1

Vs PredictionGP170/2001 #13

Gassmann’s fluid substitution requires knowledge of the bulk modulus K

Bulk Modulus ofRock w/Fluid

Bulk Modulus ofMineral Phase

Bulk Modulus ofDry Rock Bulk Modulus of

Pore Fluid

Porosity

KSat = K s

φKDry − (1 + φ)K f KDry / K s + K f

(1− φ )K f + φKs − K f KDry / Ks

Often, when only Vp is available, it is required to predict Vs from Vp to calculate K.

However, this need can be eliminated if the Vp-only fluid substitution (approximate) equation is used

K = ρb(Vp2 − 4Vs

2 / 3)

CompressionalModulus of

Rock w/Fluid

CompressionalModulus of

Mineral Phase

Compressional Modulus ofDry Rock

Bulk Modulus ofPore Fluid

Porosity

MSat = Ms

φMDry − (1+ φ )K f MDry / Ms + K f

(1 − φ)K f + φMs − K f MDry / Ms

Need 1:Fluid

Substitution

Need 2:Rock Diagnostic

Seismic Forward ModelingFluid and Lithology Prediction

1

2

3

3 4 5

Vs

(km

/s)

Vp (km/s)

Experimental Fitsfor Water-Saturated Rocks

Mudrock

WilliamsSand

WilliamsShale

Limestone

Dolomite

0

.1

.2

.3

.4

.5

3 4 5

Poi

sson

's R

atio

Vp (km/s)

Experimental Fitsfor Water-Saturated Rocks

Mudrock

WilliamsSand

WilliamsShaleLimestone

Dolomite

2

Experimental Vp/Vs Relations Often Work WellEXAMPLE: MARLS

2

3

3 4 5 6

Vs

(km

/s)

Vp (km/s)

MudrockLine

LimestoneCurves

Well A

Well B

WilliamsShale

2.0 2.2 2.4 2.6RHOB

0 0.2 0.4 0.6Porosity

PhiRHO

1 2 3 4 5Velocity (km/s)

VpVs

2 4 6 8 10 12 14P-Impedance

0.2 0.3 0.4Poisson's Ratio

Well A

25 50 75100GR

Dep

th (m

)

100 m

~ 1800 m

MARL

1 10Resistivity

2.0 2.2 2.4 2.6RHOB

0 0.2 0.4 0.6Porosity

PhiRHONPHI

1 2 3 4 5Velocity (km/s)

VpVs

2 4 6 8 10 12 14P-Impedance

0.2 0.3 0.4Poisson's Ratio

20 40 60 80GR

Dep

th (m

)

100 m

~ 1800 m

MARL?

Well B

GP170/2001 #13

3

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0 0.1 0.2 0.3 0.4 0.5

Poi

sson

's R

atio

of

Pac

k

Poisson's Ratio of Solid

Hertz-MindlinDry Sphere Pack

Effective-Medium Models and Vs (Poisson’s Ratio) Prediction

GP170/2001 #13

0.5

1.0

1.5

0 10 20 30 40 50 60 70 80

YinWinklerAzra

Vs

(km

/s)

Pressure (MPa)

DRY GLASS BEADS

HertzMindlin

0.5

1.0

1.5

2.0

2.5

0 10 20 30 40 50 60 70 80

YinWinklerAzra

Vp (km

/s)

Pressure (MPa)

DRY GLASS BEADS

HertzMindlin

SpherePack

0

0.1

0.2

0.3

0.4

0 10 20 30 40 50 60 70 80

YinWinklerAzra

Poi

sson

's R

atio

Pressure (MPa)

DRY GLASS BEADS

HertzMindlin

Hertz-Mindlin Contact Theory

KHM = [n2 (1− φ c)

2 Gs2

18π 2(1− νs )2 P]

13

GHM = 5 − 4νs

5(2 − νs )[3n2 (1− φc )2 Gs

2

2π 2 (1− ν s)2 P]

