# Part1 Fluids Rock Properties

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Basic Waterflooding & Surveillance Dr. Tarek Ahmed Tarek Ahmed & Associates Ltd www.TarekAhmedAssociates.com [email protected] Professor Emeritus of Petroleum Engineering Montana Tech of the University of Montana © 2006Tarek Ahmed & Associates, Ltd. All Rights Reserved.

Course Content I) Characterization of Reservoir Fluids  Understanding Fluid Laboratory Data  Quality Assurance of Fluid Data II) Reservoir Characterization  Porosity and Permeability  Capillary Pressure, Fluid Distribution, and Understanding the Transition Zone  Zonation and Layering System  Two and Three Phase Relative Permeability & Concept of permeability Jail  Normalizing Relative permeability Data  Characterizing Vertical and Areal Heterogeneities  Calculation of Sweep Efficiencies III) Waterflood Development & Surveillance  Stages of Oil Recovery  Factors to Consider When Planning a Waterflood  Introduction to Waterflood  Waterflood Surveillance IV) Reservoir Simulation  Gridding and Construction a Simulation Model  Simulation Studies  Developing of the Nameless Field with Waterflooding-Team Project

1

Characterization of Reservoir Fluids & Rocks Reservoir Fluid and Rock Characterization is a Scientific and Mathematical Discipline that Seeks to Define Quantitatively the Input Data Needed to Perform Predictions of Flow Through Porous Media and Subsequent Field Development Plans Understanding the Interaction Between the Reservoir Fluids and Formation is Vital in Optimal Design and Evaluating the Performance of Waterflooding Project . First Step for Implementing a Successful Secondary Recovery Process is the Proper Evaluation of the Reservoir Rock and Fluid Properties © 2006Tarek Ahmed & Associates, Ltd. All Rights Reserved.

Characterization and Classification of Reservoir Fluids

4

2

A Good Starting Point: What is the Saturation Pressure ?!! - Bubble Point - Dew Point Pressure Bubble of Gas

pb

pd

Liquid

Gas

T

Hg

T

Hg

Droplet of liquid

Understanding the Critical point is Essential Liquid

C

PC PC

C Liquid+Gas

Liquid

Gas

Gas TC

TC

3

Bubblepoint pressure curve 100% Liquid

P-T Diagram C

Liquid+Gas

7

Classification of Reservoirs  Oil Reservoirs  Gas Reservoirs  Near Critical

This Classification is Based on: 1- Initial Composition 2- Initial Pressure 3- Initial Temperature

4

Types of Reservoirs  Oil reservoirs  Gas reservoirs  Near Critical

Oil Reservoir

Gas Reservoir

5

Oil Reservoirs   

Undersaturated Oil Reservoir. Saturated Oil Reservoir. Gas-cap Reservoir.

1- undersaturated reservoir 2- saturated reservoir 3- Gas-Cap reservoir

1 2

3

6

Characterization of Crude Oils  Ordinary black oil  Low-shrinkage crude oil  High-shrinkage (volatile) crude oil  Near-critical crude oil

1- Ordinary Black Oil.

Figure 1-17. A typical p-T diagram for an ordinary black oil.

7

Liquid-shrinkage curve for black oil. 100%

Liquid Volume

E

F

Residual oil

0%

pb

Pressure

2- Low-shrinkage Oil. Liquid

E

critical point

pressure

Liquid+Gas

c

A G

85%

75%

Gas F

B

8

100% E

Liquid Volume

F

Residual oil

0%

pb

Pressure Oil-shrinkage curve for low-shrinkage oil..

17

   

Oil formation volume factor less than 1.2 bbl/STB Gas-oil ratio less than 200 scf/STB Oil gravity less than 35o API Black or deeply colored © 2006Tarek Ahmed & Associates, Ltd. All Rights Reserved.

3- Volatile Crude Oil

18

9

A typical liquid-shrinkage curve for a Volatile crude oil

Liquid Volume

100%

E

F

Residual oil

0%

19

pb

Pressure  Oil formation volume factor greater than 1.5 bbl/STB  Gas-oil ratios between 2,000-3,000 scf/STB  Oil gravities between 45-55o API  Greenish to orange in color

4- Near-critical Crude Oil

20

10

A typical liquid-shrinkage curve for near-critical crude oil 100%

Liquid Volume

E

F

0%

Residual oil Pressure

bubble-point pressure pb

Figure 1-24. A typical liquid-shrinkage curve for the near-critical crude oil. 21

Liquid shrinkage for crude oil systems 100%

Liquid Volume

E

0% Pressure

bubble-point pressure pb

Figure 1-25. Liquid shrinkage for crude oil systems. 22

11

Gas Reservoirs  Dry Gas Reservoirs  Wet Gas Reservoirs  Retrograde Gas Reservoirs  Near-critical Gas-condensate Reservoirs

1- Dry Gas Reservoirs

12

2- Wet Gas Reservoirs

26

13

A typical liquid dropout curve.

Pressure 2-phase region

Temperature 200 STB/MMscf

100% max. LDO

Condensate Yield STB/MMscf

% of liquid

28

0

0 STB/MMscf Pressure

14

4- Near-critical Gas-condensate Reservoirs

Liquid dropout curve for a near-critical gas-condensate system.

Liquid Volume

100%

2

3 1

0% Pressure Figure 1-29. Liquid-shrinkage (dropout) curve for a near-critical gas-condensate system.

15

Near Critical Systems critical point

T

critical point

31

32

16

On GOC pressure

depth GOC

pd = pb= pr

Figure 1-35. Determination of GOC from Pressure gradients 33

GOC

17

Depth Pressure Temperature Phase Sat Point Molecular Fluid Dens ft psia °F psia Weight lb/ft³ 7500 3727.52 169.27 Gas 3303.38 23.8 15.381 7600 3738.31 175.17 Gas 3392.63 24.3 15.567 7620 3740.48 176.35 Gas 3410.86 24.4 15.612 7650 3743.74 178.12 Gas 3438.55 24.6 15.684 7750 3754.78 184.02 Gas 3534.33 25.2 15.979 7786 3758.8 186.14 Gas 3570.23 25.4 16.109 7800 3760.37 186.97 Gas 3584.44 25.5 16.164 7850 3766.03 189.92 Gas 3636.67 25.9 16.385 7900 3771.78 192.87 Gas 3691.56 26.4 16.65 7930 3775.28 194.64 Gas 3726.19 26.7 16.837 7954 3778.1 196.06 Gas 3755.08 27 17.007 7960 3778.81 196.41 Gas 3762.49 27 17.052 7970 3780 197 Gas 3775.03 27.2 17.131 7972 3780.24 197.12 Gas 3777.57 27.2 17.148 7974.4 3780.52 197.26 Gas GOC 3780.57 27.2 17.167 7974.4 3780.52 197.26 Oil 3780.51 61.3 39.095 7980 3782.16 197.59 Oil 3772.82 61.7 39.239 8000 3787.64 198.77 Oil 3750.63 62.7 39.654 8080 3809.99 203.49 Oil 3673.85 66.2 40.999 8100 3815.7 204.67 Oil 3656.89 67 41.278

pd

GOR STO Oil Dens Scf/stb lb/ft³ 47382.8 58.616 40288.5 58.618 38991.5 58.621 37111.6 58.627 31337.5 58.667 29425.2 58.688 28701.7 58.697 26193.1 58.738 23783.9 58.789 22372.1 58.828 21251.9 58.863 20972.5 58.873 20506.2 58.889 20412.9 58.893 20303.2 58.897 2440.2 64.821 2414.7 64.86 2340.9 64.972 2111.6 65.326 2065.6 65.398

pr Pi>>>>Pb !!!!!!!!

?

Gas Cap GOC

pb Oil Rim

36

18

On Near-critical Reservoirs

37

Temperature

Depth

Gas Cap

Critical Region

Oil Rim

19

Pi>>>>Pb !!!!!!!! Gas Cap

Critical Mixture

Oil Rim

Fundamental of PVT Properties “A Short Review”

20

ON Gas Solubility Rs & Bo pb = 2000 psi

Oil 1.6 bbl

p = 1200 psi

p = 800 psi

p = 400 psi

200 scf

400 scf

700 scf

Oil 1.4 bbl

Oil 1.3 bbl

Oil 1.1 bbl

p = 14.7 psi

1000 scf

Oil 1 STB

Under-saturated Oil Properties 1.2048

 p   Rs   g   1.4 10 X    18.2 

X = 0.0125 API – 0.00091(T-460)

Rs Bo

µo

pb 0 co 1.705x10

7

pressure

0.69357 0.1885 Rsb g API 0.3272 T  4600.6729 P 0.5906

 Bo  0.9759  0.000120  R s 

 g    o

   

0.5

  1.25 T  460 

1.2

21

Crude Oil Viscosity  od

 

 1.8 10 7   360      0.32  API 1.53   T  260  

A 10

8.33 0.42  ( API

A

µod Rs

) µob pb 0

a = 10.715 (Rs + 100)-0.515 b = 5.44 (Rs + 150)-0.338

 ob  a  od b

1.6 0.56 o  ob  0.001( p  pb ) [ 0.024 ob  0.038 ob © 2006Tarek Ahmed & Associates, Ltd. All Rights Reserved.

