MONTANUNIVERSITÄT LEOBEN PETROLEUM ENGINEERING DEPARTMENT TEXTBOOK SERIES VOLUME 3 PETROLEUM RECOVERY

MONTANUNIVERSITÄT LEOBEN PETROLEUM ENGINEERING DEPARTMENT TEXTBOOK SERIES VOLUME 3 PETROLEUM RECOVERY by Zoltán E. HEINEMANN Professor for Reservoir Engineering Leoben, January 23

No part of this publication may be reproduced in any form. Only students of the University of Leoben may copy for studying purposes.

PREFACE This volume is the third within the series of textbooks written by Z.E.Heinemann. The previous volumes: Flow in Porous Media Reservoir Fluids deal with the properties of porous media, the one and two phase flow in reservoir rocks, phase behavior and thermodynamical properties of oil, gas and brine. To use this volume it is supposed that the reader has attended the courses mentioned above or possesses profound knowledge of these topics. All methods discussed in this volume have practical importance and are used in today work. However, it is not possible to give a complete overview about the whole range of the classical and modern reservoir engineering tools. The selection was made under the aspect of transmitting basic understandings about the reservoir processes, rather to provide recipes. This volume is followed by the textbooks Well Testing Basic Reservoir Simulation Enhanced Oil Recovery Advanced Reservoir Simulation Management The seven volumes cover the whole area of reservoir engineering normally offered in graduate university programs worldwide, and is specially used at the MMM University Leoben.

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ACKNOWLEDGEMENTS The author gratefully acknowledges permission to use material from the following sources: Table 3.1: From Accurate Formulas for Calculating the Water Influx Superposition Integral presented at the SPE Eastern Regional Meeting, Pittsburgh, October 1987. 1987 SPE-AIME. Figures 2.8, 2.9 and 2.1: From Journal of Petroleum Technology, Vol. 19, Dec ember 1967. 1967 SPE-AIME. Figure 4.3: From Transactions of the SPE of the AIME, Vol. 165, 1946. 1946 SPE-AIME. Figure 4.1: From Petroleum Production Handbook, Vol. II: Reservoir Engineering, 1962. 1962 SPE-AIME. Figures 4.12, and 4.13: From The Reservoir Engineering Aspects of Waterflooding, SPE Monograph, 1971. 1971 SPE-AIME. Figure 4.14: From Transactions of the SPE of the AIME, Vol. 27, 1956 SPE-AIME. Figure 5.1: From Transactions of the SPE of the AIME, Vol. 21, 1957. 1957 SPE-AIME. Figures 5.15, 5.16, 5.17 and 5.18: From Transactions of the SPE of the AIME, Vol. 27, 1956. 1956 SPE-AIME. Figures 5.24 and 5.25: From Effect of Flooding Rate and Permeability Ordering on Waterflooding Stratified Reservoirs, Paper 7158, 1978. 1978 SPE-AIME. Figures 5.27, 5.28 and 5.29: From The Reservoir Engineering Aspects of Waterflooding, SPE Monograph, 1971. 1971 SPE-AIME. Figures 5.3, 5.31 and 5.32: From Transactions of the SPE of the AIME, Vol. 216, 1959. 1959 SPE-AIME.

I Table of Contents Introduction 3 1.1 General Remarks...3 1.2 Classification of Reserves...7 Reserves Calculation by Volumetric Methods 9 2.1 Computation of Oil and Gas in Place...9 2.2 Recovery Factor...1 2.3 Data Distribution and Probability...11 2.3.1 Triangular Distribution...14 2.3.2 Uniform Distribution...15 2.3.3 Dependent Distribution...16 2.4 Monte Carlo Simulation Method...17 Material Balance 25 3.1 Tarner s Formulation...25 3.2 Drive Indices...29 3.3 Water Influx...3 3.3.1 Semi-Steady-State Water Influx...32 3.3.2 Steady-State Water Influx...33 3.3.3 Non-Steady-State Water Influx...35 3.3.3.1 Vogt-Wang Aquifer Model...35 3.3.3.2 Fetkovich Aquifer Model...38 3.4 Finite Difference Material Balance Equation...5 3.5 Undersaturated Oil Reservoirs...53 3.6 Gas Reservoirs...55 3.7 Calculation of Geological Reserves...56 3.8 Graphical Evaluation of Material Balance...58 3.8.1 Reservoirs Without Water Influx: W e...58 3.8.2 Reservoirs With Water Influx...6 3.9 Recovery Factor...61 Displacement Efficiency 63 4.1 Solution Gas Drive...63 4.1.1 MUSKAT s (1945) Equation of Solution Gas Drive...63 4.1.2 Calculation of the Solution Gas Drive According to PIRSON...68 4.2 Frontal Displacement...7 4.2.1 BUCKLEY-LEVERETT Theory...7 4.2.2 Oil Displacement by Water...74 4.2.3 Influence of Free Gas Saturation on Water Displacement...76 4.2.3.1 The Residual Oil Saturation...76 4.2.4 Displacement by Gas...8

II Table of Contents Sweep Efficiency 89 5.1 Mobility Ratio... 89 5.2 Stability of Displacement...9 5.3 Displacement in Dipping Layers... 92 5.3.1 Position of the Displacing Front... 92 5.3.2 Vertical Saturation Distribution... 94 5.4 Displacement in Stratified Reservoirs... 97 5.4.1 Vertical Permeability Distribution... 98 5.4.2 DYKSTRA-PARSONS Method... 11 5.4.3 JOHNSON Correlation... 15 5.4.4 Communicating Layers... 18 5.4.5 Numerical Calculation of Water Flooding in Linear Stratified Layers... 112 5.4.6 Areal Sweep Efficiency... 117 Decline Curve Analysis 125 6.1 Exponential Decline... 125 6.2 Hyperbolic Decline... 127 6.3 Harmonic Decline... 128 References

List of Figures III List of Figures Figure 1.1: Range in estimates of ultimate recovery during the life of a reservoir...4 Figure 2.1: Histogram for porosity data (Table 2.3)... 11 Figure 2.2: Cumulative frequency of porosity data (Table 2.3)... 11 Figure 2.3: Triangular probability distribution...13 Figure 2.4: Cumulative probability calculated from Fig. 2.3...13 Figure 2.5: Uniform probability distribution...14 Figure 2.6: Use of dependent distribution...15 Figure 2.7: Selecting random values from a triangular distribution (after McCray, 1975)...16 Figure 2.8: Calculated porosity (after Walstrom et al. 1967)...18 Figure 2.9: Formation factor (after Walstrom et al. 1967)...19 Figure 2.1: Calculated water saturation (after Walstrom et al. 1967)...19 Figure 2.11: Ultimate recovery calculation with Monte Carlo simulation...21 Figure 3.1: The scheme of the material balance of an oil reservoir...24 Figure 3.2: Pressure drop and production of a reservoir...25 Figure 3.3: Cumulative drive indices...28 Figure 3.4: Oil reservoir with an aquifer...29 Figure 3.5: Cumulative water influx at a constant reservoir pressure...3 Figure 3.6: Change in reservoir pressure...31 Figure 3.7: Idealized aquifers...32 Figure 3.8: Pressure-drop in a gas reservoir...54 Figure 3.9: Graphical illustration of material balance without water influx (W e )...57 Figure 3.1: Graphical illustration of material balance with balance with water influx (W e >)...59 Figure 4.1: MUSKAT s function: λ and η...64 Figure 4.2: The relative gas and oil permeabilities and their relation as a function of the oil saturation...65 Figure 4.3: Pressure and gas oil ratio histories of solution gas-drive reservoirs producing oil of different viscosities (after MUSKAT and TAYLOR, 1946)...65 Figure 4.4: Pressure, production index and GOR as a function of the production N p /N...67 Figure 4.5: Illustration of the displacing process according to Buckley and Leverett...7 Figure 4.6: The fractional flow curve and its derivative...71 Figure 4.7: The fractional flow curve and graphical method of determining the front saturation and the average saturation behind the front...72 Figure 4.8: Influence of the wettability on relative permeability and fractional curves...73 Figure 4.9: The influence of the viscosity on the fractional flow curve...73 Figure 4.1: Three-phase relative permeability functions (from Petroleum Production Handbook, Vol. 11, 1962)...76 Figure 4.11: Water displacement at free gas saturation...77

IV List of Figures Figure 4.12: Correlation between initial gas saturation and residual gas saturation (after CRAIG, 1971)... 77 Figure 4.13: Influence of initial gas saturation on the efficiency of water displacement (after CRAIG, 1971)... 78 Figure 4.14: Relation between residual gas saturation and oil saturation in case of water flooding (after KYTE et al, 1956)... 79 Figure 4.15: Illustration of gas displacement when using the BUCKLEY-LEVERETT theory... 8 Figure 4.16: Distribution of saturation in case of condensation gas drive... 8 Figure 4.17: Graphical determination of the front saturation by condensation gas drive... 82 Figure 4.18: Fractional flow curve to Example 4.1... 85 Figure 4.19: Oil recovery curve according to Example 4.1... 86 Figure 5.1: Linear water displacement demonstrated by a transparent three dimensional model (VAN MEURS 1957)... 89 Figure 5.2: Capillary forces and gravity influence the development of a viscous fingering... 89 Figure 5.3: Initial position of the water-oil contact (a) and possible changes during displacement (b)(c)... 9 Figure 5.4: Position of the interface at opportune relation of mobilities... 91 Figure 5.5: Position of the interface at an unfavorable mobility ratio... 91 Figure 5.6: Influence of capillary forces and gravity on a supercritical displacement... 93 Figure 5.7: Calculation scheme of relative permeabilities for vertical equilibrium... 94 Figure 5.8: Pseudo-capillary pressure curves... 95 Figure 5.9: Pseudo-relative permeability curves... 95 Figure 5.1: Log-normal distribution of permeability... 97 Figure 5.11: Distribution of the conductivity as a function of storage capacity... 98 Figure 5.12: Correlation between the variation coefficient and LORENZ-coefficient... 98 Figure 5.13: Stratified reservoir model... 99 Figure 5.14: Injectivity vs. injected volume... 12 Figure 5.15: JOHNSON s (1956) correlation, WOR 1... 13 Figure 5.16: JOHNSON s (1956) correlation, WOR 5... 14 Figure 5.17: JOHNSON s (1956) correlation, WOR 25... 14 Figure 5.18: JOHNSON s (1956) correlation, WOR 1... 15 Figure 5.19: Solution after application of the JOHNSON s (1956) correlation... 16 Figure 5.2: Water displacement in communicating layers... 18 Figure 5.21: Stratified reservoir with vertical communication... 19 Figure 5.22: Vertical saturation distribution depending on the sequence of layers...11 Figure 5.23: Displacement in heterogeneous reservoirs at increasing and decreasing permeability...11 Figure 5.24: Water flooding in a stratified reservoir after^berruin and MORSE, 1978...114 Figure 5.25: The position of the saturation profil S w 55 at various amounts of injected pore volume and displacement velocities after BERRUIN and MORSE, 1978...114 Figure 5.26: Definition of the areal sweep efficiency...115 Figure 5.27: Well pattern for areal flooding (after CRAIG, 1971)...116 Figure 5.28: Areal sweep efficiency at a linear pattern and uniform mobility ratios at breakthrough (after CRAIG, 1971)....116 Figure 5.29: Areal sweep efficiency with a shifted linear pattern at breakthrough, d/a 1 (after CRAIG, 1971)...117

List of Figures V Figure 5.3: Areal sweep efficiency of a five point system and the mobility ratio as a function of the fractional value (after CAUDLE and WITTE, 1959)... 118 Figure 5.31: Areal sweep efficiency of a five-spot system as a function of the injected pore volume (after CAUDLE and WITTE, 1959)... 118 Figure 5.32: Conductance ratio γ as a function of mobility ratio and pattern area sweep efficiency for five-spot pattern (after CAUDLE and WITTE, 1959)...121 Figure 6.1: Exponential decline plot...124 Figure 6.2: Harmonic decline plot...126

VI List of Figures

1 Chapter 1 Introduction The most important tasks of the reservoir engineer are to estimate the oil and gas reserves and to forecast the production. This volume describes the classic and fundamental methods that are applied for these purposes: Volumetric computation of reservoir volume Material balance calculations Estimation of displacement efficiency Estimation of sweep efficiency Production decline analysis This volume is based on the textbooks Heinemann: "Fluid Flow in Porous Media" [21.] (1991) and Heinemann and Weinhardt: "Reservoir Fluids" [2.]. 1.1 General Remarks First some commonly used notions have to be defined: PETROLEUM is the common name for all kinds of hydrocarbons existing inside of the earth, independently of its composition and aggregate stage. PETROLEUM IN PLACE is defined as the total quantity of petroleum estimated in a reservoir: O.O.I.P. - Original oil in place m 3 [bbl]. Used symbol: N O.G.I.P. - Free gas in a gas cap or in a gas reservoir m 3 [cuft]. Used symbol: G F. CUMULATIVE PRODUCTION is the accumulated production at a given date. Used symbols: N p, G p. ULTIMATE RECOVERY is the estimated ultimate production, which is expected

2 Chapter 1: Introduction during the life of the property. Used symbols: N pmax, G pmax. RESERVES are the Ultimate Recovery minus Cumulative Production: N pmax N p or G pmax G p (1.1) RECOVERY FACTOR E R CumulativeProduction ------------------------------------------------------------ O.O.I.P. N p G ----- p or E N R ------ G (1.2) ULTIMATE RECOVERY FACTOR E Rmax Ultimate ------------------------------------------------- Recovery O.O.I.P. N pmax G --------------- pmax or E N Rmax --------------- G (1.3) The amount of reserves determines the whole strategy of future activities concerning exploration, development and production. Unfortunately, reliable reserve figures are most urgently needed during the early stages of an exploration project, when only a minimum of information is available. During further activities (exploration and production), the knowledge concerning the reservoir contents becomes more and more comprehensive and naturally will reach a maximum at the end of the reservoir s life cycle. Fig.1.1 shows the life cycle of a reservoir, and the range of recovery estimates. PERIOD AB: When no well is drilled, any estimate will be supported by the analogous method based on data from similar pools. In this phase probabilities of the presence of a trap, presence of oil or gas saturation, presence of pay and amount of recoverable reservoir contents have to be estimated. No reserves is a real option. PERIOD BC: The field is being discovered and step by step developed. The production rate increases. The volumetric estimation can be made more and more precise due to the increasing

Chapter 1: Introduction 3 number of wells available. PERIOD CD: The field has been developed and the production rate achieved a maximum. The majority of possible informations from well logs core analysis and transient well testing are available. The volumetric estimation can be made more precise when analysing the reservoir performance. Early simulation studies, called reservoir modeling, make the calculation of the ultimate recovery possible. PERIOD DE: The recovery mechanisms are well known. Material balance calculations and simulation of the reservoir history provide, in most of the cases, more accurate figures of O.O.I.P. than the volumetric method. These methods are more suited to compute the reserves than the volumetric method. PERIOD EF: In addition to material balance calculations and simulation studies decline curve analysis becomes more appropriate.

