Reservoir Engineering Module - Steady State Flow - PDF

Reservoir Engineering Module Steady State Flow Dr. Mohammed Abdalla Ayoub Dr. Syed Mohammad Mahmood  Lesson’s Outcomes To explain the primary reservoir characteristics. To describe the linear and radial flow behavior of the re...

Reservoir Engineering Module Steady State Flow Dr. Mohammed Abdalla Ayoub Dr. Syed Mohammad Mahmood  Lesson’s Outcomes To explain the primary reservoir characteristics. To describe the linear and radial flow behavior of the reservoir fluids in porous media. To understand the mathematical relationships that are designed to describe the flow behavior of the reservoir fluids in porous media. Internal 4/13/2020 2  Introduction Flow in porous media is a very complex phenomenon and as such cannot be described as explicitly as flow through pipes or conduits. Unlike measuring the length and diameter of a pipe and compute its flow capacity as a function of pressure; flow in porous media is different in that there are no clear-cut flow paths which could lend themselves to measurement. Internal 4/13/2020 3  Objectives To present the mathematical relationships that are designed to describe the flow behavior of the reservoir fluids. To understand the variations in mathematical relationships depending upon the characteristics of the reservoir. To understand the primary reservoir characteristics that must be considered. Internal 4/13/2020 4  Primary Reservoir Characteristics Types of fluids in the reservoir Flow regimes Reservoir geometry Number of flowing fluids in the reservoir Internal 4/13/2020 5  Types of fluids The isothermal compressibility coefficient is essentially the controlling factor in identifying the type of the reservoir fluid. In general, reservoir fluids are classified into three groups: ❖ Incompressible fluids ❖ Slightly compressible fluids ❖ Compressible fluids Internal 4/13/2020 6  Isothermal Compressibility Coefficient The isothermal compressibility coefficient c is described mathematically by the following two equivalent expressions: In terms of fluid volume: 1  V  c =−   -------------- (1) V  P T Internal 4/13/2020 7  Isothermal Compressibility Coefficient In terms of fluid density: 1    c =   -------------- (2)   P  where V and ρ are the volume and density of the fluid, respectively. Internal 4/13/2020 8  Incompressible Fluids An incompressible fluid is defined as the fluid whose volume (or density) does not change with pressure, i.e.:  V   =0  P      =0  P  Internal 4/13/2020 9  Slightly compressible fluids ❑ These “slightly” compressible fluids exhibit small changes in volume, or density, with changes in pressure. Most of the crude oils fall in this category. ❑ The changes in the volumetric behavior of this fluid as a function of pressure p can be mathematically described by separating variables and integrating Equation (1) (on slide 7) to give: Internal 4/13/2020 10  Slightly compressible fluids P V dV − c  dP =  where; Pref Vref V P = pressure, psia V = fluid volume at pressure P, ft3 Pref = initial (reference) pressure, psia ( c Pref − P ) V e = Vref = fluid volume at initial Vref (reference) pressure, ft3 ( c Pref − P ) Vref e = V -------------- (3) Internal 4/13/2020 11  Slightly compressible fluids The ex may be represented by a series expansion as: 2 3 n x x x e = 1 + x + + +... + x -------------- (4) 2! 3! n! Because the exponent x [which represents the term c(pref−p)] is very small, the ex term can be approximated by truncating Equation (4) to: e x = 1 + x -------------- (5) Internal 4/13/2020 12  Slightly compressible fluids Combining Equation (5) with Equation (3) gives:   V = Vref 1 + c(Pref − P) -------------- (6) A similar derivation is applied to Equation (2) to give:  = ref 1 − c(Pref − P) -------------- (7) where V = volume at pressure p ρ = density at pressure p Vref = volume at initial (reference) pressure pref ρref = density at initial (reference) pressure pref Internal 4/13/2020 13  Compressible Fluids These are fluids that experience large changes in volume