Reserve Estimation Methods

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.

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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. Overexuberance 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 © 2003-2004 Petrobjects www.petrobjects.com

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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 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 © 2003-2004 Petrobjects www.petrobjects.com

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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.


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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, BAF = 7758 (1)(1)

φ S o (t ) Bo (t )

RF

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.


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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: EUR = N (t ) RF

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: N(t) =

Vbφ S o (t ) Bo (t )

Where: N(t) Vb 7758 A

= = = =

oil in place at time t, STB bulk reservoir volume, RB = 7758 A h RB/acre-ft reservoir area, acres


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H φ So(t) Bo(p)

= = = =

average reservoir thickness, ft average reservoir porosity, fraction average oil saturation, fraction oil formation volume factor at reservoir pressure p, RB/STB

Similarly, for a gas reservoir, the volumetric method is given by: EUR = G (t ) RF

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: V φ S g (t ) G(t) = b B g (t ) Where: G(t) Vb 43560 A h φ Sg(t) Bg(p)

= = = = = = = =

gas in place at time t, SCF bulk reservoir volume, CF = 43560 A h CF/acre-ft reservoir area, acres average reservoir thickness, ft average reservoir porosity, fraction average gas saturation, fraction 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 Net productive thickness Porosity Average Swi Initial reservoir pressure, pi Abandonment pressure, pa Bo at pi Bo at pa © 2003-2004 Petrobjects www.petrobjects.com

= = = = = = = =

26,700 acres 49 ft 8% 45% 2980 psia 300 psia 1.68 bbl/STB 1.15 bbl/STB 2


Sg at pa Sor after water invasion

= 34% = 20%

The following quantities will be calculated: 1. 2. 3. 4. 5. 6.

Initial oil in place Oil in place after volumetric depletion to abandonment pressure Oil in place after water invasion at initial pressure Oil reserve by volumetric depletion to abandonment pressure Oil reserve by full water drive 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: Ni=

V b φ (1 − S wi ) Boi

this yields: Ni=

10.15 x 10 9 (0.08) (1 − 0.45 ) ≈ 266 MM STB 1.68

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

N=

V b φ (1 - S w - S g ) Bo

this yields: N1=

10.15 x 109 (0.08) (1 - 0.45 - 0.34)) ≈ 148 MM STB 1.15

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


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φ N = V b S or Bo this yields: N2=

10.15 x 109 (0.08) 0.20 ≈ 97 MM STB 1.68

4. The oil reserve by volumetric depletion:

(N i - N 1) = (266 - 148 ) x 106 = 118 MM STB i.e. RF = 118/266 = 44% 5. The oil reserve by full water drive

(N i - N 2 ) = (266 - 97 ) x 106 = 169 MM STB 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:

N=

V b φ (1 - S w - S g ) Bo

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:

φ N = V b S or Bo This concludes the solution. © 2003-2004 Petrobjects www.petrobjects.com

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Example #2: Given the following data for the Bell gas field Area Net productive thickness Initial reservoir pressure Porosity Connate water Initial gas FVF Gas FVF at 2500 psia Gas FVF at 500 psia Sgr after water invasion

= = = = = = = = =

160 acres 40 ft 3250 psia 22% 23% 0.00533 ft3/SCF 0.00667 ft3/SCF 0.03623 ft3/SCF 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:

Gi =

V b φ i (1 - S wi ) B gi

this yields: © 2003-2004 Petrobjects www.petrobjects.com

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Gi =

278.784 x 106 (0.22) (1 - 0.23)) = 8860 MM SCF 0.00533

2. Gas in place after volumetric depletion to 2500 psia: G1 =

278.784 x 106 (0.22) (1 - 0.23)) = 7080 MM SCF 0.00667

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

G2 =

278.784 x 106 (0.22) (1 - 0.23)) = 1303 MM SCF 0.003623

4. Gas in place after water invasion at 3250 psia: 278.784 x 106 (0.22) (0.34) = 3912 MM SCF G3 = 0.00533

5. Gas in place after water invasion at 2500 psia: G4 =

278.784 x 106 (0.22) (0.34) = 3126 MM SCF 0.00667

6. Gas in place after water invasion at 500 psia: G5 =

278.784 x 106 (0.22) (0.34) = 576 MM SCF 0.03623

7. Gas reserve by volumetric depletion to 500 psia: 6 Gi - G 2 = (8860 - 1303) x 10 = 7557 MM SCF

i.e. RF = 7557/8860 = 85% 8. Gas reserve by water drive at 3250 psia (full water drive): 6 Gi - G 3 = (8860 - 3912 ) x 10 = 4948 MM SCF


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i.e. RF = 4948/8860 = 56% 9. Gas reserve by water drive at 2500 psia (partial water drive): 6 Gi - G 4 = (8860 - 3126 ) x 10 = 5734 MM SCF

i.e. RF = 5734/8860 = 65% 10. Gas reserve by water drive at 3250 psia if there is one un-dip well:

1 (Gi - G3 ) = 1 (8860 - 3912) x 106 = 2474 MM SCF 2 2 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:


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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. This concludes the solution.


