1

Rock Physics Reservoir Characterization

Amplitude

AcousticImpedance

ElasticImpedance

AVO (I,G)

STRUCTURES

POROSITY

SHALE CONTENT

FLUID

ROCKPHYSICSANALYSIS

Goal: 3D images of volumetric rockproperties from seismic data

Means: Rational transform functionbetween rock elastic propertiesand rock volumetric properties

Input: Controlled measurements onrock -- well logs, laboratory

GP170/2001 #1

Seismic Reflection and Impedance Inversion

2

0 0.2 0.4 0.6 0.8 1

2850

2900

2950

3000

3050

Sw and VSHALE

DR

KB

(ft)

LC 1880

Sw

VSHALE

2 2.5 3 3.5 4Vp (km/s)

0 0.1 0.2 0.3Porosity

Log Data for Designing Impedance-Reservoir Property Transform

GP170/2001 #1

3

4

5

6

7

8

9

10

0 0.2 0.4 0.6 0.8 1

LC 1880: Below 2800 ft DRKB

P-Im

peda

nce

VSHALE

SAND SHALE

0

0.1

0.2

0.3

0 0.2 0.4 0.6 0.8 1

POR

OSI

TY

VSHALE

4

5

6

7

8

9

10

11

0 0.1 0.2 0.3

P-IM

PED

AN

CE

POROSITY

Crossplots from Log Data

GP170/2001 #1

SHALE

SANDCHANNELWITH OIL

5

6

7

8

9

10

0 0.1 0.2 0.3

P-W

ave

Impe

danc

e

Porosity

SANDSHALE

Transform and Impedance Interpretation

4

Rock Physics Diagnostic

Rock physics diagnostic uses cross-plots between the rock elastic properties (e.g.,velocity, elastic modulus, and impedance) and its textural properties (e.g., porosity).Effective-medium model curves are superimposed on the well-log or core-data cross-plots.It is assumed that if the data points fall close to a theoretical model line, then the texturalproperties of the rock are those incorporated in the theoretical effective medium model.

Rock physics diagnostic allows one, for example, to discriminate between friable andslightly-cemented rocks in wells under examination.

In this example, the sands from the same formation have been diagnosed as contact-cemented in one well andcompletely unconsolidated (friable) in the other. This result is supported by the microscope image. It isconsistent with the fact that Well A is located in a high-energy depositional environment, and Well B is locatedin the lobe where clay particles cover the grains, thus preventing contact cement growth.

0.1 mm

Well A

DIAGENETICCEMENT

GRAIN

2 3 4

2.1

2.2

2.3

Vp (km/s)

Well B

Dep

th (k

m)

40 80 120GR

Well B

2 3 4

1.7

1.8

1.9

Vp (km/s)

Well A

Dep

th (k

m)

Marl

40 80 120GR

Well A

2.5

3.0

3.5

0.25 0.30 0.35 0.40

Vp

(km

/s)

Porosity

Contact CementModel

UnconsolidatedModel

ConstantCement Model

Well B

Well A

GP170/2001 #1

5

Rock physics diagnostic should be consistent with the location of the wells in the formation. The seismicamplitude map indicates that Well A is located in a high-energy depositional environment, and Well B islocated in the lobe where clay particles cover the grains, thus preventing contact cement growth.

The rock physics diagnostic allows the user to fine-tune velocity-porosity transforms to specific depositionalenvironments.

Well A Well B

Rock Physics Diagnostic

GP170/2001 #1

6

Definitions -- Measurable Rock Properties

Bulk Density (RHOB, ρb) = Total Mass / Total VolumeTotal Porosity (φ) = Pore Volume / Total VolumeSaturation (S) = Fluid Volume / Pore VolumeSw + So + Sg = 1RHOB = (1- φ) ρs + φ (Sw ρw + So ρo + Sg ρg )

GP170/2001 #1

Radius

Poro

sity

0

1

.5

Importance of Scale

7

Definitions -- Elastic Rock Properties

Tran

sdu

cer

Rec

eive

rROCK

WAVE

Ultrasonic Wave Transmission Experiment

Ppore

PconfiningA

mpl

itu

de (m

V)

Time (ns)

First-Break Travel Time

Peak-to-Peak Travel TimeINPUTSIGNAL

OUTPUTSIGNAL

V = L/τ

Pdifferential = Pconfining - Ppore

3

4

0 10 20 30 40

Vp

(km

/s)

Differential Pressure (MPa)

Sandstone9.5% Porosity

Dry

WaterSaturated

2

3

0 10 20 30 40

Vs

(km

/s)

Differential Pressure (MPa)

Sandstone9.5% Porosity

Dry

WaterSaturated

GP170/2001 #1

8

Seismic Data and Elastic Rock Properties

Ip = ρV p

ν =

1

2

(Vp / Vs )2 − 2

(Vp / Vs )2 − 1

Acoustic or P-impedance Poisson's Ratio

1200

1300

1400

1500

1600

1700

1800

1900

2000

Trav

el T

ime

(ms)

