5 IPR Saturated Gas 15

Chair of Petroleum & Geothermal Energy Recovery

Saturated Oil Reservoirs

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Saturated Oil Reservoirs Saturated reservoir: (Two phase region) 𝐩 < pb and / or pwf < pb IPR approximation Methods: (present) -

Vogel Equation Generalized Vogel Equation Composite Model Multi-rate Fetkovich Method Multi-rate Jones Model

IPR approximation Methods: (future - change of reservoir properties) -

Standing Method Multi-rate Fetkovich Method (future) Page 2

Chair of Petroleum & Geothermal Energy Recovery

Vogel Equation

Empirical relationship, based on a number of history matching simulations. (1968) -

𝐩 < pb and pwf < pb

-

Works best for solution gas drive reservoirs!

-

Does not work properly for gas wells, high viscosities and excessive skin

-

Use of the properties of only oil in a two-phase system possible

-

μ, Bo must be taken at p

-

Only for oil and gas production (no water!)

-

Depleted reservoirs can be analyzed Page 3

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Vogel equation Simplifying assumptions for the simulation: 1. reservoir is circular and completely bounded with a completely penetrating well at its center 2. porous medium is uniform and isotropic with a constant water saturation at all points 3. gravity effects are be neglected 4. compressibility of rock and water can be neglected 5. composition and equilibrium are constant for oil and gas 6. the same pressure exists in both the oil and gas phases 7. the pseudo steady-state assumption that the tank-oil desaturation rate is the same at all points at a given instant

J.V.Vogel, Inflow Performance Relationship for Solution-Gas Drive Wells

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Vogel equation

J.V.Vogel, Inflow Performance Relationship for Solution-Gas Drive Wells

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Vogel equation

J.V.Vogel, Inflow Performance Relationship for Solution-Gas Drive Wells

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Chair of Petroleum & Geothermal Energy Recovery

Vogel equation

J.V.Vogel, Inflow Performance Relationship for Solution-Gas Drive Wells

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Vogel equation

IPR Saturated Reservoir IPR Undersaturated Reservoir

qomax (saturated)

J.V.Vogel, Inflow Performance Relationship for Solution-Gas Drive Wells

qomax (undersaturated)

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Vogel equation 𝐪𝐨 𝐪𝐨,𝐦𝐚𝐱

= 𝟏 − 𝟎. 𝟐

𝐩 ∗ 𝐰𝐟 𝐩

− 𝟎. 𝟖 ∗

𝐩𝐰𝐟 𝟐 𝐩

Solution 1: Well Test Information - one test point is required

Solution 2: Use of AOF /qomax from undersaturated IPR

qo max (saturated) =

qo max (undersaturated) 1,8

qo max (undersaturated) … Oil rate from undersaturated IPR with pwf = 0 (bbl/day, m³/s) (e.g. steady state, pseudo steady state) qo … Actual oil flow-rate (bpd, m³/s, …) pwf … Well flowing pressure (psi, Pa) p … Average reservoir pressure (psi, Pa) J.V.Vogel, Inflow Performance Relationship for Solution-Gas Drive Wells

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Chair of Petroleum & Geothermal Energy Recovery

Vogel equation Example Vogel equation Develop an IPR curve for the given saturated reservoir. A well test was performed at a pressure of 3000 psi. p = 4350 psi pb = 5210 psi qo = 680 bpd

J.V.Vogel, Inflow Performance Relationship for Solution-Gas Drive Wells

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Chair of Petroleum & Geothermal Energy Recovery

Generalized Vogel Equation Combination undersaturated and saturated reservoir: 𝐩 > pb but pwf < pb 𝐉 ∗ 𝐩𝐛 𝐩𝐰𝐟 𝐩𝐰𝐟 𝐪𝐨 = 𝐪𝐛 + 𝟏 − 𝟎. 𝟐 ∗ − 𝟎. 𝟖 ∗ 𝟏. 𝟖 𝐩𝐛 𝐩𝐛

qb

𝟐

… Flow rate, where pwf = pb (with undersaturated IPR eq.) qb p−pb

PI – Index above pb :

J=

“Vogel flow” qV :

qV =

pb J 1,8

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Chair of Petroleum & Geothermal Energy Recovery

Generalized Vogel Equation

Bubble point

𝐪𝐛

𝐩𝐛 𝐉 𝐪𝐕 = 𝟏, 𝟖 𝐉 ∗ 𝐩𝐛 Page 14

Chair of Petroleum & Geothermal Energy Recovery

Generalized Vogel Equation Example Generalized Vogel Equation Develop an IPR curve for the following data. p = 4000 psi pb = 2000 psi pwf = 1200 psi qo@1200 psi = 532 bpd

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Composite Model Extension of the Vogel inflow solution that accounts for water cut.