1

3

KHM

GHM

=5(2 − ν s)

3(5 − 4νs )

νHM = 3KHM − 2GHM

2(3KHM + GHM )= 3νs

2(15 − 9νs )

2.0

2.5

0 10 20 30 40 50 60 70 80

YinWinklerAzra

Vp (km

/s)

Pressure (MPa)

GLASS BEADSw/Water

Gassmann HertzMindlin

0

0.1

0.2

0.3

0.4

0.5

0 10 20 30 40 50 60 70 80

YinWinklerAzra

Poi

sson

's R

atio

Pressure (MPa)

HertzMindlin

GLASS BEADSw/Water

Gassmann

Gassmann’sWater Saturation

Vp

Vs

=2(1− ν)

1− 2ν⇒ V s = V p

1 − 2ν2(1− ν )

4

Experimental Sphere Pack and Sand Results

GP170/2001 #13

SpherePack

0

1

2

3

4

0 10 20 30

Vp (km

/s)

Pressure (MPa)

DRY LEAD SHOT #92 mm Size

HertzMindlin

0

1

2

3

0 10 20 30

Vs

(km

/s)

Pressure (MPa)

DRY LEAD SHOT #92 mm Size

HertzMindlin

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0 10 20 30

Poi

sson

's R

atio

Pressure (MPa)

DRY LEAD SHOT #92 mm Size

HertzMindlin

0.3

0.4

0.5

0 10 20

Sand ASand B

Poi

sson

's R

atio

Pressure (MPa)

Sand w/WaterUltrasonic

Experiment

HertzMindlin

1.9

2.0

2.1

2.2

2.3

0 10 20

Sand A

Sand B

Vp (km

/s)

Pressure (MPa)

Sand w/WaterUltrasonic

Experiment

HertzMindlin

0.3

0.5

0.7

0.9

1.1

0 10 20

Sand ASand B

Vs

(km

/s)

Pressure (MPa)

Sand w/WaterUltrasonic

ExperimentHertz

Mindlin

Lead SpheresSand/Water

Ultrasonic Experiment

0

0.02

0.04

0.08

0.10

0.12

0 4 8 12 16 20

Sand ASand B1

/Q

s

Pressure (MPa)

Sand w/WaterUltrasonic

Experiment

0

0.02

0.04

0.08

0.10

0.12

0 4 8 12 16 20

Sand ASand B

1/Q

p

Pressure (MPa)

Sand w/WaterUltrasonic

Experiment

0

1

2

3

0 4 8 12 16 20

Sand A

Sand B

(1/Q

s)/(1

/Q

p)

Pressure (MPa)

Sand w/WaterUltrasonic

Experiment

5

Poisson’s Ratio in North Sea Sands

GP170/2001 #13

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.1 0.2 0.3 0.4

Poi

sson

's R

atio

Porosity

5 MPa

30 MPa

DRY

GassmannWater

TROLL SANDS

5

10

15

20

25

30

0.1 0.2 0.3 0.4

Com

pre

ssio

nal

Mod

ulu

s (G

pa)

Porosity

DRY DATA20 MPa

TROLL

OSEBERG

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.1 0.2 0.3 0.4

Poi

sson

's R

atio

Porosity

20 MPa

40 MPa

DRY

GassmannWater

OSEBERG SANDS

2

4

6

8

10

12

14

0.1 0.2 0.3 0.4

Sh

ear

Mod

ulu

s (G

pa)

Porosity

DRY DATA20 MPa

TROLL

OSEBERG

-0.1

0

0.1

0.2

0.3

0.4

0.5

-0.1 0 0.1 0.2 0.3 0.4 0.5

Poi

sson

's R

atio

30 M

Pa

Poisson's Ratio 5 MPa

DRY GassmannWater

TROLL SANDS

-0.1

0

0.1

0.2

0.3

0.4

0.5

-0.1 0 0.1 0.2 0.3 0.4 0.5

Poi

sson

's R

atio

40 M

Pa

Poisson's Ratio 20 MPa

DRY

GassmannWater

OSEBERG SANDS

Pressure-Dependent Behavior of SandsInput Experimental Data

2

3

0 0.1 0.2 0.3 0.4

Vs

(km

/s)

