Measures of Heterogeneity
Dykstra-Parson Coefficient
1)
Order data from large to small such that
k k
k
...
1 2
N
If do not have then
k1 k 2 ... k N
2)
Generate empirical CDF based on above rank
Dykstra-Parson Coefficient
3)
Plot in log scale (k/) or k versus empirical CDF in
a probability scale.
log scale
k
or k
l −1 2
CDF =
100
N
Dykstra-Parson Coefficient
4)
Fit best straight line.
VDP
or,
k
k
−
CDF =84.1%
CDF =50%
=
k
CDF =50%
( k )CDF =50% − ( k )CDF =84.1%
VDP =
( k )CDF =50%
Dykstra-Parson Coefficient
Note: If you rank k/ or k from small to large i.e
k k
k
....
1 2
N
Then plot would look like this
log scale
k
i −1 2
CDF =
100
N
Dykstra-Parson Coefficient
4)
Fit best straight line.
VDP
or,
VDP =
k
k
−
CDF =16%
CDF =50%
=
k
CDF =50%
k50% − k16%
k50%
VDP = 0
homogeneous
VDP = 1
infinitely heterogeneous
Dykstra-Parson Coefficient
Note 1: The VDP or Dykstra-Parsons coefficient is
negatively biased. It underestimates heterogeneity
Note 2: Caution should be exercised in the universal
application of VDP for data sets that are not log-normal.
Note 3: Literature values show that 0.5 VDP 0.9,
with an average of 0.7 for most reservoirs.
VDP = −
1.49 ln (1 − VDP ) (1 − VDP )
N
Lorenz Coefficient
1.
Arrange the permeability values in decreasing
order, from large to small like
k
k1 k 2
kn
...
n
1 2
2. Calculate the partial sums
j
FJ =
k h
j =1
n
j
k h
i =1
j
i i
= Flow capacity
Lorenz Coefficient
j
CJ =
h
j =1
n
j
h
i =1
j
= Storage capacity
i i
n = total number of layers
3.
Plot FJ versus CJ on a linear graph and connect the
points to form the curve BCD.
The curve must pass through (0,0) and (1,1)
Lorenz Coefficient
If A is the area between the curve and the diagonal
(shaded region, Figure below), the Lorenz coefficient
is defined as Lc = 2A
1
A
FJ
0
0
CJ
1
Lorenz Coefficient
Advantages of Lc over VDP:
1.
2.
3.
The distribution doesn’t have to be log-normal. It
can be any distribution.
Lc procedure doesn’t rely on best fit procedure.
Lc evaluation includes , h, and k.
Lorenz Coefficient
Note:
Lc = 0
homogeneous reservoir
Lc = 1
infinitely heterogeneous reservoir
Note: Field-measured values of Lc appear to range
from 0.6 to 0.9.
However, in general VDP Lc
Lorenz Coefficient
Lorenz Coefficient Example
Coefficient of Variation
CV =
It is mainly used to compare data from different
populations.
CV 0.5 homogeneous
0.5 CV 1 heterogeneous
CV 1 very hetrogeneous
I o = (10 CV )
2
I o = # of sufficient samples for statistical analysis
Koval Factor
For a miscible displacement
o
= H F 0.78 + 0.22
s
o = oil viscosity
0.25 4
s = solvent viscosity (example CO 2 = solvent)
= Koval factor
H = heterogeneity factor
F = gravity over-ride factor
Koval Factor
H = heterogeneity factor is defined as follows
VDP
log10 H =
0.2
(1 − VDP )
VDP = Dykstra − Parsons coefficient
Koval Factor
F = 0.565log10 Ck v A
+ 0.87
Qs
A = Pattern size in acres
Q = Injection rate ( reservoir bbl / day )
C = Pattern cons tan t
C = 2.5271 for five − spot pattern
C = 2.1257 for line − drive pattern
(
= Density difference between oil and solvent gm / cm3
s = Solvent vis cos ity ( cp )
k v = vertical permeability ( md )
Assume F = 1 , if do not have the means to calculate it
)
Koval Factor
The koval heterogeneity factor is used in miscible
flooding to empirically incorporate the effect of
heterogeneity on viscous fingering.
Definition: The koval factor is defined as the reciprocal
of the dimensionless breakthrough time in a unitmobility ratio displacement.
=
1
t Dbreakthrough
For homogeneous medium, =1
There is no upper limit for the koval factor
Reference for Koval Factor
J. Shaw and S. Bachu, 2002, Screening, evaluation,
and ranking of oil reservoirs suitable for CO2-flood
EOR and carbon dioxide sequestration. JCPT, vo. 41,
no. 9, p. 51-61.
