Chapter 2-reservoir System

22/02/18

SKPP 3513 – PRODUCTION ENGINEERING

Chapter 2: Reservoir System Mohd Fauzi Hamid N01a-18 07-5535616 [email protected]

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Learning Outcome

Students will able to:

 Understand the concept of IPR  Differentiate the method used for calculating single phase flow and two phase flow IPR

 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)

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.  Pressure differential (DP = Pf – Pw)  Well geometry, spacing, production area @ Pf

CHAPTER 1: INTRODUCTION

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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.

PI  J  where: PI q Ps Pwf

= = = =

q Ps  Pwf

……………………… (2-1)

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

CHAPTER 2: RESERVOIR SYSTEM

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

PI  J 

q 0.007082kh .………………. (2-2)  re Ps  Pwf Bo m ln( ) rw

 PI will be a constant if m, 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.

CHAPTER 2: RESERVOIR SYSTEM

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

q Productivity Index CHAPTER 2: RESERVOIR SYSTEM

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Productivity Index (ctd)

 For pseudosteady-state flow (from Eqn 1-4) : PI  J 

q 0.007082kh  Ps  Pwf B m (ln re  1 ) o rw 2

………… (2-3)

 In terms of average reservoir pressure, Pavg or Pr: PI  J 

CHAPTER 2: RESERVOIR SYSTEM

q 0.007082kh  Pr  Pwf B m (ln re  3 ) o rw 4

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………… (2-4)

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Productivity Index (ctd)

 Again, provided the fluid and rock properties (m, B and k) are constant, the PI should be constant, irrespective of the degree of depletion. Thus, as for the steady state case, a straight line relationship exists between P and q.

 For gas well: PI  J 

qg ( Ps 2  Pwf 2 )

CHAPTER 2: RESERVOIR SYSTEM



0.703k g h r T ( m g z ) avg ln( e ) rw

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…… (2-5)

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Productivity Index (ctd)

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

k g h( Pe 2  Pw2 ) r 1422T ( m g z ) avg ln( e ) 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) Pe2 Slope = - 1/PI

P2 Absolute Open Flow Potential (AOFP)

qo

qg

Gas well productivity relationship CHAPTER 2: RESERVOIR SYSTEM

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Productivity Index (ctd)

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

PI s  J s 

PI q  ………… (2-6) h h( Ps  Pwf )

Example 2.1 A field is drilled up on a 80-acre spacing. The reservoir pressure (Ps) is 1000 psi, k = 50 md, h = 20 ft, m = 3 cp, Bo = 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 specific PI. CHAPTER 2: RESERVOIR SYSTEM

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Productivity Index (ctd)

 If there is gas in fluid flow, the fluid is no longer incompressible, PI not constant and this situation happen when Ps < Pb. PI becomes:

PI 

dq dPwf

………… (2-7)

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

CHAPTER 2: RESERVOIR SYSTEM

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Productivity Index (ctd)

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

PI 

(qo  qw ) 0.007082h ko k  (  w ) …… (2-9) r Ps  Pwf mo Bo m w Bw ln( e ) rw

 For two production condition: ( PI )1 ( ko / mo Bo )1  ( PI ) 2 ( ko / mo Bo ) 2

……………….... (2-10)

 To estimate the PI, Lewis & Horner equation can be used:

PI  5.9*10 4 ( CHAPTER 2: RESERVOIR SYSTEM

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kh ) mo Bo

……………….... (2-11) MOHD FAUZI HAMID

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Productivity Index (ctd)

 Factors influencing PI.    

Production GOR Pressure depletion Cumulative Production Oil Saturation (So will affect the effective and relative permeability of 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.

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

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Productivity Index (ctd)

 Future PI:

kro ) mo Bo f PI f  PI p k ( ro ) p mo Bo (

………… (2-11)

where:

PI p 

CHAPTER 2: RESERVOIR SYSTEM

1.8PI P 1  0.8( wf ) Pavg

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Hydrocarbon Phase Behavior  HC reservoir fluid : complex mixture of HC molecules.  Composition depend on:  Source rock  Maturation degree  P&T  Phase change from HPHT  cool & low pressure  effect to performance  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)  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)

1. Single phase in reservoir (a) 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) 4. Evolves increasing amount of gas (d-e) until separator (f) CHAPTER 2: RESERVOIR SYSTEM

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

CHAPTER 2: RESERVOIR SYSTEM

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Inflow Performance Relationship (IPR)  The IPR is defines as the functional relationship between the 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).