IncidentP-Wave Reflected

P-Wave

ReflectedS-Wave

TransmittedS-Wave

TransmittedP-Wave

Θ1

Θ2

Φ1

Φ2

NormalReflection

T

R

R(0) =

Ip2 − Ip1

Ip2 + Ip1

=dIp

2Ip

=1

2d ln I p

R(θ ) ≡1

2d ln Ie ≈ R(0)cos2 θ + d (

1

1 − ν)s in2 θ

ln Ie = ln Ip + ( 2

1 − ν− ln Ip )sin 2 θ

Acoustic ImpedanceElastic Impedance

GP170/2001 #1

R(θ ) ≈ d(

ln I p

2) + d ( 1

1 − ν−

ln I p

2) sin 2 θ

Intercept Gradient

9

Elasticity

Ti = σ ijn j

Stress Tensor

u

x1

x2

x3

x1

Tn

x2

x3

Strain Tensor

ε ij =

1

2(∂ui

∂x j

+∂u j

∂x i

)

σ ij = σ ji i ≠ j; ε ij = ε ji i ≠ j.

Hooke's law: 21 independent constants

σ ij = cijklekl; cijkl = c jikl = cijlk = c jilk, cijkl = cklij .

Isotropic Hooke's law: 2 independent constants (elastic moduli)

σ ij = λδ ijεαα + 2µε ij; ε ij = [(1 + ν )σ ij − νδ ijσαα ] / E.

λ and µ -- Lame's constants; ν -- Poisson's ratio; E -- Young's modulus.

Bulk Modulus Compressional Modulus Young’s Modulusand

Poisson’s Ratio K = λ + 2µ / 3 M = λ + 2µ

E = µ (3λ + 2µ ) / (λ + µ )ν = 0.5λ / (λ + µ )

YoungZ

X

Y

σzz = Eεzz

ν = −εxx / εzz

σ xx = σyy =σ xy = σxz =

σ yz = 0

Compressional

σzz = Mε zz

σ yy = λε zz =

[λ / (λ + 2µ )]σzz =[ν / (1 − ν )]σzz

ε xx = εyy =

ε xy = εxz = εyz = 0

σ xz = 2µε xz

ε xx = εyy = εzz = ε xy = 0

Shear

GP170/2001 #1

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Static and Dynamic Elasticity

CompressionalExperiment

u(z)

σ(z+dz)σ(z) zdzA

Adzρ ˙ ̇ u = A[σ (z + dz ) − σ (z )] = Adz∂σ / ∂z ⇒ ρ∂ 2u / ∂ t2 = ∂σ / ∂z

σ = Mε = M∂u / ∂z ⇒ ∂ 2u / ∂ t2 = (M / ρ )∂ 2u / ∂z 2 ≡ Vp2 ∂ 2u / ∂z 2

Wave equation:

Vp = M / ρ = (K + 4G / 3) / ρ ; Vs = G / ρ ;

M = ρVp2 ; G = ρVs

2 ; K = ρ(V p2 − 4Vs

2 / 3); λ = ρ(Vp2 − 2Vs

2 )Dynamic definitions:

ε ~ 10-7

-0.005 0 0.005 0.01 0.015 0.020

10

20

30

40

50

60

70

Strain

Axi

al S

tres

s (M

Pa)

Sample 10156-58

Pc = 0

Pc = 500 psi = 3.52 MPa

Pc = 2000 psi = 14.1 MPa

RADIAL AXIAL

Static uniaxial experiment:

ε ~ 10-2

GP170/2001 #1

11

Static Moduli

0

10

20

30

Axi

al S

tres

s (M

Pa)

Axial Strain

#10118.9Pc = 0

a

-0.001

0.000

Rad

ial S

trai

n

#10118.9Pc = 0

Axial Strainb

0

2

4

6

8

You

ng's

Mod

ulu

s (G

Pa)

Axial Strain

#10118.9Pc = 0

Tangent = 3.06 GPa

c

0

0.1

0.2

0.3

0 0.002 0.004 0.006 0.008 0.01

Pois

son'

s R

atio

Axial Strain

#10118.9Pc = 0

Tangent = .253

d

GP170/2001 #1

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Static and Dynamic Moduli

0

5

10

15

20

25

You

ng's

Mod

ulu

s (G

Pa)

10198 DynamicAxial Loading

No Confinig Pressure

10197 StaticAxial Loading

3.52 MPa Confinig Pressure

10199.5 StaticAxial Loading

14.1 MPa Confinig Pressure

10197-9.5 DynamicHydrostatic

a

0

.1

.2

.3

.4

0 10 20 30

Pois

son'

s R

atio

Axial Stress (MPa)b

GP170/2001 #1

13

Effect of Porosity and Mineralogy

QUARTZ

CLAY

Load-BearingGrains

PorousClay

Void

3

4

5

0 0.1 0.2 0.3

Vp

(km

/s)

Porosity

DryClay < 35%

0 0.1 0.2 0.3Porosity

No Clay

3% < Clay < 18%

18% < Clay < 37%

0 0.1 0.2 0.32

3

4

Porosity

DryClay < 35%

Vs

(km

/s)