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Decline Curves A decline curve of a well is simply a plot of the well’s production rate on the yaxis 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. Overexuberance 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.


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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:

q = qi e − Dt Where: q qi D t

= = = =

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

The following table summarizes the equations used in exponential decline. Exponential Decline b = 0 Description Rate Cumulative Oil Production Nominal Decline Rate Effective Decline Rate Life

Equation q = qi e − Dt q −q Np = i D D = − ln (1 − De ) q −q De = i qi

De = 1 − e − D ln (qi / q ) t= D

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 after one year.

Solution: qi = 100 BOPD q = 96 BOPD t = 1 month

1. Calculate the effective decline rate per month:

De =

qi − q 100 − 96 = = 0.04 / month qi 100

2. Calculate the nominal decline rate per month:

D = − ln(1 − De ) = − ln (1 − 0.04 ) = − ln (0.96) = 0.040822/month 3. Calculate the rate after 11 more months:

q = qi e − Dt = 100e (−0.040822 x12 ) = 61.27 BOPD 4. Calculate the rate after 22.5 months:

q = qi e − Dt = 100e (−0.040822 x 22.5 ) = 39.91 POPD 5. Calculate the nominal decline rate per year:

D = 0.040822/month x 12 = 0.48986/year 6. Calculate the cumulative oil produced after one year:

Np =

qi − q (100 − 61.27 ) STB / D = * 365D / Y = 28,858 STB D 0.489864 / Y

This completes the solution.


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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:

q = qi (1 + bDi t )

−

1 b

Where: q qi Di b t

= = = = =

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

The following table summarizes the equations used in hyperbolic decline: Hyperbolic Decline b > 0, b ≠ 1 Description Rate Cumulative Oil Production

Equation

q = qi (1 + bDi t )

−

qib (qi1−b − q1−b ) Di (1 − b ) 1 −b Di = (1 − Dei ) − 1 b q −q Dei = i qi

Np =

[

Nominal Decline Rate

1 b

]

Effective Decline Rate

De = 1 − e − D

Life

b ( qi / q ) − 1 t=

bDi

Example #4: Given the following data:

qi = 100 BOPD Di = 0.5 / year b = 0.9 © 2003-2004 Petrobjects www.petrobjects.com

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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:

q = qi (1 + bDi t )

−

1 b

= (100)(1 + 0.9 x 0.5 xt )

−

1 0.9

= 100(1 + 0.45t )

−

1 0.9

BOPD

2. Calculate the cumulative oil produced at the end of each year for five years using: qib ( Np = qi1− b − q1− b ) Di (1 − b ) =

(100)0.9 (100(1−0.9 ) − q (1−0.9 ) ) 365 days 0.5(1 − 0.9 ) year

= 460598.9 * (1.584893 − q 0.1 )

3. Form the following table: Year

Rate at End of Year

0 1 2 3 4 5

100 66.176 49.009 38.699 31.854 26.992

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

Yearly Production 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:


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Harmonic Decline b = 1 Description Rate Cumulative Oil Production Nominal Decline Rate Effective Decline Rate Life


Equation qi q= 1 + bDi t q q N p = i ln i Di q Dei Di = 1 − Dei q −q Dei = i qi (q / q ) − 1 t= i Di

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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: C S  N ( B t - B ti ) + G ( B g - B gi ) + ( NB ti + GB gi )  w wi  ∆ p t  1 - S wi   Cf  + ( NB ti + GB gi )   ∆ p t + W e + W I B Iw + G I B Ig  1 - S wi 

= N p B t + N p ( R p - R soi ) B g + W p B w

Where: N Np G GI Gp We WI Wp Bti Boi Bgi Bt Bo Bg Bw BIg

= = = = = = = = = = = = = = = =

initial oil in place, STB cumulative oil produced, STB initial gas in place, SCF cumulative gas injected into reservoir, SCF cumulative gas produced, SCF water influx into reservoir, bbl cumulative water injected into reservoir, STB cumulative water produced, STB initial two-phase formation volume factor, bbl/STB = Boi initial oil formation volume factor, bbl/STB initial gas formation volume factor, bbl/SCF two-phase formation volume factor, bbl/STB = Bo + (Rsoi - Rso) Bg oil formation volume factor, bbl/STB gas formation volume factor, bbl/SCF water formation volume factor, bbl/STB injected gas formation volume factor, bbl/SCF