Class Problem: Generate and plot the PVT for a 40.7 OAPI crude oil system is characterized by a bubble point pressure of 2620 psia a reservoir temperature of 220 OF. The average specific gravity of solution gas is 0.855. 1.2048

 p   Rs   g   1.4 10 X    18.2 

X = 0.0125 API – 0.00091(T-460) 1.2

   g  0.5  Bo  0.9759  0.000120  R s    1.25 T  460    o   0.69357 0.1885 co 1.705x107 Rsb g API 0.3272 T  4600.6729 P 0.5906

 od

 

 1.8 10 7   360      0.32  API 1.53   T  260  

 ob  a  od b

A

A 10

0.42  (

8.33 API

)

a = 10.715 (Rs + 100)-0.515 b = 5.44 (Rs + 150)-0.338

1.6 0.56 o  ob  0.001( p  pb ) [ 0.024 ob  0.038 ob

22

PVT Conventional & Special Laboratory Experiments  Constant Composition Expansion “CCE”  Differential Expansion “DE”  Separator Test  Constant Volume Depletion Test “CVD”  Swelling Test  Minimum Miscibility Test “MMP”  Core Flood  Flow Assurance Tests

Constant Composition Expansion Test CCE

Pb  Co

PVT above Pb Bt

23

CCE P>>Pb

P>Pb

P
P=Pb

P<
Gas

Vt

oil

Vt

oil

oil Vsat= Vt

Y Vrel 

Vt

Gas

oil

oil

p sat  p p Vrel  1

Vt



 sat Vrel

Vt Vsat

48

24

Y=a+bP

Vrel 1 

p sat  p p a  bp 

49

Class Problem The best straight fit of the Y-function as a function of pressure for the Big-Butte oil system is given by: where:

Y = a + bp a = 1.0981 ,

b = 0.000591

Smooth the recorded relative volume data of Table 4-3. Pressure

50

1936 1930 1928 1923 1918 1911 1878 1808 1709 1600 1467 1313 1161 1035 782 600 437

Measured VreL 1.0625 1.1018 1.1611 1.2504 1.3696 1.5020 1.9283 2.4960 3.4464

25

Solution:

Vrel 1  Pressure

1936 1930 1928 1923 1918 1911 1878 1808 1709 1600 1467 1313 1161 1035 782 600 437

p sat  p p a  bp 

Measured VreL

Smoothed VreL Equation 4109

1.0625 1.1018 1.1611 1.2504 1.3696 1.5020 1.9283 2.4960 3.4464

1.0014 1.0018 1.0030 1.0042 1.0058 1.0139 1.0324 1.0630 1.1028 1.1626 1.2532 1.3741 1.5091 1.9458 2.5328 3.5290

51

What is the Bubble-Point Pressure?

26

Variable Bubble-point Pressures !!!

Gas Oil

8/13/2011

 

 S gt Pore Volume   S o  Pore Volume     RS   Bo Bg     Rsnew  So  Pore Volume Bo

 S gt  Bo  RSnew  RS     S o  Bg

   

27

Important Equations N

(porevolume) (1  S wi ) Boi

Pore Volume 

So 

N Boi 1  S wi

N  N  B p

 N Boi   1  S wi

o

  

N B  S o  (1  S wi ) 1  P  o N  Boi 

Sg = 1 – So - Swi

Class Problem: The Big Butte Field is a solution gas-drive reservoir that is under consideration for a waterflood project. The volumetric calculations of the field indicate that the areal extent of the field is 1612.6 acres. The field is characterized by the following properties: Thickness h = 25 ft Porosity = 15% Initial water saturation Swi = 20% Initial pressure pi = 2377 psi Results from the MBE in terms of cumulative oil production Np as a function of reservoir pressure are given below:

Pressure

Np

Psi

MMSTB

2377

0

2250

1.10

1950

1.76

1650

2.64

1350

3.3

28

The PVT properties of the crude oil system are tabulated below: Pressure

Bo

Rs

Bg

psi

Bbl/STB

scf/STB

bbl/scf

2377

1.706

921

--

2250

1.678

872

0.00139

1950

1.555

761

0.00162

1650

1.501

657

0.00194

1350

1.448

561

0.00240

1050

1.395

467

0.00314

750

1.336

375

0.00448

450

1.279

274

0.00754

Assume that the waterflood will commence when the reservoir pressure declines to 1650 psi; find the pressure that is required to dissolve the free gas.

Differential Expansion (Liberation) Test “DE”  Amount of gas in solution as a function of pressure “Rsd”  The shrinkage in the oil volume as a function of pressure “Bod”  Density of the oil as a function of pressure “ρo”  Properties of the evolved gas including:  the composition of the liberated gas “yi”  the gas compressibility factor “Z”  the gas specific gravity

29

Gas Solubility Rs & Bo pb = 2000 psi

p = 1200 psi

p = 800 psi

p = 400 psi

200 scf

400 scf

700 scf

Oil 1.4 bbl

Oil 1.3 bbl

Oil 1.6 bbl

p = 14.7 psi

1000 scf

Oil 1.1 bbl

Oil 1 STB

DE P1=Pb

P2
P3

P2

P3

(Vg)sc (Vg)p,T

oil (Voil)p,T

Bod 

Voil VST

Vt

Voil

oil

Voil oil (Voil)p,T

Rsd 

Gas Gas (Vg)p,T

Gas

Vb

(Vg)sc

Gas

Vt

oil

remaininggas in solution VST

Bg 

(Voil)p,T

oil

(Vg ) p ,T (Vg ) sc

30

Bod 

61

VL Vsc

Btd = Bod + (Rsdb - Rsd) Bg

V p  Z   T 

 Tsc     Vsc p sc 

 p  ZT B g   sc   Tsc  p

Table 4-5 Differential Liberation Data

Bobd

Bod

62

31

Rsd

63

Rs & GOR

Same UNITS, i.e. scf/STB, BUT they are NOT the same !!!

total gas flow rate oil flow are free gas rate  solution gas rate  oil flow rate

instantane ous GOR 

instantane ous GOR 

(Qg )free  (Qg )solution (Qg )free  Qo R s  Qo Qo

 Rs 

(Qg )free k  B  Rs  rg o o Qo kro  g Bg

64

32

You can DO so many things if you have a plot of GOR versus Time; it tells you a lot !!

GOR  Rs 

scf/STB

krg o Bo kro  g Bg

GOR

Rs = GOR

krg o Bo kro  g Bg

Sg = 0 Rs

pb 65

0 Pi

Sgc

Time Pressure

GOR

Time Observed

Rs

66

Laboratory

Pressure

33

Gas Solubility Rs & Bo pb = 2000 psi

Oil 1.6 bbl

p = 1200 psi

p = 800 psi

p = 400 psi

200 scf

400 scf

700 scf

Oil 1.4 bbl

Oil 1.3 bbl

p = 14.7 psi

Oil 1.1 bbl

1000 scf

Oil 1 STB

Separator Tests

68

34

69

Adjustment of differential data to separator conditions --Why?  Rs and Bo obtained by differential liberation are not the same as Rs and Bo obtained by flash liberation  Oil leaving reservoir is flashed to separator, therefore Rs and Bo should be determined by a flash process  Flash liberation does not cover whole range of interest, therefore differential data are corrected Composite Liberation ! 70

35

71

Rsfb

Bofb

72

36

Current adjustment method-Properties At lower pressure formation volume factor Bo might read a value less than 1 At lower pressure, the solution gas-oil ratio, Rs extrapolates to negative values.

Current adjustment method does not honor density at bubble point under reservoir conditions

73

B od B B o ofb B odb Current Adjustment of Rs

B ofb R  R  (R R ) s sfb sdb sd B odb 74

37

Total Formation Volume Factor Bt Major use in MBE, Driving Indices,…etc from DE data: Bt = (Bofb) (Vrel) from CCE data: Bt = (Btd) (Bofb) / Bodb 75

76

38

B od B B o ofb B odb

77

B ofb R  R  (R R ) s sfb sdb sd B odb

Rsdb Rsfb

78

39

CCE



 sat Vrel

DE

79

80

40

81

Adjusted oil formation volume factor vs. pressure.

Class Problem The Constant-Composition Expansion test, differential liberation test, and separator test for the Big-Butte crude oil system are given next; generate and plot Bo, Rs, Bt as a function of pressure B od B B o ofb B odb

B ofb R  R  (R  R ) s sfb sdb sd B odb

from DE data: Bt = (Bofb) (Vrel) from CCE data: Bt = (Btd) (Bofb) / Bodb

82

41

42

IMPACT OF MODELING “LIQUID SHRINKAGE” FIELD EXPERIENCE “The Nameless Field”

Modeling Strategy Identifying Problems

Updateable and History Matching Data •Location of faults •Upscaling •Static Properties •…etc

Fixed and Reliable Data •Surface Separation •Fluid Properties/EOS

43

Oil Shrinkage Fixed and Reliable Data Surface Separation Fluid Properties/EOS

1st Stage 2nd Stage 3rd Stage

OOIP Y STB Stock-Tank

Updateable and History Matching Data - Location of faults - Upscaling - Static Properties - …etc

OOIP X bbl

Reservoir Characterization and initialization

How Important is Oil Shrinkage In Modeling? Degree of shrinkage is a function of separation conditions 0% Shrinkage

25% Shrinkage

50% Shrinkage

gas gas

2 STB

1.5 STB

2 bbl

2 bbl

Bo = 1 bbl/STB

Bo = 1.333 bbl/STB

Big problem in simulation Rate constraint 100 STB/day

1 STB

2 bbl

Bo = 2 bbl/STB

100 bbl 133 bbl 200 bbl

44

The Nameless Field Schematic of the Central Processing Facilities “CPF” injection gas line

Stripper Column (19 distillation trays)

80-250oc

55oc

1500 KPA

1st Stage

Oil Composition !!! 250oc

Stock-Tank

80oc 1300 KPA

2nd Stage

changes in the temperature in the flow line

Treatment of the Temperature Profile of the Distillation Column in the FFM

Temperature profile is Described by single value of 150oc

150oc 700 KPA

55oc 1500 KPA

1st Stage 80oc 1300 KPA

Stripper Column

Stock-Tank

(19 distillation trays)

2nd Stage

The study suggests that the current treatment of the Column in the FFM overestimates the liquid shrinkage; i.e. it underestimates STOIP and ST oil produced

