4 Ipr Undersaturated 15

Chair of Petroleum & Geothermal Energy Recovery

Petroleum Engineering Summer Course 2015 Inflow Performance Relationship (IPR)

petrowiki.org

Clemens Langbauer

Chair of Petroleum & Geothermal Energy Recovery

Petroleum Production System

Beggs: Production Optimization using Nodal Analysis

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Chair of Petroleum & Geothermal Energy Recovery

Petroleum Production System

Beggs: Production Optimization using Nodal Analysis

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Chair of Petroleum & Geothermal Energy Recovery

Reservoir Types

Beggs: Production Optimization using Nodal Analysis

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Chair of Petroleum & Geothermal Energy Recovery

Reservoir Types Undersaturated Oil Reservoirs Reservoir pressure (pe , p) and well flowing pressure (pwf ) are above the bubble point pb . Oil and water are present and flowing Saturated Oil Reservoirs / Multi-phase Reservoirs pwf and / or the reservoir pressure is / are below the bubble point. Oil, water and gas are present and flowing Gas Reservoirs Gas is present and flowing (little amount of water) Page 5

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Undersaturated Oil Reservoirs

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Chair of Petroleum & Geothermal Energy Recovery

Darcy‘s Law

L.P.Drake: Fundmentals of Reservoir Engineering

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Chair of Petroleum & Geothermal Energy Recovery

Darcy‘s Law Generalized form: (frictionless & laminar flow) u=

kρ dΦ . μ dl

Φ=

pB dp pA ρ

+ g. (zA − zB )

u ρ Φ

… Velocity (m/s, ft/s) … Density (kg/m³, lb/ft³) … Potential energy per unit mass

k μ l

… Permeability (m², D) … Viscosity (Pas, cp) … Length (m, ft)

q

… Flow rate (m³/s, bbl/day)

A

… Cross-section (m², ft²)

Incompressible & horizontal flow: k dp μ dl

u=− .

Radial form: q =

Φ=

p ρ

q=−

kA dp . μ dl

2πkh dp r. μ dr

L.P.Drake: Fundmentals of Reservoir Engineering

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Chair of Petroleum & Geothermal Energy Recovery

Diffusivity Equation Basic differential equation for radial flow in a porous media -

Homogeneous rock properties and isotropic permeability Fully radial flow (entire reservoir thickness covered) Formation is completely saturated with one fluid Reservoir of cylindrical shape

L.P.Drake: Fundmentals of Reservoir Engineering

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Chair of Petroleum & Geothermal Energy Recovery

Diffusivity Equation Radial flow of any single fluid in a porous media: 1 𝜕 kρ 𝜕p r r 𝜕r μ 𝜕r

= ϕcρ

𝜕p 𝜕t

Linearization for incompressible fluids: 𝟏 𝛛 𝐫 𝛛𝐫

𝛛𝐩 𝐫 𝛛𝐫

L.P.Drake: Fundmentals of Reservoir Engineering

=

𝛟𝐜𝛍 𝛛𝐩 𝐤 𝛛𝐭

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Chair of Petroleum & Geothermal Energy Recovery

Flow Regimes Flow regimes describe the outer boundary pressure decline over time: Steady-State Flow:

𝜕p 𝜕t q=const.

=0

Constant pressure boundary

Pseudo-(Semi-) Steady-State Flow:

𝜕p 𝜕t q=const.

= const.

𝜕p 𝜕t q=const.

= f (log t)

No flow boundary

Transient Flow: Infinite acting

so far no boundary response - until pressure wave hits boundary L.P.Drake: Fundmentals of Reservoir Engineering

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Chair of Petroleum & Geothermal Energy Recovery

Steady State Solution Steady State applies after the transient period and assumes a constant pressure over time at the outer boundary. 𝜕p 𝜕t

=0 q=const.

L.P.Drake: Fundmentals of Reservoir Engineering

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Chair of Petroleum & Geothermal Energy Recovery

Steady State SI – Units (m³/s) Relationship p(r),r p − pwf = qμ ln r + S 2πkh r w

Outer boundary pressure pe

qμ

r

pe − pwf = 2πkh ln r e + S w

Average reservoir p − pwf = qμ ln 0,61.re + S 2πkh rw pressure p

Field – Units (bbl/day) p − pwf =

141,2.qμ kh

pe − pwf = p − pwf =

141,2.qμ kh

141,2.qμ kh

r

ln r + S w

r

ln r e + S w

ln

0,61.re rw

+S

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Chair of Petroleum & Geothermal Energy Recovery

Pseudo- (Semi-) Steady State Solution Pseudo Steady State applies after the transient period and assumes no fluid flow at the outer boundary and a constant pressure drop over time. 𝜕p 𝜕t

= const. q=const.

