Inflow Performance Reservoir Ipr

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Inflow Performance

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Instructional Objectives

• Calculate the IPR for oil wells • Calculate the IPR for gas wells

2 WV 12/12/2018

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Reservoir Capabilities Single phase liquid flow

• Darcy’s Law – Liquid flow in Laminar Flow through a permeable medium is described by Darcy’s Law

7.08 X 10 k h  pr  pwf  3

q

3 WV 12/12/2018

  o Bo  ln 

  re     0.75  s  a ' q   r  w 

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Required Data • • • • • • • • • 4 WV 12/12/2018

Permeability (k) Thickness of producing zone (h) Average reservoir pressure (P) Average viscosity (u) - PVT Average oil formation volume factor (Bo)-PVT Radius of drainage (re) Radius if the drilled hole (rw) Total skin (S) Turbulent flow (aq)

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Reservoir Capabilities • s = Skin Factor (dimensionless)

 k   ra  s    1 ln    ka   rw 

16 WV 12/12/2018

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Productivity Index • A common indicator of liquid reservoir behavior is PI or productivity index – Referred to as “J” in SPE nomenclature

q STB/ D / psi J p  p wf 17 WV 12/12/2018

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Productivity Index in Terms of Darcy’s Law 3

7.08 X 10 k h J   re  o Bo  ln    0.75    rw  18 WV 12/12/2018

 s  

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Calculating Flowrate • Using PI, we can calculate flowrate, q, quickly and easily from

q  J (p  p wf ) 19 WV 12/12/2018

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Exercise 1 • Given reservoir parameters: k = 30 md h = 40 ft o = 0.5 cp Bo = 1.2 RB/STB hole size = 8 ½ inches s =0

20 WV 12/12/2018

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Exercise 1 • Calculate: – J for re = 1,000 ft – q for a drawdown ( p  p wf ) of 750 psi – q for a drawdown of 1,000 psi – With p = 3,000 psia, calculate q for a complete drawdown (absolute open flow potential).

21 WV 12/12/2018

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Two-phase flow in the reservoir • Bubblepoint pressure (pb) – Pressure at which first bubble of gas is released from reservoir oils

22 WV 12/12/2018

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Multiphase Flow • Vogel’s Behavior – IPR Curve - Vogel plotted the data using the following dimensionless variables

p wf p 23 WV 12/12/2018

and

q qmax

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Multiphase Flow • Mathematical model for Vogel’s curve

 q    pwf    1  0.2   qmax    p

24 WV 12/12/2018

  pwf   0.8    p

2

    

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Vogel Curve

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1

0.8

pwf/pr

0.6

0.4

0.2

0 0 25 WV 12/12/2018

0.2

0.4

0.6

q/qmax

0.8

1

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Exercise 2 Given Data: • Pr = 2400 psi • qo=100 b/d

• Pwf=1800 psi Calculate:

• qo max • Construct IPR curve 26 WV 12/12/2018

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Combination single phase liquid and two phase flow q STB / D / psi  J p  pwf +

 q    pwf    1  0.2   qmax    p 27 WV 12/12/2018

  pwf   0.8    p

2

    

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Multiphase Flow • Combination Darcy/Vogel p

Pressure

pb

pwf

J pb

qb O 28 WV 12/12/2018

qmax

1.8 O

Rate

q

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Multiphase Flow • Mathematical relationship between Vogel (qmax) and Darcy (AOF)

qmax

29 WV 12/12/2018

J  Pb  qb  1.8

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Multiphase Flow • How to find qmax: for q  qb , Darcy's law applies : q  Jp  p wf    2     p p for q  qb then : q  qb  qmax  qb  1  0.2 wf  0.8  wf   pb   pb    

J pb qmax  qb  1.8 30 WV 12/12/2018

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Exercise 3 • Pr =3000 psia • Pb = 2000 psia (bubble point) • K = 30 md

