Reserve Estimation Methods

© 2003-2004 Petrobjects 1www.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. © 2003-2004 Petrobjects 2 www.petrobjects.com Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects 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 © 2003-2004 Petrobjects 3 www.petrobjects.com Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects 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 © 2003-2004 Petrobjects 4 www.petrobjects.com Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects 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. © 2003-2004 Petrobjects 1 www.petrobjects.com Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects 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, ( )( ) ( ) ( ) RF t B t S = BAF o o φ 1 1 7758 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. © 2003-2004 Petrobjects 1 www.petrobjects.com Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects 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: ( ) RF t N EUR = 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: ( ) ( ) t B t S V = N(t) o o b φ Where: N(t) = oil in place at time t, STB V b = bulk reservoir volume, RB = 7758 A h 7758 = RB/acre-ft A = reservoir area, acres © 2003-2004 Petrobjects 2 www.petrobjects.com Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www.petrobjects.com H = average reservoir thickness, ft φ = average reservoir porosity, fraction S o (t) = average oil saturation, fraction B o (p) = oil formation volume factor at reservoir pressure p, RB/STB Similarly, for a gas reservoir, the volumetric method is given by: ( ) RF t G EUR = 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: ( ) ( ) t B t S V = G(t) g g b φ Where: G(t) = gas in place at time t, SCF V b = 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 S g (t) = average gas saturation, fraction B g (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 S wi = 45% Initial reservoir pressure, p i = 2980 psia Abandonment pressure, p a = 300 psia B o at p i = 1.68 bbl/STB B o at p a = 1.15 bbl/STB © 2003-2004 Petrobjects 3 www.petrobjects.com Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www.petrobjects.com S g at p a = 34% S or 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: V b = 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 = N oi wi b i − 1 φ this yields: ( ) STB MM 266 1.68 5 0. (0.08) 10 x 10.15 = N 9 i ≈ − 4 1 2. The oil in place after volumetric depletion to abandonment pressure is given by: ( ) B S - S - 1 V = N o g w b φ this yields: ( ) STB MM 148 1.15 0.34) - 0.45 - 1 (0.08) 10 x 10.15 = N 9 1 ≈ 3. The oil in place after water invasion at initial reservoir pressure is given by: © 2003-2004 Petrobjects 4 www.petrobjects.com Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www.petrobjects.com B S V = N o or b φ this yields: STB MM 97 1.68 0.20 (0.08) 10 x 10.15 = N 9 2 ≈ 4. The oil reserve by volumetric depletion: ( ) ( ) STB MM 118 = 10 x 148 - 266 = N - N 6 1 i i.e. RF = 118/266 = 44% 5. The oil reserve by full water drive ( ) ( ) STB MM 169 = 10 x 97 - 266 = N - N 6 2 i 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 S g is the gas saturation and B o the oil formation volume factor at abandonment pressure, then oil in place at abandonment pressure is given by: ( ) B S - S - 1 V = N o g w b φ 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: B S V = N o or b φ This concludes the solution. © 2003-2004 Petrobjects 5 www.petrobjects.com Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects 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 ft 3 /SCF Gas FVF at 2500 psia = 0.00667 ft 3 /SCF Gas FVF at 500 psia = 0.03623 ft 3 /SCF S gr 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: V b = 43,560 x A x h = 43,560 x 160 x 40 = 278.784 MM ft 3 1. Initial gas in place is given by: ( ) B S - 1 V = G gi wi i b i φ this yields: © 2003-2004 Petrobjects 6 www.petrobjects.com Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www.petrobjects.com ( ) SCF MM 8860 = 0.00533 0.23) - 1 (0.22) 10 x 278.784 = G 6 i 2. Gas in place after volumetric depletion to 2500 psia: ( ) SCF MM 7080 = 0.00667 0.23) - 1 (0.22) 10 x 278.784 = G 6 1 3. Gas in place after volumetric depletion to 500 psia: ( ) SCF MM 1303 = 0.003623 0.23) - 1 (0.22) 10 x 278.784 = G 6 2 4. Gas in place after water invasion at 3250 psia: SCF MM 3912 = 0.00533 (0.34) (0.22) 10 x 278.784 = G 6 3 5. Gas in place after water invasion at 2500 psia: SCF MM 3126 = 0.00667 (0.34) (0.22) 10 x 278.784 = G 6 4 6. Gas in place after water invasion at 500 psia: SCF MM 576 = 0.03623 (0.34) (0.22) 10 x 278.784 = G 6 5 7. Gas reserve by volumetric depletion to 500 psia: ( ) SCF MM 7557 = 10 x 1303 - 8860 = G - G 6 2 i i.e. RF = 7557/8860 = 85% 8. Gas reserve by water drive at 3250 psia (full water drive): ( ) SCF MM 4948 = 10 x 3912 - 8860 = G - G 6 3 i © 2003-2004 Petrobjects 7 www.petrobjects.com Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www.petrobjects.com i.e. RF = 4948/8860 = 56% 9. Gas reserve by water drive at 2500 psia (partial water drive): ( ) SCF MM 5734 = 10 x 3126 - 8860 = G - G 6 4 i i.e. RF = 5734/8860 = 65% 10. Gas reserve by water drive at 3250 psia if there is one un-dip well: ( ) ( ) SCF MM 2474 = 10 x 3912 - 8860 2 1 = G - G 2 1 6 3 i 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: © 2003-2004 Petrobjects 8 www.petrobjects.com Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects 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. This