45

Treatment of the Temperature Profile of the Distillation Column

injection gas line

150oc 700 KPA

55oc 1500 KPA

1st

Stage

Temp profile is Described by single value of 150oc

Oil Composition !!! 80oc 1300 KPA

Stock-Tank

Stripper Column (19 distillation trays)

2nd Stage

Summary of Results Stage 1 Stage 2nd Stage Stripper Stripper Stripper Stripper Stripper Stripper Stripper Stripper Stripper Flow Line Flow Line Stock-Tank st

P_r88 T P

Case 1-a T P

Case 1-b T P

Case 2 T P

55 1500 80 1300 150 700

55 78 80 85 90 95 100 110 120 140

55 80 80 85 90 95 100 110 120 130

55 80 78 85 95 100 110 115 120 130 135

1500 1300 700 700 700 700 700 700 700 700

1500 1300 700 700 700 700 700 700 700 700

1500 1300 700 700 700 700 700 700 700 700 700

150 700 150 700 48 101.3 48 101.3 15.5 101.3 15.5 101.3 15.5 101.3 15.5 101.3

T

Case 3-2 P

55 80 78 85 95 100 110 115 120 130 135 150 48 15.5

Case 5 T

Case 6 P

T

Case 7 P

T

Case 8 P

T

P

1500 1300 700 700 700 700 700 700 700 700 700 700 101.3 101.3

55 1500 80 1300 80 700 83 700 90 700 95 700 100 700 110 700 120 700 130 700

65 1500 80 1300 78 700 83 700 90 700 95 700 100 700 110 700 120 700 130 700

55 1500 80 1300 80 700 130 700

55 1500 80 1300 78 700 83 700 130 700

15.5 101.3

15.5 101.3

15.5 101.3

15.5 101.3

OIIP, KSTM3

322,849.6

328,258.9

328,165.5

336,788.5

328,503.6

337,773.1

337,725.5

334,992.5

335,290.3

Cum Oil, KSTM3 % increase in cum oil Recovery Factor

188,167.0 58.28%

191,334 1.68 58.30

190,874 1.42 58.16

196,250.0 4.30 58.27%

191,487 1.76 58.29%

196,960 4.67 58.30

196,972 4.68 58.32

195,539 3.92 58.37

195,778 4.04 58.39

Increase in OIIP 4-9 MM STm3

46

Laboratory Analysis of Gas Condensate Systems In the laboratory, a standard analysis on a gas-condensate sample consists of: 1. Recombination and analysis of separator samples 2. Measuring the pressure-volume relationship, i.e., constant Composition Expansion Test. 3. Constant-Volume Depletion Test (CVD)

93

1- Recombination of Separator Samples

94

Hydrocarbon Analyses of Separator Products and Calculated Wellstream

47

Constant-Composition Test Measures:  Dewpoint Pressure  Two-Phase Compressibility Factor Z2-phase  % of LDO  p  Z  Zd   p   V rel   d 

95

Table 4-10 Pressure-Volume Relations of Reservoir Fluid at 262° F (Constant-Composition Expansion)

1.043

96

48

QC & QA Evaluation Techniques

Consistency Criteria for Composition The Hoffman Plot Component

xi %

yi %

zi %

CO2

5.04

8.82

8.60

N2

0.08

0.77

0.73

C1

15.71

74.92

71.44

C2

6.26

8.64

8.50

C3

8.33

4.50

4.73

i-C4

2.33

0.61

0.71

n-C4

4.84

0.97

1.20

i-C5

2.81

0.24

0.39

n-C5

3.09

0.22

0.39

C6s

5.11

0.12

0.41

C7s

8.31

0.10

0.58

C8s

11.46

0.07

0.74

C9s

7.53

0.02

0.46

C10+

19.10

0.00

1.12

49

Hydrocarbon Analysis of Reservoir Fluid Sample

99

 Hoffman Plots

Ki 

yi xi

P  (1 / Tbi )  (1 / T )  Fi  log( ci )   14.7  (1 / Tbi )  (1 / Tci )  N2 C1 CO2 C2 iC4

Log Ki

0 -3

C3

nC4 nC5

Fi

iC5

0

xi %

yi %

K=y/x

CO2

5.04

8.82

1.7500

N2

0.08

0.77

C1 C2 C3 i-C4 n-C4 i-C5 n-C5 C6s C7s C8s C9s C10+

15.71 6.26 8.33 2.33 4.84 2.81 3.09 5.11 8.31 11.46 7.53 19.1

74.92 8.64 4.5 0.61 0.97 0.24 0.22 0.12 0.1 0.07 0.02 0

9.6250 4.7689 1.3802 0.5402 0.2618 0.2004 0.0854 0.0712 0.0235 0.0120 0.0061 0.0027 0.0000

Compone nt

EOS Lab

+3

50

 Hoffman Plots

Ki 

yi xi

 (1 / Tbi )  (1 / T )  P Fi  log( ci )   14.7  (1 / Tbi )  (1 / Tci ) 

DISCARD N2 CO2 C1 C2 C3

Log Ki

nC4 nC5

0 -3

Fi

iC4

iC5

+3

0

 An Example of the Hoffman Plot Component

Ki= yi/xi

xi

yi

zi

%

%

%

CO2

3.65

1.29

2.46

0.353

N2

5.74

0.2

2.95

0.035

C1

10.01

57.48

33.94

5.742

C2

5.34

15.84

10.63

2.966

C3

3.64

14.04

8.88

3.857

Ki 

i-C4

2.76

2.15

2.45

0.779

n-C4

2.76

5.3

4.04

1.92

i-C5

2.23

1.29

1.76

0.578

n-C5

2.23

1.44

1.83

0.646

C6s

1.86

0.74

1.3

0.398

C7s

1.6

0.2

0.89

0.125

C8s

1.41

0.03

0.71

0.021

C9s

1.25

0

0.62

0

C10s

50.81

0

25.2

0

yi xi

P  (1 / Tbi )  (1 / T )  Fi  log( ci )   14.7  (1 / Tbi )  (1 / Tci ) 

51

Identifying the Outliers  Saturation pressure vs. depth  C1 mole% vs. depth  C7+ mole% vs. depth  MC7+ vs. depth

103

Saturation Pressure

X True Vertical Depth

X X

X

X

X

X Outlier (removed) data

52

Mol % of Methane

X

True Vertical Depth

X

X

X

X

X

X Outlier (removed) data

Mol% of Heptanes-Plus Fraction

X

True Vertical Depth

X X X

X

106

X Outlier (removed) data

53

Molecular weight of Heptanes-Plus Fraction

X

True Vertical Depth

X X

X Outlier (removed) data

Figure 2-13. Measured C7+ Molecular Weight versus Depth

107

Oil Isothermal Compressibility Coefficient “Co”

 P < Pb

Rs

Bo R  Bg s p p

Bo

 P > Pb P

pb !!!

0.00

Log(Bod/Bodb)

-0.005

Instantaneous Co : Co = A/p

-0.010 -0.015

0.00

54

Consistency Criteria for Differential Liberation Test Data

 PVT Data From the MBE:

P&T

g

Rsd & Bod

o

Bod 

Psc & Tsc

o

62.4  o  0.0136Rsd  g

o

62.4  o same as o !!!!?

55

Recombination of Separator Samples

Density =42.42 lb/ft3

111

P Gas

zi, nt

yi, nv T Oil

56

 Material Balance Tests

zi nt  xi nL  yi nv yi nt nL xi   ( ) zi nv nv zi

yi/zi

0

Calculate GOR:

GOR 

Slope = - nL/nv

0

xi/zi

Consistency Criteria for Flash Data Sample 1: Condensate System Component

xi %

yi %

zi %

CO2

5.04

8.82

8.60

N2

0.08

0.77

0.73

C1

15.71

74.92

71.44

C2

6.26

8.64

8.50

C3

8.33

4.50

4.73

i-C4

2.33

0.61

0.71

n-C4

4.84

0.97

1.20

i-C5

2.81

0.24

0.39

n-C5

3.09

0.22

0.39

C6s

5.11

0.12

0.41

C7s

8.31

0.10

0.58

C8s

11.46

0.07

0.74

C9s

7.53

0.02

0.46

C10+

19.10

0.00

1.12

Measured Data: Liquid density = 44.94 lb/ft3 Liquid MW = 92.13 Gas Gravity = 0.7702 GOR (scf/sep bbl) = 16616.8

57

Class problem Given Measured Data: Liquid density = 44.94 lb/ft3 Liquid MW = 92.13 Gas Gravity = 0.7702 GOR (scf/sep bbl) = 16616.8

Calculate GOR

Component

xi %

yi %

CO2

5.04

8.82

8.6

N2

0.08

0.77

0.73

C1

15.71

74.92

71.44

C2

6.26

8.64

8.5

C3

8.33

4.5

4.73 0.71

zi %

i-C4

2.33

0.61

n-C4

4.84

0.97

1.2

i-C5

2.81

0.24

0.39

n-C5

3.09

0.22

0.39

C6s

5.11

0.12

0.41

C7s

8.31

0.1

0.58

C8s

11.46

0.07

0.74

C9s

7.53

0.02

0.46

C10+

19.1

0

Class Problem 2: Crude Oil System Component

xi %

yi %

zi %

CO2

3.65

1.29

2.46

N2

5.74

0.20

2.95

C1

10.01

57.48

33.94

C2

5.34

15.84

10.63

C3

3.64

14.04

8.88

i-C4

2.76

2.15

2.45

n-C4

2.76

5.30

4.04

i-C5

2.23

1.29

1.76

n-C5

2.23

1.44

1.83

C6s

1.86

0.74

1.30

C7s

1.60

0.20

0.89

C8s

1.41

0.03

0.71

C9s

1.25

0.00

0.62

C10s

55.52

0.00

27.54

Measured Data: Liquid density = 44.73 lb/ft3 Liquid MW = 160.7 Gas Gravity = 0.966 GOR (scf/sep bbl) = 572

58

Reservoir Characterization

Rock Properties There are basically two main categories of core analysis tests that are performed on core samples physical properties of reservoir rocks, these are: A. Routine core analysis Tests Porosity Permeability Saturation B. Special tests Overburden pressure Capillary pressure Relative permeability Wettability Surface and interfacial tension

59

Porosity Space Between grains

Permeability Ease to flow Between grains

Low Pressure

High Pressure

Reservoir Rock

Oil getting left behind… EOR target

Trapping mechanisms: (1) Surface tension traps oil drops (2) Less viscous water “short circuits” more viscous oil

60

Sources of Error As shown below, there are several factors that must be considered as possible sources of error in determining rock properties. These factors are: 1. Core sample may not be representative of the reservoir rock because of reservoir heterogeneity. 2. Core recovery may be incomplete. 3. Permeability of the core may be altered when it is cut, or when it is cleaned and dried in preparation for analysis. This problem is likely to occur when the rock contains reactive clays. 4. Sampling process may be biased. There is a temptation to select the best parts of the core for analysis.