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

CHAPTER 2: RESERVOIR SYSTEM

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IPR (ctd)

Slope = - 1/PI

Typical IPR Curve for Incompressible Fluid CHAPTER 2: RESERVOIR SYSTEM

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IPR (ctd)  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 tan   max  PI Ps CHAPTER 2: RESERVOIR SYSTEM

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q

qmax

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Example 2-2 Given:

Permeability, k Pay thickness, h Average reservoir pressure, Pr Reservoir temperature, T Well spacing, A Drilled hole size, D Formation volume factor, B Oil viscosity, m Skin factor, St

= = = = = = = = =

30 mD 40 ft 3000 psi 200oF 160 acres 12 ¼ in (open hole) 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)  For most well, IPR curve is not a linear relationship, because 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.

CHAPTER 2: RESERVOIR SYSTEM

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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 (Pr2 – Pwf2) C = constant  For higher flow rate due non-Darcy (turbulent flow) effect  field observation  Bureau of Mines eq: q = C (Pr2 – Pwf2)n n = 0.5 – 1.0  Log-log plot of q vs (Pr2 – Pwf2)  straight line with : Slope = n Intercept = C  Standard test for gas well: measure Pwf at 4 production rate. CHAPTER 2: RESERVOIR SYSTEM

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IPR – 2 Phase (Gas-Liquid) Flow  Saturated reservoir  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 @ Pwf = 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 .

CHAPTER 2: RESERVOIR SYSTEM

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

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

  Pwf   0.8      Ps 

 Based on the equation, Vogel plot for solution gas drive reservoir.

2

 Pwf  qo vs   (qo ) max  Ps 

 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 CHAPTER 2: RESERVOIR SYSTEM

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

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

 From the equation; if and if

qo (qo ) max

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

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Example 2-3  Given qo Pwf Ps

= = =

300 BPD 1500 psi 1800 psi

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

CHAPTER 2: RESERVOIR SYSTEM

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Example 2-4  Given qo Pwf Ps

= = =

300 BPD 1500 psi 1800 psi

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

CHAPTER 2: RESERVOIR SYSTEM

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Construction of IPR 





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

CHAPTER 2: RESERVOIR SYSTEM

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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 line

CHAPTER 2: RESERVOIR SYSTEM

Slope = - 1/PI

Pwf

 Pwf = Ps  q = 0  Pwf = 0  q = qmax 

Ps

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q

qmax

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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

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(1) Ps < Pb 

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

qo Ps PI * Ps  1.8( Ps  Pwf ) 1.8

Derivation-1

q

CHAPTER 2: RESERVOIR SYSTEM

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

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Derivation - 1  Pwf   Pwf qo  1  0.2    0.8  (qo ) max  Ps   Ps Pwf   Pwf   1  1  0.8  Ps  Ps  

(qo ) max

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

  

2

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

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

     

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

qo Ps PI * Ps  1.8( Ps  Pwf ) 1.8 Back

CHAPTER 2: RESERVOIR SYSTEM

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(1) Ps < Pb (ctd) 

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



Plot graph of Pwf vs qo

Ps

 Pwf = Ps  qo = 0

Pwf

 Pwf = 0  qo = (qo)max 

The plot will produce a curve q

qmax

2   Pwf   Pwf   qo  (qo ) max 1  0.2    0.8      Ps   Ps  

CHAPTER 2: RESERVOIR SYSTEM

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(2) Ps > Pb 

IPR curve will have two parts:

 Ps > Pwf > Pb  a straight line

4000

 Pwf < Pb  a curve

3500

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.



Two situations:

Pwf(psi)

3000



500 0

 Given Pwf > Pb

0

50

100

150

200

q(BPD)

 Given Pwf < Pb CHAPTER 2: RESERVOIR SYSTEM

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(2) Ps > Pb (ctd)

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

CHAPTER 2: RESERVOIR SYSTEM

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 PI * P b   1.8 

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(2-1) Pwf > Pb 

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



Determine qb from above plot or from equation qb = PI(Ps – Pb) 



  PI * Pb    qb     qb  1.8  Ps  Pwf    1.8   

Determine qmax from: qmax  

qPb



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



Plot Pwf vs q.