concludes the solution. © 2003-2004 Petrobjects 1 www.petrobjects.com Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects 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. © 2003-2004 Petrobjects 2 www.petrobjects.com Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects 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: Dt i e q q − = Where: q = well’s production rate at time t, STB/day q i = 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 i e q q − = Cumulative Oil Production D q q N i p − = Nominal Decline Rate ( ) e D D − − = 1 ln i i e q q q D − = Effective Decline Rate D e e D − − =1 Life ( ) D q q t 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 © 2003-2004 Petrobjects 3 www.petrobjects.com Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www.petrobjects.com months and after 22.5 months. Also predict the amount of oil produced after one year. Solution: month t BOPD q BOPD q i 1 96 100 = = = 1. Calculate the effective decline rate per month: month q q q D i i e / 04 . 0 100 96 100 = − = − = 2. Calculate the nominal decline rate per month: ( ) ( ) ( ) onth 0.040822/m 96 . 0 ln 04 . 0 1 ln 1 ln = − = − − = − − = e D D 3. Calculate the rate after 11 more months: ( ) BOPD e e q q x Dt i 27 . 61 100 12 040822 . 0 = = = − − 4. Calculate the rate after 22.5 months: ( ) POPD e e q q x Dt i 91 . 39 100 5 . 22 040822 . 0 = = = − − 5. Calculate the nominal decline rate per year: ar 0.48986/ye 12 x onth 0.040822/m = = D 6. Calculate the cumulative oil produced after one year: ( ) STB Y D Y D STB D q q N i p 858 , 28 / 365 * / 489864 . 0 / 27 . 61 100 = − = − = This completes the solution. © 2003-2004 Petrobjects 4 www.petrobjects.com Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects 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: ( ) b i i t bD q q 1 1 − + = Where: q = well’s production rate at time t, STB/day q i = well’s production rate at time 0, STB/day D i = 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 ( ) b i i t bD q q 1 1 − + = Cumulative Oil Production ( ) ( ) b b i i b i p q q b D q N − − − − = 1 1 1 Nominal Decline Rate ( ) [ ] 1 1 1 − − = −b ei i D b D i i ei q q q D − = Effective Decline Rate D e e D − − =1 Life ( ) i b i bD q q t 1 / − = Example #4: Given the following data: 9 . 0 / 5 . 0 100 = = = b year D BOPD q i i © 2003-2004 Petrobjects 5 www.petrobjects.com Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects 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: ( ) ( )( ) ( ) BOPD t xt x t bD q q b i i 9 . 0 1 9 . 0 1 1 45 . 0 1 100 5 . 0 9 . 0 1 100 1 − − − + = + = + = 2. Calculate the cumulative oil produced at the end of each year for five years using: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 . 0 9 . 0 1 9 . 0 1 9 . 0 1 1 1.584893 * 460598.9 365 100 9 . 0 1 5 . 0 100 1 q year days q q q b D q N b b i i b i 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: © 2003-2004 Petrobjects 6 www.petrobjects.com Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www.petrobjects.com Harmonic Decline b = 1 Description Equation Rate t bD q q i i + = 1 Cumulative Oil Production q q D q N i i i p ln = Nominal Decline Rate ei ei i D D D − = 1 Effective Decline Rate i i ei q q q D − = Life ( ) i i D q q t 1 / − = © 2003-2004 Petrobjects 1 www.petrobjects.com Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects 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 wi t ti g gi ti gi t wi f Iw Ig ti gi e I I t wi t p soi g w p p p C S N - + G - + + p NB GB B B B B 1 - S C + + + + + p NB GB W W G B B 1 - S = + - + N N W B R R B B | | ∆ | \ . | | ∆ | \ . Where: N = initial oil in place, STB N p = cumulative oil produced, STB G = initial gas in place, SCF G I = cumulative gas injected into reservoir, SCF G p = cumulative gas produced, SCF W e = water influx into reservoir, bbl W I = cumulative water injected into reservoir, STB W p = cumulative water produced, STB B ti = initial two-phase formation volume factor, bbl/STB = B oi B oi = initial oil formation volume factor, bbl/STB B gi = initial gas formation volume factor, bbl/SCF B t = two-phase formation volume factor, bbl/STB = B o + (R soi - R so ) B g B o = oil formation volume factor, bbl/STB B g = gas formation volume factor, bbl/SCF B w = water formation volume factor, bbl/STB B Ig = injected gas formation volume factor, bbl/SCF © 2003-2004 Petrobjects 2 www.petrobjects.com Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www.petrobjects.com B Iw = injected water formation volume factor, bbl/STB R soi = initial solution gas-oil ratio, SCF/STB R so = solution gas-oil ratio, SCF/STB R p = cumulative produced gas-oil ratio, SCF/STB C f = formation compressibility, psia -1 C w = water isothermal compressibility, psia -1 S wi = initial water saturation ∆p t = reservoir pressure drop, psia = p i - 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 wi t ti g gi ti gi t wi Iw Ig e I I t p soi g w p p p + C C S N - + G - + + p NB GB B B B B 1 - S + + + W W G B B = + - + N N W B 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 GB m = NB and plugging into the equation yields: ( ) ( ) ( ) ( ) f w wi ti t ti g gi ti ti t gi wi Iw Ig e I I t p soi g w p p p + C C S B N - + Nm - + + Nm p NB B B B B B 1 - S B + + + W W G B B = + - + N N W B R R B B | | ∆ | \ . © 2003-2004 Petrobjects 3 www.petrobjects.com Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www.petrobjects.com Let: ( ) ( ) ( ) 1 o t ti ti g g gi