4 Chapter 1: Introduction A Property Status Study Method Range of Estimates Pre-drilling Period Analog B C D E F 1st Well Completed Development Period Production Operations Abandonment Volumetric Performance Simulation Studies Material Balance Studies Decline Trend Analyses Ultimate Recovery Range of Recovery Estimates Actual Recovery Production Profile Log Production Rate Cumulative Rate Cumulative Production Relative Risk High Time Risk Low A B C D E F Figure 1.1: Range in estimates of ultimate recovery during the life of a reservoir

Chapter 1: Introduction 5 1.2 Classification of Reserves Reserves are divided into classes. The definition of these classes is not common in all countries. Here the recommended nomenclature system of MARTINEZ et al. [28.] (1983) published at the 11 th World Oil Congress, London, is used. PROVED RESERVES of petroleum are the estimated quantities, as of a specific date, which analysis of geological and engineering data demonstrates, with reasonable certainty, to be recoverable in the future from known reservoirs under the economic and operational conditions at the same date. PROVED DEVELOPED RESERVES are those proved reserves that can be expected to be recovered through existing wells and facilities and by existing operating methods. Improved recovery reserves can be considered as proved developed reserves only after an improved recovery project has been installed. PROVED UNDEVELOPED RESERVES are those proved reserves that are expected to be recovered from future wells and facilities, including future improved recovery projects which are anticipated with a high degree of certainty. UNPROVED RESERVES of petroleum are the estimated quantities, as of a specific date, which analysis of geological and engineering data indicate might be economically recoverable from already discovered deposits, with a sufficient degree of certainty to suggest the likelihood or chance of their existence. Unproved reserves may be further categorized as PROBABLE RESERVES where there is a likelihood of their existence, or POSSIBLE RESERVES where there is only a chance of their existence. The estimated quantities of unproved reserves should take account of the uncertainties as to whether, and to what extent, such additional reserves may be expected to be recoverable in the future. The estimates, therefore, may be given as a range. SPECULATIVE RESERVES of petroleum are the estimated quantities, as at a specific date, which have not yet been discovered, but which general geological and engineering judgement suggests may be eventually economically obtainable. Due to the great uncertainties, they should always be given as a range.

6 Chapter 1: Introduction

7 Chapter 2 Reserves Calculation by Volumetric Methods 2.1 Computation of Oil and Gas in Place The following formulas are used: OOIP - Original oil in place: N Vφ( 1 S wi ) ----------------------------. B oi (2.1) OGIP - Free gas in a gas reservoir or in a gas cap: Vφ( 1 S wi ) G F ----------------------------. B gi (2.2) Solution gas in an oil reservoir: Vφ( 1 S wi )R si G s ------------------------------------ NR si, B oi (2.3) where V - reservoir volume [m 3 ] φ - porosity [-] S wi - initial water saturation [-] 7

8 Chapter 2: Reserves Calculation by Volumetric Methods B oi - oil formation volume factor (FVF) at initial pressure [m 3 /sm 3 ] B gi - gas formation volume factor at initial pressure [m 3 /sm 3 ] R si - initial solution (or dissolved) gas oil ratio [sm 3 /sm 3 ] The reservoir volume V can be calculated in different ways. Which method is the best depends on the available data, shape of the reservoir, etc.. 2.2 Recovery Factor Ultimate recovery is influenced by a lot of individual physical realities. It depends on the drive mechanism of the reservoir, mobility of reservoir fluids, permeability and variation of permeability, both vertically and in the area, inclination and stratification of the layers, strategy and methods of field development and exploitation. In the exploration and early exploitation stage, only analogous or statistical methods can be used to estimate the ultimate recovery factor. It is necessary to examine known reservoirs from the same region. ARPS et al. [2.] examined a large amount of oil fields with depletion drive and with water drive. The results were published by the American Petroleum Institute. The formulas are known as API formulas for estimation of the ultimate recovery factor. Depletion or gas drive: φ( 1 S w ) E R (%) 41.815 ----------------------- B ob.1611 -------- k.979 S.3722 p b.1714 w ---- µ ob p a (2.4) Water or gravitation drive: E R (%) 54.898 ----------------------- φ( 1 Sw).422 ------- k µ.77.193 p i.2159 wi S w ---- B oi µ oi p a (2.5) where p i is the initial pressure, p a the abandonment pressure, and p b the bubble point pressure. The subscripts i and b denote that the value is valid at p i or p b. The numerical results of the API examination are summarized in Table 2.1.

Chapter 2: Reserves Calculation by Volumetric Methods 9 Table 2.1: API Ultimate Recovery Factor Estimation Depletion + Gasdrive Water or Gravitationdrive Standard deviation Sand, Sandstone min. mean max. Carbonates min. mean max..229.95.213.46.155.176.27.176.131.284.579.9.218.481 2.3 Data Distribution and Probability The methods for calculation of the reservoir volume, average porosity and permeability are subjects of reservoir geology and log evaluation. These data are more or less uncertain and have to be treated as random variables. Interpretation of even moderately large amounts of data requires statistical methods. A commonly used method is to gather individual data into groups or classes. This facilitates interpretation as well as numerical computations. Porosity data are used to demonstrate this ascertainment. An example of raw data is given in Table 2.2. Table 2.2: Porosity Sample Data (n24 values).165.198.196.185.192.188.187.184.182.25.178.175.192.25.162.162.182.17.184.165.154.179.172.156 Σx 4.317 Class boundaries, as given in Table 2.3, coincide in terms of the upper boundary of one class being the same as the lower boundary of the next class. Common convention is to take the values at the boundary into the higher class. The difference between upper and lower boundaries is referred to as the class interval. Normally but not necessarily, the class intervals are equal. Average porosity can be calculated either from the data in Table 2.2 or in Table 2.3: n x j j 1 φ -------------- n 4.317 ------------.1798 24 (2.6)

1 Chapter 2: Reserves Calculation by Volumetric Methods φ * n i 1 f i x i ----------------- f i 4.33 ---------.184 24 (2.7) It is evident that φ lim φ * n (2.8) Class Boundaries Members Table 2.3: Frequency Distribution Histograms and frequency polygons are used to show the probability density. Fig. 2.1 shows both of them for the porosity data included in the foregoing tables. The frequency polygon is drawn through the midpoints of the classes. The areas below the histogram and the frequency polygon are equal. The cumulative frequency polygons are shown in Fig. 2.2. Based on this diagram, one can conclude that 6% of the samples have a porosity less than about 18.5%. Two important statistical properties of the data group are the variance Frequency f i Class Mark x i.15-.16.156,.157 2.155.311.16-.17.162,.162,.165,.165 4.165.66.17-.18.17,.172,.175,.178,.179 5.175.875.18-.19.181,.182,.182,.184,.185,.187 7.185 1.295.19-.2.192,.192,.196,.198 4.195.78.2-.21.25,.25 2.25.41 24 4.33 f i.x i n σ 2 ( x j x) j 1 ---------------------------------- n 2 (2.9) and the standard deviation σ n ( x j x) j 1 ---------------------------------- n 2 (2.1)

Chapter 2: Reserves Calculation by Volumetric Methods 11 Frequency 8 6 4 2.3.2.1 Relative frequency.14.16.18.2.22 Porosity Figure 2.1: Histogram for porosity data (Table 2.3) Cumulative frequency 24 18 12 6 1..75.5.25 Cum. relative frequency.14.16.18.2.22 Porosity Figure 2.2: Cumulative frequency of porosity data (Table 2.3) For the foregoing example, these are n 2 σ 2 ( φ j φ) j 1 -----------------------------------.2 ; σ.14. (2.11) n The local maximum of the probability density is called the modus. If the distribution is symmetrical, the modus is equal to the mean value.

12 Chapter 2: Reserves Calculation by Volumetric Methods 2.3.1 Triangular Distribution In the majority of practical cases, it is not possible to get a reliable histogram or frequency polygon. One has to be content with estimating the upper, the lower and the modus. For the data given in Table 2.2 the following estimation would be possible: φ min.15, φ max.21, φ mod.185, φ.18. (2.12) Fig. 2.3 illustrates the triangular distribution of this data. The height of the triangle is selected in a way so that the surface value becomes 1. The probability that the value (φ) will be less than the modes (φ mod ) is p φ mod φ ----------------------------- min φ max φ min (2.13) and 1 - p that it will be higher. The cumulative probability can be calculated as follows: F φ φ min 2 p ----------------------------, φ, (2.14) min φ φ mod φ max φ min φ max φ 2 F 1 ( 1 p) ----------------------------, φ. (2.15) mod φ φ max The variance is φ max φ min σ ( φ max φ min ) 2 ------------------------------------ [ 1 p( 1 p) ]. (2.16) 18 Fig. 2.4 shows the F-function for the triangular probability distribution in Fig. 2.3. It is very similar to the diagram in Fig. 2.2. In fact, cumulative relative frequency and cumulative probability have the same meaning.

Chapter 2: Reserves Calculation by Volumetric Methods 13.3 Probability.2.1.14.16.2.22 Φ min.18φ mod Φ max Porosity Figure 2.3: Triangular probability distribution 1. Cumulative probability.75.5.25.14.16.18.2.22 Porosity Figure 2.4: Cumulative probability calculated from Fig. 2.3 2.3.2 Uniform Distribution An uniform distribution is illustrated in Fig. 2.5. The randomly occurring values are evenly distributed in the range from minimum to maximum values. The cumulative probability is defined by: F φ φ φ min max p( φ) dφ ----------------------------, φ. (2.17) min φ φ max φ min φ max φ min For φ φ min, F and for φ φ max, F 1.

14 Chapter 2: Reserves Calculation by Volumetric Methods The variance is σ 2 ( φ max φ min ) 2 ------------------------------------. (2.18) 12 Probability 1..75.5.25.14.16.18.2.22 Porosity Figure 2.5: Uniform probability distribution 1..75.5.25 Cumulative probability, F 2.3.3 Dependent Distribution Reservoir data applied for computation purposes of the reservoir are not independent. For example, the water saturation may be calculated from the ARCHIE [1.] formulas: F aφ m (2.19) R t n FR w S w (2.2) thus S w ar --------- w R t 1 -- n φ m --- n (2.21) where R w - the connate water resistivity Ωm [Ωft], R t - the formation resistivity Ωm [Ωft], F - the formation factor, a, m, n - positive constants.

Chapter 2: Reserves Calculation by Volumetric Methods 15 It is evident, that S w increases if φ decreases. Fig. 2.6 shows such a dependence for a triangular distribution of the connate water saturation..3.25 Porosity.2.15 min modus max.1.5 1. Connate water saturation Figure 2.6: Use of dependent distribution 2.4 Monte Carlo Simulation Method Calculate the following formula: z xy. (2.22) x and y are stochastic variables. Their cumulative probabilities between their minimum and maximum values are known. The task is to determine the probability distribution and the expected value of z. The computation is simple, but less suitable for hand calculations than for the computer. For calculations it is necessary to use a random number generator. Values entering into Eq. 2.22 are repeatedly selected by random numbers taken from an appropriate range of values, as it is shown in Fig. 2.7. Within several hundred to several thousand trials, the number of z values for prefixed classes are counted. The result is a histogram and the cumulative relative frequency for the z variable.

16 Chapter 2: Reserves Calculation by Volumetric Methods Probability density 1 1. x min x mod x max Random numbers Uniform distribution Cumulative probability or frequency x min x max 1. Value of parameter Figure 2.7: Selecting random values from a triangular distribution (after McCray, 1975) Example 2.1 Calculating the porosity al. [39.] (1967). ( φ) The calculation steps are as follows: 1. Determine φ from the relation: and water saturation (S w ) from well logs (after Walstrom et ρ B ρ F φ+ ρ Ma ( 1 φ). 2. Use the value of φ to determine formation factor F from the Archie formula: F aφ m. 3. Use the value of F and randomly chosen values of other quantities to determine S w from the relation R t FR w S n w,

Chapter 2: Reserves Calculation by Volumetric Methods 17 where ρ B - the bulk density kg m -3 [1b/cuft], ρ F - the fluid density kg m -3 [1b/cuft], ρ Ma - the rock matrix density kg m -3 [1b/cuft], R w - the connate water resistivity Ωm [Ωft], R t - the formation resistivity F - the formation factor, a, m, n - positive constants. Ωm [Ωft], The range of parameter used in this example is given in Table 2.4. Uniform distribution was assumed for all quantities, except parameters a and ρ F, which where assumed to be constants. The results of the simulation are shown in Fig. 2.8, Fig. 2.9 and Fig. 2.1, these reflect the results of several hundred cases, wherein each case was processed through steps 1, 2 and 3. Here it may be noted that although uniform distribution was assumed for quantities entering the calculations, the resulting probability densities are not symmetrical. The functional relationship of the quantities may skew the results. Table 2.4: Ranges of Parameters Used in a Log-Interpretation Example (metric units) Parameter Lower Limit Upper Limit R t True resistivity Ωm 19. 21. R w Connate water resistivity Ωm.55.75 n Exponent in the ARCHIE equation 1.8 2.2 a Coefficient in the ARCHIE equation.62.62 m Exponent in the ARCHIE equation 2. 2.3 ρb Bulk density kg/m 3 2.36 2.38 ρ MA Rock mineral density kg/m 3 2.58 2.63 ρ F Reservoir fluid density kg/m 3.9.9

18 Chapter 2: Reserves Calculation by Volumetric Methods Table 2.5: Ranges of Parameters Used in a Log-Interpretation Example (field units) Parameter Lower Limit Upper Limit R t True resistivity Ωft 62.3 68.9 R w Connate water resistivity Ωft.18.246 n Exponent in the ARCHIE equation 1.8 2.2 a Coefficient in the ARCHIE equation.62.62 m Exponent in the ARCHIE equation 2. 2.3 ρb Bulk density lb/cu ft 147.3 148.6 ρ MA Rock mineral density lb/cu ft 161. 164.2 ρ F Reservoir fluid density lb/cu ft 56.2 56.2.5.4 Probability.3.2.1.1.12.14.16.18 Porosity Figure 2.8: Calculated porosity (after Walstrom et al. 1967)