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BIw Rsoi Rso Rp Cf Cw Swi ∆pt p(t)

= = = = = = = = =

injected water formation volume factor, bbl/STB initial solution gas-oil ratio, SCF/STB solution gas-oil ratio, SCF/STB cumulative produced gas-oil ratio, SCF/STB formation compressibility, psia-1 water isothermal compressibility, psia-1 initial water saturation reservoir pressure drop, psia = pi - 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:  C f + C w S wi  N ( B t - B ti ) + G ( B g - B gi ) + ( NB ti + GB gi )   ∆ pt  1 - S wi  + W e + W I B Iw + G I B Ig = N p B t + N p ( R p - R soi ) B g + W p B w

Defining the ratio of the initial gas cap volume to the initial oil volume as: m=

initial gas cap volume GBgi = NBti initial oil volume

and plugging into the equation yields: N ( B t - B ti ) + Nm

+   B ti ( B g - B gi ) + ( NB ti + Nm B ti )  C f1 - C w S wi  ∆ p t S wi  B gi 

+ W e + W I B Iw + G I B Ig = N p B t + N p ( R p - R soi ) B g + W p B w


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Let: E o = B t - B ti Eg=

E

F=N

p

f,w

 Bt + 

B ti ( B g - B gi ) B gi

 C f + C w S wi  = (1 + m ) B ti   ∆ pt 1 - S wi  

( R p - R soi ) B g

 +W 

p

Bw -W

I

B Iw - G I B Ig

Thus we obtain: F = NEo + mNE g + NE f ,w + W e = N ( Eo + mE g + E f ,w ) + We

The following cases are considered: 1.

No gas cap, negligible compressibilities, and no water influx

F= NEo 2.

Negligible compressibilities, and no water influx

F= NEo + NmE g Eg F = N + Nm Eo Eo 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:

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D = Eo + mE g + E f ,w Dividing through by D, we get: F W =N+ e 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: N ( B t - B ti ) + G ( B g - B gi ) + (W e - W p B w ) = N p  B t + ( R p - R soi ) B g 

Dividing through by the right hand side of the equation yields: N ( B t - B ti )

N p  B t + ( R p - R soi ) B g 

+

G ( B g - B gi )

N p  B t + ( R p - R soi ) B g 

+

(W e - W p B w ) =1 N p  B t + ( R p - R soi ) B g 

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. © 2003-2004 Petrobjects www.petrobjects.com

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Example #5: Given the following data for an oil field Volume of bulk oil zone Volume of bulk gas zone Initial reservoir pressure Initial oil FVF Initial gas FVF Initial dissolved GOR Oil produced during the interval Reservoir pressure at the end of the interval Average produced GOR Two-phase FVF at 2000 psia Volume of water encroached Volume of water produced Water FVF Gas FVF at 2000 psia

= = = = = = = = = = = = = =

112,000 acre-ft 19,600 acre-ft 2710 psia 1.340 bbl/STB 0.006266 ft3/SCF 562 SCF/STB 20 MM STB 2000 psia 700 SCF/STB 1.4954 bbl/STB 11.58 MM bbl 1.05 MM STB 1.028 bbl/STB 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: N (Bt - Bti ) + G (B g - B gi ) = N p [ Bt + (R p - R soi ) B g ] - (W e - W p B w ) Define the ratio of the initial gas cap volume to the initial oil volume as:

m=

GB gi NBti

we get:

N (Bt - Bti ) + Nm Bti (B g - B gi ) = N p [ Bt + (R p - R soi ) B g ] - (W e - W p B w ) B gi


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and solve for N, we get: N=

N p [ Bt + (R p - R soi ) B g ] - (W e - W p B w ) (Bt - Bti ) + m Bti (B g - B gi ) B gi

Since: Np Bt Rp Rsoi= Bg We Wp Bw Bti m Bgi

= 20 x 106 STB = 1.4954 bbl/STB = 700 SCF/STB 562 SCF/STB = 0.008479 ft3/SCF = 0.008479/5.6146 = 0.001510 bbl/SCF = 11.58 x 10 6 bbl = 1.05 x 106 STB = 1.028 bbl/STB = 1.34 bbl/STB = GBgi/NBti = 19,600/112,000 = 0.175 = 0.006266 ft3/SCF = 0.006266/5.6146 = 0.001116 bbl/SCF

Thus: N=

20  1.4954 + ( 700 - 562 ) 0.001510  - ( 11.58 - 1.05x1.028 ) 6 10 1.34 (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: N ( B t - B ti

)

N p  B t + ( R p - R soi ) B g 

+

G ( B g - B gi )

N p  B t + ( R p - R soi ) B g 

+

( W e - W p Bw ) =1 N p  B t + ( R p - R soi ) B g 

Thus the depletion drive index (DDI) is given by: N ( B t - B ti )

