Reservoir Flow Properties Fundamentals COPYRIGHT. Introduction

Transcription

1 Reservoir Flow Properties Fundamentals Why This Module is Important Introduction Fundamental understanding of the flow through rocks is extremely important to understand the behavior of the reservoir Permeability The property which defines the conductivity of the fluid in the rocks Darcy s Law Describes the behavior of flow, and depending on the nature of flow, can take many forms It is important to understand how to extend Darcy s law equation to account for the nature of flow Equations to calculate the rate changes based on whether it is a single phase oil, single phase gas, or a multi-phase fluids 1

2 Why This Module is Important Reservoirs can encounter: It is common to see linear flow in horizontal wells Linear Flow Radial Flow It is unusual to see radial flow in vertical wells Why This Module is Important During the drilling process: The nearby well region is often damaged The permeability near the well bore is altered How to handle the changes in the flow behavior due to damage near the wellbore is important to correctly predict the rate behavior of the well Skin factor is normally considered to properly account for this behavior When a well is stimulated or the near wellbore properties are altered, the productivity equation can be manipulated to determine how that alteration is going to impact the productivity of the well r w r e P w f P e 2

3 Why This Module is Important Reservoirs are rarely homogeneous and we need to account for the permeability heterogeneity in properly calculating the productivity of the well Depending on the distribution of the permeability heterogeneity, different sets of equations need to be used to calculate the effective flow behavior For gas wells, because of low viscosity of the gas and relatively high velocities near the wellbore, the traditional Darcy s law equation may not be applicable We need to account for additional pressure drop due to high velocity of gas by supplementing Darcy s law Why This Module is Important Darcy s law and all its deviations and extensions are crucial to understand the behavior of oil and gas wells Without understanding these extensions, no reservoir engineer would be able to correctly predict the rate at which the well is capable of producing 3

4 Learning Objectives Reservoir Flow Properties Fundamentals Darcy s Law for Linear Flow This section will cover the following learning objectives: Solve linear flow problems using Darcy s law equation Differentiate between gas and oil flows and why the equations are different for two phases Summarize the importance of gravity and pressure gradients and their influence on flow in linear systems 4

5 Recall Darcy s Law for Linear Flow For linear flow, assuming constant density: Symbol Field Unit SI Unit q Barrels per day m 3 /per day k Permeability in md Permeability in md A ft 2 m 2 cp Pa.s p psia kpa I ft m ϒ Specific gravity of fluid Specific gravity of fluid θ Dip angle of inclination measured counterclockwise from horizontal flow Dip angle of inclination measured counterclockwise from horizontal flow * For upward vertical flow θ is 90 degrees * For vertically upward flow, θ is 90 degrees

6 Example 1 A cylindrical tube 4 ft [1.22 m] long with a 3 in. [7.62 cm] diameter is packed with sand. The tube was laid horizontally and water, at a rate of 1 gallon/hour [3788 cc/hr], was flowed through the tube. The pressure at the inlet was observed to be 25 psig [172.4 kpag] and at the exit, it was 10 psig [68.9 kpag]. Assume the viscosity of water equal to 0.8 cp [ Pa.s] and the specific gravity equal to 1 What is the permeability of the sand pack? If the tube was turned vertical and water flows from the top to bottom, at what rate water will flow if the pressures at the inlet and the outlet were maintained the same as in the previous experiment? What would the flow rate be if the tube was turned at 45 and the liquid flowed in the upward direction? Solution: Example 1 Flow Rate Area Horizontal Flow = ft Substituting in Darcy s law equation Solving for k k 2,203 md k 2,203 md 6

7 Solution: Example 1 Vertical Flow θ = -90 o (downward flow) Downward flow: gravity assists the pressure drop, hence higher rate is observed. 2, / 2, Solution: Example 1 Inclined Flow θ = 45 o (upward flow) / 2, Upward flow: gravity opposes the pressure gradient, hence the rate is lower ,

8 Gas Flow Equations for Linear Flow Gas flow equation for linear flow can be written as (in terms of pressure squared method) Symbol Field Unit SI Unit q MSCFD M 3 D You can write a similar equation in the form of pseudo-real pressure as follows: Gas flow equation for linear flow can be written as (in terms of pressure squared method) You can write a similar equation in the form of pseudo-real pressure as follows: k md md A ft2 m 2 Centipoise Pa.s T Rankine L ft m p psi kpa m(p) psi 2 /cp kpa 2 /Pa.s 8

9 Example 2 A horizontal well which is 5000 ft [1525 m] long is subjected to multi-stage fractures. The total number of transverse fractures is 20 and the fracture length is 200 ft [61 m]. Example 2 A horizontal well which is 5000 ft [1525 m] long is subjected to multi-stage fractures. The total number of transverse fractures is 20 and the fracture length is 200 ft [61 m]. Assume the following: Average distance between the fractures is 250 ft [76.2 m] and the flow is linear from reservoir into the fracture. Each fracture is draining about half the distance between two fractures. Pressure in the middle of two fractures is 4,000 psia [27,579 kpa] and the pressure inside the fracture is 500 psia [3,447 kpa]. The well is producing dry gas. Assume the permeability to be 0.01 md. Height of each fracture is 100 ft [30.5 m]. The temperature of the reservoir is 180 F [82.6 C]. Average values of and z to be cp [ Pa.s ] and 0.88 respectively. What is the rate at which the horizontal well is producing? 9