Differentiate between:  Averaging Reservoir Properties  Upscaling

61

1- Porosity The porosity of a rock is a measure of the storage capacity (pore volume) that is capable of holding fluids.

=

Pore volume Bulk volume effective

absolute

a=

Bulk volume  Grain volume Bulk volume

=

Interconnected pore volume Bulk volume

PV = 43,560 (A h) 

ft3

PV = 7,758 (A h) 

bbl.

n

Arithmaticaverage :





i

i 1

n n

 ( h ) i

T hickness- weighted average :  

i

i 1 n

(h ) i

i 1

n

Areal - weighted average:



 (

i

Ai )

i 1 n

( A ) i

i 1 n

Volumetric - weighted average:  

 (

i

Ai hi )

i 1 n

( A h ) i

i

i 1

62

2- Rock Compressibility A reservoir thousands of feet underground is subjected to an overburden pressure caused by the weight of the overlying formations. Depth of the formation is the most important consideration, and a typical value of overburden pressure is approximately one psi per foot of depth.

The weight of the overburden simply applies a compressive force to the reservoir. The pressure in the rock pore spaces does not normally approach the overburden pressure. A typical pore pressure, commonly referred to as the reservoir pressure, is approximately 0.5 psi per foot of depth.

The pressure difference between overburden and internal pore pressure is referred to as the “effective overburden” pressure. During pressure depletion operations, the internal pore pressure decreases and, therefore, the “effective overburden” pressure increases. This increase causes the following effects: 1. 2.

Bulk volume of the reservoir rock is reduced. Sand grains within the pore spaces expand.

These two volume changes tend to reduce the pore space and, therefore, the porosity of the rock.

63

Types of Rock Compressibility

Rock-Matrix Compressibility

“cr” cr = -

1  Vr    V r   p T

Vr= volume of Solids

Rock-Bulk Compressibility

Pore Compressibility

“cB”

“cP or cf”

cB = -

1  V B    V B   p T

VB= Bulk Volume

cP  c f =

1  V P    VP   p T

VP= Pore Volume

cB≈ 0

cr≈ 0

cf =

1  VP 1  VP  VP  p VP  p

 VP  c f VP  p Example: Calculate the reduction in the pore volume of a reservoir due to a pressure drop of 10 psi. The reservoir original pore volume is one million barrels with an estimated formation compressibility of 10x10-6 psi-1

 VP  c f VP  p V p = (10x10- 6 ) (1x106 ) (10)= 100 bbl Hall's Correlation :  1.782  c f   0.428 106   © 2006Tarek Ahmed & Associates, Ltd. All Rights Reserved.

64

1  cf =  p

1 c f p  ( ) 

 =  o eC f (P - Po )

p

cf

p 

po

e x 1  x 



 

 ) o

c f ( p  po )  ln(

o

x2 x3   ... 2! 3!

   o [1  c f ( p  po ) ]

Example Given the following data: cf = 10x 10-6 original pressure = 5000 psi original porosity = 18% current pressure = 4500 psi

Calculate the porosity at 4500 psi.

 o [1 c f ( p  po ) ]   0.18[1  (10x106 ) (4500 5000) ]  0.179

65

3- Saturation Saturation is defined as that fraction, or percent, of the pore volume occupied by a particular fluid (oil, gas, or water). This property is expressed mathematically by the following relationship.

fluid saturation=

So =

volume of oil pore volume

& Sg=

totalvolumeof thefluid pore volume

volume of gas pore volume

& Sw=

volume of water pore volume

Sg + So + Sw = 1.0 Question: Soc or Sor are they the same????? © 2006Tarek Ahmed & Associates, Ltd. All Rights Reserved.

a) Critical saturation, Soc, Sgc, Swc b) Residual oil saturation and Trapped gas saturation, Sorw, Sorg, Sgt . How about trapped gas saturation? WAG? c) Movable oil saturation, Som & Sorm Som = 1- Swc- Soc Sorm is the Residual Oil Saturation to Miscible Flood; usually <0.10

66

Averaging the Saturation n

hS i

( S o ) avg 

oi

i=1 n

h

i

i=1

Averaging the Saturation Averaging Reservoir/Core Data

Averaging Core Data

n

n

So=

  i hi S o

i

So =

&

i=1 n

  i hi





n

n



i hi S w i

&

i=1

n

Sw =



  i hi

i=1

n

  i hi

Ai hi S wi



i

Ai hi

i=1

n

n

  i hi S g

i

i=1 n

i=1

Sg=

Ai hi

i

i=1

i=1

Sw =

Ai hi S oi

i

i=1 n

i

&

i=1

Sg=



i

Ai hi S g i

i=1 n



i

Ai hi

i=1

67

Class Problem Using the following data core samples; Calculate  Average porosity Average oil and connate water saturation

Sample 1 2 3 4 5 6

hi , ft  , % 1.0 10 1.5 12 1.0 11 2.0 13 2.1 14 1.1 10

So , % Swc , % 75 25 77 23 79 21 74 26 78 22 75 25

4- Permeability Permeability is a property of the porous medium that measures the capacity and ability of the formation to transmit fluids.

1 dp  dL k dp v=  dL v -

68

Absolute Permeability “k”  Absolute permeability is a property of the porous medium that

measures the capacity and ability of the formation to transmit fluids.  kx, ky , and kz !!!!!!!!!!!!!! P2

P1

A

q

q L

 A ( p1  p2 )  q    L    A ( p1  p2 )  q  k  L  

Permeability Tensor  K xx    K yx     K zx  

K xy K yy K zy

K xz    K yz     K zz   

69

P2

P1

A

q

q L

Liquid

Gas

A p qk( ) L

k=

k A ( p12 - p 22 ) Q gsc = 2 g L

qL A p

k=

2  g L Q gsc A ( p12 - p 22 )

WHY?

70

Effect of gas pressure on measured permeability for various gases.

1 1  0 pm 

k g = k L + (b k L)[

1 ] pm

IF one measurement with air:

6.9 k 0.64 L + pm k L - pm k g = 0 141 © 2006Tarek Ahmed & Associates, Ltd. All Rights Reserved.

Class Problem: A brine is used to measure the absolute permeability of a core plug. The rock sample is 4 cm long and 3 cm2 in cross section. The brine has a viscosity of 1.0 cp and is flowing a constant rate of 0.5 cm3/sec under a 2.0 atm pressure differential. Calculate the absolute permeability.

Class Problem: Rework the above example assuming that an oil of 2.0 cp is used to measure the permeability. Under the same differential pressure, the flow rate is 0.25 cm3/sec.

71

Averaging Permeability

143

n

hj

 (k h ) i

k

i

i 1 n

(h ) i

i 1

Li

n

( L ) i

k

i 1

n

[( L /( k )] i

i

i 1

1

k avg = ( k 1 k 2 k 3 ... k n )n

72

Parallel Layers

n

 (k h ) i

T hickness- weighted average: k 

i

i 1 n

( h ) i

i 1

Class Problem: Given the following permeability data from a core analysis report, calculate the average permeability of the reservoir.

Depth, ft

Permeability, md

3998-02 4002-04 4004-06 4006-08 4008-10

200 130 170 180 140

73

Parallel Layers

n

 (k w h ) i

wighted average: k 

i

i

i 1 n

( w h ) i

i

i 1

Series beds

n

( L / A ) i

Harmonicaverage :

k

i

i 1

n

[( L /( A k )] i

i

i

i 1

74

k avg =

l n ( re / r w ) l n ( r j / r j -1 ) [ ] k j j=1 n

Class Problem A hydrocarbon reservoir is characterized by five distinct formation segments that are connecting in series. Each segment has the same formation thickness. The length and permeability of each section of the five-bed reservoir are given below:

Length, ft 150 200 300 500 200

Permeability, md 80 50 30 20 10

75

Heterogeneous Systems  n   [ hi ln(ki ]   k  exp i 1 n   hi   i 1  

Geometric average:

1 n n

k avg = ( k 1 k 2 k 3 ... k )

If hi is constant:

Class Problem: Given the following core data, please calculate the geometric average permeability.

