reserve estimation methods

5/21/2018 Reserve Estimation Methods

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2003-2004 Petrobjects 1 www.petrobjects.com

Petroleum ReservesEstimation Methods

2003-2004 Petrobjectswww.petrobjects.com

Introduction

The process of estimating oil and gas reserves for a producing field continuesthroughout the life of the field. There is always uncertainty in making suchestimates. The level of uncertainty is affected by the following factors:

1. Reservoir type,2. Source of reservoir energy,3. Quantity and quality of the geological, engineering, and geophysical data,

4. Assumptions adopted when making the estimate,5. Available technology, and6. Experience and knowledge of the evaluator.

The magnitude of uncertainty, however, decreases with time until the economiclimit is reached and the ultimate recovery is realized, see Figure 1.

Figure 1:Magnitude of uncertainty in reserves estimates

The oil and gas reserves estimation methods can be grouped into the followingcategories:

1. Analogy,2. Volumetric,3. Decline analysis,4. Material balance calculations for oil reservoirs,5. Material balance calculations for gas reservoirs,6. Reservoir simulation.


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In the early stages of development, reserves estimates are restricted to theanalogy and volumetric calculations. The analogy method is applied by comparingfactors for the analogous and current fields or wells. A close-to-abandonmentanalogous field is taken as an approximate to the current field. This method is mostuseful when running the economics on the current field; which is supposed to be anexploratoryfield.

The volumetric method, on the other hand, entails determining the areal extentof the reservoir, the rock pore volume, and the fluid content within the porevolume. This provides an estimate of the amount of hydrocarbons-in-place. Theultimate recovery, then, can be estimated by using an appropriate recovery factor.Each of the factors used in the calculation above have inherent uncertainties that,when combined, cause significant uncertainties in the reserves estimate.

As production and pressure data from a field become available, decline analysisand material balance calculations, become the predominant methods of calculatingreserves. These methods greatly reduce the uncertainty in reserves estimates;however, during early depletion, caution should be exercised in using them. Declinecurve relationships are empirical, and rely on uniform, lengthy production periods.It is more suited to oil wells, which are usually produced against fixed bottom-holepressures. In gas wells, however, wellhead back-pressures usually fluctuate,

causing varying production trends and therefore, not as reliable.The most common decline curve relationship is the constant percentage decline

(exponential). With more and more low productivity wells coming on stream, thereis currently a swing toward decline rates proportional to production rates(hyperbolic and harmonic). Although some wells exhibit these trends, hyperbolic orharmonic decline extrapolations should only be used for these specific cases. Over-exuberance in the use of hyperbolic or harmonic relationships can result inexcessive reserves estimates.

Material balance calculation is an excellent tool for estimating gas reserves. If areservoir comprises a closed system and contains single-phase gas, the pressure in

the reservoir will decline proportionately to the amount of gas produced.Unfortunately, sometimes bottom water drive in gas reservoirs contributes to thedepletion mechanism, altering the performance of the non-ideal gas law in thereservoir. Under these conditions, optimistic reserves estimates can result.

When calculating reserves using any of the above methods, two calculationprocedures may be used: deterministic and/or probabilistic. The deterministicmethod is by far the most common. The procedure is to select a single value foreach parameter to input into an appropriate equation, to obtain a single answer.The probabilistic method, on the other hand, is more rigorous and less commonlyused. This method utilizes a distribution curve for each parameter and, through the

use of Monte Carlo Simulation; a distribution curve for the answer can bedeveloped. Assuming good data, a lot of qualifying information can be derived from


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the resulting statistical calculations, such as the minimum and maximum values,the mean (average value), the median (middle value), the mode (most likelyvalue), the standard deviation and the percentiles, see Figures 2 and 3.

Figure 1:Measures of central tendency

Figure 3:Percentiles

The probabilistic methods have several inherent problems. They are affected byall input parameters, including the most likely and maximum values for theparameters. In such methods, one can not back calculate the input parametersassociated with reserves. Only the end result is known but not the exact value ofany input parameter. On the other hand, deterministic methods calculate reserve

values that are more tangible and explainable. In these methods, all inputparameters are exactly known; however, they may sometimes ignore the variability


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and uncertainty in the input data compared to the probabilistic methods which allowthe incorporation of more variance in the data.

A comparison of the deterministic and probabilistic methods, however, canprovide quality assurance for estimating hydrocarbon reserves; i.e. reserves arecalculated both deterministically and probabilistically and the two values arecompared. If the two values agree, then confidence on the calculated reserves isincreased. If the two values are away different, the assumptions need to bereexamined.


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Analogy

The analogy method is applied by comparing the following factors for theanalogousand currentfields or wells:

Recovery Factor (RF),Barrels per Acre-Foot (BAF), andEstimated Ultimate Recovery (EUR)

The RF of a close-to-abandonmentanalogous field is taken as an approximate valuefor another field. Similarly, the BAF, which is calculated by the following equation,

( )( ) ( )

( )RF

tB

tS=BAF

o

o117758

is assumed to be the same for the analogous and current field or well. ComparingEURs is done during the exploratory phase. It is also useful when calculatingproved developed reserves.

Analogy is most useful when running the economics on a yet-to-be-drilledexploratory well. Care, however, should be taken when applying analogy technique.For example, care should be taken to make sure that the field or well being usedfor analogy is indeed analogous. That said, a dolomite reservoir with volatile crudeoil will never be analogous to a sandstone reservoir with black oil. Similarly, if yourcalculated EUR is twice as high as the EUR from the nearest 100 wells, you hadbetter check your assumptions.


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Volumetric

The volumetric method entails determining the physical size of the reservoir, thepore volume within the rock matrix, and the fluid content within the void space.This provides an estimate of the hydrocarbons-in-place, from which ultimaterecovery can be estimated by using an appropriate recovery factor. Each of thefactors used in the calculation have inherent uncertainties that, when combined,cause significant uncertainties in the reserves estimate. Figure 4 is a typical

geological net pay isopach map that is often used in the volumetric method.

Figure 4:A typical geological net pay isopach map

The estimated ultimate recovery (EUR) of an oil reservoir, STB, is given by:

( )RFtNEUR =

Where N(t) is the oil in place at time t, STB, and RF is the recovery factor, fraction.The volumetric method for calculating the amount of oil in place (N) is given by thefollowing equation:

( )( )tB

tSV=N(t)

o

ob

Where:

N(t) = oil in place at time t, STBVb = bulk reservoir volume, RB = 7758 A h

7758 = RB/acre-ftA = reservoir area, acres


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H = average reservoir thickness, ft = average reservoir porosity, fractionSo(t) = average oil saturation, fractionBo(p) = oil formation volume factor at reservoir pressure p, RB/STB

Similarly, for a gas reservoir, the volumetric method is given by:

( )RFtGEUR =

Where G(t) is the gas in place at time t, SCF, and RF is the recovery factor, fraction.The volumetric method for calculating the amount of gas in place (G) is given bythe following equation:

( )( )tB

tSV=G(t)

g

gb

Where:

G(t) = gas in place at time t, SCFVb = bulk reservoir volume, CF = 43560 A h

43560 = CF/acre-ftA = reservoir area, acresh = average reservoir thickness, ft = average reservoir porosity, fractionSg(t) = average gas saturation, fractionBg(p) = gas formation volume factor at reservoir pressure p, CF/SCF

Note that the reservoir area (A) and the recovery factor (RF) are often subject tolarge errors. They are usually determined from analogy or correlations. Thefollowing examples should clarify the errors that creep in during the calculations of

oil and gas reserves.

Example #1: Given the following data for the Hout oil field in Saudi Arabia

Area = 26,700 acresNet productive thickness = 49 ftPorosity = 8%Average Swi = 45%Initial reservoir pressure, pi = 2980 psiaAbandonment pressure, pa = 300 psiaB

oat p

i = 1.68 bbl/STB

Boat pa = 1.15 bbl/STB


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Sgat pa = 34%Sorafter water invasion = 20%

The following quantities will be calculated:

1. Initial oil in place2. Oil in place after volumetric depletion to abandonment pressure3. Oil in place after water invasion at initial pressure4. Oil reserve by volumetric depletion to abandonment pressure5. Oil reserve by full water drive6. Discussion of results

Solution:

Lets start by calculating the reservoir bulk volume:

Vb= 7758 x A x h = 7758 x 26,700 x 49 = 10.15 MMM bbl

1. The initial oil in place is given by:

( )

B

SV=N

oi

wibi

1

this yields:

( )STBMM266

1.68

50.(0.08)10x10.15=N

9

i 41

2. The oil in place after volumetric depletion to abandonment pressure is given by:

( )

B

S-S-1V=N

o

gwb

this yields:

( )STBMM148

1.15

0.34)-0.45-1(0.08)10x10.15=N

9

1

3. The oil in place after water invasion at initial reservoir pressure is given by:


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B

SV=N

o

orb

this yields:

STBMM971.68

0.20(0.08)10x10.15=N

9

2

4. The oil reserve by volumetric depletion:

( ) ( ) STBMM118=10x148-266=N-N 61i

i.e. RF = 118/266 = 44%

5. The oil reserve by full water drive

( ) ( ) STBMM169=10x97-266=N-N 62i

i.e. RF = 169/266 = 64%

6. Discussion of results: For oil reservoirs under volumetric control; i.e. no waterinflux, the produced oil must be replaced by gas the saturation of whichincreases as oil saturation decreases. If Sgis the gas saturation and Bothe oilformation volume factor at abandonment pressure, then oil in place atabandonment pressure is given by:

( )

B

S-S-1V=N

o

gwb

On the other hand, for oil reservoirs under hydraulic control, where there is noappreciable decline in reservoir pressure, water influx is either edge-water drive orbottom-water drive. In edge-water drive, water influx is inward and parallel tobedding planes. In bottom-water drive, water influx is upward where the producingoil zone is underlain by water. In this case, the oil remaining at abandonment isgiven by:

B

SV=N

o

orb

This concludes the solution.


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Example #2: Given the following data for the Bell gas field

Area = 160 acresNet productive thickness = 40 ftInitial reservoir pressure = 3250 psiaPorosity = 22%Connate water = 23%Initial gas FVF = 0.00533 ft3/SCFGas FVF at 2500 psia = 0.00667 ft3/SCFGas FVF at 500 psia = 0.03623 ft3/SCFSgrafter water invasion = 34%

The following quantities will be calculated:

1. Initial gas in place2. Gas in place after volumetric depletion to 2500 psia3. Gas in place after volumetric depletion to 500 psia

4. Gas in place after water invasion at 3250 psia5. Gas in place after water invasion at 2500 psia6. Gas in place after water invasion at 500 psia7. Gas reserve by volumetric depletion to 500 psia8. Gas reserve by full water drive; i.e. at 3250 psia9. Gas reserve by partial water drive; i.e. at 2500 psia10. Gas reserve by full water drive if there is one un-dip well11. Discussion of results

Solution:

Lets start by calculating the reservoir bulk volume:

Vb= 43,560 x A x h = 43,560 x 160 x 40 = 278.784 MM ft3

1. Initial gas in place is given by:

( )

B

S-1V=G

gi

wiibi

this yields:


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( )SCFMM8860=

0.00533

0.23)-1(0.22)10x278.784=G

6

i

2. Gas in place after volumetric depletion to 2500 psia:

( )SCFMM7080=

0.00667

0.23)-1(0.22)10x278.784=G

6

1

3. Gas in place after volumetric depletion to 500 psia:

( )SCFMM1303=

0.003623

0.23)-1(0.22)10x278.784=G

6

2

4. Gas in place after water invasion at 3250 psia:

SCFMM3912=0.00533

(0.34)(0.22)10x278.784=G

6

3


SCFMM3126=0.00667

(0.34)(0.22)10x278.784=G

6

4


SCFMM576=0.03623

(0.34)(0.22)10x278.784=G

6

5

7. Gas reserve by volumetric depletion to 500 psia:

( ) SCFMM7557=10x1303-8860=G-G 62i

i.e. RF = 7557/8860 = 85%

8. Gas reserve by water drive at 3250 psia (full water drive):

( ) SCFMM4948=10x3912-8860=G-G 63i


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i.e. RF = 4948/8860 = 56%

9. Gas reserve by water drive at 2500 psia (partial water drive):

( ) SCFMM5734=10x3126-8860=G-G 64i

i.e. RF = 5734/8860 = 65%

10.Gas reserve by water drive at 3250 psia if there is one un-dip well:

( ) ( ) SCFMM2474=10x3912-88602

1=G-G

2

1 63i

i.e. RF = 2474/8860 = 28%

11. Discussion of results

The RF for volumetric depletion to 500 psia (no water drive) is calculated to be85%. On the other hand, the RF for partial water drive is 65%, and for the full waterdrive is 56%. This can be explained as follows: As water invades the reservoir, thereservoir pressure is maintained at a higher level than if there were no waterencroachment. This leads to higher abandonment pressures for water-drivereservoirs. Recoveries, however, are lower because the main mechanism ofproduction in gas reservoirs is depletion or gas expansion. In water-drive gasreservoirs, it has been found that gas recoveries can be increased by:

1. Outrunning technique: This is accomplished by increasing gas production rates.This technique has been attempted in Bierwang Field in West Germany wherethe field production rate has been increased from 50 to 75 MM SCF/D, and theyfound that the ultimate recovery increased from 69 to 74%.

2. Co-production technique: This technique is defined as the simultaneousproduction of gas and water, see Fig. 1. In this process, as down-dip wellsbegin to be watered out, they are converted to high-rate water producers, whilethe up-dip wells are maintained on gas production. This technique enhancesproduction as follows:


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The high-rate down-dip water producers act as a pressure sink for thewater. This retards water invasion into the gas zone, thereforeprolonging its productive life.The high-rate water production lowers the average reservoir pressure,allowing for more gas expansion and therefore more gas production.When the average reservoir pressure is lowered, the immobile gas in thewater-swept portion of the reservoir could become mobile and henceproducible. It has been reported that this technique has increased gasproduction from62% to 83% in Eugene Island Field of Louisiana.



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Decline Curves

A decline curve of a wellis simply a plot of the wells production rate on the y-axis versus time on the x-axis. The plot is usually done on a semilog paper; i.e. they-axis is logarithmic and the x-axis is linear. When the data plots as a straight line,it is modeled with a constant percentage decline exponential decline. When thedata plots concave upward, it is modeled with a hyperbolic decline. A special caseof the hyperbolic decline is known as harmonic decline.

The most common decline curve relationship is the constant percentage decline(exponential). With more and more low productivity wells coming on stream, thereis currently a swing toward decline rates proportional to production rates(hyperbolic and harmonic). Although some wells exhibit these trends, hyperbolic orharmonic decline extrapolations should only be used for these specific cases. Over-exuberance in the use of hyperbolic or harmonic relationships can result inexcessive reserves estimates. Figure 5 is an example of a production graph withexponential and harmonic extrapolations.

Figure 5:Decline curve of an oil well

Decline curves are the most common means of forecasting production. Theyhave many advantages:

Data is easy to obtain,They are easy to plot,They yield results on a time basis, andThey are easy to analyze.


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If the conditions affecting the rate of production of the well are not changed byoutside influences, the curve will be fairly regular, and, if projected, will furnishuseful knowledge as to the future production of the well.

Exponential Decline

As mentioned above, in the exponential decline, the wells production dataplots as a straight line on a semilog paper. The equation of the straight line

on the semilog paper is given by:

Dt

ieqq

=

Where:

q = wells production rate at time t, STB/dayqi = wells production rate at time 0, STB/dayD = nominal exponential decline rate, 1/dayt = time, day

The following table summarizes the equations used in exponential decline.

Exponential Decline b = 0

Description Equation

RateDt

ieqq

=

Cumulative Oil ProductionD

qqN ip

=

Nominal Decline Rate

( )e

DD = 1ln

i

ie

q

qqD

=

Effective Decline RateD

e eD

=1

Life( )D

qqt i

/ln=

Example #3:A well has declined from 100 BOPD to 96 BOPD during a one

month period. Assuming exponential decline, predict the rate after 11 more


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months and after 22.5 months. Also predict the amount of oil produced afterone year.

Solution:

montht

BOPDq

BOPDqi

1

96

100

=

=

=

1. Calculate the effective decline rate per month:

monthq

qqD

i

ie /04.0

100

96100=

=

=

2. Calculate the nominal decline rate per month:

( ) ( ) ( ) onth0.040822/m96.0ln04.01ln1ln ==== eDD

3. Calculate the rate after 11 more months:

( )BOPDeeqq

xDt

i 27.6110012040822.0

===

4. Calculate the rate after 22.5 months:

( ) POPDeeqq xDti 91.391005.22040822.0===

5. Calculate the nominal decline rate per year:

ar0.48986/ye12xonth0.040822/m ==D

6. Calculate the cumulative oil produced after one year:

( )STBYD

Y

DSTB

D

qqN ip 858,28/365*

/489864.0

/27.61100=

=

=

This completes the solution.


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Hyperbolic Decline

Alternatively, if the wells production data plotted on a semilog paperconcaves upward, then it is modeled with a hyperbolic decline. The equationof the hyperbolic decline is given by:

( ) bii tbDqq1

1

+=

Where:

q = wells production rate at time t, STB/dayqi = wells production rate at time 0, STB/dayDi = initial nominal exponential decline rate (t = 0), 1/dayb = hyperbolic exponentt = time, day

The following table summarizes the equations used in hyperbolic decline:

Hyperbolic Decline b > 0, b 1Description Equation

Rate ( ) bii tbDqq1

1

+=

Cumulative Oil Production( )

( )bbii

b

ip qq

bD

qN

=11

1

Nominal Decline Rate

( )[ ]111 = beii Db

D

i

iei

q

qqD

=

Effective Decline RateD

e eD

=1

Life( )

i

b

i

bD

qqt

1/ =

Example #4:Given the following data:

9.0/5.0

100

=

=

=

byearD

BOPDq

i

i


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Assuming hyperbolic decline, predict the amount of oil produced for fiveyears.

Solution:

1. Calculate the well flow rate at the end of each year for five years using:

( ) ( )( ) ( ) BOPDtxtxtbDqq bii 9.01

9.0

11

45.011005.09.011001

+=+=+=

2. Calculate the cumulative oil produced at the end of each year for fiveyears using:

( )( )

( )( )

( ) ( )( )

( )1.0

9.019.01

9.0

11

1.584893*460598.9

3651009.015.0

100

1

q

year

daysq

qqbD

qN bbi

i

b

ip

=

=

=

3. Form the following table:

Year Rate at End of YearCum.

ProductionYearly

Production

0123

45

10066.17649.00938.699

31.85426.992

029,52450,24866,115

78,91489,606

-29,52420,72415,867

12,79910,692

This completes the solution.

Harmonic Decline

A special case of the hyperbolic decline is known as harmonic decline,where b is taken to be equal to 1. The following table summarizes theequations used in harmonic decline:


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Harmonic Decline b = 1

Description Equation

RatetbD

qq

i

i

+

=

1

Cumulative Oil Productionq

q

D

qN i

i

ip ln=

Nominal Decline Rateei

eii

D

DD

=

1

Effective Decline Ratei

iei

q

qqD

=

Life( )

i

i

D

qqt

1/ =


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Material Balance Calculations for Oil Reservoirs

A general material balance equation that can be applied to all reservoir types wasfirst developed by Schilthuis in 1936. Although it is a tank model equation, it canprovide great insight for the practicing reservoir engineer. It is written from start ofproduction to any time (t) as follows:

Expansion of oil in the oil zone +

Expansion of gas in the gas zone +Expansion of connate water in the oil and gas zones +Contraction of pore volume in the oil and gas zones +Water influx + Water injected + Gas injected =Oil produced + Gas produced + Water produced

Mathematically, this can be written as:

( ) ( ) ( )

( )

( )

w wit ti g gi ti gi t

wi

fIw Igti gi e I I t

wi

t p soi g wp p p

C SN - + G - + + pNB GBB B B B

1 - S

C+ + + + +pNB GB W W GB B1 - S

= + - +N N WB R R B B

Where:

N = initial oil in place, STBNp = cumulative oil produced, STBG = initial gas in place, SCFGI = cumulative gas injected into reservoir, SCF

Gp = cumulative gas produced, SCFWe = water influx into reservoir, bblWI = cumulative water injected into reservoir, STBWp = cumulative water produced, STBBti = initial two-phase formation volume factor, bbl/STB = BoiBoi = initial oil formation volume factor, bbl/STBBgi = initial gas formation volume factor, bbl/SCFBt = two-phase formation volume factor, bbl/STB = Bo+ (Rsoi- Rso) BgBo = oil formation volume factor, bbl/STBBg = gas formation volume factor, bbl/SCF