10 Solution: Example 2 Area of each fracture Distance gas travels for linear flow 2 x 100 x 200 = 40,000 ft 2 2 x 61 x 30.5 = 3,718 m ft (half the distance) 38.1 m (half the distance) Gas rate for linear flow Substituting Multiplying by 20, total rate at which the well will produce: , , ,685 / 11.8 MMSCFD Learning Objectives 333,703 / This section has covered the following learning objectives: Solve linear flow problems using Darcy s law equation Differentiate between gas and oil flows and why the equations are different for two phases Summarize the importance of gravity and pressure gradients and their influence on flow in linear systems 10

11 Learning Objectives Reservoir Flow Properties Fundamentals Darcy s Law for Radial Flow This section will cover the following learning objectives: Solve simple problems for radial flow across porous medium using Darcy s law for radial flows Differentiate between oil and gas flows Define and calculate productivity index Predict the inflow performance relationship for oil and gas wells 11

12 Recall Radial Flow For petroleum reservoirs, radial flow is much more common as the flow converges into the wellbore The pressure at the boundary of radius r e is p e The pressure at the wellbore radius, r w, is p wf P e > p wf and r e > r w Recall Radial Flow You can write the previous equation for radial flow in field units as: r w r e P wf Symbol Field Unit SI Unit q Barrels per day m 3 /day k Permeability in md Permeability in md h ft m p psia kpa cp Pa.s Radii ft m P e 12

13 Productivity Index Productivity index is defined as units of production per day per unit change in the bottom hole pressure. Productivity index is an important indicator of how productive the well is. For oil wells, typically, in field units, the value of J 1 (bbl/d/psi) [0.023 m 3 /d/kpa in SI units] is considered a highly productive well. If the value of productivity index, J (bbl/d/psi) [ m 3 /d/kpa], it is considered a marginal well. The rigorous definition of productivity index is. This definition tells us about the incremental oil (or gas) that can be produced per unit Why? change in BHP. For single phase oil, J is constant. Taking the derivative of rate equation with respect to BHP, the value will be constant. For radial flow it has the form where C depends on the units. Ideally it does not change with reservoir pressure. So, if indeed there is a reduction over time, it is an indication of reservoir damage. 13

14 Example 3 A well is producing in a radial reservoir at a bottom hole pressure of 5,500 psi [37,921 kpa]. The reservoir pressure is 6,000 psi [41,369 kpa]. Oil viscosity is 0.25 cp [ Pa.s]. If the permeability of the reservoir is 20 md, the thickness is 30 ft [9.15 m] and the drainage radius is 1,000 ft [304.9 m], at what rate will the well produce? The well bore radius is 6 in. [0.152 m]. If, by applying an artificial lift method, the bottom hole pressure is reduced to 3,000 psi [20,684 kpa], at what rate will the well produce? What is the value of productivity index? Plot rate vs. bottom hole pressure for this well. Solution: Example ,000 5, ,118 / ,000 3, ,707 / Use the radial flow equation and substitute all the numbers: Choosing 3,000 psia allows you to calculate the rate as: Productivity Index, J: ,368 37, / ,368 20, / / /

15 Solution: Example 3 Once the productivity index is calculated, calculate the rate at any bottom hole pressure using the equation: q J p e p wf The plot of rate vs. BHP (bottom hole pressure) indicates a straight line This graph is also called IPR or Inflow Performance Curve The slope of the line is J or productivity index p wf P wf,, psia psia p P wf wf,, kpa kpa q, q, bbl/day q, q, m 3 /day 3 15

16 Gas Wells For gas wells, write the equation for the rate as: As can be seen from this equation, the rigorous definition of productivity index, provides the following value of productivity index: For gas wells, the productivity index decreases as the bottom hole pressure (BHP) decreases. Don t expect to see the same incremental rate increase as BHP decreases further. This is an important observation and has a practical significance. 16

17 Example 4 A gas well is producing from a reservoir with permeability of 3 md and thickness of 50 ft [15.2 m]. The reservoir pressure is 5,000 psia [34,474 kpa]. The average viscosity is 0.02 cp [ Pa.s] and the z factor is The reservoir temperature is 200 F [93.7 C]. The drainage radius is 1,500 ft [457.3 m] and the wellbore radius is 4 in. [0.102 m]. Generate the inflow performance curve. Calculate the productivity index as a function of BHP. A main reason we install compressor on a gas well is to reduce the BHP. Based on your understanding of IPR curves, is it better to install compressor when the BHP is higher or lower? Solution: Example 4 Calculation of Flow Rate , ,474 17,

18 Solution: Example 4 By substituting different values of BHP, the IPR curve can be generated: / , // To calculate the productivity index at 2500 psia [17,237 kpa], use this equation Solution: Example 4 The graph shows the inflow performance curve for the gas well Unlike single phase oil well, it is not a straight line; instead, it is concave upwards P p wf,, psia p P wf wf,, kpa kpa q, MMSCF/day q, The IPR curve is gentle at the top and becomes steeper at the bottom q, Mm q, Mm 3 /day 3 18