as a function of pressure All gases are considered compressible fluids. The truncation of the series expansion, as given by Equation (5), is not valid in this category and the complete expansion as given by Equation (4) is used. Internal 4/13/2020 14  Compressible Fluids The isothermal compressibility of any compressible fluid is described by the following expression: 1 1  z  cg = −   -------------- (8) P z  P T Figures (1) and (2) show schematic illustrations of the volume and density changes as a function of pressure for the three types of fluids: Internal 4/13/2020 15  Pressure-volume relationship Internal 4/13/2020 Figure(1): Pressure-volume relationship 16  Pressure-Fluid Density Relationship Figure(2): Fluid density versus pressure for different fluid types Internal 4/13/2020 17  Flow Regimes There are basically three types of flow regimes that must be recognized in order to describe the fluid flow behavior and reservoir pressure distribution as a function of time: Steady-state flow Unsteady-state flow Pseudosteady-state flow Internal 4/13/2020 18  Steady-State Flow The pressure at every location in the reservoir remains constant and does not change with time  P    = 0 -------------- (9)  t i Internal 4/13/2020 19  When It Happens? In reservoirs, the steady-state flow condition can only occur when the reservoir is completely recharged and supported by strong aquifer or pressure maintenance operations. This condition is rarely achieved in the field. Internal 4/13/2020 20  Unsteady / Transient State Flow The fluid flowing condition at which the rate of change of pressure with respect to time at any position in the reservoir is not zero or constant The pressure derivative with respect to time is essentially a function of both position i and time t  P    = f (i, t ) -------------- (10)  t  Internal 4/13/2020 21  Pseudosteady-State Flow The pressure at any given location, i, in the reservoir is declining linearly as a function of time  P    = constant -------------- (11)  t i Internal 4/13/2020 22  Characteristic of Fluids Internal Figure (3): Volume change behavior of Fluids with change in pressure 4/13/2020 23  Reservoir Geometry The shape of a reservoir has a significant effect on its flow behavior Most reservoirs have irregular boundaries Rigorous mathematical description of geometry is often possible only with the use of numerical simulators The actual flow geometry may be represented by one of the following flow geometries: Internal 4/13/2020 24  Reservoir Geometry,… contd Linear flow Radial Flow Spherical flow Hemispherical Flow Internal 4/13/2020 25  Radial Flow structure in radial-cylindrical Flow into or away from a Three-dimensional flow wellbore will follow coordinate system radial flow lines from a substantial distance from the wellbore In the absence of severe reservoir heterogeneities, fluids move toward the well from all directions and converge at the wellbore A typical one-dimensional, Internal 4/13/2020 radial-cylindrical flow model 26  Ideal radial flow into a wellbore Figure (4) Ideal radial flow into a wellbore. Internal 4/13/2020 27  Linear Flow When flow paths are parallel and the fluid flows in a single direction (i.e. no flow perpendicular to or angular to the main flow streams) The cross sectional area to flow must be constant A common application of linear flow equations is the fluid flow into vertical hydraulic fractures Internal 4/13/2020 28  Linear Flow Figure (5) Linear flow Internal 4/13/2020 29  Ideal linear flow into vertical fracture Figure (6) Ideal linear flow into vertical fracture Internal 4/13/2020 30  Spherical flow due to limited entry Figure (7) Spherical flow due to limited entry Internal 4/13/2020 32  Spherical Flow A well which is only perforated in the central part of the formation and not in the entire pay zone could result in spherical flow in the vicinity of the perforations. The kind of limited perforation in the central part of the pay zone could be beneficial if there is a possibility of water coning from the bottom and and gas encroachment from the top.  