Ipr - Leslie Thompson

Inflow Performance Relationship IPR by Dr. Leslie Thompson

Production System

Fluid Production • Path of produced fluids – Reservoir – Perforations, gravel pack, etc. – Downhole equipment, casing, tubing. • Mixing with gas-lift gas

– Wellhead, production chokes – Flow lines, manifolds – Separator

Fluid Production • In each flow segment, the fluids interact with the production components – Pressure, temperature and flow velocity are altered. – Fluid properties constantly changing.

Pressure Losses

Driving Force • The driving force that moves fluids along the reservoir and production system is the energy stored in the form of compressed fluids in the reservoir. – As the fluids move along the system components, pressure drop occurs. The pressure in the direction of flow continuously decreases from the reservoir pressure to the final downstream pressure value at the separator.

IPR • The flow of oil, gas and water from the reservoir is characterized by the Inflow Performance relationship. – IPR is a measure of “Pressure losses” in the formation.

• The functional relationship between flow rate and bottomhole pressure is called IPR. – Indicator of well performance. q  f  p  pwf   f  p 

Productivity Index • Measure of the well’s capacity to produce fluids from reservoir to wellbore. • Definition: Fluid production rate for 1 psi pressure drop from reservoir to wellbore.

Productivity Index q t  PI   p t   pwf  t   PI

Productivity Index, STB/d/psi

p t 

Average pressure in the well’s drainage area, psi.

p wf  t  q(t)

Bottomhole flowing pressure, psi Production rate, STB/d

Parameters Controlling IPR or PI • Rock Properties – Permeability, Relative permeability

• Fluid Properties – Bo, o, Bw, w, co, cw – Bg, g, z, cg

• Flow Regime in the Reservoir – Darcy Flow – Non-Darcy (inertial, Forchheimer) flow

• Phases Flowing – Single Phase (Oil, Gas) – Two Phases (O+G, O+W, G+W) – Three Phases (O+G+W)

Parameters Controlling IPR or PI • • • •

Fluid Saturation Distribution Reservoir Drive Mechanisms Formation Damage of Stimulation Relationship between IPR/PI and other variables (k, fluid properties, relative permeability, etc.) can be very complex.

Determination of PI • Testing the Well – Measure q, pwf, p q t  PI   p t   pwf  t  

• Develop PI models (equations) – PI = f(k, kro, krg, krw, Bo, Bg, Bw, etc.) – e.g., steady-state single phase flow q PI   p  pwf

kh

  re  1     s  141.2 B ln  r 2   w 

Steady-State Radial Flow W e ll, r w R e s e r v o ir , ( r e , k )

D a m a g e d Z o n e , (r s, k s)

Radial Flow • Consider single-phase radial flow in a cylindrical reservoir of radius re with a well in the center produced at a constant rate qsc STB/D. • If the outer boundary is held at a constant pressure pe, after some time we will have steady-state flow, q(r)/B = qsc for all r and t where q(r) is the flow rate in RB/D through a cylinder of radius r. Moreover, p/t = 0. •

Darcy’s Law becomes

k p kh p qsc  r   1.127 10  2rh   r B r 141.2 B r 3

Radial Flow • In the outer zone, re > r > rs kh p qsc  r   r 141.2 B r

• Separating variables and integrating pe

 dp 

p r 

re

141.2qsc B dr r r kh

141.2qsc B  re  pe  p  r   ln  kh  r

Radial Flow • At r = rs, the pressure is given by 141.2qsc B  re  p rs   pe  ln  kh  rs 

• For the inner (damaged zone) Darcy’s law ks h p qsc  r 141.2 B r

Radial Flow • Separating variables and integrating p  rs 

r

141.2qsc B s dr  p  r dp  k h r s  r

141.2qsc B  rs  p rs   p  r   ln  ks h  r

• Eliminating p(rs)

141.2qsc B  re  141.2qsc B  rs  pe  p  r   ln   ln  kh ks h  r  rs  141.2qsc B   re  k  rs    ln   ln    kh   rs  k s  r  

Radial Flow • Expanding and simplifying 141.2qsc B   re  k  rs    rs   rs  pe  p  r   ln   ln   ln   ln    kh  r  r  ks  r     rs    rs   141.2qsc B   re   k  ln     1 ln    kh   r   ks   r  

• Evaluating at the wellbore pe  pwf

  rs   141.2qsc B   re   k       ln  1 ln   kh r k r w s w        

Skin • We define  k   rs    s  1 ln  k s   rw 

• The term s is known as the (Hawkins) skin factor • Steady state radial flow equation becomes pe  pwf

 141.2qsc B   re     s   ln   kh   rw  

Notes on Skin  k   rs    s  1 ln  k s   rw 

• Skin is a dimensionless number • If k = ks, the reservoir is not damaged, and s = 0. • If k > ks, the reservoir is damaged, and s > 0. • If k < ks, the reservoir is stimulated, and s < 0.