PI * Pb q  qb  1.8

CHAPTER 2: RESERVOIR SYSTEM

2   Pwf   Pwf   1  0.2    0.8     Pb   Pb   

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P (q  q b )  1  0.2  wf (q max  q b )  Pb q  q b  (q max

  Pwf    0.8     Pb 

2

2   Pwf   Pwf    q b ) 1  0.2    0.8     Pb   Pb   

  PI * Pb qPb but (q max  q b )  q v     1.8 P  P 1.8   s wf       Pwf qPb q  q b    1  0.2   Pb  1.8  Ps  Pwf    PI * Pb q  qb  1.8 CHAPTER 2: RESERVOIR SYSTEM

  Pwf 1  0.2   Pb  (40)

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

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

Back MOHD FAUZI HAMID

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(2-2) Pwf < Pb 

Determine qb from equation

   Pwf Pb   qb  q 1    1  0.2   1.8( Ps  Pwf )    Pb 

qmax

2   Pwf     0.8      Pb     qPb q   1.8  Ps  Pwf   b  



Determine qmax from:



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



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

CHAPTER 2: RESERVOIR SYSTEM

PI * Pb 1.8

  Pwf 1  0.2   Pb 

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

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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

Typical Composite IPR Curve CHAPTER 2: RESERVOIR SYSTEM

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Example 2-5 

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 the well in for 24 hours, the bottomhole pressure reached a static value of 1125 psi. Calculate: 1. Productivity index, PI 2. Absolute open hole flowing potential 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

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Example 2-5: Solution

1200

1125 100 800 585 400 350 200 0

0 72.2 22.9 38.0 51.0 54.6 65.1 79.2

Example 2-5

1000

Pwf(psi)

Pwf (psi) q(BPD)

800

600

400

200

0 0

CHAPTER 2: RESERVOIR SYSTEM

10

20

30

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40

50

q (STB/day)

60

70

80

90

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Example 2-6 

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 3220 psi. The estimated bubble point pressure of 4020 psi 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

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MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Example 2-6: Solution 

Vogel Equation: 2

 Pwf   Pwf  qo  1  0.2    0.8   (qo )max  Ps   Ps  qo (qo ) max  2 P  P  1  0.2  wf   0.8  wf   Ps   Ps  200 (qo )max  2  3220   3220  1  0.2   0.8     4000   4000  



624STB / day

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

CHAPTER 2: RESERVOIR SYSTEM

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Example 2-6: Solution (ctd)

(Pwf/Ps) 1 0.9375 0.8750 0.8125 0.7500 0.6875 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 0.6602 0.5625 0.4727 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

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MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

Example 2-7 Given: Static reservoir pressure, psi Bubble point pressure, psi Bottomhole flowing pressure, psi Production rate, BPD 1. 2. 3. 4.

= = = =

3620 1825 1980 108

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

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Example 2-7: Solution 

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  Pwf ) 1.8(3620  1980)

2   Pwf   Pwf   qo  (qmax  qb ) 1  0.2   0.8      qb   Pb   Pb   2   Pwf   Pwf   qo  (185.3  118.5) 1  0.2    0.8     118.5 (66.8)   1825   1825  

CHAPTER 2: RESERVOIR SYSTEM

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OR 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 2   Pwf   Pwf   1  0.2    0.8     Pb   Pb    2 (0.066) * (1825)   Pwf   Pwf     q o  118.5   0.8  1  0.2     1.8  1825   1825      2   P P       q o  118.5  66.9 1  0.2  wf   0.8  wf   1825 1825        

3) IPR: q  q b 

CHAPTER 2: RESERVOIR SYSTEM

PI * Pb 1.8

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

1 0.9041 0.8219 0.6849 0.5479 0.411 0.274 0.137 0

1.0000 0.8174 0.6755 0.4691 0.3002 0.1689 0.0751 0.0188 0.0000

0 119 130 138 151 162 171 178 182 185

Example 2-7

3500 3000

Pwf(psi)