Hemispherical flow in a partially penetrating well Figure (8) Hemispherical flow in a partially penetrating well Internal 4/13/2020 34  Hemispherical Flow A well which only partially penetrates the pay zone not in the entire pay zone could result in hemispherical flow. This kind of partial penetration could be beneficial if there is a possibility of water coning from the bottom but no chance of gas encroachment from the top.  Number of Flowing Fluids in the Reservoir Single-phase flow (oil, water, or gas) Two-phase flow (oil-water, oil-gas, or gas- water) Three-phase flow (oil, water, and gas) Internal 4/13/2020 36  Henry Darcy Was a 19th century French engineer Was designing a filter to process his town’s water demand Studied vertical flow of water through packed sand Introduced the concept of permeability (unit: mD) Internal 4/13/2020 37  Darcy’s Law What are the parameters that affect fluid flow? Internal 4/13/2020 38  L q A dx For one-dimensional, horizontal flow through a porous medium, Darcy’s Law states that: kA dp q=− -------------- (12)  dx q =Flow rate (cm /s) 3 A= Cross sectional area (cm2) μ =Viscosity of flowing fluid (cp) k =Permeability (Darcy) 𝑑𝑃/𝑑𝑥=Pressure gradient (atm/cm) Transport eqn. implying velocity is proportional to pressure gradient and reciprocal to viscosity Internal 39 4/13/2020  Pressure Profile in Linear System Internal 4/13/2020 Pressure vs. distance in a linear flow 40  Darcy’s Law for Radial Flow kA dp k (2rh) dp q=− =− -------------- (13)  dr  dr Curved surface open to flow For fluid flow to occur, a pressure gradient must be established between the inner and outer boundary of the reservoir. h Pressure gradient dp/dr Internal 4/13/2020 41  Pressure Profile in Radial System Internal 4/13/2020 Pressure gradient in radial flow 42  Conditions for Darcy’s Law Darcy’s Law applies only when the following conditions exist: Laminar (viscous) flow Steady-state flow Incompressible fluids Homogeneous formation Internal 4/13/2020 43  STEADY-STATE FLOW Represents the condition that exists when the pressure throughout the reservoir does not change with time The applications of the steady-state flow include: Linear flow of incompressible fluids Linear flow of slightly compressible fluids Linear flow of compressible fluids Radial flow of incompressible fluids Radial flow of slightly compressible fluids Radial flow of compressible fluids Multiphase flow Internal 4/13/2020 44  Section 2 In this section, students will be able to ; 1- Derive an equation for simple Linear Flow of Incompressible fluid 2- Understand the fluid potential concept 3- See how units can be converted from one system to another Internal 4/13/2020 45  Linear Flow of Incompressible Fluids ❖ The flow occurs through a constant cross-sectional area A ❖ both ends are entirely open to flow ❖ no flow crosses the sides, top, or bottom Internal 4/13/2020 Linear flow model 46  Linear Flow of Incompressible Fluids If an incompressible fluid is flowing across the element dx, then the fluid velocity v and the flow rate q are constants at all points The flow behavior in this system can be expressed by integration the differential form of Darcy’s equation within limits of Injection end (x=0, P= P1) and producing end (x=L,P=P2) q L k 2 P kA(P1 − P2 ) A0 dx = −  dP  P1 OR q= L In field units Permeability (mD) Area (ft2) Pressure (psi) 0.001127 kA(P1 − P2 ) Flowrate (bbl/d) q= -------- (14) L Viscosity (cp) Distance (ft) Internal 4/13/2020 47  Example An incompressible fluid flows in a linear porous media with the following properties: L = 2000 ft; h = 20′; width = 300′ k = 100 md; Ø= 15%; μ = 2 cp P1 = 2000 psi; P2 = 1990 psi Calculate: a. Flow rate in bbl/day b. Apparent fluid velocity in ft/day c. Actual fluid velocity in ft/day Internal 4/13/2020 48  Solution a. Calculate the flow rate Calculate the cross-sectional area A: A = (h) (width) = (20) (300) = 6000 ft2 b. Calculate the apparent velocity: c. Calculate the actual fluid velocity: Internal 4/13/2020 