Porosity

5 MPa

30 MPa

DRYSANDSTONES

2

3

4

0 0.1 0.2 0.3 0.4

Vp (km

/s)

Porosity

5 MPa

30 MPa

DRYSANDSTONES

0

0.1

0.2

0.3

0 0.1 0.2 0.3 0.4

Poi

sson

's R

atio

Porosity

5 MPa

30 MPa DRYSANDSTONES

General Sandstone Data

6

Empirical Relations and Well Log Data

GP170/2001 #13

0.1

0.2

0.3

0.4

0.5

2 3 4

Poi

sson

's R

atio

Vp (km/s)

Well ANo Hydrocarbons

MudrockWilliams

1

2

2 3 4

Vs

(km

/s)

Vp (km/s)

Well ANo Hydrocarbons

MudrockWilliams

1

2

2 3 4

Vs

(km

/s)

Vp (km/s)

Well BGas and Oil

MudrockWilliams

0.1

0.2

0.3

0.4

0.5

2 3 4

Poi

sson

's R

atio

Vp (km/s)

Well BGas and Oil

MudrockWilliams

0.1

0.2

0.3

0.4

0.5

2 3 4

Poi

sson

's R

atio

Vp (km/s)

Well CGas and Oil

MudrockWilliams

1

2

2 3 4

Vs

(km

/s)

Vp (km/s)

Well CGas and Oil

MudrockWilliams

0.1

0.2

0.3

0.4

0.5

2 3 4 5 6

Poi

sson

's R

atio

Vp (km/s)

Well EUplifted

Tight SandGas

MudrockWilliams

0.1

0.2

0.3

0.4

0.5

2 3 4 5 6

Poi

sson

's R

atio

Vp (km/s)

Well DUplifted

Tight SandGas

MudrockWilliams

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

2 3 4 5 6

Vs

(km

/s)

Vp (km/s)

Well DUplifted

Tight SandGas

MudrockWilliams

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

2 3 4 5 6

Vs

(km

/s)

Vp (km/s)

Well EUplifted

Tight SandGas

MudrockWilliams

0.5

1.0

1.5

2.0

2.5

3.0

2 3 4 5

Poi

sson

's R

atio

Vp (km/s)

Well FDeeper

Gulf CoastGas Sand

MudrockWilliams

PAY

0.1

0.2

0.3

0.4

0.5

2 3 4 5

Poi

sson

's R

atio

Vp (km/s)

Well FDeeper

Gulf CoastGas Sand

MudrockWilliams

PAY

Empirical relations do not give precise values but can be used to identify hydrocarbons (w/caution)

7

Pore Pressure and Fluid Identification

GP170/2001 #13

0 10 20 301

2

3

Diff. Pressure (MPa)

Vp

Vs

Vel

ocit

y (k

m/s)

0 10 20 30Pore Pressure (MPa)

Vp

Vs

2.0

2.5

5 10 15 20 25 30V

p (km

/s)

Effective Pressure (MPa)

GAS

OIL

BRINE

Phi = 0.35

5 10 15 20 25 30Pore Pressure (MPa)

GAS

OIL

BRINE

Phi = 0.35

5 10 15 20 25 30

2.0

2.5

Pore Pressure (MPa)

GAS

OIL

BRINE

Vp (km

/s)

WaterFlood

GasInjection

Gas out ofSolution

0

.1

.2

.3

.4

3 4 5

Poi

sson

's R

atio

P-Impedance (km/s g/cc)

BRINE

PorePressure

PorePressure

GAS

OIL

PorePressure

NORTH SEASAND

3 4 50

.1

.2

.3

.4

P-Impedance (km/s g/cc)

