Chapter 2 Reservoir System

SKPP 3513 – PRODUCTION  ENGINEERING

Chapter 2: Reservoir System Chapter 2: Reservoir System Mohd Fauzi Hamid Mohd Fauzi Hamid N01‐439 07‐5535616 [email protected]

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Learning Outcome Learning Outcome

Students will able to: Students will able to:

 Understand the concept of IPR  Differentiate the method used for calculating single phase  Differentiate the method used for calculating single phase flow and two phase flow IPR  Construct the IPR curve Construct the IPR curve  Identify the fluid phase for a particular given condition

CHAPTER 2: RESERVOIR SYSTEM

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Contents  Productivity Index (PI)  Hydrocarbon Phase Behavior  Inflow Performance Relationship (IPR) p( )

CHAPTER 2: RESERVOIR SYSTEM

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Fluid Inflow  Fluid inflow is the flow of liquid from formation to the  bottom hole.  The flow of liquids are affected by 4 main factors:  Reservoir fluid properties (physical & chemical properties).  P, T, composition, etc.  Physical & Geometrical properties of reservoir rocks. P, T,  composition, structure, over burden, cementing,  compaction etc compaction, etc.  Pressure differential (P = Pf – Pw)  Well geometry, spacing, production area @ P Well geometry spacing production area @ Pf

CHAPTER 1: INTRODUCTION

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Productivity Index (PI) Productivity Index (PI)  The PI express the ability of production of the well.  Provide a measure of the capability of a reservoir to deliver  fluids to the bottom of a wellbore for production.

 It defines the relationship between the surface production  rate and the pressure drop across the reservoir, known as  the drawdown.

where: PI q Ps Pwf CHAPTER 2: RESERVOIR SYSTEM

q PI  J  Ps  Pwf

………………………  (2‐1)

=      productivity index, STB/(day) (psi) =      producing flow rate, STB/day = reservoir static pressure psi =      reservoir static pressure, psi =      flowing bottom hole pressure, psi (5)

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Productivity Index (ctd) Productivity Index   For radial flow and strong water drive or gas cap reservoir  where flow tend to reach steady‐state flow (from Eq where flow tend to reach steady state flow (from Eqn 1‐3): 1 3)

q 0 007082kh 0.007082 .………………. (2‐2) PI  J   re Ps  Pwf Bo  ln(( ) rw  PI will be a constant if , Bo & k remain constant. A plot of P  vs q should be a straight line of slope ‐1/PI with an intercept  on the ordinate axis of Pe. on the ordinate axis of P

CHAPTER 2: RESERVOIR SYSTEM

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Productivity Index (ctd) Productivity Index  Pe Slope = 1/PI Slope = ‐ P

q Productivity Index Productivity Index CHAPTER 2: RESERVOIR SYSTEM

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Productivity Index (ctd) Productivity Index 

 For pseudosteady‐state flow (from Eqn 1‐4) : q 0.007082kh  PI  J  re 1 Ps  Pwf B  (ln  ) o (l rw 2

………… (2‐3)

 In terms of average reservoir pressure, P I t f i Pavg or P Pr: q 0 007082kh 0.007082 PI  J   Pr  Pwf B  (ln re  3 ) o rw 4

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………… (2‐4) ( )

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Productivity Index (ctd) Productivity Index 

 Again, provided the fluid and rock properties (, B and k)  are constant, the PI should be constant, irrespective of the  t t th PI h ld b t t i ti f th degree of depletion. Thus, as for the steady state case, a  straight line relationship exists between P and q straight line relationship exists between P and q.

 For gas well: PI  J 

CHAPTER 2: RESERVOIR SYSTEM

qg ( Ps 2  Pwf 2 )



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0.703k g h re T (  g z ) avg ln( ) rw

…… (2‐5) ( )

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Productivity Index (ctd) Productivity Index 

 Where the gas flow rate, qg (MSCF/day): qg 

k g h( Pe 2  Pw2 ) re 1422T (  g z ) avg ln( ) rw

 PI will remain constant if no change in the fluid and  reservoir properties. If a graph of P2 vs q is drawn, a  straight line should result of slope ‐1/PI.

CHAPTER 2: RESERVOIR SYSTEM

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Productivity Index (ctd) Productivity Index  Pe2 Slope = ‐ 1/PI

P2 Absolute Open Flow  Potential (AOFP) Potential (AOFP)

qo

qg

Gas well productivity relationship Gas well productivity relationship CHAPTER 2: RESERVOIR SYSTEM

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Productivity Index (ctd) Productivity Index 

 PI specific, PIs (STB/(day)(psi)(ft):

PI q PI s  J s   ………… (2‐6) ( ) h h( Ps  Pwf ) Example 2.1 A field is drilled up on a rectangular 80‐acre spacing. The  reservoir pressure (Ps) is 1000 psi, k = 50 md, h = 20 ft,  = 3 cp,  Bo = 1.25 bbl/STB. The wells are completed with 7‐in casing. If  = 1 25 bbl/STB The wells are completed with 7 in casing If the producing pressure at the bottom of the well is 500 psi, (a)  what is the production rate of the well, (b) the PI, and (c) the  p ,( ) , ( ) specific PI. CHAPTER 2: RESERVOIR SYSTEM

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Productivity Index (ctd) Productivity Index 

 If there is gas in fluid flow, the fluid is no longer  iincompressible, PI not constant and this situation happen  ibl PI t t t d thi it ti h when Ps < Pb. PI becomes:

ddq PI  dPwf

………… (2‐7)

 If there is water in oil flow (oil and water mix): (qo  qw ) ………… (2‐8) (2 8) PI  ( Ps  Pwf )

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Productivity Index (ctd) Productivity Index 

 For oil‐water mix & radial flow (steady‐state):

(qo  qw ) 0.007082h ko kw …… (2‐9) PI    ( ) re Ps  Pwf  B  B o o w w ln( ) rw  FFor two production condition: t d ti diti ( PI )1 ( ko / o Bo )1  ( PI ) 2 ( ko / o Bo ) 2

……………….... (2‐10) (2 10)

 To estimate the PI, Lewis & Horner equation can be used: T ti t th PI L i & H ti b d

kh PI  5 5.9*10 9*10 ( ) o Bo 4

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……………….... (2‐11) (2 11) MOHD FAUZI HAMID

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Productivity Index (ctd) Productivity Index 

 Factors influencing PI.    