gi f w wi f,w ti t wi t p soi g w Iw Ig p p I I = - E B B B = - E B B B + C C S = m B p E 1 - S F = + - + - - N W W G B R R B B B B | | + ∆ | \ . ( ¸ ¸ Thus we obtain: ( ) , , e o 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 o F= NE 2. Negligible compressibilities, and no water influx o g g o o F= NE NmE E F = N Nm E E + + Which is written as y = b + x. This would suggest that a plot of F/E o as the y coordinate versus E g /E o 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: © 2003-2004 Petrobjects 4 www.petrobjects.com Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www.petrobjects.com , o g f w D E mE E = + + Dividing through by D, we get: e F 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 W e /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 g e p p N - + G - + - = + - W W N B B B B B B R R B ( ¸ ¸ Dividing through by the right hand side of the equation yields: ( ) ( ) ( ) ( ) ( ) ( ) w g gi e p t ti t p soi g t p soi g t p soi g p p p - G - N - W W B B B B B + + = 1 + - + - + - N N N B 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. © 2003-2004 Petrobjects 5 www.petrobjects.com Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects 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 ft 3 /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 ft 3 /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: ( ) ( ) ( ) | | ( ) B W - W - B R - R + B N = B - B G + B - B N w p e g soi p t p gi g ti t Define the ratio of the initial gas cap volume to the initial oil volume as: NB GB = m ti gi we get: ( ) ( ) ( ) | | ( ) B W - W - B R - R + B N = B - B B B Nm + B - B N w p e g soi p t p gi g gi ti ti t © 2003-2004 Petrobjects 6 www.petrobjects.com Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www.petrobjects.com and solve for N, we get: ( ) | | ( ) ( ) ( ) B - B B B m + B - B B W - W - B R - R + B N = N gi g gi ti ti t w p e g soi p t p Since: N p = 20 x 10 6 STB B t = 1.4954 bbl/STB R p = 700 SCF/STB R soi = 562 SCF/STB B g = 0.008479 ft 3 /SCF = 0.008479/5.6146 = 0.001510 bbl/SCF W e = 11.58 x 10 6 bbl W p = 1.05 x 10 6 STB B w = 1.028 bbl/STB B ti = 1.34 bbl/STB m = GB gi /NB ti = 19,600/112,000 = 0.175 B gi = 0.006266 ft 3 /SCF = 0.006266/5.6146 = 0.001116 bbl/SCF Thus: ( ) ( ) ( ) ( ) 6 20 1.4954 + 700 - 562 0.001510 - 11.58 - 1.05x1.028 N = 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: ( ) ( ) ( ) ( ) ( ) ( ) w g gi e p t ti t p soi g t p soi g t p soi g p p p - G - N - W W B B B B B + + = 1 + - + - + - N N N B R R B B R R B B R R B ( ( ( ¸ ¸ ¸ ¸ ¸ ¸ Thus the depletion drive index (DDI) is given by: ( ) ( ) ( ) ( ) 6 t ti 6 t p soi g p N - 98.97x 1.4954 - 1.34 10 B B = = 0.45 20x 1.4954 + 700 - 562 0.001510 + - 10 N B R R B ( ( ¸ ¸ ¸ ¸ The segregation drive index (SDI) is given by: © 2003-2004 Petrobjects 7 www.petrobjects.com Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www.petrobjects.com ( ) ( ) ( ) ( ) ti g gi gi t p soi g p 6 6 B Nm - B B B = + - N B R R B 1.34 98.97 x x 0.175x 0.001510 - 0.001116 10 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 10 20x 10 x 1.028 x 1.05 - 10 x 11.58 = B R - R + B N B W - W 6 6 6 g soi p t p w p e 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 10 6 STB. The cumulative oil production N p and cumulative gas oil ratio R p are listed in the following table as functions of the average reservoir pressure over the first few years of production. Assume that p i = p b = 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 © 2003-2004 Petrobjects 8 www.petrobjects.com Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www.petrobjects.com Solution: Calculate the parameters F, E o , E g 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/E o versus E g /E o 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 / E o Figure 6: F/Eo vs. Eg/Eo Plot The best fit is expressed by: 108.7 58.83 g o o E F = E E + © 2003-2004 Petrobjects 9 www.petrobjects.com Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects 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 10 6 STB. The cumulative oil production N p and cumulative gas oil ratio R p 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 p i = p b = 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 © 2003-2004 Petrobjects 10 www.petrobjects.com Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www.petrobjects.com Solution: Calculate the parameters F, E o , E g , E f,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 W e /D is shown next. The best fit is expressed by: 0.0071 48.067 6 e W F = e D 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 / E o Figure 7: F/Eo vs. Eg/Eo Plot Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects 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. 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 2 including the most likely and maximum values for the parameters. however. In such methods. one can not back calculate the input parameters associated with reserves. all input parameters are exactly known. In these methods.Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www.com 3 . Only the end result is known but not the exact value of any input parameter. Figure 1: Measures of central tendency Figure 3: Percentiles The probabilistic methods have several inherent problems. deterministic methods calculate reserve values that are more tangible and explainable. They are affected by all input parameters. the median (middle value). they may sometimes ignore the variability © 2003-2004 Petrobjects www. the mean (average value).com the resulting statistical calculations.petrobjects. the standard deviation and the percentiles. such as the minimum and maximum values. On the other hand.petrobjects. see Figures 2 and 3. the mode (most likely value). com 4 . then confidence on the calculated reserves is increased. the assumptions need to be reexamined. reserves are calculated both deterministically and probabilistically and the two values are compared.Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www. © 2003-2004 Petrobjects www. If the two values are away different. i.com and uncertainty in the input data compared to the probabilistic methods which allow the incorporation of more variance in the data. however.petrobjects. can provide quality assurance for estimating hydrocarbon reserves. If the two values agree.petrobjects.e. A comparison of the deterministic and probabilistic methods. care should be taken to make sure that the field or well being used for analogy is indeed analogous. the BAF. which is calculated by the following equation.com Analogy The analogy method is applied by comparing the following factors for the analogous and current fields or wells: Recovery Factor (RF). if your calculated EUR is twice as high as the EUR from the nearest 100 wells. however. a dolomite reservoir with volatile crude oil will never be analogous to a sandstone reservoir with black oil. Analogy is most useful when running the economics on a yet-to-be-drilled exploratory well. should be taken when applying analogy technique. It is also useful when calculating proved developed reserves. Barrels per Acre-Foot (BAF). Similarly. Comparing EUR’s is done during the exploratory phase. Similarly. Care. 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. That said.petrobjects. and Estimated Ultimate Recovery (EUR) The RF of a close-to-abandonment analogous field is taken as an approximate value for another field. © 2003-2004 Petrobjects www.petrobjects. For example.com 1 . you had better check your assumptions.Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www. the pore volume within the rock matrix. STB bulk reservoir volume. when combined.Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www. Figure 4 is a typical geological net pay isopach map that is often used in the volumetric method. This provides an estimate of the hydrocarbons-in-place. and RF is the recovery factor. acres 1 Vbφ S o (t ) Bo (t ) © 2003-2004 Petrobjects www. Figure 4: A typical geological net pay isopach map The estimated ultimate recovery (EUR) of an oil reservoir. STB. Each of the factors used in the calculation have inherent uncertainties that.petrobjects. fraction. from which ultimate recovery can be estimated by using an appropriate recovery factor.com . is given by: EUR = N (t ) RF Where N(t) is the oil in place at time t. The volumetric method for calculating the amount of oil in place (N) is given by the following equation: N(t) = Where: N(t) Vb 7758 A = = = = oil in place at time t. and the fluid content within the void space. RB = 7758 A h RB/acre-ft reservoir area. STB.petrobjects.com Volumetric The volumetric method entails determining the physical size of the reservoir. cause significant uncertainties in the reserves estimate. fraction gas formation volume factor at reservoir pressure p. SCF. for a gas reservoir.com = = = = = = = = 26. pa Bo at pi Bo at pa © 2003-2004 Petrobjects www. acres average reservoir thickness. CF/SCF Note that the reservoir area (A) and the recovery factor (RF) are often subject to large errors. pi Abandonment pressure. CF = 43560 A h CF/acre-ft reservoir area. ft average reservoir porosity.petrobjects. fraction average gas saturation. ft average reservoir porosity.700 acres 49 ft 8% 45% 2980 psia 300 psia 1. and RF is the recovery factor. fraction average oil saturation. RB/STB Similarly. They are usually determined from analogy or correlations.68 bbl/STB 1.com H φ So(t) Bo(p) = = = = average reservoir thickness. fraction oil formation volume factor at reservoir pressure p. 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. Example #1: Given the following data for the Hout oil field in Saudi Arabia Area Net productive thickness Porosity Average Swi Initial reservoir pressure.Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www. The following examples should clarify the errors that creep in during the calculations of oil and gas reserves. the volumetric method is given by: EUR = G (t ) RF Where G(t) is the gas in place at time t. fraction.15 bbl/STB 2 .petrobjects. 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. 5.0.15 MMM bbl 1. 4.45 . The oil in place after water invasion at initial reservoir pressure is given by: © 2003-2004 Petrobjects www.45 ) ≈ 266 MM STB 1.68 V b φ (1 − S wi ) Boi 2. The initial oil in place is given by: Ni= this yields: Ni= 10.08) (1 .petrobjects.petrobjects.700 x 49 = 10.Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www.com 3 .15 3.0.08) (1 − 0.S w .com Sg at pa Sor after water invasion = 34% = 20% The following quantities will be calculated: 1. The oil in place after volumetric depletion to abandonment pressure is given by: N= this yields: N1= V b φ (1 .S g ) Bo 10.15 x 10 9 (0.34)) ≈ 148 MM STB 1. 2.15 x 109 (0. 3. 6. Discussion of results: For oil reservoirs under volumetric control. water influx is upward where the producing oil zone is underlain by water. water influx is either edge-water drive or bottom-water drive. where there is no appreciable decline in reservoir pressure. In bottom-water drive. If Sg is the gas saturation and Bo the oil formation volume factor at abandonment pressure.08) 0. The oil reserve by volumetric depletion: (N i .148 ) x 106 = 118 MM STB i.N 1) = (266 .e.S g ) Bo On the other hand.20 ≈ 97 MM STB 1. RF = 169/266 = 64% 6.com 4 . In edge-water drive. RF = 118/266 = 44% 5.e. the oil remaining at abandonment is given by: φ N = V b S or Bo This concludes the solution.68 4. In this case.petrobjects. for oil reservoirs under hydraulic control. the produced oil must be replaced by gas the saturation of which increases as oil saturation decreases.Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www. no water influx.15 x 109 (0.97 ) x 106 = 169 MM STB i.petrobjects. The oil reserve by full water drive (N i .com φ N = V b S or Bo this yields: N2= 10.e. water influx is inward and parallel to bedding planes. i.N 2 ) = (266 . then oil in place at abandonment pressure is given by: N= V b φ (1 . © 2003-2004 Petrobjects www.S w . Discussion of results Solution: Let’s start by calculating the reservoir bulk volume: Vb = 43.com 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. Gas reserve by full water drive if there is one un-dip well 11.e.Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www. Gas in place after water invasion at 3250 psia 5.petrobjects. Gas in place after volumetric depletion to 500 psia 4. Initial gas in place 2. Gas reserve by volumetric depletion to 500 psia 8.560 x 160 x 40 = 278. Gas in place after water invasion at 2500 psia 6. at 3250 psia 9.560 x A x h = 43. i. Gas reserve by partial water drive.784 MM ft3 1. i.e. Gas reserve by full water drive.00667 ft3/SCF 0. Gas in place after volumetric depletion to 2500 psia 3.S wi ) B gi 5 .com V b φ i (1 .03623 ft3/SCF 34% The following quantities will be calculated: 1. Gas in place after water invasion at 500 psia 7. at 2500 psia 10.petrobjects. Initial gas in place is given by: Gi = this yields: © 2003-2004 Petrobjects www. Gas in place after water invasion at 3250 psia: 278.00667 3.23)) = 8860 MM SCF 0. Gas in place after water invasion at 500 psia: G5 = 278.23)) = 1303 MM SCF 0. Gas reserve by volumetric depletion to 500 psia: 6 Gi .0.784 x 106 (0.00533 5. Gas in place after volumetric depletion to 500 psia: 278.22) (1 . Gas in place after water invasion at 2500 psia: G4 = 278.22) (1 .784 x 106 (0.G 3 = (8860 .22) (1 .34) = 576 MM SCF 0.784 x 106 (0. RF = 7557/8860 = 85% 8.784 x 106 (0.petrobjects.003623 G2 = 4.e.com 6 . Gas in place after volumetric depletion to 2500 psia: G1 = 278.0.3912 ) x 10 = 4948 MM SCF © 2003-2004 Petrobjects www.G 2 = (8860 .22) (0.22) (0.0.34) = 3912 MM SCF G3 = 0.23)) = 7080 MM SCF 0.com Gi = 278.784 x 106 (0. Gas reserve by water drive at 3250 psia (full water drive): 6 Gi .03623 7.34) = 3126 MM SCF 0.00667 6.00533 2.petrobjects.784 x 106 (0.1303) x 10 = 7557 MM SCF i.Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www.22) (0. On the other hand. however. are lower because the main mechanism of production in gas reservoirs is depletion or gas expansion. and they found that the ultimate recovery increased from 69 to 74%. RF = 4948/8860 = 56% 9. 2. Recoveries.G 4 = (8860 . This leads to higher abandonment pressures for water-drive reservoirs. In this process.Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www.e. In water-drive gas reservoirs.petrobjects.G3 ) = 1 (8860 . Gas reserve by water drive at 3250 psia if there is one un-dip well: 1 (Gi . This can be explained as follows: As water invades the reservoir.3912) x 106 = 2474 MM SCF 2 2 i. the reservoir pressure is maintained at a higher level than if there were no water encroachment.petrobjects.com 7 . RF = 5734/8860 = 65% 10. Gas reserve by water drive at 2500 psia (partial water drive): 6 Gi . it has been found that gas recoveries can be increased by: 1. 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. as down-dip wells begin to be watered out. 1. RF = 2474/8860 = 28% 11. see Fig.e. while the up-dip wells are maintained on gas production. Outrunning technique: This is accomplished by increasing gas production rates. Co-production technique: This technique is defined as the simultaneous production of gas and water. they are converted to high-rate water producers.e. the RF for partial water drive is 65%. Discussion of results The RF for volumetric depletion to 500 psia (no water drive) is calculated to be 85%.com i. and for the full water drive is 56%.3126 ) x 10 = 5734 MM SCF i. This technique enhances production as follows: © 2003-2004 Petrobjects www. com The high-rate down-dip water producers act as a pressure sink for the water. This concludes the solution. © 2003-2004 Petrobjects www.petrobjects. It has been reported that this technique has increased gas production from62% to 83% in Eugene Island Field of Louisiana. This retards water invasion into the gas zone. When the average reservoir pressure is lowered. allowing for more gas expansion and therefore more gas production. the immobile gas in the water-swept portion of the reservoir could become mobile and hence producible.Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www.petrobjects.com 8 . The high-rate water production lowers the average reservoir pressure. therefore prolonging its productive life. Figure 5: Decline curve of an oil well Decline curves are the most common means of forecasting production. With more and more low productivity wells coming on stream. hyperbolic or harmonic decline extrapolations should only be used for these specific cases. there is currently a swing toward decline rates proportional to production rates (hyperbolic and harmonic).com 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. Overexuberance in the use of hyperbolic or harmonic relationships can result in excessive reserves estimates. © 2003-2004 Petrobjects www. the y-axis is logarithmic and the x-axis is linear. When the data plots concave upward. A special case of the hyperbolic decline is known as “harmonic decline”. When the data plots as a straight line.com 1 . Figure 5 is an example of a production graph with exponential and harmonic extrapolations. They are easy to plot. The most common decline curve relationship is the constant percentage decline (exponential). it is modeled with a constant percentage decline “exponential decline”. and They are easy to analyze. They have many advantages: Data is easy to obtain. Although some wells exhibit these trends. They yield results on a time basis.petrobjects.e.petrobjects. The plot is usually done on a semilog paper. i. it is modeled with a “hyperbolic decline”.Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www. if projected. in the exponential decline. 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. will furnish useful knowledge as to the future production of the well. Assuming exponential decline.com 2 .Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www.com If the conditions affecting the rate of production of the well are not changed by outside influences. the curve will be fairly regular. 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.petrobjects. and. STB/day nominal exponential decline rate. 1/day time. the well’s production data plots as a straight line on a semilog paper.petrobjects. day The following table summarizes the equations used in exponential decline. STB/day well’s production rate at time 0. Exponential Decline As mentioned above. predict the rate after 11 more © 2003-2004 Petrobjects www. 858 STB D 0. Also predict the amount of oil produced after one year.91 POPD 5.27 ) STB / D = * 365D / Y = 28. Calculate the rate after 22.5 months: q = qi e − Dt = 100e (−0.040822/month x 12 = 0.040822/month 3. Solution: qi = 100 BOPD q = 96 BOPD t = 1 month 1. Calculate the nominal decline rate per month: D = − ln(1 − De ) = − ln (1 − 0.48986/year 6.489864 / Y This completes the solution. Calculate the effective decline rate per month: De = qi − q 100 − 96 = = 0.petrobjects. Calculate the cumulative oil produced after one year: Np = qi − q (100 − 61. © 2003-2004 Petrobjects www.com 3 . Calculate the rate after 11 more months: q = qi e − Dt = 100e (−0.96) = 0.04 / month qi 100 2.5 months.040822 x12 ) = 61.Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www.petrobjects.5 ) = 39.04 ) = − ln (0.27 BOPD 4.com months and after 22. Calculate the nominal decline rate per year: D = 0.040822 x 22. 5 / year b = 0.com 4 .petrobjects. if the well’s production data plotted on a semilog paper concaves upward. STB/day well’s production rate at time 0.com Hyperbolic Decline Alternatively.9 © 2003-2004 Petrobjects www. 1/day hyperbolic exponent time.petrobjects. day The following table summarizes the equations used in hyperbolic decline: Hyperbolic Decline b > 0.Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www. STB/day initial nominal exponential decline rate (t = 0). The equation of the hyperbolic decline is given by: q = qi (1 + bDi t ) Where: q qi Di b t = = = = = − 1 b well’s production rate at time t. then it is modeled with a hyperbolic decline. b ≠ 1 Description Rate Cumulative Oil Production Equation q = qi (1 + bDi t ) Np = − 1 b Nominal Decline Rate Effective Decline Rate Life qib (qi1−b − q1−b ) Di (1 − b ) 1 −b Di = (1 − Dei ) − 1 b q −q Dei = i qi [ ] De = 1 − e − D (qi / q )b − 1 t= bDi Example #4: Given the following data: qi = 100 BOPD Di = 0. 45t ) − 1 0.724 15. 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 BOPD 2. Solution: 1.692 This completes the solution.5 xt ) − 1 0.248 66. Production 0 29.9 x 0.com 5 .992 Cum.9 * (1.petrobjects. Form the following table: Year 0 1 2 3 4 5 Rate at End of Year 100 66.176 49.9 = 100(1 + 0.009 38.606 Yearly Production 29.524 20.914 89.799 10. predict the amount of oil produced for five years.115 78.9 ) − q (1−0.petrobjects.5(1 − 0.854 26. Calculate the cumulative oil produced at the end of each year for five years using: qib (qi1−b − q1−b ) Np = Di (1 − b ) = = 460598. where b is taken to be equal to 1. The following table summarizes the equations used in harmonic decline: © 2003-2004 Petrobjects www.1 ) (100)0.9 (100(1−0.9 ) year 3. Harmonic Decline A special case of the hyperbolic decline is known as “harmonic decline”.584893 − q 0.Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www.867 12.524 50.com Assuming hyperbolic decline.9 ) ) 365 days 0.699 31. petrobjects.petrobjects.Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www.com 6 .com 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 © 2003-2004 Petrobjects www. bbl/STB = Boi initial oil formation volume factor. STB initial gas in place.B gi ) + ( NB ti + GB gi )  w wi  ∆ p t  1 .Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www.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. bbl/SCF water formation volume factor. SCF cumulative gas injected into reservoir. bbl/STB initial gas formation volume factor. SCF water influx into reservoir. 