GAS

OIL

BRINE

Poi

sson

's R

atio

Gas out ofSolution

PorePressure

PDifferential = PConfining - PPore

2.5

3.0

3.5

4.0

4.5

0 20 40 60 80 100

Vp (km

/s)

Differential Pressure (MPa)

DRY

SATURATED

GASSMANN

1.8

2.0

2.2

2.4

2.6

2.8

3.0

0 20 40 60 80 100

Vs

(km

/s)

Differential Pressure (MPa)

Frequency Artifacts

8

Poisson’s Ratio and Pressure in Dry Rocks

GP170/2001 #13

PDifferential = PConfining - PPore

0 5 100

.1

.2

.3

Pressure (MPa)

PR

a

CascoGranite

Pore Pressure

0 10 20 30Pressure (MPa)b

Clay-FreeSandstone

18% Porosity

Pore Pressure0 10 20

Pressure (MPa)c

Ottawa Sand +5% Kaolinite33% Porosity

Pore Pressure

0 10 20Pressure (MPa)d

Sandstone13% Porosity

11% Clay

Pore Pressure

0 10 20 300

.1

.2

Pressure (MPa)

PR

e

Sand36% Porosity

Pore Pressure

0 10 20 30Pressure (MPa)f

Sand27% Porosity

Pore Pressure

0 10 20 30Pressure (MPa)g

Sand35% Porosity

Pore Pressure

0 10 20 30 40Pressure (MPa)h

Sandstone25% Porosity

Pore Pressure

0 20 40 600

.1

.2

Pressure (MPa)

PR

i

DeepTight Gas SS

4.3% Porosity31% Clay

Pore Pressure

0 20 40Pressure (MPa)j

BereaSandstone

19% Porosity

Pore Pressure

0 20 40Pressure (MPa)k

DeepSandstone

13% Porosity44% Clay

Pore Pressure

0 20 40Pressure (MPa)l

Carbonate~ 2% Porosity

Pore Pressure

0.30 0.35Porosityb

at 5 MPaat 20 MPa

0.30 0.35Porosityd

at 5 MPaat 20 MPa

0.30 0.35

1.6

1.8

2.0

2.2

Porositya

Vp (km

/s) Both at

20 MPa

0.30 0.35

.1

.2

Porosityc

Poi

sson

's R

atio

Both at20 MPa

0

.1

.2

0 .1 .2

PR

at

30 M

Pa

PR at 5 MPa

2

3

4

0.1 0.2 0.3

Vp (km

/s)

Porosity

30 MPa5 MPa

0

.1

.2

0.1 0.2 0.3

PR

Porosity

9

Modeling Poisson’s RatioChange in Cracked Rock

GP170/2001 #13

BulkModulus

ShearModulus

0 10 20 30Pressure (MPa)

Pore Pressure

Differential Pressure 35 MPa0 MPa

35 MPa 0 MPa

0

5

10

15Bulk

Modulus

ShearModulus

Ela

stic

Mod

uli (G

Pa)

1e-41e-31e-2

.1

.2

.3

Crack Porosity

PR

Model Data

AxialLoading

RadialDeformation

Physical Meaningof the Result

0

0.1

0.2

0 10 20 30 40 50

OTTAWA SAND (YIN)

GLASS BEADS (TUTUNCU)

Poi

sson

's R

atio

Pressure (MPa)

Pore Pressure

Importance of Cracks

Explanation of the Effect

1e-41e-31e-2

.1

.2

.3

.4

Crack PorosityPR

IncreasingPore Pressure

Dry

w/Water

Synthetic ExampleDry and Saturated

10

0 20 40Pressure (MPa)b

Dry

w/Water

Pore Pressure

GP170/2001 #13

Saturated Rocks

0 20 400

.1

.2

.3

Pressure (MPa)a

Dry

w/Water

PR

Pore Pressure

.20

.25

.30

.35

0 10 20 30 40

Poi

sson

's R

atio

Differential Pressure (MPa)