Sample 1 2 3 4 5 6 7 8 9 10

hi, ft 1.0 1.0 0.5 1.5 2.0 1.5 1.0 1.0 1.5 0.5

ki, md 10 30 100 40 80 70 15 50 35 20

76

Absolute Permeability Correlations 1- Timur equation

k = 8.58102

 4.4 2

S wc

2- Morris and Biggs

For an oil reservoir:

k = 62.5(

3 S wc

)2

For a gas reservoir:

k = 2.5(

3 S wc

)2

3- Kozeny’s equation

k=

153

m S wi

(

 2 ) 1 

Cementation Factor “m” m= 1.3 for unconsolidated rocks and Limestone m= 1.4-1.5 for VERY slightly cemented sands m= 1.6-1.7 for slightly cemented sands m= 1.8-1.9 for highly consolidated sands with Φ <15% m= 2.0-2.2 for low Φ sands; dolomite and chlak

Class problem: Estimate the absolute permeability of an oil zone that is characterized with a connate water saturation of 25% and an average porosity of 15%

77

UPSCALING (Averaging !!!!)

well2

Fine-Scale 173-Layer System well1

Permeability

78

Coarse Scale 19-Layer System

well2

well1

Permeability

19Layers Coarse System vs. 173 Layers Fine System TAGI Formation- Oil Rate Coarse-Scale, RC/cell

Coarse-Scale, RC/Layer

Fine-Scale

250

What is causing this difference ?

OIL PRODUCTION RATE (STM3 / DAY)

200

150 Fine-Scale

Variable RC/cell or by layer Identical performance for 32 years

100

50

0

0

1000 2000 3000 4000 5000 6000 7000 8000 9000 10000 11000 12000 13000 14000

79

19Layers Coarse System vs. 173 Layers Fine System TAGI Formation - Cum Oil Prod 19 Layers System, RC/cell

19 Layers System, RC/Layer

Fine-Scale

1600

CUMULATIVE OIL PRODUCTION (KSTM3)

1400 1200 1000 800

Fine-Scale

600 400 200 0

? 0

1000

2000

3000

4000

5000

6000

7000

8000

9000

10000 11000 12000 13000 14000

5- Wettability Wettability is defined as the tendency of one fluid to spread on or adhere to a solid surface in the presence of other immiscible fluids.

Oil

Water

Water Wet

Oil Wet

80

Wettability The tendency of a liquid to spread over the surface of a solid is an indication of the wetting characteristics of the liquid for the solid. This spreading tendency can be expressed more conveniently by measuring the angle of contact at the liquid-solid surface. “Contact Angle ϴ.” mercury retains a spherical shape oil droplet develops an approximately hemispherical shape water tends to spread over the glass surface.

The performance of a waterflood is controlled to a large extend by the wettability. © 2006Tarek Ahmed & Associates, Ltd. All Rights Reserved.

81

6- Surface and Interfacial Tension Forces acting on the interface between two immiscible fluids: surface tension Interfacial tension.

82

7- Capillary Pressure When two immiscible fluids are in contact, a discontinuity in pressure exists between the two fluids. The pressure difference between these two fluids across the interface is called the Capillary Pressure and is referred to by “pc.“ pc = pressure on the non-wetting – pressure of the wetting phase

pcwo = po -pw pcgo = pg- po pcgw = pg- pw

Oil

Pc

If all the three phases are continuous, then pcgw = pcgo + pcgo Water © 2006Tarek Ahmed & Associates, Ltd. All Rights Reserved.

83

h= Oil

2 (cos )  ow g r ( w   o) gm/cm3

pc =

pc

2  ow ( cos  ) r

Pc = (

h ) ( w   o ) 144 lb/ft3

Pc

Porous Plate

84

Porous Plate Most accurate

Others: 1- Centrifuge Technique 2- Mercury Injection

85

Pore Throat Size Distribution  2  ow cos( )  Pc    r    2  ow cos( )   r   Pc  

Pc

In the absence of measured values: for oil - water system : r 

8.7 Pc

for gas - water system: r 

10 Pc

0

r  pore radius in mincron p c  capillary pressure, psi

Sw

100%

rmax

Height of the fluid contact above the FWL is given by:  2  ow cos( )   hWOC    rmax  g 

86

Converting Laboratory Capillary Pressure Data

( pc) res = ( pc) Lab

 res  res k core  lab  core k res

Only Oil Flowing

After Correcting Pc

Only Oil

 h  Pc      144  or : Oil + Water

 144 Pc h    

  

Only water

87

Displacement pressure

WOC !!!!

1-Soc

88

 h   h  Pc    (  NW  W )      144   144  or :  144 Pc h    

h

  

Transition zone

WOC

0

Swc

FWL = WOC +

Sw

FWL

144 p d 

177

Class Problem The reservoir capillary pressure-saturation data of the Big Butte Oil reservoir as represent by an average permeability of 30 md is shown in next slide. Geophysical log interpretations and core analysis establish the WOC at 5023 ft. The following additional data are available. Oil density = 43.5 lb/ft3 Water density = 64.1 lb/ft3 Interfacial tension = 50 dynes/cm Please calculate: a) Connate Water Saturation (Swc) b) Depth to FWL c) Water saturation profile as a function of depth d) Thickness of the transition zone

89

h

1 ( k  )

90

Variable WOC Different HFU; Layers, Reservoirs,..etc

k1 k2 k3 k4

91

Important Class Problem A Five-layer oil reservoir is characterized by a set of five reservoir capillary pressure-saturation curves, as shown next. The following additional data are also available Layer

Depth, ft 1 2 3 4 5

4600-4700 4700-4750 4750-4850 4850-4950 4950-5000

Permeability, md 300 3 30 100 10

FWL = 5000 ft Water density = 65.2 llb/ft3 Oil density = 55.2 lb/ft3 Please calculate and plot water saturation versus depth for this reservoir

92

Class Problem A four-layer oil reservoir is characterized by a set of four reservoir capillary pressure-saturation curves, as shown next. The following additional data are also available Layer

Depth, ft 1 2 3 4

4000-4010 4010-4020 4020-4035 4035-4060

Permeability, md 80 100 70 90

WOC = 4060 ft Water density = 65.2 llb/ft3 Oil density = 55.2 lb/ft3 Please calculate and plot water saturation versus depth for this reservoir

93

Capillary Pressure Hystresis

Oil+Water

Only Oil

Only Water

94

Useful Capillary Pressure Applications

95

The key to characterized reservoir description is: Pore Size Distribution Index “λ” 0.00084k 0.5498 pd0.9930   1.0485  (1  S wi ) 2.2790

define :

S w*



   S 

Pc  k rw

S w  S wi  1  S wi  S or 1 *  pd S w * w

 ln(S w* ) ln( pc / pd )

Proper way of estimating λ: from curve-fit of pc vs. Sw

23

k rnw  (1  S w* ) 2

 * 1 S w 

2 

 

  

96

97

Normalizing Capillary Pressure Data The Leverett J-Function Leverett defined the dimensionless function of saturation, which he called the Jfunction, as

J ( S w ) = 0.21645 J(Sw) Pc σ k

pc

k

= J-Function = Capillary pressure, psi =Interfacial tension, dynes/cm =Permeability, md =Fractional porosity

98

J-Function can be used as Hydraulic Flow Unit Indicator Or for Zonation

J ( S w ) = 0.21645

pc

k

Class Problem A laboratory capillary pressure test was conducted on a core sample taken from the Nameless Field. The core has a porosity and permeability of 16% and 80 md, respectively. The capillary pressure-saturation data are given below. Sw 1.0 0.8 0.6 0.4 0.2

pc, psi 0.50 0.60 0.75 1.05 1.75

The interfacial tension is measured at 50 dynes/cm. Further reservoir engineering analysis indicated that the reservoir is better described at a porosity value of 19% and an absolute permeability of 120 md. Generate the capillary pressure data for the reservoir.

99

Class problem: Using the capillary pressure data as tabulated on the next slide; generate the capillary pressure data for a reservoir layer as characterized by the flowing properties: Permeability = 72 md Porosity = 19.1 % Interfacial tension = 28 dynes/cm Contact angle = 0o

Core 1 K=11.2 md ᶲ = 0.147

Core 3 K=157 md ᶲ = 0.208

Core 2 K=34 md ᶲ = 0.174

Core 4 K=569 md ᶲ = 0.275

Sw, %

Pc

Sw, %

Pc

Sw, %

Pc

Sw, %

Pc

100

2.15*

100

1.60*

100

0.93*

100

0.60*

87

2.27

82

1.79

79

1.00

65

0.67

74

2.59

54

2.28

60

1.19

47

0.87

60

2.95

43

3.25

42

1.77

40

1.15

54

3.50

35

4.91

31

3.28

33

1.63

47

3.85

29

6.55

23

5.46

29

2.31

41

4.82

28

7.92

22

7.93

22

2.80

37

5.69

27

10.25

21

10.25

19

3.59

34

7.45

16

6.52

32

10.25

16

10/25

Interfacial tension = 70 dynes/cm Contact angle = 0o

100

8- Relative Permeability

101

Water-oil relative permeability curves.

(Krw )Soc

203

Soc =20%

Gas-oil relative permeability curves.

204

102

Original Corey’s Correlation  Developed for the drainage cycle in water-wet sandstones  Applies ONLY to well-sorted homogenous rocks

kro  kro Swc

 1  S w  Sorw    1  S wc  Sorw 

 S g  S gc    Swc 1 S Lc  S gc 

 

k rg  k rg

no

 S S  krw  krw Sorw  w wc  1 S wc  Sorw 

ng

k ro  k ro Sgc

 1 S g  S Lc    1 S gc  S Lc 

nw

ngo

Accounting for Stratification Stratification Factor “m” kro  kro Swc krw

 1  S w  Sorw    1  S wc  Sorw 

 S S   krw Sorw  w wc  1 S wc  Sorw   S g  S gc    Swc 1 S Lc  S gc 

 

k rg  k rg

ng

 1 S g  S Lc  k ro  k ro Sgc   1 S gc  S Lc 

no

kro  kro Swc m S w*

nw

krw  krw Sorw m S w*

no

nw

  m S 

k rg  krg ngo

* ng g

Swc

k ro  k ro Sgc m S g*

ngo

103

104

Relative history Permeability Hysteresis

(Hysteresis describes the

of the saturation changes)

Drainage Process Oil core 100% H2O Kro Krw Imbibition Process H2O

Oil + residual H2O

0% 100%

210

SW SO

100% 0%

Soc & Sorw !!!!!