Chapter 2: Reserves Calculation by Volumetric Methods 19.2.15 Probability.1.5 18 26 34 42 5 58 66 74 82 Formation factor Figure 2.9: Formation factor (after Walstrom et al. 1967).3 Probability.2.1 24 3 36 42 48 54 6 Water saturation Figure 2.1: Calculated water saturation (after Walstrom et al. 1967) Example 2.2 Calculation of oil recovery The ultimate recovery was estimated by combination of the Eq. 2.1 and Eq. 2.5: N pmax NE R.54898V φ ( 1 S w) 1.4222 kµ ----------------------- wi.77.193 p ---------- i.2159 S w ---- (2.23) Symmetric triangular distribution was assumed for all the quantities in Eq. 2.23, except the parameter which is constant.7 cp. The limits are shown in Table 2.6 and Table 2.7. µ wi B oi The ultimate recovery from this reservoir can be characterized with the following figures µ oi p a

2 Chapter 2: Reserves Calculation by Volumetric Methods of a Monte Carlo simulation with 5 trials. Expected ultimate recovery: 16,54, m 3 or 14.3 MMbbl with standard deviation of: 1,1, m 3 or 6.91 MMbbl Probability distribution: with more than but less than 99% 13.82-19.24.1 3 m 3 [86.9-121. MMbbl] 9% 14.7-18.38.1 6 m 3 [92.4-115.6 MMbbl] 8% 15.8-18..1 6 m 3 [94.8-113.2 MMbbl] Taking for all parameters the most unfavorably and the most favorably the following realistically figures can be calculated: Worst case: Best case: 12.227x1 6 m 3 21.726x1 6 m 3 (76.9x16 bbl) (136.6x16bbl) Table 2.6: Range of Parameters Used in the Ultimate Recovery Calculation Example (metric units) Parameter Unit Lower Limit Upper Limit V Reservoir volume 1 6 m 3 285. 37. φ Porosity.165.19 S w Initial water saturation.22.28 B oi Formations volume factor 1.48 1.5 µ oi Reservoir oil viscosity cp 2.85 2.95 k Reservoir permeability darcy.8 1.5 p i Initial pressure (p i p b ) MPa 22. 22.5 p a Abandonment pressure MPa 18. 2.

Chapter 2: Reserves Calculation by Volumetric Methods 21 Table 2.7: Range of Parameters Used in the Ultimate Recovery Calculation Example (field units) Parameter Unit Lower Limit Upper Limit V Reservoir volume 1 5 ac ft 2.31 3. φ Porosity.165.19 S w Initial water saturation.22.28 B oi Formations volume factor 1.48 1.5 µ oi Reservoir oil viscosity cp 2.85 2.95 k Reservoir permeability md 8. 1.5 p i Initial pressure (p i p b ) psi 3.19 3.262 p a Abandonment pressure psi 2.61 2.9 1. Cell C1 4985.75 Probability.5.25 Frequency 13.5 15. 16.5 18. 19.5 6 3 Cumulative distribution 1 [m ] Figure 2.11: Ultimate recovery calculation with Monte Carlo simulation

22 Chapter 2: Reserves Calculation by Volumetric Methods

23 Chapter 3 Material Balance 3.1 Tarner s Formulation Fig. 3.1 shows a schematic illustration of an oil reservoir. The rock volume is V and the porosity φ. Apart from a certain saturation of connate water S wi, the rock is saturated with hydrocarbons. Thus, the effective reservoir volume is Vφ( 1 S wi ) V P ( 1 S wi ). (3.1) The reservoir temperature is defined as T, the initial pressure as p i. The virgin reservoir is in a state of hydrostatic and thermodynamic equilibrium. The original oil in place is defined as N in sm 3 or stb (stock tank barrels). At reservoir conditions gas is dissolved in the oil. The amount is expessed by the initial solution GOR (gas-oil ratio) R si sm 3 /sm 3 or scft/stb. The formation volume of the oil is NB oi. If the reservoir contains the same or a greater amount of gas than soluble at reservoir conditions (pressure and temperature), the reservoir is saturated and the surplus gas forms a gas cap. Otherwise, the reservoir is undersaturated. Should the gas cap contain an amount of G sm 3 or scft gas, then its formation volume will be GB gi. Usually the volume of the gas cap is expressed in relation to the oil volume: GB gi mnb oi. (3.2) 23

24 Chapter 3: Material Balance When regarding surface and formation volumes the following relations can be set up: Fluid Surface Formation Oil N sm 3 [stb] NB oi m 3 [bbl] Dissolved Gas NR si sm 3 [scft] -- Free Gas G F sm 3 [scft] G F B gi mn B oi m 3 [cuft] Production in reservoir volumes N(R-R)B p p s g NB p o Expanded reservoir volumes W p At reservoir pressure p Initial reservoir pressure: pi net pore volume: VF(1-S ) GB gimnboi wi GB g Reservoir content GB + g NB +NB (R -R ) o g si s +NB g(rsi-r s)- -N B (R -R ) p g si s NB oi (N-N p )B o W o W-W s p Figure 3.1: The scheme of the material balance of an oil reservoir The effective pore volume, expressed by the amounts of oil and gas, is: NB oi + GB gi V P ( 1 S wi ). (3.3) After a certain time period, an amount of Oil N p sm 3 [stb], Gas G p N p R p sm 3 [scft], Water W p sm 3 [bbl]

Chapter 3: Material Balance 25 will have been produced. R p is the cumulative production GOR. As a consequence of production, the reservoir pressure decreases from p i to p. Reservoir pressure [mpa] 26 24 22 2 18 3 3 R m/m p 1 8 6 R P Past 1 8 6 4 2 3 3 N P and W P 1 [m ] Cumulative production 16 4 24 48 72 96 12 144 Production time [month] Figure 3.2: Pressure drop and production of a reservoir W P Let us now consider the reverse situation. At first pressure drops to p. Thus, the gas cap expands and gas evolves from the oil. From the aquifer an amount of W e water will flow into the reservoir. The expanded system would have a reservoir volume at pressure p of B g NB o + mnb oi ------- + NB g ( R si R s ) + W e. (3.4) At the same pressure the produced fluids would have a total reservoir volume of N p B o + N p B g ( R p R s ) + W p N p [ B o + B g ( R p R s )] + W p. (3.5) The effective pore volume corresponding to Eq. 3.1 remains unchanged which consequently makes the following assertion valid: or B gi [expanded volume] - [initial volume] [produced volume] Eq. 3.4 - Eq. 3.3 Eq. 3.5

26 Chapter 3: Material Balance After substituting: B g NB ( o B oi ) + N mb oi ------- 1 + B g ( R si R s ) + W e B gi (3.6) N p [ B o + B g ( R p R s )] + W p From this N N p [ B o + B g ( R p R s )] ( W e W p ) -------------------------------------------------------------------------------------------------------. (3.7) B g mb oi ------- 1 + B g ( R si R s ) ( B oi B o ) B gi This is the formula of TARNER s [37.] material balance. If in addition to production, water is injected at the cumulative amount of W I and/or gas at the cumulative amount of G p I, then the term (W I + G p IB g ) has to be added to the numerator. I indicates how much of the produced gas was reinjected into the reservoir. Every specific term in Eq. 3.7 has a certain meaning: produced hydrocarbons 6444 7444 8 N N p [ B o + B g ( R p R s )] B mb g oi + 14243 B gi gas cap expansion Reservoir volume of net water influx ( ) W e B g + W I W p ( R si R s ) 14243 4 desoluted gas expansion injected gas ( ) B oi GB I g B o 14243 reservoir oil shrinkage (3.8) Eq. 3.6 is then divided by B g (3.9)

Chapter 3: Material Balance 27 Eq. 3.9 is divided by its right hand side: B o B N ----- oi R B s ------- R g B si mnb oi ------- 1 1 ----- 1 + + ----- ( W g B gi B g B e W p ) g ------------------------------------------------------------------------------------------------------------------------------------------------------------ 1 B o N p ----- R s + N p R p B g (3.1) 3.2 Drive Indices Splitting up the left side of Eq. 3.1 leaves three fractions which describe the shares of the specific drive mechanisms in reference to the whole cumulative production effected by the solution gas drive, the gas drive and the water drive. These are considered the drive indices. The solution gas drive index (a two phase expansion of the oil) is defined as I s B o B N ----- oi R B s ------- R g B si g ------------------------------------------------------------------. (3.11) B o N p ----- R s + N p R p B g The gas drive index (expansion of the gas cap) is defined as I g mnb oi ------- 1 1 ----- B gi B -------------------------------------------------- g. (3.12) B o N p ----- R s + N p R p B g The water drive index (expansion of the aquifer) is defined as I w ----- 1 ( W B e W p ) -------------------------------------------------- g. (3.13) B o N p ----- R s + N p R p B g

28 Chapter 3: Material Balance The relation between the indices is given by I s + I g + I w 1. (3.14) Cumulative oil-, gas- and water production (N p, R p, W p ) are given by production statistics. The PVT-properties (B o,b g,r s ) are determined by laboratory measurement or by correlations. The volumetric reserve calculation covers the petroleum in place (N, G). The average reservoir pressure is recorded by regular measurements of static well bottom-hole pressures. The application of these data enables a sufficient description of the water influx as a function of time. 3.3 Water Influx Operating a reservoir over years, the cumulative oil, gas and water production N p (t), G p (t), and W p (t) are naturally known. The reservoir pressure declines and the actual values p(t) will be determined by regulare pressure surveys. The fluid properties, as B o (p), B g (p) and R s (p), are messured in PVT Labs or determined from different types of charts, e.g.: from Standing correlation. Also the OOIP (N) and the gas cap factor m can be estimated by volumetric calculation (see Chapter 3.). The only quantity in Eq. 3.6, which is entierly unknown, is the water influx W e (t). The Material Balance calculation is the only method which enables to determine it as function of time. From Eq. 3.6: W e () t N p [ B o + B g ( R p R s )] (3.15) B g N B o B oi + mb -------- oi 1 + B B gi g ( R si R s ) + W p The aquifer is a water bearing formation, hydrodynamically connected to the hydrocarbon reservoir. Its form, size and permeability can vary greatly. Hydrological reflection could help to set up hypotheses. However, these can never be verified in detail since no wells will be drilled to explore an aquifer. One of the boundaries of the aquifer is the water-oil-contact (WOC). This interior boundary is usually well known, whereas the exterior boundary is an object of speculation. The exterior boundary can be considered closed if the whole amount of water flowing into the reservoir is due to the expansion of the aquifer. In this case, the aquifer is finite closed. Faults and layer pinch outs form such boundaries.

Chapter 3: Material Balance 29 The aquifer can be considered a finite open one, if the pressure at the exterior boundary is constant. A connection to the atmosphere through outcropping or hydrodynamic contact to a karstic formation are the possibilities to form such boundaries. Fig. 3.4 shows a schematic illustration of an aquifer. 1. Cumulative drive indices.8.6.4.2 I g I s Past Prognosis I w 24 48 72 96 12 144 Production time [month] Figure 3.3: Cumulative drive indices Fault (tectonic boundary) Inner WOC Outer water-oil-contact (WOC) Marl Aquifer Oil Water Reservoir Aquifer Figure 3.4: Oil reservoir with an aquifer The cumulative water influx is calculated from the rate: t W e () t qt () dt. (3.16) o Production induces pressure decline at the interior boundary of the aquifer. Let us make a theoretical consideration. We assume a unique and sudden pressure drop p p i - p at this

3 Chapter 3: Material Balance boundary, where p i is the initial reservoir pressure. The pressure drop will cause water intrusion into the reservoir. Initially, this is a consequence of the expansion of water and rock and it is independent of the distance to the exterior boundary and regardless of whether the boundary is closed or not. The depression zone stretches with time - either fast or slowly, dependent on permeability - whereby the water influx rate permanently decreases. This situation is given until the depression radius reaches the exterior boundary. We call this time interval as transient period. In case of a closed exterior boundary, water influx decreases rapidly and tends to zero, if the pressure in the whole aquifer has dropped by p. In this case, the function of the cumulative water influx W e (t) has an asymptotic value (see Fig. 3.5). Cumulative water influx, W e pp-pconstant i 1) Finite aquifer with constant pressure on the boundary pp e i r e ( ) rre p r 2) Infinite aquifer 3) Finite closed aquifer 1) 2) 3) Time Figure 3.5: Cumulative water influx at a constant reservoir pressure If there is an aquifer with constant pressure at the exterior boundary, a stabilization of the influx rate takes place and therefore the function W e (t) becomes linear with increasing time. If the aquifer is small or the permeability high, the transient period becomes short and can be neglected. Under this consideration we distinguish between three types of aquifer models or water influxes: 1. Semi-steady-state, 2. Steady-state, 3. Non-steady-state: 3.1. Transient 3.2. Pseudo-steady-state. t 3.3.1 Semi-Steady-State Water Influx Sometimes the cumulative water influx can be considered solely as a function of the

Chapter 3: Material Balance 31 pressure drop. That means, the time in which the pressure change took place has no influence on the intruded water amount, consequently Eq. 3.16 becomes: W e C 1 [ p i pt ()]. (3.17) In such a case the aquifer has always a limited size and a closed external boundary. The coefficient C 1 can be expressed by the parameters of the aquifer: C 1 Ahφc e (3.18) p i Reservoir pressure p 1 p 2 p 3 p 4 p p 1 p 2 p 3 t 1 t 2 t 3 Time, t t 4 Figure 3.6: Change in reservoir pressure 3.3.2 Steady-State Water Influx In case of a constant pressure at the exterior boundary and high aquifer permeability the transient period can be neglected. This is equivalent to the assumption of incompressible water inside the aquifer. Thus the water influx rate is proportional to the pressure difference between the two boundaries: qt () C 2 [ p i pt ()] The cumulative water influx is calculated by integration: t W e () t C 2 [ p i pt ()] dt o (3.19) (3.2) This is the SCHILTHUIS [35.] -formula (1936). The coefficient C 2 is calculated by DARCY s law with the help of the specific aquifer parameters. In the case of linear

32 Chapter 3: Material Balance aquifers (Fig. 3.7) C 2L where bhk --------, (3.21) µl b - width of the aquifer, h - thickness of the aquifer, L - length of the aquifer, k - permeability of the aquifer, µ - viscosity of the water. h r w r e b L A) B) Figure 3.7: Idealized aquifers In the case of radialsymmetric aquifers (Fig. 3.7 B), the coefficient is: C 2r -------------- 2πhk, (3.22) µ r e ln----- r w where replace 2π by 7.8 x 1-3 for field units to get [bbl/psi d]. There are r w - inner radius, r e - outer radius of the aquifer. The pressure p(t) in Eq. 3.2 is the pressure at the inner boundary. In most of the practical cases it will be replaced by the average reservoir pressure, given for discrete time points: t o, t 1, t 2,..., t n. Although this pressure can be plotted with a smooth continuous line, it is more practical to approximate it with a step function or with linear functions as shown in Fig. 3.6. For Eq. 3.2 both approximations give the same results.