19 Solution: Example 4 The graph shows the productivity index as a function of bottom hole pressure (BHP) Notice that productivity index increases with an increase in BHP For the same change of BHP, the improvement in rate is much better when BHP is higher than lower It is better to install a compressor if the BHP is higher so that improvement will be larger Learning Objectives P p wf,, psia P wf, p wf kpa, J, MSCF/day/psi J, J, mj, 3 m/day/kpa 3 /day/kpa This section has covered the following learning objectives: Solve simple problems for radial flow across porous medium using Darcy s law for radial flows Differentiate between oil and gas flows Define and calculate productivity index Predict the inflow performance relationship for oil and gas wells 19

20 Learning Objectives Reservoir Flow Properties Fundamentals Flow Regimes and Their Impact on the Performance of the Wells This section will cover the following learning objectives: Outline the differences between transient state, steady state and pseudo-steady state flow regimes Calculate the productivity of both oil and gas wells under different flow regimes Describe the importance of the shape of the reservoirs and how they impact the behavior of oil and gas wells 20

21 Equations for Other Flow Regimes Transient State Pseudo-State Transient state occurs when the production is initiated and the well is producing without seeing the influence of the boundary Pseudo-steady state occurs when the boundary effect is felt and the pressure across reservoir starts changing uniformly as a function of time due to depletion If the reservoir has different shapes than a perfect circle, different boundaries are felt at different times, resulting in a late transient period where the flow is between end of true transient and beginning of pseudo-steady state period Transient to Late Transient to PSS Boundaries Transient Late Transient Pseudo-steady State dp/dt = c pfnlogtime Time Key requirement of pseudo-steady state when well is producing: Change in the pressure with respect to time in any part of the reservoir is constant 21

22 Transient Flow Equation for Radial Flow Equation for Transient Flow: OR. Recall that time to end transient state is: : :. where t etf is in hours OR Symbol Field Unit SI Unit k md md h ft m q bbi/d m 3 /d μ cp md P psi kpa t hrs hrs c t psi -1 kpa -1 r w ft m p i represents initial pressure which will also be equal to the pressure at the boundary or p e. p wf represents the pressure at the wellbore. 22

23 Example 5 A well is producing from a reservoir with a permeability of 1 md, thickness of 50 ft [15.2 m], r w of 0.36 ft [0.11 m] and r e of 1,500 ft [457.3 m]. Porosity is 0.2, viscosity is 0.4 cp [ Pa.s], c t is 10 x 10-6 psi -1 [1.45 x 10-6 kpa -1 ], the p i is 5,000 psia [34,474 kpa] and p wf is 1,000 psia [6,895 kpa]. Calculate the time to end transient state. Calculate the rate at which the well will flow at 1, 5, 10, 20, 30, 50 and 70 days. Solution: Example 5 Calculate the time to end the transient state: = 1,706 hours 71 days = 1,706 hours 71 days 1 At all the times for which you are calculating the rate, you are in transient state. 23

24 Solution: Example 5 By substituting different values of BHP, the IPR curve can be generated: =7.59 5,000 1,000 = 527 barrels/day =330 Solution: Example 5 Rate profile as a function of time indicates rapid decline in the rate as time progresses 34,474 6,895 = 83.6 m /day q,bbl/day q,m m 3 /day To calculate the rate at 10 days, Replace this value with t, days t, For radial flow, as the time reaches 71 days, the flow regime will change to pseudo-steady state period t, days t, days 24

25 Average vs. Boundary Pressure When the transition from transient to pseudo-steady state flow begins, the rate equations can be written in terms of either boundary pressure or average pressure. Boundary Pressure Average Pressure The boundary pressure represents The average pressure represents the pressure at the boundary. volumetric average of the pressure. It will fall between p wf and p e but it is closer to p e because most of the pressure drop in radial flow occurs near the well bore and the volume in that region is very small. Denote the average pressure by. Terminology p p i, p e p p wf The difference between and p wf is smaller than the difference between p e and p wf. p - p wf p i -p wf The average pressure is a lot closer to the drainage pressure or initial pressure than the bottom hole pressure r w r r e 25

26 Pseudo-steady State For pseudo-steady state conditions, define the rate equation for radial flow as: or or It is more common to use the average pressure instead of boundary pressure Pseudo-steady State Considering oil properties (from), write the equation in terms of STB/day as: or In this equation B is formation volume factor or

27 Non-radial Reservoirs If the reservoir has a shape which is not circular, we can still write an equation for pseudo-steady conditions Shape Factors The figure shows the values of shape factors for some shapes. It includes the beginning of pseudosteady state when dimensionless time t DA reaches a certain value, defined as: C A In C A Start of PSS End of Infinite Acting Exact for t DA > <1% error Transient solution for t DA > has <1% error for t DA < Variable Definition A Area of the reservoir in ft 2 C A The shape factor which depends on the shape of the reservoir Euler s constant and has a value of Wellbore radius squared For circular reservoir, C A is substituting A as r e2, you obtain the equation on the previous slide. For Radial Flow: The end of transient period and the start of pseudo-steady state is the same Symbol Field Unit SI Unit k md md t hrs hrs Fraction Fraction cp Pa.s c t psi -1 kpa -1 A ft 2 m 2 27