K.Brown, Technology of Artificial Lift Methods, Volume 4

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Multi-rate Fetkovich Method

Saturated reservoir: 𝐩 < pb and pwf < pb -

Also called Back Pressure Equation Assuming p is known, at least two tests are required to determine C, n It can be used for high rate oil/gas wells and gas wells More accurate and flexible than Vogel-Equation Plotting (pr²-pwf²) vs. q on a log- log paper and drawing a best fit line results in a slope, equal to 1/n qo = C. p² − pwf ²

C n

n

qo qo max

= 1−

n pwf 2 p

… Curve coefficient (rock properties) (bpd/psi²) … Curve exponent (flow region, 0.5 (turbulent) < n < 1 (laminar))

M.J.Fetkovich, The Isochronal Testing of Oil Wells

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Multi-rate Fetkovich Method

M.J.Fetkovich, The Isochronal Testing of Oil Wells

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Comparison IPR - curves

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Multi-rate Fetkovich Method Example Fetkovich Method Develop Calculate and plot the IPR using the Fetkovich Approximation / compare with the Vogel equation!

Flow rate (bpd)

𝐩𝐰𝐟 (psi)

Testpoint 1

383

2897

Testpoint 2

640

2150

p= 3600 psi

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Chair of Petroleum & Geothermal Energy Recovery

Multi-rate Jones Model

Saturated reservoir: 𝐩 < pb -

Applicable in high-rate oil wells

-

Provides an indication of perforation effectiveness in normally completed wells. An abnormal high turbulence coefficient indicates too few openings

-

The laminar flow coefficient includes skin effect

-

Two or more stabilized flow tests are required

-

Based on the Forchheimer’s equation

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Chair of Petroleum & Geothermal Energy Recovery

Multi-rate Jones Method p − pwf = C + Dq q

q C D

…… Oil flow rate (bbl/day) …… Laminar flow coefficient …… Turbulence coefficient C=

hp β ρ

141,2.Bμ kh

0,472.re ln rw

+S

D=

9,08.10−13 βB²ρ 4π².h²p .rw

...... Perforation length (ft) ...... Turbulence factor (1/ft) ...... Fluid density (lb/ft³) D1 h²p2 = D2 h²p1 Page 22

Chair of Petroleum & Geothermal Energy Recovery

Multi-rate Jones Method Example Two flow tests were performed on a high rate oil well. Develop the PI – plot and compare the IPR curve for the given data and for a 20% increased perforation length Flow rate (bpd)

𝐩𝐰𝐟 (psi)

Testpoint 1

6199

5410

Testpoint 2

8115

5383

p = 5448 psi

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Multi-rate Jones Method

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Multi-rate Jones Method

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Standing Method Three parameter change as a result of the declining average reservoir pressure: k ro , μo and Bo f pp =

kro Bo μ0 P

f pf =

qo,max,F = qo,max,P . qo,F q0,max,F

p F

= 1 − 0.2 ∗

pwf,F pF

kro Bo μ0 F

f pF pF f pp pp

− 0.8 ∗

pwf,F 2 pF

… present value … future value Page 26

Chair of Petroleum & Geothermal Energy Recovery

Standing Method Example Standing Method Calculate the IPR for both, the present and the future for the following parameters: (draw the graphs) Pressure test: qo = 400 [STB/day] pwf = 1815 [psig] p o Bo So kro

Present Time 2250 [psig] 3.11 [cP] 1.173 0.768 0.815

Future Time 1800 [psig] 3.59 [cP] 1.150 0.741 0.685 Page 27

Chair of Petroleum & Geothermal Energy Recovery

Multi-rate Fetkovich Method

Multi-rate Fetkovich Method (future) This approach assumes a linear relationship between the average reservoir pressure and the coefficient C.

