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.
2Slide 13Dr Elena Pasternak
Standard deviation is compared with E(X). Coefficient of variation
)(XE=
%)6(059.0067.5
3.0)(
3.0 ==XE
3Slide 15Dr Elena Pasternak
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
=
=
Slide 16Dr Elena Pasternak
Permeability of layered rock
z Permeability parallel to the layersz Permeability normal to the layers1
2
n
v1v2
vn
Permeability Volumetric fraction
4Slide 17Dr Elena Pasternak
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
Slide 18Dr Elena Pasternak
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.
5Slide 19Dr Elena Pasternak
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.
Slide 20Dr Elena Pasternak
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
6Slide 21Dr Elena Pasternak
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.
Slide 22Dr Elena Pasternak
Permeability of heterogeneous (isotropic) rock
Permeabilities 1, 2,.. nEffective permeability
=
=
n
kkk
n
kkk vv
1
1
1
1
7Slide 23Dr Elena Pasternak
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
Slide 24Dr Elena Pasternak
Particular case
Two phase rock: 1,2
Equal volumetric fractions: v1=v2=0.5
Effective permeability
21=(Dykhne, 1970) exact solution!
(p=0)
8Slide 25Dr Elena Pasternak
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
Slide 26Dr Elena Pasternak
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.
9Slide 27Dr Elena Pasternak
Dykstra and Parsons (1950)
50=10, 84.1=3, VK=0.7 reservoir is very heterogeneous
log
Slide 28Dr Elena Pasternak
10
Slide 29Dr Elena Pasternak
ExampleGiven the permeability data in Table for well HBK5
(Chatzis et al, 1997), see sl. 9, calculate the Dykstra-Parsons coefficient
Slide 30Dr Elena Pasternak
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
Slide 31Dr Elena Pasternak
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.
Slide 32Dr Elena Pasternak
=1-cumulative frequency distribution, since permeabilitiesare arranged in descending order (not ascending as required for the cumulative distribution function)
12
Slide 33Dr Elena Pasternak
log
Slide 34Dr Elena Pasternak
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.