L.P.Drake: Fundmentals of Reservoir Engineering

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Chair of Petroleum & Geothermal Energy Recovery

Pseudo Steady State SI – Units (m³/s) Relationship p(r),r

qμ

r

Field – Units (bbl/day)

r²

p − pwf = 2πkh ln r − 2r2 + S w

e

Outer boundary pe − pwf = qμ ln 0,61.re + S 2πkh rw pressure pe Av. reservoir pressure p

qμ

p − pwf = 2πkh ln

0,472.re rw

+S

p − pwf =

141,2.qμ kh

pe − pwf = p − pwf =

141,2.qμ kh

141,2.qμ kh

r

r²

ln r − 2r2 + S w

e

0,61.re rw

+S

0,472.re rw

+S

ln ln

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Chair of Petroleum & Geothermal Energy Recovery

Transient Solution The reservoir is acting infinite – at the beginning of production and time dependent! 𝜕p = f (log t) 𝜕t q=const.

L.P.Drake: Fundmentals of Reservoir Engineering

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Chair of Petroleum & Geothermal Energy Recovery

Transient Solution SI – Units (m³/s) qμ

4kt

pi − pwf = 4πkh ln γϕμc r

t w²

re rw pe pi pwf q μ h k Φ ct γ t S

Field – Units (bbl/day) + 2. S

pi − pwf =

162,5 qμ kh

kt

log ϕμc r

t w²

− 3,23 + 0,87. S

… Outer boundary radius (m, ft) … Wellbore radius (m, ft) … Outer boundary pressure (Pa, psi) … Initial reservoir pressure (Pa, psi) … Well flowing pressure (Pa, psi) … Production rate under reservoir conditions (m³/s, STB/d) … Viscosity of the fluid (Pas, cp) … Height of the reservoir (m, ft) … Total permeability (m², mD) … Porosity (-) … Compressibility (Pa-1, psi-1) … Euler constant = 1,78 … Time (s, h) … Skin factor (-) Page 17

Chair of Petroleum & Geothermal Energy Recovery

Surface Rates from IPR Conversion of the volumes under reservoir conditions to surface conditions. Vr qr B= = Vs qs

B Vr Vs qr qs

… Formation volume factor (-) … Volume of fluid under reservoir conditions (m³, bbl, …) … Volume of fluid under stock tank conditions (m³, bbl, …) … Flow rate under reservoir conditions (m³/s, bbl/day, …) … Flow rate under stock tank conditions (m³/s, bbl/day, …)

Economides: Petroleum Production System

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Chair of Petroleum & Geothermal Energy Recovery

Surface Rates Example IPR rate calculation Calculate the surface production rate (here p = 0 MPa) and the quantities at a pressure of 5 MPa for the following reservoir under steady state conditions. p = 20 MPa pwf = 10 MPa μ = 0,003 Pas k = 100 mD h = 10 m re = 100 m rw = 0,15 m S=1 Page 19

Chair of Petroleum & Geothermal Energy Recovery

Surface Rates Homework

Calculate the production rate for steady state , pseudo steady state and transient fluid flow for the well with the following parameters: pe = pi = 17 MPa pwf = 10 MPa

μ = 1,62. 10-3 Pas B = 1,1

re = 100 m rw = 6,538” = 83 mm

h = 10 m k = 100 mD

Φ = 0,2 γ = 1,78

ct = 1,3. 10-5 Pa-1 t = 30 days = 2592000 s

L.P.Drake: Fundmentals of Reservoir Engineering

S=0

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Chair of Petroleum & Geothermal Energy Recovery

Inflow Performance Relationship

Productivity Index J or PI:

J=

qo pr −pwf

 Expression describes the deliverability of the wellbore  Constant for undersaturated reservoirs  Maximizing the PI is a major objective Page 21

Chair of Petroleum & Geothermal Energy Recovery

Productivity Index Example Productivity index The average reservoir pressure of well XY in a depth of 2000 m is 21 MPa. The test point of this undersaturated reservoir was at 18 MPa and generated 250 m³/day. Draw the IPR curve! Calculate the PI-Index and the AOF! Calculate the skin factor! μ = 1,62. 10-3 Pas B = 1,1 re = 100 m rw = 6,538” = 83 mm

h = 10 m k = 100 mD Page 22

Chair of Petroleum & Geothermal Energy Recovery

Dietz Shape Factor Irregular Drainage Pattern Up to now only cylindrical reservoirs were considered – the shape of the reservoir has a significant influence on the IPR. Dietz Shape Factor: It considers the irregularity of the drainage area. SI – Units (m³/s)

Field – Units (bbl/day)