• h = 60 ft • Bo = 1.2 • uo = 0.68 cp

• re = 2000 ft • rw = 0.4 ft • S=0

31 WV 12/12/2018

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Exercise 3 Calculate: • q(bubble point)

• qo max if it follows Vogel´s relationship below Pb • qo for flowing pressure of (a) 2500 psia

(b) 1000 psia

32 WV 12/12/2018

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Procedure for flow efficiency not Equal to 1.0

Pr  P´wf FE  Pr  Pwf Where: P´wf = Pwf = Pr = 33 WV 12/12/2018

The equivalent undamaged flowing pressure actual flowing pressure static reservoir pressure

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Pwf´

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Exercise 4 Given Data • Pr=2600 psi
• (2) qomax for FE =0.6 • (3) Find qo for Pwf=1300 psia for FE=0.6, 1.0 and 1.3 36 WV 12/12/2018

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Three or Four points tests • Fetcovich proposed that flow after flow or isochronal test as used on gas wells could also be used on oil wells

q  J ' o(Pr  Pwf ) 2

2 n

q  C (Pr  Pwf ) 2

2 n

• These equations are straight lines on log log with J’o and C representing the intercept on the q axis (where Pr2-Pwf2=1 and n = 1/slope)

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Exercise 5 • Given Data: four point oil well test: Pr = 2500 psia

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Pb=3000 psia

Test

qo

Pwf

1

880

2000

2

1320

1500

3

1595

1000

4

1752

500

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Exercise 5 Calculate: (1) value of exponent n (2) value of J’o

(3) Absolute open flow potential (AOFP) or qmax (4) qo for Pwf=2200 psia

39 WV 12/12/2018

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Three or Four points tests • Jones, Blount, and Glaze suggest that radial flow for both oil and gas could be represented to show wether near wellbore restriction exist

7.08 X 10 k h  p  pwf  3

q

40 WV 12/12/2018

  o Bo  ln 

  re     0.75  s  a ' q   r  w 

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  (ln re / rw  3 / 4  S   9.08 10 13 Bo   2  q q  Pr  Pwf   3 2   4  h r   p w   7.08 10  kh 

b

a

Pr  Pwf   (ln re / rw  3 / 4  S   9.08 10 13 Bo     q 3 2   q 4h p rw     7.08 10  kh

41 WV 12/12/2018

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 9.08 10 13 Bo    a 2   4  h r p w     (ln re / rw  3 / 4  S   b 3   7.08 10  kh

Pr  Pwf  aq  b q

b'  b  aq max 42 WV 12/12/2018

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43 WV 12/12/2018

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Conclusions based on the plot (1) if b is low -less than 0.05- no formation damage occurs in the well. The degree of damage will increase with increasing values of b (2) If the value of b’/b is low -less than 2-litlle or not turbulence is occurring in the well formation system (3) If the value of b and b’/b are low, the well has a good completion (4) If the value of b is low and b’/b is high, stimulation is not recommended. The low productivity is caused by insufficient open perforated area. Additional perforations would be recommended (5) if the value of b is high and b’/b is low, stimulation is recommended 44 WV 12/12/2018

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Exercise 6 • Given Data: well test: • Reservoir pressure= 4453 psi Test qo(b/d) Pwf (psi) 1 545 4427 2 672 4418 3 746 4412 4 822 4405 Calculate: (1) Plot (Pr-Pwf)/qo (2) Recommend ways to improve the productivity of the well 45 WV 12/12/2018

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Future IPR • Future production rate • determine when a well is to be placed on artificial lift

• Rate acceleration projects and comparing artificial lift methods

46 WV 12/12/2018

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47 WV 12/12/2018

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Future IPR • Fetcovich procedure



 Pr2  2  Pr2  Pwf 2 qo  J ' o1  Pr1 



n

From a three or four point flow test it is posible

to predict IPR curves at other static reservoir pressures 48 WV 12/12/2018

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Exercise 7 • Given Data (from exercise 5) • The equation describing this test was:  (2500) 2  Pwf 2   qo  3.906 1000  