Production GOR Pressure depletion Cumulative Production Oil Saturation (So will affect the effective and relative permeability of  th fl id) the fluid)

 Generally, PI will decrease with time & P because:  Turbulence due to increasing of flow rate.  Oil permeability decrease due to present of gas when P decrease  near the well bore.

 o increase with pressure depletion below Pb.  Permeability decrease due to formation compression. CHAPTER 2: RESERVOIR SYSTEM

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Productivity Index (ctd) Productivity Index 

 Future PI:

kro ( )f o Bo PI f  PI p kro ( )p o Bo

…………  (2‐11)

where: h

1.8 PI PI p  Pwf 1  0.8( ) Pavg

CHAPTER 2: RESERVOIR SYSTEM

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Hydrocarbon Phase Behavior Hydrocarbon Phase Behavior  HC reservoir fluid : complex mixture of HC molecules.  Composition depend on: C iti d d  Source rock  Maturation degree Maturation degree  P & T  Phase change from HPHT  Phase change from HPHT  cool & low pressure  cool & low pressure  effect to  effect to performance  Single phase in reservoir (liquid phase) Single phase in reservoir (liquid phase)  Remains single phase (liquid) at wellbore (significant reduction  in P & small change in T @ flow in reservoir (single phase  region)) ))  Starts to evolve gas as T & P reduced @ flow up in tubing  (bubble point) (bubble point)  Evolves increasing amount of gas until separator (two phase  region) CHAPTER 2: RESERVOIR SYSTEM (17) MOHD FAUZI HAMID

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Hydrocarbon Phase Behavior (ctd) Hydrocarbon Phase Behavior 

1. Single phase in reservoir (a) g p ( ) 2. Remains single phase liquid at wellbore (significant reduction in  P & small change in T @ flow in reservoir (a – b) 3 Starts to evolve gas as T & P reduced @ flow up in tubing (c) 3. St t t l T & P d d @ fl i t bi ( ) 4. Evolves increasing amount of gas (d‐e) until separator (f) CHAPTER 2: RESERVOIR SYSTEM

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Hydrocarbon Phase Behavior (ctd) Hydrocarbon Phase Behavior  Possible Phase Changes in Production Flow System

CHAPTER 2: RESERVOIR SYSTEM

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Inflow Performance Relationship (IPR) Inflow Performance Relationship (IPR)  The IPR is defines as the functional relationship between the  production rate and the flowing bottom hole pressure production rate and the flowing bottom hole pressure.

 Always termed as “IPR Curve”; a plot of Pwf vs q (Pwf on  vertical axis and q on horizontal axis). ti l i d h i t l i)

 From the definition of PI

q Pwf  Ps  PI  Ps is constant, and if PI also assumed to be constant, the  relationship between P relationship between Pwf & q is linear with a negative slope of  & q is linear with a negative slope of 1/PI.

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IPR (ctd) IPR 

Slope = ‐ 1/PI

Typical IPR Curve for Incompressible Fluid CHAPTER 2: RESERVOIR SYSTEM

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IPR (ctd) IPR   When q = 0, Pwf = Ps (no flow).  When Pwf = 0, q = Ps * PI = qmax (absolute open flow potential – AOFP)

 AOFP is the maximum rate that the  formation can deliver fluid to the  bore hole.

Ps Slope = - 1/PI

Pwf

 The figure is a typical IPR curve for 

Absolute Open flow potential

radial flow equation for single phase,  incompressible fluid.

 From the figure:

q

qmax tan    PI Ps

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qmax

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Example 2‐2 Example 2 2 Given: Permeability, k P bili k Pay thickness, h Average reservoir pressure Pr Average reservoir pressure, P Reservoir temperature, T Well spacing A Well spacing, A Drilled hole size, D Formation volume factor, B Formation volume factor, B Oil viscosity,  Skin factor, St ,

= = = = = = = = =

30 mD 30 D 40 ft 3000 psi 3000 psi 200oF 160 acres 160 acres 12 ¼ in (open hole) 1.2 bbl/STB 1.2 bbl/STB 0.8 cp 0

Calculate: Absolute Open Flow Potential and PI Plot IPR curve. CHAPTER 2: RESERVOIR SYSTEM

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IPR (ctd) IPR   For most well, IPR curve is not a linear relationship, because  PI decrease when q increase and pressure depleted PI decrease when q increase and pressure depleted.

 PI decrease with pressure depletion.  This method (PI method) of predicting IPR is only can be used  for single phase flow (Ps > Pb). For solution gas reservoir at Ps < Pb, Vogel method is applicable.

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IPR – Gas Flow IPR  Gas Flow  Compressible fluid  IPR not straight line  Can be derived from Darcy Law by using average properties  between reservoir & wellbore (low q): q = C (Pr ( 2 – Pwf2) C = constant  For higher flow rate due non‐Darcy (turbulent flow) effect  For higher flow rate due non‐Darcy (turbulent flow) effect  field observation  Bureau of Mines eq: q = C (Pr q ( 2 – Pwf2)n n  = 0.5 – 1.0  Log‐log plot of q vs (Pr2 – Pwf2)  straight line with : Slope = n l Intercept = C  Standard test for gas well: measure P Standard test for gas ell meas re Pwf at 4 production rate. at 4 prod ction rate CHAPTER 2: RESERVOIR SYSTEM

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IPR – 2 Phase (Gas IPR  2 Phase (Gas‐Liquid) Liquid) Flow Flow  Saturated reservoir  Vogel equation for solution gas‐drive wells can be used: Vogel equation for solution gas drive wells can be used: q/qmax = 1 – 0.2(Pwf/Pr) ‐0.8(Pwf/Pr)2 qmax = AOF (q @ P AOF (q @ Pwf = 0)  0)  If multirate test data available: q/qmax = {1 – (Pwf/Pr)2}n Plot log‐log q vs (Pwf/Pr)2  straight line with slope = n Plot log‐log q vs straight line with slope = n .

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Vogel’ss Equation Vogel Equation  Introduced by Vogel in 1968 and recommended for solution  gas drive reservoir where Ps < P gas drive reservoir where P < Pb .

 Vogel’s Equation;  Pwf qo  1  0.2  (qo ) max  Ps

  Pwf    0.8     Ps 

2

 Pwf  qo vs   Based on the equation, Vogel plot    (qo ) max  Ps 

for solution gas drive reservoir.

 known  as dimensionless IPR Curve of Vogel Chart, Vogel Equation  known as dimensionless IPR Curve of Vogel Chart Vogel Equation or Vogel Chart.

 can be used to predict the IPR of solution gas drive reservoir can be used to predict the IPR of solution gas drive reservoir CHAPTER 2: RESERVOIR SYSTEM

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Vogel’ss Equation  Vogel Equation (ctd)  This IPR curve can be generated if either the qmax and  reservoir pressure are known, or the reservoir pressure & a  reservoir pressure are known or the reservoir pressure & a flow rate and the corresponding bottom hole flowing  pressure are known pressure are known.

 For either case, a buildup test for the reservoir pressure, and  a flow test with bottom hole gauge are required a flow test with bottom hole gauge are required.  Pwf  qo  0, 1   (qo ) max  Ps   Pwf   0,   1  Ps 

 From the equation; if  and if 

qo (qo ) max

Pwf   Pwf   qo  0  1  1  0 0.8 8   (qo ) max Ps   Ps   CHAPTER 2: RESERVOIR SYSTEM

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Example 2‐3 Example 2 3  Given qo Pwf Ps

= = =

300 BPD 300 BPD 1500 psi 1800 psi 1800 psi

Determine: (a) q maximum ((b)) q @ Pwf = 1000 psi q p by  PI and Vogel methods.

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Example 2‐4 Example 2 4  Given qo Pwf Ps

= = =

300 BPD 300 BPD 1500 psi 1800 psi 1800 psi

Determine Pwf for q Determine P for qo = 600 BPD 600 BPD by  PI and Vogel methods. by PI and Vogel methods.

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Standing Method Standing Method  Also known as Standing’s Extension of Vogel’s IPR  Note that Vogel’s IPR is independent of skin factor, thus  applicable to undamaged well only.

 Standing extend the effect of skin on Vogel’s IPR equation and  come up with the concept of flow efficiency factor or FE  (productivity ratio).

P 'wf  Ps  FE ( Ps  Pwf ) Ideal bottom hole flowing pressure, Pwf @ S=0

 So, Vogel’s IPR can be written as S l’ b i  P 'wf qo  1  0.2 0 2 (qo ) max  Ps CHAPTER 2: RESERVOIR SYSTEM

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  P 'wf  0 8   0.8  P   s 

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Standing Method (ctd) Standing Method 

( Ps  P 'wf ) Pideal PI ideal id l id l FE    ( Ps  Pwf ) ..………. (a) Preal PI real

P 'wf  Pwf  Pskin

……….………… (b)

 Combining (a) & (b): C bi i ( ) & (b)

FE 

Pskin CHAPTER 2: RESERVOIR SYSTEM

( Ps  Pwf  Pskin ) ( Ps  Pwf )

m = slope of straight

.…. (c) line curve of PBU test (Horner)

162.6q  B   0.87 mSt  0.87  St  kh   (32)

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Standing Method (ctd) Standing Method   Equation (c) shows that if FE < 1, the well is damage. If FE > 1,  the well is stimulated. When FE = 1.0, Standing the well is stimulated When FE 1 0 Standing’ss correlation  correlation becomes Vogel’s equation.

 Standing produce IPR curve for FE = 0.5 – S di d IPR f FE 0 5 1.5, and can be  15 d b used to determine:

   

maximum flow rate for damage well (formation) maximum flow rate for repaired/stimulated well Flow rate for any value of flow pressure at different FE To construct IPR curve for damage and stimulated well.

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Standing Method (ctd) Standing Method  (Figure 9.5 page 285: Chilingarian)

CHAPTER 2: RESERVOIR SYSTEM

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Example 2‐5 Example 2 5  The following information is available from a well test: Production rate = 500 BPD P d i 500 BPD Static reservoir pressure = 3000 psig Bottomhole flowing pressure = 2500 psig flowing pressure = 2500 psig Flow efficiency = 0.7

Find: q at P Find: q at Pwf = 1700 psig, when (a) FE = 0.7, (b) FE = 1.0, and  = 1700 psig when (a) FE = 0 7 (b) FE = 1 0 and (c) FE = 1.3

 Solution: (Pwf/Ps) = (2500)/(3000) = 0.833 From Fig., at FE=1.0, q/(q)maxFE=1=0.19. Thus, (q) maxFE=1=(500)/(0.19) = 2632 BPD. F For a P Pwf of 1700 psig: f 1700 i CHAPTER 2: RESERVOIR SYSTEM

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Example 2‐5 Example 2 5 (ctd) From Figure: (a) For FE = 0.7, q/(q) (a) For FE = 0 7 q/(q)maxFE=1 = 0.47,  = 0 47 thus q = (0.47)(2632) = 1237 BPD (b) For FE = 1.0, q/(q) (b) For FE  1.0, q/(q)maxFE=1 = 0.65,  0.65, thus q = (0.65)(2632) = 1711 BPD q q maxFE=1 = 0.77,  (c) For FE = 1.3, q/(q) thus q = (0.77)(2632) = 2026 BPD

 The above example illustrates how to predict the flow rate at  various values of Pwf. Hence, as long as FE is known, an IPR  curve can be developed for a particular well.

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Skin Factor Skin Factor  Factors that increase the restriction of fluid flow to the  wellbore wellbore.  It can be mechanical or physical phenomenon.  St = SSd + SSpc + SSswp + SSp where  St Sd Spc Sswp Sp

= = = = =

ttotal dimensionless skin factor t l di i l ki f t formation damage skin factor skin factor due to partial completion skin factor due to partial completion skin factor due to well deviation perforation skin factor perforation skin factor

 The total skin can be as higher as 100 and the minimum skin  in highly stimulated well is about ‐5. in highly stimulated well is about  5. CHAPTER 2: RESERVOIR SYSTEM

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Skin Zone CHAPTER 2: RESERVOIR SYSTEM

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Skin Factor (ctd) Skin Factor   Pressure drop due to skin: 162.6q B  141.2q B  Pskin  0.87mSt  0.87   S St t    kh kh    

 Partial Completion Skin Effect, Spc  Reservoir not perforated over the entire thickness of the pay  f d h h k f h zone

Deviation from radial flow due to partial penetration or completion

CHAPTER 2: RESERVOIR SYSTEM

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MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Skin Factor (ctd) Skin Factor  



In cases where the well is partially completed, the flow can  no longer considered horizontal and well productivity will no longer considered horizontal and well productivity will  depend upon both the horizontal and vertical permeabilities. 

Formation Damage Skin Factor, Sd.  

During drilling operation and in specific completion phases,  wellbore fluid will ‘leak‐off’ into the formation.



This filtrate, together with fine particles, will cause a change  in the permeability of the reservoir rock by any one of a  range of mechanisms shown in next page range of mechanisms shown in next page. 

CHAPTER 2: RESERVOIR SYSTEM

(40)

MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Skin Factor (ctd) Skin Factor  Clay  S lli Swelling Particulate

Wettability

Plugging

Formation Damage

Fines  Migration g

Emulsion

Scale &  Inorganic  Precipitate

Water  Blocking

Formation Damage Mechanism CHAPTER 2: RESERVOIR SYSTEM

(41)

MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Skin Factor (ctd) Skin Factor  

Well Deviation Skin Effect, Sswp.   If a well enters a formation at an angle, the length of borehole  If ll f i l h l h fb h l and hence cross sectional area for flow will be increased.   The productivity will thus be increased The productivity will thus be increased

Deviated Well Geometry Deviated Well Geometry

CHAPTER 2: RESERVOIR SYSTEM

(42)

MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Skin Factor (ctd) Skin Factor  

Perforation Skin Effect, Sp. 

 The process of perforation creates a limited number of small  hole entry points for fluid to flow into the wellbore.

 This will affect the productivity of the well.

CHAPTER 2: RESERVOIR SYSTEM

(43)

MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Construction of IPR Construction of IPR 





Based on Vogel Method, the data required are:  Flow rate, q l  Bottom hole flowing pressure, Pwf  Reservoir pressure, P R i Ps. All these data can be obtained from:  Production test d  Flow test  PBU test PBU The method of constructing the curve depend upon the  condition of well, i.e. diti f ll i  Single phase flow  Two phase flow: P T h fl Ps < P < Pb and P d Ps > P > Pb

CHAPTER 2: RESERVOIR SYSTEM

(44)

MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Single Phase Flow Single Phase Flow q ( Ps  Pwf )



Determine PI: PI 



Determine qmax :  qmax = Ps * PI



Plot graph of Pwf vs q

Absolute Open flow potential

The plot will produce a straight  The plot will produce a straight line

CHAPTER 2: RESERVOIR SYSTEM

Slope = - 1/PI

Pwf

 Pwff = PPs  q  q = 0 0  Pwf = 0   q = qmax 

Ps

(45)

q

qmax

MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Two Phase Flow Two Phase Flow 

There are three methods depend upon the Ps, Pwf, and Pb. 1) Ps < Pb 2) Ps > Pwf > Pb 3) Ps > Pb > Pwf

CHAPTER 2: RESERVOIR SYSTEM

(46)

MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

((1)  P ) s < Pb 

Determine (qo)max from Vogel Equation  Pwf   Pwf qo  1  0.2    0.8  (qo ) max P  s   Ps qo (qo ) max   Pwf   Pwf 1  0.2    0.8   Ps   Ps



  

2

  

2

If production test data is not available, qo can be estimated  from Darcy equation, and  (qo)max from: (qo ) max

CHAPTER 2: RESERVOIR SYSTEM

qo Ps PI * Ps   1.8( Ps  Pwf ) 1.8

(47)

Derivation‐1

0.007082kh( Pr  Pwf ) q r 3 Bo  (ln e  ) rw 4 MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Derivation ‐ 1 qo (qo ) max

(qo ) max

2

 Pwf   Pwf  1  0.2 0 2 0 8   0.8  Ps   Ps Pwf   Pwf    1  08  1  0.8  P P s  s  

  

  qo Ps  1  Pwf ( Ps  Pwf )   1  0.8 Ps 

    1   PI * P  s   Pwf   1  0.8 Ps  

Pwf   Ps  Pwf     08  1  0.8  P P s s   

     

If Pwf to tend to Ps (with small drawdown – at initial stage) (qo ) max 

qo Ps PI * Ps  1 8( Ps  Pwf ) 1.8( 11.88 Back

CHAPTER 2: RESERVOIR SYSTEM

(48)

MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

((1)  P ) s < Pb (ctd) 



Assume a numbers of P Assume a numbers of Pwf (from 0 to  (from 0 to Ps), determine q from Vogel Eqn or  Vogel’ss Reference IPR Vogel Reference IPR Ps

Plot graph of Pwf vs qo

 Pwf = Ps  qo = 0

Pwf

 Pwf = 0   qo = (qo)max 

The plot will produce a curve q

qo  (qo ) max CHAPTER 2: RESERVOIR SYSTEM

(49)

qmax

2   Pwf   Pwf   1  0.2    0.8      Ps   Ps  

MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

((2)  P ) s > Pb 

IPR curve will have two parts:

 Ps > Pwf > Pb  a straight line

4000

 Pwf < Pb  a curve

3500 3000



Considered as generalized IPR – 2500 a combination of the PI method  2000 above the bubble point  pressure and Vogel’s IPR below  1500 1000 the bubble point pressure h b bbl i .

Pwf(psi)



500

Two situations:

0

 Given Pwf > Pb

0

50

100

150

200

q(BPD)

 Given Pwf < Pb CHAPTER 2: RESERVOIR SYSTEM

(50)

MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

((2) P ) s > Pb (ctd)

qb  PI ( Ps  Pb )  qPb  qv   1.8  Ps  Pwf 

CHAPTER 2: RESERVOIR SYSTEM

(51)

 PI * P b    1.8

MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

((2‐1) P ) wf > Pb 

Mark the point Ps @ q = 0, and Pwf @ given q. Connect the  two points with a straight line two points with a straight line.