Pore Pressure

a

b

c

Shale w/WaterUltrasonic Experiment

0

.1

.2

.3

.4

3 4 5

Poi

sson

's R

atio

P-Impedance (km/s g/cc)

BRINE

PorePressure

PorePressure

GAS

OIL

PorePressure

NORTH SEASAND

a

Seismic Diagnostic Crossplots

2

3

4

5

6

7

2 3 4

P-I

mped

ance

(km

/s

g/cc

)

S-Impedance (km/s g/cc)

BRINE

GAS

OIL

NORTH SEASAND

b

Caveats of Texture

0.25 0.30 0.35

3

4

5

6

7

8

9

Porosity

NORTH SEASANDS 20 MPa

a

P-I

mped

ance

(k

m/s

g/cc

)

Oseberg

Troll

GAS

0.25 0.30 0.35Porosity

b

NORTH SEASANDS 20 MPa

Oseberg

Troll

WATER

3 4 5 6 7 8 9

0

.1

.2

.3

.4

P-Impedance (km/s g/cc)

BRINE

ContactCementation

ContactCementation

GAS

NORTH SEASANDS 20 MPa

a

Poi

sson

's R

atio

3 4 5 6 7 8 9P-Impedance (km/s g/cc)

BRINE

PorePressure

PorePressure

GAS

NORTH SEATROLL SAND

b

11

GP170/2001 #13

More on Overpressure

5

10

15

0.2 0.3 0.4

5 MPa15 MPa

Com

pre

ssio

nal

Mod

ulu

s (G

Pa)

Porosity

TROLL SANDROOM DRY

5 MPaPhiC = .4

C = 8

15 MPaPhiC = .4C = 10

2

4

6

0.2 0.3 0.4

5 MPa

15 MPaS

hea

r M

odu

lus

(GPa)

Porosity

TROLL SANDROOM DRY

5 MPaPhiC = .4

C = 8

15 MPaPhiC = .4C = 10

0

0.1

0.2

0.2 0.3 0.4

5 MPa15 MPa

Poi

sson

's R

atio

Porosity

TROLL SANDROOM DRY

5 MPaPhiC = .4

C = 8

15 MPaPhiC = .4C = 10

Uncemented Troll Sand: Pressure Behavior

2 4 6 8 10 12M-Modulus (GPa)

12 MPa

22 MPa

Pore Pressure

0.2 0.3 0.4

1.1

1.2

1.3

Porosity

Dep

th (km

)

1 2 3 4 5G-Modulus (GPa)

12 MPa

22 MPa

Pore Pressure

0 0.1 0.2PR

12 MPa

22 MPa

Pore Pressure

Overpressured Gas Sand

2

4

6

8

10

12

2 4 6 8 10 12

M-M

odu

lus

at 5

MPa

M-Modulus at 15 MPa

1

2

3

4

5

1 2 3 4 5

Sh

ear

Mod

ulu

s at

5 M

Pa

Shear Modulus at 15 MPa

0

.1

.2

0 .1 .2

Poi

sson

's R

atio

at

5 M

Pa

Poisson's Ratio at 15 MPa

PDifferential = PConfining - PPore

12

GP170/2001 #13Two-Layer Model

0.3 0.4Porosity

2Layer

0 0.5 1Sw

2Layer

Gas

Material PropertiesMineral Bulk (GPa) Shear (GPa) Density (g/cc)--------------------------------------------------------------------------------------Quartz 36.6 45 2.65Clay 21 7 2.54Water 2.5 0 1.00Gas .02 0 0.1

Model Parameters (Uncemented Sand Model)Layer PhiC Coordination# Diff. Pressure--------------------------------------------------------------------------------------Shale (Top) 0.4 8 15 MPaSand (Normal) 0.4 8 15 MPaSand (Pressured) 0.4 5 5 MPa

1.6 1.8 2.0 2.2RHOB

2Layer

0 0.5 1

1.00

1.01

1.03

1.04

1.05

1.06

1.07

1.08

Clay (in Solid)

Dep

th (km

)