Productivity Index • For steady-state radial flow PI 

q  pe  pwf

kh 

  re     s  141.2 B ln    rw  

• Well’s productivity is increased if – – – –

s is reduced rw is increased  is reduced h is increased

Example • A reservoir with the following properties is flowed at a bottomhole pressure of 4500 psi. Calculate the flow rate. Suggest two ways of increasing the wells production rate by a factor of 2.

Properties Property

Value

Source

Permeability, md

5.2

Pressure Transient Analysis

Thickness, ft

53

Well logs

Viscosity, cp

1.7

Fluid Analysis

Formation Volume Factor, RB/STB

1.1

Fluid Analysis

Wellbore radius, ft

0.328

Well drainage area, acre

640

Well spacing

Skin factor

10

Pressure Transient Analysis

Average reservoir pressure, psi

5635

Pressure Transient Analysis

Solution • First, we convert the well’s drainage area to an 2 equivalent drainage radius using re  A • The reservoir area is 640 acre = 640 acre43560 ft2/acre = 27878400 ft2. The equivalent reservoir drainage area is re = 2979 ft. PI 

q  p  p wf



kh

  re  1    s  141.2 B ln    rw  2   5.2 53 PI   0.0561 STB/psi 141.21.11.7   ln 2979   0.5  10    0.328  

Solution • Solving for Rate

q  PI  p  p wf   0.0561  5635  4500   63.6 STB/d

• To double the rate, we could double the pressure drop. The current pressure drop is 1135 psi, so double this value is 2270 psi, which means that we must reduce the bottomhole pressure to 5635 – 2270 = 3095 psi.

Solution • Another method of doubling the well’s rate would be to double the Productivity Index to a value of 0.1122 STB/psi. • We would do this by decreasing the skin factor by stimulating the well. • To determine the new skin factor PI  0.1122 

 5.2 53 141.21.11.7   

• snew = 0.7.

  2979   0 . 5  s  new   0.328  

ln

Pseudosteady State Flow • Rate of change of pressure with time at each point in a closed reservoir is constant. – Each “point” in the reservoir contributes equally to the flow.

• Productivity Index q PI   p  p wf

kh 141.2 B

 

 re ln  rw

 3    s   4   

Non-cylindrical Drainage Area • Deitz Shape Factor”, CA PI 

q  p  p wf

kh  1  2.2458 A      s  141.2 B ln 2   2  rw C A  

5.615 N p B • AveragepPressure  t   pi  Vpc

Dietz Shape Factors

Notes • With all else equal, asymmetric well-reservoir configurations have lower PI and flow rate compared with a symmetric well-reservoir configuration • For psuedosteady state flow with constant well flowing pressure, average reservoir pressure and flow rate decline continuously due to depletion. • For single phase flow, PI does not change with changes in changes in flow rate and average reservoir pressure due to depletion.

Absolute Open Flow Potential • AOF – For a given well-reservoir pair and average reservoir pressure, AOF is the maximum theoretical flow rate that the well can provide.

q  PI  p  pwf 

q AOF  qmax  PIp • AOF is useful in analyzing IPR in terms of

pwf p

versus

q qmax

Single Phase IPR Pwf

Pr

q  PI  p  pwf 

 q PI    p wf 

 



 p

qmax  PI p q qmax

Future Linear IPR Pwf

As time t incresases, reservoir

Pr

pressure Pr decreases and cumulative production Np increases.

q  J  Pr  Pwf  q qmax

IPR for Gas Wells • Gas PVT properties are a function of pressure (g, z, Bg, cg) • If p < 2500 psi and steady-state gas flow (q Mscf/D) p p 2 e

2 wf

1424q Z T  kh



  r   ln e   s    r   w    

• Pseudosteady-state flow p p 2

2 wf

1424q Z T  kh

 1  2.2458 A    ln   s 2  2  r C   w A    

PI for Gas Wells • Pseudosteady-state flow q PI  PI   p  pwf

kh p  pwf 

 1  2.2458 A     s  1424Z T  ln 2   2  rw C A  

• PI for gas wells is a function of pressure. – When average reservoir pressure changes, gas properties change. – PI is not constant during the well’s life

Typical Gas IPR Gas IPR

Well Pressure, psi

6000 5000 4000 3000 2000 1000 0 0

10000

20000

30000

40000

Rate, MSCF/d

50000

60000

70000

IPR for non-Darcy Gas Flow • At high flow rates, Darcy’s Law is not valid. • There are many high flow rate gas wells where Darcy’s law is not obeyed in the near-wellbore region (i.e., where gas velocities are greatest). • The observed pressure drop exhibits a ratedependent skin effect

st  s  Dq

• st denotes the total apparent skin factor, and D is a non-Darcy coefficient.

