Reserve Estimation Methods

© 2003-2004 Petrobjects 1

www.petrobjects.com

Petroleum Reserves Estimation Methods

© 2003-2004 Petrobjects www.petrobjects.com

Introduction The process of estimating oil and gas reserves for a producing field continues throughout the life of the field. There is always uncertainty in making such estimates. 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, and 6. Experience and knowledge of the evaluator.

The magnitude of uncertainty, however, decreases with time until the economic limit 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 following categories:

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


www.petrobjects.com



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

The volumetric method, on the other hand, entails determining the areal extent of the reservoir, the rock pore volume, and the fluid content within the pore volume. This provides an estimate of the amount of hydrocarbons-in-place. The ultimate 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 analysis and material balance calculations, become the predominant methods of calculating reserves. These methods greatly reduce the uncertainty in reserves estimates; however, during early depletion, caution should be exercised in using them. Decline curve relationships are empirical, and rely on uniform, lengthy production periods. It is more suited to oil wells, which are usually produced against fixed bottom-hole pressures. 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, there is currently a swing toward decline rates proportional to production rates (hyperbolic and harmonic). Although some wells exhibit these trends, hyperbolic or harmonic decline extrapolations should only be used for these specific cases. Over-exuberance in the use of hyperbolic or harmonic relationships can result in excessive reserves estimates.

Material balance calculation is an excellent tool for estimating gas reserves. If a reservoir 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 the depletion mechanism, altering the performance of the non-ideal gas law in the reservoir. Under these conditions, optimistic reserves estimates can result.

When calculating reserves using any of the above methods, two calculation procedures may be used: deterministic and/or probabilistic. The deterministic method is by far the most common. The procedure is to select a single value for each parameter to input into an appropriate equation, to obtain a single answer. The probabilistic method, on the other hand, is more rigorous and less commonly used. This method utilizes a distribution curve for each parameter and, through the use of Monte Carlo Simulation; a distribution curve for the answer can be developed. Assuming good data, a lot of qualifying information can be derived from


www.petrobjects.com



the resulting statistical calculations, such as the minimum and maximum values, the mean (average value), the median (middle value), the mode (most likely value), 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 by all input parameters, including the most likely and maximum values for the parameters. In such methods, one can not back calculate the input parameters associated with reserves. Only the end result is known but not the exact value of any input parameter. On the other hand, deterministic methods calculate reserve values that are more tangible and explainable. In these methods, all input parameters are exactly known; however, they may sometimes ignore the variability


www.petrobjects.com



and uncertainty in the input data compared to the probabilistic methods which allow the incorporation of more variance in the data. A comparison of the deterministic and probabilistic methods, however, can provide quality assurance for estimating hydrocarbon reserves; i.e. reserves are calculated both deterministically and probabilistically and the two values are compared. If the two values agree, then confidence on the calculated reserves is increased. If the two values are away different, the assumptions need to be reexamined.


www.petrobjects.com



Analogy

The analogy method is applied by comparing the following factors for the analogous and current fields or wells:

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

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

( )( ) ( )( ) RFtBtS = BAF

o

oφ117758

is assumed to be the same for the analogous and current field or well. Comparing EUR’s is done during the exploratory phase. It is also useful when calculating proved developed reserves.

Analogy is most useful when running the economics on a yet-to-be-drilled exploratory 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 used for analogy is indeed analogous. That said, a dolomite reservoir with volatile crude oil will never be analogous to a sandstone reservoir with black oil. Similarly, if your calculated EUR is twice as high as the EUR from the nearest 100 wells, you had better check your assumptions.


www.petrobjects.com



Volumetric

The volumetric method entails determining the physical size of the reservoir, the pore volume within the rock matrix, and the fluid content within the void space. This provides an estimate of the hydrocarbons-in-place, from which ultimate recovery can be estimated by using an appropriate recovery factor. Each of the factors 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 the following equation:

( )( )tB

tSV = N(t)o

obφ

Where: N(t) = oil in place at time t, STB Vb = bulk reservoir volume, RB = 7758 A h 7758 = RB/acre-ft A = reservoir area, acres


www.petrobjects.com



H = average reservoir thickness, ft φ = average reservoir porosity, fraction So(t) = average oil saturation, fraction Bo(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 by the following equation:

( )( )tB

tSV = G(t)

g

gbφ

Where: G(t) = gas in place at time t, SCF Vb = bulk reservoir volume, CF = 43560 A h 43560 = CF/acre-ft A = reservoir area, acres h = average reservoir thickness, ft φ = average reservoir porosity, fraction Sg(t) = average gas saturation, fraction Bg(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 to large errors. They are usually determined from analogy or correlations. The following 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 acres Net productive thickness = 49 ft Porosity = 8% Average Swi = 45% Initial reservoir pressure, pi = 2980 psia Abandonment pressure, pa = 300 psia Bo at pi = 1.68 bbl/STB Bo at pa = 1.15 bbl/STB


www.petrobjects.com



Sg at pa = 34% Sor after water invasion = 20% The following quantities will be calculated:

1. Initial oil in place 2. Oil in place after volumetric depletion to abandonment pressure 3. Oil in place after water invasion at initial pressure 4. Oil reserve by volumetric depletion to abandonment pressure 5. Oil reserve by full water drive 6. Discussion of results

Solution: Let’s 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

S V = Noi

wibi

−1φ

this yields:

( ) STBMM 266 1.68

50. (0.08) 10 x 10.15 = N9

i ≈− 41

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

( )B

S - S - 1 V = No

gwbφ

this yields:

( ) STBMM 148 1.15

0.34) - 0.45 - 1 (0.08) 10 x 10.15 = N9

1 ≈

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


www.petrobjects.com



BS V = N

o

orbφ

this yields:

STBMM 97 1.68

0.20 (0.08) 10 x 10.15 = N9

2 ≈

4. The oil reserve by volumetric depletion:

( ) ( ) STBMM 118 = 10 x 148 - 266 = N-N 6

1i i.e. RF = 118/266 = 44%

5. The oil reserve by full water drive

( ) ( ) STBMM 169 = 10 x 97 - 266 = N-N 62i

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

6. Discussion of results: For oil reservoirs under volumetric control; i.e. no water influx, the produced oil must be replaced by gas the saturation of which increases as oil saturation decreases. If Sg is the gas saturation and Bo the oil formation volume factor at abandonment pressure, then oil in place at abandonment pressure is given by:

( )

BS - S - 1 V = N

o

gwbφ

On the other hand, for oil reservoirs under hydraulic control, where there is no appreciable decline in reservoir pressure, water influx is either edge-water drive or bottom-water drive. In edge-water drive, water influx is inward and parallel to bedding planes. In bottom-water drive, water influx is upward where the producing oil zone is underlain by water. In this case, the oil remaining at abandonment is given by:

BS V = N

o

orbφ

This concludes the solution.


www.petrobjects.com



Example #2: Given the following data for the Bell gas field Area = 160 acres Net productive thickness = 40 ft Initial reservoir pressure = 3250 psia Porosity = 22% Connate water = 23% Initial gas FVF = 0.00533 ft3/SCF Gas FVF at 2500 psia = 0.00667 ft3/SCF Gas FVF at 500 psia = 0.03623 ft3/SCF Sgr after water invasion = 34% The following quantities will be calculated:

1. Initial gas in place 2. Gas in place after volumetric depletion to 2500 psia 3. Gas in place after volumetric depletion to 500 psia 4. Gas in place after water invasion at 3250 psia 5. Gas in place after water invasion at 2500 psia 6. Gas in place after water invasion at 500 psia 7. Gas reserve by volumetric depletion to 500 psia 8. Gas reserve by full water drive; i.e. at 3250 psia 9. Gas reserve by partial water drive; i.e. at 2500 psia 10. Gas reserve by full water drive if there is one un-dip well 11. Discussion of results

Solution: Let’s 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 - 1 V = Ggi

wiibi

φ

this yields:


www.petrobjects.com



( ) SCFMM 8860 = 0.00533

0.23) - 1 (0.22) 10 x 278.784 = G6

i

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

( ) SCFMM 7080 =

0.006670.23) - 1 (0.22) 10 x 278.784 = G

6

1

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

( ) SCFMM 1303 = 0.003623

0.23) - 1 (0.22) 10 x 278.784 = G6

2

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

SCFMM 3912 = 0.00533

(0.34) (0.22) 10 x 278.784 = G6

3


SCFMM 3126 = 0.00667

(0.34) (0.22) 10 x 278.784 = G6

4


SCFMM 576 = 0.03623

(0.34) (0.22) 10 x 278.784 = G6

5

7. Gas reserve by volumetric depletion to 500 psia:

( ) SCFMM 7557 = 10 x 1303 - 8860 = G - G 6

2i i.e. RF = 7557/8860 = 85%

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

( ) SCFMM 4948 = 10 x 3912 - 8860 = G - G 63i


www.petrobjects.com



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

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

( ) SCFMM 5734 = 10 x 3126 - 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:

( ) ( ) SCFMM 2474 = 10 x 3912 - 8860 21 = G - G

21 6

3i

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

11. Discussion of results The RF for volumetric depletion to 500 psia (no water drive) is calculated to be 85%. On the other hand, the RF for partial water drive is 65%, and for the full water drive is 56%. This can be explained as follows: As water invades the reservoir, the reservoir pressure is maintained at a higher level than if there were no water encroachment. This leads to higher abandonment pressures for water-drive reservoirs. Recoveries, however, are lower because the main mechanism of production in gas reservoirs is depletion or gas expansion. In water-drive gas reservoirs, 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 where the field production rate has been increased from 50 to 75 MM SCF/D, and they found that the ultimate recovery increased from 69 to 74%.

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


www.petrobjects.com



The high-rate down-dip water producers act as a pressure sink for the water. This retards water invasion into the gas zone, therefore prolonging 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 the water-swept portion of the reservoir could become mobile and hence producible. It has been reported that this technique has increased gas production from62% to 83% in Eugene Island Field of Louisiana.



www.petrobjects.com



Decline Curves A decline curve of a well is simply a plot of the well’s production rate on the y-axis versus time on the x-axis. The plot is usually done on a semilog paper; i.e. the y-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 the data plots concave upward, it is modeled with a “hyperbolic decline”. A special case of 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, there is currently a swing toward decline rates proportional to production rates (hyperbolic and harmonic). Although some wells exhibit these trends, hyperbolic or harmonic decline extrapolations should only be used for these specific cases. Over-exuberance in the use of hyperbolic or harmonic relationships can result in excessive reserves estimates. Figure 5 is an example of a production graph with exponential and harmonic extrapolations.

Figure 5: Decline curve of an oil well Decline curves are the most common means of forecasting production. They have many advantages:

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


www.petrobjects.com



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

Exponential Decline

As mentioned above, in the exponential decline, the well’s production data plots as a straight line on a semilog paper. The equation of the straight line on the semilog paper is given by:

Dtieqq −=

Where:

q = well’s production rate at time t, STB/day qi = well’s production rate at time 0, STB/day D = nominal exponential decline rate, 1/day t = time, day

The following table summarizes the equations used in exponential decline. Exponential Decline b = 0

Description Equation Rate Dt

ieqq −=

Cumulative Oil Production DqqN i

p−

=

Nominal Decline Rate

( )eDD −−= 1ln

i

ie q

qqD −=

Effective Decline Rate De eD −−=1

Life ( )Dqqt 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


www.petrobjects.com



months and after 22.5 months. Also predict the amount of oil produced after one year. Solution:

monthtBOPDqBOPDqi

196100

===

1. Calculate the effective decline rate per month:

monthqqqD

i

ie /04.0

10096100

=−

=−

=

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 27.61100 12040822.0 === −−

4. Calculate the rate after 22.5 months:

( ) POPDeeqq xDt

i 91.39100 5.22040822.0 === −−

5. Calculate the nominal decline rate per year:

ar0.48986/ye 12x onth0.040822/m ==D

6. Calculate the cumulative oil produced after one year:

( ) STBYD

YDSTB

DqqN i

p 858,28/365*/489864.0

/27.61100=

−=

−=

This completes the solution.


www.petrobjects.com



Hyperbolic Decline

Alternatively, if the well’s production data plotted on a semilog paper concaves upward, then it is modeled with a hyperbolic decline. The equation of the hyperbolic decline is given by:

( ) bii tbDqq1

1 −+=

Where:

q = well’s production rate at time t, STB/day qi = well’s production rate at time 0, STB/day Di = initial nominal exponential decline rate (t = 0), 1/day b = hyperbolic exponent t = time, day

The following table summarizes the equations used in hyperbolic decline: Hyperbolic Decline b > 0, b ≠ 1

Description Equation

Rate ( ) bii tbDqq1

1 −+=

Cumulative Oil Production ( ) ( )bbi

i

bi

p qqbD

qN −− −−

= 11

1

Nominal Decline Rate

( )[ ]111−−= −b

eii Db

D

i

iei q

qqD −=

Effective Decline Rate De eD −−=1

Life ( )

i

bi

bDqqt 1/ −

=

Example #4: Given the following data:

9.0/5.0

100

===

byearDBOPDq

i

i


www.petrobjects.com



Assuming hyperbolic decline, predict the amount of oil produced for five years. Solution:

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

( ) ( )( ) ( ) BOPDtxtxtbDqq bii 9.01

9.011

45.011005.09.011001 −−− +=+=+=

2. Calculate the cumulative oil produced at the end of each year for five

years using:

( ) ( )

( )( )

( ) ( )( )( )1.0

9.019.019.0

11

1.584893*460598.9

3651009.015.0

100

1

qyeardaysq

qqbD

qN bbi

i

bi

p

−=

−−

=

−−

=

−−

−−

3. Form the following table:

Year Rate at End of Year Cum.

Production Yearly

Production 0 1 2 3 4 5

100 66.176 49.009 38.699 31.854 26.992

0 29,524 50,248 66,115 78,914 89,606

- 29,524 20,724 15,867 12,799 10,692

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 the equations used in harmonic decline:


www.petrobjects.com



Harmonic Decline b = 1 Description Equation

Rate tbDqq

i

i

+=

1

Cumulative Oil Production qq

DqN i

i

ip ln=

Nominal Decline Rate ei

eii D

DD−

=1

Effective Decline Rate i

iei q

qqD −=

Life ( )

i

i

Dqqt 1/ −

=


www.petrobjects.com



Material Balance Calculations for Oil Reservoirs A general material balance equation that can be applied to all reservoir types was first developed by Schilthuis in 1936. Although it is a tank model equation, it can provide great insight for the practicing reservoir engineer. It is written from start of production 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 It

wi

t p soi g wp p p

C SN - + G - + + pNB GBB B B B 1 - SC+ + + + + pNB GB W W GB B1 - S

= + - + N N WB R R B B

∆

∆

Where: N = initial oil in place, STB Np = cumulative oil produced, STB G = initial gas in place, SCF GI = cumulative gas injected into reservoir, SCF Gp = cumulative gas produced, SCF We = water influx into reservoir, bbl WI = cumulative water injected into reservoir, STB Wp = cumulative water produced, STB Bti = initial two-phase formation volume factor, bbl/STB = Boi Boi = initial oil formation volume factor, bbl/STB Bgi = initial gas formation volume factor, bbl/SCF Bt = two-phase formation volume factor, bbl/STB = Bo + (Rsoi - Rso) Bg Bo = oil formation volume factor, bbl/STB Bg = gas formation volume factor, bbl/SCF Bw = water formation volume factor, bbl/STB BIg = injected gas formation volume factor, bbl/SCF


www.petrobjects.com



BIw = injected water formation volume factor, bbl/STB Rsoi = initial solution gas-oil ratio, SCF/STB Rso = solution gas-oil ratio, SCF/STB Rp = cumulative produced gas-oil ratio, SCF/STB Cf = formation compressibility, psia-1 Cw = water isothermal compressibility, psia-1 Swi = initial water saturation ∆pt = 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 the corresponding production data is considered as being a separate point from other pressure values. From each separate point, a calculation is made and the results of these calculations are averaged. However, a method is required to make use of all data points with the requirement that these points must yield solutions to the material balance equation that behave linearly to obtain values of the independent variable. The straight-line method begins with the material balance 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 ratio of the initial gas cap volume to the initial oil volume as:

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 - + + Nm pNBB B B B B 1 - SB+ + + W W GB B

= + - + N N WB R R B B

∆


www.petrobjects.com



Let:

( )

( )

( )

1

o t ti

tig g gi

gi

f w wif,w ti t

wi

t p soi g w Iw Igp p I I

= - E B B

B= - E B BB

+ 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 water influx

oF= NE

2. Negligible compressibilities, and no water influx

o g

g

o o

F= NE NmE

EF = N NmE E

+

+

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

3. Including compressibilities and water influx, let:


www.petrobjects.com



,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 as the 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), and 3. Water drive (water zone water influx).

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

( ) ( ) ( ) ( )t ti g gi w t p soi ge 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 gp 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 drive index (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 the calculations of oil and gas reserves.


www.petrobjects.com



Example #5: Given the following data for an oil field Volume of bulk oil zone = 112,000 acre-ft Volume of bulk gas zone = 19,600 acre-ft Initial reservoir pressure = 2710 psia Initial oil FVF = 1.340 bbl/STB Initial gas FVF = 0.006266 ft3/SCF Initial dissolved GOR = 562 SCF/STB Oil produced during the interval = 20 MM STB Reservoir pressure at the end of the interval = 2000 psia Average produced GOR = 700 SCF/STB Two-phase FVF at 2000 psia = 1.4954 bbl/STB Volume of water encroached = 11.58 MM bbl Volume of water produced = 1.05 MM STB Water FVF = 1.028 bbl/STB Gas 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 - B R - R + B N = B - BG + B - BN wpegsoiptpgigtit

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

NBGB = m

ti

gi

we get:

( ) ( ) ( )[ ] ( )BW - W - B R - R + B N = B - B BBNm + B - BN wpegsoiptpgig

gi

titit


www.petrobjects.com



and solve for N, we get:

( )[ ] ( )( ) ( )B - B

BBm + B - B

BW - W - B R - R + B N = Ngig

gi

titit

wpegsoiptp

Since:

Np = 20 x 106 STB Bt = 1.4954 bbl/STB Rp = 700 SCF/STB Rsoi= 562 SCF/STB Bg = 0.008479 ft3/SCF = 0.008479/5.6146 = 0.001510 bbl/SCF We = 11.58 x 10 6 bbl Wp = 1.05 x 106 STB Bw = 1.028 bbl/STB Bti = 1.34 bbl/STB m = GBgi/NBti = 19,600/112,000 = 0.175 Bgi = 0.006266 ft3/SCF = 0.006266/5.6146 = 0.001116 bbl/SCF

Thus:

( ) ( )

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

N = 101.341.4954 - 1.34 + 0.175 0.001510 - 0.0011160.001116

= 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 gp 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 gp

N - 98.97x 1.4954 - 1.3410B B = = 0.4520x 1.4954 + 700 - 562 0.001510 + - 10N B R R B

The segregation drive index (SDI) is given by:


www.petrobjects.com



( )

( )

( )( )

tig gi

gi

t p soi gp

6

6

BNm - B BB =

+ - N B R R B

1.3498.97 x x 0.175x 0.001510 - 0.00111610 0.001116 = 0.24

20x 1.4954 + 700 - 562 0.001510 10

The water drive index (WDI) is given by:

( )( )[ ]

( )( )[ ] 0.31 =

0.001510 562 - 700 + 1.4954 1020x 10 x 1.028 x 1.05 - 10 x 11.58 =

B R - R + B NB W - W

6

66

gsoiptp

wpe

3. The drive mechanisms as calculated in part (2) indicate that when the

reservoir pressure has declined from 2710 psia to 2000 psia, 45% of the total production was by oil expansion, 31% was by water drive, and 24% was by gas cap expansion.


Example #6: Given the following data for an oil field A gas cap reservoir is estimated, from volumetric calculations, to have an initial oil volume N of 115 x 106 STB. The cumulative oil production Np and cumulative gas oil ratio Rp are listed in the following table as functions of the average reservoir pressure over the first few years of production. Assume that 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 this figure confirmed by the production and pressure history? If not, what is the correct 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 0.00113 2400 17.73 1300 1.1822 352 0.00120


www.petrobjects.com



Solution: Calculate the parameters F, Eo, Eg as 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 4.28810302 1.31883 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/Eo versus Eg/Eo is 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

+


www.petrobjects.com



Therefore, N = 108.7 MM STB and m = 58.83/108.7 = 0.54. This concludes the solution of this problem. Example #7: Given the following data for an oil field A gas cap reservoir is estimated, from volumetric calculations, to have an initial oil volume N of 47 x 106 STB. The cumulative oil production Np and cumulative gas oil ratio Rp are listed in the following table as functions of the average reservoir pressure over the first few years of production. Other pertinent data are also supplied. Assume pi = pb = 3640 psia. The size of the gas cap is uncertain with the best estimate, based on geological information, giving the value of m = 0.0. Is this figure confirmed by the production history? If not, what is the correct value of m?

pi = 3640 psia Cf = 0.000004 psia-1 Cw = 0.000003 psia-1 Swi = 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 1.491 825 3300 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 2.5753 0.001014 93.211 1.38 3.417647059 3.5294


www.petrobjects.com



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

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 3626.017847 0.027000 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

Documents

Reserve Estimation Methods