Dykstra parson

1Slide 11Dr Elena Pasternak

2)ln(ln i

Slide 12Dr Elena Pasternak

The effective permeability:Calculate the geometric mean of the natural log of the core-derived permeability values

The arithmetic average of the natural log of the 14 permeabilityvalues (5.067 mD) is practically equal to the geometric mean of the same permeability values (5.058 mD). This further indicates that this particular formation is practically homogeneous.


Standard deviation is compared with E(X). Coefficient of variation

)(XE=

%)6(059.0067.5

3.0)(

3.0 ==XE


http://www.answers.com/topic/darcy-s-law?cat=technology

permeability (m2~1012darcy)Q discharge or flow rate (m3/sec)Pa, Pb - pressures (Pa)A area (m2) dynamic viscosity (Pa sec)

Pressure increment P= Pa-Pb>0, Px (pressure drop)Flow occurs from high pressure Pa to low pressure Pb

xxPkAQ

LPPkAQ ba

=

=


Permeability of layered rock

z Permeability parallel to the layersz Permeability normal to the layers1

2

n

v1v2

vn

Permeability Volumetric fraction


Permeability parallel to the layers

12

n

Q Q

Pa Pbv1v2

vn

Q1Q2

Qn

===

=

==

n

k

bakkn

k

bakkn

k

bakk

LPPvA

LPP

ALLAA

LPPAQ

111

=

=n

kkkv

1

hk

=

== nk

k

kkk

h

hALLA

1

-volume fraction - effective permeability


pk=kpk probability that a layer k has a permeability k; the same as the volume fraction (geometric probability)

=

= nk

k

kk

h

hp

1

hk width of the layer k.


In the case of linear flow parallel to a stratified medium, typical of shallow marine sheet sands for example, the effective permeability of the region is the expected value of the layer permeabilities.Clearly, permeability is additive in this case.


Permeability normal to the layers

Q

Q

Pa

Pb12

n

v1v2

vn

P1P2

Pn

=

==

=

=

==

=

n

k k

k

n

k kk

n

k k

kk

n

kk

ba

vAQ

AQh

HhPh

H

PHH

PP

1

11

1

11

1

=

=n

kkkv

1

11

=

=

=n

kk

kk

hH

Hh

1

-volume fraction

- effective permeability

hk


When linear flow is orthogonal to the layers (eg, dune crossbeds), permeability is no longer additive. However, its inverse is additive. (Resistance to flow is additive.)The effective resistance to flow -1 is the expected value of the layer resistances to flow.


Permeability of heterogeneous (isotropic) rock

Permeabilities 1, 2,.. nEffective permeability

=

=

n

kkk

n

kkk vv

1

1

1

1


Representation through general averaging

( )

=

=

=

=

0lnexp

0

1

1

1

pv

pv

n

kkk

pn

k

pkk

p

Effective permeability

11, = ppGenerally, p depends upon the permeability distribution


Particular case

Two phase rock: 1,2

Equal volumetric fractions: v1=v2=0.5

Effective permeability

21=(Dykhne, 1970) exact solution!

(p=0)


Dykstra-Parsons coefficientz Dykstra and Parsons (1950) used the normal

distribution of permeability to define the coefficient of permeability variation VK

)(

)( 2

=

==

E

ns

sV

i

K

n is the total number of data points, i is the permeability of individual core samples.In a normal distribution, the value of is such that 84.1% of the permeability values are less than E()+s and 15.9% of the values are less than E()-s.

Standard deviationThe mean value of , arithmetic average of permeability


The Dykstra-Parsons coefficient of permeability variation can be obtained graphically by plotting permeability values on log-probability paper and using the following eqn

50

1.8450

=KV

where50=permeability value with 50% probability84.1=permeability at 84.1% of the cumulative sampleThe Dykstra-Parsons coefficient of permeability variation is an excellent tool for characterising the degree of reservoirs heterogeneity. The term VK is also called the Reservoir Heterogeneity Index.


Dykstra and Parsons (1950)

50=10, 84.1=3, VK=0.7 reservoir is very heterogeneous

log


10


ExampleGiven the permeability data in Table for well HBK5

(Chatzis et al, 1997), see sl. 9, calculate the Dykstra-Parsons coefficient


The procedure for graphically determining the Dykstra-Parsons

coefficienta) Arrange permeability data in descending order (column 2, sl.

32)b) Determine the frequency of each permeability value (column

3)c) Find the number of samples with larger permeability

(column 4)d) Calculate the cumulative frequency distribution by

dividing values in column 4 with the total number of permeability points, n (n=14 in this example) (column 5)

e) Plot permeability data (column 2) versus cumulative frequency data on a log-probability paper (natural log of permeabilities vs. percent of samples with larger probability)

11


f) Draw the best straight line through the data, with more weight placed on points in the central portion where the cumulative frequency is close to 50%. This straight line reflects a quantitative, as well as a qualitative, measure of the heterogeneity of the reservoir rock.

g) From graph (sl. 33) read the values 50=158.7 mD and 84.1=117.2 mD. Otherwise these values can be interpolated from Table (sl. 32).

h) Calculate the Dykstra-Parsons coefficient

26.07.158

22.1177.158

50

1.8450 ===KV

This formation is slightly heterogeneous, but it can be treated as homogeneous for reservoir simulation purposes.


=1-cumulative frequency distribution, since permeabilitiesare arranged in descending order (not ascending as required for the cumulative distribution function)

12


log


log-probability paper

Notes Vertical axis - Plot natural log of permeabilities Horizontal axis - cumulative frequency data (percent of

samples with larger probability). Observe that the ends (0, 0.1) and (0.9, 1) (or (0, 10%) and (90%, 100)) are stretched and the middle (0.1, 0.9) (or (10%, 90%)) is shrunk.

Log-probability paper is used to make graph look like a straight line in the middle. If standard coordinates are used we will get a curve! Interpolation of values is required to get 50 and 84.1.

Dykstra parson

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