28 Example 6 A well is located as shown in the figure here. The area of the reservoir is 4,000,000 ft 2 [3.718x10 5 m 2 ], permeability is 1 md, thickness is 70 ft [21.3 m], viscosity is 0.5 cp [ Pa.s], c t is 13 x10-6 psi -1 [1.89 x10-6 kpa -1 ] porosity is 0.15, r w is 0.45 ft [0.137 m], is 4,000 psia [27,579 kpa] and p wf is 1,500 psia [10,342 kpa]. The value of B is 1.2 bbl/stb [1.2 m 3 /Sm 3 ]. Calculate the time to reach pseudo-steady state. Calculate the rate at which the well will produce under pseudo-steady state conditions. Calculate the value of the productivity index. Solution: Example 6 The time to reach pseudo-steady state is calculated by knowing that the value of t DA = , , It will take approximately one year to reach pseudo steady state 28

29 Solution: Example 6 Calculate the rate as: Productivity Index, J = 256/(4,000 1,500) = STB/d/psi ,000 1, STB/day Productivity Index, J = 40.8/(27,579 10,342) = Sm 3 /d/kpa ,579 10, = 40.8 Sm3 /day

30 Equations for Gas for Other Flow Regimes For transient state: where Equations for Gas for Other Flow Regimes Pressure Squared Method where where Symbol Field Unit SI Unit k md md h ft m t hrs hrs cp Pa.s c t psi -1 kpa -1 r w ft m Fraction Fraction P psia kpa m(p) psi 2 /cp kpa 2 /Pa.s Symbol Field Unit SI Unit k md md h ft m t hrs hrs cp Pa.s c t psi -1 kpa -1 r w ft m Fraction Fraction P psia kpa m(p) psi 2 /cp kpa 2 /Pa.s 30

31 Gas Flow: Pseudo-steady State Pseudo-Steady State Rate Equation / /4 Gas Flow: Pseudo-steady State Pressure Squared / /4 Symbol Field Unit SI Unit k md md h ft m m(p) psi 2 /cp. kpa 2 /Pa.s T R K Symbol Field Unit SI Unit k md md h ft m t hrs hrs cp Pa.s c t psi -1 kpa -1 r w ft m Fraction Fraction P psia kpa m(p) psi 2 /cp kpa 2 /Pa.s 31

32 Example 7 A gas well is producing from a reservoir with initial reservoir pressure of 5,000 psia [34,474 kpa] and bottom hole pressure of 1,500 psia [10,342 kpa]. The permeability is 0.2 md, h is 50 ft [15.24 m], r e is 1,500 ft [457.3 m], and r w is 0.4 ft [0.122 m]. The reservoir temperature is 200 F [93.6 C]. The gas gravity is The porosity is Calculate the time required to reach pseudo-steady state. Calculate the rate at which the well will produce at different times until it reaches pseudo-steady state. Assume that the average reservoir pressure is reduced to 4,970 psia [34,267 kpa] at the beginning of pseudo-steady state period. Calculate the rate at the beginning of pseudo-steady state period and compare it to the end of transient period. What do you observe? Solution: Example 7 Calculate time to reach pseudo-steady state: ,500 = 4,392 hours = 183 days ,392 hours 183 days

33 Solution: Example 7 Calculate a g for 5 days as follows: = 576, x Solution: Example = 9.67x108 Calculate the value of a for pressure squared method: =1.52x =2.55x104 33

34 Solution: Example 7 For using m(p) method, you will need to calculate m(p) as a function of pressure by numerically integrating p/(z) as a function of pressure. P wf, psia m(p) 1.E+09 8.E+08 6.E+08 4.E+08 2.E+08 0.E q, bbl/day p, psia Solution: Example 7 Calculate rate using m(p) method as: P wf, m(p) kpa = =1,362 MSCFD 576,533 5.E+13 4.E+13 3.E+13 2.E+13 1.E+13 0.E q, mp, 3 /day kpa q, MSCFD q, 1,800 1,600 1,400 1,200 1, m(p) p squared t, days t, days /4 q, q, Sm 3 /D m(p) p squared 55,000 50,000 45,000 40,000 35,000 30,000 25,000 20, t, days t, days 34

35 Solution: Example 7 Calculate rate using pressure squared method as: / = 1,496 MSCFD /4 The match between m(p) and p 2 method is not exact since the initial pressure is significantly greater than 3,000 psia [21,000 kpa] Solution: Example 7 Calculate the rate under pseudo steady state conditions as: /4 q, MSCFD q, q, q, Sm 3 /D 1,800 1,600 1,400 1,200 1, m(p) t, days m(p) p squared t, days p squared 55,000 50,000 45,000 40,000 35,000 30,000 25,000 20, t, days t, days = = 1,109 MSCFD 3/ /4 = =31,422 Sm 3 /D /4 35

36 Solution: Example 7 For pressure squared method, the rate is: /4 = / ,203 MSCFD There is some difference between the two answers because the reservoir pressure higher than 3,000 psia [21,000 kpa] = ,267 10,342 3/ = 34,061 Sm 3 /D /4 Learning Objectives This section has covered the following learning objectives: Outline the differences between transient state, steady state and pseudo-steady state flow regimes Calculate the productivity of both oil and gas wells under different flow regimes Describe the importance of the shape of the reservoirs and how they impact the behavior of oil and gas wells 36