Determine qb from above plot or from equation qb = PI(P ( s – Pb)  qPb   1.8  1 8  Ps  Pwf

  PI * Pb    qb    qb  1.88   1  



Determine qmax from: qmax



Assume a numbers of Pwf ((0 to Pb)) and determine q from: q 2 P P  wf   wf  (q  qb ) Other Form  1  0.2    0.8   (qmax  qb )  Pb   Pb  Plot Pwf vs q. 2  



q  qb 

CHAPTER 2: RESERVOIR SYSTEM

(52)

PI * Pb 1.8

 Pwf  1  0.2    Pb 

  Pwf    0.8     P   b   MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

 Pwf (q  q b )  1  0.2  ( max  q b ) (q  Pb

  Pwf    0.8     Pb 

2

2   Pwf   Pwf   q  q b  (q ( max  q b ) 1  00.2 2 8   00.8    Pb   Pb      PI * Pb qP Pb but (q max  q b )  q v     1.8  1.8  Ps  Pwf  

    Pwf qPb q  q b    1  0.2   Pb  1.8  Ps  Pwf   

  Pwf       0.8  P   b   2   Pwf   Pwf   PI * Pb  q  qb  2 8 1  00.2   00.8   1.8   Pb   Pb    CHAPTER 2: RESERVOIR SYSTEM

(53)

2

Back

MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

((2‐2) P ) wf < Pb 



Determine qb from equation 2     P P  wf   wf   Pb   qb  q 1    1  0.2    0.8     1.8( Ps  Pwf )    Pb   Pb      qPb   qb Determine qmax from:  qmax    1.8   1 8  Ps  Pwff  



Assume a numbers of Pwf ((0 to Pb)) and determine q from: q 2  Pwf   Pwf  (q  qb )  1  0.2    0.8   (qmax  qb ) P P  b   b 



Plot Pwf vs q + a straight line from Pb to Ps. PI * Pb q  qb  1.8

CHAPTER 2: RESERVOIR SYSTEM

(54)

  Pwf 1  0.2    Pb 

2   Pwf     0.8    P   b   MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

4000

Ps 3500

Pwf(psi)

3000

2500

2000

Pb 1500

1000

500

qb

0 0

20

40

60

80

100

120

140

160

q(BPD)

180

qmax

200

T i lC Typical Composite IPR Curve it IPR C CHAPTER 2: RESERVOIR SYSTEM

(55)

MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Example 2‐5 p 

The Pulai No 1 was tested for eight hours at a rate of about 38  STB/day Wellbore flowing pressure is 585 psi After shutting STB/day. Wellbore flowing pressure is 585 psi. After shutting  the well in for 24 hours, the bottomhole pressure reached a  static value of 1125 psi Calculate: static value of 1125 psi. Calculate: 1. Productivity index, PI 2 Absolute openhole 2. Ab l h l flowing potential fl i i l 3. Oil flow rate for a wellbore flowing pressure of 350 psi  4. Wellbore flowing pressure required to produce 60 STB/day

Draw the IPR curve, indicating the calculated quantities.

CHAPTER 2: RESERVOIR SYSTEM

(56)

MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Example 2‐5: Solution p

1200

1125 100 800 585 400 350 200 0

0 72.2 22.9 38.0 51 0 51.0 54.6 65.1 79 2 79.2

1000

Pwf(psi)

Pwf (psi) q(BPD)

Example 2‐5

800

600

400

200

0 0

CHAPTER 2: RESERVOIR SYSTEM

10

20

30

(57)

40

50

q (STB/day)

60

70

80

90

MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Example 2‐6 p 

A discovery well, Tanjung No. 3, was tested in the T‐9 Sand at  a rate of 200 STB/day with a bottomhole flowing pressure of  a rate of 200 STB/day with a bottomhole flowing pressure of 3220 psi. The estimated bubble point pressure of 4020 psi  indicates that the well is draining saturated oil since the indicates that the well is draining saturated oil, since the  static pressure  was 4000 psi.  Plot the IPR using the Vogel equation.

CHAPTER 2: RESERVOIR SYSTEM

(58)

MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Example 2‐6: Solution p 

Vogel Equation:  qo (qo ) max

 Pwf  1  0.2   Ps

(qo ) max 

(qo ) max 



 Pwf 1  0.2   Ps

  Pwf    0.8   P   s  qo

  Pwf    0.8   P   s  200

2

2

 3220   3220  1  0.2   0.8    4000 4000    

2



624 STB / day

Calculate several rates at specific drawdown to have enough  p g points to plot IPR

CHAPTER 2: RESERVOIR SYSTEM

(59)

MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Example 2‐6: Solution  p ((ctd))

(Pwf/Ps) 1 0.9375 0 8750 0.8750 0.8125 0.7500 0.6875 0 6250 0.6250 0.5625 0.5000 0.4375 0.3750 0.3125 0.2500 0.1875 0.125 0.0625 0

(Pwf/Ps)2 1 0.8789 0 7656 0.7656 0.6602 0.5625 0.4727 0 3906 0.3906 0.3164 0.2500 0.1914 0.1406 0.0977 0.0625 0.0352 0.0156 0.0039 0.0000

q(BPD) 0 68 133 193 250 302 351 396 437 474 507 536 562 583 601 614 624

4500

Example 2-6

4000 3500 3000

Pwf(psi)

Pwf (psi) 4000 3750 3500 3250 3000 2750 2500 2250 2000 1750 1500 1250 1000 750 500 250 0

2500 2000 1500 1000 500 0 0

100

200

300

400

500

600

700

q(BPD)

2   Pwf   Pwf   qo  624 1  0.2    0.8    4000 4000      

CHAPTER 2: RESERVOIR SYSTEM

(60)

MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Example 2‐7 p Given: Static reservoir pressure, psi            =   3620  Bubble point pressure, psi                =   1825 Bottomhole flowing pressure, psi   =   1980 Production rate, BPD                         =   108 1. 2. 3. 4.

Determine the Productivity Index that is valid if Pwf > Pb. Calculate the oil rate if Pwf = Pb. Calculate the maximum oil rate. Plot IPR

CHAPTER 2: RESERVOIR SYSTEM

(61)

MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Example 2‐7: Solution p 

Ps > Pwf > Pb:

1) PI:

PI  (

q 108 )  0.066 BPD / psi Ps  Pwf 3620  1980

2) qb = PI(Ps – Pb) = 0.066(3620 – 1825) = 118.5 BPD 3) qmax: 4) IPR:

qmax 

qPb (108)(1825)  qb   118.5  185.3BPD 1 8( Ps  Pwff ) 1.8( 1.8(3620 1 8(3620  1980)

2   Pwf   Pwf   qo  (qmax  qb ) 1  0.2    0.8     qb P P   b   b   2   Pwf   Pwf   qo  (185.3  118.5) 1  0.2    0.8     118.5 1825 1825      

CHAPTER 2: RESERVOIR SYSTEM

(62)

MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Example 2‐7: Solution  p ((ctd)) Pwf (psi) (Pwf/Pb) (Pwf/Pb)2 q(BPD)

1 0.9041 0.8219 0.6849 0.5479 0.411 0.274 0 3 0.137 0

1.0000 1 0000 0.8174 0.6755 0.4691 0.3002 0.1689 0.0751 0 0 88 0.0188 0.0000

 P qo  (qmax  qb ) 1  0.2  wf   Pb

0 119 130 138 151 162 171 178 182 82 185

Example 2-7

3500 3000 2500

Pwff(psi)

3620 1825 1650 1500 1250 1000 750 500 2 0 250 0

4000

2000 1500 1000 500 0

  Pwf     0.8     qb P   b   2

0

20

40

60

80

100

120

140

160

180

200

q(BPD)

2   Pwf   Pwf   qo  (185.3 (185 3  118.5) 118 5) 1  0.2 0 2 0 8 118 5   0.8    118.5 1825 1825      

CHAPTER 2: RESERVOIR SYSTEM

(63)

MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Example 2‐8 p Given: Static reservoir pressure, psi Bubble point pressure, psi Bottomhole flowing pressure, psi Production rate, BPD

=    1750  =    1200 =     900 =     600

1. Plot IPR 2. Determine well potential 3. Determine Pwf required if qo is 800 BPD

CHAPTER 2: RESERVOIR SYSTEM

(64)

MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Example 2‐8: Solution p 

Ps > Pb > Pwf: Determine qb:    Pwf Pb  qb  q 1  1  0.2   1.8( Ps  Pwf )   Pb 

2   Pwf     0.8      Pb  

2   1200  900   900   qb  600 1  1  0.2    0.8     412 BPD (  900))   1200   1200    1.8(1750



Determine qmax: qmax 



qPb 600*1200  qb   412  883BPD 1.8( Ps  Pwf ) 1.8(1750  900)

Calculate several rates at specific drawdown to have enough  points to plot IPR

CHAPTER 2: RESERVOIR SYSTEM

(65)

MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Example 2‐8: Solution  p ((ctd)) 2000

Example 2-8 28

1800

Pwf (psi) (Pwf/Pb) (Pwf/Pb)2 q(BPD)

1600 1400

1 0.8333 0.6667 0.5 0.3333 0.1667 0

1 0.6944 0.4444 0.2500 0.1111 0.0278 0.0000

412 543 653 742 810 857 883

1200

Pwf(psi)

1200 1000 800 600 400 200 0

1000 800 600 400 200 0

 P qo  (qmax  qb ) 1  0.2  wf   Pb

  Pwf     0.8     qb P   b   2

2   Pwf   Pwf   qo  (883  412) 1  0.2    0.8     412   1200   1200  

CHAPTER 2: RESERVOIR SYSTEM

0

200

400

q(BPD)

600

800

1000

(ii)  Well Potential  =  qmax =  883 BPD (iii)  Pwf @ q  =  800 BPD: from IPR, Pwf =  (66)

MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Fetkovich Method 

Multipoint back pressure test of gas wells is a common  procedure to establish the performance curve of gas wells or procedure to establish the performance curve of gas wells or  deliverability.



FFetkovich k i h applied these tests on oil well and found that oil  li d h il ll d f d h il well producing below the bubble point pressure behave just  like gas well like gas well.



The general conclusion from the backpressure tests:

q  PI '( Ps2  Pwf2 ) n where:  PI’ is the productivity index, the exponent “n” was found to be  between 0.5 and 1.0. CHAPTER 2: RESERVOIR SYSTEM

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MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Fetkovich Method (ctd) ( )  A plot of q vs A l f ( s2 – Pwf2) on log‐log paper results in a straight line  (P ) l l l i i h li with slope 1/n.

 This equation often referred to as the backpressure equation. Thi ti ft f dt th b k ti 

The equation was derived from Evinger and Muskat for two  phase radial flow: h di l fl

(7.08kh) q f ( P)dP re ln( ) rw kro f ( P)  o B0 CHAPTER 2: RESERVOIR SYSTEM

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MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Example 2‐8‐1 p Given: Static reservoir pressure, psi

=

2000 

Bottomhole flowing pressure, psi =

1400

Production rate, BPD Production rate, BPD

1020

=

1. Plot IPR when n = 1.0 1 Plot IPR when n = 1 0 2. Plot IPR when n = 0.5

CHAPTER 2: RESERVOIR SYSTEM

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MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Example 2‐8‐1: Solution p (a) n = 1.0: [(b) n = 0.5]:

q  PI '( Ps2  Pwf2 ) n q 1020 BPD PI '  2  2 1.0 2 1.0 2 ( Ps  Pwf ) (2000 ( ppsi )  1400 ppsi      5 x 104 bbl /(day )( psi ) 2   0.7141  2 2   q  5 x 10 bbl /(day )( psi )  2000 psi   Pwf    5 x 104  4 x 106   Pwf2  BPD 4

2

 0.7141  4 x 10   P  6

CHAPTER 2: RESERVOIR SYSTEM

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2 wf

05 0.5

BPD MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Example 2‐8‐1: Solution  p ((ctd)) 2500

Example 2-8-1

Pwf(psi) 2000

Pwf (psi) q(BPD) 2000 0 1500 875 1200 1280 1000 1500 800 1680 600 1820 400 1920 200 1980 0 2000

Pwf (psi) 2000 1500 1200 1000 800 600 400 200 0

q(BPD) 0 944.67 1142.56 1236.86 1308.97 1362.42 1399.34 1421.04 1428.20

1500

1000

500

0 0

CHAPTER 2: RESERVOIR SYSTEM

500

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1000

q(BPD)

1500

2000

2500

MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Prediction of Future IPR 

One of the method used to predict future IPR is Standing Method Develop by Standing based on Vogel Equation Method. Develop by Standing based on Vogel Equation.



Standing rewrote Vogel Equation as: Pwf Pwf q )(1  0.8 )  (1  qmax Ps Ps Pwf qmax q )(1  0.8 ) ( Ps  Pwff Ps Ps Pwf qmax PI  ( )(1  0.8 ) Ps Ps



……………. (1)

Let PI* be the initial value of PI (the PI for small drawdown.  Allowing P g wf to tend to Ps:

CHAPTER 2: RESERVOIR SYSTEM

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MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Prediction of Future IPR (ctd) ( ) 1.8qmax ………….. (2) PI *  ( ) Ps 

Equation (1) & (2): 



1 8PI 1.8 PI PI *  Pwf (1  0.8 ) Ps Vogel equation & (2): produce an equation for future IPR  PI *f *( Ps ) f qf    18 1.8 

CHAPTER 2: RESERVOIR SYSTEM

 Pwf   1  0.2     ( Ps ) f

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  Pwf   0.8    ( Ps ) f

  

2

   

MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Prediction of Future IPR (ctd) ( )  where: 

 (kro / o Bo ) f  PI  PI    (kro / o Bo ) p  * f

* p

 so  sor  kro    1  sor  swi 

CHAPTER 2: RESERVOIR SYSTEM

n

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MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Example 2‐9 p 

Production test data for a well are as follows: Parameter SStatic reservoir pressure, psig i i i Bottomhole flowing pressure, psig Buble point pressure, psig P d i Production rate, BPD BPD Oil saturation, % Residual oil saturation, % I iti l t Initial water saturation, % t ti % Oil formation volume factor, bbl/STB Oil viscosity cp Oil viscosity, cp Gas liquid ratio, MCF/STB Oil specific gravity, oAPI Gas specific gravity Gas specific gravity

CHAPTER 2: RESERVOIR SYSTEM

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2009

2015

3000 2500 3500 5000 60 20 10

2000 ‐ ‐ ‐ 45 15 30

1.3 08 0.8 0.8 30 06 0.6

1.01 08 0.8 0.8 30 06 0.6

MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Example 2‐9  p ((ctd)) i.

Construct IPR Curve for the year of 2009 and 2015.

ii. Determine well potential for the year of 2009 and 2015. iii. Determine the bottomhole flowing pressure if the well should  produced 14000 BPD in 2009. iv. Determine production rate in 2015 if the bottomhole p flowing  g pressure is 750 psig.

CHAPTER 2: RESERVOIR SYSTEM

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MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Example 2‐9: Solution p 

For 2009: Ps
(qo ) max 



qo P 1  0.2  wf  Ps

  Pwf   0.8    P   s 

2

5000  2500   2500  1  0.2   0.8     3000   3000 

2

 18, 000 BPD

Calculate several present rates at specific drawdown to have  enough points to plot 2009 IPR.

CHAPTER 2: RESERVOIR SYSTEM

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MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Example 2‐9: Solution  p ((ctd)) Present IPR Curve Present IPR Curve (Pwf/Ps) 1 0 9167 0.9167 0.8333 0.75 0.6667 0 5833 0.5833 0.5 0.4167 0.3333 0.25 0.1667 0.0833 0 0333 0.0333 0

(Pwf/Ps)2 1 0 8403 0.8403 0.6944 0.5625 0.4444 0 3403 0.3403 0.2500 0.1736 0.1111 0.0625 0.0278 0.0069 0 0011 0.0011 0

 P qo  (qo ) max 1  0.2 0 2  wf   Ps

q(BPD) 0 2600 5000 7200 9200 11000 12600 14000 15200 16200 17000 17600 17864 18000

2009 IPR 3500 3000 2500 2000 1500

2009 IPR

Pwf(psi)

Pwf (psi) 3000 2750 2500 2250 2000 1750 1500 1250 1000 750 500 250 100 0

1000 500 0 0

2000

4000

8000

10000

12000

14000

16000

18000

20000

q(BPD)

2   Pwf   0 8   0.8   P   s  

CHAPTER 2: RESERVOIR SYSTEM

6000

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MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Example 2‐9: Solution  p ((ctd)) 

To predict future IPR, use Standing Approach:  (q )    Pwf   18000   2500   PI   o max  1  0.8  1 0.8        10   3000    Ps    Ps   3000  PI *p 

1.8* PI 1.8*10   10.8 Pwf    (2500)  1  0.8 1  0.8     (3000) P   s  

 (kro / o Bo ) f  PI  PI    11.53 ( k /  B )  ro o o p  * f

* p

 Pwf PI *( Ps ) f   1  0.2  qf   (P ) 1.8   s f  * f

  Pwf   0.8    ( Ps ) f

  

2

   

2  P P  wf   wf   11.53* 2000 1  0.2 qf  0 2 0 8   0.8   1.8   2000   2000   CHAPTER 2: RESERVOIR SYSTEM

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MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Example 2‐9: Solution  p ((ctd)) 

Calculate several future rates at specific drawdown to have enough points to plot 2015 IPR IPR. Future IPR Curve 3500

Pwf (psi) (Pwf/Ps) (Pwf/Ps)2 qf(BPD) 2000 1750 1500 1250 1000 750 500 250 0

1 0.875 0.75 0.625 05 0.5 0.375 0.25 0.125 0

1 0.7656 0.5625 0.3906 0 2500 0.2500 0.1406 0.0625 0.0156 0.0000

0 2722 5124 7206 8968 10409 11530 12331 12811

Pwf(psi)

Example 2 2-9 9 3000

2500

2009 IPR

2000

2015 IPR 1500

1000

500

2  Pwf   Pwf   PI *( Ps ) f  1  0.2    0.8  qf   ( P )   ( P )   0 1.8   s f   s f    23060 1  1*104 ( Pwf )  2 *107 ( Pwf ) 2  qf  1.8 * f

CHAPTER 2: RESERVOIR SYSTEM

0

2000

4000

6000

8000

10000

12000

14000

16000

18000

20000

q(BPD) (80)

MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Predicting Future IPR by Fetkovich g y Method 

From Muskat: PI at any specific time is related to the PI at other  time:

PI1  (kro / o Bo )1    PI 2  (kro / o Bo ) 2 



Fetkovich: kro is linear towards the pressure: p

kro @ Ps Ps  kroi Psi 

From backpressure equation:

q  PI '( Ps2  Pwf2 ) n …………. (1)

CHAPTER 2: RESERVOIR SYSTEM

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MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Predicting Future IPR by Fetkovich g y Method ((ctd)) 

Based on these three equations:

Ps 2 q  PI ' ( Ps  Pwf2 ) n Psi OR: q  PI ' Ps 2 ( P 2  P 2 ) n 1 s2 wf Ps1

………… (2)

 Determine PI’ by backpressure equation.  Find the relationship q vs Pwf by Eqn. (2)  Assume a numbers of Pwf, and use Eqn (2) to find q vs Pwf at  specific Ps.

 Plot Pwf vs qq. CHAPTER 2: RESERVOIR SYSTEM

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MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Predicting Future IPR by Simple Method g y p 

Collect several PI data from wells which known Ps.



Plot PI/(PI)i vs Ps.  (PI)i = PI at early stage of production PI @ Pi).



Find the best straight line.



For any specific value of Ps, find PI.



Based on the obtained PI, future IPR curve can be plotted using  Based on the obtained PI future IPR curve can be plotted using previous methods.

CHAPTER 2: RESERVOIR SYSTEM

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MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Determination of PI or IPR at Field 

Field determination is depend on the conditions of the well.

a) Simple Case.  Data required are P q q wf, q and P s.  Shut‐in well for 24 – 72 hours   PBU  With pressure gauge at bottom, flow the well at lowest q and  p g g , q record Pwf (stable – may be 24 hours)  Flow the well at highest rate and record Pwf (as step 2)  Repeat for another flow rate  minimum 3 reading  Ps from pressure buildup test.  PI can be determine by method explained earlier.

CHAPTER 2: RESERVOIR SYSTEM

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MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Determination of PI or IPR at Field (ctd) ( ) b) Special Case  For well without packer. F ll i h k  During production, some of the gas will come out and  accumulated at the top of the annulus accumulated at the top of the annulus.  At equilibrium,  Pwf = P Pc + pressure due to gas column + d t l  Gilbert: 1 5/100 psia P Pressure due to gas column = P d t l PcD1.5 /100 i Pwf = Pc(1+D1.5/100 ) psia where Pc = D = CHAPTER 2: RESERVOIR SYSTEM

casing head pressure, psia tubing depth per 1000 ft tubing depth, per 1000 ft (85)

MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Example 2‐10 p Given: Tubing depth b d h q

= 3000 ft f = 42 BPD @ Pc = 550 psig = 66 BPD @ Pc = 320 psig By assuming constant PI, determine: 1. PI 2. Ps 3. Well potential

CHAPTER 2: RESERVOIR SYSTEM

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MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Example 2‐10: Solution p  D1.5   31.5  Pwf  Pc 1    565 1    595 psia  580 psig  100   100  

Pwf = 580 psig @ q = 42 BPD 580 i @ 42 BPD  D1.5   31.5  Pwf  Pc 1    335 1    353 psia  338 psig  100   100 



Pwf = 338 psig @ q = 66 BPD

CHAPTER 2: RESERVOIR SYSTEM

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MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Pwf (psig g)

Example 2‐10: Solution  p ((ctd))

1000



 PI = 100/1000 = 0.1 BPD/psi  Ps = 1000 psig  qmax = 100 BPD 100 BPD

580 338

0

CHAPTER 2: RESERVOIR SYSTEM

From the plot:

42

66

100 q (BPD)

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MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

IPR – Method Selection



Dissolved Gas Drive Reservoir  Above P Above Pb: • Constant GOR ‐ No gas breaking out of solution and no  interfering with oil flow. g • PI method is applicable



Below P B l Pb: • Gas break out of solution & impedes the oil flow • Vogel method V l h d

CHAPTER 2: RESERVOIR SYSTEM

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MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

IPR – Method Selection (ctd) ( )



Water Drive Reservoir  Has: H • Constant producing GOR • Pr is maintained by water influx is maintained by water influx • No gas breaking out of solution • PI method can be used.  However, if water influx rate is exceeded, then the reservoir  will perform like a dissolved gas drive reservoir, therefore  Vogel method will be applicable Vogel method will be applicable .

CHAPTER 2: RESERVOIR SYSTEM

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MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

IPR – Method Selection (ctd) ( )



Gas Cap Drive Reservoir  Wells producing from the oil column below the GOC  W ll d i f th il l b l th GOC • Produce with a constant GOR • PI method can be used. PI method can be used.  But, after gas breaks through to a well, neither the PI nor a  vogel method is applicable. In most instances, the well  should be shut in after the gas cap break through in order to  conserve gas cap energy to produce down‐structure well.

CHAPTER 2: RESERVOIR SYSTEM

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MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

IPR – Method Selection Summaryy





PI Method  Use to evaluate dissolved gas‐drive wells above the bubble  U t l t di l d di ll b th b bbl point  Use to evaluate gas cap expansion drive wells prior to gas  g p p p g cap break through  Use to evaluate water drive wells before exceeding water  influx rate influx rate Vogel Method  Use to evaluate dissolved gas drive wells below the bubble  U t l t di l d di ll b l th b bbl point  Use to evaluate water drive wells after water influx rate is  exceeded

CHAPTER 2: RESERVOIR SYSTEM

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MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Gas Well Performance



Gas well productivity determined with deliverability testing  Two basic relationship in used: Two basic relationship in used: • Rawlins & Schellhardt (empirical backpressure method) • Houpeurt (theoretical ) (theoretical )



Deliverability test method: • Flow‐after‐flow test (four‐point test) • Isochronal test • Modified isochronal test

CHAPTER 2: RESERVOIR SYSTEM

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MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Gas Well Performance (ctd) ( )



Rawlins & Schellhardt’s deliverability eq: qg = C (Ps2 – Pwf2)n where: h

C = flow coefficient C fl ffi i t n = deliverability exponent non‐Darcy: n = 0.5 – 1.0 Darcy flow: n = 1.0

 A plot of q A l t f g vs (Ps2 – Pwf2) on log‐log paper results in a straight  ) l l lt i t i ht line with slope 1/n.

 This equation often referred to as the backpressure equation Thi ti ft f dt th b k ti  IPR can be plotted by: qg/qg max = [ 1 – (Pwf/Ps)2]n CHAPTER 2: RESERVOIR SYSTEM

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MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Gas Well Performance (ctd) ( )



Houpeurt relationship: Deliverability eq:  Ps2 – Pwf2 = aqg + bqg2 where; a = laminar flow coefficient b = turbulence coefficient From plot of (P p ( s2 – Pwf2 )) / q / qg vs qg ((straight line) g ) a = intercept with y axis b = slope

CHAPTER 2: RESERVOIR SYSTEM

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MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Gas Well Performance (ctd) ( )



OR: a 1422zT[ln (re/rw) – a = 1422 ) ¾ + s] / kh ¾ + s] / kh b = (1422zT/kh)D D = non‐Darcy flow coefficient, D/Mscf D = non Darcy flow coefficient D/Mscf D = [2.715x10‐15  kMPsc] / hgrwTsc

 = turbulence factor, ft‐1:       = 1.88x1010k‐1.47‐0.53 k      



Psc Tsc M M     



s        h CHAPTER 2: RESERVOIR SYSTEM

=   md =  cp cp =   standard pressure, psia =   standard temperature, oR =   molecular weight, lbm/lbm‐mole l l i ht lb /lb l =   porosity, fraction =   skin factor =   formation thickness, ft (96)

MOHD FAUZI HAMID