105

Gas-oil relative permeability curves.

Sgc & Sgr , Trapped Gas!!!!! 211

k ro /krw as a function of saturation.

kro  a ebSw krw

106

One of the most practical applications of the MBE is its ability to generate the field relative permeability ratio as a function of gas saturation that can be used to adjust the laboratory core relative permeability data. The main advantage of the field or well generated relative permeability ratio is that it incorporates some of the complexities of reservoir heterogeneity and degree of the segregation of the oil and the evolved gas.

8/13/2011

N = 7758 A h φ (1 – Swi)/Boi So 

N  N  B p

o

 N Boi     1 S wi 

N p  Bo   So  1 S wi  1  N  Boi  Sg = 1 – So - Swi

krg WOR 

k rw   o Bo    k ro   w Bw 

 B  a e b S w  WOR  w w    o Bo    B  WOR  a e b S w / ( w w )   o Bo  

kro

  g Bg  GOR  Rs    o Bo

  

  g Bg   a e b So  GOR  Rs     o Bo   g Bg   GOR  Rs  a e b So / ( )  o Bo  

8/13/2011

107

Class Problem: A volumetric undersaturated oil reservoir has a bubble-point pressure of 4500. The initial reservoir pressure is 7150 psia and the volumetric calculations indicate the reservoir contains 750MM STB of oil initially in place. The field is a tight naturally fractured chalk reservoir and was developed without pressure support by water injection. The following additional data is available: Swi = 43% , Bw = 1.0 bbl/STB

cw = 3.00x10-6 psi-1 , pb = 4500 psi

,

cf = 3.3x10-6 psi-1

The field production and PVT data are summarized below:

p psia

Qo STB/day

7150 6600 5800 4950 4500 4350 4060 3840 3600 3480 3260 3100 2940 2800

44230 79326 75726 70208 50416 35227 26027 27452 20975 15753 14268 13819

Qg Bo MMscf/day bbl/STB 64.110 115.616 110.192 134.685 147.414 135.282 115.277 151.167 141.326 125.107 116.970 111.792

1.743 1.760 1.796 1.830 1.850 1.775 1.670 1.611 1.566 1.523 1.474 1.440 1.409 1.382

8/13/2011

Rs Scf/STB

Bg BBL/scf

μo/ μg

Np MMSTB

RP Scf/STB

1450 1450 1450 1450 1450 1323 1143 1037 958 882 791 734 682 637

0.000797 0.000840 0.000881 0.000916 0.000959 0.001015 0.001065 0.001121 0.001170

5.60 6.02 7.24 8.17 9.35 9.95 11.1 11.9 12.8 13.5

0 8.072 22.549 36.369 43.473 49.182 58.383 64.812 69.562 74.572 78.400 81.275 83.879 86.401

1450 1450 1455 1455 1447 1576 1788 1992 2158 2383 2596 2785 2953 3103

8/13/2011

108

3- Phase Relative Permeability kro

Sg krg

krw

So kro Sw

Sw krw krg

kro

Sg

 k    krog  kro  kro Swc  row  krw    krg   krw  krg   kro Swc    kro Swc 

217

Data and Observations Rock Properties Critical water saturation in low-perm rocks

Typical reservoir rock

109

Poor quality reservoir rock

Poor quality reservoir rock

Critical Saturations increase

Critical Saturations increase

Worse quality reservoir rock

110

Data and Observations Rock Properties All show and no flow!!

Worse quality reservoir rock

“Permeability Jail”

Near-Well Upscaling ( K rf ) I , J 1/ 2 ,K  coarse scale

I

 f qIC, J 1,K TI , J 1/ 2 , K ( PI , J 1,K  PI , J , K )

J-1 I-1

fine scale

I+1

West

J

East

J+1

South Figure 1. Diagram to illustrate the near-well upscaling method

111

Near-Well Approach f q q k   k f w

w

t

t

t

t

k ro   o (1  f w ) k rw   w f w

t k

t k

Analytical representations for individual-phase relative permeabilities are commonly used in numerical simulators. The most frequently used functional forms used for relative-permeability and capillary-pressure data are given below: Oil-Water Systems:  1  S w  Sorw  kro  kro Swc   1  S wc  Sorw 

no

krw

 S S   krw Sorw  w wc  1  S  S wc orw  

nw

 1 S w  Sorw   Pcwo  Pc Swc   1 S wc  Sorw 

np

Gas-Oil Systems:  S g  S gc  krg  krg Swc   1 Slc  S gc 

ng

 1 S g  Slc  kro  kro Sgc   1 S gc  Slc 

ngo

 S g  S gc  Pcgo  Pc Slc   1 Slc  S gc 

n pg

SLc = Swc + Sorg where: SLc = Total critical liquid saturation no, nw, ng, ngo = Exponents on relative permeability curves np = Exponent of the capillary pressure curve in gas-oil system (Pc)Sl = Capillary pressure at critical liquid saturation. © 2006Tarek Ahmed & Associates, Ltd. All Rights Reserved.

112

Class Problem. Generate the relative permeability and capillary pressure data using the following information on the water-oil and gas-oil systems: Swc = 0.25,

Sorw = 0.35,

Sgc = 0.05,

(kro)Swc= 0.85,

(krw)Sorw = 0.4,

(Pc)Swc= 20 psi

kro)Sgc= 0.60,

(krg)Swc

no = 0.9,

nw = 1.5,

(Pc)slc

Sorg = 0.23

= 0.95 np= 0.71

,ngo= 1.2,ng= 0.6 ,

= 30 psi, npg= 0.51

Analytical Expressions So* 

So 1  S wc

S w* 

S w  S wc 1  S wc

S g* 

Sg 1  S wc

Drainage Oil-Water Relative Permeabilities Type of formation kro krw Equati on Unconsolidated sand, 1  S * * 3 Sw w well sorted (5-4) Unconsolidated sand, 1 S * 2 1  S *1.5  * 3.5 w w Sw poorly sorted (5-5) 2 4 Cemented sandstone, * *2 * 1 S w 1 S w Sw oolitic limestone (5-6)



     

113

Drainage Gas-Oil Relative Permeabilities Type of formation kro krg

S 

Unconsolidated sand, well sorted Unconsolidated sand, poorly sorted Cemented Sandstone, oolitic limestone, rocks with vugular porosity

Equatio n

1 S 

* 3 o

* 3 o

(5-7)

S 

1 S  1 S 

(5-8)

S 

1  S  1 S 

(5-9)

* 3.5 o

* 2 o

* 4 o

* 2 o

*1.5 o

*2 o

Relative Permeability from Capillary Pressure Data 1- Oil-Water System k ro

 1 S w      1  S wc 

2

 

 S  S wc   k rw   w  1  S wc 

1

Sw 1

(1 / Pc2 )dSw

Swc

2

(1 / Pc2 )dSw

 

Sw

Swc 1 Swc

(1 Pc2 )dSw

Pc

(1 Pc2 )dSw

Sw

2- Gas-Oil System  S  S or   k ro   o  1  S or 

k rg

2

So

 (1 P )dS  (1 P )dS

 S  S or   1  o   S g  S gc  

2 c

o 1 o

2

o

2 c

1

1/P2c

o

 (1 P )dS  (1 P )dS So 1 o

2 c

o

2 c

o

Sw

114

Normalization and Averaging Relative Permeability Data Results of relative permeability tests performed on several core samples of a reservoir rock often vary. Therefore, it is necessary to average the relative permeability data obtained on individual rock samples. Prior to usage for oil recovery prediction, the relative permeability curves should first be normalized to remove the effect of different initial water and critical oil saturations. The relative permeability can then be de-normalized and assigned to different regions of the reservoir based on the existing critical fluid saturation for each reservoir region.

To perform the normalization procedure, it is helpful to set up the calculation steps for each core sample i in a tabulated form as shown below:

Relative Permeability Data for Core Sample i (1) (2) (3) (4) (5) (6) k rw k ro * * S w  S wc * k ro  k rw  Sw k ro @ S wc 1 S wc  Soc k rw @ S oc Sw kro krw The following normalization methodology describes the necessary steps for a water-oil system as outlined in the above table. Step 1. Select several values of Sw starting at Swc (column 1), and list the corresponding values of kro and krw in columns 2 and 3.

115

Relative Permeability Data for Core Sample i (1) (2) (3) (4) (5) (6) k rw k ro * * S  S wc k ro  k rw  S *w  w k ro @ S wc 1 S wc  Soc k rw @ S oc Sw kro krw Step 2. Calculate the normalized water saturation S*w for each set of relative permeability curves and lists the calculated values in column 4 by using the following expression: S *w 

S w  S wc 1 S wc  Soc

Step 3. Calculate the normalized relative permeability for the oil and water phase at different water saturations by using the following relationships and list results in column 5 and 6:

k ro* 

k ro

k ro Swc

* krw 

krw krw  Soc

Relative Permeability Data for Core Sample i (1) (2) (3) (4) (5) (6) k rw k ro * * S w  S wc * k ro  k rw  Sw k @ S 1 S wc  Soc k rw @ S oc Sw kro krw ro wc Step 4. Using regular Cartesian coordinate, plot the normalized k*ro and k*rw versus S*w for ALL core samples on the same graph.