37 Learning Objectives Reservoir Flow Properties Fundamentals Vertically Fractured and Horizontal Wells This section will cover the following learning objectives: Outline the difference between vertically fractured and horizontal wells and their advantages and disadvantages Calculate the rates and productivity indices for vertically fractured and horizontal wells using the concept of effective well bore radius Describe different flow regimes encountered by vertically fractured and horizontal wells Evaluate efficacy of horizontal wells and compare the performance to vertically fractured wells 37

38 Other Types of Wells Hydraulically Fractured Vertical Well: A vertical well drilled in tight formations but is hydraulically fractured to improve the productivity Horizontal Well: Instead of vertical, a horizontally drilled well to improve the connectivity with reservoir Fractured Vertical Well Vertical Well Assume that the fracture is vertical and contained within formation L f r e 2 Half fracture length is also denoted as x f L f Fracture has significantly higher conductivity (permeability) than the formation 38

39 Flow Regimes Similar to vertical wells, vertically fractured as well as horizontal wells also exhibit different flow regimes during the production Unlike vertical wells, the flow regimes are more complex and many variations in transient flow regimes exist Flow Regimes (Transient Flow) Fracture Linear Flow There will be a demonstration on the physical reasons for the existence of such flow regimes but will also concentrate on pseudo-radial transient flow Unless the permeability is extremely low, the existence of other flow regimes is short lived, and pseudo-radial flow starts within days When the well is open, the first thing which drains into the well bore Bi Linear is the Flow fluid inside the fracture. And because we assume that the fracture is linear, the flow will be linear inside the fracture. Formation Linear Flow Pseudo radial Linear Flow 39

40 Importance of Permeability on Flow Regimes k, md ft X f m Linear Flow Start t, Days Linear Flow Ends Radial Flow , ,316 6, , ,825 47, The lower the permeability, the more likely is the possibility that linear or bi-linear flow can extend over a long time. 40

41 Vertically Fractured vs. Horizontal Well When comparing a vertically fractured with a horizontal well, it is observed that the vertically fractured well with the same length as the horizontal well will be superior compared to the horizontal well. Because the horizontal well only contacts a limited portion of the reservoir in a horizontal direction; whereas, the vertically fractured well contacts the entire thickness in a vertical direction. Horizontal well performance is controlled by vertical permeability but this is not the case for vertically fractured wells. We still drill horizontal wells because we have much better geological control over horizontal wells compared to vertical fractures. Horizontal Well Z X L h z w The assumption is that once the horizontal well reaches a pseudoradial flow the performance of the horizontal well can be estimated. h We need to know: The distance from the bottom of the formation where the center of the horizontal well is drilled The length of the horizontal well L h r e 41

42 What About Horizontal Wells that are Vertically Fractured? These wells overcome the biggest problem of horizontal wells; i.e., vertical communication To determine the performance of these wells analytically is difficult However, as an approximation, consider each vertical fracture as less efficient vertically fractured well The efficiency can be as low as 10% to as much as 75% 42

43 Example 8 A vertically fractured well is producing from a gas reservoir. The average pressure is 2,800 psia [19,305 kpa] and the BHP is 700 psia [4,826 kpa]. The reservoir has a permeability of 0.1 md, thickness of 60 ft [18.3 m], r e is 1,700 ft [518.3 m], and r w is 0.38 ft [0.116 m]. The half fracture length is 100 ft [30.5 m]. Assume the fracture to be infinite conductivity. The viscosity of gas is cp [ Pa.s], z is and the reservoir temperature is 130 F [54.8 C]. Calculate the rate with and without hydraulic fracture under pseudo-steady state conditions. Use pressure squared method. Generate the IPR curve for both the conditions. Solution: Example 8 Rate without Fracture / =.... / = 433 MSCFD. /...,, =... = 12,253 Sm 3 /D./ 43

44 Solution: Example 8 Rate with Fracture The effective wellbore radius is 100/2=50 ft [30.5/2 = m] /4 = ,193 MSCFD 3/ = ,305 4,826 3/ = 33,788 Sm 3 /D /4 Solution: Example 8 The IPR curve can be generated by calculating the rate at other BHP s Examining the IPR curve, one can clearly see the difference between the productivity of the well with and without a hydraulic fracture P p wf,, psia P wf, p wf kpa, kpa w/o Fracture ,000 1,500 q, q, MSCFD w/ Fracture w/o Fracture w/ Fracture ,000 20,000 30,000 40,000 50,000 q, Mm q, Sm 3 /day 3 /D 44

45 Example 9 A horizontal well is planned to be drilled in an oil reservoir. The permeability of the formation is 1 md, vertical to horizontal permeability ratio is 0.1, thickness is 60 ft [18.3 m], the oil viscosity is 0.6 cp [ Pa.s], the compressibility is 20 x 10-6 psi -1 [2.81 x 10-5 kpa -1 ], the porosity is 0.12, the r w is 0.4 ft [0.122 m], and r e is 2000 ft [609.8 m]. The length of the well is proposed to be 1,000 ft [304.9 m]. The initial reservoir pressure is 2,800 psia [19,305 kpa] and the initial formation volume factor is 1.15 bbl/stb [1.15 m 3 /Sm 3 ]. Assuming that permeability in x and y directions are identical and are equal to 1 md. The distance from bottom of the formation to the well is 30 ft [9.15 m]. Calculate the inflow performance of the horizontal well Compare the performance with the inflow performance of a vertical well drilled in the same reservoir Solution: Example 9 To determine the performance of the horizontal well: First determine the value of F:

46 Solution: Example ln ft 2 ln =.. Solution: Example 9 = 34 m The rate at any well flowing pressure using standard rate equation for radial flow can be calculated. Rate at 1,000 psia = ,800 1, = STB/day Rate at 6,895 kpa = ,305 6, = 82 Sm 3 /day 46

47 Solution: Example 9 Contrast this to standard vertical well, where the rate is equal to: ,800 1, STB/day = ,305 6, = 22.7 Sm 3 /day Solution: Example 9 The figure on the right compares the IPR curves for vertical vs. horizontal wells Horizontal wells are more than three times prolific than vertical wells p P wf wf,, psia psia p P wf wf,, kpa kpa Vertical q, q, STB/D Vertical Horizontal If the cost of drilling horizontal wells are significantly smaller than vertical wells, horizontal wells are clearly a more effective method of producing this formation Horizontal q, q, Sm 3 /D 47

48 Learning Objectives This section has covered the following learning objectives: Outline the difference between vertically fractured and horizontal wells and their advantages and disadvantages Calculate the rates and productivity indices for vertically fractured and horizontal wells using the concept of effective well bore radius Describe different flow regimes encountered by vertically fractured and horizontal wells Evaluate efficacy of horizontal wells and compare the performance to vertically fractured wells 48

49 Learning Objectives Reservoir Flow Properties Fundamentals Heterogeneous Systems and Skin Factor This section will cover the following learning objectives: Calculate the effective permeability of heterogeneous systems when the layers are either in parallel or in series Distinguish between effective permeability calculations for linear and radial flows Describe the concept of skin factor and its influence on the performance of the well Evaluate the well performance with limited production data 49

50 Application of Darcy s Law for Heterogeneous Systems Layers in Parallel Layers in Series Flow direction is parallel to the layers Layers in Parallel k 1,hk 1,h 1 k 2,hk 2,h 2 K n 1 k,h n 1 n 1,h n 1 k n,h K n n,h n k 1 k,l 1 1 k 2 k,l 2 2 k k n,l n Flow direction is perpendicular to the layers Assume that we have n layers with different permeabilities and different thicknesses. The flow is parallel to the layers as shown by the direction of the arrows. k 1,h 1 k 2,h 2 k n 1,h n 1 k n,h n 50

51 Layers in Parallel Basic Derivation If we define the total thickness of all the layers as: The effective permeability is calculated as: It is easy to show that the same equation can also be used for radial flow when the flow is through parallel layers 51

52 Example 10 Assume that we have n layers with different permeabilities and different thicknesses. The flow is parallel to the layers as shown by the direction of the arrows. Solution: Example 10.. k = 100 md, h = 5 k = 10 md, h = 10 k = 1 md, h = 60.. Calculate the effective permeability as: Assuming that the highest permeability is 1000 instead of 100 (an order of magnitude higher), the effective permeability will be: Assuming that the lowest permeability is 0.1 instead of 1 (an order of magnitude smaller), the effective permeability will be: Notice that the effective permeability is much more sensitive to the largest value but not as sensitive to the smallest value. This is an important characteristics of arithmetic average. 52

53 Layers in Series By applying Darcy s law and making similar assumptions, we can also derive an effective permeability value for layers in series. The flow direction is shown for both linear and radial flow below. k 1,L 1 k 2,L 2 k n,l n Linear Flow Layers in Series Linear vs. Radial Flow r e k n k 1 Radial Flow If we define the total length of all the layers as: We can write the expression for effective permeability as: Radial Flow We can write the expression for effective permeability as: 53

54 Example 11 Following system consists of three layers with different permeabilities and thicknesses. Calculate the effective permeability of the system if the flow is perpendicular to the layers. Solution: Example k = 100 md, L = 5 k = 10 md, L = 10 k = 1 md, L = 60 Calculate the effective permeability If we assume that the highest permeability is 1000 instead of 100 (an order of magnitude higher), the effective permeability will be On the other hand, if we assume that the lowest permeability is 0.1 instead of 1 (an order of magnitude smaller), the effective permeability will be Notice that the effective permeability is much more sensitive to the smallest value but not as sensitive to the largest value. This is an important characteristic of harmonic average 54

55 Near Wellbore Damage Earlier there was a discussion on how a small region near the wellbore can alter the productivity of a well This damage can also be explained in an alternate equation using a concept of skin factor. The equation can be written as: 0.75 In the equation, C is a constant depending on the units, and S is called a skin factor which is a dimensionless quantity and represents the near wellbore alteration in permeability. Physical Meaning of Skin Factor If we assume that up to a radius of r d, the original permeability k is altered to k d, we can write an equation for effective permeability as: 1 1 It is known that the rate calculated using effective permeability or using original permeability with skin factor has to be the same. The equation can be written as:

56 Physical Meaning of Skin Factor Substituting the value of k eff in the previous equation, and with some algebraic manipulation, you can write: This expression allows us to calculate the value of skin factor in terms of damaged zone. Thick Skin The assumption that the damaged zone has a finite thickness. Thin Skin Where we assume that very close to the wellbore, there is an instantaneous pressure drop due to damage. 56

57 Example 12 A well is currently producing at a rate of 130 bbl/day [20.7 m 3 /day]. As an operator you suspect that there is a near wellbore damage where the original permeability of 20 md has been altered to 5 md up to a distance of 3 ft [0.91 m]. The drainage radius is 900 ft [274.4 m], and the wellbore radius is 0.35 ft [0.107 m]. Calculate the skin factor under the current conditions. If the permeability in the damaged zone is restored to original value, what would be the new rate? If by using stimulation, the permeability up to a distance of 3 ft [0.91 m] is increased to 100 md, what would be the skin factor? How much would be the rate under these conditions? Solution: Example 12 Using the equation for skin factor, you can calculate the value as the following:

58 Solution: Example 12 Altering the near wellbore permeability to 20, the skin factor would be zero. Examining the rate equation for the rate in terms of skin factor, write the following: and Solution: Example 12 Substituting..... / /.. 58

59 Solution: Example 12 By restoring the permeability to the original value, there is a significant improvement in the rate The negative skin factor implies that the near wellbore alteration has resulted in improved permeability which exceeds the reservoir permeability. Solution: Example 12 Using the same logic, calculate the new rate for negative skin factor as, substituting: / / Big improvement in the rate is obtained by changing the permeability over a very small interval. 59

60 Rate Equation Approximation For a typical vertical well, the value of can be approximated by a constant 7. Although both the drainage radius and wellbore radius can vary, the ratio is such that the log term is close to 7. Assuming this value, write the rate equation for different skin factors as: 7 7 The advantage of using this equation is that without knowing reservoir parameters, you can still approximate the rate improvement by altering the near wellbore permeability. Learning Objectives This section has covered the following learning objectives: Calculate the effective permeability of heterogeneous systems when the layers are either in parallel or in series Distinguish between effective permeability calculations for linear and radial flows Describe the concept of skin factor and its influence on the performance of the well Evaluate the well performance with limited production data 60

61 Learning Objectives Reservoir Flow Properties Fundamentals Deviations from Single Phase Darcy s Law Equations This section will cover the following learning objectives: Evaluate the well performance in the presence of non-darcy flow using both pressure squared and pseudo-real pressure methods Evaluate the multi-rate test and generate the inflow performance curve for gas wells Use the single rate test to predict inflow performance of oil wells producing below bubble point using both Fetkovich and Vogel methods 61

62 Non-Darcy Flow Darcy law is applicable for wells when the velocity at which the fluid is moving is very slow. Non-Darcy Flow Darcy law is applicable for wells when the velocity at which the fluid is moving is very slow. If the velocity is higher, the pressure drop cannot be calculated using Darcy s law equation. Instead we will need to use Forchheimer s equation. If the velocity is higher, the pressure drop cannot be calculated using Darcy s law equation. Instead we will need to use Forchheimer s equation. Gas expands as it approaches the The wellbore. non-darcy flow is important near At the lower well bore and mostly for pressures gas wells. the gas will expand, and hence the velocity will increase. Gas expands as it approaches the The wellbore. non-darcy flow is important near At the lower well bore and mostly for pressures gas wells. the gas will expand, and hence the velocity will increase. For oil wells, non- Darcy flow is rare and need not be considered. For GAS wells, non-darcy flow is more of a RULE, whereas for OIL wells non-darcy flow is more of an EXCEPTION. For oil wells, non- Darcy flow is rare and need not be considered. 62

63 Forchheimer s Equation Forchheimer s equation is written as: This equation requires the knowledge of which is called high velocity coefficient. Firoozabadi and Katz empirical equation provides us with a value of for different reservoirs Gas Flow Rewrite Forchheimer s equation For gas flow rate and, for radial flow, 2 Substituting these two equations into the first equation In the above equation, you can simplify further by assuming that viscosity can be defined at near wellbore conditions on the right hand side 63

64 Non-Darcy Flow When non-darcy flow is important, a quadratic equation is governing the flow. For given pressures, solve the quadratic equation to calculate the rate. That equation is written as: or If non-darcy term is negligible, the equation reduces to: or = This is the same equation we have seen before. Note that the values of a and b are different for pressure squared and m(p) methods. 64

65 Example 13 A gas well is producing from a reservoir with a reservoir pressure of 3,500 psia [24,132 kpa]. The drainage radius is 1,500 ft [457.3 m], the wellbore radius is 0.4 ft [0.122 m], the permeability is 16 md, and the thickness is 45 ft [13.7 m]. The average viscosity is 0.02 cp [ Pas], z factor is 0.87, and the reservoir temperature is 180 F [82.6 C]. The perforated interval is 30 ft [9.15 m]. The specific gravity of gas is Assume skin factor to be zero. Assume consolidated formation. Calculate the inflow performance of the well with and without non-darcy flow. Use pressure squared method. Solution: Example ft m , First calculate the value of

66 For Calculating Rate without Non-Darcy Flow BHP of 2000 psia [13,789 kpa], the rate is For Calculating Rate without Non-Darcy Flow For calculating rate without non-darcy flow, use: ,420 3 / / The rate at other BHP s can be calculated using similar equation. 66

67 Solution: Example 13 (continued) The IPR relationship shows large differences with and without non-darcy flow P p wf,, psia non Darcy w/o non Darcy If we ignore non-darcy flow for this well, the difference in the rate prediction can be off by 40% At low rates, the two curves converge, indicating that non-darcy flow becomes important at high rates P wf, p kpa wf,kpa q, MMSCF/day q, non Darcy w/o non Darcy ,000 1,500 2,000 2,500 q, Mm 3 3 /day 67

68 Well Productivity for Multiphase Cases So far, we have considered well productivity for single phase oil and single phase gas. What happens if we have oil producing below bubble point? If oil produces below bubble point, both free gas and oil flow in the reservoir The well is producing under multiphase flow conditions As the bottom hole pressure changes, so does the saturation surrounding the wellbore; hence, the productivity of both oil and gas will be impacted 68

69 Equation for Below Bubble Point ln 0.75 The relative permeability of oil is changing as the saturation changes. Write the equation for rate under pseudosteady state as the following: If single phase oil is present throughout the reservoir, k ro is 1 and the equation will reduce to standard single phase equation. Equation for Below Bubble Point ln Based on numerical studies and field observations, we can reasonably assume that where a and b are constants Substituting in the above equation, integrate to obtain: ln We can also write the equation for maximum flow rate which corresponds to a case when BHP is zero If we assume that b is zero and take ratios of the two rates, we obtain Fetkovich s equation There is also another equation called Vogel s equation which has a slightly different form Both Fetkovich and Vogel s equations require that to generate IPR, we will need a single rate and the corresponding BHP. 69

70 Example 14 A well is producing at a rate of 72 STB/day [11.5 Sm 3 /day] at BHP of 1,000 psia [6,895 kpa]. The reservoir pressure is 2,000 psia [13,790 kpa.]. If by installing rod pump, the BHP can be reduced to 100 psia [690 kpa]. What would be the incremental production from this well? Use both Fetkovich and Vogel methods. Solution: Example 14 Fetkovich Method Calculate q max as: 1 = = 96 STB/day 1 = = 15.3 Sm3 /day 70

71 Solution: Example 14 Fetkovich Method Calculate the rate at 100 psia (690 kpa) as: 1 = Therefore, q = 96 STB/day , // The improvement would be 24 STB/day (3.7 Sm 3 /day) Solution: Example 14 Vogel Method Calculate q max as: STB/day Sm3 /day 71

72 Solution: Example 14 Vogel Method Calculate the rate at 100 psia (690 kpa) as: The improvement would be 30 STB/day (4.7 Sm 3 /day) Therefore, q 102 STB/day 2000 Vogel's equation is always going to be a little bit more optimistic than Fetkovich method Therefore, q 16.2 Sm /day 72

73 Back to Work Suggestions Reservoir Flow Properties Fundamentals Leverage the skills you ve learned by discussing the skill module objectives with your supervisor to develop a personalized plan to implement on the job. Some suggestions are provided. Back to Work Suggestions Reservoir Flow Properties Fundamentals Leverage the skills you ve learned by discussing the skill module objectives with your supervisor to develop a personalized plan to implement on the job. Some suggestions are provided. How can you better predict the performance of an existing well? Identify if the well is vertical, horizontal, or fractured. What type of flow regime is it producing from? Is the well producing under single phase or multi phase conditions? If the well is a gas well, consider non Darcy effect in predicting the performance of the well. Compare the performance of the well with the predicted performance. Have you made adjustments in the reservoir parameters to match the performance? What does the history of well performance tell us? Do you have sufficient information about the reservoir pressure as a function of time? Is the well performance consistent with what is expected from that well? Does the past performance indicate potential skin damage? What possible solutions you can implement to improve the performance of the well in the future? 73

74 Learning Objectives This section has covered the following learning objectives: Evaluate the well performance in the presence of non-darcy flow using both pressure squared and pseudo-real pressure methods Evaluate the multi-rate test and generate the inflow performance curve for gas wells Use the single rate test to predict inflow performance of oil wells producing below bubble point using both Fetkovich and Vogel methods This is Reservoir Engineering Core Reservoir Rock Properties Core Reservoir Rock Properties Fundamentals Reservoir Fluid Core Reservoir Fluid Fundamentals Reservoir Flow Properties Core Reservoir Flow Properties Fundamentals Reservoir Fluid Displacement Core Reservoir Fluid Displacement Fundamentals Reservoir Material Balance Core Applied Reservoir Engineering Properties Analysis Management Reservoir Material Balance Fundamentals Decline Curve Analysis and Empirical Approaches Core Decline Curve Analysis and Empirical Approaches Fundamentals Pressure Transient Analysis Core Rate Transient Analysis Core Enhanced Oil Recovery Core Enhanced Oil Recovery Fundamentals Reservoir Simulation Core Reserves and Resources Core Reservoir Surveillance Core Reservoir Surveillance Fundamentals Reservoir Management Core Reservoir Management Fundamentals 74