Bw = water formation volume factor, bbl/STBBIg = injected gas formation volume factor, bbl/SCF


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BIw = injected water formation volume factor, bbl/STBRsoi = initial solution gas-oil ratio, SCF/STBRso = solution gas-oil ratio, SCF/STBRp = cumulative produced gas-oil ratio, SCF/STBCf = formation compressibility, psia

-1Cw = water isothermal compressibility, psia

-1Swi = initial water saturationpt = reservoir pressure drop, psia = pi- p(t)p(t) = current reservoir pressure, psia

The MBE as a Straight Line

Normally, when using the material balance equation, each pressure and thecorresponding production data is considered as being a separate point fromother pressure values. From each separate point, a calculation is made and theresults of these calculations are averaged. However, a method is required tomake use of all data points with the requirement that these points must yieldsolutions to the material balance equation that behave linearly to obtain valuesof the independent variable. The straight-line method begins with the materialbalance written as:

( ) ( ) ( )

( )

f w wit ti g gi ti gi t

wi

Iw Ige I I

t p soi g wp p p

+C C SN - + G - + + pNB GBB B B B

1 - S

+ + +W W GB B

= + - +N N WB R R B B

Defining the ratioof the initial gas cap volumeto the initial oil volumeas:

initial gas cap volume=

initial oil volume

gi

ti

GBm =

NB

and plugging into the equation yields:

( ) ( ) ( )

( )

f w witit ti g gi titi t

gi wi

Iw Ige I I

t p soi g wp p p

+C C SBN - + Nm - + + NmNBB B B B B

1 - SB

+ + +W W GB B

= + - +N N WB R R B B


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Let:

( )

( )

( )

1

o t ti

tig g gi

gi

f w wi

f,w ti twi

t p soi g w Iw Ig p p I I

= -E B B

B= -E B B

B

+C C S

= m B pE 1 - S

F = + - + - -N W W GB R R B B B B

+

Thus we obtain:

( )

,

,

eo g f w

o g f w e

F = NE + mNE + NE +W

N E mE E W= + + +

The following cases are considered:

1. No gas cap, negligible compressibilities, and no waterinflux

oF= NE

2. Negligible compressibilities, and no water influx

o g

g

o o

F= NE NmE

EF= N Nm

E E

+

+

Which is written as y = b + x. This would suggest that a plot of F/Eoasthe y coordinate versus Eg/Eoas the x coordinate would yield a straightline with slope equal to mN and intercept equal to N.

3. Including compressibilities and water influx, let:


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,o g f wD E mE E= + +

Dividing through by D, we get:

eF W= N +D D

Which is written as y = b + x. This would suggest that a plot of F/D asthe y coordinate and We/D as the x coordinate would yield a straight line

with slope equal to 1 and intercept equal to N.

Drive Indexes from the MBE

The three major driving mechanisms are:

1. Depletion drive (oil zone oil expansion),2. Segregation drive (gas zone gas expansion), and3. Water drive (water zone water influx).

To determine the relative magnitude of each of these driving mechanisms, thecompressibility term in the material balance equation is neglected and theequation is rearranged as follows:

( ) ( ) ( ) ( )t ti g gi w t p soi g e p pN - + G - + - = + -W W NB B B B B B R R B

Dividing through by the right hand side of the equation yields:

( )

( )( )( )

( )( )

wg gi e pt ti

t p soi g t p soi g t p soi g p p p

-G -N - W W BB BB B+ + = 1

+ - + - + -N N NB R R B B R R B B R R B

The terms on the left hand side of equation (3) represent the depletion driveindex (DDI), the segregation drive index (SDI), and the water drive index(WDI) respectively. Thus, using Pirson's abbreviations, we write:

DDI + SDI + WDI = 1

The following examples should clarify the errors that creep in during thecalculations of oil and gas reserves.


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Example #5: Given the following data for an oil field

Volume of bulk oil zone = 112,000 acre-ftVolume of bulk gas zone = 19,600 acre-ftInitial reservoir pressure = 2710 psiaInitial oil FVF = 1.340 bbl/STBInitial gas FVF = 0.006266 ft3/SCF

Initial dissolved GOR = 562 SCF/STBOil produced during the interval = 20 MM STBReservoir pressure at the end of the interval = 2000 psiaAverage produced GOR = 700 SCF/STBTwo-phase FVF at 2000 psia = 1.4954 bbl/STBVolume of water encroached = 11.58 MM bblVolume of water produced = 1.05 MM STBWater FVF = 1.028 bbl/STBGas FVF at 2000 psia = 0.008479 ft3/SCF

The following values will be calculated:

1. The stock tank oil initially in place.2. The driving indexes.3. Discussion of results.

Solution:

1. The material balance equation is written as:

( ) ( ) ( )[ ] ( )BW-W-BR-R+BN=B-BG+B-BN wpegsoiptpgigtit

Define the ratio of the initial gas cap volume to the initial oil volume as:

NB

GB=m

ti

gi

we get:

( ) ( ) ( )[ ] ( )BW-W-BR-R+BN=B-BB

BNm+B-BN wpegsoiptpgig

gi

titit


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and solve for N, we get:

( )[ ] ( )

( ) ( )B-BB

Bm+B-B

BW-W-BR-R+BN=N

gig

gi

titit

wpegsoiptp

Since:

Np = 20 x 106

STBBt = 1.4954 bbl/STBRp = 700 SCF/STBRsoi= 562 SCF/STBBg = 0.008479 ft

3/SCF = 0.008479/5.6146 = 0.001510 bbl/SCFWe = 11.58 x 10

6bblWp = 1.05 x 10

6STBBw = 1.028 bbl/STBBti = 1.34 bbl/STBm = GBgi/NBti= 19,600/112,000 = 0.175

Bgi = 0.006266 ft

3

/SCF = 0.006266/5.6146 = 0.001116 bbl/SCF

Thus:

( ) ( )

( ) ( )

620 1.4954 + 700 - 562 0.001510 - 11.58 - 1.05x1.028

N = 101.34

1.4954 - 1.34 + 0.175 0.001510 - 0.001116 0.001116

= 98.97 MM STB

2. In terms of drive indexes, the material balance equation is written as:

( )

( )

( )

( )( )

( )wg gi e pt ti

t p soi g t p soi g t p soi g p p p

-G -N - W W BB BB B+ + = 1

+ - + - + -N N NB R R B B R R B B R R B

Thus the depletion drive index (DDI) is given by:

( )

( )( )

( )

6t ti

6t p soi g p

N - 98.97x 1.4954 - 1.3410B B= = 0.45

20x 1.4954 + 700 - 562 0.001510+ - 10N B R R B

The segregation drive index (SDI) is given by:


26/29




( )

( )

( )

( )

tig gi

gi

t p soi g p

6

6

BNm -B B

B=

+ -N B R R B

1.3498.97 x x 0.175x 0.001510 - 0.00111610

0.001116 = 0.2420x 1.4954 + 700 - 562 0.00151010

The water drive index (WDI) is given by:

( )( )[ ]

( )( )[ ]

0.31=0.001510562-700+1.49541020x

10x1.028x1.05-10x11.58=

BR-R+BN

BW-W6

66

gsoiptp

wpe

3. The drive mechanisms as calculated in part (2) indicate that when thereservoir pressure has declined from 2710 psia to 2000 psia, 45% of thetotal production was by oil expansion, 31% was by water drive, and 24%was by gas cap expansion.



A gas cap reservoir is estimated, from volumetric calculations, to have aninitial oil volume N of 115 x 106STB. The cumulative oil production Npandcumulative gas oil ratio Rpare listed in the following table as functions of theaverage reservoir pressure over the first few years of production. Assumethat pi= pb= 3330 psia. The size of the gas cap is uncertain with the best

estimate, based on geological information, giving the value of m = 0.4. Is thisfigure confirmed by the production and pressure history? If not, what is thecorrect value of m?