Chapter 3: Material Balance 33 n W e () t C 2 [ p i pt ()] dt j 1 n t j t j 1 p j + p j 1 W e () t C 2 p i ---------------------- ( t 2 j t j 1 ) j 1 n W e () t C p 2 ( i p ) j ( t j t j 1 ) j 1 (3.23) 3.3.3 Non-Steady-State Water Influx Usually the water influx has non-steady-state character. We distinquish between transient and pseudo-steady state flow regime. Note that both are non-steady state flow. The theoretical funded and general applicable method, covering transient, pseudo-steady-state and steady-state flow as well, was published by van Everdingen and Hurst [16.] (1949). The method was slighly modified by Vogt and Wang [41.] (1988), making it more convenient for computer programming. The derivation was discussed in the Volume 1 of this Textbook series (Heinemann, Z.E.: "Fluid Flow in Porous Media", Chapter 3). We use the Vogt-Wang formulation as our standard method. Under pseudo steady state conditions the water influx results from the uniform expansion of the aquifer, which means that the rate of pressure change is equal in the whole aquifer domain. 3.3.3.1 Vogt-Wang Aquifer Model It was assumed that the reservoir area can be approximated by a segment of a circle. The radius is r w 2A ------ 1 2, (3.24) ω where ω is the arc in radian ( 2π for complete circle) and r e is the outer radius of the aquifer. The dimensionless outer radius is r ed r e -----. (3.25) r w

34 Chapter 3: Material Balance The dimensionless time is t D --------------------- k t αt φµ w c e r2 w (x.634 for field units, t in days). (3.26) The cumulative water influx at the time t Dj is p o p 1 p W ej C 3 ----------------Q 1 p 2 p ( t t Dj ) --------------------- o p 1 + ---------------- Q ( tdj t D1 t D2 t D1 t D1 )... D1 p j 1 p j p + --------------------------- j 2 p j 1 ---------------------------------- Q ( t Dj t Dj 1 t Dj 1 t tdj Dj 2 t Dj )} 1 (3.27) The Q ( t D ) functions are given in Table 3.1 only for an infinite acting radial aquifer. In case of finite aquifers we refer to the paper of VOGT and WANG [41.] (1988). In radial symmetrical homogeneous case the coefficient C 3 can be calculated as C 3 id ωφhc e r w 2 (x.1781 for bbl/psi). (3.28) Example 3.2 - Example 3.4 demonstrate how the coefficients C 1 -C 3 and the function W e (t) are calculated. Usually the aquifer parameters are unknown. It is possible however (as shown in Example 3.1) to determine the water influx of the past with help of the material balance equation. First it is essential to ascertain the semi-steady-state, steady state or non-steady-state character of the water influx. Example 3.1 is continued by Example 3.5. If either C 1 nor C 2 is a constant but the coefficient C 1 calculated at various times increase steadily whereas C 2 continuously decreases, coefficient C 3 can be determined with sufficient accuracy. The dispersion of the C 3 -values indicates how appropriate r ed and α were chosen. If α is too small, the function W e f -------- p t Q D (3.29) is not linear, but has an upward curvature. If α is too big, the curvature is directed downwards. Only numerous repetitions of the calculation using various values of r ed provide a favorable solution.

Chapter 3: Material Balance 35 Table 3.1: Dimensionless Functions for an Infinite Radial Aquifer (Q(t D ) after van EVERDINGEN and HURST, 1949) ( Q ( t D ) after VOGT and WANG, 1988 t D Q(t D ) t D Q ( t D ).1E-1.112E+.1E-1.7522528E-3.25E-1.17425E+.178E-1.18364557E-2.5E-1.278E+.316E-1.4432129E-2.75E-1.341E+.562E-1.1722197E-1.1E+.44E+.1E+.2643625E-1.25E+.689E+.178E+.63654354E-1.5E+.12E+1.316E+.15612493E+.75E+.261E+1.562E+.38757698E+.1E+1.1569E+1.1E+1.9731597E+.25E+1.5649E+1.178E+1.2473688E+1.5E+1.4539E+1.316E+1.63361924E+1.75E+1.6285E+1.562E+1.1649726E+2.1E+2.7411E+1.1E+2.43576673E+2.25E+2.14573E+2.178E+2.1167682E+3.5E+2.24855E+2.316E+2.3155979E+3.75E+2.34247E+2.562E+2.86662947E+3.1E+3.43129E+2.1E+3.2412943E+4.25E+3.9184E+2.178E+3.6781498E+4.5E+3.162698E+3.316E+3.19176731E+5.75E+3.229515E+3.562E+3.5488684E+5.1E+4.293514E+3.1E+4.15847778E+6.25E+4.648781E+3.178E+4.4618726E+6.5E+4.1192198E+4.316E+4.13431358E+7.75E+4.176688E+4.562E+4.394725E+7.1E+5.223861E+4.1E+5.11665744E+8.25E+5.55726E+4.178E+5.3465488E+8.5E+5.936399E+4.316E+5.1279585E+9.75E+5.1353145E+5.562E+5.37431E+9.1E+6.1758628E+5.1E+6.9278873E+9.25E+6.466E+5.178E+6.277871E+1.5E+6.7699E+5.316E+6.83166975E+1.75E+6.112E+6.562E+6.2515398E+11.1E+7.1462E+6.1E+7.7614114E+11.25E+7.3427E+6.178E+7.2375497E+12.5E+7.6544E+6.316E+7.69814848E+12.75E+7.95625E+6.562E+7.2123343E+13.1E+8.1252E+7.1E+8.64719162E+13.25E+8.2961E+7.178E+8.19772497E+14.5E+8.5689E+7.316E+8.6187517E+14.75E+8.8341E+7.562E+8.18414932E+15.1E+9.195E+8.1E+9.56428423E+15

36 Chapter 3: Material Balance The VAN EVERDINGEN-HURST solution requires the calculation of the sum at every time t D for every j. This is for limited aquifers not necessary if the early transient period is over. 3.3.3.2 Fetkovich Aquifer Model Fetkovich [17.] presented a simplified approach for such cases that utilizes the pseudo-steady-state aquifer productivity index and an aquifer material balance to represent a finite compressibility system. Assuming that the flow obeys DARCY s law and is at pseudo-steady-state or steady-state, the generalized rate equation for an aquifer without regarding the geometry can be written: q w J w ( p p wf ). (3.3) J w is defined as the productivity index of the aquifer. p wf is the pressure at the inner radius and p is the average aquifer pressure. The later value can be calculated from a material balance for a constant compressibility: p W e --------- + p, (3.31) c e W i where W is the water content and c e c w + c φ the total compressibility of the aquifer. The maximum encroachable water at p is: W ei c e Wp i (3.32) After substituting Eq. 3.32 into Eq. 3.31: p p i -------- W e + p i W ei (3.33) The calculation is reduced to the following steps: For a time interval j+1 t t j+1 - t j the constant influx rate would be: q w J w ( p j p wfj + 1 ), (3.34) where p j is the average aquifer pressure for the time t j and p wfj+1 is the average inner boundary pressure during the period j+1 t.

Chapter 3: Material Balance 37 The total efflux during the timer interval j+1 t would be W ej + 1 W ej + 1 W ej q w j 1 + t (3.35) The cumulative efflux to the time j+1 t would be: W ej 1 j + 1 + W ej + W ej + 1 W en n + 1 (3.36) The aquifer average pressure for the next time interval: p i p j + 1 W ei --------W ej 1 + + p i (3.37) The efflux of the aquifer is naturally the water influx for the reservoir. The VAN EVERDINGEN-HURST [16.] method in the original form and in the modified form by VOGT and WANG [41.] as well uses three parameters: C 3, α and r ed. The first two are numerical constants, the third relates to the mathematical assumption of a radial symmetrical aquifer. The Fetkovich [17.] method uses only two parameters (J w and W ei ) and no relation was made to any geometrical form. In spite of that J w and W ei can be calculated for the radial symmetric case easily as described in the volume "Fluid Flow in Porous Media": J w ----------------------------- 2πhkf r e µ ----- 3 ln -- 4 r w (3.38) 2 W ei fπ( r e r 2 w)hc e p i (3.39) where f1 if the circle is in full size. For field units replace 2π by 7.8x1-3 in Eq. 3.38 to get [bbl/psi d]. There Eq. 3.34 - Eq. 3.37 give the exact solution of the time step go to zero. For practical cases a time step of some months gives results accurate enough. Example 3.1 An oil reservoir contains N 3.6 x 1 6 m 3 [2.264x1 7 bbl] oil. The ratio of the gas-/oil-pore volume is m.3. The change of the reservoir pressure and the cumulative productions are given in columns (1)-(5) in Table 3.4 and Table 3.5. Columns (6)-(8) tabulate the values for B g, B o and R s at the corresponding reservoir pressure. The task is to calculate water influx.

38 Chapter 3: Material Balance Solution From Eq. 3.6: W e N p [ B o + B g ( R p R s )] B g N B o B oi + mb oi ------- 1 + B g ( R si R s ) + W p B gi The routine of the calculation procedure can be set up as follows (the numbers indicate the column numbers): ( 9) ( 3) [( 6) + ( 7) (( 4) ( 8) )] N [( 6) B oi + mb oi (( 7) B gi 1) + 7 ( R si ( 8) )] + ( 5) The results are written in Column (9) Example 3.2 Figure 3.4 shows an oil reservoir with a semicircular aquifer. The parameters of the aquifer are r w - 1 m [328 ft] r e - 5 m [164 ft] h - 7.2 m [23.6 ft] φ -.23 k -.225x1-12 m 2 [~ 22.2 md] µ w -.25 Pas [.25 cp] At the time t reservoir pressure is reduced by 1 MPa [145 psi] and is then kept constant. The water influx can be considered steady-state. The cumulative water influx after 3 years is to be calculated.

Chapter 3: Material Balance 39 Solution From Eq. 3.22 for the semicircular aquifer: 1 C 2r -- -------------- 2πhk 2 µ r e ln----- r w π7.2.225 1 12 -----------------------------------------------------.25x1 3 1.265 1 9 m 3 Pa 1 ln5 s 1 C 19.3 m 3 In field units: MPa 1 d 1 1 C 2r -- 2 3 7.8 1 ------------------------------------------------------------- 23.6 22.2.25 ln5 4.695 bbl psi d The cumulative water influx after 1 days totals W e C 2r pt 19.3 1 1 19 1 3 m 3 In field units: W e 4.695 145 1 67741 bbl Example 3.3 The task is to determine cumulative water influx after a production time of t 1 days at non-steady-state conditions (Fig. 3.4). During this time the reservoir pressure is reduced linear by 1 MPa [145 psi]. The aquifer acting infinite. The parameters of the aquifer are: A 14x1 6 m 2 [3459 ac] h 7.2 m [23.6 ft] φ.23 k.225x1-12 m 2 [~ 222 md] c φ 6x1-1 Pa -1 [4.13685x1-6 1/psi] c w 5x1-1 Pa -1 [3.4473x1-6 1/psi] µ w.25x1-3 Pas [.25 cp] A is the area of the aquifer.

4 Chapter 3: Material Balance Solution The radii of the reservoir are r w 1 -- 2A 2 ------ 28 1 6 -------------------- π π 2985 m In field units: 2 3459 4356 r w ----------------------------------------- π 1 -- 2 9794 ft The effective compressibility of the aquifer is c e c w + c φ ( 6 + 5) 1 1 1.1 1 9 Pa 1 In field units: c e 3.448 1 6 + 4.137 1 6 7.585 1 6 1 psi According to Eq. 3.26: α k -------------------- 2 µφc e r w.225 1 12 ------------------------------------------------------------------------------------------------.25 1 3.23 1.1 1 9 2985 2 α.39924 1 6 s 1.345 d 1 In field units: α 6.34 1 3 ------------------------------------------------------------------------------------------------------ 222.25 1 3.23 7.585 1 6 9794 2.345 d 1 From Table 3.1, at t D αt 34.5 we get: Q ( t D ) 38.5562

Chapter 3: Material Balance 41 Eq. 3.28, since the reservoir is semicircular C 3 id C 3 id 2 πφhc e r w π.23 7.2 1.1 1 9 2985 2.51 m 3 Pa 1 51 m 3 MPa 1 In field units: C 3 id.1781π.23 23.6 7.585 1 6 9794 2 228 bbl psi From Eq. 3.26: W e id p C 3 ------Q ( t t D ) 51 1 38.5562 34.5 568 m 3 D In field units: W e 228 145 38.5562 34.5 3.53246 1 6 bbl Example 3.4 A reservoir produces three years. Reservoir pressure has decreased from 3 MPa [435 psi] to 27 MPa [3915 psi]. The data known are tabulated in columns (1)-(3) in Table 3.4 and Table 3.5. The cumulative water influx after 3 years is to be calculated. The parameters of the aquifer are the same as in Example 3.3. Solution Constants α.345 d -1 and C 3 51 m 3 MPa -1 [2233 bbl/psi] were already calculated in Example 3.3. The water influx results in p W e C 3 j + 1 -------- t Q ( tdn t Dj )51 26.189 1.3356 1 6 m 3 D In field units: n 1 j W e 228 379.74 8.38465 1 6 bbl

42 Chapter 3: Material Balance Example 3.5 A water influx equation for the reservoir in Example 3.1 has to be determined. The radius of the reservoir is ~ 1 m [328 ft], the data of the aquifer are: φ.23 k 8x1-15 m 2 [8 md] c e 1.1x1-9 Pa -1 [7.585x1-6 1/psi] µ w.25x1-3 Pas [.25 cp] Solution First, it is essential to determine whether the water influx is steady-state or non-steady-state. The procedure is comprised in Table 3.6 and Table 3.7. The coefficients C 1 and C 2 are calculated for all t j by using Eq. 3.17 and Eq. 3.23. Due to the fact that C 1 increases and C 2 decreases, the water influx can be considered non-steady-state. The aquifer is assumed infinite and α is calculated as follows: α k -------------------- 2 µφc e r w 8 1 15 ------------------------------------------------------------------------------------------.25 1 3.23 1.1 1 9 1 6 α 1.2648 1 7 s 1.19 d 1 1 3month 1 In field units: α 6.34 1 3 ------------------------------------------------------------------------------------- 8.23.25 7.585 1 6 328 2.19d 1 ( 1 3)month 1 In Table 3.8 and Table 3.9, the terms n 1 j j + 1 -------- p t Q ( tdn t Dj ), n 1, 6 D are calculated for the last three time points. The coefficients C 3 are calculated in Table 3.6 and Table 3.7 using Eq. 3.26. The values C 3 can be regarded with fair accuracy as constants during the last 6 months. The value C 3 53 m 3 MPa -1 [23 bbl/psi] can be accepted for prediction purposes.

Chapter 3: Material Balance 43 t month Table 3.2: Calculation of Water Influx into an Oil-Reservoir - Example 3.1 (metric units) p MPa N p W 1 3 m 3 R p W p 1 3 m 3 B o B g R e s 1 3 m 3 (1) (2) (3) (4) (5) (6) (7) (8) (9) 23 1.332*.48* 97.8*. 6 22 87.8 94.. 1.2957.52 93.9 5.8 12 21 183.7 92..1 1.2879.526 9. 11.1 27 2 38.9 89. 1.1 1.28.552 86. 39.5 48 19 52.8 95. 3.3 1.2719.581 82. 166.8 75 18 693.2 99. 13.2 1.2636.613 77.9 292.2 18 17 95. 15. 21.8 1.255.649 73.8 468.5 t month Table 3.3: Calculation of Water Influx into an Oil-Reservoir - Example 3.1 (field units) N p W p W e 1 3 /bbl p R p R psi 1 3 bbl scf/bbl 1 3 B o B s g /bbl scf/bbl (1) (2) (3) (4) (5) (6) (7) (8) (9) 3335 1.332*.48* 549.2 6 319 552 527. 1.2957.52 527. 36 12 345 1155 516.6 1.2879.526 55. 7 27 29 1943 499 6.9 1.28.552 483. 248 48 2755 3162 532 2.7 1.2719.581 46. 149 75 261 4359 555 82.9 1.2636.613 437. 1837 18 2465 5691 589 136.9 1.255.649 414. 2946

44 Chapter 3: Material Balance Date Reservoir pressure MPa α.345 t D const. 6.3 α.345 t D const. 6.3 Table 3.4: Calculation of Water Influx into an Oil-Reservoir - Example 3.1 (metric units p MPa t days t Dj αt t D -t Dj 37.8-t Dj p t D Table 3.5: Calculation of Water Influx into an Oil-Reservoir - Example 3.1 (field units) (8)x(9) (1) (2) (3) (4) (5) (6) (7) (8) (9) (1) 76.7.1 3..1 37.8.158.158 454.475 7.181 77.1.1 29.8.25 183 6.3 31.5.397.239 314.8 7.56 77.7.1 29.5.4 365 12.6 25.2.635.238 223.422 5.317 78.1.1 29..7 548 18.9 18.9.1111.476 132.683 6.316 78.7.1 28.1.75 73 25.2 12.6.119.79 67.573.538 79.1.1 27.5.55 913 31.5 6.3.873 -.323 2.74 -.669 79.9.1 27. 195 37.8 Σ 26.189 Date Reservoir pressure psi p psi t days t Dj αt t D -t Dj 37.8-t Dj p t D p --------- j + 1 t D p --------- j + 1 t D Q ( t Dn t Dj ) Q ( t Dn t Dj ) (8)x(9) (1) (2) (3) (4) (5) (6) (7) (8) (9) (1) 76.7.1 435. 14.5 37.8.229.229 454.475 14.12 77.1.1 4321. 36.3 183 6.3 31.5.576.347 314.8 18.84 77.7.1 4277.5 58. 365 12.6 25.2.921.345 223.422 77.9 78.1.1 425. 11.5 548 18.9 18.9 1.611.69 132.683 91.58 78.7.1 474. 18.8 73 25.2 12.6 1.725.114 67.573 7.8 79.1.1 3987.5 79.8 913 31.5 6.3 1.266 -.459 2.74-9.7 79.9.1 3915. 195 37.8 Σ 379.74

Chapter 3: Material Balance 45 Table 3.6: Determination of the Aquifer Type (metric units) No t p W e p i -p C 1 p j Σ p C 2 Σ j+1 C 3 (t-t j ) p -------- Q t D month MPa 1 3 m 3 MPa (3)/(4) MPa (3)/(7) (3)/(9) (1) (2) (3) (4) (5) (6) (7) (8) (9) 23. 1 6 22. 5.8 1. 5.8.5 3. 1.93 2 12 21. 11.1 2. 5.6 1. 12..93 3 27 2. 39.5 3. 13.2 1. 49.5.8 4 48 19. 166.8 4. 41.7 1. 123. 1.36 3.444 548 5 75 18. 292.2 5. 58.5 1. 244.5 1.19 54.46 536 6 18 17. 468.5 6. 78.1 1. 426. 1.1 88.29 531 N 36 m 3 Table 3.7: Determination of the Aquifer Type (field units) No t p W e p i -p C 1 p j Σ p C 2 Σ j+1 C 3 (t-t j ) p -------- Q t D month psi 1 3 bbl psi (3)/(4) psi (3)/(7) (3)/(9) (1) (2) (3) (4) (5) (6) (7) (8) (9) 3335 1 6 319 36.5 145 2.52 72.5 435 83.9 2 12 345 69.8 29 2.42 145. 174 4.1 3 27 29 248.4 435 5.72 145. 7177 34.6 4 48 2755 149. 58 18.9 145. 17835 58.8 4413 237.6 5 75 261 1838. 725 25.38 145. 35452 51.8 7897 232.6 6 18 2465 2946. 87 33.88 145. 6177 47.7 1279 23.5 N 2.26x1 7 bbl

46 Chapter 3: Material Balance Table 3.8: Calculation of the p j + 1 -------- Q ( td t t Dj ) D Example 3.5 (field units) Function for p -------- p t j + 1 -------- D t D psi 145 72.5 72.5 1 2 145 72.5. 2 4 145 29. -43.5 3 9 145 2.76-8.294 4 16 145 16.195-4.5965 5 25 145 13.185-2.929 No t D p j For 6 th time point: t Dn 36 No t Dn -t Dj Q ( t D t Dj ) p j + 1 -------- Q t D 36 414.157 3 26.31 1 34 378.317. 2 32 326.557-14 25.215 3 27 249.154-2 66.54 4 21 148.294-681.645 5 11 96.536-282.75 Σ 12 79.16 For 5 th time point: t Dn 25 No t Dn -t Dj Q ( t D t Dj ) p j + 1 -------- Q t D 25 22.337 15 974.55 1 23 191.521. 2 21 162.74-7.77.595 3 16 99.872-828.385 4 9 37.394-171.825 Σ 7 896.7 For 4 th time point: t Dn 16 No t Dn -t Dj Q ( t D t Dj ) p j + 1 -------- Q t D 25 22.337 15 974.55 1 23 191.521. 2 21 162.74-7.77.595 3 16 99.872-828.385 4 9 37.394-171.825 Σ 7 896.7

Chapter 3: Material Balance 47 Table 3.9: Calculation of p j + 1 -------- Q ( td t t Dj ) D Example 3.5 (metric units) Function for p -------- p t j + 1 -------- D t D MPa 1.5.5 1 2 1.5. 2 4 1.2-3. 3 9 1.1428 -.572 4 16 1.1111 -.317 5 25 1.99 -.22 No t D p j For 6 th time point: t Dn 36 No t Dn -t Dj Q ( t D t Dj ) p j + 1 -------- Q t D 36 414.157 27.78 1 34 378.317. 2 32 326.557-97.967 3 27 249.154-14.252 4 21 148.294-4.71 5 11 96.536-1.95 Σ 88.28 For 5 th time point: t Dn 25 No t Dn -t Dj Q ( t D t Dj ) p j + 1 -------- Q t D 25 22.337 11.169 1 23 191.521. 2 21 162.74-48.811 3 16 99.872-5.713 4 9 37.394-1.185 Σ 54.46 For 4 th time point: t Dn 16 No t Dn -t Dj Q ( t D t Dj ) p j + 1 -------- Q t D 16 99.872 49.936 1 14 81.17. 2 12 62.342-18.73 3 7 25.29 -.793 Σ 3.44

48 Chapter 3: Material Balance 3.4 Finite Difference Material Balance Equation The time is divided into a finite number of optional intervals. At time j, Eq. 3.9 becomes N B o ----- R s B g B o N p ----- R s B g j B oi ------- R si mnb oi ------- 1 1 ------- 1 + + ------- ( W ej W pj ) j B gj + G pj B gi B gj B gj (3.4) For time j+1 N pj 1 + N pj + j + 1 N p N pj + q o j 1 + t G pj 1 + G pj + R j 1 G pj + q o R j 1 + N p + t, (3.41) W pj 1 + W pj + j + 1 W p W pj + q w j 1 + t where q o - oil production rate m 3 /d [bbl/d], q w - water production rate m 3 /d [bbl/d], R - the average production GOR. In order to simplify calculations, the water influx is assumed steady-state. (Calculations are similar, but more complicated for non-steady-state water influx). According to Fetkovich [17.] equation Eq. 3.35, W ej + 1 W ej + J w ( p aj p) j + 1 t, (3.42) where p aj is the avarage aquifer pressure at time j and p p j + 1 + p j ---------------------- 2 (3.43) is the average reservoir pressure in a time period (t j, t j+1 ). Thus, Eq. 3.4 at time j+1 leads to

Chapter 3: Material Balance 49 N B o ----- R s B g j + 1 B oi ------- R si mnb oi ------- 1 1 --------------- 1 + + --------------- ( W ej W pj ) B gj 1 + --------------- [ J w ( p aj p) q w ] j 1 B gj + 1 B o N pj ----- R s B g j + 1 + t B gi B gj + 1 B o + G pj + j + 1 N p ----- R s B g j + 1 + R B gj + 1 (3.44) From Eq. 3.44 either the production for the time intervall j,j+1 j + 1 N p B oi mnb 1 oi ------- 1 --------------- B gj B N B o + ----- R gj + 1 B s --------------- R g B si j + 1 gj + 1 ----------------------------------------------------------------------------------------------------------------------------------------------- B ----- o R s + R B g j + 1 (3.45) B o ( W N pj ----- e W p ) J R B s G pj ------------------------- w ( p aj p) q w + + --------------------------------------- g B j + 1 gj + 1 B j + 1 t gj + 1 + N p + -------------------------------------------------------------------------------------------------------------------------------------------------------- B ----- o R s + R j 1 B g j + 1 or the duration j + 1 t B oi mnb 1 1 oi ------- --------------- N B o + ----- R B gj B gj + 1 B s --------------- R g B si j + 1 gj + 1 ----------------------------------------------------------------------------------------------------------------------------------------------- B o J q o ----- w ( p aj p) q w R s + R ------------------------------------- B g j + 1 B gj + 1 (3.46) B W o e W p j N pj ----- R B s G pj + ---------------------- g j + 1 B gj + 1 + t + ---------------------------------------------------------------------------------- B o J q o ----- w p aj p q w R B s + R -------------------------------- g j + 1 B gj + 1 j 1 can be expressed. Application of the differential form of MB equation:

5 Chapter 3: Material Balance 1. Eq. 3.44 is applied for the production forecast j+1 N p, if the reservoir pressure drops from p j to p j+1 in a time period j+1 t. 2. Eq. 3.46 enables the calculation of the time period j+1 t corresponding to a given pressure change j+1 p and production j+1 N p. 3. For the third case, where the pressure change in a certain time interval j+1 t at a given production rate has to determined, a NEWTON- RAPHSON iteration must be applied to calculate j+1 N p and j+1 t according to j+1 p. Subtraction of Eq. 3.4 from Eq. 3.44 leads to B o N j + 1 B g ----- R s NBoi 1 j + 1 ----- mnb oi 1 j + 1 ----- 1 + ( W e W p ) j j + 1 ----- B g B g B g 1 + --------------- [ J w ( p aj p) q w ] j + 1 t B gj + 1 B o B N pj j + 1 ----- o R s + j + 1 N p ----- R s B g B g j + 1 + R (3.47) where 1 j + 1 ----- B g B o j + 1 B g 1 --------------- B gj + 1 1 -------, B gj B ----- o R s ----- R s B g j + 1 B o ----- R s. B g j (3.48) After reordering: B o 1 ( N N p ) j + 1 ----- R B s ( 1 + m)nb oi j + 1 ----- g B g ------------------------------------------------------------------------------------------------------------------------- B o j + 1 N p ----- R s + R B g j + 1 ( W e W p ) j 1 ----- 1 + --------------- [ J B g B w ( p aj p) q w ] j + 1 t gj + 1 + -------------------------------------------------------------------------------------------------------------------------------------- 1 B o j + 1 N p ----- R s + R + 1 B g j + 1 (3.49) Again, it is possible to split this term into three fractions, each representing one specific drive index at a production of j+1 N p :

Chapter 3: Material Balance 51 The solution gas drive index: B o 1 ( N N p ) j + 1 ----- R B s NB oi j + 1 ----- g B g i s ------------------------------------------------------------------------------------------------------ B o B g ----- R s j + 1 + R j + 1 N p (3.5) The gas drive index: mnb oi j 1 ----- B g i g --------------------------------------------------------------------, (3.51) B ----- o R s + R j + 1 N p B g j + 1 The water drive index: + 1 ( W e W p ) j + 1 ----- 1 1 B + --------------- [ J g B w ( p aj p) q w ] j + 1 t gj + 1 i w ------------------------------------------------------------------------------------------------------------------------------------ B ----- o R s + R j + 1 N p B g j + 1 (3.52) Again i s + i g + i w 1 (3.53) is valid. In contrast to the indices applied in Eq. 3.11 - Eq. 3.13, which refer to the total cumulative production, these are only valid for the production in the time interval j+1 t. 3.5 Undersaturated Oil Reservoirs Up to this point, the compressibility of both the connate water and the reservoir rock has been neglected due to the compressibility of gas and the two phase compressibility of oil. In case of an unsaturated reservoir, oil is in single phase state if pressure ranges from initial pressure p i to bubble point pressure p b. Since the compressibility of water is in its order of magnitude comparable to that of oil, it has to be taken into consideration also.