0.70192

Calculate:

• (1) qo max when Pr lowers to 1800 psia • (2) qo for Pwf=800 psia when Pr=1800 psia 49 WV 12/12/2018

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Gas Reservoirs • Pseudosteady State – The behavior of gas flowing in laminar flow through a porous medium (Darcy’s Law)





7.03X10 4 k h p 2  p2wf q   re     0.75  s  g T z  ln    rw   50 WV 12/12/2018

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Calculation of IPR Curve 4000

pwfs, psia

3000

2000

1000

0 0

2000

4000

6000

q, Mscf/day 51 WV 12/12/2018

8000

10000

12000

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Using the Real Gas Pseudopressure (pp (p))





7.03 x 10 4 kh pp p   pp p wf  q   re   T  ln    0.75  s    rw  

and p

p pp p   2  dp p z o

52 WV 12/12/2018

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Exercise 8 Plot the IPR Curve

Given the following data and using the pressure squared relationship: k

53 WV 12/12/2018

= 100 md

ug = 0.02 cp

h = 20 ft

T = 610R

re = 1,500 ft

Z = 0.9

rw = 0.33 ft

P = 4,000 psig

s

= 0

g = 0.65

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IPR in Gas Reservoirs • Jones’ Gas IPR – Problem • Darcy’s law valid for laminar flow only

• High permeability gas wells turbulent flow near the wellbore

2

2 2 p  pwf  aq  bq 54 WV 12/12/2018

produce

in

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Definitions    re  3 1.424x10 g z T  ln    0.75     rw  a   kh      3.16 x1012   g T z b   hp2 rw  55 WV 12/12/2018

 s  

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Exercise 9 • Given data: four point gas well test: • Pr=4750 psia Pwf (psia) Gas flow rate MMScf/d 4213 9.45 3806 12.37 3243 15.21 2763 16.98 Recommended a way to improve the productivity of this well 56 WV 12/12/2018

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Transient IPR Curves • Assumptions – Darcy’s and Jones’ laws assume that the average pressure P is constant – Drainage radius, re, is constant and that

• These assumptions are true in pseudosteady state only, i.e. when all of the outer boundaries of the reservoir are reached.

57 WV 12/12/2018

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Transient IPR Curves  The time to reach pseudo-steady state (pss), tstab, can be calculated with the following equation

948   c t re t stab  k 58 WV 12/12/2018

2

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Exercise 10 Oil Well Stabilization Time

• Find tstab with the following data – – – –

f = 0.1 o = 0.5 cp ct = 2 X 10-5 psi-1 re = 1,500 ft

For the following values of k: 0.1 md, 1.0 md, 10 md, and 100 md 59 WV 12/12/2018

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Exercise 11 Gas Well Stabilization Time

• Find tstab with the following data – – – –

f = 0.1 g = 0.02 cp ct = 2 X 10-4 psi-1 re = 1,500 ft

For the following values of k: 0.1 md, 1.0 md, 10 md, and 100 md 60 WV 12/12/2018

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Transient Flow pwf

p

tstab > t3 > t2 > t1 t1 t2 tsta b

t3

q 61 WV 12/12/2018

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Transient IPR Curves • Transient Flow Equation – for oil wells

k h p  p wf  qo      k t    3.23  0.87s  162.6o Bo log     c r2    o t w   

– for gas wells (low pressure only)



62 WV 12/12/2018



k h p 2  p2wf qg      k t   3.23  0.87s  1638g T z  log      c r2  g t w   

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Summary • Purpose---> to show the various procedures used in the construction of IPR curves for oil and gas • if reservoirs models are readily available, they may be used in place of a less rigorous procedure • IPR curves for gas condensate reser voirs and many wells producing from secondary and tertiary recovery projects are good examples where more sophisticated techniques are needed 63 WV 12/12/2018

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References

64 WV 12/12/2018