116

S*w Step 5. Determine the average normalized relative permeability values for oil and water as a function of the normalized water saturation by select arbitrary values of S*w and calculate the average of k*ro and k*rw by applying the following relationships:

 h k k  n

k 

* ro avg

* ro i

i 1

n

 h k i

 h k k  n

k 

* rw avg

* rw i

i 1

n

 h k  i 1

i 1

i

where n= Total number of core samples hi= Thickness of sample i ki= Absolute permeability of sample i

Step 6. The last step in this methodology involves de-normalizing the average curve to reflect actual reservoir and conditions of Swc and Soc. These parameters are the most critical part of the methodology and, therefore, a major effort should be spent in determining representative values. When representative critical saturations have been estimated, it is again convenient to perform the de-normalization calculations in a tabular form as illustrated below:

(1)

(2)

(3)

(4)

S*w (k*ro)avg (k*rw)avg

(5)

(6)

 

* Sw  Sw* 1  Swc  Soc   Swc kro  (kro )avg k ro

Swc

 

* krw  (krw ) avg k rw

Soc

Where average (kro)Swc and average (krw)Soc are the average relative permeability of oil and water at connate water and critical oil, respectively, and n n given by: h k krw Soc i h k k ro Swc i k rw Soc i 1 n k ro Swc  i 1 n h k i h k 

 

 i 1

 

i

 i 1

117

The Swc and Soc are usually determined by averaging the core data, log analysis, or correlations. Versus graphs, such as: (kro)Swc vs. Swc, (krw)Soc vs. Soc, and Soc vs. Swc should be constructed to determine if a significant correlation exists. Often plots of Swc and Sor versus log k  may demonstrate a reliable correlation to determine end-point saturations as shown schematically below :

Class Problem Relative permeability measurements are made on three core samples. The measured data are summarized below:

Sw 0.20 0.25 0.30 0.40 0.50 0.60 0.65 0.72

Core Sample #1

Core Sample #2

Core Sample #3

h = 1ft k = 100 md Soc = 0.35 Swc = 0.25

h = 1ft k = 80 md Soc = 0.28 Swc = 0.30

h=1 k = 150 md Soc = 0.35 Swc = 0.20

kro 0.850* 0.754 0.557 0.352 0.131 0.000 -

krw 0.000 0.018 0.092 0.198 0.327 0.400* -

kro 0.800 0.593 0.393 0.202 0.111 0.000

krw 0 0.077 0.191 0.323 0.394 0.500*

kro 1.000* 0.872 0.839 0.663 0.463 0.215 0.000 -

krw 0.000 0.008 0.027 0.088 0.176 0.286 0.350* -

*values at critical saturations

It is believed that a connate water saturation of 0.27 and a critical oil saturation of 30% better describe the formation. Generate the oil and water relative permeability data using the new critical saturations.

118

UPSCALING

Absolute Permeability Upscaling  Arithmetic (Volumetric) Average  Root-Mean-Square “RMS” Average  Tensor Averaging

 Renormalization (on-going)

238

119

Power Averaging Procedures 1/ 

1  (ki ) coarse    [ki ]fine dV   Vb V  b   (ki ) coarse  upscaled permeabili ty [ki ]fine

 fine - scale permeabili ty

Vb

 coarse block bulk volume

 power average exponent !!! Choice of  is very tedious

ω lies between -1 and 1: ω = 1; Arithmetic average (parallel flow) ω = -1; harmonic average (Series flow) ω = 0; geometric average (kw is replaced by ln(k)) ω = 1/3; average k will lie between geometric and arithmetic averages ω = 0.57, for horizontal flow in shale-sand environments ω = 0.12, vertical flow 239

Root Mean Square “RMS”

n

(k IJK ) coarse 

m

z

 (k i

j

ijk

) 2fine

k

nm z

240

120

Permeability Tensor Permeability tensors allow a more rigorous representation of permeability. The method uses the numerical solution of the PDE that governs single-phase steady state flow conditions to calculate kxx, kyy,and kzz. The equation for the system consisting of fine-scale blocks that fall inside the coarse black is:  p  p  p (k xx )  (k yy )  (k zz )  0 x x y y z z

This equation is solved with constant pressure conditions at inlet and outlet faces, which are normal to the direction in which coarse-scale permeability is being calculated; with closed boundaries in the other directions; k1

k2

. . . .

k3

p ( x  0)  p o

k4

p ( x  L)  p L p y p z

dp/dy = 0

0 y 0, Ly

0 z  0 , Lz

The numerical solutions to the above equations provides the pressure distribution And flow rates in the fine scale system; with coarse grid permeability as given by: 241

Kxx = µ Lx (ΣQ)fine) / [ A (po – pL)]

Renormalization The Method uses the analogy between flow in Porous media and electric circuits to calculate Kcoarse. The approach based on averaging the permeability over small regions (2x2x2 of the fine-scale block) to form a new “average permeability” with the equation as derived from circuit analogs.

k1

k2

k3

k4

KA 

KA

KB

KC

KD

KCoarse

4 A (k1  k 3 ) (k 2  k 4 ) 3 B  (k1  k 2  k 3  k 4 ) A

A  k 2 k 4 (k1  k 3 )  k1 k 3 (k 2  k 4 ) B  (k1  k 2 ) (k 3  k 4 ) (k 1  k 3 ) (k 2  k 4 )

242

121

Identifying Hydraulic Flow Units & Zonation

Prediction of Hydraulic Flow Units “HFU” HFU is commonly characterized by the following three functions:

1) Reservoir quality Index (RQI) :

k 

RQI  0.0314*

2) Pore volume to grain volume ratio

   z    1    3) Flow Zone Indicator (FZI) : FZI 

RQI z

Linear equation with a unit slope

log RQI  logz  log FZI

122

Proposed Approach for Defining HFU: The proposed approach is based on utilizing to all available core data and determining : calculate RQI and ᶲz for each sample core data Plot RQI vs. ᶲz on loglog-log coordinates. The plotting data will form a straight line with a 45o degree angle (unit slope) if FZI is constant for all samples . Data samples with similar but not identical FZI values will be located around the single unit slope straight line. Samples with significantly different FZI will for another parallel unitunit-slope straight line Each line is a FLOW UNIT that is characterized with an average FZI FZI.. The average FZI for for each FU is the intercept of a unitunit-slope straight line with the coordinates ᶲz =1 Permeability of a sample point can be calculated from a pertinent HFU using the average FZI values and the corresponding sample porosity:

k 1014( FZI ) 3

3 (1   ) 2

Log-log X (FZI)1

RQI

X (FZI)2

X (FZI)3 45o

z

1

123

Identifying Hydraulic Flow Units 

RQI vs PHIZ z Ourhoud Field Wells 10.00 HFU12 HFU11 HFU10 HFU9 HFU8 HFU7 HFU6 HFU5 HFU4 HFU3 HFU2 HFU1

RQI

1.00

0.10

0.01 0.01

0.10 PHIz

z

1.00

Class Problem: The following tabulated data shows properties of core plugs from different field locations. Determine the numbers of HFU .

124

Solution:

The fitting parameters a&b of the straight for the Nameless Field:

log(S w )  log(a)  log[ J (S w ) b ] Sw 

a J (S w )b

Sw 

0.207 J ( S w ) 0.5852

125

Applied Mathematical Manipulations for each HFU Sw  Sw=

Sw=

a J (S w )b

J ( S w ) = 0.21645

pc

k

a  pc 0.21645   a

k  

 h  0.21645 144 

b

pc = (

k  

b

h )  144

Sw above FWL

FOIL Function The FOIL function (Cuddy and co-workers(1993) proposed a simple mathematical function that relates the Bulk Water Volume “BWV” to the height “h” above the free water level by using 2-constant mathematical expression:

A hB A 1 Sw  B h 

BWV   S w 

A & B are fitting parameters Independed of k

It should be noted that a log-log plot of the J(Sw) vs. Sw (as shown on next slide) will yield a linear relationship of the form:

S w  a J (S w )

b

log(S w )  log(a)  log[ J (S w ) b ]

126

The fitting parameters a&b of the straight for the Nameless Field:

log(S w )  log(a)  log[ J (S w ) b ] Sw 

a J (S w )b

Sw 

0.207 J ( S w ) 0.5852

Reservoir Heterogeneity The reservoir heterogeneity is then defined as a variation in reservoir properties as a function of space. Ideally, if the reservoir is homogeneous, measuring a reservoir property at any location will allow us to fully describe the reservoir. The task of reservoir description is very simple for homogeneous reservoirs. On the other hand, if the reservoir is heterogeneous, the reservoir properties vary as a function of a spatial location. These properties may include permeability, porosity, thickness, saturation, faults and fractures, rock facies and rock characteristics. For a proper reservoir description, we need to predict the variation in these reservoir properties as a function of spatial locations. There are essentially two types of heterogeneity: Vertical Heterogeneity Areal Heterogeneity

127

Reservoir Heterogeneity •Vertical Heterogeneity Dykstra and Parsons permeability variation “V” Lorenz coefficient “L” •Areal Heterogeneity The Polygon Method The Inverse Distance Method The Inverse Distance Squared Method Triangulation Method Delaunay Triangulation

255

Vertical Heterogeneity Perhaps the area of the greatest uncertainty in designing a waterflood is the quantitative knowledge of the permeability variation within the reservoir. The degree of permeability variation is considered by far the most significant parameter influencing the vertical sweep efficiency. To calculate the vertical sweep efficiency, the engineer must be able to address the following three problems: 1. How to describe and define the permeability variation in mathematical terms 2. How to determine the minimum number of layers that are sufficient to model the performance of the fluid 3. How to assign the proper average rock properties for each layer (called the Zonation Problem) 8/13/2011

128

It is appropriate to be able to describe the degree of heterogeneity within a particular system in quantitative terms. The degree of homogeneity of a reservoir property is a number that characterizes the departure from uniformity or constancy of that particular measured property through the thickness of reservoir. A formation is said to have a uniformity coefficient of zero in a specified property when that property is constant throughout the formation thickness. A completely heterogeneous formation has a uniformity coefficient of unity. Between the two extremes, formations have uniformity coefficients comprised between zero and one. The following are the two most widely used descriptors of the vertical heterogeneity of the formation: Dykstra and Parsons permeability variation “V” Lorenz coefficient “L”

Dykstra-Parsons “D&P” Permeability Variation Dykstra and Parsons (1950) introduced the concept of the permeability variation V, which is designed to describe the degree of heterogeneity within the reservoir. The value of this uniformity coefficient ranges between zero for a completely homogeneous system and one for a completely heterogeneous system.