Spatial Statistics
Covariance
Covariance:
For bivariate distributions (two variables), we define
another expected value. Covariance
C ( X , Y ) = E ( X − x ) (Y − y )
= E ( X Y ) − E ( X ) E (Y )
Note:
X = Y C ( X ,Y ) = 2
In practice, the covariance can be defined as
n
1 n
1 n
C ( x, y ) = xi yi − 2 xi yi
n i=1
n i=1 i=1
⚫
n is the total number of sample pairs.
Covariance
➢
The covariance quantifies a relationship between
the two variables X and Y.
➢
If the variables are closely related to each other, the
absolute value of covariance is high.
➢
If the two variables are independent of each other,
the covariance is equal to zero.
➢
Covariance can be either positive, or negative.
Correlation Coefficient
( x, y ) =
where
C ( x, y )
x y
( x, y ) = correlation coefficient
C ( x, y ) = covariance between x and y
x = standard deviation of x
y = standard deviation of y
Correlation Coefficient
The correlation coefficient is a dimensionless quantity.
The value of the correlation coefficient varies between
-1 and 1.
If two variables are strongly, positively related, the
correlation coefficient is close to +1
If two variables are strongly, negatively related, the
correlation coefficient is close to -1.
If two variables are independent of each other, the
correlation coefficient is zero.
Example 1
The following table lists the values of core permeability
and core porosity values for a sandstone reservoir
1)
2)
Calculate the covariance between the porosity and
the log of permeability
Calculate the correlation coefficient.
Example 1
(%)
K (md)
Log (k)
29.49
1156
3.063
26.79
531
2.725
28.74
1059
3.025
27.65
822
2.915
27.69
1014
3.006
22.69
109
2.037
23.30
138
2.140
23.81
1656
2.22
25.54
362
2.559
=
=
Example 1
n
1 n
1 n
C ( x, y ) = xi yi − 2 xi yi
n i=1
n i=1 i=1
let
x = log k and
y =
Example 1 Solution
( 3.063 29.49 ) + ( 2.723 26.79 ) + ( 3.025 28.74 )
+ ( 2.915 27.65 ) + ( 3.006 27.69 ) + ( 2.037 22.69 )
+ 2.14 23.3 + 2.22 23.81 + 2.559 25.54 +
)
) (
) (
1 (
C ( X ,Y ) = 1
9 − [3.063 + 2.725 + 3.025 + 2.915 + 3.006 + 2.037
9
+ 2.14 + 2.22 + 2.559][29.49 + 26.79 + 28.74 +
25.54]
+
23.81
+
23.3
+
2.69
2
+
27.69
+
27.65
= 0.8875
Example 1 solution
S x = 0.3859
S y = 2.3291
C ( x, y )
0.8875
( x, y ) =
=
= 0.987
SxS y
( 0.3859 )( 2.3291)
Strong relationship between log k and
Spatial Covariance
1
1
(
)
C x, y = xi yi − 2 xi yi
n
n
E ( X , Y ) − E ( X )E (Y )
n
1 n
1 n
C Z ( xi ) , Z ( xi + h ) = Z ( xi ) Z ( xi + h ) − 2 Z ( xi ) Z ( xi + h )
n i=1
n i=1
i =1
Z ( xi ) = is the measured value of a variable at location xi
Z ( xi + h ) = is the measured value of the variable at location ( xi + h )
n = is the number of pairs located h distance apart
h = is the distance between two values.
C ( Z ( xi ) , Z ( xi + h ) ) = is covariance of variable Z , h distance apart
Covariance is a measure of similarity
Note (1):
Here we are extending the concept of covariance
where a given variable is a function of distance.
Example:
Consider wellbore data collected at every one foot
interval. Intuitively we can assume that the values
which are closer to each other are similar to each
other. On the other hand, as the distance between the
measured values increases, the similarity will also
decrease.
Quantifying such relationship can be accomplished by
covariance.
Spatial Covariance
Hence the covariance is defined between the same
variable measured at different locations.
COV (Z i , Z i + h ) = E (Z i Z i + h ) − E (Z i )E (Z i + h )
Note (3): COV ( X i , X i + h ) = E ( X i X i + h ) − E ( X i )E ( X i + h )
Note (4): The special relationship is also expressed
using variograms:
h
Z1
Z2
Z3
Z4
Z5
Z6
Z7
Spatial Statistics
Suppose that we have the variable Z example porosity
measured at different locations.
Z1 = Z ( X 1 )
Z2 = Z (X 2 )
.
.