3620 1825 1650 1500 1250 1000 750 500 250 0

4000

2500 2000 1500 1000 500 0

2 20 40   Pwf   Pwf   0 qo  (qmax  qb ) 1  0.2    0.8     qb   Pb   Pb   2   Pwf   Pwf   qo  (185.3  118.5) 1  0.2    0.8     118.5  1825   1825   (66.8) 

CHAPTER 2: RESERVOIR SYSTEM

60

80

100

120

140

160

180

200

q(BPD)

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Example 2-8 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

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Example 2-8: Solution 

Ps > Pb > Pwf: Determine qb: 2    Pwf   Pwf   Pb qb  q 1  1  0.2    0.8     1.8( Ps  Pwf )   Pb   Pb   

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



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 2 points to plot IPR  P  P   wf wf qo  (qmax  qb ) 1  0.2    0.8     qb P P   b   b  

CHAPTER 2: RESERVOIR SYSTEM

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OR 1) PI:

PI  (

q 600 )  0.706 BPD / psi Ps  Pwf 1750  900   Pwf 1  0.2   Pb 

2   Pwf   2) qb : q b   0.8      Pb   2 0.706 *1200     900   900   q b  600   0.8 1  0.2       411.73BPD 1.8  1200   1200     

PI * Pb  q 1.8

3) IPR:

q o  411.73 

2 (0.706) * (1200)   Pwf   Pwf     1  0.2  0.8   1200   1200   1.8        

2   P   P     q o  412  471 1  0.2  wf   0.8  wf   1200 1200        

CHAPTER 2: RESERVOIR SYSTEM

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Example 2-8: Solution (ctd) 2000

Example 2-8

1800 1600

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

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

1000 800

Pwf(psi)

1200

1200 1000 800 600 400 200 0

600 400 200 0 0

100

200

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

2   Pwf   Pwf   qo  (883  412) 1  0.2    0.8     412 (471)   1200   1200  

CHAPTER 2: RESERVOIR SYSTEM

300

400

500

600

700

800

900

1000

q(BPD)

(ii) Well Potential = qmax = 883 BPD (iii) Pwf @ q = 800 BPD: from IPR, Pwf = (55)

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 deliverability.



Fetkovich applied these tests on oil well and found that oil well producing below the bubble point pressure behave just 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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Fetkovich Method (ctd)  A plot of (Ps2 – Pwf2) vs q on log-log paper results in a straight line with slope 1/n. [q vs (Ps2 – Pwf2) with slope n]

 This equation often referred to as the backpressure equation. 

The equation was derived from Evinger and Muskat for two phase radial flow:

q

(7.08kh) f ( P )dP re ln( ) rw

f ( P) 

kro mo B0

CHAPTER 2: RESERVOIR SYSTEM

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Example 2-8-1 Given: Static reservoir pressure, psi

=

2000

Bottomhole flowing pressure, psi =

1400

Production rate, BPD

1020

=

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

CHAPTER 2: RESERVOIR SYSTEM

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

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

PI ' 

2 s

2 q  5 x 104 bbl /( day )( psi) 2  2000 psi   Pw2f   

 5 x 104  4 x 106   Pwf2  BPD  0.7141  4 x 106   Pw2f 

CHAPTER 2: RESERVOIR SYSTEM

0.5

BPD

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Example 2-8-1: Solution (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 1500 1142.56 1236.86 1308.97 1000 1362.42 1399.34 1421.04 500 1428.20

0 0

CHAPTER 2: RESERVOIR SYSTEM

500

(60)

1000

q(BPD)

1500

2000

2500

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Prediction of Future IPR 

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



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



……………. (1)

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

CHAPTER 2: RESERVOIR SYSTEM

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

Equation (1) & (2):



1.8PI P (1  0.8 wf ) Ps Vogel equation & (2): produce an equation for future IPR PI * 

 PI *f *( Ps ) f qf    1.8 

CHAPTER 2: RESERVOIR SYSTEM

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

(62)

  Pwf   0.8    ( Ps ) f

  

2

   

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Prediction of Future IPR (ctd)  where:

 ( kro / mo Bo ) f  PI *f  PI *p    ( kro / mo Bo ) p   s s  kro   o or  1  sor  swi 

CHAPTER 2: RESERVOIR SYSTEM

n

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Prediction of Future IPR (ctd)  procedure: i. PI: PI  ii. PI*p :

iii. PI*f :

PI *p 

 Pwf  (q )   q   o max  1  0.8  Ps  Pwf  Ps    Ps 1.8* PI Pwf   1  0.8  Ps  

 (kro / mo Bo ) f  PI  PI    (kro / mo Bo ) p 

iv. IPR :

CHAPTER 2: RESERVOIR SYSTEM

* f

   

* p

 s  sor  (kro )   o  1  sor  swi 

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

(64)

  Pwf   0.8    ( Ps ) f

  

2

n

   

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Example 2-9 

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

CHAPTER 2: RESERVOIR SYSTEM

2017

2027

3000 2500 3500 5000 60 20 10

2000 45 15 30

1.3 0.8 0.8 30 0.6

1.01 0.8 0.8 30 0.6

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SKPP 3513: PRODUCTION ENGINEERING

Example 2-9 (ctd) i.

Construct IPR Curve for the year of 2017 and 2027.

ii. Determine well potential for the year of 2017 and 2027. iii. Determine the bottomhole flowing pressure if the well should produced 14000 BPD in 2017. iv. Determine production rate in 2027 if the bottomhole flowing pressure is 750 psig.

CHAPTER 2: RESERVOIR SYSTEM

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Example 2-9: Solution 

For 2017: Ps
(qo ) max 



qo P 1  0.2  wf  Ps

  Pwf    0.8     Ps 

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 2017 IPR. 2   Pwf   Pwf   qo  (qo ) max 1  0.2    0.8      Ps   Ps   2   Pwf   Pwf   qo  18000 1  0.2    0.8      3000   3000  

CHAPTER 2: RESERVOIR SYSTEM

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MOHD FAUZI HAMID

SKPP 3513: PRODUCTION ENGINEERING

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

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

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

2017 IPR 3500 3000 2500 2000 1500

2017 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

CHAPTER 2: RESERVOIR SYSTEM

6000

8000

10000

12000

14000

16000

18000

20000

q(BPD)

2   Pwf   Pwf   qo  (qo ) max 1  0.2    0.8      Ps   Ps  

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Example 2-9: Solution (ctd) 

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

where

  18000   2500     1  0.8  3000    10     3000 

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

 (kro / mo Bo ) f  PI *f  PI *p    (kro / mo Bo ) p  n 4  s  sor   0.45  0.15  (kro ) f   o      0.0885 1  sor  swi  f 1  0.15  0.3  n

4

 s  sor   0.60  0.20  (kro ) p   o     0.1066 1  sor  swi  p 1  0.20  0.1

CHAPTER 2: RESERVOIR SYSTEM

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MOHD FAUZI HAMID

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Example 2-9: Solution (ctd)  (kro / mo Bo ) f   (0.0885 /(0.8 x1.01))  PI *f  PI *p    10.8    11.53  (kro / mo Bo ) p   (0.1066 /(0.8 x1.3)) 

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

  Pwf   0.8    ( Ps ) f

  

2

   

 Pwf   Pwf  11.53 * 2000  1  0.2    0.8   1.8   2000   2000 

2

  

2   Pwf   Pwf   q f  12,811 1  0.2    0.8      2000   2000  

q f  12,8111  1*104 (Pwf )  2 *107 (Pwf )2 

CHAPTER 2: RESERVOIR SYSTEM

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Example 2-9: Solution (ctd) 

Calculate several future rates at specific drawdown to have enough points to plot 2027 IPR. Future IPR Curve 3500

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

1 0.875 0.75 0.625 0.5 0.375 0.25 0.125 0

1 0.7656 0.5625 0.3906 0.2500 0.1406 0.0625 0.0156 0.0000

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

qf(BPD) 0 2722 5124 7206 8968 10409 11530 12331 12811

Pwf(psi)

Example 2-9

)2

3000

2500

2017 IPR

2000

2027 IPR 1500

1000

  Pwf   0.8    ( Ps ) f

  

500

2

  0 

0

2000

4000

6000

q f  12,8111  1*104 (Pwf )  2 *107 (Pwf )2  CHAPTER 2: RESERVOIR SYSTEM

8000

10000

12000

14000

16000

18000

20000

q(BPD) (71)

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Predicting Future IPR by Fetkovich Method 

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

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



Fetkovich: kro is linear towards the pressure:

kro @ Ps Ps  kroi Psi 

From backpressure equation:

q  PI '( Ps2  Pwf2 )n …………. (1)

CHAPTER 2: RESERVOIR SYSTEM

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Predicting Future IPR by Fetkovich Method (ctd) 

Based on these three equations:

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

………… (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 q. CHAPTER 2: RESERVOIR SYSTEM

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Example 2-10 Given: Static reservoir pressure Bubble point pressure Flowing bottom hole pressure Production rate

= = = =

3000 psi 3000 psi 2000 psi 500 BPD

If n=1.0, construct IPR when static reservoir pressure decline to 2000 psi.

CHAPTER 2: RESERVOIR SYSTEM

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Predicting Future IPR by Simple Method 

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 previous methods.

CHAPTER 2: RESERVOIR SYSTEM

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Determination of PI or IPR at Field 

Field determination is depend on the conditions of the well.

a) Simple Case.  Data required are Pwf, q and Ps.  Shut-in well for 24 – 72 hours  PBU  With pressure gauge at bottom, flow the well at lowest q and 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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Determination of PI or IPR at Field (ctd) b) Special Case  For well without packer.  During production, some of the gas will come out and accumulated at the top of the annulus.  At equilibrium, Pwf = Pc + pressure due to gas column  Gilbert: Pressure due to gas column = PcD1.5/100 psia Pwf = Pc(1+D1.5/100 ) psia where Pc = D = CHAPTER 2: RESERVOIR SYSTEM

casing head pressure, psia tubing depth, per 1000 ft (77)

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Example 2-11 Given: Tubing depth q

= 3000 ft = 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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Example 2-11: Solution  D1.5   31.5  Pwf  Pc 1   565  1    595 psia  580 psig  100   100  

Pwf = 580 psig @ q = 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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Pwf (psig)

Example 2-11: Solution (ctd)

1000



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

580 338

0

CHAPTER 2: RESERVOIR SYSTEM

From the plot:

42

66

100 q (BPD)

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IPR – Method Selection



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



Below Pb: • Gas break out of solution & impedes the oil flow • Vogel method

CHAPTER 2: RESERVOIR SYSTEM

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IPR – Method Selection (ctd)



Water Drive Reservoir  Has: • Constant producing GOR • Pr 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 .

CHAPTER 2: RESERVOIR SYSTEM

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IPR – Method Selection (ctd)



Gas Cap Drive Reservoir  Wells producing from the oil column below the GOC • Produce with a constant GOR • 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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IPR – Method Selection Summary





PI Method  Use to evaluate dissolved gas-drive wells above the bubble point  Use to evaluate gas cap expansion drive wells prior to gas cap break through  Use to evaluate water drive wells before exceeding water influx rate Vogel Method

 Use to evaluate dissolved gas drive wells below the bubble point

 Use to evaluate water drive wells after water influx rate is exceeded

CHAPTER 2: RESERVOIR SYSTEM

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Gas Well Performance



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



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

CHAPTER 2: RESERVOIR SYSTEM

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Gas Well Performance (ctd)



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

C = flow coefficient n = deliverability exponent non-Darcy: n = 0.5 – 1.0 Darcy flow: n = 1.0

 A plot of qg vs (Ps2 – Pwf2) on log-log paper results in a straight line with slope 1/n.

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

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Gas Well Performance (ctd)



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

CHAPTER 2: RESERVOIR SYSTEM

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Gas Well Performance (ctd)



OR: a = 1422mzT[ln (re/rw) – ¾ + s] / kh b = (1422mzT/kh)D D = non-Darcy flow coefficient, D/Mscf D = [2.715x10-15 b kMPsc] / hmgrwTsc

b = turbulence factor, ft-1: k

m Psc Tsc M

f

s h CHAPTER 2: RESERVOIR SYSTEM

= = = = = = = =

b = 1.88x1010k-1.47f-0.53

md cp standard pressure, psia standard temperature, oR molecular weight, lbm/lbm-mole porosity, fraction skin factor formation thickness, ft (88)

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