V

8/13/2011

k50  k84.1 k50

129

Step 1. Step 2.

Arrange the core samples in decreasing permeability sequence, i.e. descending order. For each sample, calculate the percentage of thickness with permeability greater than this sample.

Step 3

Using a Log-probability graph paper, plot permeability values on the log scale and the % of thickness on the probability scale.

Step 4

Draw the best straight line through the points

Step 5.

Read the corresponding permeability values at 84.1% and 50% of thickness. these two values are designated as k84.1 and k50.

Step 6.

The Dykstra and Parsons permeability variation is defined by the following expression:

V 

k 50  k 84.1 k 50

It should be noted that if all the permeabilities are equal, the numerator of the above equation would be zero, and the “V” would also be zero. This would be the case for a completely homogeneous system. The Dykstra and Parsons method is commonly referred to as a “Permeability Ordering Technique”

260

130

Class Problem: The following conventional core analysis data are available from three wells. Depth ft

Well #1 k md

5389-5391 --5393 -5395 -5397 -5399 -5401 -5403 -5405 -5406 -5409

166 435 147 196 254 105 158 153 128 172

Well #2 Porosit Depth k y ft md % 17.4 5397-5398.5 72 18.0 -539.95 100 16.7 -5402 49 17.4 -5404.5 90 19.2 -5407 91 16.8 -5409 44 16.8 -5411 62 15.9 -5413 49 17.6 -5415 49 17.2 -5417 83

porosi ty % 15.7 15.6 15.2 15.4 16.1 14.1 15.6 14.9 14.8 15.2

Dept ft

Well #3 k md

5401-5403 -5405 -5407 -5409 -5411 -5413 -5415 -5417 -5419

28 40 20 32 35 27 27 9 30

poros ity % 14.0 13.7 12.2 13.6 14.2 12.6 12.3 10.6 14.1

Minimum Number of Layers Based on a computer study, Craig (1971) outlined some guidelines for selecting the minimum number of layers needed to predict the performance of a reservoir under waterflooding operation. The author simulated the performance of a waterflood five-spot pattern that is composed of 100 layers with permeability variations ranging from 0.4 to 0.8. The minimum number of layers required to match results of the 100-layer model was determined as a function of mobility ratio M and permeability variation V.

131

The Zonation Problem In waterflooding calculations, it is frequently desirable to divide the reservoir into a number of layers that have equal thickness but different permeabilities and porosities. Traditionally, two methods are used in the industry to assign the proper average permeability for each layer: (1) The positional method (2) the permeability ordering method.

132

Positional Method The positional method describes layers according to their relative location within the vertical rock column. This method assumes that the injected fluid remains in the same elevation (layer) as it moves from the injector to the producer. The method have been successfully demonstrated this in predicting the performance in several Projects. The average permeability in a selected layer (elevation) should be calculated by applying the geometric-average permeability as given by:

k avg

 n  hi ln(ki )     exp  i 1 n   hi   i 1  

If the all thicknesses are equal, then: 1

k avg  (k1 k 2 k 3 . . . k n ) n

k avg  (k1 k 2 k 3 . . . k n )

1 n

133

Class Problem Using the core analysis data given below for ten wells system, assign the proper average permeability for each layer if the reservoir is divided into: 10 equal-thickness layers; each with 1 ft. thickness 5 equal-thickness layers; each with 2 ft. thickness

Layering using D&P Method In water flooding calculations, it is frequently desired to divide the reservoir into layers that have equal thickness and different permeability. The log-probability scale can be used in this case to assign the permeability scale into equal percent increments and reading the corresponding permeability at the midpoint of each interval.

134

Class Problem Using the data given in the previous problem, determine the average layer permeability for a 10 layered system, assuming a uniform porosity. Well #1 Depth ft

5389-5391 --5393 -5395 -5397 -5399 -5401 -5403 -5405 -5406 -5409

Well #2

Well #3

k md

Porosity %

Depth ft

k md

porosity %

Depth ft

k md

porosity %

166 435 147 196 254 105 158 153 128 172

17.4 18.0 16.7 17.4 19.2 16.8 16.8 15.9 17.6 17.2

5397-5398.5 -539.95 -5402 -5404.5 -5407 -5409 -5411 -5413 -5415 -5417

72 100 49 90 91 44 62 49 49 83

15.7 15.6 15.2 15.4 16.1 14.1 15.6 14.9 14.8 15.2

5401-5403 -5405 -5407 -5409 -5411 -5413 -5415 -5417 -5419

28 40 20 32 35 27 27 9 30

14.0 13.7 12.2 13.6 14.2 12.6 12.3 10.6 14.1

Lorenz Coefficient “L” Schmalz and Rahme (1950) introduced a single parameter that describes the degree of heterogeneity within a pay zone section. The term is called Lorenz Coefficient and varies between zero, for a completely homogeneous system, to one for a completely heterogeneous system. The following steps summarize the methodology of calculating Lorenz Coefficient: Step 1. Arrange all the available permeability values in a descending order. Step 2. Calculate the cumulative permeability capacity ∑kh and cumulative volume capacity ∑Фh. Step 3. Normalize both cumulative capacities such that each cumulative capacity ranges from 0 to 1. Step 4. Plot the normalized cumulative permeability capacity versus the normalized cumulative volume capacity on a Cartesian scale.

135

L

Area above the straight line A  Area below the straight line B

271

A B

L

Area above the straight line A  Area below the straight line B

136

This relationship can be expressed mathematically by the following two expressions: Lorenz Coefficient in Terms of Permeability Variation:

L  0.0116356 0.339794V  1.066405V 2  0.3852407V 3 Permeability Variation in Terms of Lorenz Coefficient:

V  5.05971 (10 4 )  1.747525 L  1.468847 L2  0.701023 L3 The above two expressions are applicable between 0
Class Problem Using the data given in last Example, calculate Lorenz Coefficient assuming a uniform porosity. Well #1 Depth ft

5389-5391 --5393 -5395 -5397 -5399 -5401 -5403 -5405 -5406 -5409

Well #2

Well #3

k md

Porosity %

Depth ft

k md

porosity %

166 435 147 196 254 105 158 153 128 172

17.4 18.0 16.7 17.4 19.2 16.8 16.8 15.9 17.6 17.2

5397-5398.5 -539.95 -5402 -5404.5 -5407 -5409 -5411 -5413 -5415 -5417

72 100 49 90 91 44 62 49 49 83

15.7 15.6 15.2 15.4 16.1 14.1 15.6 14.9 14.8 15.2

Dept ft

5401-5403 -5405 -5407 -5409 -5411 -5413 -5415 -5417 -5419

k md

porosity %

28 40 20 32 35 27 27 9 30

14.0 13.7 12.2 13.6 14.2 12.6 12.3 10.6 14.1

137

Layering using Lorenz Plot A plot of the cumulative permeability capacity “∑kh” vs. ∑h (without normalization) is commonly constructed, as shown, and used to assign average permeability values for a selected number of reservoir layers. If the intervals of the thickness are chosen, then the average values of permeability for each thickness interval (layer) can be calculated by dividing the incremental (∑kh) by the incremental thickness ∑h . It should be noted that it is not necessary that equal thickness sections be chosen. They may be selected at irregular increments as desired. There are also some advantages of selecting layer properties so that each layer has the same permeability thickness product.

Class problem Using Lorenz approach and the data in D&P example, calculate the average permeability for a ten-layered system reservoir. Compare the results with those of Dykstra and Parsons method.

138

Areal Heterogeneity   ( x) 

n

 i 1

n

 i 1

i  xi 

i  1

Where: Z*(x) = estimate of the regionalized variable at location x Z (xi) = measured value of the regionalized variable at position xi λi = weight factor n = number of nearby data points

277

The Polygon Method This technique is essentially based on assigning the nearest measured value of the regionalized variable to the designated location. This implies that all the weighting factors, λ i are set equal to zero except the corresponding λi for the nearest point is set equal to one.

The Inverse Distance Method

1

1  i 1  i  n

  d

i     di 

The Inverse Distance Squared Method

1 i    di

  

2

1   i 1  d i n

  

2

278

139

Class Problem: The illustration below shows the locations of four wells and distance between the wells and point “x”. The average permeability in each well location is given below

Well # 1 2 3 4

Permeability, md 73 110 200 140

Well #3 410 ft Well #4

380 ft

x

Well #2

170 ft Well #1

Estimate the permeability at location “x” by the polygon and the two inverse distance methods.

Delaunay Triangulation Method

k k 280

(k1 ) (area I )  (k1 ) (area II )  (k3 ) (area III ) (area I )  (area II )  (area III )

(696) ( 22.25)  ( 227) (12.)  (606) (9.5)  548.7 md ( 22.25)  (12)  (9.5)

140

Triangulation Method

Z=ax+by+c k1 = a x1 + b y1 + c k2 = a x2 + b y2 + c k3 = a x3 + b y3 + c

63 a + 140 b + c =696 64 a + 129 b + c =227 71 a + 140 b + c =606

a = -11.25, b = 41.614 , c = - 4421.159 k = -11.25 (65) + 41.614 (137) - 4421.159 = 548.7 md 281

End of 1st part

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