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.com .S wi  = N p B t + N p ( R p . bbl/STB = Bo + (Rsoi .B ti ) + G ( B g . Although it is a tank model equation.S wi   Cf  + ( NB ti + GB gi )   ∆ p t + W e + W I B Iw + G I B Ig  1 . bbl/STB gas formation volume factor.petrobjects.petrobjects. this can be written as: C S  N ( B t . bbl cumulative water injected into reservoir.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.Rso) Bg oil formation volume factor. bbl/STB injected gas formation volume factor. STB cumulative oil produced. SCF cumulative gas produced. STB cumulative water produced. bbl/SCF two-phase formation volume factor. bbl/SCF 1 © 2003-2004 Petrobjects www. STB initial two-phase formation volume factor. it can provide great insight for the practicing reservoir engineer. B ti ) + Nm +   B ti ( B g . each pressure and the corresponding production data is considered as being a separate point from other pressure values.petrobjects. SCF/STB formation compressibility. The straight-line method begins with the material balance written as:  C f + C w S wi  N ( B t . psia = pi .B gi ) + ( NB ti + Nm B ti )  C f1 . SCF/STB cumulative produced gas-oil ratio.com BIw Rsoi Rso Rp Cf Cw Swi ∆pt p(t) = = = = = = = = = injected water formation volume factor. However.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 ) + G ( B g . SCF/STB solution gas-oil ratio.Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www.petrobjects.R soi ) B g + W p B w © 2003-2004 Petrobjects www.p(t) current reservoir pressure. bbl/STB initial solution gas-oil ratio. when using the material balance equation.B gi ) + ( NB ti + GB gi )   ∆ pt  1 . a calculation is made and the results of these calculations are averaged. psia-1 water isothermal compressibility.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 . psia-1 initial water saturation reservoir pressure drop.com 2 . From each separate point.S wi  + W e + W I B Iw + G I B Ig = N p B t + N p ( R p . psia The MBE as a Straight Line Normally. 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. and no water F= NEo 2.w + W e = N ( Eo + mE g + E f . Negligible compressibilities. 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.com 3 .petrobjects.G I B Ig Thus we obtain: F = NEo + mNE g + NE f . negligible compressibilities.petrobjects. Including compressibilities and water influx.w ) + We The following cases are considered: 1.Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www. 3. influx No gas cap.com Let: E o = B t . let: © 2003-2004 Petrobjects www.B ti Eg= B ti ( B g .B gi ) B gi E f.S wi   F=N p  Bt +  ( R p .w  C f + C w S wi  = (1 + m ) B ti   ∆ pt 1 .R soi ) B g  +W  p Bw -W I B Iw . and no water influx F= NEo + NmE g Eg F = N + Nm Eo Eo Which is written as y = b + x. com 4 . we write: DDI + SDI + WDI = 1 The following examples should clarify the errors that creep in during the calculations of oil and gas reserves. and the water drive index (WDI) respectively.Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www. 2.w Dividing through by D. Segregation drive (gas zone gas expansion).R soi ) B g    (W e . Water drive (water zone water influx). using Pirson's abbreviations.R soi ) B g    The terms on the left hand side of equation (3) represent the depletion drive index (DDI).petrobjects.B gi ) + N p  B t + ( R p . we get: F W =N+ e D D Which is written as y = b + x.B ti ) + G ( B g .B ti ) + G ( B g .R soi ) B g    N p  B t + ( R p .petrobjects. © 2003-2004 Petrobjects www.W p B w ) = N p  B t + ( R p . the compressibility term in the material balance equation is neglected and the equation is rearranged as follows: N ( B t .R soi ) B g    Dividing through by the right hand side of the equation yields: N ( B t . Depletion drive (oil zone oil expansion).W p B w ) =1 N p  B t + ( R p . 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.B gi ) + (W e . Thus. and 3.com D = Eo + mE g + E f . the segregation drive index (SDI). Drive Indexes from the MBE The three major driving mechanisms are: 1. To determine the relative magnitude of each of these driving mechanisms. 600 acre-ft 2710 psia 1.58 MM bbl 1.W p B w ) B gi © 2003-2004 Petrobjects www.Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www.Bti ) + Nm Bti (B g . Discussion of results.com 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 The following values will be calculated: 1.W p B w ) Define the ratio of the initial gas cap volume to the initial oil volume as: m= we get: GB gi NBti N (Bt .05 MM STB 1.006266 ft3/SCF 562 SCF/STB 20 MM STB 2000 psia 700 SCF/STB 1.R soi ) B g ] .028 bbl/STB 0.008479 ft3/SCF Solution: 1. The material balance equation is written as: N (Bt .petrobjects.(W e .B gi ) = N p [ Bt + (R p .com 5 .000 acre-ft 19.340 bbl/STB 0. 2. The driving indexes.4954 bbl/STB 11. The stock tank oil initially in place.B gi ) = N p [ Bt + (R p .R soi ) B g ] . = = = = = = = = = = = = = = 112. 3.Bti ) + G (B g .petrobjects.(W e . 001116 = 98.4954 + (700 .008479 ft3/SCF = 0.1.W p B w ) (Bt .175 (0. we get: N= N p [ Bt + (R p .4954 .R soi ) B g    G ( B g .34 ) = 0.001510    98. In terms of drive indexes.4954 .B gi ) B gi Since: Np Bt Rp Rsoi= Bg We Wp Bw Bti m Bgi Thus: N= 20  1.petrobjects.6146 = 0. the material balance equation is written as: N p  B t + ( R p .001510 bbl/SCF = 11.34 ) + 0.001510 .com 6 .58 .006266/5.