0 0.1 0.2 0.3Porosity

3% < Clay < 18%

18% < Clay < 37%

No Clay

GP170/2001 #1

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Data Sources in Rock Physics -- History

Source Time Period Scale______________________________________________________________________Core Data 1950's - current 1 inchLog Data 1990's - current 1 ftSeismic Data current - future 100 ft

Raymer's Equation

Vp = (1 − φ )2 V ps + φV pf

Rock Type Mineral Velocity (km/s)

Sandstone 5.480 to 5.950

Limestone 6.400 to 7.000

Dolomite 7.000 to 7.925

GP170/2001 #1

2

3

4

5

6

0 0.1 0.2 0.3 0.4

Han'sOsebergTroll

Vp

(km

/s)

Porosity

Raymer

Wyllie

Core-Data Based Equations -- Wyllie's Time Average

τ p = τps + τpf ⇒1

Vp

=1 − φVps

+φ

Vpf

Mineral Fluid

15

Calculating Solid-Phase Properties

Hill's Elastic Moduli Average (ad-hoc)

MSolid = 0.5(MV + MR ),

MV = XiMii =1

n

∑ , MR = ( Xi

Mii=1

n

∑ )−1

M is either bulk or shear modulus; Xi is the volume fraction of i-th mineral in thesolid phase; Mi is the corresponding elastic modulus

ρS = Xiρii =1

n

∑Density of the solid phase

QUARTZ

CLAY

CALCITE

MICA

Mineral Bulk Modulus (GPa) Shear Modulus(GPa) Density (g/cc)--------------------------------------------------------------------------------------Quartz 36.6 45 2.65Clay 21 7 2.54

Vp = (K +4

3G) /ρ

Vs = G / ρ

Solid Phase

GP170/2001 #1

Mineral's volume in whole rock vs. volume in the solid

Whole rock: f Clay + f Quartz + φ = 1

f Solid = fClay + f Quartz

Solid phase: XClay + XQuartz = 1

XClay =fClay

f Clay + f Quartz

=f Clay

f Solid

=f Clay

1 − φ

QUARTZ

PORE

CLAY

Whole Rock

16

Core database -- Han '86

2

3

4

5

6

0 0.1 0.2 0.3

Vp

(km

/s)

Porosity

5 MPa

0 0.1 0.2 0.3Porosity

20 MPa

0 0.1 0.2 0.3Porosity

40 MPa

GP170/2001 #1

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Core database -- Han '86

1

2

3

4

0 0.1 0.2 0.3

Vs

(km

/s)

Porosity

5 MPa

0 0.1 0.2 0.3Porosity

20 MPa

0 0.1 0.2 0.3Porosity

40 MPa

GP170/2001 #1

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Core database -- Han '86 -- Effect of ClayGP170/2001 #1

0 0.05 0.1 0.15 0.2 0.25 0.32.5

3

3.5

4

4.5

5

5.5

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

Clay

Porosity

Vp (km/s)

40 MPa

2 4 6 8 10 12 140.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

0.22

0.24

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

Clay

P-Impedance

Poisson'sRatio

40 MPa

Pressure and Porosity

.1

.2

.3

.1 .2 .3Porosity at 5 MPa

40 MPa

Poro

sity

at

40 M

Pa

0

.01

.1 .2 .3Porosity at 5 MPa

Poro

sity

Incr

ease

Left: Porosity at40 MPa versus porosityat 5 MPa.

Right: Increase of porositybetween 40 MPa and 5 MPaversus porosity at 5 MPa.

All samples are sandstone.

0

.1

.2

.1 .2 .3Porosity at 5 MPa

Rel

ativ

e C

hang

e

5 MPa to 40 MPaRelative Reduction

19

Home Assignment -- Matlab

1. Load Han.datThe data are given for room-dry rock samples --Phi ClayWholeRock BulkDensity Vp5MPa Vp20MPa Vp40MPa Vs5MPa Vs20MPa Vs40MPa

2. Plot Vp and Vs at 40 MPa versus porosity, color by clay content (2 plots)

3. Calculate Ip, Nu at all pressures

4. Plot Nu versus Ip at all pressures, color by clay content (3 plots)

5. For use in Wyllie's and Raymer's equations, calculate VpSolid versus ClayWholeRock

6. For fluid use air at normal conditions -- bulk modulus ~ 1 bar = 0.1 MPa; density ~ 1 kg/cu m= 0.001 g/cc

7. Apply Wyllie's and Raymer's equations to predict Vp in Han's data, sample-by-sample; plot Vppredicted versus Vp measured for all pressures -- altogether 6 plots.

3

4

5

6

3 4 5 6

Vp

Wyl

lie (

km/s

)

Vp True (km/s)

Consolidated SandstonesWGG Equation

3

4

5

6

3 4 5 6

Vp

Ray

mer

(km

/s)

Vp True (km/s)

Consolidated SandstonesRHG Equation

GP170/2001 #1

Example