49  Fluid Potential (Φ) The fluid potential at any point in the reservoir is the pressure at that point plus the pressure that would be exerted by a fluid head extending to an arbitrarily assigned datum level. Δzi is the vertical distance from a point i in the reservoir to this datum level -------------- (15) -------------- (16) where ρ is the density in lb/ft3 Internal 4/13/2020 50  Darcy’s units In terms of Darcy units -------------- (17) v= apparent fluid velocity, cm/sec; k= permeability, darcy; μ= fluid viscosity, cp; p= pressure, atm l= length, cm; ρ= fluid density, gm/cm3; g= Acceleration due to gravity, cm/sec2; and Z= elevation, cm. The vertical distance Δzi is assigned as a positive value when the point i is below the datum level and as a negative when it is above the datum level Internal 4/13/2020 51  In fields units -------------- (18) where Φi = fluid potential at point i, psi pi = pressure at point i, psi Δzi = vertical distance from point i to the selected datum level ρ = fluid density, lb/ft3 γ = fluid specific gravity (water=1) v= apparent fluid velocity, res bbl/day/ft2; A= total cross-sectional area, ft2 B= formation volume factor, RB/STB. k= permeability, md θ= dip angle of the reservoir or formation measured counterclockwise from the horizontal to the positive flow path The negative sign in Eq. (16) accounts for the sign convention that flow is considered positive in the positive direction of the flow path length, and pressure decreases in the direction of flow Internal 4/13/2020 52  Applying the previous-generalized concept to Darcy’s equation gives: -------------- (19) The fluid potential drop (Φ1 − Φ2) is equal to the pressure drop (p1 − p2) only when the flow system is horizontal Internal 4/13/2020 53  Linear Flow of Slightly Compressible Fluids The relationship that exists between pressure and volume for slightly compressible fluid is:   V = Vref 1 + c(Pref − P) -------------- (6) The above equation can be modified and written in terms of flow rate as:   q = qref 1 + c(Pref − P ) -------------- (20) where qref is the flow rate at some reference pressure Pref. Internal 4/13/2020 54  Substituting the previous relationship in Darcy’s equation gives: Separating the variables and arranging: Integrating gives: -------------- (21) where qref = flow rate at a reference pressure pref, bbl/day p1 = upstream pressure, psi p2 = downstream pressure, psi k = permeability, md μ = viscosity, cp Internal 4/13/2020 c = average liquid compressibility, psi−1 55  Selecting the upstream pressure P1 as the reference pressure Pref and substituting in Equation (21) gives the flow rate at Point 1 as: ---------- (22) Choosing the downstream pressure P2 as the reference pressure and substituting in Equation (19) gives: ---------- (23) where q1 and q2 are the flow rates at point 1 and 2, respectively Internal 4/13/2020 56  Darcy’s Unit Conversion Internal 4/13/2020 57  You hate extra homework given by the tutors or instructors??? Good news, you are not alone Internal 4/13/2020 58  Linear Flow of Compressible Fluids (Gases) For a viscous (laminar) gas flow in a homogeneous-linear system The real-gas equation-of-state can be applied to calculate the number of gas moles n at pressure p, temperature T, and volume V PV n = ZRT At standard conditions, the volume occupied by the above n moles is given by: nZsc RTsc Vsc = Psc Internal 4/13/2020 59  Combining the two expressions and assuming zsc = 1 gives: PV PscVsc = ZT Tsc Equivalently, the above relation can be expressed in terms of the flow rate as: 5.615 Pq PscQsc = ZT Tsc Internal 4/13/2020 60  Rearranging:  Psc  ZT  Qsc   T   P  5.615  = q ---------- (24)  sc    where: q = gas flow rate at pressure p in bbl/day Qsc = gas flow rate at standard conditions, scf/day Z = gas compressibility factor Tsc, psc = standard temperature and pressure in ◦R and psia, respectively. Replacing the gas flow rate q with that of Darcy’s Law gives: q  Psc  ZT  Qsc  1  K dP =        = − 0. 