(W e .028 bbl/STB = 1.petrobjects.006266 ft3/SCF = 0.562 ) 0.Bti ) + m Bti (B g .175 = 0.97 MM STB = 20 x 106 STB = 1.05 x 106 STB = 1.( 11.000 = 0.B ti ) + N p  B t + ( R p .028 ) 6   10 1.R soi ) B g ] .4954 bbl/STB = 700 SCF/STB 562 SCF/STB = 0.600/112.R soi ) B g    Thus the depletion drive index (DDI) is given by: N p  B t + ( R p .58 x 10 6 bbl = 1.0.6146 = 0.W p Bw ) =1 N p  B t + ( R p .001116 ) 0.B ti ) = 20x10 6 1.com and solve for N.34 (1.008479/5.001116 bbl/SCF 2.1.001510  .4954 + ( 700 .R soi ) B g    N ( B t .R soi ) B g    N ( B t .Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www.34 bbl/STB = GBgi/NBti = 19.97x10 6 ( 1.45 The segregation drive index (SDI) is given by: © 2003-2004 Petrobjects www.1.562 ) 0.05x1.B gi ) + ( W e . 2353 1.001116 ) 0.2511 1.503 14.31 6 N p [ Bt + (R p .com Nm B ti ( B g .1.001510    The water drive index (WDI) is given by: ( W e . The size of the gas cap is uncertain with the best estimate.1922 1.24 20x10 6 1.001116 = 0.1822 Rso SCF/STB 510 477 450 425 401 375 352 Bg bbl/SCF 0.00087 0.petrobjects.R soi ) B g  Np  98.4.com 7 .028 x 106 ) = 0.00101 0.175x 1.73 Rp SCF/STB 0 1050 1060 1160 1235 1265 1300 Bo BBL/STB 1. This concludes the solution. 31% was by water drive.4954 + (700 . The drive mechanisms as calculated in part (2) indicate that when the reservoir pressure has declined from 2710 psia to 2000 psia.00113 0.513 17. Example #6: Given the following data for an oil field A gas cap reservoir is estimated.B gi ) B gi =  B t + ( R p . giving the value of m = 0. what is the correct value of m? Pressure psia 3330 3150 3000 2850 2700 2550 2400 Np MM STB 0 3.34 ( 0. based on geological information.562 ) 0.97 x 10 6 x 0.00092 0.petrobjects.2122 1.58 x 106 .W p Bw ) = ( 11. 45% of the total production was by oil expansion. to have an initial oil volume N of 115 x 106 STB.00107 0.001510 .001510 ] 3.2222 1.00096 0.0.R soi ) B g ] 20x 10 [1.852 11.00120 © 2003-2004 Petrobjects www.295 5.562 ) 0.Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www. Is this figure confirmed by the production and pressure history? If not. 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.05 x 1.4954 + (700 . Assume that pi = pb = 3330 psia. from volumetric calculations. and 24% was by gas cap expansion.903 8.2022 1. 29805 1.509551844 4.246407728 3.201326437 0.83x + 108.2 4.7 + 58.8 5 5.6099594 340.8 4 4.com Solution: Calculate the parameters F.5128136 355.067730 0.8073 10.6714 17.petrobjects.373891954 0.31883 1.28810302 4.046950 0.Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www. Eo.7626678 Eg/Eo 4.2511 1.34475 1.014560 0.7 320 300 3.4 Eg/Eo 4.26566 1.129424138 0.071902299 0.0940 31.093650 0.3017 24.2798 1.petrobjects.3718 F MM/RB 5.938344701 4.992439445 3. Eg as given by the above equations: Bt BBL/STB 1. Eg/Eo Plot The best fit is expressed by: Eg F = 108.120700 Eg RB/SCF 0.7353276 340.83 Eo Eo © 2003-2004 Petrobjects www.474555172 F/Eo MM/STB 398.1301 Eo RB/STB 0.8534135 371.028700 0.8981 41.2 Figure 6: F/Eo vs.6 4.8272962 368.931691569 The plot of F/Eo versus Eg/Eo is shown next: Chart Title 420 400 380 F/Eo 360 340 y = 58.com 8 .287609195 0. 000918 0.000957 0. to have an initial oil volume N of 47 x 106 STB.26 0.com Therefore.41 9.32 80.08 2. giving the value of m = 0.25 1.215189873 4. based on geological information.476 1. The size of the gas cap is uncertain with the best estimate.08 0.petrobjects.6114 1. from volumetric calculations. N = 108.043269231 3.21 1.187 71.000936 0.763565891 3.000905 0.41 0.7 MM STB and m = 58.491 1.4 We MM BBL 0 48.001014 Np MM STB 0 0.9567 2.62 Wp MM STB 0 0.025 0 Gp MM SCF 0 4.com 9 .6 0. what is the correct value of m? pi = Cf = Cw = Swi = Bw = m= Pressure psia 3640 3585 3530 3460 3385 3300 3200 Bg bbl/SCF 0.545454545 4.0713 1.71 11.000003 0.92 1.5753 3. Assume pi = pb = 3640 psia.000892 0.564 93. Other pertinent data are also supplied.58 3.38 psia psia-1 psia-1 psia Bt BBL/STB 1.519 Rp SCF/STB 0 5. 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.83/108.Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www. Example #7: Given the following data for an oil field A gas cap reservoir is estimated.417647059 Rso SCF/STB 888 874 860 846 825 804 779 F MM/RB 0.81 61.12 5.469 1.79 1.000004 0.482 1.7 = 0.4291 1.54 2.464 1.0.68 7 8.5294 © 2003-2004 Petrobjects www.694214876 4. Is this figure confirmed by the production history? If not. This concludes the solution of this problem.211 3640 0.petrobjects.293 87.501 1.54.000982 0. 027000 0.147713004 0.00236436 0.00166896 0. The best fit is expressed by: W F = 0.042672646 0.021336323 0.00101992 0.369147 2180. This concludes the solution of this problem. Eg/Eo Plot © 2003-2004 Petrobjects www.9559779 82.28173445 72. N = 48 MM STB and m = 0.01966896 0.012000 0.055000 Eg RB/SCF 0.00407968 D F/D MM/STB 110.02936436 0.05907968 8858.005000 0.50351 4699.1383531 59.018000 0.00315248 0. Eo. Ef. Eg.067e6 e D D Therefore.petrobjects.200233184 Ef. as given by the above equations: Eo RB/STB 0.00550996 0.037000 0.01301992 0.65677901 66.0071 + 48.106681614 0.072215247 0.com Solution: Calculate the parameters F.017847 2734.Petroleum Reserves Estimation Methods © 2003-2004 Petrobjects www.491241 3626.w.716738 The plot of F/D versus We/D is shown next. and D.739895 We/D 0.00050996 0.0071x + 48.04015248 0.63557762 64.0071.com 10 .786841 1577. Chart Title 119 109 99 F/Eo 89 79 69 y = 0.067 59 0 1000 2000 3000 4000 5000 Eg/Eo 6000 7000 8000 9000 10000 Figure 7: F/Eo vs.petrobjects.w RB/SCF 0.