2Layer

SH

ALE

SA

ND

4

5

6

0.2 0.3 0.4

P-I

mped

ance

Porosity

Top ShaleModel

Ip = 7.32 - 8.88 Phi.37

.38

.39

.40

.41

0.2 0.3 0.4

Poi

sson

's R

atio

Porosity

Top ShaleModel

PR = 0.34 + 0.17 Phi

3

4

5

0.2 0.3 0.4

P-I

mped

ance

Porosity

Gas Sand15 MPa

Ip = 7.84 - 12.76 Phi.01

.03

.05

.07

.09

0.2 0.3 0.4

Poi

sson

's R

atio

Porosity

PR = 0.176 - 0.38 Phi

Gas Sand15 MPa

3

4

5

0.2 0.3 0.4

P-I

mped

ance

Porosity

Gas Sand5 MPa

Ip = 5.83 - 9.73 Phi

.01

.03

.05

.07

.09

0.2 0.3 0.4

Poi

sson

's R

atio

Porosity

PR = 0.19 - 0.39 Phi

Gas Sand5 MPa

Model Results

2 3 4 5Ip

2Layer

15 M

Pa

5 M

Pa

0 0.2 0.4PR

2Layer

15 M

Pa

5 M

Pa

From Model

0 0.2 0.4PR

2Layer

5 M

Pa

15 M

Pa

Rea

list

ic

PR Corrected

SHALEIp = 4.42RP = 0.4

SAND NORMALIp = 3.37RP = 0.15

SAND OVERPRESSUREDIp = 2.42RP = 0.06

PDifferential = PConfining - PPore

R(0 )=Ip2 − Ip1

Ip2 + Ip1

; R(θ1 ) ≈ R(0)cos2 θ1 + (1

1− ν2

−1

1− ν1

)sin 2 θ1.

-0.5

-0.4

-0.3

-0.2

-0.1

0 10 20 30 40

Rpp

Angle

Normal

Overpressure

Overpressurew/Normal PR

13

GP170/2001 #13Backus Average

AndWyllie’s Average

Frequency

Wavelength

Vel

oci

ty

This velocity dispersion mechanism is related to the heterogeneity of natural rocks and reservoirs. The high-frequency wave travels faster than the low-frequencywave.

Consider a layered 1D medium made out of two types of materials: one material is soft and has density ρ1 and the elastic modulus M1; the other has density ρ2

and the elastic modulus M2. If the wavelength is short, the wave sees every layer as an infinite space. Then the travel time through two layers made of differentmaterials but having the same thickness (one length unit) is (Wyllie's time average):

Let now the wavelength be much larger than the individual layers. Then the wave sees the layers as a single elastic body. The effective elastic modulus of thisbody is given by the iso-stress (Reuss) average:

The effective density of the new elastic body is ; the resulting effective velocity is , and the travel time through two layersis

For N layers:

τ Short =

1

V1

+1

V2

=1

M1 / ρ1

+1

M2 / ρ2

= τ1 + τ2 .

1 / MEff = (1/ M1 + 1/ M2 )/ 2.

ρEff = (ρ1 + ρ2 )/ 2 VEff = MEff /ρEff

τ Long = 2 / VEff

τ Backus =hi

i =1

N

∑VBackus

=hi

i =1

N

∑MEff / ρEff

,1

MEff

=1

hii=1

N

∑hi

Mii=1

N

∑ , ρEff =1

hii =1

N

∑hiρi

i =1

N

∑ .

τ Short =

1

M1 / ρ1

+1

M2 / ρ2

< 2(ρ1 + ρ2 )

2

1

2(

1

M1

+1

M2

) = τLong

SHALEVp = 2.147 km/sRHOB = 2.06 g/cc

M = 9.5 GPa

SAND NORMALVp = 1.92 km/s

RHOB = 1.76 g/ccM = 6.5 GPa

Vp_Wyllie = 2.027 m/sVp_Backus = 2.010 m/s

Example: