5 IPR Saturated Gas 15

8/16/2019 5 IPR Saturated Gas 15

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Chair of Petroleum & Geothermal Energy Recovery

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Saturated Oil Reservoirs


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Saturated reservoir: (Two phase region)

< 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)

Saturated Oil Reservoirs


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Empirical relationship, based on a number of history matching simulations.

(1968)

- < 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

Vogel Equation


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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 rateis the same at all points at a given instant

Vogel equation

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


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Vogel equation



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Vogel equation



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Vogel equation



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IPR Undersaturated Reservoir

IPR Saturated Reservoir

qomax (saturated) qomax (undersaturated)

Vogel equation



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, . ∗

. ∗

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)

,

qo max (undersaturated) … Oil rate from undersaturated IPR with pw = 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)

Vogel equation



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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 psipb = 5210 psiqo = 680 bpd


Vogel equation


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Combination undersaturated and saturated reservoir: > pb but pwf < pb

∗ .

. ∗

. ∗

qb … Flow rate, where pw pb (with undersaturated IPR eq.)

PI – Index above pb: J

−

“Vogel flow” qV: qV

,

Generalized Vogel Equation


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Bubblepoint

∗

,



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Example


Develop an IPR curve for the following data.

p = 4000 psipb = 2000 psipw = 1200 psiqo@ si = 532 bpd


C f & G

C f & G


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

Composite Model

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

Ch i f P t l & G th l E R



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Saturated reservoir: < 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 (pr ²-pwf ²) vs. q on a log- log paper and drawing a best fit line

results in a slope, equal to 1/n

qo C. p² pw ² 1

C … Curve coefficient (rock properties) (bpd/psi²)

n … Curve exponent (flow region, 0.5 (turbulent) < n < 1 (laminar))

Multi-rate Fetkovich Method

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




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




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Comparison IPR - curves




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Example

Fetkovich Method

Develop Calculate and plot the IPR using the Fetkovich Approximation /

compare with the Vogel equation!

p= 3600 psi

Flow rate (bpd) (psi)

Testpoint 1 383 2897






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Saturated reservoir: < pb

- Applicable in high-rate oil wells

- Provides an indication of perforation effectiveness in normallycompleted 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

Multi-rate Jones Model

Chair of Petrole m & Geothermal Energ Reco er

Chair of Petrole m & Geothermal Energ Reco er


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p pw

q

C D q

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

C ,.

ln ,.

S D ,.²

π².².

h ...... Perforation length (ft)

β ...... Turbulence factor (1/ft)ρ ...... Fluid density (lb/ft³)

DD

h²h²

Multi-rate Jones Method




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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

p = 5448 psi

Flow rate (bpd) (psi)Testpoint 1 6199 5410






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Three parameter change as a result of the declining average reservoir

pressure: ko, μo and Bo

f p

P f p

F

qo,ax,F qo,ax,P.

,

,, 1 0 . 2 ∗

,

0 . 8 ∗

,

p … present value

F … future value

Standing Method




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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]

Present Time Future Time

p 2250 [psig] 1800 [psig]

o 3.11 [cP] 3.59 [cP]

Bo 1.173 1.150

So 0.768 0.741

kro 0.815 0.685

Standing Method




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This approach assumes a linear relationship between the average

reservoir pressure and the coefficient C.

CF C.

qo,F CF pF² pw,F²

p … present value

F … future value


Multi-rate Fetkovich Method (future)




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Example

Fetkovich Method (future)

Calculate the future IPR curve for the following parameters: (use the

properties from the Fetkovich example)

p = 3600 psipF = 2950 psi




Multi-rate Fetkovich Method (future)




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Model Fluids Conditions

PI - Method 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

Darcy Equation Oil, Water Above pb

Composite Model Oil, Gas, Water Above and below pb

Summary




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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 pb = 7,6 MPa μo = 1,5. 10-3 PasBw = 1,04 Sw = 55 % μw = 1. 10-3 Pas

re = 200 m h = 10 m S = 0 rw = 0,083 m k = 200 mD

Deadline: 26.08.2015 09:00 Homework




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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!










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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 µ = 1,7 cp rw = 0,328 fth = 50 ft B = 1,1 re = 2500 ft









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Gas Reservoirs




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Specifics of gas:

- μ and the isothermal compressibility of real gas are highly pressuredependent

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

units

Z – Factor:

p

pc … Critical pressure of the gas (psi, Pa)

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

p … Pseudo-reduced pressure (-)

T … Pseudo-reduced temperature (-)

p … Pressure of interest (psi, Pa)

T … Temperature of interest (°R, °C)

Gas Properties




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know gas composition:

pc yi. pciNi=

Tc yi. TciNi=

unknown gas composition:

pc 709,6 58. γg

Tc 170,5 307,3. γg

y … Mole fraction (-)

γg … Gas gravity (-)

Gas Properties




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Gas Properties




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gy y

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gy y

Viscosity – Approximation:

Gas Properties

B.C.Craft, Applied Petroleum Reservoir Engineering




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gy y

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gy y

Gas Properties

B.C.Craft, Applied Petroleum Reservoir Engineering




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gy y

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gy y

Pseudo-steady State Solution:

p² p²w .

ln

,.

S

μ μ+

Z Z

+

Q … Gas flow rate (Mscf/day, at 60°F and 14,7 psi)

k … Permeability (mD)h, r … Distances (ft)

Z … Factor (-)

T … Temperature (°R = 460 + °F)

p … Pressure (psi)

μ … Viscosity (cp)

Russell and Goodrich Solution




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gy y

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gy y





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gy y

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gy y

Example

Gas IPR

Generate the IPR curve for the following gas reservoir:

γg = 0,9 p = 4000 psiT = 240°F (700°R) S = 3

k = 10 mD h = 5 ft

rw = 0,3125 ft re = 500 ft





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gy y

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gy y

Z Q

4000 - - - - 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 56840 2000 3,0 0,8 0,0177 5899





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Pseudo-steady State Solution:

“real gas pseudo pressure”: m p m pw 2. .d

m p m pw 1422. QT

khln

0,472. rerw

S

Q … Gas flow rate (Mscf/day, at 60°F and 14,7 psi)

k … Permeability (mD)h, r … Distances (ft)

Z … Factor (-)

T … Temperature (°R = 460 + °F)

p … Pressure (psi)

μ … Viscosity (cp)

Al Hussainy, Ramey and Crawford Solution




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Example

Gas IPR

Create the IPR curve and evaluate pw for a production rate of Q = 3865Mscf/day for the following gas reservoir:

γg = 0,85 p = 3600 psiT = 200°F (660°R) S = 2,5

k = 10 mD h = 5 ft

rw = 0,3125 ft re = 1000 ft





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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





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Gas Reservoirs

5 IPR Saturated Gas 15

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