001127 A  Tsc  P  5.615  A   dx Internal 4/13/2020 61  The constant 0.001127 is to convert from Darcy’s units to field units. Separating variables and arranging yields:  Qsc PscT L p2 P    dx = −  dP p1 Z  0.006328 Tsc Ak  0 g For simplification purposes, we assume that the product of Z*μg is constant over the specified pressure range between p1 and p2, and integrating, gives: Qsc = gas flow rate at standard conditions, scf/day 0.003164 Tsc Ak ( p12 − p22 ) k = permeability, md Qsc = T = temperature, °R PscTZ g L μg = gas viscosity, cp A = cross-sectional area, ft2 L = total length of the linear system, ft Internal 4/13/2020 62  It is essential to notice that those gas properties Z and μg are very strong functions of pressure, but they have been removed from the integral to simplify the final form of the gas flow equation. The above equation is valid for applications when the pressure is less than 2000 psi. The gas properties must be evaluated at the average pressure p as defined below: Setting psc = 14. 7 psi and Tsc = 520◦R in the above expression gives: Qsc = ( 0. 111924 Ak p12 − p22 ) ---------- (25) TZ g L p12 + p22 p= 2 ---------------- (26) Internal 4/13/2020 63  Example A linear system with a specific gravity of 0.72 is flowing in linear porous media at 140◦F. The upstream and downstream pressures are 2100 psi and 1894.73 psi, respectively. The cross-sectional area is constant at 4500 ft2. The total length is 2500 ft with an absolute permeability of 60 md. Calculate the gas flow rate in scf/day (psc = 14. 7 psia, Tsc = 520◦R). Solution: Step 1. Calculate average pressure by using Equation (26) 2100 2 + 1894.37 2 p= = 2000 psi 2 Internal 4/13/2020 64  Step 2. Using the specific gravity of the gas, calculate its pseudo-critical properties Tpc = 168 + 325γg − 12. 5γg2 = 168 + 325(0. 72) − 12. 5(0. 72)2 = 395. 5◦R ppc = 677 + 15. 0γg − 37. 5 γg2 = 677 + 15. 0(0. 72) − 37. 5(0. 72)2 = 668. 4 psia Step 3. Calculate the pseudo-reduced pressure and temperature. P 2000 T 140 + 460 p pr = = = 2.99 Tpr = = = 1.52 p pc 668.4 Tpc 395.5 Internal 4/13/2020 65  Step 4. Determine the Z- factor from a Standing– Katz chart to give: Z = 0. 78 Internal 4/13/2020 66  Step 5. Solve for the viscosity of the gas by applying the Lee– Gonzales–Eakin method and using the following sequence of calculations: Ma = 28. 96γg = 28. 96(0. 72) = 20. 85 ρg =pMa/ZRT = (2000)(20. 85)/(0. 78)(10. 73)(600) = 8. 30 lb/ft3 K = (9. 4 + 0. 02Ma)T1.5/(209 + 19Ma + T) = (9. 4 + 0. 02(20. 96)(600)1.5/(209 + 19(20. 96) + 600) = 119. 72 X = 3. 5 +986/T + 0. 01Ma = 3. 5 +986/600 + 0. 01(20. 85) = 5. 35 Y = 2. 4 − 0. 2X = 2. 4 − (0. 2)(5. 35) = 1. 33 μg = 10−4K exp[ X(ρg /62. 4)Y ] = 0. 0173 cp Internal 4/13/2020 67  Step 6. Calculate the gas flow rate by applying equation (25) Qsc = ( 0. 111924 Ak p12 − p22 ) TZ g L 0. 111924 (4300 )(60 )(2100 2 − 1894.732 ) Qsc = = 1,224,242 scf/day (600 )(2500 )(0.78)(0.0173 ) 1,224,242 scf/day Internal 4/13/2020 68 Internal 4/13/2020 69  Lesson Outcomes- “Reminder” i. To explain the primary reservoir characteristics. ii. To describe the linear and radial flow behavior of the reservoir fluids in porous media. iii. To understand the mathematical relationships that are designed to describe the flow behavior of the reservoir fluids in porous media. Internal 4/13/2020 70  Radial Flow Section Outcome ❖Radial Flow of Incompressible Fluids ❖Average Reservoir Pressure Pr ❖Radial Flow of Slightly Compressible Fluids ❖Radial Flow of Compressible Gases ❖Approximation of the Gas Flow Rate ❖Horizontal Multiple-Phase Flow Internal 4/13/2020 71  Radial Flow of Incompressible Fluids All fluids move toward the producing well from all directions The pressure in the formation at the wellbore must be less than the pressure in the formation at some distance from the well for flow to occur The pressure in the formation at the wellbore of a producing well is know as the bottom-hole flowing pressure (flowing BHP, pwf) Internal 4/13/2020 72  The formation is considered to have a uniform thickness h and a constant permeability k. The flow rate q must be