Pressure Np Rp Bo Rso Bg

psia MM STB SCF/STB BBL/STB SCF/STB bbl/SCF

3330 0 0 1.2511 510 0.00087

3150 3.295 1050 1.2353 477 0.00092

3000 5.903 1060 1.2222 450 0.00096

2850 8.852 1160 1.2122 425 0.00101

2700 11.503 1235 1.2022 401 0.00107

2550 14.513 1265 1.1922 375 0.001132400 17.73 1300 1.1822 352 0.00120


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Solution:

Calculate the parameters F, Eo, Egas given by the above equations:

Bt F Eo Eg F/Eo Eg/Eo

BBL/STB MM/RB RB/STB RB/SCF MM/STB

1.2511

1.26566 5.8073 0.014560 0.071902299 398.8534135 4.938344701

1.2798 10.6714 0.028700 0.129424138 371.8272962 4.509551844

1.29805 17.3017 0.046950 0.201326437 368.5128136 4.288103021.31883 24.0940 0.067730 0.287609195 355.7353276 4.246407728

1.34475 31.8981 0.093650 0.373891954 340.6099594 3.992439445

1.3718 41.1301 0.120700 0.474555172 340.7626678 3.931691569

The plot of F/Eoversus Eg/Eois shown next:

Chart Title

y = 58.83x + 108.7

300

320

340

360

380

400

420

3.8 4 4.2 4.4 4.6 4.8 5 5.2

Eg/Eo

F/Eo

Figure 6: F/Eo vs. Eg/Eo Plot

The best fit is expressed by:

108.7 58.83 g

o o

EF=E E

+


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Therefore, N = 108.7 MM STB and m = 58.83/108.7 = 0.54.

This concludes the solution of this problem.


A gas cap reservoir is estimated, from volumetric calculations, to have aninitial oil volume N of 47 x 106STB. The cumulative oil production Np andcumulative gas oil ratio Rpare listed in the following table as functions of theaverage reservoir pressure over the first few years of production. Otherpertinent data are also supplied. Assume pi= pb= 3640 psia. The size of thegas cap is uncertain with the best estimate, based on geological information,giving the value of m = 0.0. Is this figure confirmed by the productionhistory? If not, what is the correct value of m?

pi = 3640 psia

Cf = 0.000004 psia-1

Cw = 0.000003 psia-1Swi = 0.25

Bw = 1.025 psia

m = 0

Pressure Np Gp Bt Rso

psia MM STB MM SCF BBL/STB SCF/STB

3640 0 0 1.464 888

3585 0.79 4.12 1.469 874

3530 1.21 5.68 1.476 860

3460 1.54 7 1.482 846

3385 2.08 8.41 1.491 8253300 2.58 9.71 1.501 804

3200 3.4 11.62 1.519 779

Bg We Wp Rp F

bbl/SCF MM BBL MM STB SCF/STB MM/RB

0.000892 0 0 0

0.000905 48.81 0.08 5.215189873 0.6114

0.000918 61.187 0.26 4.694214876 1.0713

0.000936 71.32 0.41 4.545454545 1.4291

0.000957 80.293 0.6 4.043269231 1.9567

0.000982 87.564 0.92 3.763565891 2.57530.001014 93.211 1.38 3.417647059 3.5294


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Solution:Calculate the parameters F, Eo, Eg, Ef,w, and D, as given by the aboveequations:

Eo Eg Ef,w D F/D We/D

RB/STB RB/SCF RB/SCF MM/STB

0.005000 0.021336323 0.00050996 0.00550996 110.9559779 8858.50351

0.012000 0.042672646 0.00101992 0.01301992 82.28173445 4699.491241

0.018000 0.072215247 0.00166896 0.01966896 72.65677901 3626.0178470.027000 0.106681614 0.00236436 0.02936436 66.63557762 2734.369147

0.037000 0.147713004 0.00315248 0.04015248 64.1383531 2180.786841

0.055000 0.200233184 0.00407968 0.05907968 59.739895 1577.716738

The plot of F/D versus We/D is shown next. The best fit is expressed by:

0.0071 48.067 6 eWF

= eD D

+

Therefore, N = 48 MM STB and m = 0.0071.

This concludes the solution of this problem.

Chart Title

y = 0.0071x + 48.067

59

69

79

89

99

109

119

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000

Eg/Eo

F/Eo

Figure 7: F/Eo vs. Eg/Eo Plot

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