Pseudosteady State Gas Flow • For Pseudosteady state flow p p 2

2 wf

 1424qZ T  1  2.2458 A     ln  s  Dq  2   kh  2  rw C A  

• Back Pressure Equation (Rawlins – Schellhardt)

p 2  p wf2  aq  bq 2 1424Z T a kh b

 1  2.2458 A    ln   s 2  2  r C   w A    

1424Z TD kh

Limiting Cases • When non-Darcy flow is negligible (b << a) p 2  pwf2  aq



1 2 q  p  pwf2 a



• When non-Darcy flow is dominant (b >> a)

p 2  pwf2  bq 2



1 q p 2  pwf2 b



1 2

Generalized Gas Flow • Limiting forms of the gas flow equation can be generalized as



qC p  p 2

2 wf



n

– Negligible non-Darcy flow, n = 1 – Purely non-Darcy flow, n = 0.5 – Both components play a role, 0.5 < n < 1

• Equation can be rearranged as 1 1 log p 2  pwf2  log q   log C  n n





Back Pressure Analysis • At constant reservoir pressure, flow at four different flow rates. – Measure stabilized bottomhole pressure at each rate.

• Plot log p 2  pwf2  versus log q  • Straight line with slope 1/n. • With n, can construct the IPR.

Multiphase Flow • Need rates of oil, gas, water – PIo, PIg, PIw PI o 

qo  p  p wf

ko h

 1  2.2458 A      141.2 Bo  o ln  s 2   2   rw C A    qw kwh PI w   p  p wf  1  2.2458 A      141.2 Bw  w ln  s 2   2   rw C A    qg kg h PI g   p  pwf    2.2458 A  * * 1   s  141.2 Bg  g  ln 2   2  rw C A  

k o  kk ro k w  kk rw k g  kk rg

Multiphase IPR • IPR under multiphase flow conditions cannot be easily calculated. • The most accurate method is by solving the equations governing the flow in the porous media through a reservoir simulator. • The IPR is so important to Production Engineers that simplified or empirical methods to estimate it are necessary. • The most common correlations are Vogel and Fetkovich

Vogel IPR • Vogel used a numerical reservoir simulator to generate the IPR. He studied several cases for a specific condition: – Mechanism of production – Solution Gas Drive – No water production – Reservoir pressure below bubble point Saturated conditions

–

• He changed several other conditions such as fluid and rock properties

P w fr q m ax Vogel IPR

1

0.8

0.6

0.4

 pwf   pwf  q   0.8   1  0.2  qmax  p   p 

0.2

2

0

0

0.2

0.4

0.6

0.8

1

Vogel IPR • Vogel IPR can be obtained from well tests. • Although the method was developed for solution gas drive reservoirs, the equation is generally accepted for other drive mechanisms as well. • It is found to give excellent results for any well with a reservoir pressure below the oil bubble point, i.e., saturated reservoirs.

Undersaturated Reservoirs P

q  J  Pr  Pwf 

Pr Pb

?

qb

qmax

q

Undersaturated Reservoirs P

Pr Pb

Pr’ = Pb

q’ = q - qb qb

qmax

q

Undersaturated Reservoirs P

q  J  Pr  Pwf 

Pr

  Pwf  q  qb   0.8   1  0.2   qmax  qb  Pb 

Pwf  Pb

Pb

qb

qmax

q







2

Fetkovich IPR • Based on observations of hundreds of sets of field production data • C and n are unique to each well • The values of the constants C and n are determined from at two sets of rate and bottomhole pressure data at a given average reservoir pressure.

qo  C  p  p 2

q o ,max  Cp qo q o ,max





2 n wf

2n

p

2 wf

  1 2    p  



n

Example Given data,

p = 2,400 psia qo = 100 STB/d pwf = 1800 psia Generate inflow performance curve using both Vogel's and Fetkovich's (n = 1) equations.

Example Solution •Vogel's Equation Determine qo,max 100  1800   1800   1  0.2   0.8  q o ,max  2400   2400 

q o ,max  250 STB/d

So Vogel’s equation becomes  p wf   p wf  qo   0.8   1  0.2 250  2400   2400 

2

2

Example •Fetkovich's Equation Determine qo,max 100  1800   1   q o ,max  2400 

2

q o ,max  228.6 STB/d

Example Vogel pwf, psia qo, STB/d 0 250.0 600 225.0 1200 175.0 1800 100.0 2400 0.0

Fetkovich pwf, psia qo, STB/d 0 228.6 600 214.3 1200 171.5 1800 100.0* 2400 0.0

Multiphase IPR 300

250

Rate, STB/D

200

150

100

50

0 0

500

1000

1500 Pressure, psia

2000

2500

3000

Multi-layer inflow Performance

Cross-Flow between 2 Layers