constant at all radii Due to the steady-state flowing condition, the pressure profile around the wellbore is maintained constant with time ---------------- (27) where v = apparent fluid velocity, bbl/day-ft2 q = flow rate at radius r, bbl/day k = permeability, md μ = viscosity, cp 0.001127 = conversion factor to express the equation in field units Ar = cross-sectional area at radius r Internal 4/13/2020 73  Radial flow model Internal 74 4/13/2020  At any point in the reservoir the cross-sectional area across which flow occurs will be the surface area of a cylinder, which is 2πrh, The flow rate for a crude oil system is customarily expressed in surface units. Change it to reservoir conditions. where Bo = the oil formation volume factor bbl/STB Qo = the oil flow in STB/day Internal 4/13/2020 75  The flow rate in Darcy’s equation can be expressed in Rb/day to give: Integrating the above equation between two radii, r1 and r2, when the pressures are p1 and p2 yields: ------------ (28) Internal 4/13/2020 76  For incompressible system in a uniform formation, Equation (26) can be simplified to: Performing the integration, gives: Internal 4/13/2020 77  The steady state radial flow equation In field units: External pressure (psi) Permeability (mD) Height (ft) Bottom hole flowing Pressure at 0.00708 kh(Pe − Pw ) (psi) q=  re  ---------------- (29) o Bo ln  r   w External or drainage radius(ft) Oil flow rate (STB/d) Viscosity Oil formation volume Wellbore (cp) factor (Bbl/STB) radius(ft) Internal 78 4/13/2020  The external (drainage) radius re is usually determined from the well spacing by equating the area of the well spacing with that of a circle OR ------ (30) Equation (29) can be arranged to solve for the pressure p at any radius r to give: -------------- (31) Internal 4/13/2020 79  Example An oil well in the Nameless Field is producing at a stabilized rate of 600 STB/day at a stabilized bottom-hole flowing pressure of 1800 psi. Analysis of the pressure buildup test data indicates that the pay zone is characterized by a permeability of 120 md and a uniform thickness of 25 ft. The well drains an area of approximately 40 acres. The following additional data is available: rw = 0.25 ft A = 40 acres Bo = 1.25 bbl/STB μo = 2.5 cp Calculate the pressure profile (distribution) and list the pressure drop across 1 ft intervals from rw to 1.25 ft, 4 to 5 ft, 19 to 20 ft, 99 to 100 ft, and 744 to 745 ft. Internal 4/13/2020 80  Solution Step 1. Rearrange Equation (29) and solve for the pressure p at radius r. Internal 4/13/2020 81  Step 2. Calculate the pressure at the designated radii. r, ft p, psi Radius Interval Pressure drop 0.25 1800 1.25 1942 0.25 – 1.25 1942 − 1800 = 142 psi 4 2045 5 2064 4–5 2064 − 2045 = 19 psi 19 2182 20 2186 19 – 20 2186 − 2182 = 4 psi 99 2328 100 2329 99 – 100 2329 − 2328 = 1 psi 744 2506.1 745 2506.2 744 – 745 2506.2 − 2506.1 = 0.1 psi Figure (1) shows the pressure profile on a function of radius for the calculated data. Internal 4/13/2020 82  Figure (1) Pressure profile around the wellbore Internal 4/13/2020 83  Average Reservoir Pressure Pr Should be used in performing material balance calculations and flow rate prediction. Craft and Hawkins (1959) showed that the average pressure is located at about 61% of the drainage radius re for a steady-state flow condition. Substitute 0.61 re in Equation (29) to give: Internal 4/13/2020 84  in terms of flow rate: ----------------- (32) -------------- (33) Internal 4/13/2020 85  Golan and Whitson (1986) suggesteded a method for approximating drainage area of wells producing from a common reservoir. They assume that the volume drained by a single well is proportional to its rate of flow. Assuming constant reservoir properties and a uniform thickness, the approximate drainage area of a single well, Aw, is: -------------- (34) Where Aw = drainage area AT = total area of the field qT = total flow rate of the field Internal 4/13/2020 qw = well flow rate 86  Radial Flow of Slightly Compressible Fluids where qref is the flow