Steady state qμB

p − pwf = 2πkh

1 4A ln + 2 0,61.γCA rw ²

S

p − pwf =

141,2.qμB 1 4A ln + kh 2 0,61.γCA rw ²

p − pwf =

141,2.qμB 1 4A ln + kh 2 γCA rw ²

S

Pseudo steady state qμB

p − pwf = 2πkh CA A

1 4A ln + 2 γCA rw ²

S

S

… Dietz shape factor … Drainage area (m², ft²) Page 23

Chair of Petroleum & Geothermal Energy Recovery

Dietz Shape Factor

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Chair of Petroleum & Geothermal Energy Recovery

Dietz Shape Factor Example

1

2

Dietz shape factor Compare the surface production rates under steady state conditions for the two different drainage patterns and the given reservoir data. Draw the IPR curves and calculate the productivity index. B = 1,1 μ = 1,5 cP k = 9 mD rw = 0,3281 ft p = 3000 psi

A = 570 acre = 24829200 ft² S=0 h = 35 ft γ = 1,78 pwf = 1200 psi

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Chair of Petroleum & Geothermal Energy Recovery

Relative Permeabilities Separation between oil and water production e.g. steady state solution: 2πkk ro h p − pwf qo = 0,61. re μo Bo ln +S rw qw =

k ro k rw

2πkk rw h p − pwf 0,61. re μw Bw ln +S rw

… Relative permeability oil (-) … Relative permeability water (-) Page 26

Chair of Petroleum & Geothermal Energy Recovery

Relative Permeabilities Example Water and Oil Production Calculate the oil and water production rate under steady state conditions. The water saturation is 55 %. p = 15 MPa re = 100 m h = 10 m S=3

pwf = 10 MPa rw = 83 mm k = 100 mD

μoil = 1,62. 10-3 Pas μwater = 0,8. 10-3 Pas

Boil = 1,12 Bwater = 1,04

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Chair of Petroleum & Geothermal Energy Recovery

Horizontal Wells The drainage shape for horizontal wells is ellipsoidal. One must distinguish between horizontal and vertical permeability.

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Chair of Petroleum & Geothermal Energy Recovery

Horizontal Wells Steady state flow in the horizontal plane and pseudo steady state flow in the vertical plane are assumed: L

a = 2 0,5 + 0,25 +

reH 4

0,5 0,5

L

for 2 ≤ 0,9 . reH

L 2

Iani =

kH kV

k H . h. ∆p

q= 141,2. Bμ ln

L a + a² − 2 ² L 2

I h Iani h + ani ln L rw (Iani + 1)

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Chair of Petroleum & Geothermal Energy Recovery

Example IPR Horizontal Wells Calculate the production rate for the given horizontal well.

k H = 10 mD

k V = 3 mD

pi = 5000 psi h = 55 ft

pwf = 2750 psi

reH = 500 ft

B = 1,15 μ = 1,55 cP rw = 0,351 ft

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Chair of Petroleum & Geothermal Energy Recovery

Homework Deadline: 25.08.2015

09:00

The following examples should be calculated using the following undersaturated reservoir properties:

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Chair of Petroleum & Geothermal Energy Recovery

Homework 1) Infinite – acting oil well: Develop a production rate profile (q – t diagram) for 1 year assuming that no boundary effects emerge. Do this in increments of 2 months and use a pwf of 3500 psi. 2) Steady state production: Assume that the well has a drainage area equal to 640 acre (re = 2980ft) and is producing at steady state with an outer boundary (constant) pressure equal to 5651 psi. Calculate the steady-state production rate if the pwf is equal to 4500 psi. Use a skin factor of 10. Describe two mechanisms to increase the flow rate by 50%. (Show calculations)

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Chair of Petroleum & Geothermal Energy Recovery

Homework 3) No-flow boundary reservoir: What would be the average reservoir pressure if the outer boundary pressure is 6000 psi, the pwf is 3000 psi, the drainage area 640 acres and the well radius is 0,328 ft ? What would be the ratio of the flow rates before and after the average reservoir pressure drops by 1000 psi? Assume skin = 0. 4) Irregular well positioning: Assume that two wells each drain 640 acre. Furthermore, assume that p =5651 psi (same as pi ) and that S = 0. The pwf in both wells is 3500 psi. Well A is placed at the center of a square, whereas well B is at the center of the upper right quadrant of a square drainage shape. Calculate the production rates from the two wells for pseudo steady state!

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Chair of Petroleum & Geothermal Energy Recovery

Homework 5) Transient IPR: Construct transient IPR curves for 1, 6 and 24 months. Assume zero skin! 6) Steady state IPR: Assume that the initial reservoir pressure of the well is also the constant pressure of the outer boundary. Draw IPR curves for skin effects equal to 0, 5, 10 and 50. Use a drainage radius of 2980 ft (A = 640 acre).

7) Pseudo steady state IPR: Calculate the IPR curves for zero skin effect but for average reservoir pressures in increments of 500 psi from initial 5651 psi to 3500 psi. drainage radius of 2980 ft Page 34