52 Chapter 3: Material Balance A simplification can be achieved by introducing an apparent oil compressibility c oe representing all three phases: S o c oe c o S o + c w S w + c φ. (3.54) Since S o 1 S w, then S w c oe c o + c w. (3.55) 1 -------------- + S c -------------- 1 φ w 1 S w The volume of the reservoir oil at initial pressure p i is V p ( 1 S wi ) NB oi. (3.56) If the reservoir pressure drops to p as a consequence of the production N p, the volume of the total reservoir contents taking the water influx into account would be NB oi [ 1 + c oe ( p i p) ] + W e. (3.57) Due to the fact that the compressibility of the pore volume is conveyed to the oil, the pore volume is considered constant as in Chapter 3.1. The difference between the volumes according Eq. 3.57 and Eq. 3.56 lies in the volume of cumulative production at pressure p: N p B o + Thus, W p. (3.58) NB oi c oe ( p i p) + W e N p B o + W p (3.59) Substitution of Eq. 3.55 leads to B oi ( c φ + S wi c w )( p i p) N B oi c o ( p i p) + ---------------------------------------------------------- 1 S wi + W e N p B o + W p. (3.6) Since B o B oi [ 1 + c o ( p i p) ], (3.61) Eq. 3.6 can be written as follows: B oi ( c φ + S wi c w )( p i p) N B o B oi + ---------------------------------------------------------- 1 S wi + W e N p B o + W p. (3.62) Analogous forms can be set up for Eq. 3.45, Eq. 3.46 and Eq..

Chapter 3: Material Balance 53 The equations of material balance for unsaturated reservoirs can be applied solely between initial and bubble point pressure. In case of pressure being below bubble point pressure, the equations for saturated reservoirs must be applied. 3.6 Gas Reservoirs The compressibility of rock and connate water are again neglected. The formation volume at initial presure p i is defined as V p ( 1 S wi ) GB gi (3.63) The production G p effects a drop of the reservoir pressure. The total reservoir content including the water influx then adds up to GB g + W e. (3.64) The difference between Eq. 3.63 and Eq. 3.64 is conditioned by the volume of cumulative production at pressure p j : GB g + W e GB gi G p B g + W p (3.65) and thus G G p B g W ------------------- e W p --------------------. (3.66) B g B gi B g B gi The formation volume factor can be calculated from the real gas equation. B g C ZT ------ C p p o ----------- Z o T o (3.67) Since C and T (reservoir temperature) are constant, Eq. 3.66 can be written as follows: Z -- p W G G p ------------------------ e W p -------------------- Z -- p Z -- p B g B gi i (3.68) If no water influx is present, a gas reservoir is called volumetric. Thus, the second term on the right side of the equatin vanishes and Eq. 3.68 can be transformed to -- p Z p -- Z i G p 1 ------. (3.69) G

54 Chapter 3: Material Balance A specific property of a volumetric reservoir is that the fraction p/z is a linear function of the cumulative production. Therefore, the presence of a water influx is easily observed by this function not being linear (see Fig. 3.8). Figure 3.8: Pressure-drop in a gas reservoir 3.7 Calculation of Original Fluid in Place The material balance formulas as used in Chapters 3.1, 3.4, 3.5 can be regarded as specific cases. The general form of the material balance equation is B oi ( c φ + S wi c w )( p i p) B N B o B oi B g ( R si R s ) ---------------------------------------------------------- g + + + GB 1 S gi ------- 1 + W wi B e gi (3.7) N p [ B o + B g ( R p R s )] + W p ( W I + G I B g ) where W I is the cumulative injected water and G I the cumulative injected gas. Eq. 3.7 includes the material balance equations derivated earlier: The compressibility of the pore volume and the connate water can be neglected if the reservoir contains free gas. Then R si > R s and/or GB gi mnb oi. Disregarding gas and water injection means that Eq. 3.7 is identical with Eq. 3.6. If the reservoir pressure is above bubble point pressure R s R si, R p R si, and GB gi. Therefore, Eq. 3.7 becomes identical with Eq. 3.62.

Chapter 3: Material Balance 55 In a gas reservoir, N, N p. Thus Eq. 3.7 is identical with Eq. 3.65. The water influx can be calculated by the modified VAN EVERDINGEN-HURST method (see Eq. 3.26): p W e C j + 1 --------, (3.71) t Q ( tdn t Dj ) Cη w D where n 1 j n 1 p η w j + 1 -------- t Q ( tdn t Dj ) D j (3.72) Further, the following notions are defined: ε o B o B oi + B g ( R si R s ), (3.73) B oi ( c φ + S w c w )( p i p) ε w --------------------------------------------------------, (3.74) 1 S wi ε ow ε o + ε w ε g B g B gi, (3.75), (3.76) Q F N p [ B o + B g ( R p R s )] + W p ( W I + G I B g ). (3.77) Substituting Eq. 3.71 - Eq. 3.77 into Eq. 3.7 the following simple form is obtained: Nε ow + Gε g + Cη w Q F. (3.78) In order to determine the functions ε, η and Q F, it is necessary to have knowledge of both the average reservoir pressure as a function of time and cumulative production. If the corresponding values are known at n dates, it is possible to set up a system of n linear equations: Nε ow1 + Gε g1 + Cη w1 Q F1 Nε ow2 + Gε g2 + Cη w2 Q F2 (3.79)... Nε own + Gε gn + Cη wn Q Fn Solutions for N, G and C could be obtained by the method of the GAUSS s normal equations, whereby Eq. 3.79 is gradually multiplied with the coefficients of the first,

56 Chapter 3: Material Balance second and third unknown and than summed up: N ε 2 + owj G ε gj ε + owj C η wj ε Q owj Fj ε owj N ε owj ε + 2 G ε gj + gj C η wj ε Q gj Fj ε gj N ε owj η + wj G ε gj η + wj C η 2 wj Q Fj η wj (3.8) where n a. (3.81) j 1 3.8 Graphical Evaluation of Material Balance Let us consider Eq. 3.78: Nε ow + Gε g + Cη w Q F (3.82) or with G mnb ---------------- oi B gi (3.83) in the form: mnb oi N ε ow + ----------------ε g + Cη w Q F B gi (3.84) HAVLENA and ODEH [19.] used this equation for graphical evaluation of reservoir performances. We investigate several special cases illustrated in Fig. 3.9 and Fig. 3.1. 3.8.1 Reservoirs Without Water Influx: W e a) For an oil reservoirs without a gas cap m and Eq. 3.84 becomes: Q F Nε ow. (3.85)

Chapter 3: Material Balance 57 Illustrated in the coordinate system ε ow νs Q F, Eq. 3.85 results in a straight line through the origin of the coordinate system with a slope N. b) For an oil reservoir with a gas cap and supposing that m is known and Eq. 3.84 becomes: Q F N ε ow m B o + ----- B g ε g i. (3.86) Illustrated in the coordinate system ε ow + m(b o /B g ) i ε g νs. Q F, Eq. 3.86 appears as a straight line through the origin of the coordinate system with a slope N. c) For an oil reservoir with a gas cap of unknown size must one select a value for m. A diagram is drawn according to the case b). If m is too large, the series of points will bend downwards and upwards if m is too small. After several attempts, the correct m will result in a straight line. d) For a gas reservoir Eq. 3.82 becomes: G p B g Gε g. (3.87) Illustrated in the coordinate system ε g νs G p B g, Eq. 3.87 is straight line through the origin of the coordinate system with a slope N. 3.8.2 Reservoirs With Water Influx a) For an oil reservoir without a gas cap: m. Eq. 3.84 can then be written as Q F -------- N C η w + --------. (3.88) ε ow ε ow Illustrated in the coordinate system Q F /ε ow νs. η m /ε ow, Eq. 3.88 is a straight line with a slope C. The intersection of the straight line with the axis of coordinates gives N.

58 Chapter 3: Material Balance Q F Q F N N AA) W : m e ε o ε o +m(b o /B g ) i e g AB) W : mknown e Q F m too small m too large GB p g G ε o AC) W : munknown e AD) Gas reservoir Figure 3.9: Graphical illustration of material balance without water influx (W e ) b) For an oil reservoir with a gas cap of known size Eq. 3.84 can be written as ε g Q F ----------------------------------- ε o m B N+ C----------------------------------- o + ----- ε g ε o m B o + ----- ε g B g i η w B g i (3.89) The analogy to case a) is obvious. c) Gas reservoir Eq. 3.82 can be written in the following form: G p B g + W --------------------------- p G C η w + ------. (3.9) ε g ε g Evaluation is made as in case a). In order to calculate n 1 p η w j + 1 -------- t Q ( tdn t Dj ) D j (3.91) one must estimate the parameters r ed r e ----- and α r w -------------------- k 2 µc e φr w (3.92)

Chapter 3: Material Balance 59 correctly. If η w has a large value, Eq. 3.88 will be a rather flat curve. η w too small Q F ε ow N C η w too large Q F ε m N C η/ε w ow η w/εm BB) Q F ε ow η w ε ow N+C ; m BC) Q F ε m η w ε m N+C ; m> GB+W p g e ε g G C BD) GB+W p g p ε g G+C η w ε g η/ε w g Figure 3.1: Graphical illustration of material balance with balance with water influx (W e >) 3.9 Recovery Factor The ultimate recovery factor is defined as the relation between the ultimate recovery and the original oil in place, short OOIP: E R N --------------- pmax. (3.93) N The recovery factor can be applied either for the whole or selected parts of the reservoir. The recovery factor at a specific point is considered equal to that of the displacement efficiency inferred by the existing oil recovery mechanisms at just this point. If several oil recovery mechanisms exist in a reservoir, they will result in an overall oil recovery (various possible combinations of elementary mechanisms are considered extra): E R n j 1 E E vj Dj, (3.94) where E vj is the volumetric efficiency of recovery mechanism j.

6 Chapter 3: Material Balance Of course, n E vj 1. (3.95) j 1 It is customary to split E v into two factors: E A - areal sweep efficiency E I - vertical sweep efficiency. The contribution of recovery mechanism j to the recovery factor is defined by the multiplication of the following three factors: j E R E Dj E Aj E Ij. (3.96)

61 Chapter 4 Displacement Efficiency 4.1 Solution Gas Drive Imagine a part of the reservoir with 1 m 3 [bbl] pore volume, an immobile water saturation S wi and oil saturation S oi 1 - S wi. The oil is saturated with gas at initial reservoir pressure. In case of a pressure drop, the fluids will expand and gas will liberate from the oil. The total fluid volume will increase. Since the pore volume remains constant, surplus fluid must be displaced. An essential precondition for displacement is the mobility of the phase. In this case, the immobility of the water is inferred from the constant value of water saturation (expansion of the water was neglected). The involved gas develops a disperse, free gas phase consisting of small immobile gas bubbles. The oil is the only phase capable of flowing out of the regarded volume. A further pressure drop enlarges the number and size of the gas bubbles. The bubbles start to connect themselves and consequently develop continuous channels. The gas begins to move, more and more gas rather than oil will flow out. Pressure drops rapidly and as a result only oil with lost interior energy will remain. 4.1.1 MUSKAT s (1945) Equation of Solution Gas Drive A pressure below the bubble point p b is assumed. Water, oil and gas saturations represent the reservoir fluid volumes: S o + S g + S w 1 (4.1) An unit pore volume at this pressure contains S o B o sm 3 [stb] stock tank oil and 61

62 Chapter 4: Displacement Efficiency S o sb o ----- + ( 1 S o S w )----- sm 1 3 [ scf] B g (4.2) gas. The first term in this formula is the amount of dissolved, the second term the amount of free gas. As pressure drops from p i to p, Q o sm 3 [stb] oil and Q g sm 3 [scf] gas are produced from this volume unit. Further pressure drops of a differently small dp cause an increase of production by dq o and dq g. Naturally these terms must coincide with the changes of the oil and gas contents: dq --------- o dp d ----- S o ----- dp B o 1 ----- ds o S -------- ----- db o o + ------- B o dp 2 B dp o (4.3) and dq --------- o dp d S ----- o R dp s ----- + ( 1 S B w S o )----- 1 o B g R ----- ds s o R -------- s S ----------- db o o S -------- ----- dr o s -------- d + ( 1 S dp dp dp w S o )----- 1 ----- ----- 1 ds o + -------- dp dp B o B o 2 B o B g B g (4.4) The momentary GOR results in: R dq g B --------- R. (4.5) dq s ----- k o g ---- µ o + ----- o B g k o µ g Eq. 4.5 demands a reduction of saturation both of oil and gas in such a way that the equation for the GOR is fulfilled when regarding the present saturations. A combination of Eq. 4.3, Eq. 4.4 and Eq. 4.5 thus leads to R s R s R ----- ds s o R -------- s S ----------- db o o S -------- ----- dr o s + -------- B ----- k o g ---- µ B o dp 2 o B dp B o dp + ----- o --------------------------------------------------------------------------- B g k o µ g ----- 1 ds o S -------- -------- db o o -------- dp dp B o B 2 o B ----- k o g ---- µ ( 1 S w S o ) ----- d ----- 1 ----- 1 ds o o dp B -------- + ----- g B g dp + --------------------------------------------------------------------------------- B g k o µ g ----- 1 ds o S -------- o B ----- o ----- B o dp 2 B dp o (4.6)

Chapter 4: Displacement Efficiency 63 Thus ds -------- o dp can be defined as ds -------- o dp µ o B g S o ----- dr s k -------- g S B o dp o ---- ------------ db o -------- d ( 1 S k o B o µ g dp w S o )B g ----- 1 + + ----- dp B ------------------------------------------------------------------------------------------------------------------------------------------------- g k g 1 ---- µ o + ----- k o µ g (4.7) This is a nonlinear first order differential equation. The oil saturation at pressure p is calculated by integration from p i to p. The following functions are introduced: λ( p) B g ----- dr s -------- dp B o, d ε( p) B g ----- dp ----- 1 B g (4.8) η( p) 1 ----- µ o ----- db o --------, B o µ g dp ω( ps, o ) 1 + k g ---- µ o ----- k o µ g. These are the MUSKAT functions and are given either graphically (Fig. 4.1) or in tables, but never analytically. Therefore integration can only be carried out numerically. The finite difference form of Eq. 4.7 is defined as S o p k g S o λ( p) + S o ---- η( p) + ( 1 S k w S o )ε( p) ---------------------------------------------------------------------------------------------------- o ω( ps, o ) (4.9) and the cumulative amount of produced oil as p N p V p S o S ----- o V p ----- pi B o B o i S o ----- B o. (4.1)

64 Chapter 4: Displacement Efficiency 2 η(p) λ(p) 1.5 14 13 11 λ(p)*1 4 1 λ(p) 1..5 η(p)*1 2 9 7 5 3 λ(p)*1 4 25 1 5 75 1 125 15 175 Pressure [bar] Figure 4.1: MUSKAT s function: λ and η The method of MUSKAT is advantageous if a greater number of calculations are to be executed, whereby the effects of various factors on the solution gas drive are to be analysed. The fluid and rock characteristics remain unchanged. The functions λ( p), η( p), ε( p), ω( p, S o ) are calculated only once. The corresponding values are obtained from the given curves. Nevertheless there are various methods to solve Eq. 4.7 or Eq. 4.9. Several authors have analysed the effects of various factors influencing the progress of expansion and recovery. MUSKAT and TAYLOR [32.] (1946) using Eq. 4.7, analysed the influence of µ o - viscosity p i - initial pressure R si - solution gas-oil ratio k rg /k ro - relative permeability ratio, see Fig. 4.2 S wi - connate water saturation Recovery as a function of the oil viscosity µ o in case of solution gas drive is illustrated in Fig. 4.3. Essential for production planning of a volumetric reservoir is the calculation of the pressure decrease p(n p ), the gas -oil ratio R(N p ) and the production index J(N p ), each as a function of the cumulative oil production N p. In case the initial production rates of each well are known, N p may be calculated with the help of the productivity index as a function of time.

Chapter 4: Displacement Efficiency 65 k rg Intersection k ro point A Slope b 1*2.33 1 3 S o 1 2 k r 1 S or k rg kro S gc Sgc 1 1 1 1-1 1-2 S o 1 Cycle 1. S or S gc S o S o Figure 4.2: The relative gas and oil permeabilities and their relation as a function of the oil saturation 3 3 Gas - oil ratio [ft /bbl]* 1 13 12 11 1 9 8 7 6 5 4 3 2 1 Pressure [psi]*1 2 26 24 Res v 22 2 18 16 14 12 1 8 6 I II III IV V 4 2 rat I II III IV V 1 2 3 4 5 6 7 8 9 1 11 12 13 14 15 16 17 18 19 Cumulative recovery in per cent of pore space Gas er o r e i - oil p sure r s io Curve I: atmospheric-pressure viscosity 11.4 cp. Curve II: atmos. pres. viscosity 5.52 cp. Curve III: atmos. pres. viscosity 2.76 cp. Curve IV: atmos. pres. viscosity 1.38 cp. Curve V: atmos. pres. viscosity.69 cp. Gas solubility at 25 lb. per sq. in. 534 cu. ft. per bbl. in all cases Figure 4.3: Pressure and gas oil ratio histories of solution gas-drive reservoirs producing oil of different viscosities (after MUSKAT and TAYLOR, 1946)

66 Chapter 4: Displacement Efficiency 4.1.2 Calculation of the Solution Gas Drive According to PIRSON The task is again the calculation of the pressure drop and GOR as a function of the cumulative produced oil. The equations will be written for N m 3 O.O.I.P.. Fluid saturation equation: ( N N p )B o S L S w + --------------------------- ( 1 S w ) NB oi. (4.11) GOR equation: B R R s ----- k o rg ------ µ o + -----. (4.12) B g k ro µ g PIRSON [33.] (1958) applied the material balance form in its difference form which is advantageous because it converges and the number of numerical operations is minimal. Since only the solution gas drive is acting the drive index i s 1 can be set in Eq. 3.53 and the production increase can be calculated as follows: where j + 1 N p B o 1 ( N N pj ) j + 1 ----- R B s NB oi j + 1 ----- g B g ------------------------------------------------------------------------------------------------------------- B ----- o R s + R B g j + 1 (4.13) R R j + R ----------------------- j + 1. 2 j or j + 1 refer to two subsequent dates. Further, the production capacity is defined by the productivity index: J lim [ q o ( p p wf )] p wf p (4.14) where p - the reservoir pressure

Chapter 4: Displacement Efficiency 67 p wf - the bottom hole flowing pressure q o - the oil production rate J i J(p i ) is the productivity index at initial reservoir pressure. The change of the productivity index during recovery may be calculated with the help of the two phase DARCY-equation: ( B o µ o ) i J J i k ro ------------------ B o µ o. (4.15) In order to avoid cumulative errors, it is advisable to check this method using the finite form of the material balance equation. For this, the following formula is applied: B o j + 1 N pj + 1 ----- R B s + N p R g j + 1 j 1 -------------------------------------------------------------------------------------- 1 B o B N ----- oi R s --------------- R si B g j + 1 B gj + 1 (4.16) Fig. 4.4 shows the characteristic functions p(n p ), R(N p ) and J(N p ) of a solution gas drive reservoir with initial pressure p i p b. pp i b p R J R si.1.2 Recovery Final pressure.3 N /N p Figure 4.4: Pressure, production index and GOR as a function of the production N p /N

68 Chapter 4: Displacement Efficiency 4.2 Frontal Displacement Due to pressure drop in the oil zone of the reservoir, the aquifer and gas cap expand, whereby the oil is displaced front-like by gas and water. This front is considered solely a transition zone between the displaced and displacing phase. Its size is very small in relation to the reservoir. Not only natural depletion, but also artificial methods such as gas and water injection perform frontal displacement. Gas injection wells are situated inside the gas cap, water injection wells at the exterior boundary of the reservoir. If reservoirs have a minor thickness and inclination, the displacing phase can also be injected inside the oil zone. However, direction of flow and distribution of saturation are in this case much more complicated. In modern reservoir engineering, these processes are calculated with numerical solutions of very complicated systems of differential equations. But this does not imply that analytical solutions of more simple and idealized models are superfluous. An essential aspect of these models is the possibility to make universal assertions and to have a quick overall picture of the considered case. This chapter analyses several idealized models in order to describe frontal displacement. 4.2.1 BUCKLEY-LEVERETT Theory This theory proves that displacement can proceed as a front and provides a clear description of how phase mobility and gravity influence this procedure. The calculation of front saturation and average saturation behind the front are additional aspects underlining the importance of this theory. The BUCKLEY-LEVERETT [4.] (1942) theory has proven successful for solutions of more complicated problems such as condensation gas drive, CO 2 gas drive or polymer flooding. Explanations considering this theory were already given in the volume "Fluid Flow in Porous Media" (HEINEMANN [21.] (1995)). Therefore it is only necessary to recall the most important aspects. BUCKLEY and LEVERETT [4.] (1942) assumed that: 1. the two fluids are non-compressible and immiscible, 2. the porous medium is homogeneous, 3. the displacement is one dimensional and stable,

Chapter 4: Displacement Efficiency 69 4. the filtration can be described by the multiphase DARCY-law, 5. the capillary force is neglectable. The equations for a one dimensional two phase filtration are u 1 kk r1 p --------- 1 ------- + ρ, (4.17) x 1 gsinα µ 1 u 2 kk r2 p --------- 2 ------- + ρ, (4.18) x 2 gsinα µ 2 where α is the inclination of the direction of displacement. Phase 1 is the displacing phase, and can either be water or gas. Phase 2 is the displaced phase: oil or gas. The difference between the two phase pressures p 1 and p 2 is regarded as the capillary pressure: p 2 p 1 P c ( S 1 ). (4.19) This is neglected according to condition (5). Therefore, p 2 p 1 p. (4.2) Let u u 1 + u 2 (4.21) be the overall filtration velocity, which is according to condition (1) independent of x. The portion of the displacing phase f 1 referring to the overall flow is calculated with the help of Eq. 4.17 and Eq. 4.18. Therefore f 1 ( S 1 ) u 1 ---- u µ 2 k( ρ ------ 1 ρ 2 )gsinα ---------------------------------------- k r2 u ------------------------------------------------------ µ 1 ------ + ------ k r1 µ 2 k r2 (4.22) or f 1 ( S 1 ) 1 kk ------------------------- r2 ( ρ 1 ρ 2 )gsinα k r2 1 ------ µ 1 ------------------------------------------------ 1 µ + ----- 2 u k r1 µ 2. (4.23)

7 Chapter 4: Displacement Efficiency The function f 1 depends - due to the relative permeabilities - on the saturation S 1. Fig. 4.6 shows this function at a constant displacement velocity. In the volume "Fluid Flow in porous Media" (HEINEMANN [21.] (1995)) it was shown that every saturation value moves with the constant velocity: w -- u df 1 -------- φ ds 1. (4.24) This term is only a function of u and S i. Thus, it is only necessary to know the initial saturation distribution and the displacement velocity in order to calculate the saturation distribution during the whole displacing process. This discontinuity is regarded as the front of displacement. Fig. 4.5 illustrates the displacement process. The initial saturation can be zero (for example in case of a gas displacement) or it can be so small that fluid 1 becomes immobile (S 1i < S1 c, S1 c is defined as the critical saturation). Otherwise it is large enough to be mobile. Two periods are to be distinguished. The first period lasts until the front has reached L. This is regarded as the breakthrough. The front saturation S 1F can be determined graphically with the f i -curve as shown in Fig. 4.7 The procedure is as follows: 1 S or S 1M S 1F S 1F S 1m x L Figure 4.5: Illustration of the displacing process according to Buckley and Leverett First a tangent is drawn from point S 1i to the curve. The tangential point indicates directly the front saturation and the fraction value of the front. In case the initial saturation (S 1i ) is higher than the critical saturation (S 1c ), the tangent is drawn from this initial value situated on the fractional flow curve to the curve itself.

Chapter 4: Displacement Efficiency 71 Evidently the front saturation (S 1F ) will be smaller than before. The same is valid for the average saturation behind the front (S 1F < S 1F ). Thus, the efficiency of the frontal displacement is smaller for S 1i > S1 c than for S 1i < S 1c. The average values of saturation behind the front (S 1F < S 1F ) remain constant during the progression of the front until breakthrough is achieved. On the right hand side of Fig. 4.7 a part of the fractional flow curve is drawn in magnification. The part of the curve between f 1F and f 1 1 describes the displacement conditions after breakthrough. The proportion of the displacing phase f 1L increases continuously at the exit of the medium. Saturation at x L is S 1L and average saturation is S 1L. The last value is the intersection on the f 1 lines with the tangent drawn from the point f 1L (S 1L ). Eq. 4.24 enables the calculation of the date of breakthrough: t d ---- L w f φl( S 1F S 1i ) ---------------------------------. (4.25) u 1. 5. f 1 df 1 ds 1 df 1 ds 1 f 1 S 1m S 1M 1. S 1 Figure 4.6: The fractional flow curve and its derivative The displacing efficiency at and after breakthrough is defined as S 1 S 1i E D ------------------. (4.26) 1 S 1i

72 Chapter 4: Displacement Efficiency 4.2.2 Oil Displacement by Water Fig. 4.8 left shows the relative permeability curves of a highly water wet and highly oil wet sandstone. Corresponding to this diagram is Fig. 4.9 right, which displays the fractional flow curves at the same viscosity ratio and horizontal displacement. In case of a given relative permeability function, the viscosity ratio has a deciding influence on the fractional flow curve and therefore on displacement efficiency. Fig. 4.9 was calculated on the basis of Fig. 4.8. The influence of gravity is considered as an advantage if the displacing phase is the heavier one and displacement progresses from bottom to top, or if it is the lighter phase and displacement progresses from top to bottom. The value f w describes the fraction of water in the overall flow at reservoir conditions. The notion q o and q w are defined as the oil and water production in reference to the volume at surface conditions, therefore: B w q -------------------------------- w f. (4.27) B w q w + B o q w o S 1F S 1F.3 S 1F S 1L S 1.6.7 1. f 1F S or.9 f f 1L 1F.8.5 f 1 f 1.7 f 1F.6 f 1i S 1i S 1c S 1i S 1F S 1F Saturation Figure 4.7: The fractional flow curve and graphical method of determining the front saturation and the average saturation behind the front..5

Chapter 4: Displacement Efficiency 73 k r 1..8.6.4.2 STRONG OIL-WET k ro k ro k rw STRONG WATER-WET k rw.2.4.6.8 1. S w A) RELATIVE PERMEABILITY CURVES OIL-WET WATER-WET µ o 1.2 cp µ w.8 cp ρ.2.4.6.8 1. S w Figure 4.8: Influence of the wettability on relative permeability and fractional curves f w 1..8.6.4.2 B) FRACTION FLOW CURVES 1..8.6 µ o 1 5 2 1.8 µ o 1 5 2 1 µ w 1..6 µ w f w.4 f w.4.2.2.2.4.6.8 1. S w.2.4.6.8 1. S w A) Oil-wetted B) Water-wetted Figure 4.9: The influence of the viscosity on the fractional flow curve In production statistics, the notions water cut (WC) or water oil ratio (WOR) are applied to describe water production WC q w q w q o -----------------, (4.28) + WOR q w -----. (4.29) q o

74 Chapter 4: Displacement Efficiency From Eq. 4.27 and Eq. 4.29: WOR B o f w ------ ------------- B w 1 f w (4.3) 4.2.3 Influence of Free Gas Saturation on Water Displacement Often water displacement takes place when a gas saturation has developed as a consequence of solution gas drive or gas displacement. Gas is mobile if S g > S gc. Now the relative permeabilities are a function of two variables: k rw k rw ( S w, S o ), k ro k ro ( S w, S o ), k rg k rg ( S w, S o ). (4.31) These are illustrated by a triangular diagram (Fig. 4.1). Fig. 4.11 shows a schematic illustration of the displacing process. The first section shows the existence of two mobile phases: oil and gas. The connate water is immobile and the free gas is displaced by the accumulating oil. Due to the fact that gas has a greater mobility than oil, the remaining gas saturation behind the front will be practically immobile. The gas saturation is also diminished by pressure increase effecting compression and solution in the oil. A three phase displacement can only be calculated numerically. An approximation is possible by applying the two phase flow equations because there are only two mobile phases present at a specific section. The BUCKLEY-LEVERETT [4.] (1942) solution is satisfactory for the calculation of the oil-gas and water-oil front saturations and material balance can be used to calculate the extension of the oil bank. According to laboratory experiments, the gas saturation in an oil bank (S gt ) depends on the initial gas saturation (S gi ). Fig. 4.12 shows a summary of several results achieved by CRAIG [8.] (1971). 4.2.3.1 The Residual Oil Saturation The residual oil saturation is defined by the relative permeability function. This is the nonreducible oil saturation achieved during a steady-state-measurement of relative permeability at an increasing water saturation (imbibition). Therefore, this saturation does not coincide with the oil saturation after displacement by water since it depends on several other parameters such as velocity of displacement, oil viscosity, capillary pressure function, injected pore volume. Fig. 4.13 demonstrates the effects of gas saturation on oil saturation after a complete water

Chapter 4: Displacement Efficiency 75 flooding in the case of water wet reservoir rock. This free gas saturation can have a beneficial effect on the displacing efficiency (assuming that no other factors oppose). In case of an oil wet reservoir, no specific associations between initial gas saturation and oil saturation after a water flooding are detectable. The influence depends on pore structure, viscosity ratio, injected amount of water and displacing velocity. The presence of gas diminishes the mobility of water and thereby can be considered beneficial for displacement. This facts results, as shown in Fig. 4.14, in a smaller amount of water needed.