N p  B t + ( R p - R soi ) B g 

=

98.97x10 6 ( 1.4954 - 1.34 )

20x10 6 1.4954 + (700 - 562 ) 0.001510 

= 0.45

The segregation drive index (SDI) is given by:


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Nm B ti ( B g - B gi ) B gi =  N p  B t + ( R p - R soi ) B g  1.34 ( 0.001510 - 0.001116 ) 0.001116 = 0.24 20x10 6 1.4954 + (700 - 562 ) 0.001510 

98.97 x 10 6 x 0.175x

The water drive index (WDI) is given by:

( W e - W p Bw ) = ( 11.58 x 106 - 1.05 x 1.028 x 106 ) = 0.31 6 N p [ Bt + (R p - R soi ) B g ] 20x 10 [1.4954 + (700 - 562 ) 0.001510 ] 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. This concludes the solution.

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 psia 3330 3150 3000 2850 2700 2550 2400

Np MM STB 0 3.295 5.903 8.852 11.503 14.513 17.73


Rp SCF/STB 0 1050 1060 1160 1235 1265 1300

Bo BBL/STB 1.2511 1.2353 1.2222 1.2122 1.2022 1.1922 1.1822

Rso SCF/STB 510 477 450 425 401 375 352

Bg bbl/SCF 0.00087 0.00092 0.00096 0.00101 0.00107 0.00113 0.00120

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Solution: Calculate the parameters F, Eo, Eg as given by the above equations: Bt BBL/STB 1.2511 1.26566 1.2798 1.29805 1.31883 1.34475 1.3718

F MM/RB

Eo RB/STB

Eg RB/SCF

F/Eo MM/STB

Eg/Eo

5.8073 10.6714 17.3017 24.0940 31.8981 41.1301

0.014560 0.028700 0.046950 0.067730 0.093650 0.120700

0.071902299 0.129424138 0.201326437 0.287609195 0.373891954 0.474555172

398.8534135 371.8272962 368.5128136 355.7353276 340.6099594 340.7626678

4.938344701 4.509551844 4.28810302 4.246407728 3.992439445 3.931691569

The plot of F/Eo versus Eg/Eo is shown next: Chart Title 420

400

F/Eo

380

360

340 y = 58.83x + 108.7 320

300 3.8

4

4.2

4.4

4.6

4.8

5

5.2

Eg/Eo

Figure 6: F/Eo vs. Eg/Eo Plot The best fit is expressed by: Eg F = 108.7 + 58.83 Eo Eo


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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.

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?

Pressure psia 3640 3585 3530 3460 3385 3300 3200 Bg bbl/SCF 0.000892 0.000905 0.000918 0.000936 0.000957 0.000982 0.001014


pi = Cf = Cw = Swi = Bw = m=

3640 0.000004 0.000003 0.25 1.025 0

psia psia-1 psia-1

Np MM STB 0 0.79 1.21 1.54 2.08 2.58 3.4

Gp MM SCF 0 4.12 5.68 7 8.41 9.71 11.62

Bt BBL/STB 1.464 1.469 1.476 1.482 1.491 1.501 1.519

We MM BBL 0 48.81 61.187 71.32 80.293 87.564 93.211

Wp MM STB 0 0.08 0.26 0.41 0.6 0.92 1.38

psia

Rp SCF/STB 0 5.215189873 4.694214876 4.545454545 4.043269231 3.763565891 3.417647059

Rso SCF/STB 888 874 860 846 825 804 779 F MM/RB 0.6114 1.0713 1.4291 1.9567 2.5753 3.5294

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

Eg RB/SCF

Ef,w RB/SCF

D

F/D MM/STB

We/D

0.005000 0.012000 0.018000 0.027000 0.037000 0.055000

0.021336323 0.042672646 0.072215247 0.106681614 0.147713004 0.200233184

0.00050996 0.00101992 0.00166896 0.00236436 0.00315248 0.00407968

0.00550996 0.01301992 0.01966896 0.02936436 0.04015248 0.05907968

110.9559779 82.28173445 72.65677901 66.63557762 64.1383531 59.739895

8858.50351 4699.491241 3626.017847 2734.369147 2180.786841 1577.716738

The plot of F/D versus We/D is shown next. The best fit is expressed by: W F = 0.0071 + 48.067e6 e D D

Therefore, N = 48 MM STB and m = 0.0071. This concludes the solution of this problem. Chart Title 119

109

F/Eo

99

89

79

69 y = 0.0071x + 48.067 59 0

1000

2000

3000

4000

5000

6000

7000

8000

9000

10000

Eg/Eo

Figure 7: F/Eo vs. Eg/Eo Plot © 2003-2004 Petrobjects www.petrobjects.com

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Reserve Estimation Methods

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