CF = Cp .

pF pp

qo,F = CF pF ² − pwf,F ²

p F

n

… present value … future value

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Chair of Petroleum & Geothermal Energy Recovery

Multi-rate Fetkovich Method (future) Example Fetkovich Method (future) Calculate the future IPR curve for the following parameters: (use the properties from the Fetkovich example)

Flow rate (bpd)

𝐩𝐰𝐟 (psi)

Testpoint 1

383

2897

Testpoint 2

640

2150

p = 3600 psi pF = 2950 psi

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Chair of Petroleum & Geothermal Energy Recovery

Summary Model PI - Method

Fluids

Conditions

Oil

Above pb

Vogel Equation

Oil, Gas

Below pb

Generalized Vogel Equation

Oil, Gas

Above and below pb

Multi-rate Fetkovich Method

Oil, Gas

Below pb

Multi-rate Jones Model

Oil, Gas

Below pb

Standing Method

Oil, Gas

Below pb

Oil, Water

Above pb

Darcy Equation Composite Model

Oil, Gas, Water

Above and below pb

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Chair of Petroleum & Geothermal Energy Recovery Deadline: 26.08.2015 09:00

Homework

1) Draw the IPR curve for the given reservoir and calculate the surface quantities (oil, water and gas) for a well flowing pressure of 3 MPa! Vogel’s Method p = 7,5 MPa Bw = 1,04 re = 200 m rw = 0,083 m

pb = 7,6 MPa Sw = 55 % h = 10 m k = 200 mD

μo = 1,5. 10-3 Pas μw = 1. 10-3 Pas S=0

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Chair of Petroleum & Geothermal Energy Recovery Deadline: 26.08.2015 09:00

Homework

2) Use the Fetkovich equation to generate the IPR curve for the following reservoir data! p= 3600 psi Calculate the future IPR for a new average reservoir pressure of 3000 psi!

Flow rate (bpd)

𝐩𝐰𝐟 (psi)

Testpoint 1

383

2897

Testpoint 2

640

2150

Testpoint 3

263

3200

Testpoint 4

497

2530

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Chair of Petroleum & Geothermal Energy Recovery Deadline: 26.08.2015 09:00

Homework

3) You are give 3 test points from an oil well. Use the Jones Method to evaluate the average reservoir pressure and the skin factor S. k = 10 mD h = 50 ft

µ = 1,7 cp B = 1,1

rw = 0,328 ft re = 2500 ft

Flow rate (bpd)

𝐩𝐰𝐟 (psi)

Testpoint 1

6599

6610

Testpoint 2

8515

6583

Testpoint 3

21400

6256

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Gas Reservoirs

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Chair of Petroleum & Geothermal Energy Recovery

Gas Properties Specifics of gas: -

μ and the isothermal compressibility of real gas are highly pressure dependent

-

Gas flow equations using surface rates at standard conditions in field units

Z – Factor: ppr = Tpr =

p ppc T Tpc

ppc

… Critical pressure of the gas (psi, Pa)

Tpc

… Critical temperature of the gas (°R, °C)

ppr

… Pseudo-reduced pressure (-)

Tpr p T

… Pseudo-reduced temperature (-) … Pressure of interest (psi, Pa) … Temperature of interest (°R, °C) Page 35

Chair of Petroleum & Geothermal Energy Recovery

Gas Properties know gas composition: ppc =

N i=1 yi . pci

Tpc =

N i=1 yi . Tci

unknown gas composition:

ppc = 709,6 − 58. γg Tpc = 170,5 + 307,3. γg y … Mole fraction (-) γg … Gas gravity (-) Page 36

Chair of Petroleum & Geothermal Energy Recovery

Gas Properties

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Chair of Petroleum & Geothermal Energy Recovery

Gas Properties Viscosity – Approximation:

B.C.Craft, Applied Petroleum Reservoir Engineering

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Chair of Petroleum & Geothermal Energy Recovery

Gas Properties

B.C.Craft, Applied Petroleum Reservoir Engineering

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Chair of Petroleum & Geothermal Energy Recovery