76 Chapter 4: Displacement Efficiency A 1% gas Gas isoperms berea core Relative permeability [%] 6 5 4 3 2 1 5 1% B water A 1% gas 1 C 1% oil Oil isoperms berea core Irreducible saturation Relative permeability [%] 1 5 1 2 3 1% B water A 1% gas 4 5 6 C 1% oil Water isoperms berea core B 1% water 6 5 4 3 2 1 5 1 RELATIVE PERMEABILITY [%] C 1% oil Figure 4.1: Three-phase relative permeability functions (from Petroleum Production Handbook, Vol. 11, 1962)

S gt S gi Chapter 4: Displacement Efficiency 77 p 1. x S or S So Oil bank Initial saturation conditions Water front S or Figure 4.11: Water displacement at free gas saturation x S gt, trapped gas saturation, % pore volume 2 18 16 14 12 1 8 6 4 2 2 4 6 8 1 12 14 16 18 2 22 24 26 28 3 S, initial flowing gas saturation, % pore volume gi Figure 4.12: Correlation between initial gas saturation and residual gas saturation (after CRAIG, 1971)

78 Chapter 4: Displacement Efficiency S or, reduction in residual oil satn., % pore volume 18 16 14 12 1 8 6 4 2 2 4 6 8 1 12 14 16 18 2 22 24 26 28 3 S gi, INITIAL FLOWING GAS SATURATION, % PORE VOLUME Figure 4.13: Influence of initial gas saturation on the efficiency of water displacement (after CRAIG, 1971) 4.2.4 Displacement by Gas By assuming that the phases are incompressible and the gas will not dissolve in oil, the BUCKLEY-LEVERETT [4.] (1942) theory can be applied for frontal gas displacement. This is the case when displacement occurs at constant pressure with very small depression. The critical gas saturation at which the gas becomes immobile lies between and.12. Fig. 4.15 shows for the same relative permeability functions for different fractional flow curves. µ g is assumed 2 x 1-5 Pas [.2 cp] and ρ o - ρ g 7 kg/m 3 [43.71 lbm/cuft]. Diagram b) displays the influence of the oil viscosity in case of horizontal displacement and diagram c) the influence of gravity. At low velocities, gas displacement in a vertical direction can also achieve favorable results also with viscous oils. The influence of the displacement velocity is illustrated in diagram d). The BUCKLEY-LEVERETT [4.] (1942) theory may be applied for calculation of the displacement caused by the expanding gas cap or better by a gas injection into the gas cap. In these cases, one can assume that the gas does not condensate in the oil (i.e. it does not dissolve). Gas injection into the oil zone is sometimes associated (i.e. it does not dissolve). Gas injection into the oil zone is sometimes associated with a pressure higher than a reservoir pressure, which means that gas dissolved in the oil enlarges the volume and diminishes the viscosity of the oil. This process is called condensation gas drive. Fig. 4.16 shows a linear gas displacement at constant connate water saturation S wi. At the front, the gas saturation is S gf which increases at x to the value S gmax 1 - S wi - S or.

Chapter 4: Displacement Efficiency 79 1 Oil saturation in % of pore volume 9 8 7 6 5 4 Water breakthrough 1 P.V. water injected 22 CP. Oil 1.4 CP. Oil - Gas initially trapped - Gas initially mobile Water breakthrough 1 P.V. water injected 3 P.V. water injected 3 3 P.V. water injected 5 1 15 2 25 3 Trapped gas saturation in % of pore volume Figure 4.14: Relation between residual gas saturation and oil saturation in case of water flooding (after KYTE et al, 1956)

8 Chapter 4: Displacement Efficiency k r 1. k ro S gc k rg 1-Swi-Swo f g 1. 1-Swi-Swo S gc µ µ µ α 1 cp 5 cp 2 cp S g 1. A) Rel. perm. curves S g 1. B) Influence of oil viscosity f g 1. 1-Swi-Swo S gc α 9 α µ 1 cp u.1 m/day f g 1. 1-Swi-Swo S gc u.1 m/day u 1. m/day µ 1 cp α 9 S g 1. C) Influence of gravity S g 1. D) Influence of velocity Figure 4.15: Illustration of gas displacement when using the BUCKLEY-LEVERETT theory 1. S or Saturation Gas S gf Oil x Figure 4.16: Distribution of saturation in case of condensation gas drive For a specific date t, gas balance can be defined as follows: Injected gas free gas + additional dissolved gas. S wi Water

Chapter 4: Displacement Efficiency 81 q g t Aφ ------ S B gf x f + xs ( g ) ds g g ( +Aφ R s R si ) 1 S wi S or ----------------------- ( 1 S wi S gf )x f xs ( g ) ds g B o 1 S wi S or S gf S gf (4.32) q g is the gas injection rate, A the surface and R s - R si the increase of dissolved gas. The distance x f or x(s g ) are calculated according to the BUCKLEY-LEVERETT [4.] (1942) theory from Eq. 4.25: x f ut w f t ---- f φ gf B g q g t ------------- f Aφ gf (4.33) and xs ( g ) B g q g t -------------f A gf (4.34) x f and x(s g ) are inserted into Eq. 4.32: q g t Aφ ------ B g q g t ------------- S B g Aφ gf f gf + 1 f gf q g t + B g( R s R si ) ------------------------------ [( 1 S B wi S gf )f gf 1 + f gf ] o (4.35) or f gf where + a f gf ( S gf + b), (4.36) a B g ( R s R si ) ------------------------------------------, (4.37) B o B g ( R s R si ) B g ( R s R si ) b ( 1 S w )------------------------------------------. (4.38) B o B g ( R s R si ) Comparison of Eq. 4.35 with the common BUCKLEY-LEVERETT [4.] (1942) equation for S gi leads to f gf f gf S gf. (4.39)

82 Chapter 4: Displacement Efficiency This proves that the condensation gas drive can be designed graphically by coordinate transformation (see Fig. 4.17). The tangent to the f g curve is drawn from the point (-b,a), left from the origin. 1. S gf f g b a S gf S g 1. Figure 4.17: Graphical determination of the front saturation by condensation gas drive Example 4.1 A linear water displacement is to be calculated. The following reservoir parameters are available: Length m[ft] L 5 [164] Width m[ft] b 2 [656.2] Thickness m[ft] h 1 [32.8] Inclination [Grad] a 6 Porosity [-] f.2 Initial water saturation [-] S wi.27 Permeability m 2 [md] k 5.x1-13 [56.6] Oil: Formation volume factor B oi [-] B oi 1.22 Viscosity Pas [cp] µ o 2x1-3 [2] Density kgm -3 [lbm/cuft] ρ o 7 [43.69] Water: Formation volume factor B wi [-] B wi 1.1 Viscosity Pas [cp] µ w.7x1-3 [.7]

Chapter 4: Displacement Efficiency 83 Density kgm -3 [lbm/cuft] ρ w 98 [61.17] Filtration velocity ms -1 [ft/s] u 1.16x1-6 [3.85x1-6 ] Relative permeability [-] k r (see Tab.3.3.1) Solution The reservoir contains L b h φ( 1 S wi ) ----------------------------------------------- 11967 m 3 OOIP B oi In field units: ----------------------------------------------------------------------------------------------------------------------------------------------- 7758 164 656.2 32.8 ( 1 4356).2 ( 1.27) 1.22 5 7.5233 1 bbl OOIP The fractional flow curve is calculated with Eq. 4.24 in columns (4) and (5) in Table 4.1. Table 4.1: Calculation of the Fractional Flow Curve S w (1) k rw (2) k ro (3) (4) f w (5).27. 1....35.13.56.66.47.4.5.44.314.244.45.13.33.112.841.5.25.235.2331.25.55.5.165.464.4248.6.8.115.6653.6261.65.12.7.834.86.7.165.45.9129.8918.75.22.23.9647.9533.82.31. 1. 1. Fig. 4.18 shows the f w -curve. The tangent drawn from S wi to the f w -curve enables the determination of the front saturation: S wf.65, S wf.74.

84 Chapter 4: Displacement Efficiency At breakthrough, the oil recovery (now identical with the displacement efficiency) is calculated using Eq. 4.27: Column( 4) ------------------------- 1 µ w 1 ------- k ro + ------ k rw µ o kk ro ( ρ w ρ o )gsinα Column( 5) 1 ------------------------------------------------- ( 4) µ o u The displacement efficiency is S wf S wi E D -----------------------.74.27 1 S -------------------------- wi 1.27.644 (4.4) I is defined as the produced (or injected) pore volume. It is obvious that 1 f w de D -------------di. (4.41) S wi Until breakthrough, f w and therefore E Dd I d ----------------. (4.42) 1 S wi The function E D (I) is linear. After breakthrough, one must proceed as follows: 1. Values for S wf < S w1 < S w2 <... are assumed at regular distances. 2. At every specific S wj the corresponding E Dj is determined from Eq. 4.4. 3. A tangent is drawn from the middle of the considered interval (S wj, S wj+1 ) to the f w -curve. The tangent points are the f wj+1 values. 4. From the finite form of Eq. 4.41 I is obtained: j 1 + E D E Dj + 1 E Dj 1 f ---------------------- wj + 1 1 S j 1 wi + I (4.43) and m I I d + j + 1 I j (4.44) Calculations (1) - (4) are demonstrated in Table 4.2. Results are displayed in Fig. 4.19.

Chapter 4: Displacement Efficiency 85 Table 4.2: Calculation of the Recovery Curve S w F w E D E D I I.74.81.6438...47.75.815.6575.137.541.5241.76.86.6712.137.714.5955.77.89.6849.137.99.6864.78.91.6986.137.1111.7975.79.932.7123.137.1471.9446.8.952.726.137.284 1.153.81.97.7397.137.3334 1.4863.82.98.7534.137.5 1.9864 1..8 f w.6.4 S wf.65 S wf.74 f.81 w.2.2.4.6.8 1. Water saturation Figure 4.18: Fractional flow curve to Example 4.1

86 Chapter 4: Displacement Efficiency 1. Recovery factor, E d.8.6.4.2 1. Injected pore volumes, PV 2. 3. Figure 4.19: Oil recovery curve according to Example 4.1

87 Chapter 5 Sweep Efficiency 5.1 Mobility Ratio In the theory of piston like displacement, it is assumed that ahead of the front only the displaced fluid - for example the oil - is flowing and the oil saturation is presumed as S oi 1 S wi, where S wi is the initial immobile water saturation. Behind the front, the displacing fluid is flowing - for example water - and the water saturation is S w 1 S or, where S or is the immobile residual oil saturation. Further, the mobility ratio is defined as M ow λ w ------ λ o k rwm µ w k rom ------------ ----------- µ o (5.1) where λ w,λ o - are the water and oil mobilities k rwm - is the relative permeability of the water behind the front k rom - is the relative permeability of the oil before the front The mobility ratio is essential for determining the characteristics of the displacement. If M ow 1 then the displaced fluid - the oil - is more mobile than the displacing fluid. This is beneficial for the displacement. If M ow > 1 then the reverse situation is given. The above definition is valid in a waterflooding in which there is no saturation gradient behind the waterflood front and consequently there is no ambiguity about the value of 87

88 Chapter 5: Sweep Efficiency displacing phase relative permeability. In case of a saturation gradient, the displacing phase mobility is defined at the average displacing fluid saturation - e.g. water saturation - in the displaced pore volume of the reservoir. This definition has been proved by S w CRAIG et al. [7.] (1955). 5.2 Stability of Displacement The theory of BUCKLEY-LEVERETT [4.] (1942) presumes that the displacing front proceeds in an uniform way. ENGELBERTS and KLINKENBERG [15.] (1951) observed that in most cases so-called viscous fingers proceed ahead of the front. If the front catches up with the fingers then the displacing front can be regarded as stable, in the reverse case as unstable. In Fig. 5.1 illustrated cases the capillary force is zero and the fluid densities are equal. In case a) the Mobility Ratio M 1. As it can be seen, the front is stable and the majority of the oil is displaced before breakthrough. In case b) M 8. Early breakthrough and low displacement efficiency is manifested due to the viscous fingering. Another aspect is that the capillary force and gravity can have a favorable effect on the stability of the front. Fig. 5.2 shows a displacement of oil from bottom to top by the heavier water which is the wetting fluid. Capillary forces tend to widen the finger and gravity tends to segregate the phases vertically. Both effects can be considered to be independent of the displacement velocity. The tendency to instability increases with rising displacement velocity. If displacement proceeds slowly enough, time remains for the stabilization of the front through capillary forces and gravity. Reasons for the development of such a finger can be caused by local irregularities of the porous medium. The width of the finger depends directly on the extension of this inhomogeneity. If a porous medium is macroscopically considered homogeneous, this appearance will result in a flattening of the saturation profile. If the development of the finger is due to a macroscopically detectable change of properties (in most cases the permeability) then the whole problem can be regarded as the movement of a front progressing in a stratified medium. Reservoirs are more or less stratified, which infers that viscous fingering is due to heterogeneity.

Chapter 5: Sweep Efficiency 89 NW11% p i NW42% p i NW6% p i NW34% p i NW32% p i NW85% p i NW12% p i N p52%; W65% i NW12% p i NW6% p i NW6.6% p i N p27%; W146% i NW37% p i N p64%; W24% i N p15%; W28% i N p44%; W547% i Figure 5.1: Linear water displacement demonstrated by a transparent three dimensional model (VAN MEURS 1957) Gravity Capillary pressure Displacement Front M>1 Figure 5.2: Capillary forces and gravity influence the development of a viscous fingering