875 (24)

Differentiating with respect to pressure at constant tempera-ture

(120597V120597119901

)

119879

= 119899119877119879[1119901

(120597119911

120597119901)minus

119911

1199012 ] (25)

Substituting (23) into (25)

119862119892=

1119901

minus1119911[119889119911

119889119901] (26)

In terms of density

119862119892=

1119901

minus1120588

[120597120588

120597119901] (27)

This equation is applicable for real gas conditionFor compressible fluid

119862119892=

1119901

minus1120588

[120597120588

120597119901]

1120588

[120597120588

120597119901] =

1119901

minus119862119892

int

120588

1205880

120597120588

120588= int

119901

1199010

[120597119901

119901minus 120597119901119862

119892]

ln [120588

1205880] = ln [

119901

1199010]minus [119901minus1199010] 119862119892

(28)

Applying the power series for ln119901

ln [119901] = [119901 minus 1] minus[119901 minus 1]2

2+ sdot sdot sdot +

[minus1]119899 [119901 minus 1]119899

119899

+ sdot sdot sdot 0 lt 119901 le 2(29)

Limit ln119909 to the 1st term only

[120588

1205880]minus 1 = [

119901

119901119900

]minus 1minus [119901minus1199010] 119862119892

[120588

1205880] = [

119901

119901119900

]minus [119901minus1199010] 119862119892

(30)

[120588

1205880] =

119901 minus 119901120588119900119888119892minus 119901119900

2119888119892

119901119900

119901119900120588

120588119900

minus119901119900

2119888119892= 119901 [1minus119901

119900119888119892]

(31)

119875 =119875119900120588 minus 119875119900

2120588119900119888119892

120588119900(1 minus 119875

119900119888119892) (32)

For compressible fluids such as gas (32) is used to generatethe pressure equivalent from the fluid density obtained fromreservoir simulation or PDG tool at the well and then thepressure is analyzed in any available well test softwares

Also outside the available welltest software the pressurederivative can be generated from the pressure equivalentobtained from the oil gas and water densities at the well byapplying (1) formulated by Horne (1995) [12] or (2) to (4) byBiu and Zheng (2015) [11]

4 Density Weighted Average (DWA)

While (19) or (22) and (32) give the pressure equivalentfor independent fluid phases such as gas oil and waterthe weighted average method is used to obtain the densityequivalent for a two- or three-phase combinationThe equiv-alent pressure derived from the density for all three fluidcomponents such as gas oil and water is given as

119875119894=

120588119892119901119892+ 120588119900119901119900+ 120588119908119901119908

120588119892+ 120588119900+ 120588119908

(33)

Equation (33) comprises all the fluid phases in the system andas such will be comparable to the conventional bottom-holepressure measurement during the derivative analysis

41 Empirical Model Correlation between Fluid Phases119870 Anempirical model integrating the fluid phasersquos permeabilitiesfor a given system is formulated to determine the averagereservoir permeability The mathematical model is given as

119896ave = 4radic1198961199001198961199081198962119892 (34)

where 119896119900= oil phase permeability 119896

119892= gas phase permeabil-

ity and 119896119908= water phase permeability

With the estimation of the phasersquos permeabilities it istherefore possible to estimate the possible relative permeabil-ity to each phase and the percentage contribution to flow byeach phase at one point analysis hence at several points therelative 119896 can be generated

To illustrate applicability of this approach 6 scenarios inconventional oil reservoir and gas condensate reservoir are

Journal of Petroleum Engineering 5

Local grid refinement

Pressure (psia) Pressure (barsa)

PROD1

PROD1

35591 35630 35669 35708 35747 3093430899 30970 31006 31041

Figure 2 Simulation model for gas cap + oil + water reservoir and gas condensate + water showing local grid refinement around the well

Table 1 Reservoir and fluid data for Example 1

Parameters Design valueEclipse model Black oilModel dimension 10 times 5 times 5Length by width ft by ft 500 times 400Thickness ℎ ft 250Permeability 119870119909 by 119870119910 mD 500 by 500Porosity 20Well diameter ft 060Initial water saturation 119878119908

119894 22

Permeability 119870 mD 50Gas oil contact (GOC) ft 8820Oil water contact (OWC) ft 90000Initial pressure 119875

119894 psia 40000

Formation temperature 119879 oF 2000

investigated using numerical model built with a commercialsimulator Local grid refinement (LGR) is imposed aroundthe well to capture sharp change in fluid densities as shownin Figure 2

The simulation software keywords LBPR LDENOLDENW LDENG AND WBHP were outputs to obtain thedensity and pressure change around the well and as far as theperturbation could extend

Example 1 Table 1 presents a summary of the well andreservoir synthetic data used for the buildup and drawdownsimulated scenarios with additional information given belowIt is required to generate the pressure equivalent and deriva-tive for each phase compare their diagnostic signaturesand also determine the phases permeabilities and averagereservoir permeability

Assumption (i) Oil reservoir + gas cap is completed with onewell

(ii) LGR is imposed around the well and far across toaccount for pressure and density changes

Gas oil and water densities around the local grid refine-ment (wellbore) andWBHP were output using the simulatorkeywords The following scenarios were evaluated

(a) Flowing+ buildup sequence well perforated ℎ119901= 30 ft

between oil and water layer Net sand thickness ℎ =250 ft

(b) Flowing+buildup sequence well perforated ℎ119901= 30 ft

inside the oil layer Net sand thickness ℎ = 250 ft(c) Flowing+ buildup sequence well perforated ℎ

119901= 30 ft

between gas and oil layer Net sand thickness ℎ =250 ft

(d) Falloff test flowing + buildup sequence well perfo-rated ℎ

119901= 30 ft inside the oil layer Net sand thickness

ℎ = 250 ftThe schematic of the 4 scenarios is depicted in Figures 3and 4 Figure 5 shows the production shut-in and injectionsequence for the 4 scenarios in Example 1

In scenario (a) the well is completed between the oil andwater layer to mimic multiphase conditions at the wellboreand some distance away from the well and also see the effecton pressure distribution fluid densities changes around thewellbore and estimate fluid phase permeabilities 119896 using thespecialised plot and 119896ave from the empirical model for three-phase conditions

The derivative in Figure 6 shows a good radial flow butdrop in derivative at late time due to constant pressuresupport (likely aquifer support) A continuous drop is seenfrom 10 hrs in the model parameters The derivatives forall output parameters display the same well and reservoirsignatures (3 flow periods early to late time response)but with different 119889119901

1015840 stabilisation The well bottom-holepressure (BPRrarrBHP) response shows good overlay withpressure equivalent of density weighted average (PDENDWArarr PDENA) and pressure equivalent of water density (PDEN-WAT rarr PDENW) while pressure equivalent of gas den-sity (PDENGAS rarr PDENG) and pressure equivalent ofoil density (PDENOIL rarr PDENO) differ completely ThePDENDWA gives a better fingerprint that is less noisy

A permeability value of 508mD is estimated from thebottom-hole pressure BPR where 119896 = 1627119902119861120583119898ℎ and119898 is

6 Journal of Petroleum Engineering

Gas

Oil

Water

Case 1 production test

Net sand thickness h = 250ftwater layerWell perforated hp = 30ft between oil andFlowing + buildup sequence

(a)

Gas

Oil

Water

Case 2 production test Flowing + buildup sequence

Net sand thickness h = 250ftWell perforated hp = 30ft inside the oil layer

(b)

Gas

Oil

Water

Case 3 production test Flowing + buildup sequence

Net sand thickness h = 250ftoil layerWell perforated hp = 30ft between gas and

(c)

Gas

Oil

Water

Case 4 falloff test Flowing + buildup sequence

Net sand thickness h = 250ftwater layerWell perforated hp = 30ft between oil and

(d)

Figure 3 Schematics of well perforation interval and sand thickness for oil + gas cap + water reservoir

obtained from the specialised plot This is an approximate ofthe input value in the simulation model Also the simulatoroutputs PDENDWA and PDENWAT in the simulation givethe same 119896 value while PDENGAS and PDENOIL differgiving 158 and 4037mD respectively At ℎ = 50 f t the best119896 estimate is obtained depicting ℎ = 50 f t as the thicknesscontributing to flowAt ℎ gt 50ft 119896 drops below the 119896 imputedin the model

Using (34) 119896ave = 473mD is obtained which isapproximately close to that of BPR hence a good estimateof the fluid phase permeabilities A summary of the result isshown in Tables 2 and 5

In scenario (b) thewell is completedwithin the oil sectionto capture the pressure and fluid densities changes around thewell The derivative in Figure 7 shows a good radial flow butwith noisy numerical artefact It is likely that the boundaryresponse is masked by numerical artefact Also a continuousdrop is seen after 10 hrs in the derivative

Using the estimated fluid phase permeabilities 119896ave from(34) is 472mD as shown in Table 3 which is the same asscenario (a) This is in line with the uniform 119896 used in thesimulation model Also the BPR gives a permeability value of500mD which is the same for PDENDWA and PDENWATbut differs with PDENGAS and PDENOIL that give 121 and

Journal of Petroleum Engineering 7

Water

Case 1 production test

Gas + condensate

Flowing + buildup sequence

condensate and water layerNet sand thickness h = 45m

Well perforated hp = 90m between gas

(a)

Water

Case 2 production test

Gas + condensate

Net sand thickness h = 45m

Flowing + buildup sequence

condensate layerWell perforated hp = 90m inside gas

(b)

Figure 4 Schematics of well perforation interval and sand thickness for gas condensate + water reservoir

00

5000

0 50 100 150 200 250 300 350

Prod

uctio

nin

ject

ion

rate

bop

d

Duration

Test design sequence

Case dCases a to c

minus25000

minus20000

minus15000

minus10000

minus5000

Figure 5 Production and injection test sequence for scenarios (a)to (d) (buildup and falloff)

Table 2 119870 estimates for new approach versus conventionalapproach for scenario (a)

Parameters Numerical density119896 (mD)

Equivalentℎ (ft)

BHP 508PDENA 508PDENG 158 50PDENO 4037PDENW 508

4844mD respectively At ℎ = 150ft (60 of sand thicknessand 83 of oil thickness) the estimated 119896 is still within range

00

00

01

10

100

0000 0001 0010 0100 1000 10000 100000Log time (h)

BHPPENDAPDENO

PDENGPDENW

508mD

4037mD

158mD

k =1627qB120583

mh

Case a derivatives for conventional and numerical density models

logdp998400

Figure 6 Derivative and 119870 estimation for scenario (a)

(119896 = 489mD)This indicates ℎ = 50ft or 83 of oil thicknesscontributing to flow

To test this approach in the gas column the well wascompleted in between the gas and oil layer which is consid-ered as scenario (c) and scenario (d) with the well completedwithin the oil layer but with water injection after flowingand shut-in sequence In both scenarios the multiphasefluid distribution is triggered at the wellbore in order tocapture the density changes for each phase and calculate fluidphase permeabilities First the well fluid densities equivalent

8 Journal of Petroleum Engineering

Case b derivatives for conventional and numerical density models

00

01

10

100

0000 0001 0010 0100 1000 10000 100000Log time (h)

BHPPENDAPDENO

PDENGPDENW

4842mDk =1627qB120583

mh

120mD

489mD

logdp998400

Figure 7 Derivative and 119870 estimation for scenario (b)

0001

0010

0100

1000

10000

0000 0001 0010 0100 1000 10000 100000Log time (h)

BHPPENDAPDENO

PDENGPDENW

412mD

209mD

Case c derivatives for conventional and numerical density models

4842mD

k =1627qB120583

mh

logdp998400

Figure 8 Derivative and 119870 estimation for scenario (c)

Table 3 119870 estimates for new approach versus conventionalapproach for scenario (b)

Parameters Numerical density119896 (mD)

Equivalentℎ (ft)

BHP 500PDENA 500PDENG 121 50PDENO 4842PDENW 500

pressures and pressures at bottom-hole for flowing andbuildup test are generated

The derivative for both scenarios declines after 3 hoursprobably due to fluid redistribution but is noisy (numericalartefact) in scenario (c) as shown in Figure 8 A good radial

Table 4 119870 estimates for new approach versus conventionalapproach for scenario (d)

Parameters Numerical Density119896 (mD) Equivalent ℎ (ft)

Scenario (d)

BHP 500 100PDENA 500PDENG 175 50PDENO 490 250PDENW 500 100

stabilisation for both the conventional method BPR and thedensity outputs PDENDWA PDENWAT PDENGAS andPDENOIL is observed Likewise as obtained in scenarios (a)and (b) a good 119870ave value of 57 and 52mD from (34) isobtained for scenarios (c) and (d) respectivelyThis is slightlyhigher than 119896 = 50mD imputed in the simulation model

For each of the fluid phase permeabilities estimates 119896value of 50mD is obtained using the conventional BPR forscenario (c) which is in line with the uniform 119896 in thesimulation model This is the same for the PDENDWA andPDENWAT Also at ℎ = 150ft (60 of sand thickness and83 of oil thickness) the estimated 119896 is still within range(119896 = 412mD)This also indicates that ℎ = 150ft or 83 of oilthickness is contributing to flowThe 119896 values for PDENGASandPDENOIL are 209 and 4842mD respectively Summaryof the result is shown in Table 4

However in scenario (d) the permeability value of500mD using the conventional BPR which is the samefor PDENDWA and PDENWAT is only achieved if ℎ =

100ft (40 of sand thickness and 56 of oil thickness)PDENGAS and PDENOIL differ giving 350mD at ℎ =

50ft and 1960mD at ℎ = 250ft respectively This indicatesthe impact of water injection on densities and pressureschanges around the well and consequently its impact on sandthickness contributing to flow The drop in derivative curvein Figure 9 depicts the impact of the injected water

In summary for scenarios (a) to (d) the 119896 value of500mD is achieved if the thickness contributing to flowranges from 50 to 150 ft Generally results indicate that in all4 scenarios investigated the heavier fluid such as water andthe weighted average pressure-density equivalent of all fluidgive exact effective 119896 as the BPR likewise the density outputsPDENDWA and PDENWAT Also results from Table 5 showthat the empirical model from (34) with all fluid phasepermeabilities gives an average effective 119896ave of 47ndash57mDwhich is within that used in the simulationmodel and also thesame as the estimated conventional approach This approachprovides an estimate of the possible fluid phase permeabilitiesand theof each phase contribution to flow hence at severalpoints the relative 119896 can be generated as shown in Table 5

Example 2 To capture the influence of highly compressiblefluid on estimated fluid phase permeabilities an example ongas condensate reservoir (volatile system) was tested Table 6presents a summary of the well and reservoir synthetic dataused for the buildup and drawdown simulated scenarios with

Journal of Petroleum Engineering 9

Table 5 Comparison of 119870 estimates between conventional and numerical density parameters

Scenario Numerical density method Calculated119870 (mD)

Conventional119896 (mD)

Simulationmodel 119870 (mD) Relative 119870

phasecontribution to flowPhases 119870 (mD)

(a)Gas 158 003 30Oil 4037 473 508 500 086 860

Water 508 011 110

(b)Gas 12 003 20Oil 4844 432 500 500 113 890

Water 50 012 90

(c)Gas 209 004 40Oil 4842 570 500 500 086 87

Water 50 009 90

(d)Gas 175 003 30Oil 490 523 500 500 095 880

Water 50 010 90

1000

Case d derivatives for conventional and numerical density models

0010

0100

10000

BHPPENDAPDENO

PDENGPDENW

0100 1000 10000 100000Log time (h)

k =1627qB120583

mh

209mD

500mD

4842mD

logdp998400

Figure 9 Derivative and 119870 estimation for scenario (d)

additional information given below It is required to generatethe pressure equivalent and derivative for each fluid phasecompare their diagnostic signatures and also determine thephases permeabilities and average reservoir permeability

Assumption (i) Condensate reservoir is completed with onewell

(ii) LGR is imposed around the well and far across toaccount for pressure and density changes

Gas condensate andwater densities around the local gridrefinement (wellbore) andWBHPwere output using the sim-ulatorrsquos keywords The following scenarios were evaluated

(a) Flowing + buildup sequence well perforated ℎ119901

=

30ft between gas condensate and water layer Netsand thickness ℎ = 150ft

Table 6 Summary of reservoir simulations data

Parameters Design valueEclipse model Black oilModel dimension 9 times 3 times 3Length by width ft by ft 1312 times 984Thickness ℎ ft 150Permeability 119870119909 by 119870119910 mD 4000 by 3000Porosity 30Well diameter ft 115Initial water saturation 119878119908

119894 60

Permeability 119870 mD 400Gas oil contact (GOC) ft 6890Oil water contact (OWC) ft 6890Initial pressure 119875

119894 psi 4495

Formation temperature 119879 ∘C 120 2000

(b) Flowing + buildup sequence well perforated ℎ119901

=

30ft inside gas condensate layer Net sand thicknessℎ = 150ft

Figure 10 shows the production and shut-in sequence for 2scenarios in Example 2

As in Example 1 the derivative in scenario (a) ofExample 2 also shows good radial flowwith no late time effectas seen in Figure 11 A negative half-slope fingerprint is seen at03 to 30 hrs of buildup and a good stabilisation from 40 hrsin the model parameters

The derivatives for all output parameters display the samewell and reservoir fingerprint (2 flow periods early to middletime response) but with different 1198891199011015840 stabilisation

Permeability value of 3707mD was obtained from theconventionalmethod BPRwhere 119896 = 1637119902119879119898ℎ is the samefor the PDENDWA and PDENWAT and 119898 is obtained fromthe specialised plot This is in line with the uniform 119896 (119896119909 =

400mD and 119896119910 = 300mD) imputed in the simulationmodel

10 Journal of Petroleum Engineering

00

5000000

10000000

15000000

20000000

25000000

0 40 80 120 160

Prod

uctio

n ra

te (s

cfd

)

Duration

Test design sequence

Cases a to b

Figure 10 Production test sequence (buildup) for Example 2

10000

Case a derivatives for conventional and numerical density models

00100

01000

BHPPENDAPDENO

PDENGPDENW

1422mD

3707mD

35747mD

0000 0001 0010 0100 1000 10000

100000

100000Log time (h)

k =1627qB120583

mh

logdp998400

Figure 11 Derivative and 119870 estimation for scenario (a)

However PDENGAS and PDENOIL differ giving 350mDand 35747mD respectively if ℎ = 148ft

Also the derivative in Figure 12 shows good radial flowand a negative half-slope fingerprint is seen at 05 to 20 hrsof buildup and a good stabilisation from 30 hrs in the modelparameters After 3 hrs of shut-in somenoisy data (numericalartefact) is seen in all the fluid phase derivatives but still goodenough to identify 119889119901

1015840 stabilisationFor each of the fluid phase permeabilities 119896 value of

3409mD at ℎ = 148ft is obtained using BPR which is thesame for the PDENDWA and PDENWAT PDENGAS andPDENOIL differ giving 1490 and 35141mD at ℎ = 148ftrespectively

For scenarios (a) and (b) a good 119896ave value of 4050and 4030mD is obtained respectively from the empiricalmodel (34) integrating all fluid phase permeabilities whichis within that used in the simulationmodel and the estimated

0001000

0010000

0100000

1000000

10000000

0000 0001 0010 0100 1000 10000 100000

Case b derivatives for conventional and numerical density models

BHPPENDAPDENO

PDENGPDENW

Log time (h)

k =1627qB120583

mh

1490mD

3410mD

35140mD

logdp998400

Figure 12 Derivative and119870 estimation for scenario (b)

Table 7 Summary of reservoir modelling properties imputed ineclipse model for case study 2

Parameters Numerical density119896 (mD)

Equivalentℎ (ft)

Scenario(a)

BHP 371PDENA 371PDENG 142 148PDENO 3575PDENW 371

Scenario(b)

BHP 341PDENA 341PDENG 149 148PDENO 3514PDENW 341

conventional approach From the 2 scenarios investigated ithas been demonstrated that the heavier fluid such as waterand the weighted average pressure-density equivalent of allfluid give exact effective 119896 as the conventional method BPRA summary of the result is shown in Table 7

This approach estimates the fluid phase permeabilitiesand the of each phase contribution to flow at a given pointhence at several points the relative 119896 can be generated asshown in Table 8

5 Density Related Radial Flow EquationDerivation for Each Fluid Phase

Radial diffusivity equation is given as

1119903

120597120588

120597119903[119903

120597120588

120597119903] =

120601120583119888

119896

120597120588

120597119905 (35)

Journal of Petroleum Engineering 11

Table 8 Comparison of 119870 estimates between conventional and numerical density parameters

Scenario Numerical density method Calculated119870 (mD)

Conventional119896 (mD)

Simulationmodel 119870 (mD) Relative 119870

phasecontribution to flowPhases 119870 (mD)

(a)Gas 1422 004 30Oil 35747 4046 3707 400 089 870

Water 3707 009 90

(b)Gas 149 004 40Oil 35141 4038 3409 400 089 880

Water 3409 008 90

For oil

1119903

120597120588

120597119903[119903

120597120588119900

120597119903] =

120601120583119900119888

119896119900

120597120588119900

120597119905 (36a)

For water phase

1119903

120597120588

120597119903[119903

120597120588119908

120597119903] =

120601120583119908119888

119896119908

120597120588119908

120597119905 (36b)

For gas phase

1119903

120597120588

120597119903[119903

120597120588119892

120597119903] =

120601120583119892119888

119896119892

120597120588119892

120597119905 (36c)

51 From Equation for Slight and Small Compressibility Suchas Water and Oil From (22)

119875 = 119875119900minus

120588120588119900+ 1

119888119900

(37)

Differentiating with respect to 120588

120597119875

120597120588= minus

1119888119900120588119900

120597119875 = minus120597120588

119888119900120588119900

(38)

From Darcy flow equation

119902 = minus2120587119896ℎ120583

119903120597120588

120597119903 (39)

Substitute for 120597119875

119902 =2120587119896ℎ120583

119903

119888119900120588119900

120597120588

120597119903 (40)

To derive the analytical density transient equation for eachphase the following assumptionconditions are applicable

For oil phase from (35)

1119903

120597120588

120597119903[119903

120597120588

120597119903] =

120601120583119888

119896

120597120588

120597119905 (41)

Initial condition is as follows

120588 (119903 119905 = 0) = 120588119894 (42a)

BC at the wellbore is as follows

lim119903rarr 0

2120587119896ℎ120583

119903

119888119900120588119900

120597120588

120597119903= 119876 (42b)

BC at infirmity is as follows

lim119903rarrinfin

120588 (119903 119905) = 120588119894 (42c)

Using the Boltzmann transformation consider the followingAssuming 120578 = 120601120583119888119903

2119896119905

120597120588

120597119903=

120597120588

120597120578

120597120578

120597119903=2120601120583119888119903119896119905

120597120588

120597120578=

1206011205831198881199032

119896119905

2119903

120597120588

120597120578=2120578119903

120597120588

120597120578 (43)

Therefore1119903

120597

120597119903[119903

120597120588

120597119903] =

21205781199032

120597

120597120578[2120578

120597120588

120597120578] =

41205781199032

120597

120597120578[120578

120597120588

120597120578] (44)

Differentiating with respect to 119903 is equal to differentiatingwith respect to 120578 multiply by 2120578119903

From (35) the LHS is resolved as follows

120597120588

120597119905=

120597120588

120597120578

120597120578

120597119905=

120601120583119888119903

1198961199052120597120588

120597120578= minus

120578

119905

120597120588

120597120578 (45)

120601120583119888

119896

120597120588

120597119905=

minus120601120583119888

119896

120578

119905

120597120588

120597119905=

minus1206011205831198881199032

119896119905

120578

1199032120597120588

120597120578=

minus1205782

1199032120597120588

120597120578 (46)

Equating (44) and (46) we have

120597

120597120578[120578

120597120588

120597120578] =

minus120578

4120597120588

120597120578 (47)

This is the simplified ordinary differential equation of 120588 as afunction of 120578

Apply boundary conditions as followsInitial condition is as follows

lim119903rarrinfin

120588 (120578) = 120588119894 (48a)

Substituting (43) into (40)

lim119903rarr 0

2120587119896ℎ120583

119903

119888119900120588119900

120597120588

120597119903= 119876 (48b)

lim119903rarr 0

[120578120597120588

120597120578] =

120601120583119888119900120588119900

4120587119896ℎ (49)

Assume that119898 = 120578(120597120588120597120578)

12 Journal of Petroleum Engineering

From (47)

120597119898

120597120578=

minus119898

4 (50)

Integrating from 120578 = 0 to 120578

997904rArr int

119898(120578)

119898(0)

120597119898

120597120578= minusint

120578

0

120597120578

4

997904rArr ln[119898 (120578)

119898 (0)] =

minus120578

4

997904rArr 119898(120578) = 119898 (0) 119890minus1205784

(51)

Recall boundary condition

lim119903rarr 0

[120578120597120588

120597120578] =

119876120583119888119900120588119900

4120587119896ℎ

119898 (0) =119876120583119888119900120588119900

4120587119896ℎ

119898 (120578) =119876120583119888119900120588119900

4120587119896ℎ119890minus1205784

(52)

Recall that119898 = 120578(120597120588120597120578)Therefore

120597120588 (120578)

120597120578=

120583119876119888119900120588119900

4120587119896ℎ119890minus1205784

120578 (53)

Integrating from 120578 = infin where 120588 = 120588119894

int

120588(120578)

120588119894

120597120588 = minusint

120578

infin

120583119876119888119900120588119900

4120587119896ℎ119890minus1205784

120578120597120578

120588 (120578) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎminusint

infin

120578

119890minus1205784

120578120597120578

(54)

where

120578 =120601120583119888119903

2

119896119905 (55)

Assuming 119906 = 1205784 then 120597120578120578 = 120597119906119906 and 119906 = 12060112058311988811990324119896119905

Hence

120588 = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎminusint

infin

12060112058311988811990324119896119905

119890minus1205784

120578120597120578 (56)

where

minus119864119894 (minus119909) = int

infin

119909

119890minus119906

119906120597119906 (57)

Known as the exponential integral function defined by [13]

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎ119864119894 (minus119909) (58)

where 119909 = 12060112058311988811990324119896119905 and

minus119864119894 (minus119909) = int

infin

119909

119890minus119906

119906120597119906 (59)

At large times 119909 will be small applying Taylorrsquos series andintegrating

minus119864119894 (minus119909) = ln (120574)

ln (120574) = ln (1781) = 05772(60)

where 120574 is known as Eulerrsquos numberFor 14119909 gt 25 consider the followingThen

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎ[ln119909+ ln 120574]

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎln[

1206011205831198881199032120574

4119896119905]

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎln [

22461198961199051206011205831198881199032

]

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎ[ln [

22461198961199051206011205831198881199032

]+ 080907]

(61)

Plotting 120588119908(119905) versus ln(119905) will yield a straight line at longer

time and the slope of the line is given as120597120588119908

120597 ln 119905= 119898oil =

120583119876119888119900120588119900

4120587119896ℎ (62)

Therefore

119896ℎ =120583119876119888119900120588119900

4120587119898oil (63)

52 For Water Phase The radial density equation is given as

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎ[ln [

22461198961199051206011205831198881199032

]+ 080907] (64)

Plotting 120588119908(119905) versus ln(119905) will yield a straight line at longer

time and the slope of the line is given as120597120588119908

120597 ln 119905= 119898water =

120583119876119888119900120588119900

4120587119896ℎ (65)

where

119896ℎ =120583119876119888119900120588119900

4120587119898water (66)

53 For Gas Phase From equation for compressible fluidsuch as gas consider the following

From (32)

119875 =119875119900120588 minus 119875119900

2120588119900119888119892

120588119900(1 minus 119875

119900119888119892)

120597119875

120597120588=

119875119900

120588119900(1 minus 119875

119900119888119892)

120597119875 =119875119900

120588119900(1 minus 119875

119900119888119892)120597120588

(67)

Journal of Petroleum Engineering 13

From the fundamental

119899 =119875119881

119911119877119879 (68)

At standard condition

119875119881

119911119879=

119875sc119881sc119879sc

119875 (5615119902)119911119879

=119875sc119876sc119879sc

119902 =119875sc119879sc

(119911119879

119875)

119876sc5615

119902 =119875sc119879sc

119876sc5615

(119911119879

119875) =

2120587119896ℎ120583

119903120597119901

120597119903

119902 =119875sc119879sc

119876sc5615

(119911119879) =2120587119896ℎ120583

119903119875120597119901

120597119903

(69)

where

120597119875 =119875119900

120588119900(1 minus 119875

119900119888119892)120597120588

120572 =119875119900

120588119900(1 minus 119875

119900119888119892)

120597119875 = 120572120597120588

119875 = 120588119877119911119879

(70)

Substitute into the equation where

119898(120588119894) minus119898 (120588

119908119891) =

2120583

int

120588119894

120588119908119891

120588120597120588

119898 (120588119894) minus119898 (120588

119908119891) =

120588119894

2minus 120588119908119891

2

120583

120588 = radic120588119894

2+ 120588119908119891

2

2

(71)

From the diffusivity equation for gas

1119903

120597120588

120597119903[119903

120597120588119892

120597119903] =

120601120583119892119888

119896119892

120597120588119892

120597119905 (72)

Initial condition is as follows

120588 (119903 119905 = 0) = 120588119894 (73)

BC at the wellbore is as follows

lim119903rarr 0

4120587119896ℎ120572119877120583

(5615119879sc

119875sc) 119903120588

120597120588

120597119903= 119876 (74)

BC at infirmity is as follows

lim119903rarrinfin

120588 (119903 119905) = 120588119894 (75)

where119877119875sc and119879sc are known as temperature pressure andgas constant at standard condition

Therefore integrating from 120578 = infin where 120588 = 120588119894

int

120588(120578)

120588119894

120588120597120588 = minusint

120578

infin

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)119890minus1205784

120578120597120578

1205882(120578) = 120588

119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

minusint

infin

120578

119890minus1205784

120578120597120578

(76)

where

120578 =120601120583119888119903

2

119896119905 (77)

Assume that 119906 = (1205784) (120597120578120578) = 120597119906119906 and 119906 = 12060112058311988811990324119896119905

Hence

1205882= 120588119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)minusint

infin

12060112058311988811990324119896119905

119890minus1205784

120578 (78)

where minus119864119894(minus119909) = intinfin

119909(119890minus119906

119906)120597119906Known as the exponential integral function defined by

[13]

1205882= 120588119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)minus119864119894 (minus119909) (79)

where 119909 = 12060112058311988811990324119896119905 and minus119864119894(minus119909) = int

infin

119909(119890minus119906

119906)120597119906At large times 119909will be small applying Taylorrsquos series and

integrating

minus119864119894 (minus119909) = ln (120574)

ln (120574) = ln (1781) = 05772(80)

where 120574 is known as Eulerrsquos numberFor 14119909 gt 25 consider the followingThen

1205882(119903 119905) = 120588

119894

2+

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) [ln119909+ ln 120574] (81)

1205882(119903 119905) = 120588

119894

2+

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) ln[120601120583119888119903

2120574

4119896119905]

1205882(119903 119905) = 120588

119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) ln [22461198961199051206011205831198881199032

]

119898 (120588119894) minus119898 (120588

119908119891) =

2120583

int

120588119894

120588119908119891

120588 120597120588

119898 (120588119908119891

) = 119898 (120588119894) minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

sdot [ln [22461198961199051206011205831198881199032

]+ 080907]

120597119898 (120588119908)

120597 ln 119905= 119898gas =

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

(82)

14 Journal of Petroleum Engineering

4160041650417004175041800418504190041950420004205042100

560856095609561056105611561156125612

ODENGDEN

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

1205880

(Ibm

ft)

120597120588w120597lnt

= moil =1627120583Qc01205880

kh= minus0009

120597m(120588w)

120597lnt= mgas =

1637120583Q

kh120572R= minus12195

120588g2

(Ibm

ft)

Figure 13 Oil and gas phase density versus log Horner time

6279

6279

6279

6279

6279

6279

6279

6279

01 1 10 100 1000 10000 1000001000000Horner time (h)

Density versus log Horner time

WDEN

120588w

(Ibm

ft)

120597120588w120597lnt

= mwater =1627120583Qc01205880

kh= minus00003

Figure 14 Water phase density versus log Horner time

Plotting 120588119908

2(119905) or 119898(120588

119908) versus ln(119905) will yield a straight line

at longer time and the slope of the line is given as follows

119896ℎ =120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)1

119898gas (83)

Example 3 Data from Example 1(a) was used for this caseand the semilog plot of density for each phase (oil water andgas) is plotted against log of Horner timelowastlowast

119876 used for 119896 calculation is the average rate for allflowing periods

From Figures 13 and 14 the calculated slopes of the radialflow for oil and water phases are given as

120597120588119908

120597 ln 119905= 119898oil =

1627120583119876119888119900120588119900

119896ℎ= minus 0009

120597120588119908

120597 ln 119905= 119898water =

1627120583119876119888119900120588119900

119896ℎ= minus 00003

(84)

The estimated phase permeabilities are

119896oil =1627120583119876119888

119900120588119900

ℎ119898oil= 445mD

119896water =1627120583119876119888

119900120588119900

ℎ119898water= 50mD

(85)

Also for the gas phase consider the following

ODENGDEN

422004225042300423504240042450425004255042600

5613561356145614561556155616

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

1205880

(Ibm

ft)

120597120588w120597lnt

= moil =1627120583Qc01205880

kh= minus0009

120597m(120588w)

120597lnt= mgas =

1637120583Q

kh120572R= minus12502

120588g2

(Ibm

ft)

Figure 15 Oil and gas phase density versus log Horner time

WDEN

6279627962796279627962796280628062806280

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

120588w

(Ibm

ft)

120597120588w120597lnt

= mwater =1627120583Qc01205880

kh= minus00003

Figure 16 Water phase density versus log Horner time

The calculated slope from the semilog plot shown inFigure 13 is given as

120597119898 (120588119908)

120597 ln 119905= 119898gas =

1637120583119876119896ℎ120572119877

= minus 12195 (86)

And estimated gas permeability is

119896gas =1637120583119876

119900

ℎ120572119877119898gas= 170mD (87)

Using the empirical model for all three phases the averageestimated permeability is as follows

119896ave =4radic50 times 445 times 1702

119896ave = 50mD

(88)

Example 4 Data from Example 1(c) was used for this caseand the semilog plot of density for each phase (oil water andgas) is plotted against log of Horner time

From Figures 15 and 16 the calculated slopes of the radialflow for oil and water phases are given as

120597120588119908

120597 ln 119905= 119898oil =

1627120583119876119888119900120588119900

119896ℎ= minus 0009

120597120588119908

120597 ln 119905= 119898water =

1627120583119876119888119900120588119900

119896ℎ= minus 00003

(89)

Journal of Petroleum Engineering 15

The estimated phase permeabilities are

119896oil =1627120583119876119888

119900120588119900

ℎ119898oil= 4352mD

119896water =1627120583119876119888

119900120588119900

ℎ119898water= 49mD

(90)

Also for the gas phase consider the following

The calculated slope from the semilog plot shown inFigure 15 is given as

120597119898 (120588119908)

120597 ln 119905= 119898gas =

1637120583119876119896ℎ120572119877

= minus 12502 (91)

And estimated gas permeability is

119896gas =1637120583119876

119900

ℎ120572119877119898gas= 179mD (92)

Using the empirical model for all three phases the averageestimated permeability is as follows

119896ave =4radic49 times 4352 times 1792

119896ave = 51mD

(93)

6 Conclusion

The following inferences were drawn from the six scenariosreviewed

(i) Results from the three-phase fluid empirical modeldeveloped for average effective 119896ave estimation rangebetween 47 and 57mD for Example 1 and give404mD for Example 2 which is within that used inthe simulation model and also estimated from BPR

(ii) The BPR PDENDWA and PDENWAT give the samestabilisation and the same 119896 estimates making it suit-able for interpretation of pressure transient analysis

(iii) It has been demonstrated in all 6 scenarios inves-tigated that the heavier fluid such as water andthe weighted average pressure-density equivalent ofall fluid give exact effective 119896 as the conventionalmethod

(iv) For oil reservoir system (scenarios (a) to (d)) 119896 =

500mD is achieved if the thickness contributing toflow ranges from 50 to 150 ft while for gas condensatereservoir (scenarios (a) to (b)) 119896 ranges between340 and 371mD which is achieved if the net sandthickness ℎ = 148ft is contributing to flow

(v) This approach also provides an estimate of the pos-sible fluid phase permeabilities and the of eachphase contribution to flow at a given point hence atseveral 1198891199011015840 stabilisation points the relative 119896 can begenerated

(vi) The derivatives for all output parameters display thesame wellbore and reservoir fingerprint as the BPRmethod

(vii) Generally where output parameters (BPR PDE-NOIL PDENGAS PDENWAT and PDENDW)depict different fingerprint the density derivative willserve as support to distinguish between reservoir andnonreservoir response

Nomenclature

119875 Pressure psi119879 Temperature oF119903 Radius ft119896 Permeability mDOslash Porosity fraction120583 Viscosity cp119905 Time hrs119902 Production rate bblday119861 Formation volume factor rbStb119862119905 Total compressibility psiminus1119903119908 Wellbore radius ftΔ119901 Change in pressure psiaℎ Formation thickness ft119860 Drainage area acres119875119908119891 Bottom-hole flowing pressure psi119875119894 Initial pressure psi

119885 Difference between two pointtime series119894 Subscript of an observed variable119888 Subscript of a calculated variableSTEYX SSE of data point119899 Number of data pointsCov Covariance of data point120575 Standard deviation119905119901 Cumulative production time119862119904 Wellbore storage constant119909 Mean of data point

Abbreviations

LBPR Local grid bottom-hole pressureLDENO Local grid oil densityLDENW Local grid water densityLDENG Local grid gas densityWBHP Well bottom-hole pressureBPR Well bottom-hole pressurePDENOIL Pressure equivalent of LDENOPDENGAS Pressure equivalent of LDENGPDENWAT Pressure equivalent of LDENWPDENDWA Pressure equivalent of density weighted

average (LDENO LDENG and LDENW)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Jalali S D Mohaghegh and R Gaskari ldquoIdentifying infilllocations and underperformer wells in mature fields usingmonthly production rate data Carthage Field Cotton Valley

16 Journal of Petroleum Engineering

Formation Texasrdquo in Proceedings of the SPE Eastern RegionalMeeting SPE Paper 104550 Canton Ohio USA October 2006

[2] D Tiab and A Kumar ldquoApplication of PD function to inter-ference analysisrdquo in Proceedings of the 51st Annual Fall Meetingof the Society of Petroleum Engineers of AIME Paper SPE 6053New Orleans La USA 1976

[3] D Tiab and A Kumar ldquoDetection and location of two parallelsealing fault around a wellrdquo in Proceedings of the 51st AnnualFall Meeting of the Society of Petroleum Engineers of AIME SPEPaper 6056 New Orleans La USA 1976

[4] D Bourdet T M Whittle A A Douglas and Y M Pirard ldquoAnew set of type curves simplifies well test analysisrdquo World Oilvol 196 no 6 pp 95ndash106 1983

[5] S Y Zheng ldquoFighting against non-unique solution problemsin heterogeneous reservoirs through numerical well testingrdquo inProceedings of the SPE Asia Pacific Oil and Gas Conference andExhibition SPE 100951 Adelaide Australia September 2006

[6] R N Horne and K O Temeng ldquoRecognition and location ofpinch-out boundaries by pressure transient analysisrdquo SPE 9905Society of Petroleum Engineers 1981

[7] L C Ayestaran R D Nurmi G A K Shehab and W S ElsildquoWelltest design and interpretation improved by integrated welltesting and geological effortsrdquo SPE 17945 1989

[8] G J Massonnat and D Bandiziol ldquoInterdependence betweengeology and well test interpretationrdquo SPE 22740 1991

[9] K F Du and G Stewart ldquoReservoir description from well testinterpretationrdquo in North Sea Oil and Gas ReservoirmdashIII pp339ndash356 Norwegian Institute of Technology (NTH) 1994

[10] S Y Zheng Well testing and characterisation of meanderingfluvial channel reservoirs [PhD thesis] Heriot-Watt University1997

[11] V T Biu and S Y Zheng ldquoA new approach in pressure transientanalysis part I improved diagnosis of flow regimes in oil and gaswellsrdquo in Proceedings of the 3rd EAGEAAPGWorkshop on TightReservoirs in theMiddle East AbuDhabi UnitedArab Emirates2015

[12] R N Horne Modern Welltest Analysis Petroway Palo AltoCalif USA 1995

[13] C S Matthews and D G Russell Pressure Buildup and FlowTests in Wells vol 1 ofMonograph Series SPE Richardson TexUSA 1967

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Journal of Petroleum Engineering 5

Local grid refinement

Pressure (psia) Pressure (barsa)

PROD1

PROD1

35591 35630 35669 35708 35747 3093430899 30970 31006 31041

Figure 2 Simulation model for gas cap + oil + water reservoir and gas condensate + water showing local grid refinement around the well

Table 1 Reservoir and fluid data for Example 1

Parameters Design valueEclipse model Black oilModel dimension 10 times 5 times 5Length by width ft by ft 500 times 400Thickness ℎ ft 250Permeability 119870119909 by 119870119910 mD 500 by 500Porosity 20Well diameter ft 060Initial water saturation 119878119908

119894 22

Permeability 119870 mD 50Gas oil contact (GOC) ft 8820Oil water contact (OWC) ft 90000Initial pressure 119875

119894 psia 40000

Formation temperature 119879 oF 2000

investigated using numerical model built with a commercialsimulator Local grid refinement (LGR) is imposed aroundthe well to capture sharp change in fluid densities as shownin Figure 2

The simulation software keywords LBPR LDENOLDENW LDENG AND WBHP were outputs to obtain thedensity and pressure change around the well and as far as theperturbation could extend

Example 1 Table 1 presents a summary of the well andreservoir synthetic data used for the buildup and drawdownsimulated scenarios with additional information given belowIt is required to generate the pressure equivalent and deriva-tive for each phase compare their diagnostic signaturesand also determine the phases permeabilities and averagereservoir permeability

Assumption (i) Oil reservoir + gas cap is completed with onewell

(ii) LGR is imposed around the well and far across toaccount for pressure and density changes

Gas oil and water densities around the local grid refine-ment (wellbore) andWBHP were output using the simulatorkeywords The following scenarios were evaluated

(a) Flowing+ buildup sequence well perforated ℎ119901= 30 ft

between oil and water layer Net sand thickness ℎ =250 ft

(b) Flowing+buildup sequence well perforated ℎ119901= 30 ft

inside the oil layer Net sand thickness ℎ = 250 ft(c) Flowing+ buildup sequence well perforated ℎ

119901= 30 ft

between gas and oil layer Net sand thickness ℎ =250 ft

(d) Falloff test flowing + buildup sequence well perfo-rated ℎ

119901= 30 ft inside the oil layer Net sand thickness

ℎ = 250 ftThe schematic of the 4 scenarios is depicted in Figures 3and 4 Figure 5 shows the production shut-in and injectionsequence for the 4 scenarios in Example 1

In scenario (a) the well is completed between the oil andwater layer to mimic multiphase conditions at the wellboreand some distance away from the well and also see the effecton pressure distribution fluid densities changes around thewellbore and estimate fluid phase permeabilities 119896 using thespecialised plot and 119896ave from the empirical model for three-phase conditions

The derivative in Figure 6 shows a good radial flow butdrop in derivative at late time due to constant pressuresupport (likely aquifer support) A continuous drop is seenfrom 10 hrs in the model parameters The derivatives forall output parameters display the same well and reservoirsignatures (3 flow periods early to late time response)but with different 119889119901

1015840 stabilisation The well bottom-holepressure (BPRrarrBHP) response shows good overlay withpressure equivalent of density weighted average (PDENDWArarr PDENA) and pressure equivalent of water density (PDEN-WAT rarr PDENW) while pressure equivalent of gas den-sity (PDENGAS rarr PDENG) and pressure equivalent ofoil density (PDENOIL rarr PDENO) differ completely ThePDENDWA gives a better fingerprint that is less noisy

A permeability value of 508mD is estimated from thebottom-hole pressure BPR where 119896 = 1627119902119861120583119898ℎ and119898 is

6 Journal of Petroleum Engineering

Gas

Oil

Water

Case 1 production test

Net sand thickness h = 250ftwater layerWell perforated hp = 30ft between oil andFlowing + buildup sequence

(a)

Gas

Oil

Water

Case 2 production test Flowing + buildup sequence

Net sand thickness h = 250ftWell perforated hp = 30ft inside the oil layer

(b)

Gas

Oil

Water

Case 3 production test Flowing + buildup sequence

Net sand thickness h = 250ftoil layerWell perforated hp = 30ft between gas and

(c)

Gas

Oil

Water

Case 4 falloff test Flowing + buildup sequence

Net sand thickness h = 250ftwater layerWell perforated hp = 30ft between oil and

(d)

Figure 3 Schematics of well perforation interval and sand thickness for oil + gas cap + water reservoir

obtained from the specialised plot This is an approximate ofthe input value in the simulation model Also the simulatoroutputs PDENDWA and PDENWAT in the simulation givethe same 119896 value while PDENGAS and PDENOIL differgiving 158 and 4037mD respectively At ℎ = 50 f t the best119896 estimate is obtained depicting ℎ = 50 f t as the thicknesscontributing to flowAt ℎ gt 50ft 119896 drops below the 119896 imputedin the model

Using (34) 119896ave = 473mD is obtained which isapproximately close to that of BPR hence a good estimateof the fluid phase permeabilities A summary of the result isshown in Tables 2 and 5

In scenario (b) thewell is completedwithin the oil sectionto capture the pressure and fluid densities changes around thewell The derivative in Figure 7 shows a good radial flow butwith noisy numerical artefact It is likely that the boundaryresponse is masked by numerical artefact Also a continuousdrop is seen after 10 hrs in the derivative

Using the estimated fluid phase permeabilities 119896ave from(34) is 472mD as shown in Table 3 which is the same asscenario (a) This is in line with the uniform 119896 used in thesimulation model Also the BPR gives a permeability value of500mD which is the same for PDENDWA and PDENWATbut differs with PDENGAS and PDENOIL that give 121 and

Journal of Petroleum Engineering 7

Water

Case 1 production test

Gas + condensate

Flowing + buildup sequence

condensate and water layerNet sand thickness h = 45m

Well perforated hp = 90m between gas

(a)

Water

Case 2 production test

Gas + condensate

Net sand thickness h = 45m

Flowing + buildup sequence

condensate layerWell perforated hp = 90m inside gas

(b)

Figure 4 Schematics of well perforation interval and sand thickness for gas condensate + water reservoir

00

5000

0 50 100 150 200 250 300 350

Prod

uctio

nin

ject

ion

rate

bop

d

Duration

Test design sequence

Case dCases a to c

minus25000

minus20000

minus15000

minus10000

minus5000

Figure 5 Production and injection test sequence for scenarios (a)to (d) (buildup and falloff)

Table 2 119870 estimates for new approach versus conventionalapproach for scenario (a)

Parameters Numerical density119896 (mD)

Equivalentℎ (ft)

BHP 508PDENA 508PDENG 158 50PDENO 4037PDENW 508

4844mD respectively At ℎ = 150ft (60 of sand thicknessand 83 of oil thickness) the estimated 119896 is still within range

00

00

01

10

100

0000 0001 0010 0100 1000 10000 100000Log time (h)

BHPPENDAPDENO

PDENGPDENW

508mD

4037mD

158mD

k =1627qB120583

mh

Case a derivatives for conventional and numerical density models

logdp998400

Figure 6 Derivative and 119870 estimation for scenario (a)

(119896 = 489mD)This indicates ℎ = 50ft or 83 of oil thicknesscontributing to flow

To test this approach in the gas column the well wascompleted in between the gas and oil layer which is consid-ered as scenario (c) and scenario (d) with the well completedwithin the oil layer but with water injection after flowingand shut-in sequence In both scenarios the multiphasefluid distribution is triggered at the wellbore in order tocapture the density changes for each phase and calculate fluidphase permeabilities First the well fluid densities equivalent

8 Journal of Petroleum Engineering

Case b derivatives for conventional and numerical density models

00

01

10

100

0000 0001 0010 0100 1000 10000 100000Log time (h)

BHPPENDAPDENO

PDENGPDENW

4842mDk =1627qB120583

mh

120mD

489mD

logdp998400

Figure 7 Derivative and 119870 estimation for scenario (b)

0001

0010

0100

1000

10000

0000 0001 0010 0100 1000 10000 100000Log time (h)

BHPPENDAPDENO

PDENGPDENW

412mD

209mD

Case c derivatives for conventional and numerical density models

4842mD

k =1627qB120583

mh

logdp998400

Figure 8 Derivative and 119870 estimation for scenario (c)

Table 3 119870 estimates for new approach versus conventionalapproach for scenario (b)

Parameters Numerical density119896 (mD)

Equivalentℎ (ft)

BHP 500PDENA 500PDENG 121 50PDENO 4842PDENW 500

pressures and pressures at bottom-hole for flowing andbuildup test are generated

The derivative for both scenarios declines after 3 hoursprobably due to fluid redistribution but is noisy (numericalartefact) in scenario (c) as shown in Figure 8 A good radial

Table 4 119870 estimates for new approach versus conventionalapproach for scenario (d)

Parameters Numerical Density119896 (mD) Equivalent ℎ (ft)

Scenario (d)

BHP 500 100PDENA 500PDENG 175 50PDENO 490 250PDENW 500 100

stabilisation for both the conventional method BPR and thedensity outputs PDENDWA PDENWAT PDENGAS andPDENOIL is observed Likewise as obtained in scenarios (a)and (b) a good 119870ave value of 57 and 52mD from (34) isobtained for scenarios (c) and (d) respectivelyThis is slightlyhigher than 119896 = 50mD imputed in the simulation model

For each of the fluid phase permeabilities estimates 119896value of 50mD is obtained using the conventional BPR forscenario (c) which is in line with the uniform 119896 in thesimulation model This is the same for the PDENDWA andPDENWAT Also at ℎ = 150ft (60 of sand thickness and83 of oil thickness) the estimated 119896 is still within range(119896 = 412mD)This also indicates that ℎ = 150ft or 83 of oilthickness is contributing to flowThe 119896 values for PDENGASandPDENOIL are 209 and 4842mD respectively Summaryof the result is shown in Table 4

However in scenario (d) the permeability value of500mD using the conventional BPR which is the samefor PDENDWA and PDENWAT is only achieved if ℎ =

100ft (40 of sand thickness and 56 of oil thickness)PDENGAS and PDENOIL differ giving 350mD at ℎ =

50ft and 1960mD at ℎ = 250ft respectively This indicatesthe impact of water injection on densities and pressureschanges around the well and consequently its impact on sandthickness contributing to flow The drop in derivative curvein Figure 9 depicts the impact of the injected water

In summary for scenarios (a) to (d) the 119896 value of500mD is achieved if the thickness contributing to flowranges from 50 to 150 ft Generally results indicate that in all4 scenarios investigated the heavier fluid such as water andthe weighted average pressure-density equivalent of all fluidgive exact effective 119896 as the BPR likewise the density outputsPDENDWA and PDENWAT Also results from Table 5 showthat the empirical model from (34) with all fluid phasepermeabilities gives an average effective 119896ave of 47ndash57mDwhich is within that used in the simulationmodel and also thesame as the estimated conventional approach This approachprovides an estimate of the possible fluid phase permeabilitiesand theof each phase contribution to flow hence at severalpoints the relative 119896 can be generated as shown in Table 5

Example 2 To capture the influence of highly compressiblefluid on estimated fluid phase permeabilities an example ongas condensate reservoir (volatile system) was tested Table 6presents a summary of the well and reservoir synthetic dataused for the buildup and drawdown simulated scenarios with

Journal of Petroleum Engineering 9

Table 5 Comparison of 119870 estimates between conventional and numerical density parameters

Scenario Numerical density method Calculated119870 (mD)

Conventional119896 (mD)

Simulationmodel 119870 (mD) Relative 119870

phasecontribution to flowPhases 119870 (mD)

(a)Gas 158 003 30Oil 4037 473 508 500 086 860

Water 508 011 110

(b)Gas 12 003 20Oil 4844 432 500 500 113 890

Water 50 012 90

(c)Gas 209 004 40Oil 4842 570 500 500 086 87

Water 50 009 90

(d)Gas 175 003 30Oil 490 523 500 500 095 880

Water 50 010 90

1000

Case d derivatives for conventional and numerical density models

0010

0100

10000

BHPPENDAPDENO

PDENGPDENW

0100 1000 10000 100000Log time (h)

k =1627qB120583

mh

209mD

500mD

4842mD

logdp998400

Figure 9 Derivative and 119870 estimation for scenario (d)

additional information given below It is required to generatethe pressure equivalent and derivative for each fluid phasecompare their diagnostic signatures and also determine thephases permeabilities and average reservoir permeability

Assumption (i) Condensate reservoir is completed with onewell

(ii) LGR is imposed around the well and far across toaccount for pressure and density changes

Gas condensate andwater densities around the local gridrefinement (wellbore) andWBHPwere output using the sim-ulatorrsquos keywords The following scenarios were evaluated

(a) Flowing + buildup sequence well perforated ℎ119901

=

30ft between gas condensate and water layer Netsand thickness ℎ = 150ft

Table 6 Summary of reservoir simulations data

Parameters Design valueEclipse model Black oilModel dimension 9 times 3 times 3Length by width ft by ft 1312 times 984Thickness ℎ ft 150Permeability 119870119909 by 119870119910 mD 4000 by 3000Porosity 30Well diameter ft 115Initial water saturation 119878119908

119894 60

Permeability 119870 mD 400Gas oil contact (GOC) ft 6890Oil water contact (OWC) ft 6890Initial pressure 119875

119894 psi 4495

Formation temperature 119879 ∘C 120 2000

(b) Flowing + buildup sequence well perforated ℎ119901

=

30ft inside gas condensate layer Net sand thicknessℎ = 150ft

Figure 10 shows the production and shut-in sequence for 2scenarios in Example 2

As in Example 1 the derivative in scenario (a) ofExample 2 also shows good radial flowwith no late time effectas seen in Figure 11 A negative half-slope fingerprint is seen at03 to 30 hrs of buildup and a good stabilisation from 40 hrsin the model parameters

The derivatives for all output parameters display the samewell and reservoir fingerprint (2 flow periods early to middletime response) but with different 1198891199011015840 stabilisation

Permeability value of 3707mD was obtained from theconventionalmethod BPRwhere 119896 = 1637119902119879119898ℎ is the samefor the PDENDWA and PDENWAT and 119898 is obtained fromthe specialised plot This is in line with the uniform 119896 (119896119909 =

400mD and 119896119910 = 300mD) imputed in the simulationmodel

10 Journal of Petroleum Engineering

00

5000000

10000000

15000000

20000000

25000000

0 40 80 120 160

Prod

uctio

n ra

te (s

cfd

)

Duration

Test design sequence

Cases a to b

Figure 10 Production test sequence (buildup) for Example 2

10000

Case a derivatives for conventional and numerical density models

00100

01000

BHPPENDAPDENO

PDENGPDENW

1422mD

3707mD

35747mD

0000 0001 0010 0100 1000 10000

100000

100000Log time (h)

k =1627qB120583

mh

logdp998400

Figure 11 Derivative and 119870 estimation for scenario (a)

However PDENGAS and PDENOIL differ giving 350mDand 35747mD respectively if ℎ = 148ft

Also the derivative in Figure 12 shows good radial flowand a negative half-slope fingerprint is seen at 05 to 20 hrsof buildup and a good stabilisation from 30 hrs in the modelparameters After 3 hrs of shut-in somenoisy data (numericalartefact) is seen in all the fluid phase derivatives but still goodenough to identify 119889119901

1015840 stabilisationFor each of the fluid phase permeabilities 119896 value of

3409mD at ℎ = 148ft is obtained using BPR which is thesame for the PDENDWA and PDENWAT PDENGAS andPDENOIL differ giving 1490 and 35141mD at ℎ = 148ftrespectively

For scenarios (a) and (b) a good 119896ave value of 4050and 4030mD is obtained respectively from the empiricalmodel (34) integrating all fluid phase permeabilities whichis within that used in the simulationmodel and the estimated

0001000

0010000

0100000

1000000

10000000

0000 0001 0010 0100 1000 10000 100000

Case b derivatives for conventional and numerical density models

BHPPENDAPDENO

PDENGPDENW

Log time (h)

k =1627qB120583

mh

1490mD

3410mD

35140mD

logdp998400

Figure 12 Derivative and119870 estimation for scenario (b)

Table 7 Summary of reservoir modelling properties imputed ineclipse model for case study 2

Parameters Numerical density119896 (mD)

Equivalentℎ (ft)

Scenario(a)

BHP 371PDENA 371PDENG 142 148PDENO 3575PDENW 371

Scenario(b)

BHP 341PDENA 341PDENG 149 148PDENO 3514PDENW 341

conventional approach From the 2 scenarios investigated ithas been demonstrated that the heavier fluid such as waterand the weighted average pressure-density equivalent of allfluid give exact effective 119896 as the conventional method BPRA summary of the result is shown in Table 7

This approach estimates the fluid phase permeabilitiesand the of each phase contribution to flow at a given pointhence at several points the relative 119896 can be generated asshown in Table 8

5 Density Related Radial Flow EquationDerivation for Each Fluid Phase

Radial diffusivity equation is given as

1119903

120597120588

120597119903[119903

120597120588

120597119903] =

120601120583119888

119896

120597120588

120597119905 (35)

Journal of Petroleum Engineering 11

Table 8 Comparison of 119870 estimates between conventional and numerical density parameters

Scenario Numerical density method Calculated119870 (mD)

Conventional119896 (mD)

Simulationmodel 119870 (mD) Relative 119870

phasecontribution to flowPhases 119870 (mD)

(a)Gas 1422 004 30Oil 35747 4046 3707 400 089 870

Water 3707 009 90

(b)Gas 149 004 40Oil 35141 4038 3409 400 089 880

Water 3409 008 90

For oil

1119903

120597120588

120597119903[119903

120597120588119900

120597119903] =

120601120583119900119888

119896119900

120597120588119900

120597119905 (36a)

For water phase

1119903

120597120588

120597119903[119903

120597120588119908

120597119903] =

120601120583119908119888

119896119908

120597120588119908

120597119905 (36b)

For gas phase

1119903

120597120588

120597119903[119903

120597120588119892

120597119903] =

120601120583119892119888

119896119892

120597120588119892

120597119905 (36c)

51 From Equation for Slight and Small Compressibility Suchas Water and Oil From (22)

119875 = 119875119900minus

120588120588119900+ 1

119888119900

(37)

Differentiating with respect to 120588

120597119875

120597120588= minus

1119888119900120588119900

120597119875 = minus120597120588

119888119900120588119900

(38)

From Darcy flow equation

119902 = minus2120587119896ℎ120583

119903120597120588

120597119903 (39)

Substitute for 120597119875

119902 =2120587119896ℎ120583

119903

119888119900120588119900

120597120588

120597119903 (40)

To derive the analytical density transient equation for eachphase the following assumptionconditions are applicable

For oil phase from (35)

1119903

120597120588

120597119903[119903

120597120588

120597119903] =

120601120583119888

119896

120597120588

120597119905 (41)

Initial condition is as follows

120588 (119903 119905 = 0) = 120588119894 (42a)

BC at the wellbore is as follows

lim119903rarr 0

2120587119896ℎ120583

119903

119888119900120588119900

120597120588

120597119903= 119876 (42b)

BC at infirmity is as follows

lim119903rarrinfin

120588 (119903 119905) = 120588119894 (42c)

Using the Boltzmann transformation consider the followingAssuming 120578 = 120601120583119888119903

2119896119905

120597120588

120597119903=

120597120588

120597120578

120597120578

120597119903=2120601120583119888119903119896119905

120597120588

120597120578=

1206011205831198881199032

119896119905

2119903

120597120588

120597120578=2120578119903

120597120588

120597120578 (43)

Therefore1119903

120597

120597119903[119903

120597120588

120597119903] =

21205781199032

120597

120597120578[2120578

120597120588

120597120578] =

41205781199032

120597

120597120578[120578

120597120588

120597120578] (44)

Differentiating with respect to 119903 is equal to differentiatingwith respect to 120578 multiply by 2120578119903

From (35) the LHS is resolved as follows

120597120588

120597119905=

120597120588

120597120578

120597120578

120597119905=

120601120583119888119903

1198961199052120597120588

120597120578= minus

120578

119905

120597120588

120597120578 (45)

120601120583119888

119896

120597120588

120597119905=

minus120601120583119888

119896

120578

119905

120597120588

120597119905=

minus1206011205831198881199032

119896119905

120578

1199032120597120588

120597120578=

minus1205782

1199032120597120588

120597120578 (46)

Equating (44) and (46) we have

120597

120597120578[120578

120597120588

120597120578] =

minus120578

4120597120588

120597120578 (47)

This is the simplified ordinary differential equation of 120588 as afunction of 120578

Apply boundary conditions as followsInitial condition is as follows

lim119903rarrinfin

120588 (120578) = 120588119894 (48a)

Substituting (43) into (40)

lim119903rarr 0

2120587119896ℎ120583

119903

119888119900120588119900

120597120588

120597119903= 119876 (48b)

lim119903rarr 0

[120578120597120588

120597120578] =

120601120583119888119900120588119900

4120587119896ℎ (49)

Assume that119898 = 120578(120597120588120597120578)

12 Journal of Petroleum Engineering

From (47)

120597119898

120597120578=

minus119898

4 (50)

Integrating from 120578 = 0 to 120578

997904rArr int

119898(120578)

119898(0)

120597119898

120597120578= minusint

120578

0

120597120578

4

997904rArr ln[119898 (120578)

119898 (0)] =

minus120578

4

997904rArr 119898(120578) = 119898 (0) 119890minus1205784

(51)

Recall boundary condition

lim119903rarr 0

[120578120597120588

120597120578] =

119876120583119888119900120588119900

4120587119896ℎ

119898 (0) =119876120583119888119900120588119900

4120587119896ℎ

119898 (120578) =119876120583119888119900120588119900

4120587119896ℎ119890minus1205784

(52)

Recall that119898 = 120578(120597120588120597120578)Therefore

120597120588 (120578)

120597120578=

120583119876119888119900120588119900

4120587119896ℎ119890minus1205784

120578 (53)

Integrating from 120578 = infin where 120588 = 120588119894

int

120588(120578)

120588119894

120597120588 = minusint

120578

infin

120583119876119888119900120588119900

4120587119896ℎ119890minus1205784

120578120597120578

120588 (120578) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎminusint

infin

120578

119890minus1205784

120578120597120578

(54)

where

120578 =120601120583119888119903

2

119896119905 (55)

Assuming 119906 = 1205784 then 120597120578120578 = 120597119906119906 and 119906 = 12060112058311988811990324119896119905

Hence

120588 = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎminusint

infin

12060112058311988811990324119896119905

119890minus1205784

120578120597120578 (56)

where

minus119864119894 (minus119909) = int

infin

119909

119890minus119906

119906120597119906 (57)

Known as the exponential integral function defined by [13]

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎ119864119894 (minus119909) (58)

where 119909 = 12060112058311988811990324119896119905 and

minus119864119894 (minus119909) = int

infin

119909

119890minus119906

119906120597119906 (59)

At large times 119909 will be small applying Taylorrsquos series andintegrating

minus119864119894 (minus119909) = ln (120574)

ln (120574) = ln (1781) = 05772(60)

where 120574 is known as Eulerrsquos numberFor 14119909 gt 25 consider the followingThen

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎ[ln119909+ ln 120574]

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎln[

1206011205831198881199032120574

4119896119905]

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎln [

22461198961199051206011205831198881199032

]

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎ[ln [

22461198961199051206011205831198881199032

]+ 080907]

(61)

Plotting 120588119908(119905) versus ln(119905) will yield a straight line at longer

time and the slope of the line is given as120597120588119908

120597 ln 119905= 119898oil =

120583119876119888119900120588119900

4120587119896ℎ (62)

Therefore

119896ℎ =120583119876119888119900120588119900

4120587119898oil (63)

52 For Water Phase The radial density equation is given as

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎ[ln [

22461198961199051206011205831198881199032

]+ 080907] (64)

Plotting 120588119908(119905) versus ln(119905) will yield a straight line at longer

time and the slope of the line is given as120597120588119908

120597 ln 119905= 119898water =

120583119876119888119900120588119900

4120587119896ℎ (65)

where

119896ℎ =120583119876119888119900120588119900

4120587119898water (66)

53 For Gas Phase From equation for compressible fluidsuch as gas consider the following

From (32)

119875 =119875119900120588 minus 119875119900

2120588119900119888119892

120588119900(1 minus 119875

119900119888119892)

120597119875

120597120588=

119875119900

120588119900(1 minus 119875

119900119888119892)

120597119875 =119875119900

120588119900(1 minus 119875

119900119888119892)120597120588

(67)

Journal of Petroleum Engineering 13

From the fundamental

119899 =119875119881

119911119877119879 (68)

At standard condition

119875119881

119911119879=

119875sc119881sc119879sc

119875 (5615119902)119911119879

=119875sc119876sc119879sc

119902 =119875sc119879sc

(119911119879

119875)

119876sc5615

119902 =119875sc119879sc

119876sc5615

(119911119879

119875) =

2120587119896ℎ120583

119903120597119901

120597119903

119902 =119875sc119879sc

119876sc5615

(119911119879) =2120587119896ℎ120583

119903119875120597119901

120597119903

(69)

where

120597119875 =119875119900

120588119900(1 minus 119875

119900119888119892)120597120588

120572 =119875119900

120588119900(1 minus 119875

119900119888119892)

120597119875 = 120572120597120588

119875 = 120588119877119911119879

(70)

Substitute into the equation where

119898(120588119894) minus119898 (120588

119908119891) =

2120583

int

120588119894

120588119908119891

120588120597120588

119898 (120588119894) minus119898 (120588

119908119891) =

120588119894

2minus 120588119908119891

2

120583

120588 = radic120588119894

2+ 120588119908119891

2

2

(71)

From the diffusivity equation for gas

1119903

120597120588

120597119903[119903

120597120588119892

120597119903] =

120601120583119892119888

119896119892

120597120588119892

120597119905 (72)

Initial condition is as follows

120588 (119903 119905 = 0) = 120588119894 (73)

BC at the wellbore is as follows

lim119903rarr 0

4120587119896ℎ120572119877120583

(5615119879sc

119875sc) 119903120588

120597120588

120597119903= 119876 (74)

BC at infirmity is as follows

lim119903rarrinfin

120588 (119903 119905) = 120588119894 (75)

where119877119875sc and119879sc are known as temperature pressure andgas constant at standard condition

Therefore integrating from 120578 = infin where 120588 = 120588119894

int

120588(120578)

120588119894

120588120597120588 = minusint

120578

infin

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)119890minus1205784

120578120597120578

1205882(120578) = 120588

119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

minusint

infin

120578

119890minus1205784

120578120597120578

(76)

where

120578 =120601120583119888119903

2

119896119905 (77)

Assume that 119906 = (1205784) (120597120578120578) = 120597119906119906 and 119906 = 12060112058311988811990324119896119905

Hence

1205882= 120588119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)minusint

infin

12060112058311988811990324119896119905

119890minus1205784

120578 (78)

where minus119864119894(minus119909) = intinfin

119909(119890minus119906

119906)120597119906Known as the exponential integral function defined by

[13]

1205882= 120588119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)minus119864119894 (minus119909) (79)

where 119909 = 12060112058311988811990324119896119905 and minus119864119894(minus119909) = int

infin

119909(119890minus119906

119906)120597119906At large times 119909will be small applying Taylorrsquos series and

integrating

minus119864119894 (minus119909) = ln (120574)

ln (120574) = ln (1781) = 05772(80)

where 120574 is known as Eulerrsquos numberFor 14119909 gt 25 consider the followingThen

1205882(119903 119905) = 120588

119894

2+

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) [ln119909+ ln 120574] (81)

1205882(119903 119905) = 120588

119894

2+

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) ln[120601120583119888119903

2120574

4119896119905]

1205882(119903 119905) = 120588

119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) ln [22461198961199051206011205831198881199032

]

119898 (120588119894) minus119898 (120588

119908119891) =

2120583

int

120588119894

120588119908119891

120588 120597120588

119898 (120588119908119891

) = 119898 (120588119894) minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

sdot [ln [22461198961199051206011205831198881199032

]+ 080907]

120597119898 (120588119908)

120597 ln 119905= 119898gas =

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

(82)

14 Journal of Petroleum Engineering

4160041650417004175041800418504190041950420004205042100

560856095609561056105611561156125612

ODENGDEN

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

1205880

(Ibm

ft)

120597120588w120597lnt

= moil =1627120583Qc01205880

kh= minus0009

120597m(120588w)

120597lnt= mgas =

1637120583Q

kh120572R= minus12195

120588g2

(Ibm

ft)

Figure 13 Oil and gas phase density versus log Horner time

6279

6279

6279

6279

6279

6279

6279

6279

01 1 10 100 1000 10000 1000001000000Horner time (h)

Density versus log Horner time

WDEN

120588w

(Ibm

ft)

120597120588w120597lnt

= mwater =1627120583Qc01205880

kh= minus00003

Figure 14 Water phase density versus log Horner time

Plotting 120588119908

2(119905) or 119898(120588

119908) versus ln(119905) will yield a straight line

at longer time and the slope of the line is given as follows

119896ℎ =120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)1

119898gas (83)

Example 3 Data from Example 1(a) was used for this caseand the semilog plot of density for each phase (oil water andgas) is plotted against log of Horner timelowastlowast

119876 used for 119896 calculation is the average rate for allflowing periods

From Figures 13 and 14 the calculated slopes of the radialflow for oil and water phases are given as

120597120588119908

120597 ln 119905= 119898oil =

1627120583119876119888119900120588119900

119896ℎ= minus 0009

120597120588119908

120597 ln 119905= 119898water =

1627120583119876119888119900120588119900

119896ℎ= minus 00003

(84)

The estimated phase permeabilities are

119896oil =1627120583119876119888

119900120588119900

ℎ119898oil= 445mD

119896water =1627120583119876119888

119900120588119900

ℎ119898water= 50mD

(85)

Also for the gas phase consider the following

ODENGDEN

422004225042300423504240042450425004255042600

5613561356145614561556155616

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

1205880

(Ibm

ft)

120597120588w120597lnt

= moil =1627120583Qc01205880

kh= minus0009

120597m(120588w)

120597lnt= mgas =

1637120583Q

kh120572R= minus12502

120588g2

(Ibm

ft)

Figure 15 Oil and gas phase density versus log Horner time

WDEN

6279627962796279627962796280628062806280

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

120588w

(Ibm

ft)

120597120588w120597lnt

= mwater =1627120583Qc01205880

kh= minus00003

Figure 16 Water phase density versus log Horner time

The calculated slope from the semilog plot shown inFigure 13 is given as

120597119898 (120588119908)

120597 ln 119905= 119898gas =

1637120583119876119896ℎ120572119877

= minus 12195 (86)

And estimated gas permeability is

119896gas =1637120583119876

119900

ℎ120572119877119898gas= 170mD (87)

Using the empirical model for all three phases the averageestimated permeability is as follows

119896ave =4radic50 times 445 times 1702

119896ave = 50mD

(88)

Example 4 Data from Example 1(c) was used for this caseand the semilog plot of density for each phase (oil water andgas) is plotted against log of Horner time

From Figures 15 and 16 the calculated slopes of the radialflow for oil and water phases are given as

120597120588119908

120597 ln 119905= 119898oil =

1627120583119876119888119900120588119900

119896ℎ= minus 0009

120597120588119908

120597 ln 119905= 119898water =

1627120583119876119888119900120588119900

119896ℎ= minus 00003

(89)

Journal of Petroleum Engineering 15

The estimated phase permeabilities are

119896oil =1627120583119876119888

119900120588119900

ℎ119898oil= 4352mD

119896water =1627120583119876119888

119900120588119900

ℎ119898water= 49mD

(90)

Also for the gas phase consider the following

The calculated slope from the semilog plot shown inFigure 15 is given as

120597119898 (120588119908)

120597 ln 119905= 119898gas =

1637120583119876119896ℎ120572119877

= minus 12502 (91)

And estimated gas permeability is

119896gas =1637120583119876

119900

ℎ120572119877119898gas= 179mD (92)

Using the empirical model for all three phases the averageestimated permeability is as follows

119896ave =4radic49 times 4352 times 1792

119896ave = 51mD

(93)

6 Conclusion

The following inferences were drawn from the six scenariosreviewed

(i) Results from the three-phase fluid empirical modeldeveloped for average effective 119896ave estimation rangebetween 47 and 57mD for Example 1 and give404mD for Example 2 which is within that used inthe simulation model and also estimated from BPR

(ii) The BPR PDENDWA and PDENWAT give the samestabilisation and the same 119896 estimates making it suit-able for interpretation of pressure transient analysis

(iii) It has been demonstrated in all 6 scenarios inves-tigated that the heavier fluid such as water andthe weighted average pressure-density equivalent ofall fluid give exact effective 119896 as the conventionalmethod

(iv) For oil reservoir system (scenarios (a) to (d)) 119896 =

500mD is achieved if the thickness contributing toflow ranges from 50 to 150 ft while for gas condensatereservoir (scenarios (a) to (b)) 119896 ranges between340 and 371mD which is achieved if the net sandthickness ℎ = 148ft is contributing to flow

(v) This approach also provides an estimate of the pos-sible fluid phase permeabilities and the of eachphase contribution to flow at a given point hence atseveral 1198891199011015840 stabilisation points the relative 119896 can begenerated

(vi) The derivatives for all output parameters display thesame wellbore and reservoir fingerprint as the BPRmethod

(vii) Generally where output parameters (BPR PDE-NOIL PDENGAS PDENWAT and PDENDW)depict different fingerprint the density derivative willserve as support to distinguish between reservoir andnonreservoir response

Nomenclature

119875 Pressure psi119879 Temperature oF119903 Radius ft119896 Permeability mDOslash Porosity fraction120583 Viscosity cp119905 Time hrs119902 Production rate bblday119861 Formation volume factor rbStb119862119905 Total compressibility psiminus1119903119908 Wellbore radius ftΔ119901 Change in pressure psiaℎ Formation thickness ft119860 Drainage area acres119875119908119891 Bottom-hole flowing pressure psi119875119894 Initial pressure psi

119885 Difference between two pointtime series119894 Subscript of an observed variable119888 Subscript of a calculated variableSTEYX SSE of data point119899 Number of data pointsCov Covariance of data point120575 Standard deviation119905119901 Cumulative production time119862119904 Wellbore storage constant119909 Mean of data point

Abbreviations

LBPR Local grid bottom-hole pressureLDENO Local grid oil densityLDENW Local grid water densityLDENG Local grid gas densityWBHP Well bottom-hole pressureBPR Well bottom-hole pressurePDENOIL Pressure equivalent of LDENOPDENGAS Pressure equivalent of LDENGPDENWAT Pressure equivalent of LDENWPDENDWA Pressure equivalent of density weighted

average (LDENO LDENG and LDENW)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Jalali S D Mohaghegh and R Gaskari ldquoIdentifying infilllocations and underperformer wells in mature fields usingmonthly production rate data Carthage Field Cotton Valley

16 Journal of Petroleum Engineering

Formation Texasrdquo in Proceedings of the SPE Eastern RegionalMeeting SPE Paper 104550 Canton Ohio USA October 2006

[2] D Tiab and A Kumar ldquoApplication of PD function to inter-ference analysisrdquo in Proceedings of the 51st Annual Fall Meetingof the Society of Petroleum Engineers of AIME Paper SPE 6053New Orleans La USA 1976

[3] D Tiab and A Kumar ldquoDetection and location of two parallelsealing fault around a wellrdquo in Proceedings of the 51st AnnualFall Meeting of the Society of Petroleum Engineers of AIME SPEPaper 6056 New Orleans La USA 1976

[4] D Bourdet T M Whittle A A Douglas and Y M Pirard ldquoAnew set of type curves simplifies well test analysisrdquo World Oilvol 196 no 6 pp 95ndash106 1983

[5] S Y Zheng ldquoFighting against non-unique solution problemsin heterogeneous reservoirs through numerical well testingrdquo inProceedings of the SPE Asia Pacific Oil and Gas Conference andExhibition SPE 100951 Adelaide Australia September 2006

[6] R N Horne and K O Temeng ldquoRecognition and location ofpinch-out boundaries by pressure transient analysisrdquo SPE 9905Society of Petroleum Engineers 1981

[7] L C Ayestaran R D Nurmi G A K Shehab and W S ElsildquoWelltest design and interpretation improved by integrated welltesting and geological effortsrdquo SPE 17945 1989

[8] G J Massonnat and D Bandiziol ldquoInterdependence betweengeology and well test interpretationrdquo SPE 22740 1991

[9] K F Du and G Stewart ldquoReservoir description from well testinterpretationrdquo in North Sea Oil and Gas ReservoirmdashIII pp339ndash356 Norwegian Institute of Technology (NTH) 1994

[10] S Y Zheng Well testing and characterisation of meanderingfluvial channel reservoirs [PhD thesis] Heriot-Watt University1997

[11] V T Biu and S Y Zheng ldquoA new approach in pressure transientanalysis part I improved diagnosis of flow regimes in oil and gaswellsrdquo in Proceedings of the 3rd EAGEAAPGWorkshop on TightReservoirs in theMiddle East AbuDhabi UnitedArab Emirates2015

[12] R N Horne Modern Welltest Analysis Petroway Palo AltoCalif USA 1995

[13] C S Matthews and D G Russell Pressure Buildup and FlowTests in Wells vol 1 ofMonograph Series SPE Richardson TexUSA 1967

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6 Journal of Petroleum Engineering

Gas

Oil

Water

Case 1 production test

Net sand thickness h = 250ftwater layerWell perforated hp = 30ft between oil andFlowing + buildup sequence

(a)

Gas

Oil

Water

Case 2 production test Flowing + buildup sequence

Net sand thickness h = 250ftWell perforated hp = 30ft inside the oil layer

(b)

Gas

Oil

Water

Case 3 production test Flowing + buildup sequence

Net sand thickness h = 250ftoil layerWell perforated hp = 30ft between gas and

(c)

Gas

Oil

Water

Case 4 falloff test Flowing + buildup sequence

Net sand thickness h = 250ftwater layerWell perforated hp = 30ft between oil and

(d)

Figure 3 Schematics of well perforation interval and sand thickness for oil + gas cap + water reservoir

obtained from the specialised plot This is an approximate ofthe input value in the simulation model Also the simulatoroutputs PDENDWA and PDENWAT in the simulation givethe same 119896 value while PDENGAS and PDENOIL differgiving 158 and 4037mD respectively At ℎ = 50 f t the best119896 estimate is obtained depicting ℎ = 50 f t as the thicknesscontributing to flowAt ℎ gt 50ft 119896 drops below the 119896 imputedin the model

Using (34) 119896ave = 473mD is obtained which isapproximately close to that of BPR hence a good estimateof the fluid phase permeabilities A summary of the result isshown in Tables 2 and 5

In scenario (b) thewell is completedwithin the oil sectionto capture the pressure and fluid densities changes around thewell The derivative in Figure 7 shows a good radial flow butwith noisy numerical artefact It is likely that the boundaryresponse is masked by numerical artefact Also a continuousdrop is seen after 10 hrs in the derivative

Using the estimated fluid phase permeabilities 119896ave from(34) is 472mD as shown in Table 3 which is the same asscenario (a) This is in line with the uniform 119896 used in thesimulation model Also the BPR gives a permeability value of500mD which is the same for PDENDWA and PDENWATbut differs with PDENGAS and PDENOIL that give 121 and

Journal of Petroleum Engineering 7

Water

Case 1 production test

Gas + condensate

Flowing + buildup sequence

condensate and water layerNet sand thickness h = 45m

Well perforated hp = 90m between gas

(a)

Water

Case 2 production test

Gas + condensate

Net sand thickness h = 45m

Flowing + buildup sequence

condensate layerWell perforated hp = 90m inside gas

(b)

Figure 4 Schematics of well perforation interval and sand thickness for gas condensate + water reservoir

00

5000

0 50 100 150 200 250 300 350

Prod

uctio

nin

ject

ion

rate

bop

d

Duration

Test design sequence

Case dCases a to c

minus25000

minus20000

minus15000

minus10000

minus5000

Figure 5 Production and injection test sequence for scenarios (a)to (d) (buildup and falloff)

Table 2 119870 estimates for new approach versus conventionalapproach for scenario (a)

Parameters Numerical density119896 (mD)

Equivalentℎ (ft)

BHP 508PDENA 508PDENG 158 50PDENO 4037PDENW 508

4844mD respectively At ℎ = 150ft (60 of sand thicknessand 83 of oil thickness) the estimated 119896 is still within range

00

00

01

10

100

0000 0001 0010 0100 1000 10000 100000Log time (h)

BHPPENDAPDENO

PDENGPDENW

508mD

4037mD

158mD

k =1627qB120583

mh

Case a derivatives for conventional and numerical density models

logdp998400

Figure 6 Derivative and 119870 estimation for scenario (a)

(119896 = 489mD)This indicates ℎ = 50ft or 83 of oil thicknesscontributing to flow

To test this approach in the gas column the well wascompleted in between the gas and oil layer which is consid-ered as scenario (c) and scenario (d) with the well completedwithin the oil layer but with water injection after flowingand shut-in sequence In both scenarios the multiphasefluid distribution is triggered at the wellbore in order tocapture the density changes for each phase and calculate fluidphase permeabilities First the well fluid densities equivalent

8 Journal of Petroleum Engineering

Case b derivatives for conventional and numerical density models

00

01

10

100

0000 0001 0010 0100 1000 10000 100000Log time (h)

BHPPENDAPDENO

PDENGPDENW

4842mDk =1627qB120583

mh

120mD

489mD

logdp998400

Figure 7 Derivative and 119870 estimation for scenario (b)

0001

0010

0100

1000

10000

0000 0001 0010 0100 1000 10000 100000Log time (h)

BHPPENDAPDENO

PDENGPDENW

412mD

209mD

Case c derivatives for conventional and numerical density models

4842mD

k =1627qB120583

mh

logdp998400

Figure 8 Derivative and 119870 estimation for scenario (c)

Table 3 119870 estimates for new approach versus conventionalapproach for scenario (b)

Parameters Numerical density119896 (mD)

Equivalentℎ (ft)

BHP 500PDENA 500PDENG 121 50PDENO 4842PDENW 500

pressures and pressures at bottom-hole for flowing andbuildup test are generated

The derivative for both scenarios declines after 3 hoursprobably due to fluid redistribution but is noisy (numericalartefact) in scenario (c) as shown in Figure 8 A good radial

Table 4 119870 estimates for new approach versus conventionalapproach for scenario (d)

Parameters Numerical Density119896 (mD) Equivalent ℎ (ft)

Scenario (d)

BHP 500 100PDENA 500PDENG 175 50PDENO 490 250PDENW 500 100

stabilisation for both the conventional method BPR and thedensity outputs PDENDWA PDENWAT PDENGAS andPDENOIL is observed Likewise as obtained in scenarios (a)and (b) a good 119870ave value of 57 and 52mD from (34) isobtained for scenarios (c) and (d) respectivelyThis is slightlyhigher than 119896 = 50mD imputed in the simulation model

For each of the fluid phase permeabilities estimates 119896value of 50mD is obtained using the conventional BPR forscenario (c) which is in line with the uniform 119896 in thesimulation model This is the same for the PDENDWA andPDENWAT Also at ℎ = 150ft (60 of sand thickness and83 of oil thickness) the estimated 119896 is still within range(119896 = 412mD)This also indicates that ℎ = 150ft or 83 of oilthickness is contributing to flowThe 119896 values for PDENGASandPDENOIL are 209 and 4842mD respectively Summaryof the result is shown in Table 4

However in scenario (d) the permeability value of500mD using the conventional BPR which is the samefor PDENDWA and PDENWAT is only achieved if ℎ =

100ft (40 of sand thickness and 56 of oil thickness)PDENGAS and PDENOIL differ giving 350mD at ℎ =

50ft and 1960mD at ℎ = 250ft respectively This indicatesthe impact of water injection on densities and pressureschanges around the well and consequently its impact on sandthickness contributing to flow The drop in derivative curvein Figure 9 depicts the impact of the injected water

In summary for scenarios (a) to (d) the 119896 value of500mD is achieved if the thickness contributing to flowranges from 50 to 150 ft Generally results indicate that in all4 scenarios investigated the heavier fluid such as water andthe weighted average pressure-density equivalent of all fluidgive exact effective 119896 as the BPR likewise the density outputsPDENDWA and PDENWAT Also results from Table 5 showthat the empirical model from (34) with all fluid phasepermeabilities gives an average effective 119896ave of 47ndash57mDwhich is within that used in the simulationmodel and also thesame as the estimated conventional approach This approachprovides an estimate of the possible fluid phase permeabilitiesand theof each phase contribution to flow hence at severalpoints the relative 119896 can be generated as shown in Table 5

Example 2 To capture the influence of highly compressiblefluid on estimated fluid phase permeabilities an example ongas condensate reservoir (volatile system) was tested Table 6presents a summary of the well and reservoir synthetic dataused for the buildup and drawdown simulated scenarios with

Journal of Petroleum Engineering 9

Table 5 Comparison of 119870 estimates between conventional and numerical density parameters

Scenario Numerical density method Calculated119870 (mD)

Conventional119896 (mD)

Simulationmodel 119870 (mD) Relative 119870

phasecontribution to flowPhases 119870 (mD)

(a)Gas 158 003 30Oil 4037 473 508 500 086 860

Water 508 011 110

(b)Gas 12 003 20Oil 4844 432 500 500 113 890

Water 50 012 90

(c)Gas 209 004 40Oil 4842 570 500 500 086 87

Water 50 009 90

(d)Gas 175 003 30Oil 490 523 500 500 095 880

Water 50 010 90

1000

Case d derivatives for conventional and numerical density models

0010

0100

10000

BHPPENDAPDENO

PDENGPDENW

0100 1000 10000 100000Log time (h)

k =1627qB120583

mh

209mD

500mD

4842mD

logdp998400

Figure 9 Derivative and 119870 estimation for scenario (d)

additional information given below It is required to generatethe pressure equivalent and derivative for each fluid phasecompare their diagnostic signatures and also determine thephases permeabilities and average reservoir permeability

Assumption (i) Condensate reservoir is completed with onewell

(ii) LGR is imposed around the well and far across toaccount for pressure and density changes

Gas condensate andwater densities around the local gridrefinement (wellbore) andWBHPwere output using the sim-ulatorrsquos keywords The following scenarios were evaluated

(a) Flowing + buildup sequence well perforated ℎ119901

=

30ft between gas condensate and water layer Netsand thickness ℎ = 150ft

Table 6 Summary of reservoir simulations data

Parameters Design valueEclipse model Black oilModel dimension 9 times 3 times 3Length by width ft by ft 1312 times 984Thickness ℎ ft 150Permeability 119870119909 by 119870119910 mD 4000 by 3000Porosity 30Well diameter ft 115Initial water saturation 119878119908

119894 60

Permeability 119870 mD 400Gas oil contact (GOC) ft 6890Oil water contact (OWC) ft 6890Initial pressure 119875

119894 psi 4495

Formation temperature 119879 ∘C 120 2000

(b) Flowing + buildup sequence well perforated ℎ119901

=

30ft inside gas condensate layer Net sand thicknessℎ = 150ft

Figure 10 shows the production and shut-in sequence for 2scenarios in Example 2

As in Example 1 the derivative in scenario (a) ofExample 2 also shows good radial flowwith no late time effectas seen in Figure 11 A negative half-slope fingerprint is seen at03 to 30 hrs of buildup and a good stabilisation from 40 hrsin the model parameters

The derivatives for all output parameters display the samewell and reservoir fingerprint (2 flow periods early to middletime response) but with different 1198891199011015840 stabilisation

Permeability value of 3707mD was obtained from theconventionalmethod BPRwhere 119896 = 1637119902119879119898ℎ is the samefor the PDENDWA and PDENWAT and 119898 is obtained fromthe specialised plot This is in line with the uniform 119896 (119896119909 =

400mD and 119896119910 = 300mD) imputed in the simulationmodel

10 Journal of Petroleum Engineering

00

5000000

10000000

15000000

20000000

25000000

0 40 80 120 160

Prod

uctio

n ra

te (s

cfd

)

Duration

Test design sequence

Cases a to b

Figure 10 Production test sequence (buildup) for Example 2

10000

Case a derivatives for conventional and numerical density models

00100

01000

BHPPENDAPDENO

PDENGPDENW

1422mD

3707mD

35747mD

0000 0001 0010 0100 1000 10000

100000

100000Log time (h)

k =1627qB120583

mh

logdp998400

Figure 11 Derivative and 119870 estimation for scenario (a)

However PDENGAS and PDENOIL differ giving 350mDand 35747mD respectively if ℎ = 148ft

Also the derivative in Figure 12 shows good radial flowand a negative half-slope fingerprint is seen at 05 to 20 hrsof buildup and a good stabilisation from 30 hrs in the modelparameters After 3 hrs of shut-in somenoisy data (numericalartefact) is seen in all the fluid phase derivatives but still goodenough to identify 119889119901

1015840 stabilisationFor each of the fluid phase permeabilities 119896 value of

3409mD at ℎ = 148ft is obtained using BPR which is thesame for the PDENDWA and PDENWAT PDENGAS andPDENOIL differ giving 1490 and 35141mD at ℎ = 148ftrespectively

For scenarios (a) and (b) a good 119896ave value of 4050and 4030mD is obtained respectively from the empiricalmodel (34) integrating all fluid phase permeabilities whichis within that used in the simulationmodel and the estimated

0001000

0010000

0100000

1000000

10000000

0000 0001 0010 0100 1000 10000 100000

Case b derivatives for conventional and numerical density models

BHPPENDAPDENO

PDENGPDENW

Log time (h)

k =1627qB120583

mh

1490mD

3410mD

35140mD

logdp998400

Figure 12 Derivative and119870 estimation for scenario (b)

Table 7 Summary of reservoir modelling properties imputed ineclipse model for case study 2

Parameters Numerical density119896 (mD)

Equivalentℎ (ft)

Scenario(a)

BHP 371PDENA 371PDENG 142 148PDENO 3575PDENW 371

Scenario(b)

BHP 341PDENA 341PDENG 149 148PDENO 3514PDENW 341

conventional approach From the 2 scenarios investigated ithas been demonstrated that the heavier fluid such as waterand the weighted average pressure-density equivalent of allfluid give exact effective 119896 as the conventional method BPRA summary of the result is shown in Table 7

This approach estimates the fluid phase permeabilitiesand the of each phase contribution to flow at a given pointhence at several points the relative 119896 can be generated asshown in Table 8

5 Density Related Radial Flow EquationDerivation for Each Fluid Phase

Radial diffusivity equation is given as

1119903

120597120588

120597119903[119903

120597120588

120597119903] =

120601120583119888

119896

120597120588

120597119905 (35)

Journal of Petroleum Engineering 11

Table 8 Comparison of 119870 estimates between conventional and numerical density parameters

Scenario Numerical density method Calculated119870 (mD)

Conventional119896 (mD)

Simulationmodel 119870 (mD) Relative 119870

phasecontribution to flowPhases 119870 (mD)

(a)Gas 1422 004 30Oil 35747 4046 3707 400 089 870

Water 3707 009 90

(b)Gas 149 004 40Oil 35141 4038 3409 400 089 880

Water 3409 008 90

For oil

1119903

120597120588

120597119903[119903

120597120588119900

120597119903] =

120601120583119900119888

119896119900

120597120588119900

120597119905 (36a)

For water phase

1119903

120597120588

120597119903[119903

120597120588119908

120597119903] =

120601120583119908119888

119896119908

120597120588119908

120597119905 (36b)

For gas phase

1119903

120597120588

120597119903[119903

120597120588119892

120597119903] =

120601120583119892119888

119896119892

120597120588119892

120597119905 (36c)

51 From Equation for Slight and Small Compressibility Suchas Water and Oil From (22)

119875 = 119875119900minus

120588120588119900+ 1

119888119900

(37)

Differentiating with respect to 120588

120597119875

120597120588= minus

1119888119900120588119900

120597119875 = minus120597120588

119888119900120588119900

(38)

From Darcy flow equation

119902 = minus2120587119896ℎ120583

119903120597120588

120597119903 (39)

Substitute for 120597119875

119902 =2120587119896ℎ120583

119903

119888119900120588119900

120597120588

120597119903 (40)

To derive the analytical density transient equation for eachphase the following assumptionconditions are applicable

For oil phase from (35)

1119903

120597120588

120597119903[119903

120597120588

120597119903] =

120601120583119888

119896

120597120588

120597119905 (41)

Initial condition is as follows

120588 (119903 119905 = 0) = 120588119894 (42a)

BC at the wellbore is as follows

lim119903rarr 0

2120587119896ℎ120583

119903

119888119900120588119900

120597120588

120597119903= 119876 (42b)

BC at infirmity is as follows

lim119903rarrinfin

120588 (119903 119905) = 120588119894 (42c)

Using the Boltzmann transformation consider the followingAssuming 120578 = 120601120583119888119903

2119896119905

120597120588

120597119903=

120597120588

120597120578

120597120578

120597119903=2120601120583119888119903119896119905

120597120588

120597120578=

1206011205831198881199032

119896119905

2119903

120597120588

120597120578=2120578119903

120597120588

120597120578 (43)

Therefore1119903

120597

120597119903[119903

120597120588

120597119903] =

21205781199032

120597

120597120578[2120578

120597120588

120597120578] =

41205781199032

120597

120597120578[120578

120597120588

120597120578] (44)

Differentiating with respect to 119903 is equal to differentiatingwith respect to 120578 multiply by 2120578119903

From (35) the LHS is resolved as follows

120597120588

120597119905=

120597120588

120597120578

120597120578

120597119905=

120601120583119888119903

1198961199052120597120588

120597120578= minus

120578

119905

120597120588

120597120578 (45)

120601120583119888

119896

120597120588

120597119905=

minus120601120583119888

119896

120578

119905

120597120588

120597119905=

minus1206011205831198881199032

119896119905

120578

1199032120597120588

120597120578=

minus1205782

1199032120597120588

120597120578 (46)

Equating (44) and (46) we have

120597

120597120578[120578

120597120588

120597120578] =

minus120578

4120597120588

120597120578 (47)

This is the simplified ordinary differential equation of 120588 as afunction of 120578

Apply boundary conditions as followsInitial condition is as follows

lim119903rarrinfin

120588 (120578) = 120588119894 (48a)

Substituting (43) into (40)

lim119903rarr 0

2120587119896ℎ120583

119903

119888119900120588119900

120597120588

120597119903= 119876 (48b)

lim119903rarr 0

[120578120597120588

120597120578] =

120601120583119888119900120588119900

4120587119896ℎ (49)

Assume that119898 = 120578(120597120588120597120578)

12 Journal of Petroleum Engineering

From (47)

120597119898

120597120578=

minus119898

4 (50)

Integrating from 120578 = 0 to 120578

997904rArr int

119898(120578)

119898(0)

120597119898

120597120578= minusint

120578

0

120597120578

4

997904rArr ln[119898 (120578)

119898 (0)] =

minus120578

4

997904rArr 119898(120578) = 119898 (0) 119890minus1205784

(51)

Recall boundary condition

lim119903rarr 0

[120578120597120588

120597120578] =

119876120583119888119900120588119900

4120587119896ℎ

119898 (0) =119876120583119888119900120588119900

4120587119896ℎ

119898 (120578) =119876120583119888119900120588119900

4120587119896ℎ119890minus1205784

(52)

Recall that119898 = 120578(120597120588120597120578)Therefore

120597120588 (120578)

120597120578=

120583119876119888119900120588119900

4120587119896ℎ119890minus1205784

120578 (53)

Integrating from 120578 = infin where 120588 = 120588119894

int

120588(120578)

120588119894

120597120588 = minusint

120578

infin

120583119876119888119900120588119900

4120587119896ℎ119890minus1205784

120578120597120578

120588 (120578) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎminusint

infin

120578

119890minus1205784

120578120597120578

(54)

where

120578 =120601120583119888119903

2

119896119905 (55)

Assuming 119906 = 1205784 then 120597120578120578 = 120597119906119906 and 119906 = 12060112058311988811990324119896119905

Hence

120588 = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎminusint

infin

12060112058311988811990324119896119905

119890minus1205784

120578120597120578 (56)

where

minus119864119894 (minus119909) = int

infin

119909

119890minus119906

119906120597119906 (57)

Known as the exponential integral function defined by [13]

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎ119864119894 (minus119909) (58)

where 119909 = 12060112058311988811990324119896119905 and

minus119864119894 (minus119909) = int

infin

119909

119890minus119906

119906120597119906 (59)

At large times 119909 will be small applying Taylorrsquos series andintegrating

minus119864119894 (minus119909) = ln (120574)

ln (120574) = ln (1781) = 05772(60)

where 120574 is known as Eulerrsquos numberFor 14119909 gt 25 consider the followingThen

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎ[ln119909+ ln 120574]

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎln[

1206011205831198881199032120574

4119896119905]

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎln [

22461198961199051206011205831198881199032

]

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎ[ln [

22461198961199051206011205831198881199032

]+ 080907]

(61)

Plotting 120588119908(119905) versus ln(119905) will yield a straight line at longer

time and the slope of the line is given as120597120588119908

120597 ln 119905= 119898oil =

120583119876119888119900120588119900

4120587119896ℎ (62)

Therefore

119896ℎ =120583119876119888119900120588119900

4120587119898oil (63)

52 For Water Phase The radial density equation is given as

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎ[ln [

22461198961199051206011205831198881199032

]+ 080907] (64)

Plotting 120588119908(119905) versus ln(119905) will yield a straight line at longer

time and the slope of the line is given as120597120588119908

120597 ln 119905= 119898water =

120583119876119888119900120588119900

4120587119896ℎ (65)

where

119896ℎ =120583119876119888119900120588119900

4120587119898water (66)

53 For Gas Phase From equation for compressible fluidsuch as gas consider the following

From (32)

119875 =119875119900120588 minus 119875119900

2120588119900119888119892

120588119900(1 minus 119875

119900119888119892)

120597119875

120597120588=

119875119900

120588119900(1 minus 119875

119900119888119892)

120597119875 =119875119900

120588119900(1 minus 119875

119900119888119892)120597120588

(67)

Journal of Petroleum Engineering 13

From the fundamental

119899 =119875119881

119911119877119879 (68)

At standard condition

119875119881

119911119879=

119875sc119881sc119879sc

119875 (5615119902)119911119879

=119875sc119876sc119879sc

119902 =119875sc119879sc

(119911119879

119875)

119876sc5615

119902 =119875sc119879sc

119876sc5615

(119911119879

119875) =

2120587119896ℎ120583

119903120597119901

120597119903

119902 =119875sc119879sc

119876sc5615

(119911119879) =2120587119896ℎ120583

119903119875120597119901

120597119903

(69)

where

120597119875 =119875119900

120588119900(1 minus 119875

119900119888119892)120597120588

120572 =119875119900

120588119900(1 minus 119875

119900119888119892)

120597119875 = 120572120597120588

119875 = 120588119877119911119879

(70)

Substitute into the equation where

119898(120588119894) minus119898 (120588

119908119891) =

2120583

int

120588119894

120588119908119891

120588120597120588

119898 (120588119894) minus119898 (120588

119908119891) =

120588119894

2minus 120588119908119891

2

120583

120588 = radic120588119894

2+ 120588119908119891

2

2

(71)

From the diffusivity equation for gas

1119903

120597120588

120597119903[119903

120597120588119892

120597119903] =

120601120583119892119888

119896119892

120597120588119892

120597119905 (72)

Initial condition is as follows

120588 (119903 119905 = 0) = 120588119894 (73)

BC at the wellbore is as follows

lim119903rarr 0

4120587119896ℎ120572119877120583

(5615119879sc

119875sc) 119903120588

120597120588

120597119903= 119876 (74)

BC at infirmity is as follows

lim119903rarrinfin

120588 (119903 119905) = 120588119894 (75)

where119877119875sc and119879sc are known as temperature pressure andgas constant at standard condition

Therefore integrating from 120578 = infin where 120588 = 120588119894

int

120588(120578)

120588119894

120588120597120588 = minusint

120578

infin

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)119890minus1205784

120578120597120578

1205882(120578) = 120588

119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

minusint

infin

120578

119890minus1205784

120578120597120578

(76)

where

120578 =120601120583119888119903

2

119896119905 (77)

Assume that 119906 = (1205784) (120597120578120578) = 120597119906119906 and 119906 = 12060112058311988811990324119896119905

Hence

1205882= 120588119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)minusint

infin

12060112058311988811990324119896119905

119890minus1205784

120578 (78)

where minus119864119894(minus119909) = intinfin

119909(119890minus119906

119906)120597119906Known as the exponential integral function defined by

[13]

1205882= 120588119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)minus119864119894 (minus119909) (79)

where 119909 = 12060112058311988811990324119896119905 and minus119864119894(minus119909) = int

infin

119909(119890minus119906

119906)120597119906At large times 119909will be small applying Taylorrsquos series and

integrating

minus119864119894 (minus119909) = ln (120574)

ln (120574) = ln (1781) = 05772(80)

where 120574 is known as Eulerrsquos numberFor 14119909 gt 25 consider the followingThen

1205882(119903 119905) = 120588

119894

2+

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) [ln119909+ ln 120574] (81)

1205882(119903 119905) = 120588

119894

2+

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) ln[120601120583119888119903

2120574

4119896119905]

1205882(119903 119905) = 120588

119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) ln [22461198961199051206011205831198881199032

]

119898 (120588119894) minus119898 (120588

119908119891) =

2120583

int

120588119894

120588119908119891

120588 120597120588

119898 (120588119908119891

) = 119898 (120588119894) minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

sdot [ln [22461198961199051206011205831198881199032

]+ 080907]

120597119898 (120588119908)

120597 ln 119905= 119898gas =

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

(82)

14 Journal of Petroleum Engineering

4160041650417004175041800418504190041950420004205042100

560856095609561056105611561156125612

ODENGDEN

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

1205880

(Ibm

ft)

120597120588w120597lnt

= moil =1627120583Qc01205880

kh= minus0009

120597m(120588w)

120597lnt= mgas =

1637120583Q

kh120572R= minus12195

120588g2

(Ibm

ft)

Figure 13 Oil and gas phase density versus log Horner time

6279

6279

6279

6279

6279

6279

6279

6279

01 1 10 100 1000 10000 1000001000000Horner time (h)

Density versus log Horner time

WDEN

120588w

(Ibm

ft)

120597120588w120597lnt

= mwater =1627120583Qc01205880

kh= minus00003

Figure 14 Water phase density versus log Horner time

Plotting 120588119908

2(119905) or 119898(120588

119908) versus ln(119905) will yield a straight line

at longer time and the slope of the line is given as follows

119896ℎ =120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)1

119898gas (83)

Example 3 Data from Example 1(a) was used for this caseand the semilog plot of density for each phase (oil water andgas) is plotted against log of Horner timelowastlowast

119876 used for 119896 calculation is the average rate for allflowing periods

From Figures 13 and 14 the calculated slopes of the radialflow for oil and water phases are given as

120597120588119908

120597 ln 119905= 119898oil =

1627120583119876119888119900120588119900

119896ℎ= minus 0009

120597120588119908

120597 ln 119905= 119898water =

1627120583119876119888119900120588119900

119896ℎ= minus 00003

(84)

The estimated phase permeabilities are

119896oil =1627120583119876119888

119900120588119900

ℎ119898oil= 445mD

119896water =1627120583119876119888

119900120588119900

ℎ119898water= 50mD

(85)

Also for the gas phase consider the following

ODENGDEN

422004225042300423504240042450425004255042600

5613561356145614561556155616

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

1205880

(Ibm

ft)

120597120588w120597lnt

= moil =1627120583Qc01205880

kh= minus0009

120597m(120588w)

120597lnt= mgas =

1637120583Q

kh120572R= minus12502

120588g2

(Ibm

ft)

Figure 15 Oil and gas phase density versus log Horner time

WDEN

6279627962796279627962796280628062806280

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

120588w

(Ibm

ft)

120597120588w120597lnt

= mwater =1627120583Qc01205880

kh= minus00003

Figure 16 Water phase density versus log Horner time

The calculated slope from the semilog plot shown inFigure 13 is given as

120597119898 (120588119908)

120597 ln 119905= 119898gas =

1637120583119876119896ℎ120572119877

= minus 12195 (86)

And estimated gas permeability is

119896gas =1637120583119876

119900

ℎ120572119877119898gas= 170mD (87)

Using the empirical model for all three phases the averageestimated permeability is as follows

119896ave =4radic50 times 445 times 1702

119896ave = 50mD

(88)

Example 4 Data from Example 1(c) was used for this caseand the semilog plot of density for each phase (oil water andgas) is plotted against log of Horner time

From Figures 15 and 16 the calculated slopes of the radialflow for oil and water phases are given as

120597120588119908

120597 ln 119905= 119898oil =

1627120583119876119888119900120588119900

119896ℎ= minus 0009

120597120588119908

120597 ln 119905= 119898water =

1627120583119876119888119900120588119900

119896ℎ= minus 00003

(89)

Journal of Petroleum Engineering 15

The estimated phase permeabilities are

119896oil =1627120583119876119888

119900120588119900

ℎ119898oil= 4352mD

119896water =1627120583119876119888

119900120588119900

ℎ119898water= 49mD

(90)

Also for the gas phase consider the following

The calculated slope from the semilog plot shown inFigure 15 is given as

120597119898 (120588119908)

120597 ln 119905= 119898gas =

1637120583119876119896ℎ120572119877

= minus 12502 (91)

And estimated gas permeability is

119896gas =1637120583119876

119900

ℎ120572119877119898gas= 179mD (92)

Using the empirical model for all three phases the averageestimated permeability is as follows

119896ave =4radic49 times 4352 times 1792

119896ave = 51mD

(93)

6 Conclusion

The following inferences were drawn from the six scenariosreviewed

(i) Results from the three-phase fluid empirical modeldeveloped for average effective 119896ave estimation rangebetween 47 and 57mD for Example 1 and give404mD for Example 2 which is within that used inthe simulation model and also estimated from BPR

(ii) The BPR PDENDWA and PDENWAT give the samestabilisation and the same 119896 estimates making it suit-able for interpretation of pressure transient analysis

(iii) It has been demonstrated in all 6 scenarios inves-tigated that the heavier fluid such as water andthe weighted average pressure-density equivalent ofall fluid give exact effective 119896 as the conventionalmethod

(iv) For oil reservoir system (scenarios (a) to (d)) 119896 =

500mD is achieved if the thickness contributing toflow ranges from 50 to 150 ft while for gas condensatereservoir (scenarios (a) to (b)) 119896 ranges between340 and 371mD which is achieved if the net sandthickness ℎ = 148ft is contributing to flow

(v) This approach also provides an estimate of the pos-sible fluid phase permeabilities and the of eachphase contribution to flow at a given point hence atseveral 1198891199011015840 stabilisation points the relative 119896 can begenerated

(vi) The derivatives for all output parameters display thesame wellbore and reservoir fingerprint as the BPRmethod

(vii) Generally where output parameters (BPR PDE-NOIL PDENGAS PDENWAT and PDENDW)depict different fingerprint the density derivative willserve as support to distinguish between reservoir andnonreservoir response

Nomenclature

119875 Pressure psi119879 Temperature oF119903 Radius ft119896 Permeability mDOslash Porosity fraction120583 Viscosity cp119905 Time hrs119902 Production rate bblday119861 Formation volume factor rbStb119862119905 Total compressibility psiminus1119903119908 Wellbore radius ftΔ119901 Change in pressure psiaℎ Formation thickness ft119860 Drainage area acres119875119908119891 Bottom-hole flowing pressure psi119875119894 Initial pressure psi

119885 Difference between two pointtime series119894 Subscript of an observed variable119888 Subscript of a calculated variableSTEYX SSE of data point119899 Number of data pointsCov Covariance of data point120575 Standard deviation119905119901 Cumulative production time119862119904 Wellbore storage constant119909 Mean of data point

Abbreviations

LBPR Local grid bottom-hole pressureLDENO Local grid oil densityLDENW Local grid water densityLDENG Local grid gas densityWBHP Well bottom-hole pressureBPR Well bottom-hole pressurePDENOIL Pressure equivalent of LDENOPDENGAS Pressure equivalent of LDENGPDENWAT Pressure equivalent of LDENWPDENDWA Pressure equivalent of density weighted

average (LDENO LDENG and LDENW)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Jalali S D Mohaghegh and R Gaskari ldquoIdentifying infilllocations and underperformer wells in mature fields usingmonthly production rate data Carthage Field Cotton Valley

16 Journal of Petroleum Engineering

Formation Texasrdquo in Proceedings of the SPE Eastern RegionalMeeting SPE Paper 104550 Canton Ohio USA October 2006

[2] D Tiab and A Kumar ldquoApplication of PD function to inter-ference analysisrdquo in Proceedings of the 51st Annual Fall Meetingof the Society of Petroleum Engineers of AIME Paper SPE 6053New Orleans La USA 1976

[3] D Tiab and A Kumar ldquoDetection and location of two parallelsealing fault around a wellrdquo in Proceedings of the 51st AnnualFall Meeting of the Society of Petroleum Engineers of AIME SPEPaper 6056 New Orleans La USA 1976

[4] D Bourdet T M Whittle A A Douglas and Y M Pirard ldquoAnew set of type curves simplifies well test analysisrdquo World Oilvol 196 no 6 pp 95ndash106 1983

[5] S Y Zheng ldquoFighting against non-unique solution problemsin heterogeneous reservoirs through numerical well testingrdquo inProceedings of the SPE Asia Pacific Oil and Gas Conference andExhibition SPE 100951 Adelaide Australia September 2006

[6] R N Horne and K O Temeng ldquoRecognition and location ofpinch-out boundaries by pressure transient analysisrdquo SPE 9905Society of Petroleum Engineers 1981

[7] L C Ayestaran R D Nurmi G A K Shehab and W S ElsildquoWelltest design and interpretation improved by integrated welltesting and geological effortsrdquo SPE 17945 1989

[8] G J Massonnat and D Bandiziol ldquoInterdependence betweengeology and well test interpretationrdquo SPE 22740 1991

[9] K F Du and G Stewart ldquoReservoir description from well testinterpretationrdquo in North Sea Oil and Gas ReservoirmdashIII pp339ndash356 Norwegian Institute of Technology (NTH) 1994

[10] S Y Zheng Well testing and characterisation of meanderingfluvial channel reservoirs [PhD thesis] Heriot-Watt University1997

[11] V T Biu and S Y Zheng ldquoA new approach in pressure transientanalysis part I improved diagnosis of flow regimes in oil and gaswellsrdquo in Proceedings of the 3rd EAGEAAPGWorkshop on TightReservoirs in theMiddle East AbuDhabi UnitedArab Emirates2015

[12] R N Horne Modern Welltest Analysis Petroway Palo AltoCalif USA 1995

[13] C S Matthews and D G Russell Pressure Buildup and FlowTests in Wells vol 1 ofMonograph Series SPE Richardson TexUSA 1967

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Journal of Petroleum Engineering 7

Water

Case 1 production test

Gas + condensate

Flowing + buildup sequence

condensate and water layerNet sand thickness h = 45m

Well perforated hp = 90m between gas

(a)

Water

Case 2 production test

Gas + condensate

Net sand thickness h = 45m

Flowing + buildup sequence

condensate layerWell perforated hp = 90m inside gas

(b)

Figure 4 Schematics of well perforation interval and sand thickness for gas condensate + water reservoir

00

5000

0 50 100 150 200 250 300 350

Prod

uctio

nin

ject

ion

rate

bop

d

Duration

Test design sequence

Case dCases a to c

minus25000

minus20000

minus15000

minus10000

minus5000

Figure 5 Production and injection test sequence for scenarios (a)to (d) (buildup and falloff)

Table 2 119870 estimates for new approach versus conventionalapproach for scenario (a)

Parameters Numerical density119896 (mD)

Equivalentℎ (ft)

BHP 508PDENA 508PDENG 158 50PDENO 4037PDENW 508

4844mD respectively At ℎ = 150ft (60 of sand thicknessand 83 of oil thickness) the estimated 119896 is still within range

00

00

01

10

100

0000 0001 0010 0100 1000 10000 100000Log time (h)

BHPPENDAPDENO

PDENGPDENW

508mD

4037mD

158mD

k =1627qB120583

mh

Case a derivatives for conventional and numerical density models

logdp998400

Figure 6 Derivative and 119870 estimation for scenario (a)

(119896 = 489mD)This indicates ℎ = 50ft or 83 of oil thicknesscontributing to flow

To test this approach in the gas column the well wascompleted in between the gas and oil layer which is consid-ered as scenario (c) and scenario (d) with the well completedwithin the oil layer but with water injection after flowingand shut-in sequence In both scenarios the multiphasefluid distribution is triggered at the wellbore in order tocapture the density changes for each phase and calculate fluidphase permeabilities First the well fluid densities equivalent

8 Journal of Petroleum Engineering

Case b derivatives for conventional and numerical density models

00

01

10

100

0000 0001 0010 0100 1000 10000 100000Log time (h)

BHPPENDAPDENO

PDENGPDENW

4842mDk =1627qB120583

mh

120mD

489mD

logdp998400

Figure 7 Derivative and 119870 estimation for scenario (b)

0001

0010

0100

1000

10000

0000 0001 0010 0100 1000 10000 100000Log time (h)

BHPPENDAPDENO

PDENGPDENW

412mD

209mD

Case c derivatives for conventional and numerical density models

4842mD

k =1627qB120583

mh

logdp998400

Figure 8 Derivative and 119870 estimation for scenario (c)

Table 3 119870 estimates for new approach versus conventionalapproach for scenario (b)

Parameters Numerical density119896 (mD)

Equivalentℎ (ft)

BHP 500PDENA 500PDENG 121 50PDENO 4842PDENW 500

pressures and pressures at bottom-hole for flowing andbuildup test are generated

The derivative for both scenarios declines after 3 hoursprobably due to fluid redistribution but is noisy (numericalartefact) in scenario (c) as shown in Figure 8 A good radial

Table 4 119870 estimates for new approach versus conventionalapproach for scenario (d)

Parameters Numerical Density119896 (mD) Equivalent ℎ (ft)

Scenario (d)

BHP 500 100PDENA 500PDENG 175 50PDENO 490 250PDENW 500 100

stabilisation for both the conventional method BPR and thedensity outputs PDENDWA PDENWAT PDENGAS andPDENOIL is observed Likewise as obtained in scenarios (a)and (b) a good 119870ave value of 57 and 52mD from (34) isobtained for scenarios (c) and (d) respectivelyThis is slightlyhigher than 119896 = 50mD imputed in the simulation model

For each of the fluid phase permeabilities estimates 119896value of 50mD is obtained using the conventional BPR forscenario (c) which is in line with the uniform 119896 in thesimulation model This is the same for the PDENDWA andPDENWAT Also at ℎ = 150ft (60 of sand thickness and83 of oil thickness) the estimated 119896 is still within range(119896 = 412mD)This also indicates that ℎ = 150ft or 83 of oilthickness is contributing to flowThe 119896 values for PDENGASandPDENOIL are 209 and 4842mD respectively Summaryof the result is shown in Table 4

However in scenario (d) the permeability value of500mD using the conventional BPR which is the samefor PDENDWA and PDENWAT is only achieved if ℎ =

100ft (40 of sand thickness and 56 of oil thickness)PDENGAS and PDENOIL differ giving 350mD at ℎ =

50ft and 1960mD at ℎ = 250ft respectively This indicatesthe impact of water injection on densities and pressureschanges around the well and consequently its impact on sandthickness contributing to flow The drop in derivative curvein Figure 9 depicts the impact of the injected water

In summary for scenarios (a) to (d) the 119896 value of500mD is achieved if the thickness contributing to flowranges from 50 to 150 ft Generally results indicate that in all4 scenarios investigated the heavier fluid such as water andthe weighted average pressure-density equivalent of all fluidgive exact effective 119896 as the BPR likewise the density outputsPDENDWA and PDENWAT Also results from Table 5 showthat the empirical model from (34) with all fluid phasepermeabilities gives an average effective 119896ave of 47ndash57mDwhich is within that used in the simulationmodel and also thesame as the estimated conventional approach This approachprovides an estimate of the possible fluid phase permeabilitiesand theof each phase contribution to flow hence at severalpoints the relative 119896 can be generated as shown in Table 5

Example 2 To capture the influence of highly compressiblefluid on estimated fluid phase permeabilities an example ongas condensate reservoir (volatile system) was tested Table 6presents a summary of the well and reservoir synthetic dataused for the buildup and drawdown simulated scenarios with

Journal of Petroleum Engineering 9

Table 5 Comparison of 119870 estimates between conventional and numerical density parameters

Scenario Numerical density method Calculated119870 (mD)

Conventional119896 (mD)

Simulationmodel 119870 (mD) Relative 119870

phasecontribution to flowPhases 119870 (mD)

(a)Gas 158 003 30Oil 4037 473 508 500 086 860

Water 508 011 110

(b)Gas 12 003 20Oil 4844 432 500 500 113 890

Water 50 012 90

(c)Gas 209 004 40Oil 4842 570 500 500 086 87

Water 50 009 90

(d)Gas 175 003 30Oil 490 523 500 500 095 880

Water 50 010 90

1000

Case d derivatives for conventional and numerical density models

0010

0100

10000

BHPPENDAPDENO

PDENGPDENW

0100 1000 10000 100000Log time (h)

k =1627qB120583

mh

209mD

500mD

4842mD

logdp998400

Figure 9 Derivative and 119870 estimation for scenario (d)

additional information given below It is required to generatethe pressure equivalent and derivative for each fluid phasecompare their diagnostic signatures and also determine thephases permeabilities and average reservoir permeability

Assumption (i) Condensate reservoir is completed with onewell

(ii) LGR is imposed around the well and far across toaccount for pressure and density changes

Gas condensate andwater densities around the local gridrefinement (wellbore) andWBHPwere output using the sim-ulatorrsquos keywords The following scenarios were evaluated

(a) Flowing + buildup sequence well perforated ℎ119901

=

30ft between gas condensate and water layer Netsand thickness ℎ = 150ft

Table 6 Summary of reservoir simulations data

Parameters Design valueEclipse model Black oilModel dimension 9 times 3 times 3Length by width ft by ft 1312 times 984Thickness ℎ ft 150Permeability 119870119909 by 119870119910 mD 4000 by 3000Porosity 30Well diameter ft 115Initial water saturation 119878119908

119894 60

Permeability 119870 mD 400Gas oil contact (GOC) ft 6890Oil water contact (OWC) ft 6890Initial pressure 119875

119894 psi 4495

Formation temperature 119879 ∘C 120 2000

(b) Flowing + buildup sequence well perforated ℎ119901

=

30ft inside gas condensate layer Net sand thicknessℎ = 150ft

Figure 10 shows the production and shut-in sequence for 2scenarios in Example 2

As in Example 1 the derivative in scenario (a) ofExample 2 also shows good radial flowwith no late time effectas seen in Figure 11 A negative half-slope fingerprint is seen at03 to 30 hrs of buildup and a good stabilisation from 40 hrsin the model parameters

The derivatives for all output parameters display the samewell and reservoir fingerprint (2 flow periods early to middletime response) but with different 1198891199011015840 stabilisation

Permeability value of 3707mD was obtained from theconventionalmethod BPRwhere 119896 = 1637119902119879119898ℎ is the samefor the PDENDWA and PDENWAT and 119898 is obtained fromthe specialised plot This is in line with the uniform 119896 (119896119909 =

400mD and 119896119910 = 300mD) imputed in the simulationmodel

10 Journal of Petroleum Engineering

00

5000000

10000000

15000000

20000000

25000000

0 40 80 120 160

Prod

uctio

n ra

te (s

cfd

)

Duration

Test design sequence

Cases a to b

Figure 10 Production test sequence (buildup) for Example 2

10000

Case a derivatives for conventional and numerical density models

00100

01000

BHPPENDAPDENO

PDENGPDENW

1422mD

3707mD

35747mD

0000 0001 0010 0100 1000 10000

100000

100000Log time (h)

k =1627qB120583

mh

logdp998400

Figure 11 Derivative and 119870 estimation for scenario (a)

However PDENGAS and PDENOIL differ giving 350mDand 35747mD respectively if ℎ = 148ft

Also the derivative in Figure 12 shows good radial flowand a negative half-slope fingerprint is seen at 05 to 20 hrsof buildup and a good stabilisation from 30 hrs in the modelparameters After 3 hrs of shut-in somenoisy data (numericalartefact) is seen in all the fluid phase derivatives but still goodenough to identify 119889119901

1015840 stabilisationFor each of the fluid phase permeabilities 119896 value of

3409mD at ℎ = 148ft is obtained using BPR which is thesame for the PDENDWA and PDENWAT PDENGAS andPDENOIL differ giving 1490 and 35141mD at ℎ = 148ftrespectively

For scenarios (a) and (b) a good 119896ave value of 4050and 4030mD is obtained respectively from the empiricalmodel (34) integrating all fluid phase permeabilities whichis within that used in the simulationmodel and the estimated

0001000

0010000

0100000

1000000

10000000

0000 0001 0010 0100 1000 10000 100000

Case b derivatives for conventional and numerical density models

BHPPENDAPDENO

PDENGPDENW

Log time (h)

k =1627qB120583

mh

1490mD

3410mD

35140mD

logdp998400

Figure 12 Derivative and119870 estimation for scenario (b)

Table 7 Summary of reservoir modelling properties imputed ineclipse model for case study 2

Parameters Numerical density119896 (mD)

Equivalentℎ (ft)

Scenario(a)

BHP 371PDENA 371PDENG 142 148PDENO 3575PDENW 371

Scenario(b)

BHP 341PDENA 341PDENG 149 148PDENO 3514PDENW 341

conventional approach From the 2 scenarios investigated ithas been demonstrated that the heavier fluid such as waterand the weighted average pressure-density equivalent of allfluid give exact effective 119896 as the conventional method BPRA summary of the result is shown in Table 7

This approach estimates the fluid phase permeabilitiesand the of each phase contribution to flow at a given pointhence at several points the relative 119896 can be generated asshown in Table 8

5 Density Related Radial Flow EquationDerivation for Each Fluid Phase

Radial diffusivity equation is given as

1119903

120597120588

120597119903[119903

120597120588

120597119903] =

120601120583119888

119896

120597120588

120597119905 (35)

Journal of Petroleum Engineering 11

Table 8 Comparison of 119870 estimates between conventional and numerical density parameters

Scenario Numerical density method Calculated119870 (mD)

Conventional119896 (mD)

Simulationmodel 119870 (mD) Relative 119870

phasecontribution to flowPhases 119870 (mD)

(a)Gas 1422 004 30Oil 35747 4046 3707 400 089 870

Water 3707 009 90

(b)Gas 149 004 40Oil 35141 4038 3409 400 089 880

Water 3409 008 90

For oil

1119903

120597120588

120597119903[119903

120597120588119900

120597119903] =

120601120583119900119888

119896119900

120597120588119900

120597119905 (36a)

For water phase

1119903

120597120588

120597119903[119903

120597120588119908

120597119903] =

120601120583119908119888

119896119908

120597120588119908

120597119905 (36b)

For gas phase

1119903

120597120588

120597119903[119903

120597120588119892

120597119903] =

120601120583119892119888

119896119892

120597120588119892

120597119905 (36c)

51 From Equation for Slight and Small Compressibility Suchas Water and Oil From (22)

119875 = 119875119900minus

120588120588119900+ 1

119888119900

(37)

Differentiating with respect to 120588

120597119875

120597120588= minus

1119888119900120588119900

120597119875 = minus120597120588

119888119900120588119900

(38)

From Darcy flow equation

119902 = minus2120587119896ℎ120583

119903120597120588

120597119903 (39)

Substitute for 120597119875

119902 =2120587119896ℎ120583

119903

119888119900120588119900

120597120588

120597119903 (40)

To derive the analytical density transient equation for eachphase the following assumptionconditions are applicable

For oil phase from (35)

1119903

120597120588

120597119903[119903

120597120588

120597119903] =

120601120583119888

119896

120597120588

120597119905 (41)

Initial condition is as follows

120588 (119903 119905 = 0) = 120588119894 (42a)

BC at the wellbore is as follows

lim119903rarr 0

2120587119896ℎ120583

119903

119888119900120588119900

120597120588

120597119903= 119876 (42b)

BC at infirmity is as follows

lim119903rarrinfin

120588 (119903 119905) = 120588119894 (42c)

Using the Boltzmann transformation consider the followingAssuming 120578 = 120601120583119888119903

2119896119905

120597120588

120597119903=

120597120588

120597120578

120597120578

120597119903=2120601120583119888119903119896119905

120597120588

120597120578=

1206011205831198881199032

119896119905

2119903

120597120588

120597120578=2120578119903

120597120588

120597120578 (43)

Therefore1119903

120597

120597119903[119903

120597120588

120597119903] =

21205781199032

120597

120597120578[2120578

120597120588

120597120578] =

41205781199032

120597

120597120578[120578

120597120588

120597120578] (44)

Differentiating with respect to 119903 is equal to differentiatingwith respect to 120578 multiply by 2120578119903

From (35) the LHS is resolved as follows

120597120588

120597119905=

120597120588

120597120578

120597120578

120597119905=

120601120583119888119903

1198961199052120597120588

120597120578= minus

120578

119905

120597120588

120597120578 (45)

120601120583119888

119896

120597120588

120597119905=

minus120601120583119888

119896

120578

119905

120597120588

120597119905=

minus1206011205831198881199032

119896119905

120578

1199032120597120588

120597120578=

minus1205782

1199032120597120588

120597120578 (46)

Equating (44) and (46) we have

120597

120597120578[120578

120597120588

120597120578] =

minus120578

4120597120588

120597120578 (47)

This is the simplified ordinary differential equation of 120588 as afunction of 120578

Apply boundary conditions as followsInitial condition is as follows

lim119903rarrinfin

120588 (120578) = 120588119894 (48a)

Substituting (43) into (40)

lim119903rarr 0

2120587119896ℎ120583

119903

119888119900120588119900

120597120588

120597119903= 119876 (48b)

lim119903rarr 0

[120578120597120588

120597120578] =

120601120583119888119900120588119900

4120587119896ℎ (49)

Assume that119898 = 120578(120597120588120597120578)

12 Journal of Petroleum Engineering

From (47)

120597119898

120597120578=

minus119898

4 (50)

Integrating from 120578 = 0 to 120578

997904rArr int

119898(120578)

119898(0)

120597119898

120597120578= minusint

120578

0

120597120578

4

997904rArr ln[119898 (120578)

119898 (0)] =

minus120578

4

997904rArr 119898(120578) = 119898 (0) 119890minus1205784

(51)

Recall boundary condition

lim119903rarr 0

[120578120597120588

120597120578] =

119876120583119888119900120588119900

4120587119896ℎ

119898 (0) =119876120583119888119900120588119900

4120587119896ℎ

119898 (120578) =119876120583119888119900120588119900

4120587119896ℎ119890minus1205784

(52)

Recall that119898 = 120578(120597120588120597120578)Therefore

120597120588 (120578)

120597120578=

120583119876119888119900120588119900

4120587119896ℎ119890minus1205784

120578 (53)

Integrating from 120578 = infin where 120588 = 120588119894

int

120588(120578)

120588119894

120597120588 = minusint

120578

infin

120583119876119888119900120588119900

4120587119896ℎ119890minus1205784

120578120597120578

120588 (120578) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎminusint

infin

120578

119890minus1205784

120578120597120578

(54)

where

120578 =120601120583119888119903

2

119896119905 (55)

Assuming 119906 = 1205784 then 120597120578120578 = 120597119906119906 and 119906 = 12060112058311988811990324119896119905

Hence

120588 = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎminusint

infin

12060112058311988811990324119896119905

119890minus1205784

120578120597120578 (56)

where

minus119864119894 (minus119909) = int

infin

119909

119890minus119906

119906120597119906 (57)

Known as the exponential integral function defined by [13]

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎ119864119894 (minus119909) (58)

where 119909 = 12060112058311988811990324119896119905 and

minus119864119894 (minus119909) = int

infin

119909

119890minus119906

119906120597119906 (59)

At large times 119909 will be small applying Taylorrsquos series andintegrating

minus119864119894 (minus119909) = ln (120574)

ln (120574) = ln (1781) = 05772(60)

where 120574 is known as Eulerrsquos numberFor 14119909 gt 25 consider the followingThen

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎ[ln119909+ ln 120574]

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎln[

1206011205831198881199032120574

4119896119905]

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎln [

22461198961199051206011205831198881199032

]

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎ[ln [

22461198961199051206011205831198881199032

]+ 080907]

(61)

Plotting 120588119908(119905) versus ln(119905) will yield a straight line at longer

time and the slope of the line is given as120597120588119908

120597 ln 119905= 119898oil =

120583119876119888119900120588119900

4120587119896ℎ (62)

Therefore

119896ℎ =120583119876119888119900120588119900

4120587119898oil (63)

52 For Water Phase The radial density equation is given as

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎ[ln [

22461198961199051206011205831198881199032

]+ 080907] (64)

Plotting 120588119908(119905) versus ln(119905) will yield a straight line at longer

time and the slope of the line is given as120597120588119908

120597 ln 119905= 119898water =

120583119876119888119900120588119900

4120587119896ℎ (65)

where

119896ℎ =120583119876119888119900120588119900

4120587119898water (66)

53 For Gas Phase From equation for compressible fluidsuch as gas consider the following

From (32)

119875 =119875119900120588 minus 119875119900

2120588119900119888119892

120588119900(1 minus 119875

119900119888119892)

120597119875

120597120588=

119875119900

120588119900(1 minus 119875

119900119888119892)

120597119875 =119875119900

120588119900(1 minus 119875

119900119888119892)120597120588

(67)

Journal of Petroleum Engineering 13

From the fundamental

119899 =119875119881

119911119877119879 (68)

At standard condition

119875119881

119911119879=

119875sc119881sc119879sc

119875 (5615119902)119911119879

=119875sc119876sc119879sc

119902 =119875sc119879sc

(119911119879

119875)

119876sc5615

119902 =119875sc119879sc

119876sc5615

(119911119879

119875) =

2120587119896ℎ120583

119903120597119901

120597119903

119902 =119875sc119879sc

119876sc5615

(119911119879) =2120587119896ℎ120583

119903119875120597119901

120597119903

(69)

where

120597119875 =119875119900

120588119900(1 minus 119875

119900119888119892)120597120588

120572 =119875119900

120588119900(1 minus 119875

119900119888119892)

120597119875 = 120572120597120588

119875 = 120588119877119911119879

(70)

Substitute into the equation where

119898(120588119894) minus119898 (120588

119908119891) =

2120583

int

120588119894

120588119908119891

120588120597120588

119898 (120588119894) minus119898 (120588

119908119891) =

120588119894

2minus 120588119908119891

2

120583

120588 = radic120588119894

2+ 120588119908119891

2

2

(71)

From the diffusivity equation for gas

1119903

120597120588

120597119903[119903

120597120588119892

120597119903] =

120601120583119892119888

119896119892

120597120588119892

120597119905 (72)

Initial condition is as follows

120588 (119903 119905 = 0) = 120588119894 (73)

BC at the wellbore is as follows

lim119903rarr 0

4120587119896ℎ120572119877120583

(5615119879sc

119875sc) 119903120588

120597120588

120597119903= 119876 (74)

BC at infirmity is as follows

lim119903rarrinfin

120588 (119903 119905) = 120588119894 (75)

where119877119875sc and119879sc are known as temperature pressure andgas constant at standard condition

Therefore integrating from 120578 = infin where 120588 = 120588119894

int

120588(120578)

120588119894

120588120597120588 = minusint

120578

infin

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)119890minus1205784

120578120597120578

1205882(120578) = 120588

119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

minusint

infin

120578

119890minus1205784

120578120597120578

(76)

where

120578 =120601120583119888119903

2

119896119905 (77)

Assume that 119906 = (1205784) (120597120578120578) = 120597119906119906 and 119906 = 12060112058311988811990324119896119905

Hence

1205882= 120588119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)minusint

infin

12060112058311988811990324119896119905

119890minus1205784

120578 (78)

where minus119864119894(minus119909) = intinfin

119909(119890minus119906

119906)120597119906Known as the exponential integral function defined by

[13]

1205882= 120588119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)minus119864119894 (minus119909) (79)

where 119909 = 12060112058311988811990324119896119905 and minus119864119894(minus119909) = int

infin

119909(119890minus119906

119906)120597119906At large times 119909will be small applying Taylorrsquos series and

integrating

minus119864119894 (minus119909) = ln (120574)

ln (120574) = ln (1781) = 05772(80)

where 120574 is known as Eulerrsquos numberFor 14119909 gt 25 consider the followingThen

1205882(119903 119905) = 120588

119894

2+

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) [ln119909+ ln 120574] (81)

1205882(119903 119905) = 120588

119894

2+

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) ln[120601120583119888119903

2120574

4119896119905]

1205882(119903 119905) = 120588

119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) ln [22461198961199051206011205831198881199032

]

119898 (120588119894) minus119898 (120588

119908119891) =

2120583

int

120588119894

120588119908119891

120588 120597120588

119898 (120588119908119891

) = 119898 (120588119894) minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

sdot [ln [22461198961199051206011205831198881199032

]+ 080907]

120597119898 (120588119908)

120597 ln 119905= 119898gas =

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

(82)

14 Journal of Petroleum Engineering

4160041650417004175041800418504190041950420004205042100

560856095609561056105611561156125612

ODENGDEN

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

1205880

(Ibm

ft)

120597120588w120597lnt

= moil =1627120583Qc01205880

kh= minus0009

120597m(120588w)

120597lnt= mgas =

1637120583Q

kh120572R= minus12195

120588g2

(Ibm

ft)

Figure 13 Oil and gas phase density versus log Horner time

6279

6279

6279

6279

6279

6279

6279

6279

01 1 10 100 1000 10000 1000001000000Horner time (h)

Density versus log Horner time

WDEN

120588w

(Ibm

ft)

120597120588w120597lnt

= mwater =1627120583Qc01205880

kh= minus00003

Figure 14 Water phase density versus log Horner time

Plotting 120588119908

2(119905) or 119898(120588

119908) versus ln(119905) will yield a straight line

at longer time and the slope of the line is given as follows

119896ℎ =120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)1

119898gas (83)

Example 3 Data from Example 1(a) was used for this caseand the semilog plot of density for each phase (oil water andgas) is plotted against log of Horner timelowastlowast

119876 used for 119896 calculation is the average rate for allflowing periods

From Figures 13 and 14 the calculated slopes of the radialflow for oil and water phases are given as

120597120588119908

120597 ln 119905= 119898oil =

1627120583119876119888119900120588119900

119896ℎ= minus 0009

120597120588119908

120597 ln 119905= 119898water =

1627120583119876119888119900120588119900

119896ℎ= minus 00003

(84)

The estimated phase permeabilities are

119896oil =1627120583119876119888

119900120588119900

ℎ119898oil= 445mD

119896water =1627120583119876119888

119900120588119900

ℎ119898water= 50mD

(85)

Also for the gas phase consider the following

ODENGDEN

422004225042300423504240042450425004255042600

5613561356145614561556155616

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

1205880

(Ibm

ft)

120597120588w120597lnt

= moil =1627120583Qc01205880

kh= minus0009

120597m(120588w)

120597lnt= mgas =

1637120583Q

kh120572R= minus12502

120588g2

(Ibm

ft)

Figure 15 Oil and gas phase density versus log Horner time

WDEN

6279627962796279627962796280628062806280

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

120588w

(Ibm

ft)

120597120588w120597lnt

= mwater =1627120583Qc01205880

kh= minus00003

Figure 16 Water phase density versus log Horner time

The calculated slope from the semilog plot shown inFigure 13 is given as

120597119898 (120588119908)

120597 ln 119905= 119898gas =

1637120583119876119896ℎ120572119877

= minus 12195 (86)

And estimated gas permeability is

119896gas =1637120583119876

119900

ℎ120572119877119898gas= 170mD (87)

Using the empirical model for all three phases the averageestimated permeability is as follows

119896ave =4radic50 times 445 times 1702

119896ave = 50mD

(88)

Example 4 Data from Example 1(c) was used for this caseand the semilog plot of density for each phase (oil water andgas) is plotted against log of Horner time

From Figures 15 and 16 the calculated slopes of the radialflow for oil and water phases are given as

120597120588119908

120597 ln 119905= 119898oil =

1627120583119876119888119900120588119900

119896ℎ= minus 0009

120597120588119908

120597 ln 119905= 119898water =

1627120583119876119888119900120588119900

119896ℎ= minus 00003

(89)

Journal of Petroleum Engineering 15

The estimated phase permeabilities are

119896oil =1627120583119876119888

119900120588119900

ℎ119898oil= 4352mD

119896water =1627120583119876119888

119900120588119900

ℎ119898water= 49mD

(90)

Also for the gas phase consider the following

The calculated slope from the semilog plot shown inFigure 15 is given as

120597119898 (120588119908)

120597 ln 119905= 119898gas =

1637120583119876119896ℎ120572119877

= minus 12502 (91)

And estimated gas permeability is

119896gas =1637120583119876

119900

ℎ120572119877119898gas= 179mD (92)

Using the empirical model for all three phases the averageestimated permeability is as follows

119896ave =4radic49 times 4352 times 1792

119896ave = 51mD

(93)

6 Conclusion

The following inferences were drawn from the six scenariosreviewed

(i) Results from the three-phase fluid empirical modeldeveloped for average effective 119896ave estimation rangebetween 47 and 57mD for Example 1 and give404mD for Example 2 which is within that used inthe simulation model and also estimated from BPR

(ii) The BPR PDENDWA and PDENWAT give the samestabilisation and the same 119896 estimates making it suit-able for interpretation of pressure transient analysis

(iii) It has been demonstrated in all 6 scenarios inves-tigated that the heavier fluid such as water andthe weighted average pressure-density equivalent ofall fluid give exact effective 119896 as the conventionalmethod

(iv) For oil reservoir system (scenarios (a) to (d)) 119896 =

500mD is achieved if the thickness contributing toflow ranges from 50 to 150 ft while for gas condensatereservoir (scenarios (a) to (b)) 119896 ranges between340 and 371mD which is achieved if the net sandthickness ℎ = 148ft is contributing to flow

(v) This approach also provides an estimate of the pos-sible fluid phase permeabilities and the of eachphase contribution to flow at a given point hence atseveral 1198891199011015840 stabilisation points the relative 119896 can begenerated

(vi) The derivatives for all output parameters display thesame wellbore and reservoir fingerprint as the BPRmethod

(vii) Generally where output parameters (BPR PDE-NOIL PDENGAS PDENWAT and PDENDW)depict different fingerprint the density derivative willserve as support to distinguish between reservoir andnonreservoir response

Nomenclature

119875 Pressure psi119879 Temperature oF119903 Radius ft119896 Permeability mDOslash Porosity fraction120583 Viscosity cp119905 Time hrs119902 Production rate bblday119861 Formation volume factor rbStb119862119905 Total compressibility psiminus1119903119908 Wellbore radius ftΔ119901 Change in pressure psiaℎ Formation thickness ft119860 Drainage area acres119875119908119891 Bottom-hole flowing pressure psi119875119894 Initial pressure psi

119885 Difference between two pointtime series119894 Subscript of an observed variable119888 Subscript of a calculated variableSTEYX SSE of data point119899 Number of data pointsCov Covariance of data point120575 Standard deviation119905119901 Cumulative production time119862119904 Wellbore storage constant119909 Mean of data point

Abbreviations

LBPR Local grid bottom-hole pressureLDENO Local grid oil densityLDENW Local grid water densityLDENG Local grid gas densityWBHP Well bottom-hole pressureBPR Well bottom-hole pressurePDENOIL Pressure equivalent of LDENOPDENGAS Pressure equivalent of LDENGPDENWAT Pressure equivalent of LDENWPDENDWA Pressure equivalent of density weighted

average (LDENO LDENG and LDENW)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Jalali S D Mohaghegh and R Gaskari ldquoIdentifying infilllocations and underperformer wells in mature fields usingmonthly production rate data Carthage Field Cotton Valley

16 Journal of Petroleum Engineering

Formation Texasrdquo in Proceedings of the SPE Eastern RegionalMeeting SPE Paper 104550 Canton Ohio USA October 2006

[2] D Tiab and A Kumar ldquoApplication of PD function to inter-ference analysisrdquo in Proceedings of the 51st Annual Fall Meetingof the Society of Petroleum Engineers of AIME Paper SPE 6053New Orleans La USA 1976

[3] D Tiab and A Kumar ldquoDetection and location of two parallelsealing fault around a wellrdquo in Proceedings of the 51st AnnualFall Meeting of the Society of Petroleum Engineers of AIME SPEPaper 6056 New Orleans La USA 1976

[4] D Bourdet T M Whittle A A Douglas and Y M Pirard ldquoAnew set of type curves simplifies well test analysisrdquo World Oilvol 196 no 6 pp 95ndash106 1983

[5] S Y Zheng ldquoFighting against non-unique solution problemsin heterogeneous reservoirs through numerical well testingrdquo inProceedings of the SPE Asia Pacific Oil and Gas Conference andExhibition SPE 100951 Adelaide Australia September 2006

[6] R N Horne and K O Temeng ldquoRecognition and location ofpinch-out boundaries by pressure transient analysisrdquo SPE 9905Society of Petroleum Engineers 1981

[7] L C Ayestaran R D Nurmi G A K Shehab and W S ElsildquoWelltest design and interpretation improved by integrated welltesting and geological effortsrdquo SPE 17945 1989

[8] G J Massonnat and D Bandiziol ldquoInterdependence betweengeology and well test interpretationrdquo SPE 22740 1991

[9] K F Du and G Stewart ldquoReservoir description from well testinterpretationrdquo in North Sea Oil and Gas ReservoirmdashIII pp339ndash356 Norwegian Institute of Technology (NTH) 1994

[10] S Y Zheng Well testing and characterisation of meanderingfluvial channel reservoirs [PhD thesis] Heriot-Watt University1997

[11] V T Biu and S Y Zheng ldquoA new approach in pressure transientanalysis part I improved diagnosis of flow regimes in oil and gaswellsrdquo in Proceedings of the 3rd EAGEAAPGWorkshop on TightReservoirs in theMiddle East AbuDhabi UnitedArab Emirates2015

[12] R N Horne Modern Welltest Analysis Petroway Palo AltoCalif USA 1995

[13] C S Matthews and D G Russell Pressure Buildup and FlowTests in Wells vol 1 ofMonograph Series SPE Richardson TexUSA 1967

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8 Journal of Petroleum Engineering

Case b derivatives for conventional and numerical density models

00

01

10

100

0000 0001 0010 0100 1000 10000 100000Log time (h)

BHPPENDAPDENO

PDENGPDENW

4842mDk =1627qB120583

mh

120mD

489mD

logdp998400

Figure 7 Derivative and 119870 estimation for scenario (b)

0001

0010

0100

1000

10000

0000 0001 0010 0100 1000 10000 100000Log time (h)

BHPPENDAPDENO

PDENGPDENW

412mD

209mD

Case c derivatives for conventional and numerical density models

4842mD

k =1627qB120583

mh

logdp998400

Figure 8 Derivative and 119870 estimation for scenario (c)

Table 3 119870 estimates for new approach versus conventionalapproach for scenario (b)

Parameters Numerical density119896 (mD)

Equivalentℎ (ft)

BHP 500PDENA 500PDENG 121 50PDENO 4842PDENW 500

pressures and pressures at bottom-hole for flowing andbuildup test are generated

The derivative for both scenarios declines after 3 hoursprobably due to fluid redistribution but is noisy (numericalartefact) in scenario (c) as shown in Figure 8 A good radial

Table 4 119870 estimates for new approach versus conventionalapproach for scenario (d)

Parameters Numerical Density119896 (mD) Equivalent ℎ (ft)

Scenario (d)

BHP 500 100PDENA 500PDENG 175 50PDENO 490 250PDENW 500 100

stabilisation for both the conventional method BPR and thedensity outputs PDENDWA PDENWAT PDENGAS andPDENOIL is observed Likewise as obtained in scenarios (a)and (b) a good 119870ave value of 57 and 52mD from (34) isobtained for scenarios (c) and (d) respectivelyThis is slightlyhigher than 119896 = 50mD imputed in the simulation model

For each of the fluid phase permeabilities estimates 119896value of 50mD is obtained using the conventional BPR forscenario (c) which is in line with the uniform 119896 in thesimulation model This is the same for the PDENDWA andPDENWAT Also at ℎ = 150ft (60 of sand thickness and83 of oil thickness) the estimated 119896 is still within range(119896 = 412mD)This also indicates that ℎ = 150ft or 83 of oilthickness is contributing to flowThe 119896 values for PDENGASandPDENOIL are 209 and 4842mD respectively Summaryof the result is shown in Table 4

However in scenario (d) the permeability value of500mD using the conventional BPR which is the samefor PDENDWA and PDENWAT is only achieved if ℎ =

100ft (40 of sand thickness and 56 of oil thickness)PDENGAS and PDENOIL differ giving 350mD at ℎ =

50ft and 1960mD at ℎ = 250ft respectively This indicatesthe impact of water injection on densities and pressureschanges around the well and consequently its impact on sandthickness contributing to flow The drop in derivative curvein Figure 9 depicts the impact of the injected water

In summary for scenarios (a) to (d) the 119896 value of500mD is achieved if the thickness contributing to flowranges from 50 to 150 ft Generally results indicate that in all4 scenarios investigated the heavier fluid such as water andthe weighted average pressure-density equivalent of all fluidgive exact effective 119896 as the BPR likewise the density outputsPDENDWA and PDENWAT Also results from Table 5 showthat the empirical model from (34) with all fluid phasepermeabilities gives an average effective 119896ave of 47ndash57mDwhich is within that used in the simulationmodel and also thesame as the estimated conventional approach This approachprovides an estimate of the possible fluid phase permeabilitiesand theof each phase contribution to flow hence at severalpoints the relative 119896 can be generated as shown in Table 5

Example 2 To capture the influence of highly compressiblefluid on estimated fluid phase permeabilities an example ongas condensate reservoir (volatile system) was tested Table 6presents a summary of the well and reservoir synthetic dataused for the buildup and drawdown simulated scenarios with

Journal of Petroleum Engineering 9

Table 5 Comparison of 119870 estimates between conventional and numerical density parameters

Scenario Numerical density method Calculated119870 (mD)

Conventional119896 (mD)

Simulationmodel 119870 (mD) Relative 119870

phasecontribution to flowPhases 119870 (mD)

(a)Gas 158 003 30Oil 4037 473 508 500 086 860

Water 508 011 110

(b)Gas 12 003 20Oil 4844 432 500 500 113 890

Water 50 012 90

(c)Gas 209 004 40Oil 4842 570 500 500 086 87

Water 50 009 90

(d)Gas 175 003 30Oil 490 523 500 500 095 880

Water 50 010 90

1000

Case d derivatives for conventional and numerical density models

0010

0100

10000

BHPPENDAPDENO

PDENGPDENW

0100 1000 10000 100000Log time (h)

k =1627qB120583

mh

209mD

500mD

4842mD

logdp998400

Figure 9 Derivative and 119870 estimation for scenario (d)

additional information given below It is required to generatethe pressure equivalent and derivative for each fluid phasecompare their diagnostic signatures and also determine thephases permeabilities and average reservoir permeability

Assumption (i) Condensate reservoir is completed with onewell

(ii) LGR is imposed around the well and far across toaccount for pressure and density changes

Gas condensate andwater densities around the local gridrefinement (wellbore) andWBHPwere output using the sim-ulatorrsquos keywords The following scenarios were evaluated

(a) Flowing + buildup sequence well perforated ℎ119901

=

30ft between gas condensate and water layer Netsand thickness ℎ = 150ft

Table 6 Summary of reservoir simulations data

Parameters Design valueEclipse model Black oilModel dimension 9 times 3 times 3Length by width ft by ft 1312 times 984Thickness ℎ ft 150Permeability 119870119909 by 119870119910 mD 4000 by 3000Porosity 30Well diameter ft 115Initial water saturation 119878119908

119894 60

Permeability 119870 mD 400Gas oil contact (GOC) ft 6890Oil water contact (OWC) ft 6890Initial pressure 119875

119894 psi 4495

Formation temperature 119879 ∘C 120 2000

(b) Flowing + buildup sequence well perforated ℎ119901

=

30ft inside gas condensate layer Net sand thicknessℎ = 150ft

Figure 10 shows the production and shut-in sequence for 2scenarios in Example 2

As in Example 1 the derivative in scenario (a) ofExample 2 also shows good radial flowwith no late time effectas seen in Figure 11 A negative half-slope fingerprint is seen at03 to 30 hrs of buildup and a good stabilisation from 40 hrsin the model parameters

The derivatives for all output parameters display the samewell and reservoir fingerprint (2 flow periods early to middletime response) but with different 1198891199011015840 stabilisation

Permeability value of 3707mD was obtained from theconventionalmethod BPRwhere 119896 = 1637119902119879119898ℎ is the samefor the PDENDWA and PDENWAT and 119898 is obtained fromthe specialised plot This is in line with the uniform 119896 (119896119909 =

400mD and 119896119910 = 300mD) imputed in the simulationmodel

10 Journal of Petroleum Engineering

00

5000000

10000000

15000000

20000000

25000000

0 40 80 120 160

Prod

uctio

n ra

te (s

cfd

)

Duration

Test design sequence

Cases a to b

Figure 10 Production test sequence (buildup) for Example 2

10000

Case a derivatives for conventional and numerical density models

00100

01000

BHPPENDAPDENO

PDENGPDENW

1422mD

3707mD

35747mD

0000 0001 0010 0100 1000 10000

100000

100000Log time (h)

k =1627qB120583

mh

logdp998400

Figure 11 Derivative and 119870 estimation for scenario (a)

However PDENGAS and PDENOIL differ giving 350mDand 35747mD respectively if ℎ = 148ft

Also the derivative in Figure 12 shows good radial flowand a negative half-slope fingerprint is seen at 05 to 20 hrsof buildup and a good stabilisation from 30 hrs in the modelparameters After 3 hrs of shut-in somenoisy data (numericalartefact) is seen in all the fluid phase derivatives but still goodenough to identify 119889119901

1015840 stabilisationFor each of the fluid phase permeabilities 119896 value of

3409mD at ℎ = 148ft is obtained using BPR which is thesame for the PDENDWA and PDENWAT PDENGAS andPDENOIL differ giving 1490 and 35141mD at ℎ = 148ftrespectively

For scenarios (a) and (b) a good 119896ave value of 4050and 4030mD is obtained respectively from the empiricalmodel (34) integrating all fluid phase permeabilities whichis within that used in the simulationmodel and the estimated

0001000

0010000

0100000

1000000

10000000

0000 0001 0010 0100 1000 10000 100000

Case b derivatives for conventional and numerical density models

BHPPENDAPDENO

PDENGPDENW

Log time (h)

k =1627qB120583

mh

1490mD

3410mD

35140mD

logdp998400

Figure 12 Derivative and119870 estimation for scenario (b)

Table 7 Summary of reservoir modelling properties imputed ineclipse model for case study 2

Parameters Numerical density119896 (mD)

Equivalentℎ (ft)

Scenario(a)

BHP 371PDENA 371PDENG 142 148PDENO 3575PDENW 371

Scenario(b)

BHP 341PDENA 341PDENG 149 148PDENO 3514PDENW 341

conventional approach From the 2 scenarios investigated ithas been demonstrated that the heavier fluid such as waterand the weighted average pressure-density equivalent of allfluid give exact effective 119896 as the conventional method BPRA summary of the result is shown in Table 7

This approach estimates the fluid phase permeabilitiesand the of each phase contribution to flow at a given pointhence at several points the relative 119896 can be generated asshown in Table 8

5 Density Related Radial Flow EquationDerivation for Each Fluid Phase

Radial diffusivity equation is given as

1119903

120597120588

120597119903[119903

120597120588

120597119903] =

120601120583119888

119896

120597120588

120597119905 (35)

Journal of Petroleum Engineering 11

Table 8 Comparison of 119870 estimates between conventional and numerical density parameters

Scenario Numerical density method Calculated119870 (mD)

Conventional119896 (mD)

Simulationmodel 119870 (mD) Relative 119870

phasecontribution to flowPhases 119870 (mD)

(a)Gas 1422 004 30Oil 35747 4046 3707 400 089 870

Water 3707 009 90

(b)Gas 149 004 40Oil 35141 4038 3409 400 089 880

Water 3409 008 90

For oil

1119903

120597120588

120597119903[119903

120597120588119900

120597119903] =

120601120583119900119888

119896119900

120597120588119900

120597119905 (36a)

For water phase

1119903

120597120588

120597119903[119903

120597120588119908

120597119903] =

120601120583119908119888

119896119908

120597120588119908

120597119905 (36b)

For gas phase

1119903

120597120588

120597119903[119903

120597120588119892

120597119903] =

120601120583119892119888

119896119892

120597120588119892

120597119905 (36c)

51 From Equation for Slight and Small Compressibility Suchas Water and Oil From (22)

119875 = 119875119900minus

120588120588119900+ 1

119888119900

(37)

Differentiating with respect to 120588

120597119875

120597120588= minus

1119888119900120588119900

120597119875 = minus120597120588

119888119900120588119900

(38)

From Darcy flow equation

119902 = minus2120587119896ℎ120583

119903120597120588

120597119903 (39)

Substitute for 120597119875

119902 =2120587119896ℎ120583

119903

119888119900120588119900

120597120588

120597119903 (40)

To derive the analytical density transient equation for eachphase the following assumptionconditions are applicable

For oil phase from (35)

1119903

120597120588

120597119903[119903

120597120588

120597119903] =

120601120583119888

119896

120597120588

120597119905 (41)

Initial condition is as follows

120588 (119903 119905 = 0) = 120588119894 (42a)

BC at the wellbore is as follows

lim119903rarr 0

2120587119896ℎ120583

119903

119888119900120588119900

120597120588

120597119903= 119876 (42b)

BC at infirmity is as follows

lim119903rarrinfin

120588 (119903 119905) = 120588119894 (42c)

Using the Boltzmann transformation consider the followingAssuming 120578 = 120601120583119888119903

2119896119905

120597120588

120597119903=

120597120588

120597120578

120597120578

120597119903=2120601120583119888119903119896119905

120597120588

120597120578=

1206011205831198881199032

119896119905

2119903

120597120588

120597120578=2120578119903

120597120588

120597120578 (43)

Therefore1119903

120597

120597119903[119903

120597120588

120597119903] =

21205781199032

120597

120597120578[2120578

120597120588

120597120578] =

41205781199032

120597

120597120578[120578

120597120588

120597120578] (44)

Differentiating with respect to 119903 is equal to differentiatingwith respect to 120578 multiply by 2120578119903

From (35) the LHS is resolved as follows

120597120588

120597119905=

120597120588

120597120578

120597120578

120597119905=

120601120583119888119903

1198961199052120597120588

120597120578= minus

120578

119905

120597120588

120597120578 (45)

120601120583119888

119896

120597120588

120597119905=

minus120601120583119888

119896

120578

119905

120597120588

120597119905=

minus1206011205831198881199032

119896119905

120578

1199032120597120588

120597120578=

minus1205782

1199032120597120588

120597120578 (46)

Equating (44) and (46) we have

120597

120597120578[120578

120597120588

120597120578] =

minus120578

4120597120588

120597120578 (47)

This is the simplified ordinary differential equation of 120588 as afunction of 120578

Apply boundary conditions as followsInitial condition is as follows

lim119903rarrinfin

120588 (120578) = 120588119894 (48a)

Substituting (43) into (40)

lim119903rarr 0

2120587119896ℎ120583

119903

119888119900120588119900

120597120588

120597119903= 119876 (48b)

lim119903rarr 0

[120578120597120588

120597120578] =

120601120583119888119900120588119900

4120587119896ℎ (49)

Assume that119898 = 120578(120597120588120597120578)

12 Journal of Petroleum Engineering

From (47)

120597119898

120597120578=

minus119898

4 (50)

Integrating from 120578 = 0 to 120578

997904rArr int

119898(120578)

119898(0)

120597119898

120597120578= minusint

120578

0

120597120578

4

997904rArr ln[119898 (120578)

119898 (0)] =

minus120578

4

997904rArr 119898(120578) = 119898 (0) 119890minus1205784

(51)

Recall boundary condition

lim119903rarr 0

[120578120597120588

120597120578] =

119876120583119888119900120588119900

4120587119896ℎ

119898 (0) =119876120583119888119900120588119900

4120587119896ℎ

119898 (120578) =119876120583119888119900120588119900

4120587119896ℎ119890minus1205784

(52)

Recall that119898 = 120578(120597120588120597120578)Therefore

120597120588 (120578)

120597120578=

120583119876119888119900120588119900

4120587119896ℎ119890minus1205784

120578 (53)

Integrating from 120578 = infin where 120588 = 120588119894

int

120588(120578)

120588119894

120597120588 = minusint

120578

infin

120583119876119888119900120588119900

4120587119896ℎ119890minus1205784

120578120597120578

120588 (120578) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎminusint

infin

120578

119890minus1205784

120578120597120578

(54)

where

120578 =120601120583119888119903

2

119896119905 (55)

Assuming 119906 = 1205784 then 120597120578120578 = 120597119906119906 and 119906 = 12060112058311988811990324119896119905

Hence

120588 = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎminusint

infin

12060112058311988811990324119896119905

119890minus1205784

120578120597120578 (56)

where

minus119864119894 (minus119909) = int

infin

119909

119890minus119906

119906120597119906 (57)

Known as the exponential integral function defined by [13]

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎ119864119894 (minus119909) (58)

where 119909 = 12060112058311988811990324119896119905 and

minus119864119894 (minus119909) = int

infin

119909

119890minus119906

119906120597119906 (59)

At large times 119909 will be small applying Taylorrsquos series andintegrating

minus119864119894 (minus119909) = ln (120574)

ln (120574) = ln (1781) = 05772(60)

where 120574 is known as Eulerrsquos numberFor 14119909 gt 25 consider the followingThen

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎ[ln119909+ ln 120574]

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎln[

1206011205831198881199032120574

4119896119905]

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎln [

22461198961199051206011205831198881199032

]

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎ[ln [

22461198961199051206011205831198881199032

]+ 080907]

(61)

Plotting 120588119908(119905) versus ln(119905) will yield a straight line at longer

time and the slope of the line is given as120597120588119908

120597 ln 119905= 119898oil =

120583119876119888119900120588119900

4120587119896ℎ (62)

Therefore

119896ℎ =120583119876119888119900120588119900

4120587119898oil (63)

52 For Water Phase The radial density equation is given as

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎ[ln [

22461198961199051206011205831198881199032

]+ 080907] (64)

Plotting 120588119908(119905) versus ln(119905) will yield a straight line at longer

time and the slope of the line is given as120597120588119908

120597 ln 119905= 119898water =

120583119876119888119900120588119900

4120587119896ℎ (65)

where

119896ℎ =120583119876119888119900120588119900

4120587119898water (66)

53 For Gas Phase From equation for compressible fluidsuch as gas consider the following

From (32)

119875 =119875119900120588 minus 119875119900

2120588119900119888119892

120588119900(1 minus 119875

119900119888119892)

120597119875

120597120588=

119875119900

120588119900(1 minus 119875

119900119888119892)

120597119875 =119875119900

120588119900(1 minus 119875

119900119888119892)120597120588

(67)

Journal of Petroleum Engineering 13

From the fundamental

119899 =119875119881

119911119877119879 (68)

At standard condition

119875119881

119911119879=

119875sc119881sc119879sc

119875 (5615119902)119911119879

=119875sc119876sc119879sc

119902 =119875sc119879sc

(119911119879

119875)

119876sc5615

119902 =119875sc119879sc

119876sc5615

(119911119879

119875) =

2120587119896ℎ120583

119903120597119901

120597119903

119902 =119875sc119879sc

119876sc5615

(119911119879) =2120587119896ℎ120583

119903119875120597119901

120597119903

(69)

where

120597119875 =119875119900

120588119900(1 minus 119875

119900119888119892)120597120588

120572 =119875119900

120588119900(1 minus 119875

119900119888119892)

120597119875 = 120572120597120588

119875 = 120588119877119911119879

(70)

Substitute into the equation where

119898(120588119894) minus119898 (120588

119908119891) =

2120583

int

120588119894

120588119908119891

120588120597120588

119898 (120588119894) minus119898 (120588

119908119891) =

120588119894

2minus 120588119908119891

2

120583

120588 = radic120588119894

2+ 120588119908119891

2

2

(71)

From the diffusivity equation for gas

1119903

120597120588

120597119903[119903

120597120588119892

120597119903] =

120601120583119892119888

119896119892

120597120588119892

120597119905 (72)

Initial condition is as follows

120588 (119903 119905 = 0) = 120588119894 (73)

BC at the wellbore is as follows

lim119903rarr 0

4120587119896ℎ120572119877120583

(5615119879sc

119875sc) 119903120588

120597120588

120597119903= 119876 (74)

BC at infirmity is as follows

lim119903rarrinfin

120588 (119903 119905) = 120588119894 (75)

where119877119875sc and119879sc are known as temperature pressure andgas constant at standard condition

Therefore integrating from 120578 = infin where 120588 = 120588119894

int

120588(120578)

120588119894

120588120597120588 = minusint

120578

infin

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)119890minus1205784

120578120597120578

1205882(120578) = 120588

119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

minusint

infin

120578

119890minus1205784

120578120597120578

(76)

where

120578 =120601120583119888119903

2

119896119905 (77)

Assume that 119906 = (1205784) (120597120578120578) = 120597119906119906 and 119906 = 12060112058311988811990324119896119905

Hence

1205882= 120588119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)minusint

infin

12060112058311988811990324119896119905

119890minus1205784

120578 (78)

where minus119864119894(minus119909) = intinfin

119909(119890minus119906

119906)120597119906Known as the exponential integral function defined by

[13]

1205882= 120588119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)minus119864119894 (minus119909) (79)

where 119909 = 12060112058311988811990324119896119905 and minus119864119894(minus119909) = int

infin

119909(119890minus119906

119906)120597119906At large times 119909will be small applying Taylorrsquos series and

integrating

minus119864119894 (minus119909) = ln (120574)

ln (120574) = ln (1781) = 05772(80)

where 120574 is known as Eulerrsquos numberFor 14119909 gt 25 consider the followingThen

1205882(119903 119905) = 120588

119894

2+

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) [ln119909+ ln 120574] (81)

1205882(119903 119905) = 120588

119894

2+

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) ln[120601120583119888119903

2120574

4119896119905]

1205882(119903 119905) = 120588

119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) ln [22461198961199051206011205831198881199032

]

119898 (120588119894) minus119898 (120588

119908119891) =

2120583

int

120588119894

120588119908119891

120588 120597120588

119898 (120588119908119891

) = 119898 (120588119894) minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

sdot [ln [22461198961199051206011205831198881199032

]+ 080907]

120597119898 (120588119908)

120597 ln 119905= 119898gas =

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

(82)

14 Journal of Petroleum Engineering

4160041650417004175041800418504190041950420004205042100

560856095609561056105611561156125612

ODENGDEN

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

1205880

(Ibm

ft)

120597120588w120597lnt

= moil =1627120583Qc01205880

kh= minus0009

120597m(120588w)

120597lnt= mgas =

1637120583Q

kh120572R= minus12195

120588g2

(Ibm

ft)

Figure 13 Oil and gas phase density versus log Horner time

6279

6279

6279

6279

6279

6279

6279

6279

01 1 10 100 1000 10000 1000001000000Horner time (h)

Density versus log Horner time

WDEN

120588w

(Ibm

ft)

120597120588w120597lnt

= mwater =1627120583Qc01205880

kh= minus00003

Figure 14 Water phase density versus log Horner time

Plotting 120588119908

2(119905) or 119898(120588

119908) versus ln(119905) will yield a straight line

at longer time and the slope of the line is given as follows

119896ℎ =120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)1

119898gas (83)

Example 3 Data from Example 1(a) was used for this caseand the semilog plot of density for each phase (oil water andgas) is plotted against log of Horner timelowastlowast

119876 used for 119896 calculation is the average rate for allflowing periods

From Figures 13 and 14 the calculated slopes of the radialflow for oil and water phases are given as

120597120588119908

120597 ln 119905= 119898oil =

1627120583119876119888119900120588119900

119896ℎ= minus 0009

120597120588119908

120597 ln 119905= 119898water =

1627120583119876119888119900120588119900

119896ℎ= minus 00003

(84)

The estimated phase permeabilities are

119896oil =1627120583119876119888

119900120588119900

ℎ119898oil= 445mD

119896water =1627120583119876119888

119900120588119900

ℎ119898water= 50mD

(85)

Also for the gas phase consider the following

ODENGDEN

422004225042300423504240042450425004255042600

5613561356145614561556155616

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

1205880

(Ibm

ft)

120597120588w120597lnt

= moil =1627120583Qc01205880

kh= minus0009

120597m(120588w)

120597lnt= mgas =

1637120583Q

kh120572R= minus12502

120588g2

(Ibm

ft)

Figure 15 Oil and gas phase density versus log Horner time

WDEN

6279627962796279627962796280628062806280

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

120588w

(Ibm

ft)

120597120588w120597lnt

= mwater =1627120583Qc01205880

kh= minus00003

Figure 16 Water phase density versus log Horner time

The calculated slope from the semilog plot shown inFigure 13 is given as

120597119898 (120588119908)

120597 ln 119905= 119898gas =

1637120583119876119896ℎ120572119877

= minus 12195 (86)

And estimated gas permeability is

119896gas =1637120583119876

119900

ℎ120572119877119898gas= 170mD (87)

Using the empirical model for all three phases the averageestimated permeability is as follows

119896ave =4radic50 times 445 times 1702

119896ave = 50mD

(88)

Example 4 Data from Example 1(c) was used for this caseand the semilog plot of density for each phase (oil water andgas) is plotted against log of Horner time

From Figures 15 and 16 the calculated slopes of the radialflow for oil and water phases are given as

120597120588119908

120597 ln 119905= 119898oil =

1627120583119876119888119900120588119900

119896ℎ= minus 0009

120597120588119908

120597 ln 119905= 119898water =

1627120583119876119888119900120588119900

119896ℎ= minus 00003

(89)

Journal of Petroleum Engineering 15

The estimated phase permeabilities are

119896oil =1627120583119876119888

119900120588119900

ℎ119898oil= 4352mD

119896water =1627120583119876119888

119900120588119900

ℎ119898water= 49mD

(90)

Also for the gas phase consider the following

The calculated slope from the semilog plot shown inFigure 15 is given as

120597119898 (120588119908)

120597 ln 119905= 119898gas =

1637120583119876119896ℎ120572119877

= minus 12502 (91)

And estimated gas permeability is

119896gas =1637120583119876

119900

ℎ120572119877119898gas= 179mD (92)

Using the empirical model for all three phases the averageestimated permeability is as follows

119896ave =4radic49 times 4352 times 1792

119896ave = 51mD

(93)

6 Conclusion

The following inferences were drawn from the six scenariosreviewed

(i) Results from the three-phase fluid empirical modeldeveloped for average effective 119896ave estimation rangebetween 47 and 57mD for Example 1 and give404mD for Example 2 which is within that used inthe simulation model and also estimated from BPR

(ii) The BPR PDENDWA and PDENWAT give the samestabilisation and the same 119896 estimates making it suit-able for interpretation of pressure transient analysis

(iii) It has been demonstrated in all 6 scenarios inves-tigated that the heavier fluid such as water andthe weighted average pressure-density equivalent ofall fluid give exact effective 119896 as the conventionalmethod

(iv) For oil reservoir system (scenarios (a) to (d)) 119896 =

500mD is achieved if the thickness contributing toflow ranges from 50 to 150 ft while for gas condensatereservoir (scenarios (a) to (b)) 119896 ranges between340 and 371mD which is achieved if the net sandthickness ℎ = 148ft is contributing to flow

(v) This approach also provides an estimate of the pos-sible fluid phase permeabilities and the of eachphase contribution to flow at a given point hence atseveral 1198891199011015840 stabilisation points the relative 119896 can begenerated

(vi) The derivatives for all output parameters display thesame wellbore and reservoir fingerprint as the BPRmethod

(vii) Generally where output parameters (BPR PDE-NOIL PDENGAS PDENWAT and PDENDW)depict different fingerprint the density derivative willserve as support to distinguish between reservoir andnonreservoir response

Nomenclature

119875 Pressure psi119879 Temperature oF119903 Radius ft119896 Permeability mDOslash Porosity fraction120583 Viscosity cp119905 Time hrs119902 Production rate bblday119861 Formation volume factor rbStb119862119905 Total compressibility psiminus1119903119908 Wellbore radius ftΔ119901 Change in pressure psiaℎ Formation thickness ft119860 Drainage area acres119875119908119891 Bottom-hole flowing pressure psi119875119894 Initial pressure psi

119885 Difference between two pointtime series119894 Subscript of an observed variable119888 Subscript of a calculated variableSTEYX SSE of data point119899 Number of data pointsCov Covariance of data point120575 Standard deviation119905119901 Cumulative production time119862119904 Wellbore storage constant119909 Mean of data point

Abbreviations

LBPR Local grid bottom-hole pressureLDENO Local grid oil densityLDENW Local grid water densityLDENG Local grid gas densityWBHP Well bottom-hole pressureBPR Well bottom-hole pressurePDENOIL Pressure equivalent of LDENOPDENGAS Pressure equivalent of LDENGPDENWAT Pressure equivalent of LDENWPDENDWA Pressure equivalent of density weighted

average (LDENO LDENG and LDENW)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Jalali S D Mohaghegh and R Gaskari ldquoIdentifying infilllocations and underperformer wells in mature fields usingmonthly production rate data Carthage Field Cotton Valley

16 Journal of Petroleum Engineering

Formation Texasrdquo in Proceedings of the SPE Eastern RegionalMeeting SPE Paper 104550 Canton Ohio USA October 2006

[2] D Tiab and A Kumar ldquoApplication of PD function to inter-ference analysisrdquo in Proceedings of the 51st Annual Fall Meetingof the Society of Petroleum Engineers of AIME Paper SPE 6053New Orleans La USA 1976

[3] D Tiab and A Kumar ldquoDetection and location of two parallelsealing fault around a wellrdquo in Proceedings of the 51st AnnualFall Meeting of the Society of Petroleum Engineers of AIME SPEPaper 6056 New Orleans La USA 1976

[4] D Bourdet T M Whittle A A Douglas and Y M Pirard ldquoAnew set of type curves simplifies well test analysisrdquo World Oilvol 196 no 6 pp 95ndash106 1983

[5] S Y Zheng ldquoFighting against non-unique solution problemsin heterogeneous reservoirs through numerical well testingrdquo inProceedings of the SPE Asia Pacific Oil and Gas Conference andExhibition SPE 100951 Adelaide Australia September 2006

[6] R N Horne and K O Temeng ldquoRecognition and location ofpinch-out boundaries by pressure transient analysisrdquo SPE 9905Society of Petroleum Engineers 1981

[7] L C Ayestaran R D Nurmi G A K Shehab and W S ElsildquoWelltest design and interpretation improved by integrated welltesting and geological effortsrdquo SPE 17945 1989

[8] G J Massonnat and D Bandiziol ldquoInterdependence betweengeology and well test interpretationrdquo SPE 22740 1991

[9] K F Du and G Stewart ldquoReservoir description from well testinterpretationrdquo in North Sea Oil and Gas ReservoirmdashIII pp339ndash356 Norwegian Institute of Technology (NTH) 1994

[10] S Y Zheng Well testing and characterisation of meanderingfluvial channel reservoirs [PhD thesis] Heriot-Watt University1997

[11] V T Biu and S Y Zheng ldquoA new approach in pressure transientanalysis part I improved diagnosis of flow regimes in oil and gaswellsrdquo in Proceedings of the 3rd EAGEAAPGWorkshop on TightReservoirs in theMiddle East AbuDhabi UnitedArab Emirates2015

[12] R N Horne Modern Welltest Analysis Petroway Palo AltoCalif USA 1995

[13] C S Matthews and D G Russell Pressure Buildup and FlowTests in Wells vol 1 ofMonograph Series SPE Richardson TexUSA 1967

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Journal of Petroleum Engineering 9

Table 5 Comparison of 119870 estimates between conventional and numerical density parameters

Scenario Numerical density method Calculated119870 (mD)

Conventional119896 (mD)

Simulationmodel 119870 (mD) Relative 119870

phasecontribution to flowPhases 119870 (mD)

(a)Gas 158 003 30Oil 4037 473 508 500 086 860

Water 508 011 110

(b)Gas 12 003 20Oil 4844 432 500 500 113 890

Water 50 012 90

(c)Gas 209 004 40Oil 4842 570 500 500 086 87

Water 50 009 90

(d)Gas 175 003 30Oil 490 523 500 500 095 880

Water 50 010 90

1000

Case d derivatives for conventional and numerical density models

0010

0100

10000

BHPPENDAPDENO

PDENGPDENW

0100 1000 10000 100000Log time (h)

k =1627qB120583

mh

209mD

500mD

4842mD

logdp998400

Figure 9 Derivative and 119870 estimation for scenario (d)

additional information given below It is required to generatethe pressure equivalent and derivative for each fluid phasecompare their diagnostic signatures and also determine thephases permeabilities and average reservoir permeability

Assumption (i) Condensate reservoir is completed with onewell

(ii) LGR is imposed around the well and far across toaccount for pressure and density changes

Gas condensate andwater densities around the local gridrefinement (wellbore) andWBHPwere output using the sim-ulatorrsquos keywords The following scenarios were evaluated

(a) Flowing + buildup sequence well perforated ℎ119901

=

30ft between gas condensate and water layer Netsand thickness ℎ = 150ft

Table 6 Summary of reservoir simulations data

Parameters Design valueEclipse model Black oilModel dimension 9 times 3 times 3Length by width ft by ft 1312 times 984Thickness ℎ ft 150Permeability 119870119909 by 119870119910 mD 4000 by 3000Porosity 30Well diameter ft 115Initial water saturation 119878119908

119894 60

Permeability 119870 mD 400Gas oil contact (GOC) ft 6890Oil water contact (OWC) ft 6890Initial pressure 119875

119894 psi 4495

Formation temperature 119879 ∘C 120 2000

(b) Flowing + buildup sequence well perforated ℎ119901

=

30ft inside gas condensate layer Net sand thicknessℎ = 150ft

Figure 10 shows the production and shut-in sequence for 2scenarios in Example 2

As in Example 1 the derivative in scenario (a) ofExample 2 also shows good radial flowwith no late time effectas seen in Figure 11 A negative half-slope fingerprint is seen at03 to 30 hrs of buildup and a good stabilisation from 40 hrsin the model parameters

The derivatives for all output parameters display the samewell and reservoir fingerprint (2 flow periods early to middletime response) but with different 1198891199011015840 stabilisation

Permeability value of 3707mD was obtained from theconventionalmethod BPRwhere 119896 = 1637119902119879119898ℎ is the samefor the PDENDWA and PDENWAT and 119898 is obtained fromthe specialised plot This is in line with the uniform 119896 (119896119909 =

400mD and 119896119910 = 300mD) imputed in the simulationmodel

10 Journal of Petroleum Engineering

00

5000000

10000000

15000000

20000000

25000000

0 40 80 120 160

Prod

uctio

n ra

te (s

cfd

)

Duration

Test design sequence

Cases a to b

Figure 10 Production test sequence (buildup) for Example 2

10000

Case a derivatives for conventional and numerical density models

00100

01000

BHPPENDAPDENO

PDENGPDENW

1422mD

3707mD

35747mD

0000 0001 0010 0100 1000 10000

100000

100000Log time (h)

k =1627qB120583

mh

logdp998400

Figure 11 Derivative and 119870 estimation for scenario (a)

However PDENGAS and PDENOIL differ giving 350mDand 35747mD respectively if ℎ = 148ft

Also the derivative in Figure 12 shows good radial flowand a negative half-slope fingerprint is seen at 05 to 20 hrsof buildup and a good stabilisation from 30 hrs in the modelparameters After 3 hrs of shut-in somenoisy data (numericalartefact) is seen in all the fluid phase derivatives but still goodenough to identify 119889119901

1015840 stabilisationFor each of the fluid phase permeabilities 119896 value of

3409mD at ℎ = 148ft is obtained using BPR which is thesame for the PDENDWA and PDENWAT PDENGAS andPDENOIL differ giving 1490 and 35141mD at ℎ = 148ftrespectively

For scenarios (a) and (b) a good 119896ave value of 4050and 4030mD is obtained respectively from the empiricalmodel (34) integrating all fluid phase permeabilities whichis within that used in the simulationmodel and the estimated

0001000

0010000

0100000

1000000

10000000

0000 0001 0010 0100 1000 10000 100000

Case b derivatives for conventional and numerical density models

BHPPENDAPDENO

PDENGPDENW

Log time (h)

k =1627qB120583

mh

1490mD

3410mD

35140mD

logdp998400

Figure 12 Derivative and119870 estimation for scenario (b)

Table 7 Summary of reservoir modelling properties imputed ineclipse model for case study 2

Parameters Numerical density119896 (mD)

Equivalentℎ (ft)

Scenario(a)

BHP 371PDENA 371PDENG 142 148PDENO 3575PDENW 371

Scenario(b)

BHP 341PDENA 341PDENG 149 148PDENO 3514PDENW 341

conventional approach From the 2 scenarios investigated ithas been demonstrated that the heavier fluid such as waterand the weighted average pressure-density equivalent of allfluid give exact effective 119896 as the conventional method BPRA summary of the result is shown in Table 7

This approach estimates the fluid phase permeabilitiesand the of each phase contribution to flow at a given pointhence at several points the relative 119896 can be generated asshown in Table 8

5 Density Related Radial Flow EquationDerivation for Each Fluid Phase

Radial diffusivity equation is given as

1119903

120597120588

120597119903[119903

120597120588

120597119903] =

120601120583119888

119896

120597120588

120597119905 (35)

Journal of Petroleum Engineering 11

Table 8 Comparison of 119870 estimates between conventional and numerical density parameters

Scenario Numerical density method Calculated119870 (mD)

Conventional119896 (mD)

Simulationmodel 119870 (mD) Relative 119870

phasecontribution to flowPhases 119870 (mD)

(a)Gas 1422 004 30Oil 35747 4046 3707 400 089 870

Water 3707 009 90

(b)Gas 149 004 40Oil 35141 4038 3409 400 089 880

Water 3409 008 90

For oil

1119903

120597120588

120597119903[119903

120597120588119900

120597119903] =

120601120583119900119888

119896119900

120597120588119900

120597119905 (36a)

For water phase

1119903

120597120588

120597119903[119903

120597120588119908

120597119903] =

120601120583119908119888

119896119908

120597120588119908

120597119905 (36b)

For gas phase

1119903

120597120588

120597119903[119903

120597120588119892

120597119903] =

120601120583119892119888

119896119892

120597120588119892

120597119905 (36c)

51 From Equation for Slight and Small Compressibility Suchas Water and Oil From (22)

119875 = 119875119900minus

120588120588119900+ 1

119888119900

(37)

Differentiating with respect to 120588

120597119875

120597120588= minus

1119888119900120588119900

120597119875 = minus120597120588

119888119900120588119900

(38)

From Darcy flow equation

119902 = minus2120587119896ℎ120583

119903120597120588

120597119903 (39)

Substitute for 120597119875

119902 =2120587119896ℎ120583

119903

119888119900120588119900

120597120588

120597119903 (40)

To derive the analytical density transient equation for eachphase the following assumptionconditions are applicable

For oil phase from (35)

1119903

120597120588

120597119903[119903

120597120588

120597119903] =

120601120583119888

119896

120597120588

120597119905 (41)

Initial condition is as follows

120588 (119903 119905 = 0) = 120588119894 (42a)

BC at the wellbore is as follows

lim119903rarr 0

2120587119896ℎ120583

119903

119888119900120588119900

120597120588

120597119903= 119876 (42b)

BC at infirmity is as follows

lim119903rarrinfin

120588 (119903 119905) = 120588119894 (42c)

Using the Boltzmann transformation consider the followingAssuming 120578 = 120601120583119888119903

2119896119905

120597120588

120597119903=

120597120588

120597120578

120597120578

120597119903=2120601120583119888119903119896119905

120597120588

120597120578=

1206011205831198881199032

119896119905

2119903

120597120588

120597120578=2120578119903

120597120588

120597120578 (43)

Therefore1119903

120597

120597119903[119903

120597120588

120597119903] =

21205781199032

120597

120597120578[2120578

120597120588

120597120578] =

41205781199032

120597

120597120578[120578

120597120588

120597120578] (44)

Differentiating with respect to 119903 is equal to differentiatingwith respect to 120578 multiply by 2120578119903

From (35) the LHS is resolved as follows

120597120588

120597119905=

120597120588

120597120578

120597120578

120597119905=

120601120583119888119903

1198961199052120597120588

120597120578= minus

120578

119905

120597120588

120597120578 (45)

120601120583119888

119896

120597120588

120597119905=

minus120601120583119888

119896

120578

119905

120597120588

120597119905=

minus1206011205831198881199032

119896119905

120578

1199032120597120588

120597120578=

minus1205782

1199032120597120588

120597120578 (46)

Equating (44) and (46) we have

120597

120597120578[120578

120597120588

120597120578] =

minus120578

4120597120588

120597120578 (47)

This is the simplified ordinary differential equation of 120588 as afunction of 120578

Apply boundary conditions as followsInitial condition is as follows

lim119903rarrinfin

120588 (120578) = 120588119894 (48a)

Substituting (43) into (40)

lim119903rarr 0

2120587119896ℎ120583

119903

119888119900120588119900

120597120588

120597119903= 119876 (48b)

lim119903rarr 0

[120578120597120588

120597120578] =

120601120583119888119900120588119900

4120587119896ℎ (49)

Assume that119898 = 120578(120597120588120597120578)

12 Journal of Petroleum Engineering

From (47)

120597119898

120597120578=

minus119898

4 (50)

Integrating from 120578 = 0 to 120578

997904rArr int

119898(120578)

119898(0)

120597119898

120597120578= minusint

120578

0

120597120578

4

997904rArr ln[119898 (120578)

119898 (0)] =

minus120578

4

997904rArr 119898(120578) = 119898 (0) 119890minus1205784

(51)

Recall boundary condition

lim119903rarr 0

[120578120597120588

120597120578] =

119876120583119888119900120588119900

4120587119896ℎ

119898 (0) =119876120583119888119900120588119900

4120587119896ℎ

119898 (120578) =119876120583119888119900120588119900

4120587119896ℎ119890minus1205784

(52)

Recall that119898 = 120578(120597120588120597120578)Therefore

120597120588 (120578)

120597120578=

120583119876119888119900120588119900

4120587119896ℎ119890minus1205784

120578 (53)

Integrating from 120578 = infin where 120588 = 120588119894

int

120588(120578)

120588119894

120597120588 = minusint

120578

infin

120583119876119888119900120588119900

4120587119896ℎ119890minus1205784

120578120597120578

120588 (120578) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎminusint

infin

120578

119890minus1205784

120578120597120578

(54)

where

120578 =120601120583119888119903

2

119896119905 (55)

Assuming 119906 = 1205784 then 120597120578120578 = 120597119906119906 and 119906 = 12060112058311988811990324119896119905

Hence

120588 = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎminusint

infin

12060112058311988811990324119896119905

119890minus1205784

120578120597120578 (56)

where

minus119864119894 (minus119909) = int

infin

119909

119890minus119906

119906120597119906 (57)

Known as the exponential integral function defined by [13]

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎ119864119894 (minus119909) (58)

where 119909 = 12060112058311988811990324119896119905 and

minus119864119894 (minus119909) = int

infin

119909

119890minus119906

119906120597119906 (59)

At large times 119909 will be small applying Taylorrsquos series andintegrating

minus119864119894 (minus119909) = ln (120574)

ln (120574) = ln (1781) = 05772(60)

where 120574 is known as Eulerrsquos numberFor 14119909 gt 25 consider the followingThen

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎ[ln119909+ ln 120574]

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎln[

1206011205831198881199032120574

4119896119905]

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎln [

22461198961199051206011205831198881199032

]

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎ[ln [

22461198961199051206011205831198881199032

]+ 080907]

(61)

Plotting 120588119908(119905) versus ln(119905) will yield a straight line at longer

time and the slope of the line is given as120597120588119908

120597 ln 119905= 119898oil =

120583119876119888119900120588119900

4120587119896ℎ (62)

Therefore

119896ℎ =120583119876119888119900120588119900

4120587119898oil (63)

52 For Water Phase The radial density equation is given as

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎ[ln [

22461198961199051206011205831198881199032

]+ 080907] (64)

Plotting 120588119908(119905) versus ln(119905) will yield a straight line at longer

time and the slope of the line is given as120597120588119908

120597 ln 119905= 119898water =

120583119876119888119900120588119900

4120587119896ℎ (65)

where

119896ℎ =120583119876119888119900120588119900

4120587119898water (66)

53 For Gas Phase From equation for compressible fluidsuch as gas consider the following

From (32)

119875 =119875119900120588 minus 119875119900

2120588119900119888119892

120588119900(1 minus 119875

119900119888119892)

120597119875

120597120588=

119875119900

120588119900(1 minus 119875

119900119888119892)

120597119875 =119875119900

120588119900(1 minus 119875

119900119888119892)120597120588

(67)

Journal of Petroleum Engineering 13

From the fundamental

119899 =119875119881

119911119877119879 (68)

At standard condition

119875119881

119911119879=

119875sc119881sc119879sc

119875 (5615119902)119911119879

=119875sc119876sc119879sc

119902 =119875sc119879sc

(119911119879

119875)

119876sc5615

119902 =119875sc119879sc

119876sc5615

(119911119879

119875) =

2120587119896ℎ120583

119903120597119901

120597119903

119902 =119875sc119879sc

119876sc5615

(119911119879) =2120587119896ℎ120583

119903119875120597119901

120597119903

(69)

where

120597119875 =119875119900

120588119900(1 minus 119875

119900119888119892)120597120588

120572 =119875119900

120588119900(1 minus 119875

119900119888119892)

120597119875 = 120572120597120588

119875 = 120588119877119911119879

(70)

Substitute into the equation where

119898(120588119894) minus119898 (120588

119908119891) =

2120583

int

120588119894

120588119908119891

120588120597120588

119898 (120588119894) minus119898 (120588

119908119891) =

120588119894

2minus 120588119908119891

2

120583

120588 = radic120588119894

2+ 120588119908119891

2

2

(71)

From the diffusivity equation for gas

1119903

120597120588

120597119903[119903

120597120588119892

120597119903] =

120601120583119892119888

119896119892

120597120588119892

120597119905 (72)

Initial condition is as follows

120588 (119903 119905 = 0) = 120588119894 (73)

BC at the wellbore is as follows

lim119903rarr 0

4120587119896ℎ120572119877120583

(5615119879sc

119875sc) 119903120588

120597120588

120597119903= 119876 (74)

BC at infirmity is as follows

lim119903rarrinfin

120588 (119903 119905) = 120588119894 (75)

where119877119875sc and119879sc are known as temperature pressure andgas constant at standard condition

Therefore integrating from 120578 = infin where 120588 = 120588119894

int

120588(120578)

120588119894

120588120597120588 = minusint

120578

infin

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)119890minus1205784

120578120597120578

1205882(120578) = 120588

119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

minusint

infin

120578

119890minus1205784

120578120597120578

(76)

where

120578 =120601120583119888119903

2

119896119905 (77)

Assume that 119906 = (1205784) (120597120578120578) = 120597119906119906 and 119906 = 12060112058311988811990324119896119905

Hence

1205882= 120588119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)minusint

infin

12060112058311988811990324119896119905

119890minus1205784

120578 (78)

where minus119864119894(minus119909) = intinfin

119909(119890minus119906

119906)120597119906Known as the exponential integral function defined by

[13]

1205882= 120588119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)minus119864119894 (minus119909) (79)

where 119909 = 12060112058311988811990324119896119905 and minus119864119894(minus119909) = int

infin

119909(119890minus119906

119906)120597119906At large times 119909will be small applying Taylorrsquos series and

integrating

minus119864119894 (minus119909) = ln (120574)

ln (120574) = ln (1781) = 05772(80)

where 120574 is known as Eulerrsquos numberFor 14119909 gt 25 consider the followingThen

1205882(119903 119905) = 120588

119894

2+

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) [ln119909+ ln 120574] (81)

1205882(119903 119905) = 120588

119894

2+

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) ln[120601120583119888119903

2120574

4119896119905]

1205882(119903 119905) = 120588

119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) ln [22461198961199051206011205831198881199032

]

119898 (120588119894) minus119898 (120588

119908119891) =

2120583

int

120588119894

120588119908119891

120588 120597120588

119898 (120588119908119891

) = 119898 (120588119894) minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

sdot [ln [22461198961199051206011205831198881199032

]+ 080907]

120597119898 (120588119908)

120597 ln 119905= 119898gas =

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

(82)

14 Journal of Petroleum Engineering

4160041650417004175041800418504190041950420004205042100

560856095609561056105611561156125612

ODENGDEN

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

1205880

(Ibm

ft)

120597120588w120597lnt

= moil =1627120583Qc01205880

kh= minus0009

120597m(120588w)

120597lnt= mgas =

1637120583Q

kh120572R= minus12195

120588g2

(Ibm

ft)

Figure 13 Oil and gas phase density versus log Horner time

6279

6279

6279

6279

6279

6279

6279

6279

01 1 10 100 1000 10000 1000001000000Horner time (h)

Density versus log Horner time

WDEN

120588w

(Ibm

ft)

120597120588w120597lnt

= mwater =1627120583Qc01205880

kh= minus00003

Figure 14 Water phase density versus log Horner time

Plotting 120588119908

2(119905) or 119898(120588

119908) versus ln(119905) will yield a straight line

at longer time and the slope of the line is given as follows

119896ℎ =120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)1

119898gas (83)

Example 3 Data from Example 1(a) was used for this caseand the semilog plot of density for each phase (oil water andgas) is plotted against log of Horner timelowastlowast

119876 used for 119896 calculation is the average rate for allflowing periods

From Figures 13 and 14 the calculated slopes of the radialflow for oil and water phases are given as

120597120588119908

120597 ln 119905= 119898oil =

1627120583119876119888119900120588119900

119896ℎ= minus 0009

120597120588119908

120597 ln 119905= 119898water =

1627120583119876119888119900120588119900

119896ℎ= minus 00003

(84)

The estimated phase permeabilities are

119896oil =1627120583119876119888

119900120588119900

ℎ119898oil= 445mD

119896water =1627120583119876119888

119900120588119900

ℎ119898water= 50mD

(85)

Also for the gas phase consider the following

ODENGDEN

422004225042300423504240042450425004255042600

5613561356145614561556155616

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

1205880

(Ibm

ft)

120597120588w120597lnt

= moil =1627120583Qc01205880

kh= minus0009

120597m(120588w)

120597lnt= mgas =

1637120583Q

kh120572R= minus12502

120588g2

(Ibm

ft)

Figure 15 Oil and gas phase density versus log Horner time

WDEN

6279627962796279627962796280628062806280

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

120588w

(Ibm

ft)

120597120588w120597lnt

= mwater =1627120583Qc01205880

kh= minus00003

Figure 16 Water phase density versus log Horner time

The calculated slope from the semilog plot shown inFigure 13 is given as

120597119898 (120588119908)

120597 ln 119905= 119898gas =

1637120583119876119896ℎ120572119877

= minus 12195 (86)

And estimated gas permeability is

119896gas =1637120583119876

119900

ℎ120572119877119898gas= 170mD (87)

Using the empirical model for all three phases the averageestimated permeability is as follows

119896ave =4radic50 times 445 times 1702

119896ave = 50mD

(88)

Example 4 Data from Example 1(c) was used for this caseand the semilog plot of density for each phase (oil water andgas) is plotted against log of Horner time

From Figures 15 and 16 the calculated slopes of the radialflow for oil and water phases are given as

120597120588119908

120597 ln 119905= 119898oil =

1627120583119876119888119900120588119900

119896ℎ= minus 0009

120597120588119908

120597 ln 119905= 119898water =

1627120583119876119888119900120588119900

119896ℎ= minus 00003

(89)

Journal of Petroleum Engineering 15

The estimated phase permeabilities are

119896oil =1627120583119876119888

119900120588119900

ℎ119898oil= 4352mD

119896water =1627120583119876119888

119900120588119900

ℎ119898water= 49mD

(90)

Also for the gas phase consider the following

The calculated slope from the semilog plot shown inFigure 15 is given as

120597119898 (120588119908)

120597 ln 119905= 119898gas =

1637120583119876119896ℎ120572119877

= minus 12502 (91)

And estimated gas permeability is

119896gas =1637120583119876

119900

ℎ120572119877119898gas= 179mD (92)

Using the empirical model for all three phases the averageestimated permeability is as follows

119896ave =4radic49 times 4352 times 1792

119896ave = 51mD

(93)

6 Conclusion

The following inferences were drawn from the six scenariosreviewed

(i) Results from the three-phase fluid empirical modeldeveloped for average effective 119896ave estimation rangebetween 47 and 57mD for Example 1 and give404mD for Example 2 which is within that used inthe simulation model and also estimated from BPR

(ii) The BPR PDENDWA and PDENWAT give the samestabilisation and the same 119896 estimates making it suit-able for interpretation of pressure transient analysis

(iii) It has been demonstrated in all 6 scenarios inves-tigated that the heavier fluid such as water andthe weighted average pressure-density equivalent ofall fluid give exact effective 119896 as the conventionalmethod

(iv) For oil reservoir system (scenarios (a) to (d)) 119896 =

500mD is achieved if the thickness contributing toflow ranges from 50 to 150 ft while for gas condensatereservoir (scenarios (a) to (b)) 119896 ranges between340 and 371mD which is achieved if the net sandthickness ℎ = 148ft is contributing to flow

(v) This approach also provides an estimate of the pos-sible fluid phase permeabilities and the of eachphase contribution to flow at a given point hence atseveral 1198891199011015840 stabilisation points the relative 119896 can begenerated

(vi) The derivatives for all output parameters display thesame wellbore and reservoir fingerprint as the BPRmethod

(vii) Generally where output parameters (BPR PDE-NOIL PDENGAS PDENWAT and PDENDW)depict different fingerprint the density derivative willserve as support to distinguish between reservoir andnonreservoir response

Nomenclature

119875 Pressure psi119879 Temperature oF119903 Radius ft119896 Permeability mDOslash Porosity fraction120583 Viscosity cp119905 Time hrs119902 Production rate bblday119861 Formation volume factor rbStb119862119905 Total compressibility psiminus1119903119908 Wellbore radius ftΔ119901 Change in pressure psiaℎ Formation thickness ft119860 Drainage area acres119875119908119891 Bottom-hole flowing pressure psi119875119894 Initial pressure psi

119885 Difference between two pointtime series119894 Subscript of an observed variable119888 Subscript of a calculated variableSTEYX SSE of data point119899 Number of data pointsCov Covariance of data point120575 Standard deviation119905119901 Cumulative production time119862119904 Wellbore storage constant119909 Mean of data point

Abbreviations

LBPR Local grid bottom-hole pressureLDENO Local grid oil densityLDENW Local grid water densityLDENG Local grid gas densityWBHP Well bottom-hole pressureBPR Well bottom-hole pressurePDENOIL Pressure equivalent of LDENOPDENGAS Pressure equivalent of LDENGPDENWAT Pressure equivalent of LDENWPDENDWA Pressure equivalent of density weighted

average (LDENO LDENG and LDENW)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Jalali S D Mohaghegh and R Gaskari ldquoIdentifying infilllocations and underperformer wells in mature fields usingmonthly production rate data Carthage Field Cotton Valley

16 Journal of Petroleum Engineering

Formation Texasrdquo in Proceedings of the SPE Eastern RegionalMeeting SPE Paper 104550 Canton Ohio USA October 2006

[2] D Tiab and A Kumar ldquoApplication of PD function to inter-ference analysisrdquo in Proceedings of the 51st Annual Fall Meetingof the Society of Petroleum Engineers of AIME Paper SPE 6053New Orleans La USA 1976

[3] D Tiab and A Kumar ldquoDetection and location of two parallelsealing fault around a wellrdquo in Proceedings of the 51st AnnualFall Meeting of the Society of Petroleum Engineers of AIME SPEPaper 6056 New Orleans La USA 1976

[4] D Bourdet T M Whittle A A Douglas and Y M Pirard ldquoAnew set of type curves simplifies well test analysisrdquo World Oilvol 196 no 6 pp 95ndash106 1983

[5] S Y Zheng ldquoFighting against non-unique solution problemsin heterogeneous reservoirs through numerical well testingrdquo inProceedings of the SPE Asia Pacific Oil and Gas Conference andExhibition SPE 100951 Adelaide Australia September 2006

[6] R N Horne and K O Temeng ldquoRecognition and location ofpinch-out boundaries by pressure transient analysisrdquo SPE 9905Society of Petroleum Engineers 1981

[7] L C Ayestaran R D Nurmi G A K Shehab and W S ElsildquoWelltest design and interpretation improved by integrated welltesting and geological effortsrdquo SPE 17945 1989

[8] G J Massonnat and D Bandiziol ldquoInterdependence betweengeology and well test interpretationrdquo SPE 22740 1991

[9] K F Du and G Stewart ldquoReservoir description from well testinterpretationrdquo in North Sea Oil and Gas ReservoirmdashIII pp339ndash356 Norwegian Institute of Technology (NTH) 1994

[10] S Y Zheng Well testing and characterisation of meanderingfluvial channel reservoirs [PhD thesis] Heriot-Watt University1997

[11] V T Biu and S Y Zheng ldquoA new approach in pressure transientanalysis part I improved diagnosis of flow regimes in oil and gaswellsrdquo in Proceedings of the 3rd EAGEAAPGWorkshop on TightReservoirs in theMiddle East AbuDhabi UnitedArab Emirates2015

[12] R N Horne Modern Welltest Analysis Petroway Palo AltoCalif USA 1995

[13] C S Matthews and D G Russell Pressure Buildup and FlowTests in Wells vol 1 ofMonograph Series SPE Richardson TexUSA 1967

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10 Journal of Petroleum Engineering

00

5000000

10000000

15000000

20000000

25000000

0 40 80 120 160

Prod

uctio

n ra

te (s

cfd

)

Duration

Test design sequence

Cases a to b

Figure 10 Production test sequence (buildup) for Example 2

10000

Case a derivatives for conventional and numerical density models

00100

01000

BHPPENDAPDENO

PDENGPDENW

1422mD

3707mD

35747mD

0000 0001 0010 0100 1000 10000

100000

100000Log time (h)

k =1627qB120583

mh

logdp998400

Figure 11 Derivative and 119870 estimation for scenario (a)

However PDENGAS and PDENOIL differ giving 350mDand 35747mD respectively if ℎ = 148ft

Also the derivative in Figure 12 shows good radial flowand a negative half-slope fingerprint is seen at 05 to 20 hrsof buildup and a good stabilisation from 30 hrs in the modelparameters After 3 hrs of shut-in somenoisy data (numericalartefact) is seen in all the fluid phase derivatives but still goodenough to identify 119889119901

1015840 stabilisationFor each of the fluid phase permeabilities 119896 value of

3409mD at ℎ = 148ft is obtained using BPR which is thesame for the PDENDWA and PDENWAT PDENGAS andPDENOIL differ giving 1490 and 35141mD at ℎ = 148ftrespectively

For scenarios (a) and (b) a good 119896ave value of 4050and 4030mD is obtained respectively from the empiricalmodel (34) integrating all fluid phase permeabilities whichis within that used in the simulationmodel and the estimated

0001000

0010000

0100000

1000000

10000000

0000 0001 0010 0100 1000 10000 100000

Case b derivatives for conventional and numerical density models

BHPPENDAPDENO

PDENGPDENW

Log time (h)

k =1627qB120583

mh

1490mD

3410mD

35140mD

logdp998400

Figure 12 Derivative and119870 estimation for scenario (b)

Table 7 Summary of reservoir modelling properties imputed ineclipse model for case study 2

Parameters Numerical density119896 (mD)

Equivalentℎ (ft)

Scenario(a)

BHP 371PDENA 371PDENG 142 148PDENO 3575PDENW 371

Scenario(b)

BHP 341PDENA 341PDENG 149 148PDENO 3514PDENW 341

conventional approach From the 2 scenarios investigated ithas been demonstrated that the heavier fluid such as waterand the weighted average pressure-density equivalent of allfluid give exact effective 119896 as the conventional method BPRA summary of the result is shown in Table 7

This approach estimates the fluid phase permeabilitiesand the of each phase contribution to flow at a given pointhence at several points the relative 119896 can be generated asshown in Table 8

5 Density Related Radial Flow EquationDerivation for Each Fluid Phase

Radial diffusivity equation is given as

1119903

120597120588

120597119903[119903

120597120588

120597119903] =

120601120583119888

119896

120597120588

120597119905 (35)

Journal of Petroleum Engineering 11

Table 8 Comparison of 119870 estimates between conventional and numerical density parameters

Scenario Numerical density method Calculated119870 (mD)

Conventional119896 (mD)

Simulationmodel 119870 (mD) Relative 119870

phasecontribution to flowPhases 119870 (mD)

(a)Gas 1422 004 30Oil 35747 4046 3707 400 089 870

Water 3707 009 90

(b)Gas 149 004 40Oil 35141 4038 3409 400 089 880

Water 3409 008 90

For oil

1119903

120597120588

120597119903[119903

120597120588119900

120597119903] =

120601120583119900119888

119896119900

120597120588119900

120597119905 (36a)

For water phase

1119903

120597120588

120597119903[119903

120597120588119908

120597119903] =

120601120583119908119888

119896119908

120597120588119908

120597119905 (36b)

For gas phase

1119903

120597120588

120597119903[119903

120597120588119892

120597119903] =

120601120583119892119888

119896119892

120597120588119892

120597119905 (36c)

51 From Equation for Slight and Small Compressibility Suchas Water and Oil From (22)

119875 = 119875119900minus

120588120588119900+ 1

119888119900

(37)

Differentiating with respect to 120588

120597119875

120597120588= minus

1119888119900120588119900

120597119875 = minus120597120588

119888119900120588119900

(38)

From Darcy flow equation

119902 = minus2120587119896ℎ120583

119903120597120588

120597119903 (39)

Substitute for 120597119875

119902 =2120587119896ℎ120583

119903

119888119900120588119900

120597120588

120597119903 (40)

To derive the analytical density transient equation for eachphase the following assumptionconditions are applicable

For oil phase from (35)

1119903

120597120588

120597119903[119903

120597120588

120597119903] =

120601120583119888

119896

120597120588

120597119905 (41)

Initial condition is as follows

120588 (119903 119905 = 0) = 120588119894 (42a)

BC at the wellbore is as follows

lim119903rarr 0

2120587119896ℎ120583

119903

119888119900120588119900

120597120588

120597119903= 119876 (42b)

BC at infirmity is as follows

lim119903rarrinfin

120588 (119903 119905) = 120588119894 (42c)

Using the Boltzmann transformation consider the followingAssuming 120578 = 120601120583119888119903

2119896119905

120597120588

120597119903=

120597120588

120597120578

120597120578

120597119903=2120601120583119888119903119896119905

120597120588

120597120578=

1206011205831198881199032

119896119905

2119903

120597120588

120597120578=2120578119903

120597120588

120597120578 (43)

Therefore1119903

120597

120597119903[119903

120597120588

120597119903] =

21205781199032

120597

120597120578[2120578

120597120588

120597120578] =

41205781199032

120597

120597120578[120578

120597120588

120597120578] (44)

Differentiating with respect to 119903 is equal to differentiatingwith respect to 120578 multiply by 2120578119903

From (35) the LHS is resolved as follows

120597120588

120597119905=

120597120588

120597120578

120597120578

120597119905=

120601120583119888119903

1198961199052120597120588

120597120578= minus

120578

119905

120597120588

120597120578 (45)

120601120583119888

119896

120597120588

120597119905=

minus120601120583119888

119896

120578

119905

120597120588

120597119905=

minus1206011205831198881199032

119896119905

120578

1199032120597120588

120597120578=

minus1205782

1199032120597120588

120597120578 (46)

Equating (44) and (46) we have

120597

120597120578[120578

120597120588

120597120578] =

minus120578

4120597120588

120597120578 (47)

This is the simplified ordinary differential equation of 120588 as afunction of 120578

Apply boundary conditions as followsInitial condition is as follows

lim119903rarrinfin

120588 (120578) = 120588119894 (48a)

Substituting (43) into (40)

lim119903rarr 0

2120587119896ℎ120583

119903

119888119900120588119900

120597120588

120597119903= 119876 (48b)

lim119903rarr 0

[120578120597120588

120597120578] =

120601120583119888119900120588119900

4120587119896ℎ (49)

Assume that119898 = 120578(120597120588120597120578)

12 Journal of Petroleum Engineering

From (47)

120597119898

120597120578=

minus119898

4 (50)

Integrating from 120578 = 0 to 120578

997904rArr int

119898(120578)

119898(0)

120597119898

120597120578= minusint

120578

0

120597120578

4

997904rArr ln[119898 (120578)

119898 (0)] =

minus120578

4

997904rArr 119898(120578) = 119898 (0) 119890minus1205784

(51)

Recall boundary condition

lim119903rarr 0

[120578120597120588

120597120578] =

119876120583119888119900120588119900

4120587119896ℎ

119898 (0) =119876120583119888119900120588119900

4120587119896ℎ

119898 (120578) =119876120583119888119900120588119900

4120587119896ℎ119890minus1205784

(52)

Recall that119898 = 120578(120597120588120597120578)Therefore

120597120588 (120578)

120597120578=

120583119876119888119900120588119900

4120587119896ℎ119890minus1205784

120578 (53)

Integrating from 120578 = infin where 120588 = 120588119894

int

120588(120578)

120588119894

120597120588 = minusint

120578

infin

120583119876119888119900120588119900

4120587119896ℎ119890minus1205784

120578120597120578

120588 (120578) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎminusint

infin

120578

119890minus1205784

120578120597120578

(54)

where

120578 =120601120583119888119903

2

119896119905 (55)

Assuming 119906 = 1205784 then 120597120578120578 = 120597119906119906 and 119906 = 12060112058311988811990324119896119905

Hence

120588 = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎminusint

infin

12060112058311988811990324119896119905

119890minus1205784

120578120597120578 (56)

where

minus119864119894 (minus119909) = int

infin

119909

119890minus119906

119906120597119906 (57)

Known as the exponential integral function defined by [13]

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎ119864119894 (minus119909) (58)

where 119909 = 12060112058311988811990324119896119905 and

minus119864119894 (minus119909) = int

infin

119909

119890minus119906

119906120597119906 (59)

At large times 119909 will be small applying Taylorrsquos series andintegrating

minus119864119894 (minus119909) = ln (120574)

ln (120574) = ln (1781) = 05772(60)

where 120574 is known as Eulerrsquos numberFor 14119909 gt 25 consider the followingThen

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎ[ln119909+ ln 120574]

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎln[

1206011205831198881199032120574

4119896119905]

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎln [

22461198961199051206011205831198881199032

]

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎ[ln [

22461198961199051206011205831198881199032

]+ 080907]

(61)

Plotting 120588119908(119905) versus ln(119905) will yield a straight line at longer

time and the slope of the line is given as120597120588119908

120597 ln 119905= 119898oil =

120583119876119888119900120588119900

4120587119896ℎ (62)

Therefore

119896ℎ =120583119876119888119900120588119900

4120587119898oil (63)

52 For Water Phase The radial density equation is given as

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎ[ln [

22461198961199051206011205831198881199032

]+ 080907] (64)

Plotting 120588119908(119905) versus ln(119905) will yield a straight line at longer

time and the slope of the line is given as120597120588119908

120597 ln 119905= 119898water =

120583119876119888119900120588119900

4120587119896ℎ (65)

where

119896ℎ =120583119876119888119900120588119900

4120587119898water (66)

53 For Gas Phase From equation for compressible fluidsuch as gas consider the following

From (32)

119875 =119875119900120588 minus 119875119900

2120588119900119888119892

120588119900(1 minus 119875

119900119888119892)

120597119875

120597120588=

119875119900

120588119900(1 minus 119875

119900119888119892)

120597119875 =119875119900

120588119900(1 minus 119875

119900119888119892)120597120588

(67)

Journal of Petroleum Engineering 13

From the fundamental

119899 =119875119881

119911119877119879 (68)

At standard condition

119875119881

119911119879=

119875sc119881sc119879sc

119875 (5615119902)119911119879

=119875sc119876sc119879sc

119902 =119875sc119879sc

(119911119879

119875)

119876sc5615

119902 =119875sc119879sc

119876sc5615

(119911119879

119875) =

2120587119896ℎ120583

119903120597119901

120597119903

119902 =119875sc119879sc

119876sc5615

(119911119879) =2120587119896ℎ120583

119903119875120597119901

120597119903

(69)

where

120597119875 =119875119900

120588119900(1 minus 119875

119900119888119892)120597120588

120572 =119875119900

120588119900(1 minus 119875

119900119888119892)

120597119875 = 120572120597120588

119875 = 120588119877119911119879

(70)

Substitute into the equation where

119898(120588119894) minus119898 (120588

119908119891) =

2120583

int

120588119894

120588119908119891

120588120597120588

119898 (120588119894) minus119898 (120588

119908119891) =

120588119894

2minus 120588119908119891

2

120583

120588 = radic120588119894

2+ 120588119908119891

2

2

(71)

From the diffusivity equation for gas

1119903

120597120588

120597119903[119903

120597120588119892

120597119903] =

120601120583119892119888

119896119892

120597120588119892

120597119905 (72)

Initial condition is as follows

120588 (119903 119905 = 0) = 120588119894 (73)

BC at the wellbore is as follows

lim119903rarr 0

4120587119896ℎ120572119877120583

(5615119879sc

119875sc) 119903120588

120597120588

120597119903= 119876 (74)

BC at infirmity is as follows

lim119903rarrinfin

120588 (119903 119905) = 120588119894 (75)

where119877119875sc and119879sc are known as temperature pressure andgas constant at standard condition

Therefore integrating from 120578 = infin where 120588 = 120588119894

int

120588(120578)

120588119894

120588120597120588 = minusint

120578

infin

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)119890minus1205784

120578120597120578

1205882(120578) = 120588

119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

minusint

infin

120578

119890minus1205784

120578120597120578

(76)

where

120578 =120601120583119888119903

2

119896119905 (77)

Assume that 119906 = (1205784) (120597120578120578) = 120597119906119906 and 119906 = 12060112058311988811990324119896119905

Hence

1205882= 120588119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)minusint

infin

12060112058311988811990324119896119905

119890minus1205784

120578 (78)

where minus119864119894(minus119909) = intinfin

119909(119890minus119906

119906)120597119906Known as the exponential integral function defined by

[13]

1205882= 120588119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)minus119864119894 (minus119909) (79)

where 119909 = 12060112058311988811990324119896119905 and minus119864119894(minus119909) = int

infin

119909(119890minus119906

119906)120597119906At large times 119909will be small applying Taylorrsquos series and

integrating

minus119864119894 (minus119909) = ln (120574)

ln (120574) = ln (1781) = 05772(80)

where 120574 is known as Eulerrsquos numberFor 14119909 gt 25 consider the followingThen

1205882(119903 119905) = 120588

119894

2+

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) [ln119909+ ln 120574] (81)

1205882(119903 119905) = 120588

119894

2+

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) ln[120601120583119888119903

2120574

4119896119905]

1205882(119903 119905) = 120588

119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) ln [22461198961199051206011205831198881199032

]

119898 (120588119894) minus119898 (120588

119908119891) =

2120583

int

120588119894

120588119908119891

120588 120597120588

119898 (120588119908119891

) = 119898 (120588119894) minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

sdot [ln [22461198961199051206011205831198881199032

]+ 080907]

120597119898 (120588119908)

120597 ln 119905= 119898gas =

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

(82)

14 Journal of Petroleum Engineering

4160041650417004175041800418504190041950420004205042100

560856095609561056105611561156125612

ODENGDEN

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

1205880

(Ibm

ft)

120597120588w120597lnt

= moil =1627120583Qc01205880

kh= minus0009

120597m(120588w)

120597lnt= mgas =

1637120583Q

kh120572R= minus12195

120588g2

(Ibm

ft)

Figure 13 Oil and gas phase density versus log Horner time

6279

6279

6279

6279

6279

6279

6279

6279

01 1 10 100 1000 10000 1000001000000Horner time (h)

Density versus log Horner time

WDEN

120588w

(Ibm

ft)

120597120588w120597lnt

= mwater =1627120583Qc01205880

kh= minus00003

Figure 14 Water phase density versus log Horner time

Plotting 120588119908

2(119905) or 119898(120588

119908) versus ln(119905) will yield a straight line

at longer time and the slope of the line is given as follows

119896ℎ =120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)1

119898gas (83)

Example 3 Data from Example 1(a) was used for this caseand the semilog plot of density for each phase (oil water andgas) is plotted against log of Horner timelowastlowast

119876 used for 119896 calculation is the average rate for allflowing periods

From Figures 13 and 14 the calculated slopes of the radialflow for oil and water phases are given as

120597120588119908

120597 ln 119905= 119898oil =

1627120583119876119888119900120588119900

119896ℎ= minus 0009

120597120588119908

120597 ln 119905= 119898water =

1627120583119876119888119900120588119900

119896ℎ= minus 00003

(84)

The estimated phase permeabilities are

119896oil =1627120583119876119888

119900120588119900

ℎ119898oil= 445mD

119896water =1627120583119876119888

119900120588119900

ℎ119898water= 50mD

(85)

Also for the gas phase consider the following

ODENGDEN

422004225042300423504240042450425004255042600

5613561356145614561556155616

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

1205880

(Ibm

ft)

120597120588w120597lnt

= moil =1627120583Qc01205880

kh= minus0009

120597m(120588w)

120597lnt= mgas =

1637120583Q

kh120572R= minus12502

120588g2

(Ibm

ft)

Figure 15 Oil and gas phase density versus log Horner time

WDEN

6279627962796279627962796280628062806280

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

120588w

(Ibm

ft)

120597120588w120597lnt

= mwater =1627120583Qc01205880

kh= minus00003

Figure 16 Water phase density versus log Horner time

The calculated slope from the semilog plot shown inFigure 13 is given as

120597119898 (120588119908)

120597 ln 119905= 119898gas =

1637120583119876119896ℎ120572119877

= minus 12195 (86)

And estimated gas permeability is

119896gas =1637120583119876

119900

ℎ120572119877119898gas= 170mD (87)

Using the empirical model for all three phases the averageestimated permeability is as follows

119896ave =4radic50 times 445 times 1702

119896ave = 50mD

(88)

Example 4 Data from Example 1(c) was used for this caseand the semilog plot of density for each phase (oil water andgas) is plotted against log of Horner time

From Figures 15 and 16 the calculated slopes of the radialflow for oil and water phases are given as

120597120588119908

120597 ln 119905= 119898oil =

1627120583119876119888119900120588119900

119896ℎ= minus 0009

120597120588119908

120597 ln 119905= 119898water =

1627120583119876119888119900120588119900

119896ℎ= minus 00003

(89)

Journal of Petroleum Engineering 15

The estimated phase permeabilities are

119896oil =1627120583119876119888

119900120588119900

ℎ119898oil= 4352mD

119896water =1627120583119876119888

119900120588119900

ℎ119898water= 49mD

(90)

Also for the gas phase consider the following

The calculated slope from the semilog plot shown inFigure 15 is given as

120597119898 (120588119908)

120597 ln 119905= 119898gas =

1637120583119876119896ℎ120572119877

= minus 12502 (91)

And estimated gas permeability is

119896gas =1637120583119876

119900

ℎ120572119877119898gas= 179mD (92)

Using the empirical model for all three phases the averageestimated permeability is as follows

119896ave =4radic49 times 4352 times 1792

119896ave = 51mD

(93)

6 Conclusion

The following inferences were drawn from the six scenariosreviewed

(i) Results from the three-phase fluid empirical modeldeveloped for average effective 119896ave estimation rangebetween 47 and 57mD for Example 1 and give404mD for Example 2 which is within that used inthe simulation model and also estimated from BPR

(ii) The BPR PDENDWA and PDENWAT give the samestabilisation and the same 119896 estimates making it suit-able for interpretation of pressure transient analysis

(iii) It has been demonstrated in all 6 scenarios inves-tigated that the heavier fluid such as water andthe weighted average pressure-density equivalent ofall fluid give exact effective 119896 as the conventionalmethod

(iv) For oil reservoir system (scenarios (a) to (d)) 119896 =

500mD is achieved if the thickness contributing toflow ranges from 50 to 150 ft while for gas condensatereservoir (scenarios (a) to (b)) 119896 ranges between340 and 371mD which is achieved if the net sandthickness ℎ = 148ft is contributing to flow

(v) This approach also provides an estimate of the pos-sible fluid phase permeabilities and the of eachphase contribution to flow at a given point hence atseveral 1198891199011015840 stabilisation points the relative 119896 can begenerated

(vi) The derivatives for all output parameters display thesame wellbore and reservoir fingerprint as the BPRmethod

(vii) Generally where output parameters (BPR PDE-NOIL PDENGAS PDENWAT and PDENDW)depict different fingerprint the density derivative willserve as support to distinguish between reservoir andnonreservoir response

Nomenclature

119875 Pressure psi119879 Temperature oF119903 Radius ft119896 Permeability mDOslash Porosity fraction120583 Viscosity cp119905 Time hrs119902 Production rate bblday119861 Formation volume factor rbStb119862119905 Total compressibility psiminus1119903119908 Wellbore radius ftΔ119901 Change in pressure psiaℎ Formation thickness ft119860 Drainage area acres119875119908119891 Bottom-hole flowing pressure psi119875119894 Initial pressure psi

119885 Difference between two pointtime series119894 Subscript of an observed variable119888 Subscript of a calculated variableSTEYX SSE of data point119899 Number of data pointsCov Covariance of data point120575 Standard deviation119905119901 Cumulative production time119862119904 Wellbore storage constant119909 Mean of data point

Abbreviations

LBPR Local grid bottom-hole pressureLDENO Local grid oil densityLDENW Local grid water densityLDENG Local grid gas densityWBHP Well bottom-hole pressureBPR Well bottom-hole pressurePDENOIL Pressure equivalent of LDENOPDENGAS Pressure equivalent of LDENGPDENWAT Pressure equivalent of LDENWPDENDWA Pressure equivalent of density weighted

average (LDENO LDENG and LDENW)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Jalali S D Mohaghegh and R Gaskari ldquoIdentifying infilllocations and underperformer wells in mature fields usingmonthly production rate data Carthage Field Cotton Valley

16 Journal of Petroleum Engineering

Formation Texasrdquo in Proceedings of the SPE Eastern RegionalMeeting SPE Paper 104550 Canton Ohio USA October 2006

[2] D Tiab and A Kumar ldquoApplication of PD function to inter-ference analysisrdquo in Proceedings of the 51st Annual Fall Meetingof the Society of Petroleum Engineers of AIME Paper SPE 6053New Orleans La USA 1976

[3] D Tiab and A Kumar ldquoDetection and location of two parallelsealing fault around a wellrdquo in Proceedings of the 51st AnnualFall Meeting of the Society of Petroleum Engineers of AIME SPEPaper 6056 New Orleans La USA 1976

[4] D Bourdet T M Whittle A A Douglas and Y M Pirard ldquoAnew set of type curves simplifies well test analysisrdquo World Oilvol 196 no 6 pp 95ndash106 1983

[5] S Y Zheng ldquoFighting against non-unique solution problemsin heterogeneous reservoirs through numerical well testingrdquo inProceedings of the SPE Asia Pacific Oil and Gas Conference andExhibition SPE 100951 Adelaide Australia September 2006

[6] R N Horne and K O Temeng ldquoRecognition and location ofpinch-out boundaries by pressure transient analysisrdquo SPE 9905Society of Petroleum Engineers 1981

[7] L C Ayestaran R D Nurmi G A K Shehab and W S ElsildquoWelltest design and interpretation improved by integrated welltesting and geological effortsrdquo SPE 17945 1989

[8] G J Massonnat and D Bandiziol ldquoInterdependence betweengeology and well test interpretationrdquo SPE 22740 1991

[9] K F Du and G Stewart ldquoReservoir description from well testinterpretationrdquo in North Sea Oil and Gas ReservoirmdashIII pp339ndash356 Norwegian Institute of Technology (NTH) 1994

[10] S Y Zheng Well testing and characterisation of meanderingfluvial channel reservoirs [PhD thesis] Heriot-Watt University1997

[11] V T Biu and S Y Zheng ldquoA new approach in pressure transientanalysis part I improved diagnosis of flow regimes in oil and gaswellsrdquo in Proceedings of the 3rd EAGEAAPGWorkshop on TightReservoirs in theMiddle East AbuDhabi UnitedArab Emirates2015

[12] R N Horne Modern Welltest Analysis Petroway Palo AltoCalif USA 1995

[13] C S Matthews and D G Russell Pressure Buildup and FlowTests in Wells vol 1 ofMonograph Series SPE Richardson TexUSA 1967

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Journal of Petroleum Engineering 11

Table 8 Comparison of 119870 estimates between conventional and numerical density parameters

Scenario Numerical density method Calculated119870 (mD)

Conventional119896 (mD)

Simulationmodel 119870 (mD) Relative 119870

phasecontribution to flowPhases 119870 (mD)

(a)Gas 1422 004 30Oil 35747 4046 3707 400 089 870

Water 3707 009 90

(b)Gas 149 004 40Oil 35141 4038 3409 400 089 880

Water 3409 008 90

For oil

1119903

120597120588

120597119903[119903

120597120588119900

120597119903] =

120601120583119900119888

119896119900

120597120588119900

120597119905 (36a)

For water phase

1119903

120597120588

120597119903[119903

120597120588119908

120597119903] =

120601120583119908119888

119896119908

120597120588119908

120597119905 (36b)

For gas phase

1119903

120597120588

120597119903[119903

120597120588119892

120597119903] =

120601120583119892119888

119896119892

120597120588119892

120597119905 (36c)

51 From Equation for Slight and Small Compressibility Suchas Water and Oil From (22)

119875 = 119875119900minus

120588120588119900+ 1

119888119900

(37)

Differentiating with respect to 120588

120597119875

120597120588= minus

1119888119900120588119900

120597119875 = minus120597120588

119888119900120588119900

(38)

From Darcy flow equation

119902 = minus2120587119896ℎ120583

119903120597120588

120597119903 (39)

Substitute for 120597119875

119902 =2120587119896ℎ120583

119903

119888119900120588119900

120597120588

120597119903 (40)

To derive the analytical density transient equation for eachphase the following assumptionconditions are applicable

For oil phase from (35)

1119903

120597120588

120597119903[119903

120597120588

120597119903] =

120601120583119888

119896

120597120588

120597119905 (41)

Initial condition is as follows

120588 (119903 119905 = 0) = 120588119894 (42a)

BC at the wellbore is as follows

lim119903rarr 0

2120587119896ℎ120583

119903

119888119900120588119900

120597120588

120597119903= 119876 (42b)

BC at infirmity is as follows

lim119903rarrinfin

120588 (119903 119905) = 120588119894 (42c)

Using the Boltzmann transformation consider the followingAssuming 120578 = 120601120583119888119903

2119896119905

120597120588

120597119903=

120597120588

120597120578

120597120578

120597119903=2120601120583119888119903119896119905

120597120588

120597120578=

1206011205831198881199032

119896119905

2119903

120597120588

120597120578=2120578119903

120597120588

120597120578 (43)

Therefore1119903

120597

120597119903[119903

120597120588

120597119903] =

21205781199032

120597

120597120578[2120578

120597120588

120597120578] =

41205781199032

120597

120597120578[120578

120597120588

120597120578] (44)

Differentiating with respect to 119903 is equal to differentiatingwith respect to 120578 multiply by 2120578119903

From (35) the LHS is resolved as follows

120597120588

120597119905=

120597120588

120597120578

120597120578

120597119905=

120601120583119888119903

1198961199052120597120588

120597120578= minus

120578

119905

120597120588

120597120578 (45)

120601120583119888

119896

120597120588

120597119905=

minus120601120583119888

119896

120578

119905

120597120588

120597119905=

minus1206011205831198881199032

119896119905

120578

1199032120597120588

120597120578=

minus1205782

1199032120597120588

120597120578 (46)

Equating (44) and (46) we have

120597

120597120578[120578

120597120588

120597120578] =

minus120578

4120597120588

120597120578 (47)

This is the simplified ordinary differential equation of 120588 as afunction of 120578

Apply boundary conditions as followsInitial condition is as follows

lim119903rarrinfin

120588 (120578) = 120588119894 (48a)

Substituting (43) into (40)

lim119903rarr 0

2120587119896ℎ120583

119903

119888119900120588119900

120597120588

120597119903= 119876 (48b)

lim119903rarr 0

[120578120597120588

120597120578] =

120601120583119888119900120588119900

4120587119896ℎ (49)

Assume that119898 = 120578(120597120588120597120578)

12 Journal of Petroleum Engineering

From (47)

120597119898

120597120578=

minus119898

4 (50)

Integrating from 120578 = 0 to 120578

997904rArr int

119898(120578)

119898(0)

120597119898

120597120578= minusint

120578

0

120597120578

4

997904rArr ln[119898 (120578)

119898 (0)] =

minus120578

4

997904rArr 119898(120578) = 119898 (0) 119890minus1205784

(51)

Recall boundary condition

lim119903rarr 0

[120578120597120588

120597120578] =

119876120583119888119900120588119900

4120587119896ℎ

119898 (0) =119876120583119888119900120588119900

4120587119896ℎ

119898 (120578) =119876120583119888119900120588119900

4120587119896ℎ119890minus1205784

(52)

Recall that119898 = 120578(120597120588120597120578)Therefore

120597120588 (120578)

120597120578=

120583119876119888119900120588119900

4120587119896ℎ119890minus1205784

120578 (53)

Integrating from 120578 = infin where 120588 = 120588119894

int

120588(120578)

120588119894

120597120588 = minusint

120578

infin

120583119876119888119900120588119900

4120587119896ℎ119890minus1205784

120578120597120578

120588 (120578) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎminusint

infin

120578

119890minus1205784

120578120597120578

(54)

where

120578 =120601120583119888119903

2

119896119905 (55)

Assuming 119906 = 1205784 then 120597120578120578 = 120597119906119906 and 119906 = 12060112058311988811990324119896119905

Hence

120588 = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎminusint

infin

12060112058311988811990324119896119905

119890minus1205784

120578120597120578 (56)

where

minus119864119894 (minus119909) = int

infin

119909

119890minus119906

119906120597119906 (57)

Known as the exponential integral function defined by [13]

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎ119864119894 (minus119909) (58)

where 119909 = 12060112058311988811990324119896119905 and

minus119864119894 (minus119909) = int

infin

119909

119890minus119906

119906120597119906 (59)

At large times 119909 will be small applying Taylorrsquos series andintegrating

minus119864119894 (minus119909) = ln (120574)

ln (120574) = ln (1781) = 05772(60)

where 120574 is known as Eulerrsquos numberFor 14119909 gt 25 consider the followingThen

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎ[ln119909+ ln 120574]

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎln[

1206011205831198881199032120574

4119896119905]

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎln [

22461198961199051206011205831198881199032

]

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎ[ln [

22461198961199051206011205831198881199032

]+ 080907]

(61)

Plotting 120588119908(119905) versus ln(119905) will yield a straight line at longer

time and the slope of the line is given as120597120588119908

120597 ln 119905= 119898oil =

120583119876119888119900120588119900

4120587119896ℎ (62)

Therefore

119896ℎ =120583119876119888119900120588119900

4120587119898oil (63)

52 For Water Phase The radial density equation is given as

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎ[ln [

22461198961199051206011205831198881199032

]+ 080907] (64)

Plotting 120588119908(119905) versus ln(119905) will yield a straight line at longer

time and the slope of the line is given as120597120588119908

120597 ln 119905= 119898water =

120583119876119888119900120588119900

4120587119896ℎ (65)

where

119896ℎ =120583119876119888119900120588119900

4120587119898water (66)

53 For Gas Phase From equation for compressible fluidsuch as gas consider the following

From (32)

119875 =119875119900120588 minus 119875119900

2120588119900119888119892

120588119900(1 minus 119875

119900119888119892)

120597119875

120597120588=

119875119900

120588119900(1 minus 119875

119900119888119892)

120597119875 =119875119900

120588119900(1 minus 119875

119900119888119892)120597120588

(67)

Journal of Petroleum Engineering 13

From the fundamental

119899 =119875119881

119911119877119879 (68)

At standard condition

119875119881

119911119879=

119875sc119881sc119879sc

119875 (5615119902)119911119879

=119875sc119876sc119879sc

119902 =119875sc119879sc

(119911119879

119875)

119876sc5615

119902 =119875sc119879sc

119876sc5615

(119911119879

119875) =

2120587119896ℎ120583

119903120597119901

120597119903

119902 =119875sc119879sc

119876sc5615

(119911119879) =2120587119896ℎ120583

119903119875120597119901

120597119903

(69)

where

120597119875 =119875119900

120588119900(1 minus 119875

119900119888119892)120597120588

120572 =119875119900

120588119900(1 minus 119875

119900119888119892)

120597119875 = 120572120597120588

119875 = 120588119877119911119879

(70)

Substitute into the equation where

119898(120588119894) minus119898 (120588

119908119891) =

2120583

int

120588119894

120588119908119891

120588120597120588

119898 (120588119894) minus119898 (120588

119908119891) =

120588119894

2minus 120588119908119891

2

120583

120588 = radic120588119894

2+ 120588119908119891

2

2

(71)

From the diffusivity equation for gas

1119903

120597120588

120597119903[119903

120597120588119892

120597119903] =

120601120583119892119888

119896119892

120597120588119892

120597119905 (72)

Initial condition is as follows

120588 (119903 119905 = 0) = 120588119894 (73)

BC at the wellbore is as follows

lim119903rarr 0

4120587119896ℎ120572119877120583

(5615119879sc

119875sc) 119903120588

120597120588

120597119903= 119876 (74)

BC at infirmity is as follows

lim119903rarrinfin

120588 (119903 119905) = 120588119894 (75)

where119877119875sc and119879sc are known as temperature pressure andgas constant at standard condition

Therefore integrating from 120578 = infin where 120588 = 120588119894

int

120588(120578)

120588119894

120588120597120588 = minusint

120578

infin

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)119890minus1205784

120578120597120578

1205882(120578) = 120588

119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

minusint

infin

120578

119890minus1205784

120578120597120578

(76)

where

120578 =120601120583119888119903

2

119896119905 (77)

Assume that 119906 = (1205784) (120597120578120578) = 120597119906119906 and 119906 = 12060112058311988811990324119896119905

Hence

1205882= 120588119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)minusint

infin

12060112058311988811990324119896119905

119890minus1205784

120578 (78)

where minus119864119894(minus119909) = intinfin

119909(119890minus119906

119906)120597119906Known as the exponential integral function defined by

[13]

1205882= 120588119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)minus119864119894 (minus119909) (79)

where 119909 = 12060112058311988811990324119896119905 and minus119864119894(minus119909) = int

infin

119909(119890minus119906

119906)120597119906At large times 119909will be small applying Taylorrsquos series and

integrating

minus119864119894 (minus119909) = ln (120574)

ln (120574) = ln (1781) = 05772(80)

where 120574 is known as Eulerrsquos numberFor 14119909 gt 25 consider the followingThen

1205882(119903 119905) = 120588

119894

2+

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) [ln119909+ ln 120574] (81)

1205882(119903 119905) = 120588

119894

2+

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) ln[120601120583119888119903

2120574

4119896119905]

1205882(119903 119905) = 120588

119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) ln [22461198961199051206011205831198881199032

]

119898 (120588119894) minus119898 (120588

119908119891) =

2120583

int

120588119894

120588119908119891

120588 120597120588

119898 (120588119908119891

) = 119898 (120588119894) minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

sdot [ln [22461198961199051206011205831198881199032

]+ 080907]

120597119898 (120588119908)

120597 ln 119905= 119898gas =

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

(82)

14 Journal of Petroleum Engineering

4160041650417004175041800418504190041950420004205042100

560856095609561056105611561156125612

ODENGDEN

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

1205880

(Ibm

ft)

120597120588w120597lnt

= moil =1627120583Qc01205880

kh= minus0009

120597m(120588w)

120597lnt= mgas =

1637120583Q

kh120572R= minus12195

120588g2

(Ibm

ft)

Figure 13 Oil and gas phase density versus log Horner time

6279

6279

6279

6279

6279

6279

6279

6279

01 1 10 100 1000 10000 1000001000000Horner time (h)

Density versus log Horner time

WDEN

120588w

(Ibm

ft)

120597120588w120597lnt

= mwater =1627120583Qc01205880

kh= minus00003

Figure 14 Water phase density versus log Horner time

Plotting 120588119908

2(119905) or 119898(120588

119908) versus ln(119905) will yield a straight line

at longer time and the slope of the line is given as follows

119896ℎ =120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)1

119898gas (83)

Example 3 Data from Example 1(a) was used for this caseand the semilog plot of density for each phase (oil water andgas) is plotted against log of Horner timelowastlowast

119876 used for 119896 calculation is the average rate for allflowing periods

From Figures 13 and 14 the calculated slopes of the radialflow for oil and water phases are given as

120597120588119908

120597 ln 119905= 119898oil =

1627120583119876119888119900120588119900

119896ℎ= minus 0009

120597120588119908

120597 ln 119905= 119898water =

1627120583119876119888119900120588119900

119896ℎ= minus 00003

(84)

The estimated phase permeabilities are

119896oil =1627120583119876119888

119900120588119900

ℎ119898oil= 445mD

119896water =1627120583119876119888

119900120588119900

ℎ119898water= 50mD

(85)

Also for the gas phase consider the following

ODENGDEN

422004225042300423504240042450425004255042600

5613561356145614561556155616

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

1205880

(Ibm

ft)

120597120588w120597lnt

= moil =1627120583Qc01205880

kh= minus0009

120597m(120588w)

120597lnt= mgas =

1637120583Q

kh120572R= minus12502

120588g2

(Ibm

ft)

Figure 15 Oil and gas phase density versus log Horner time

WDEN

6279627962796279627962796280628062806280

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

120588w

(Ibm

ft)

120597120588w120597lnt

= mwater =1627120583Qc01205880

kh= minus00003

Figure 16 Water phase density versus log Horner time

The calculated slope from the semilog plot shown inFigure 13 is given as

120597119898 (120588119908)

120597 ln 119905= 119898gas =

1637120583119876119896ℎ120572119877

= minus 12195 (86)

And estimated gas permeability is

119896gas =1637120583119876

119900

ℎ120572119877119898gas= 170mD (87)

Using the empirical model for all three phases the averageestimated permeability is as follows

119896ave =4radic50 times 445 times 1702

119896ave = 50mD

(88)

Example 4 Data from Example 1(c) was used for this caseand the semilog plot of density for each phase (oil water andgas) is plotted against log of Horner time

From Figures 15 and 16 the calculated slopes of the radialflow for oil and water phases are given as

120597120588119908

120597 ln 119905= 119898oil =

1627120583119876119888119900120588119900

119896ℎ= minus 0009

120597120588119908

120597 ln 119905= 119898water =

1627120583119876119888119900120588119900

119896ℎ= minus 00003

(89)

Journal of Petroleum Engineering 15

The estimated phase permeabilities are

119896oil =1627120583119876119888

119900120588119900

ℎ119898oil= 4352mD

119896water =1627120583119876119888

119900120588119900

ℎ119898water= 49mD

(90)

Also for the gas phase consider the following

The calculated slope from the semilog plot shown inFigure 15 is given as

120597119898 (120588119908)

120597 ln 119905= 119898gas =

1637120583119876119896ℎ120572119877

= minus 12502 (91)

And estimated gas permeability is

119896gas =1637120583119876

119900

ℎ120572119877119898gas= 179mD (92)

Using the empirical model for all three phases the averageestimated permeability is as follows

119896ave =4radic49 times 4352 times 1792

119896ave = 51mD

(93)

6 Conclusion

The following inferences were drawn from the six scenariosreviewed

(i) Results from the three-phase fluid empirical modeldeveloped for average effective 119896ave estimation rangebetween 47 and 57mD for Example 1 and give404mD for Example 2 which is within that used inthe simulation model and also estimated from BPR

(ii) The BPR PDENDWA and PDENWAT give the samestabilisation and the same 119896 estimates making it suit-able for interpretation of pressure transient analysis

(iii) It has been demonstrated in all 6 scenarios inves-tigated that the heavier fluid such as water andthe weighted average pressure-density equivalent ofall fluid give exact effective 119896 as the conventionalmethod

(iv) For oil reservoir system (scenarios (a) to (d)) 119896 =

500mD is achieved if the thickness contributing toflow ranges from 50 to 150 ft while for gas condensatereservoir (scenarios (a) to (b)) 119896 ranges between340 and 371mD which is achieved if the net sandthickness ℎ = 148ft is contributing to flow

(v) This approach also provides an estimate of the pos-sible fluid phase permeabilities and the of eachphase contribution to flow at a given point hence atseveral 1198891199011015840 stabilisation points the relative 119896 can begenerated

(vi) The derivatives for all output parameters display thesame wellbore and reservoir fingerprint as the BPRmethod

(vii) Generally where output parameters (BPR PDE-NOIL PDENGAS PDENWAT and PDENDW)depict different fingerprint the density derivative willserve as support to distinguish between reservoir andnonreservoir response

Nomenclature

119875 Pressure psi119879 Temperature oF119903 Radius ft119896 Permeability mDOslash Porosity fraction120583 Viscosity cp119905 Time hrs119902 Production rate bblday119861 Formation volume factor rbStb119862119905 Total compressibility psiminus1119903119908 Wellbore radius ftΔ119901 Change in pressure psiaℎ Formation thickness ft119860 Drainage area acres119875119908119891 Bottom-hole flowing pressure psi119875119894 Initial pressure psi

119885 Difference between two pointtime series119894 Subscript of an observed variable119888 Subscript of a calculated variableSTEYX SSE of data point119899 Number of data pointsCov Covariance of data point120575 Standard deviation119905119901 Cumulative production time119862119904 Wellbore storage constant119909 Mean of data point

Abbreviations

LBPR Local grid bottom-hole pressureLDENO Local grid oil densityLDENW Local grid water densityLDENG Local grid gas densityWBHP Well bottom-hole pressureBPR Well bottom-hole pressurePDENOIL Pressure equivalent of LDENOPDENGAS Pressure equivalent of LDENGPDENWAT Pressure equivalent of LDENWPDENDWA Pressure equivalent of density weighted

average (LDENO LDENG and LDENW)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Jalali S D Mohaghegh and R Gaskari ldquoIdentifying infilllocations and underperformer wells in mature fields usingmonthly production rate data Carthage Field Cotton Valley

16 Journal of Petroleum Engineering

Formation Texasrdquo in Proceedings of the SPE Eastern RegionalMeeting SPE Paper 104550 Canton Ohio USA October 2006

[2] D Tiab and A Kumar ldquoApplication of PD function to inter-ference analysisrdquo in Proceedings of the 51st Annual Fall Meetingof the Society of Petroleum Engineers of AIME Paper SPE 6053New Orleans La USA 1976

[3] D Tiab and A Kumar ldquoDetection and location of two parallelsealing fault around a wellrdquo in Proceedings of the 51st AnnualFall Meeting of the Society of Petroleum Engineers of AIME SPEPaper 6056 New Orleans La USA 1976

[4] D Bourdet T M Whittle A A Douglas and Y M Pirard ldquoAnew set of type curves simplifies well test analysisrdquo World Oilvol 196 no 6 pp 95ndash106 1983

[5] S Y Zheng ldquoFighting against non-unique solution problemsin heterogeneous reservoirs through numerical well testingrdquo inProceedings of the SPE Asia Pacific Oil and Gas Conference andExhibition SPE 100951 Adelaide Australia September 2006

[6] R N Horne and K O Temeng ldquoRecognition and location ofpinch-out boundaries by pressure transient analysisrdquo SPE 9905Society of Petroleum Engineers 1981

[7] L C Ayestaran R D Nurmi G A K Shehab and W S ElsildquoWelltest design and interpretation improved by integrated welltesting and geological effortsrdquo SPE 17945 1989

[8] G J Massonnat and D Bandiziol ldquoInterdependence betweengeology and well test interpretationrdquo SPE 22740 1991

[9] K F Du and G Stewart ldquoReservoir description from well testinterpretationrdquo in North Sea Oil and Gas ReservoirmdashIII pp339ndash356 Norwegian Institute of Technology (NTH) 1994

[10] S Y Zheng Well testing and characterisation of meanderingfluvial channel reservoirs [PhD thesis] Heriot-Watt University1997

[11] V T Biu and S Y Zheng ldquoA new approach in pressure transientanalysis part I improved diagnosis of flow regimes in oil and gaswellsrdquo in Proceedings of the 3rd EAGEAAPGWorkshop on TightReservoirs in theMiddle East AbuDhabi UnitedArab Emirates2015

[12] R N Horne Modern Welltest Analysis Petroway Palo AltoCalif USA 1995

[13] C S Matthews and D G Russell Pressure Buildup and FlowTests in Wells vol 1 ofMonograph Series SPE Richardson TexUSA 1967

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12 Journal of Petroleum Engineering

From (47)

120597119898

120597120578=

minus119898

4 (50)

Integrating from 120578 = 0 to 120578

997904rArr int

119898(120578)

119898(0)

120597119898

120597120578= minusint

120578

0

120597120578

4

997904rArr ln[119898 (120578)

119898 (0)] =

minus120578

4

997904rArr 119898(120578) = 119898 (0) 119890minus1205784

(51)

Recall boundary condition

lim119903rarr 0

[120578120597120588

120597120578] =

119876120583119888119900120588119900

4120587119896ℎ

119898 (0) =119876120583119888119900120588119900

4120587119896ℎ

119898 (120578) =119876120583119888119900120588119900

4120587119896ℎ119890minus1205784

(52)

Recall that119898 = 120578(120597120588120597120578)Therefore

120597120588 (120578)

120597120578=

120583119876119888119900120588119900

4120587119896ℎ119890minus1205784

120578 (53)

Integrating from 120578 = infin where 120588 = 120588119894

int

120588(120578)

120588119894

120597120588 = minusint

120578

infin

120583119876119888119900120588119900

4120587119896ℎ119890minus1205784

120578120597120578

120588 (120578) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎminusint

infin

120578

119890minus1205784

120578120597120578

(54)

where

120578 =120601120583119888119903

2

119896119905 (55)

Assuming 119906 = 1205784 then 120597120578120578 = 120597119906119906 and 119906 = 12060112058311988811990324119896119905

Hence

120588 = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎminusint

infin

12060112058311988811990324119896119905

119890minus1205784

120578120597120578 (56)

where

minus119864119894 (minus119909) = int

infin

119909

119890minus119906

119906120597119906 (57)

Known as the exponential integral function defined by [13]

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎ119864119894 (minus119909) (58)

where 119909 = 12060112058311988811990324119896119905 and

minus119864119894 (minus119909) = int

infin

119909

119890minus119906

119906120597119906 (59)

At large times 119909 will be small applying Taylorrsquos series andintegrating

minus119864119894 (minus119909) = ln (120574)

ln (120574) = ln (1781) = 05772(60)

where 120574 is known as Eulerrsquos numberFor 14119909 gt 25 consider the followingThen

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎ[ln119909+ ln 120574]

120588 (119903 119905) = 120588119894+

120583119876119888119900120588119900

4120587119896ℎln[

1206011205831198881199032120574

4119896119905]

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎln [

22461198961199051206011205831198881199032

]

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎ[ln [

22461198961199051206011205831198881199032

]+ 080907]

(61)

Plotting 120588119908(119905) versus ln(119905) will yield a straight line at longer

time and the slope of the line is given as120597120588119908

120597 ln 119905= 119898oil =

120583119876119888119900120588119900

4120587119896ℎ (62)

Therefore

119896ℎ =120583119876119888119900120588119900

4120587119898oil (63)

52 For Water Phase The radial density equation is given as

120588 (119903 119905) = 120588119894minus

120583119876119888119900120588119900

4120587119896ℎ[ln [

22461198961199051206011205831198881199032

]+ 080907] (64)

Plotting 120588119908(119905) versus ln(119905) will yield a straight line at longer

time and the slope of the line is given as120597120588119908

120597 ln 119905= 119898water =

120583119876119888119900120588119900

4120587119896ℎ (65)

where

119896ℎ =120583119876119888119900120588119900

4120587119898water (66)

53 For Gas Phase From equation for compressible fluidsuch as gas consider the following

From (32)

119875 =119875119900120588 minus 119875119900

2120588119900119888119892

120588119900(1 minus 119875

119900119888119892)

120597119875

120597120588=

119875119900

120588119900(1 minus 119875

119900119888119892)

120597119875 =119875119900

120588119900(1 minus 119875

119900119888119892)120597120588

(67)

Journal of Petroleum Engineering 13

From the fundamental

119899 =119875119881

119911119877119879 (68)

At standard condition

119875119881

119911119879=

119875sc119881sc119879sc

119875 (5615119902)119911119879

=119875sc119876sc119879sc

119902 =119875sc119879sc

(119911119879

119875)

119876sc5615

119902 =119875sc119879sc

119876sc5615

(119911119879

119875) =

2120587119896ℎ120583

119903120597119901

120597119903

119902 =119875sc119879sc

119876sc5615

(119911119879) =2120587119896ℎ120583

119903119875120597119901

120597119903

(69)

where

120597119875 =119875119900

120588119900(1 minus 119875

119900119888119892)120597120588

120572 =119875119900

120588119900(1 minus 119875

119900119888119892)

120597119875 = 120572120597120588

119875 = 120588119877119911119879

(70)

Substitute into the equation where

119898(120588119894) minus119898 (120588

119908119891) =

2120583

int

120588119894

120588119908119891

120588120597120588

119898 (120588119894) minus119898 (120588

119908119891) =

120588119894

2minus 120588119908119891

2

120583

120588 = radic120588119894

2+ 120588119908119891

2

2

(71)

From the diffusivity equation for gas

1119903

120597120588

120597119903[119903

120597120588119892

120597119903] =

120601120583119892119888

119896119892

120597120588119892

120597119905 (72)

Initial condition is as follows

120588 (119903 119905 = 0) = 120588119894 (73)

BC at the wellbore is as follows

lim119903rarr 0

4120587119896ℎ120572119877120583

(5615119879sc

119875sc) 119903120588

120597120588

120597119903= 119876 (74)

BC at infirmity is as follows

lim119903rarrinfin

120588 (119903 119905) = 120588119894 (75)

where119877119875sc and119879sc are known as temperature pressure andgas constant at standard condition

Therefore integrating from 120578 = infin where 120588 = 120588119894

int

120588(120578)

120588119894

120588120597120588 = minusint

120578

infin

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)119890minus1205784

120578120597120578

1205882(120578) = 120588

119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

minusint

infin

120578

119890minus1205784

120578120597120578

(76)

where

120578 =120601120583119888119903

2

119896119905 (77)

Assume that 119906 = (1205784) (120597120578120578) = 120597119906119906 and 119906 = 12060112058311988811990324119896119905

Hence

1205882= 120588119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)minusint

infin

12060112058311988811990324119896119905

119890minus1205784

120578 (78)

where minus119864119894(minus119909) = intinfin

119909(119890minus119906

119906)120597119906Known as the exponential integral function defined by

[13]

1205882= 120588119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)minus119864119894 (minus119909) (79)

where 119909 = 12060112058311988811990324119896119905 and minus119864119894(minus119909) = int

infin

119909(119890minus119906

119906)120597119906At large times 119909will be small applying Taylorrsquos series and

integrating

minus119864119894 (minus119909) = ln (120574)

ln (120574) = ln (1781) = 05772(80)

where 120574 is known as Eulerrsquos numberFor 14119909 gt 25 consider the followingThen

1205882(119903 119905) = 120588

119894

2+

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) [ln119909+ ln 120574] (81)

1205882(119903 119905) = 120588

119894

2+

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) ln[120601120583119888119903

2120574

4119896119905]

1205882(119903 119905) = 120588

119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) ln [22461198961199051206011205831198881199032

]

119898 (120588119894) minus119898 (120588

119908119891) =

2120583

int

120588119894

120588119908119891

120588 120597120588

119898 (120588119908119891

) = 119898 (120588119894) minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

sdot [ln [22461198961199051206011205831198881199032

]+ 080907]

120597119898 (120588119908)

120597 ln 119905= 119898gas =

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

(82)

14 Journal of Petroleum Engineering

4160041650417004175041800418504190041950420004205042100

560856095609561056105611561156125612

ODENGDEN

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

1205880

(Ibm

ft)

120597120588w120597lnt

= moil =1627120583Qc01205880

kh= minus0009

120597m(120588w)

120597lnt= mgas =

1637120583Q

kh120572R= minus12195

120588g2

(Ibm

ft)

Figure 13 Oil and gas phase density versus log Horner time

6279

6279

6279

6279

6279

6279

6279

6279

01 1 10 100 1000 10000 1000001000000Horner time (h)

Density versus log Horner time

WDEN

120588w

(Ibm

ft)

120597120588w120597lnt

= mwater =1627120583Qc01205880

kh= minus00003

Figure 14 Water phase density versus log Horner time

Plotting 120588119908

2(119905) or 119898(120588

119908) versus ln(119905) will yield a straight line

at longer time and the slope of the line is given as follows

119896ℎ =120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)1

119898gas (83)

Example 3 Data from Example 1(a) was used for this caseand the semilog plot of density for each phase (oil water andgas) is plotted against log of Horner timelowastlowast

119876 used for 119896 calculation is the average rate for allflowing periods

From Figures 13 and 14 the calculated slopes of the radialflow for oil and water phases are given as

120597120588119908

120597 ln 119905= 119898oil =

1627120583119876119888119900120588119900

119896ℎ= minus 0009

120597120588119908

120597 ln 119905= 119898water =

1627120583119876119888119900120588119900

119896ℎ= minus 00003

(84)

The estimated phase permeabilities are

119896oil =1627120583119876119888

119900120588119900

ℎ119898oil= 445mD

119896water =1627120583119876119888

119900120588119900

ℎ119898water= 50mD

(85)

Also for the gas phase consider the following

ODENGDEN

422004225042300423504240042450425004255042600

5613561356145614561556155616

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

1205880

(Ibm

ft)

120597120588w120597lnt

= moil =1627120583Qc01205880

kh= minus0009

120597m(120588w)

120597lnt= mgas =

1637120583Q

kh120572R= minus12502

120588g2

(Ibm

ft)

Figure 15 Oil and gas phase density versus log Horner time

WDEN

6279627962796279627962796280628062806280

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

120588w

(Ibm

ft)

120597120588w120597lnt

= mwater =1627120583Qc01205880

kh= minus00003

Figure 16 Water phase density versus log Horner time

The calculated slope from the semilog plot shown inFigure 13 is given as

120597119898 (120588119908)

120597 ln 119905= 119898gas =

1637120583119876119896ℎ120572119877

= minus 12195 (86)

And estimated gas permeability is

119896gas =1637120583119876

119900

ℎ120572119877119898gas= 170mD (87)

Using the empirical model for all three phases the averageestimated permeability is as follows

119896ave =4radic50 times 445 times 1702

119896ave = 50mD

(88)

Example 4 Data from Example 1(c) was used for this caseand the semilog plot of density for each phase (oil water andgas) is plotted against log of Horner time

From Figures 15 and 16 the calculated slopes of the radialflow for oil and water phases are given as

120597120588119908

120597 ln 119905= 119898oil =

1627120583119876119888119900120588119900

119896ℎ= minus 0009

120597120588119908

120597 ln 119905= 119898water =

1627120583119876119888119900120588119900

119896ℎ= minus 00003

(89)

Journal of Petroleum Engineering 15

The estimated phase permeabilities are

119896oil =1627120583119876119888

119900120588119900

ℎ119898oil= 4352mD

119896water =1627120583119876119888

119900120588119900

ℎ119898water= 49mD

(90)

Also for the gas phase consider the following

The calculated slope from the semilog plot shown inFigure 15 is given as

120597119898 (120588119908)

120597 ln 119905= 119898gas =

1637120583119876119896ℎ120572119877

= minus 12502 (91)

And estimated gas permeability is

119896gas =1637120583119876

119900

ℎ120572119877119898gas= 179mD (92)

Using the empirical model for all three phases the averageestimated permeability is as follows

119896ave =4radic49 times 4352 times 1792

119896ave = 51mD

(93)

6 Conclusion

The following inferences were drawn from the six scenariosreviewed

(i) Results from the three-phase fluid empirical modeldeveloped for average effective 119896ave estimation rangebetween 47 and 57mD for Example 1 and give404mD for Example 2 which is within that used inthe simulation model and also estimated from BPR

(ii) The BPR PDENDWA and PDENWAT give the samestabilisation and the same 119896 estimates making it suit-able for interpretation of pressure transient analysis

(iii) It has been demonstrated in all 6 scenarios inves-tigated that the heavier fluid such as water andthe weighted average pressure-density equivalent ofall fluid give exact effective 119896 as the conventionalmethod

(iv) For oil reservoir system (scenarios (a) to (d)) 119896 =

500mD is achieved if the thickness contributing toflow ranges from 50 to 150 ft while for gas condensatereservoir (scenarios (a) to (b)) 119896 ranges between340 and 371mD which is achieved if the net sandthickness ℎ = 148ft is contributing to flow

(v) This approach also provides an estimate of the pos-sible fluid phase permeabilities and the of eachphase contribution to flow at a given point hence atseveral 1198891199011015840 stabilisation points the relative 119896 can begenerated

(vi) The derivatives for all output parameters display thesame wellbore and reservoir fingerprint as the BPRmethod

(vii) Generally where output parameters (BPR PDE-NOIL PDENGAS PDENWAT and PDENDW)depict different fingerprint the density derivative willserve as support to distinguish between reservoir andnonreservoir response

Nomenclature

119875 Pressure psi119879 Temperature oF119903 Radius ft119896 Permeability mDOslash Porosity fraction120583 Viscosity cp119905 Time hrs119902 Production rate bblday119861 Formation volume factor rbStb119862119905 Total compressibility psiminus1119903119908 Wellbore radius ftΔ119901 Change in pressure psiaℎ Formation thickness ft119860 Drainage area acres119875119908119891 Bottom-hole flowing pressure psi119875119894 Initial pressure psi

119885 Difference between two pointtime series119894 Subscript of an observed variable119888 Subscript of a calculated variableSTEYX SSE of data point119899 Number of data pointsCov Covariance of data point120575 Standard deviation119905119901 Cumulative production time119862119904 Wellbore storage constant119909 Mean of data point

Abbreviations

LBPR Local grid bottom-hole pressureLDENO Local grid oil densityLDENW Local grid water densityLDENG Local grid gas densityWBHP Well bottom-hole pressureBPR Well bottom-hole pressurePDENOIL Pressure equivalent of LDENOPDENGAS Pressure equivalent of LDENGPDENWAT Pressure equivalent of LDENWPDENDWA Pressure equivalent of density weighted

average (LDENO LDENG and LDENW)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Jalali S D Mohaghegh and R Gaskari ldquoIdentifying infilllocations and underperformer wells in mature fields usingmonthly production rate data Carthage Field Cotton Valley

16 Journal of Petroleum Engineering

Formation Texasrdquo in Proceedings of the SPE Eastern RegionalMeeting SPE Paper 104550 Canton Ohio USA October 2006

[2] D Tiab and A Kumar ldquoApplication of PD function to inter-ference analysisrdquo in Proceedings of the 51st Annual Fall Meetingof the Society of Petroleum Engineers of AIME Paper SPE 6053New Orleans La USA 1976

[3] D Tiab and A Kumar ldquoDetection and location of two parallelsealing fault around a wellrdquo in Proceedings of the 51st AnnualFall Meeting of the Society of Petroleum Engineers of AIME SPEPaper 6056 New Orleans La USA 1976

[4] D Bourdet T M Whittle A A Douglas and Y M Pirard ldquoAnew set of type curves simplifies well test analysisrdquo World Oilvol 196 no 6 pp 95ndash106 1983

[5] S Y Zheng ldquoFighting against non-unique solution problemsin heterogeneous reservoirs through numerical well testingrdquo inProceedings of the SPE Asia Pacific Oil and Gas Conference andExhibition SPE 100951 Adelaide Australia September 2006

[6] R N Horne and K O Temeng ldquoRecognition and location ofpinch-out boundaries by pressure transient analysisrdquo SPE 9905Society of Petroleum Engineers 1981

[7] L C Ayestaran R D Nurmi G A K Shehab and W S ElsildquoWelltest design and interpretation improved by integrated welltesting and geological effortsrdquo SPE 17945 1989

[8] G J Massonnat and D Bandiziol ldquoInterdependence betweengeology and well test interpretationrdquo SPE 22740 1991

[9] K F Du and G Stewart ldquoReservoir description from well testinterpretationrdquo in North Sea Oil and Gas ReservoirmdashIII pp339ndash356 Norwegian Institute of Technology (NTH) 1994

[10] S Y Zheng Well testing and characterisation of meanderingfluvial channel reservoirs [PhD thesis] Heriot-Watt University1997

[11] V T Biu and S Y Zheng ldquoA new approach in pressure transientanalysis part I improved diagnosis of flow regimes in oil and gaswellsrdquo in Proceedings of the 3rd EAGEAAPGWorkshop on TightReservoirs in theMiddle East AbuDhabi UnitedArab Emirates2015

[12] R N Horne Modern Welltest Analysis Petroway Palo AltoCalif USA 1995

[13] C S Matthews and D G Russell Pressure Buildup and FlowTests in Wells vol 1 ofMonograph Series SPE Richardson TexUSA 1967

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International Journal of

Journal of Petroleum Engineering 13

From the fundamental

119899 =119875119881

119911119877119879 (68)

At standard condition

119875119881

119911119879=

119875sc119881sc119879sc

119875 (5615119902)119911119879

=119875sc119876sc119879sc

119902 =119875sc119879sc

(119911119879

119875)

119876sc5615

119902 =119875sc119879sc

119876sc5615

(119911119879

119875) =

2120587119896ℎ120583

119903120597119901

120597119903

119902 =119875sc119879sc

119876sc5615

(119911119879) =2120587119896ℎ120583

119903119875120597119901

120597119903

(69)

where

120597119875 =119875119900

120588119900(1 minus 119875

119900119888119892)120597120588

120572 =119875119900

120588119900(1 minus 119875

119900119888119892)

120597119875 = 120572120597120588

119875 = 120588119877119911119879

(70)

Substitute into the equation where

119898(120588119894) minus119898 (120588

119908119891) =

2120583

int

120588119894

120588119908119891

120588120597120588

119898 (120588119894) minus119898 (120588

119908119891) =

120588119894

2minus 120588119908119891

2

120583

120588 = radic120588119894

2+ 120588119908119891

2

2

(71)

From the diffusivity equation for gas

1119903

120597120588

120597119903[119903

120597120588119892

120597119903] =

120601120583119892119888

119896119892

120597120588119892

120597119905 (72)

Initial condition is as follows

120588 (119903 119905 = 0) = 120588119894 (73)

BC at the wellbore is as follows

lim119903rarr 0

4120587119896ℎ120572119877120583

(5615119879sc

119875sc) 119903120588

120597120588

120597119903= 119876 (74)

BC at infirmity is as follows

lim119903rarrinfin

120588 (119903 119905) = 120588119894 (75)

where119877119875sc and119879sc are known as temperature pressure andgas constant at standard condition

Therefore integrating from 120578 = infin where 120588 = 120588119894

int

120588(120578)

120588119894

120588120597120588 = minusint

120578

infin

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)119890minus1205784

120578120597120578

1205882(120578) = 120588

119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

minusint

infin

120578

119890minus1205784

120578120597120578

(76)

where

120578 =120601120583119888119903

2

119896119905 (77)

Assume that 119906 = (1205784) (120597120578120578) = 120597119906119906 and 119906 = 12060112058311988811990324119896119905

Hence

1205882= 120588119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)minusint

infin

12060112058311988811990324119896119905

119890minus1205784

120578 (78)

where minus119864119894(minus119909) = intinfin

119909(119890minus119906

119906)120597119906Known as the exponential integral function defined by

[13]

1205882= 120588119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)minus119864119894 (minus119909) (79)

where 119909 = 12060112058311988811990324119896119905 and minus119864119894(minus119909) = int

infin

119909(119890minus119906

119906)120597119906At large times 119909will be small applying Taylorrsquos series and

integrating

minus119864119894 (minus119909) = ln (120574)

ln (120574) = ln (1781) = 05772(80)

where 120574 is known as Eulerrsquos numberFor 14119909 gt 25 consider the followingThen

1205882(119903 119905) = 120588

119894

2+

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) [ln119909+ ln 120574] (81)

1205882(119903 119905) = 120588

119894

2+

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) ln[120601120583119888119903

2120574

4119896119905]

1205882(119903 119905) = 120588

119894

2minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

) ln [22461198961199051206011205831198881199032

]

119898 (120588119894) minus119898 (120588

119908119891) =

2120583

int

120588119894

120588119908119891

120588 120597120588

119898 (120588119908119891

) = 119898 (120588119894) minus

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

sdot [ln [22461198961199051206011205831198881199032

]+ 080907]

120597119898 (120588119908)

120597 ln 119905= 119898gas =

120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)

(82)

14 Journal of Petroleum Engineering

4160041650417004175041800418504190041950420004205042100

560856095609561056105611561156125612

ODENGDEN

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

1205880

(Ibm

ft)

120597120588w120597lnt

= moil =1627120583Qc01205880

kh= minus0009

120597m(120588w)

120597lnt= mgas =

1637120583Q

kh120572R= minus12195

120588g2

(Ibm

ft)

Figure 13 Oil and gas phase density versus log Horner time

6279

6279

6279

6279

6279

6279

6279

6279

01 1 10 100 1000 10000 1000001000000Horner time (h)

Density versus log Horner time

WDEN

120588w

(Ibm

ft)

120597120588w120597lnt

= mwater =1627120583Qc01205880

kh= minus00003

Figure 14 Water phase density versus log Horner time

Plotting 120588119908

2(119905) or 119898(120588

119908) versus ln(119905) will yield a straight line

at longer time and the slope of the line is given as follows

119896ℎ =120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)1

119898gas (83)

Example 3 Data from Example 1(a) was used for this caseand the semilog plot of density for each phase (oil water andgas) is plotted against log of Horner timelowastlowast

119876 used for 119896 calculation is the average rate for allflowing periods

From Figures 13 and 14 the calculated slopes of the radialflow for oil and water phases are given as

120597120588119908

120597 ln 119905= 119898oil =

1627120583119876119888119900120588119900

119896ℎ= minus 0009

120597120588119908

120597 ln 119905= 119898water =

1627120583119876119888119900120588119900

119896ℎ= minus 00003

(84)

The estimated phase permeabilities are

119896oil =1627120583119876119888

119900120588119900

ℎ119898oil= 445mD

119896water =1627120583119876119888

119900120588119900

ℎ119898water= 50mD

(85)

Also for the gas phase consider the following

ODENGDEN

422004225042300423504240042450425004255042600

5613561356145614561556155616

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

1205880

(Ibm

ft)

120597120588w120597lnt

= moil =1627120583Qc01205880

kh= minus0009

120597m(120588w)

120597lnt= mgas =

1637120583Q

kh120572R= minus12502

120588g2

(Ibm

ft)

Figure 15 Oil and gas phase density versus log Horner time

WDEN

6279627962796279627962796280628062806280

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

120588w

(Ibm

ft)

120597120588w120597lnt

= mwater =1627120583Qc01205880

kh= minus00003

Figure 16 Water phase density versus log Horner time

The calculated slope from the semilog plot shown inFigure 13 is given as

120597119898 (120588119908)

120597 ln 119905= 119898gas =

1637120583119876119896ℎ120572119877

= minus 12195 (86)

And estimated gas permeability is

119896gas =1637120583119876

119900

ℎ120572119877119898gas= 170mD (87)

Using the empirical model for all three phases the averageestimated permeability is as follows

119896ave =4radic50 times 445 times 1702

119896ave = 50mD

(88)

Example 4 Data from Example 1(c) was used for this caseand the semilog plot of density for each phase (oil water andgas) is plotted against log of Horner time

From Figures 15 and 16 the calculated slopes of the radialflow for oil and water phases are given as

120597120588119908

120597 ln 119905= 119898oil =

1627120583119876119888119900120588119900

119896ℎ= minus 0009

120597120588119908

120597 ln 119905= 119898water =

1627120583119876119888119900120588119900

119896ℎ= minus 00003

(89)

Journal of Petroleum Engineering 15

The estimated phase permeabilities are

119896oil =1627120583119876119888

119900120588119900

ℎ119898oil= 4352mD

119896water =1627120583119876119888

119900120588119900

ℎ119898water= 49mD

(90)

Also for the gas phase consider the following

The calculated slope from the semilog plot shown inFigure 15 is given as

120597119898 (120588119908)

120597 ln 119905= 119898gas =

1637120583119876119896ℎ120572119877

= minus 12502 (91)

And estimated gas permeability is

119896gas =1637120583119876

119900

ℎ120572119877119898gas= 179mD (92)

Using the empirical model for all three phases the averageestimated permeability is as follows

119896ave =4radic49 times 4352 times 1792

119896ave = 51mD

(93)

6 Conclusion

The following inferences were drawn from the six scenariosreviewed

(i) Results from the three-phase fluid empirical modeldeveloped for average effective 119896ave estimation rangebetween 47 and 57mD for Example 1 and give404mD for Example 2 which is within that used inthe simulation model and also estimated from BPR

(ii) The BPR PDENDWA and PDENWAT give the samestabilisation and the same 119896 estimates making it suit-able for interpretation of pressure transient analysis

(iii) It has been demonstrated in all 6 scenarios inves-tigated that the heavier fluid such as water andthe weighted average pressure-density equivalent ofall fluid give exact effective 119896 as the conventionalmethod

(iv) For oil reservoir system (scenarios (a) to (d)) 119896 =

500mD is achieved if the thickness contributing toflow ranges from 50 to 150 ft while for gas condensatereservoir (scenarios (a) to (b)) 119896 ranges between340 and 371mD which is achieved if the net sandthickness ℎ = 148ft is contributing to flow

(v) This approach also provides an estimate of the pos-sible fluid phase permeabilities and the of eachphase contribution to flow at a given point hence atseveral 1198891199011015840 stabilisation points the relative 119896 can begenerated

(vi) The derivatives for all output parameters display thesame wellbore and reservoir fingerprint as the BPRmethod

(vii) Generally where output parameters (BPR PDE-NOIL PDENGAS PDENWAT and PDENDW)depict different fingerprint the density derivative willserve as support to distinguish between reservoir andnonreservoir response

Nomenclature

119875 Pressure psi119879 Temperature oF119903 Radius ft119896 Permeability mDOslash Porosity fraction120583 Viscosity cp119905 Time hrs119902 Production rate bblday119861 Formation volume factor rbStb119862119905 Total compressibility psiminus1119903119908 Wellbore radius ftΔ119901 Change in pressure psiaℎ Formation thickness ft119860 Drainage area acres119875119908119891 Bottom-hole flowing pressure psi119875119894 Initial pressure psi

119885 Difference between two pointtime series119894 Subscript of an observed variable119888 Subscript of a calculated variableSTEYX SSE of data point119899 Number of data pointsCov Covariance of data point120575 Standard deviation119905119901 Cumulative production time119862119904 Wellbore storage constant119909 Mean of data point

Abbreviations

LBPR Local grid bottom-hole pressureLDENO Local grid oil densityLDENW Local grid water densityLDENG Local grid gas densityWBHP Well bottom-hole pressureBPR Well bottom-hole pressurePDENOIL Pressure equivalent of LDENOPDENGAS Pressure equivalent of LDENGPDENWAT Pressure equivalent of LDENWPDENDWA Pressure equivalent of density weighted

average (LDENO LDENG and LDENW)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Jalali S D Mohaghegh and R Gaskari ldquoIdentifying infilllocations and underperformer wells in mature fields usingmonthly production rate data Carthage Field Cotton Valley

16 Journal of Petroleum Engineering

Formation Texasrdquo in Proceedings of the SPE Eastern RegionalMeeting SPE Paper 104550 Canton Ohio USA October 2006

[2] D Tiab and A Kumar ldquoApplication of PD function to inter-ference analysisrdquo in Proceedings of the 51st Annual Fall Meetingof the Society of Petroleum Engineers of AIME Paper SPE 6053New Orleans La USA 1976

[3] D Tiab and A Kumar ldquoDetection and location of two parallelsealing fault around a wellrdquo in Proceedings of the 51st AnnualFall Meeting of the Society of Petroleum Engineers of AIME SPEPaper 6056 New Orleans La USA 1976

[4] D Bourdet T M Whittle A A Douglas and Y M Pirard ldquoAnew set of type curves simplifies well test analysisrdquo World Oilvol 196 no 6 pp 95ndash106 1983

[5] S Y Zheng ldquoFighting against non-unique solution problemsin heterogeneous reservoirs through numerical well testingrdquo inProceedings of the SPE Asia Pacific Oil and Gas Conference andExhibition SPE 100951 Adelaide Australia September 2006

[6] R N Horne and K O Temeng ldquoRecognition and location ofpinch-out boundaries by pressure transient analysisrdquo SPE 9905Society of Petroleum Engineers 1981

[7] L C Ayestaran R D Nurmi G A K Shehab and W S ElsildquoWelltest design and interpretation improved by integrated welltesting and geological effortsrdquo SPE 17945 1989

[8] G J Massonnat and D Bandiziol ldquoInterdependence betweengeology and well test interpretationrdquo SPE 22740 1991

[9] K F Du and G Stewart ldquoReservoir description from well testinterpretationrdquo in North Sea Oil and Gas ReservoirmdashIII pp339ndash356 Norwegian Institute of Technology (NTH) 1994

[10] S Y Zheng Well testing and characterisation of meanderingfluvial channel reservoirs [PhD thesis] Heriot-Watt University1997

[11] V T Biu and S Y Zheng ldquoA new approach in pressure transientanalysis part I improved diagnosis of flow regimes in oil and gaswellsrdquo in Proceedings of the 3rd EAGEAAPGWorkshop on TightReservoirs in theMiddle East AbuDhabi UnitedArab Emirates2015

[12] R N Horne Modern Welltest Analysis Petroway Palo AltoCalif USA 1995

[13] C S Matthews and D G Russell Pressure Buildup and FlowTests in Wells vol 1 ofMonograph Series SPE Richardson TexUSA 1967

International Journal of

AerospaceEngineeringHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

RoboticsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Active and Passive Electronic Components

Control Scienceand Engineering

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

RotatingMachinery

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporation httpwwwhindawicom

Journal ofEngineeringVolume 2014

Submit your manuscripts athttpwwwhindawicom

VLSI Design

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Shock and Vibration

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Civil EngineeringAdvances in

Acoustics and VibrationAdvances in

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Electrical and Computer Engineering

Journal of

Advances inOptoElectronics

Hindawi Publishing Corporation httpwwwhindawicom

Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

SensorsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Modelling amp Simulation in EngineeringHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Chemical EngineeringInternational Journal of Antennas and

Propagation

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Navigation and Observation

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

DistributedSensor Networks

International Journal of

14 Journal of Petroleum Engineering

4160041650417004175041800418504190041950420004205042100

560856095609561056105611561156125612

ODENGDEN

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

1205880

(Ibm

ft)

120597120588w120597lnt

= moil =1627120583Qc01205880

kh= minus0009

120597m(120588w)

120597lnt= mgas =

1637120583Q

kh120572R= minus12195

120588g2

(Ibm

ft)

Figure 13 Oil and gas phase density versus log Horner time

6279

6279

6279

6279

6279

6279

6279

6279

01 1 10 100 1000 10000 1000001000000Horner time (h)

Density versus log Horner time

WDEN

120588w

(Ibm

ft)

120597120588w120597lnt

= mwater =1627120583Qc01205880

kh= minus00003

Figure 14 Water phase density versus log Horner time

Plotting 120588119908

2(119905) or 119898(120588

119908) versus ln(119905) will yield a straight line

at longer time and the slope of the line is given as follows

119896ℎ =120583119876

4120587119896ℎ120572119877(

119875sc5615119879sc

)1

119898gas (83)

Example 3 Data from Example 1(a) was used for this caseand the semilog plot of density for each phase (oil water andgas) is plotted against log of Horner timelowastlowast

119876 used for 119896 calculation is the average rate for allflowing periods

From Figures 13 and 14 the calculated slopes of the radialflow for oil and water phases are given as

120597120588119908

120597 ln 119905= 119898oil =

1627120583119876119888119900120588119900

119896ℎ= minus 0009

120597120588119908

120597 ln 119905= 119898water =

1627120583119876119888119900120588119900

119896ℎ= minus 00003

(84)

The estimated phase permeabilities are

119896oil =1627120583119876119888

119900120588119900

ℎ119898oil= 445mD

119896water =1627120583119876119888

119900120588119900

ℎ119898water= 50mD

(85)

Also for the gas phase consider the following

ODENGDEN

422004225042300423504240042450425004255042600

5613561356145614561556155616

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

1205880

(Ibm

ft)

120597120588w120597lnt

= moil =1627120583Qc01205880

kh= minus0009

120597m(120588w)

120597lnt= mgas =

1637120583Q

kh120572R= minus12502

120588g2

(Ibm

ft)

Figure 15 Oil and gas phase density versus log Horner time

WDEN

6279627962796279627962796280628062806280

01 1 10 100 1000 10000 100000 1000000Horner time (h)

Density versus log Horner time

120588w

(Ibm

ft)

120597120588w120597lnt

= mwater =1627120583Qc01205880

kh= minus00003

Figure 16 Water phase density versus log Horner time

The calculated slope from the semilog plot shown inFigure 13 is given as

120597119898 (120588119908)

120597 ln 119905= 119898gas =

1637120583119876119896ℎ120572119877

= minus 12195 (86)

And estimated gas permeability is

119896gas =1637120583119876

119900

ℎ120572119877119898gas= 170mD (87)

Using the empirical model for all three phases the averageestimated permeability is as follows

119896ave =4radic50 times 445 times 1702

119896ave = 50mD

(88)

Example 4 Data from Example 1(c) was used for this caseand the semilog plot of density for each phase (oil water andgas) is plotted against log of Horner time

From Figures 15 and 16 the calculated slopes of the radialflow for oil and water phases are given as

120597120588119908

120597 ln 119905= 119898oil =

1627120583119876119888119900120588119900

119896ℎ= minus 0009

120597120588119908

120597 ln 119905= 119898water =

1627120583119876119888119900120588119900

119896ℎ= minus 00003

(89)

Journal of Petroleum Engineering 15

The estimated phase permeabilities are

119896oil =1627120583119876119888

119900120588119900

ℎ119898oil= 4352mD

119896water =1627120583119876119888

119900120588119900

ℎ119898water= 49mD

(90)

Also for the gas phase consider the following

The calculated slope from the semilog plot shown inFigure 15 is given as

120597119898 (120588119908)

120597 ln 119905= 119898gas =

1637120583119876119896ℎ120572119877

= minus 12502 (91)

And estimated gas permeability is

119896gas =1637120583119876

119900

ℎ120572119877119898gas= 179mD (92)

Using the empirical model for all three phases the averageestimated permeability is as follows

119896ave =4radic49 times 4352 times 1792

119896ave = 51mD

(93)

6 Conclusion

The following inferences were drawn from the six scenariosreviewed

(i) Results from the three-phase fluid empirical modeldeveloped for average effective 119896ave estimation rangebetween 47 and 57mD for Example 1 and give404mD for Example 2 which is within that used inthe simulation model and also estimated from BPR

(ii) The BPR PDENDWA and PDENWAT give the samestabilisation and the same 119896 estimates making it suit-able for interpretation of pressure transient analysis

(iii) It has been demonstrated in all 6 scenarios inves-tigated that the heavier fluid such as water andthe weighted average pressure-density equivalent ofall fluid give exact effective 119896 as the conventionalmethod

(iv) For oil reservoir system (scenarios (a) to (d)) 119896 =

500mD is achieved if the thickness contributing toflow ranges from 50 to 150 ft while for gas condensatereservoir (scenarios (a) to (b)) 119896 ranges between340 and 371mD which is achieved if the net sandthickness ℎ = 148ft is contributing to flow

(v) This approach also provides an estimate of the pos-sible fluid phase permeabilities and the of eachphase contribution to flow at a given point hence atseveral 1198891199011015840 stabilisation points the relative 119896 can begenerated

(vi) The derivatives for all output parameters display thesame wellbore and reservoir fingerprint as the BPRmethod

(vii) Generally where output parameters (BPR PDE-NOIL PDENGAS PDENWAT and PDENDW)depict different fingerprint the density derivative willserve as support to distinguish between reservoir andnonreservoir response

Nomenclature

119875 Pressure psi119879 Temperature oF119903 Radius ft119896 Permeability mDOslash Porosity fraction120583 Viscosity cp119905 Time hrs119902 Production rate bblday119861 Formation volume factor rbStb119862119905 Total compressibility psiminus1119903119908 Wellbore radius ftΔ119901 Change in pressure psiaℎ Formation thickness ft119860 Drainage area acres119875119908119891 Bottom-hole flowing pressure psi119875119894 Initial pressure psi

119885 Difference between two pointtime series119894 Subscript of an observed variable119888 Subscript of a calculated variableSTEYX SSE of data point119899 Number of data pointsCov Covariance of data point120575 Standard deviation119905119901 Cumulative production time119862119904 Wellbore storage constant119909 Mean of data point

Abbreviations

LBPR Local grid bottom-hole pressureLDENO Local grid oil densityLDENW Local grid water densityLDENG Local grid gas densityWBHP Well bottom-hole pressureBPR Well bottom-hole pressurePDENOIL Pressure equivalent of LDENOPDENGAS Pressure equivalent of LDENGPDENWAT Pressure equivalent of LDENWPDENDWA Pressure equivalent of density weighted

average (LDENO LDENG and LDENW)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Jalali S D Mohaghegh and R Gaskari ldquoIdentifying infilllocations and underperformer wells in mature fields usingmonthly production rate data Carthage Field Cotton Valley

16 Journal of Petroleum Engineering

Formation Texasrdquo in Proceedings of the SPE Eastern RegionalMeeting SPE Paper 104550 Canton Ohio USA October 2006

[2] D Tiab and A Kumar ldquoApplication of PD function to inter-ference analysisrdquo in Proceedings of the 51st Annual Fall Meetingof the Society of Petroleum Engineers of AIME Paper SPE 6053New Orleans La USA 1976

[3] D Tiab and A Kumar ldquoDetection and location of two parallelsealing fault around a wellrdquo in Proceedings of the 51st AnnualFall Meeting of the Society of Petroleum Engineers of AIME SPEPaper 6056 New Orleans La USA 1976

[4] D Bourdet T M Whittle A A Douglas and Y M Pirard ldquoAnew set of type curves simplifies well test analysisrdquo World Oilvol 196 no 6 pp 95ndash106 1983

[5] S Y Zheng ldquoFighting against non-unique solution problemsin heterogeneous reservoirs through numerical well testingrdquo inProceedings of the SPE Asia Pacific Oil and Gas Conference andExhibition SPE 100951 Adelaide Australia September 2006

[6] R N Horne and K O Temeng ldquoRecognition and location ofpinch-out boundaries by pressure transient analysisrdquo SPE 9905Society of Petroleum Engineers 1981

[7] L C Ayestaran R D Nurmi G A K Shehab and W S ElsildquoWelltest design and interpretation improved by integrated welltesting and geological effortsrdquo SPE 17945 1989

[8] G J Massonnat and D Bandiziol ldquoInterdependence betweengeology and well test interpretationrdquo SPE 22740 1991

[9] K F Du and G Stewart ldquoReservoir description from well testinterpretationrdquo in North Sea Oil and Gas ReservoirmdashIII pp339ndash356 Norwegian Institute of Technology (NTH) 1994

[10] S Y Zheng Well testing and characterisation of meanderingfluvial channel reservoirs [PhD thesis] Heriot-Watt University1997

[11] V T Biu and S Y Zheng ldquoA new approach in pressure transientanalysis part I improved diagnosis of flow regimes in oil and gaswellsrdquo in Proceedings of the 3rd EAGEAAPGWorkshop on TightReservoirs in theMiddle East AbuDhabi UnitedArab Emirates2015

[12] R N Horne Modern Welltest Analysis Petroway Palo AltoCalif USA 1995

[13] C S Matthews and D G Russell Pressure Buildup and FlowTests in Wells vol 1 ofMonograph Series SPE Richardson TexUSA 1967

International Journal of

AerospaceEngineeringHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

RoboticsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Active and Passive Electronic Components

Control Scienceand Engineering

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

RotatingMachinery

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporation httpwwwhindawicom

Journal ofEngineeringVolume 2014

Submit your manuscripts athttpwwwhindawicom

VLSI Design

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Shock and Vibration

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Civil EngineeringAdvances in

Acoustics and VibrationAdvances in

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Electrical and Computer Engineering

Journal of

Advances inOptoElectronics

Hindawi Publishing Corporation httpwwwhindawicom

Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

SensorsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Modelling amp Simulation in EngineeringHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Chemical EngineeringInternational Journal of Antennas and

Propagation

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Navigation and Observation

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

DistributedSensor Networks

International Journal of

Journal of Petroleum Engineering 15

The estimated phase permeabilities are

119896oil =1627120583119876119888

119900120588119900

ℎ119898oil= 4352mD

119896water =1627120583119876119888

119900120588119900

ℎ119898water= 49mD

(90)

Also for the gas phase consider the following

The calculated slope from the semilog plot shown inFigure 15 is given as

120597119898 (120588119908)

120597 ln 119905= 119898gas =

1637120583119876119896ℎ120572119877

= minus 12502 (91)

And estimated gas permeability is

119896gas =1637120583119876

119900

ℎ120572119877119898gas= 179mD (92)

Using the empirical model for all three phases the averageestimated permeability is as follows

119896ave =4radic49 times 4352 times 1792

119896ave = 51mD

(93)

6 Conclusion

The following inferences were drawn from the six scenariosreviewed

(i) Results from the three-phase fluid empirical modeldeveloped for average effective 119896ave estimation rangebetween 47 and 57mD for Example 1 and give404mD for Example 2 which is within that used inthe simulation model and also estimated from BPR

(ii) The BPR PDENDWA and PDENWAT give the samestabilisation and the same 119896 estimates making it suit-able for interpretation of pressure transient analysis

(iii) It has been demonstrated in all 6 scenarios inves-tigated that the heavier fluid such as water andthe weighted average pressure-density equivalent ofall fluid give exact effective 119896 as the conventionalmethod

(iv) For oil reservoir system (scenarios (a) to (d)) 119896 =

500mD is achieved if the thickness contributing toflow ranges from 50 to 150 ft while for gas condensatereservoir (scenarios (a) to (b)) 119896 ranges between340 and 371mD which is achieved if the net sandthickness ℎ = 148ft is contributing to flow

(v) This approach also provides an estimate of the pos-sible fluid phase permeabilities and the of eachphase contribution to flow at a given point hence atseveral 1198891199011015840 stabilisation points the relative 119896 can begenerated

(vi) The derivatives for all output parameters display thesame wellbore and reservoir fingerprint as the BPRmethod

(vii) Generally where output parameters (BPR PDE-NOIL PDENGAS PDENWAT and PDENDW)depict different fingerprint the density derivative willserve as support to distinguish between reservoir andnonreservoir response

Nomenclature

119875 Pressure psi119879 Temperature oF119903 Radius ft119896 Permeability mDOslash Porosity fraction120583 Viscosity cp119905 Time hrs119902 Production rate bblday119861 Formation volume factor rbStb119862119905 Total compressibility psiminus1119903119908 Wellbore radius ftΔ119901 Change in pressure psiaℎ Formation thickness ft119860 Drainage area acres119875119908119891 Bottom-hole flowing pressure psi119875119894 Initial pressure psi

119885 Difference between two pointtime series119894 Subscript of an observed variable119888 Subscript of a calculated variableSTEYX SSE of data point119899 Number of data pointsCov Covariance of data point120575 Standard deviation119905119901 Cumulative production time119862119904 Wellbore storage constant119909 Mean of data point

Abbreviations

LBPR Local grid bottom-hole pressureLDENO Local grid oil densityLDENW Local grid water densityLDENG Local grid gas densityWBHP Well bottom-hole pressureBPR Well bottom-hole pressurePDENOIL Pressure equivalent of LDENOPDENGAS Pressure equivalent of LDENGPDENWAT Pressure equivalent of LDENWPDENDWA Pressure equivalent of density weighted

average (LDENO LDENG and LDENW)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Jalali S D Mohaghegh and R Gaskari ldquoIdentifying infilllocations and underperformer wells in mature fields usingmonthly production rate data Carthage Field Cotton Valley

16 Journal of Petroleum Engineering

Formation Texasrdquo in Proceedings of the SPE Eastern RegionalMeeting SPE Paper 104550 Canton Ohio USA October 2006

[2] D Tiab and A Kumar ldquoApplication of PD function to inter-ference analysisrdquo in Proceedings of the 51st Annual Fall Meetingof the Society of Petroleum Engineers of AIME Paper SPE 6053New Orleans La USA 1976

[3] D Tiab and A Kumar ldquoDetection and location of two parallelsealing fault around a wellrdquo in Proceedings of the 51st AnnualFall Meeting of the Society of Petroleum Engineers of AIME SPEPaper 6056 New Orleans La USA 1976

[4] D Bourdet T M Whittle A A Douglas and Y M Pirard ldquoAnew set of type curves simplifies well test analysisrdquo World Oilvol 196 no 6 pp 95ndash106 1983

[5] S Y Zheng ldquoFighting against non-unique solution problemsin heterogeneous reservoirs through numerical well testingrdquo inProceedings of the SPE Asia Pacific Oil and Gas Conference andExhibition SPE 100951 Adelaide Australia September 2006

[6] R N Horne and K O Temeng ldquoRecognition and location ofpinch-out boundaries by pressure transient analysisrdquo SPE 9905Society of Petroleum Engineers 1981

[7] L C Ayestaran R D Nurmi G A K Shehab and W S ElsildquoWelltest design and interpretation improved by integrated welltesting and geological effortsrdquo SPE 17945 1989

[8] G J Massonnat and D Bandiziol ldquoInterdependence betweengeology and well test interpretationrdquo SPE 22740 1991

[9] K F Du and G Stewart ldquoReservoir description from well testinterpretationrdquo in North Sea Oil and Gas ReservoirmdashIII pp339ndash356 Norwegian Institute of Technology (NTH) 1994

[10] S Y Zheng Well testing and characterisation of meanderingfluvial channel reservoirs [PhD thesis] Heriot-Watt University1997

[11] V T Biu and S Y Zheng ldquoA new approach in pressure transientanalysis part I improved diagnosis of flow regimes in oil and gaswellsrdquo in Proceedings of the 3rd EAGEAAPGWorkshop on TightReservoirs in theMiddle East AbuDhabi UnitedArab Emirates2015

[12] R N Horne Modern Welltest Analysis Petroway Palo AltoCalif USA 1995

[13] C S Matthews and D G Russell Pressure Buildup and FlowTests in Wells vol 1 ofMonograph Series SPE Richardson TexUSA 1967

International Journal of

AerospaceEngineeringHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

RoboticsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Active and Passive Electronic Components

Control Scienceand Engineering

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

RotatingMachinery

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporation httpwwwhindawicom

Journal ofEngineeringVolume 2014

Submit your manuscripts athttpwwwhindawicom

VLSI Design

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Shock and Vibration

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Civil EngineeringAdvances in

Acoustics and VibrationAdvances in

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Electrical and Computer Engineering

Journal of

Advances inOptoElectronics

Hindawi Publishing Corporation httpwwwhindawicom

Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

SensorsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Modelling amp Simulation in EngineeringHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Chemical EngineeringInternational Journal of Antennas and

Propagation

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Navigation and Observation

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

DistributedSensor Networks

International Journal of

16 Journal of Petroleum Engineering

Formation Texasrdquo in Proceedings of the SPE Eastern RegionalMeeting SPE Paper 104550 Canton Ohio USA October 2006

[2] D Tiab and A Kumar ldquoApplication of PD function to inter-ference analysisrdquo in Proceedings of the 51st Annual Fall Meetingof the Society of Petroleum Engineers of AIME Paper SPE 6053New Orleans La USA 1976

[3] D Tiab and A Kumar ldquoDetection and location of two parallelsealing fault around a wellrdquo in Proceedings of the 51st AnnualFall Meeting of the Society of Petroleum Engineers of AIME SPEPaper 6056 New Orleans La USA 1976

[4] D Bourdet T M Whittle A A Douglas and Y M Pirard ldquoAnew set of type curves simplifies well test analysisrdquo World Oilvol 196 no 6 pp 95ndash106 1983

[5] S Y Zheng ldquoFighting against non-unique solution problemsin heterogeneous reservoirs through numerical well testingrdquo inProceedings of the SPE Asia Pacific Oil and Gas Conference andExhibition SPE 100951 Adelaide Australia September 2006

[6] R N Horne and K O Temeng ldquoRecognition and location ofpinch-out boundaries by pressure transient analysisrdquo SPE 9905Society of Petroleum Engineers 1981

[7] L C Ayestaran R D Nurmi G A K Shehab and W S ElsildquoWelltest design and interpretation improved by integrated welltesting and geological effortsrdquo SPE 17945 1989

[8] G J Massonnat and D Bandiziol ldquoInterdependence betweengeology and well test interpretationrdquo SPE 22740 1991

[9] K F Du and G Stewart ldquoReservoir description from well testinterpretationrdquo in North Sea Oil and Gas ReservoirmdashIII pp339ndash356 Norwegian Institute of Technology (NTH) 1994

[10] S Y Zheng Well testing and characterisation of meanderingfluvial channel reservoirs [PhD thesis] Heriot-Watt University1997

[11] V T Biu and S Y Zheng ldquoA new approach in pressure transientanalysis part I improved diagnosis of flow regimes in oil and gaswellsrdquo in Proceedings of the 3rd EAGEAAPGWorkshop on TightReservoirs in theMiddle East AbuDhabi UnitedArab Emirates2015

[12] R N Horne Modern Welltest Analysis Petroway Palo AltoCalif USA 1995

[13] C S Matthews and D G Russell Pressure Buildup and FlowTests in Wells vol 1 ofMonograph Series SPE Richardson TexUSA 1967

International Journal of

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RotatingMachinery

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VLSI Design

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Shock and Vibration

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Acoustics and VibrationAdvances in

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Electrical and Computer Engineering

Journal of

Advances inOptoElectronics

Hindawi Publishing Corporation httpwwwhindawicom

Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

SensorsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Modelling amp Simulation in EngineeringHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Chemical EngineeringInternational Journal of Antennas and

Propagation

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Navigation and Observation

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

DistributedSensor Networks

International Journal of

International Journal of

AerospaceEngineeringHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

RoboticsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Active and Passive Electronic Components

Control Scienceand Engineering

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

RotatingMachinery

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporation httpwwwhindawicom

Journal ofEngineeringVolume 2014

Submit your manuscripts athttpwwwhindawicom

VLSI Design

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Shock and Vibration

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Civil EngineeringAdvances in

Acoustics and VibrationAdvances in

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Electrical and Computer Engineering

Journal of

Advances inOptoElectronics

Hindawi Publishing Corporation httpwwwhindawicom

Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

SensorsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Modelling amp Simulation in EngineeringHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Chemical EngineeringInternational Journal of Antennas and

Propagation

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Navigation and Observation

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

DistributedSensor Networks

International Journal of