Russell and Goodrich Solution

Pseudo-steady State Solution: p² − p²wf =

μ=μ

Q k h, r Z T p μ

p+pwf 2

1422.QμZT kh

ln

0,472.re rw

+S

Z=Z

p+pwf 2

… Gas flow rate (Mscf/day, at 60°F and 14,7 psi) … Permeability (mD) … Distances (ft) … Factor (-) … Temperature (°R = 460 + °F) … Pressure (psi) … Viscosity (cp) Page 40

Chair of Petroleum & Geothermal Energy Recovery

Russell and Goodrich Solution

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Chair of Petroleum & Geothermal Energy Recovery

Russell and Goodrich Solution Example Gas IPR Generate the IPR curve for the following gas reservoir:

γg = 0,9 T = 240°F (700°R) k = 10 mD rw = 0,3125 ft

p = 4000 psi S=3 h = 5 ft re = 500 ft

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Chair of Petroleum & Geothermal Energy Recovery

Russell and Goodrich Solution 𝐩𝐩𝐫

Z

𝛍

Q

4000

𝐩 + 𝐩𝐰𝐟 𝟐 -

-

-

-

0

3500

3750

5,7

0,86

0,0244

932

3000

3500

5,3

0,84

0,0232

1876

2500

3250

4,9

0,825

0,0226

2732

2000

3000

4,6

0,815

0,0220

3498

1500

2750

4,2

0,805

0,0201

4247

1000

2500

3,8

0,795

0,0195

5043

500

2250

3,4

0,79

0,0183

5684

0

2000

3,0

0,8

0,0177

5899

𝐩𝐰𝐟

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Russell and Goodrich Solution

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Chair of Petroleum & Geothermal Energy Recovery

Al Hussainy, Ramey and Crawford Solution

Pseudo-steady State Solution: “real gas pseudo pressure”: m p − m pwf = 2. m p − m pwf Q k h, r Z T p μ

p p.dp pwf μZ

1422. QT 0,472. re = ln +S kh rw

… Gas flow rate (Mscf/day, at 60°F and 14,7 psi) … Permeability (mD) … Distances (ft) … Factor (-) … Temperature (°R = 460 + °F) … Pressure (psi) … Viscosity (cp) Page 45

Chair of Petroleum & Geothermal Energy Recovery

Al Hussainy, Ramey and Crawford Solution

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Chair of Petroleum & Geothermal Energy Recovery

Al Hussainy, Ramey and Crawford Solution Example Gas IPR Create the IPR curve and evaluate pwf for a production rate of Q = 3865 Mscf/day for the following gas reservoir: γg = 0,85 T = 200°F (660°R) k = 10 mD rw = 0,3125 ft

p = 3600 psi S = 2,5 h = 5 ft re = 1000 ft

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Chair of Petroleum & Geothermal Energy Recovery

Al Hussainy, Ramey and Crawford Solution p

Z

*106

*106

400

0,61

1,05

0,0122

0,937

200

34898

400

14,0

14,0

800

1,21

1,13

0,0132

0,890

600

102420

“

41,0

54,9

1200

1,82

1,25

0,0146

0,840

1000

163499

“

65,4

120,3

1600

2,42

1,35

0,0157

0,795

1400

223940

“

89,6

209,9

2000

3,03

1,50

0,0175

0,770

1800

267544

“

107,0

316,9

2400

3,63

1,72

0,0200

0,763

2200

287789

“

115,1

432,0

2800

4,24

1,88

0,0219

0,780

2600

304386

“

121,8

553,8

3200

4,85

1,98

0,0231

0,805

3000

323120

“

129,2

683,0

3600

5,45

2,15

0,0250

0,835

3400

325131

“

130,1

813,1

4000

6,06

2,20

0,0256

0,870

3800

340836

“

136,3

949,4

4400

6,66

2,35

0,0274

0,900

4200

340913

“

136,4

1085,8 Page 48

Chair of Petroleum & Geothermal Energy Recovery

Al Hussainy, Ramey and Crawford Solution

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Chair of Petroleum & Geothermal Energy Recovery

Al Hussainy, Ramey and Crawford Solution

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Chair of Petroleum & Geothermal Energy Recovery

Gas Reservoirs

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