rate at some reference pressure pref Separating the variables in the above equation and integrating over the length of the porous medium gives: OR   1 + c ( pref − pe )      0.00708 kh  = ln    re   1 + c( pref − pwf ) qref   c ln    where qref is oil flow   rw   rate at a reference pressure pref Internal 4/13/2020 87  Choosing the bottom- hole flow pressure pwf as the reference pressure and expressing the flow rate in STB/day gives:     Qo =   0.00708 kh   re   o   ln 1 + c ( p − p ) -------------- (35) wf e  o Bo co ln      rw   Where Qo = oil flow rate, STB/day k = permeability, md co = isothermal compressibility coefficient, psi−1 Internal 4/13/2020 88  Example The following data are available on a well in the Red River Field: pe = 2506 psi pwf = 1800 re = 745′ rw = 0.25 Bo = 1.25 μo = 2.5 co = 25 × 10−6 psi−1 k = 0.12 Darcy h = 25 ft. Assuming a slightly compressible fluid, calculate the oil flow rate. Compare the result with that of incompressible fluid. Solution: For a slightly compressible fluid, the oil flow rate can be calculated by applying Equation (35):    0.00708 (120 )(25)  Qo =    (2.5)(1.25)(25  10 −6 )ln  745     0.25     ln 1 + (25  10 −6 )(1800 − 2506 ) = 595 STB / day Internal 4/13/2020 89  Assuming an incompressible fluid, the flow rate can be estimated by applying Darcy’s equation, i.e., Equation (29): Internal 4/13/2020 90  Radial Flow of Compressible Gases The basic differential form of Darcy’s Law for a horizontal laminar flow is valid for describing the flow of both gas and liquid systems For a radial gas flow, the Darcy’s equation takes the form: -------------- (36) where qgr = gas flow rate at radius r, bbl/day r = radial distance, ft h = zone thickness, ft μg = gas viscosity, cp p = pressure, psi 0.001127 = conversion constant from Darcy units to field units Internal 4/13/2020 91  The gas flow rate is usually expressed in Scf/day Applying the real gas equation-of-state to both conditions to get: OR -------------- (37) where psc = standard pressure, psia Tsc = standard temperature, °R Qg = gas flow rate, scf/day qgr = gas flow rate at radius r, bbl/day p = pressure at radius r, psia T = reservoir temperature, °R z = gas compressibility factor at p and T zsc = gas compressibility factor at standard condition ≅ 1.0 Internal 4/13/2020 92  Combining Equations 36 and 37 yields: Assuming that Tsc = 520 °R and psc = 14.7 psia: ----------------- (38) Integrating Equation 6-36 from the wellbore conditions (rw and pwf) to any point in the reservoir (r and p) gives: ------------ (39) Internal 4/13/2020 93  Imposing Darcy’s Law conditions on Equation (39) – Steady-state flow which requires that Qg is constant at all radii – Homogeneous formation which implies that k and h are constant Internal 4/13/2020 94  Combining the above relationships yields: --------- (40) The integral is called the real gas potential or realgas pseudo-pressure and it is usually represented by m(p) or ψ -------------------------- (41) Internal 4/13/2020 95  in terms of the real gas potential OR ------------ (42) Internal 4/13/2020 96  Equation (41) indicates that a graph of ψ vs. ln r/rw yields a straight line of slope (QgT/0.703kh) and intercepts ψw (Figure 2). The flow rate is given exactly by ---------------- (43) In the case when r = re, then: ---------------- (44) Internal 4/13/2020 97  Figure 2. Graph of Ψ vs. ln (r/rw). where ψe = real gas potential as evaluated from 0 to pe, psi2/cp ψw = real gas potential as evaluated from 0 to Pwf, psi2/cp k = permeability, md h = thickness, ft re = drainage radius, ft rw = wellbore radius, ft Qg = gas flow rate, scf/day Internal 4/13/2020 98  The gas flow rate is commonly expressed in Mscf/day ---------------- (45) Where Qg = gas flow rate, Mscf/day, expressed in terms of the average reservoir pressure pr instead of the initial reservoir pressure pe as: -------------- (46) Internal 4/13/2020 99  Approximation of the Gas Flow Rate ❑Removing the term 2 g z outside the integral as a constant. ❑Zg is considered constant over a pressure range