Z6 = Z (X 6 )
The separation distance between two consecutive
locations is h, Therefore, we can define the variogram
as such:
Semi-Variogram
( )
1 n
2
(
)
(
)
(
)
h = Z Xi − Z Xi + h
2n i =1
( h ) is the semi variaogram at a distance h
Z ( X i ) and Z ( X i + h ) are values of the variable distance h apart
n is the numbr of pairs h distance apart.
Note 1: The semi-variogram is a function of lag
distance h. Similarly we can estimate (2h), (3h),
(4h)…
Semi-Variogram
1 n
(2h ) = Z ( X i ) − Z ( X i + 2h )2
Then plot (h):
2n
i =1
h 2h 3h 4h − −
Note : You can think of variogram as a variance. It is a
measure of differences.
1 n
2
Var ( X ) =
X
n
i =1
i
− X
Spatial Statistics
If we allow h to be a variable distance say h = 1m then
h = 2m then h = 3m etc…, Then
1 n
(h ) = Z ( X i ) − Z ( X i + h )2
2n i =1
and we can plot the variogram as such:
COV (0)
COV (h )
h
(h ) = COV (0) − COV (h )
h
Spatial Statistics
( h ) = COV ( 0 ) − COV ( h )
COV ( h ) = covariance at lag distance h
γ ( h ) = semi variogram at a lag distance h
C ( 0 ) = is the covariance at lag distance zero = sill
The covariance at log distance zero is equivalent to
the variance of the data.
Note: This relationship (h) = COV(0) – COV(h) is very
important in geostatistical analysis.
cov (0)= sill = plateau value of semi-variogram
Range is defined as lag distance when have 95% of maximum variogram value
Nugget
Definition:
Value of semi-variogram at zero lag distance. This
is also the estimate on the measurement error on
Z.
Nugget Effect:
1.
2.
The nugget effects (i.e, the higher the nugget
value) makes weights become more similar to
each other and results in higher kriging
variance.
A pure nugget effect model entails a complete
lack of spatial correlation.
Hole Effect Model
h
sinh
(
)
(
)
h = COV 0 1 −
h
h in radians
This model has been used to represent fairly continuous processes that show a
periodic behavior such as successions of rich and poor mining zones.
MonteCarlo Simulation
Ghawar field in Saudi Arabia has been
estimated to contain 75 billion bbl of oil.
How much uncertainty is associated with
that?
What is MonteCarlo Simulation?
Algorithm of the MonteCarlo Simulation
Definition
Process of
probabilities.
finding
estimates
associated
with
Example:
Let’s say we have data for H, , and Swc of a particular
reservoir.
Objective: Estimate reserves
probabilities, or with confidence.
associated
7758 AH (1 − S wc )
N=
Boi
with
Procedure
Given either data for H, , and Swc or pdf of H, , and
Swc
f
h
g
H
Swc
Procedure Continue
Step 1: From data of , or pdf of construct a CDF
plot say in a probability paper.
Do the same for H, and Swc.
1
CDF
H
0
1
0
H
Procedure Continue
Step 2: Using a random number generator, generate 3
numbers ( , )
Step 3: Use , as a CDF and enter CDF plots for
H, , and Sw and read values of H, , and Sw
Calculate (in field units)
7758 AH (1 − S wc )
N=
Boi
1
1
CDF
CDF
Procedure Continue
0
1
1
0
Sw
Procedure Continue
Step 4: Repeat Step 3 say a fixed number of times
(example 1000 times).
Step 5: Now you have generated data for N, or you
have generated a distribution for N.
Step 6: The reserve distribution (pdf of N) can be
estimated as follows
*
x
f =
*
XN
X = total # of N values
x*= number of N* values within a class interval
Procedure Continue
f is a pdf ____
f (N )dN = 1
You can also find empirical CDF of N.
From empirical CDF of N, can get
(1) the mean N value
(2) the standard deviation.
If N is normally distributed N 2s is within 95%
confidence interval.
Procedure Continue
If have analytic function for pdf may estimate
probability for reserves N to be within a certain interval
b
P(a N b ) = f ( N )dN
a
Reserve Classification
UPSCALING
Static and Dynamic Properties
Up-Scaling
1
Module Objectives
2
Volumes of Investigation
3
UPSCALING
Definition: Up-scaling techniques
transform the reservoir properties from
the fine-scale grid into effective properties
for a grid with coarser blocks, such as the
simulation results are the same at both
scales.
4
Example of UpScaling
5
Why Do We UpScale?
6
Upgridding vs. Upscaling?
7
Upgridding involves averaging
8
Re-cap: Average Calculations
9
Exercice on Upgridding
10
Exercise on Upgridding
➢
➢
➢
➢
➢
Data heterogeneity is better preserved with
3 layers.
Apply geometric average for uncorrelated
perm.
Use harmonic average for vertical perm, or
series vertical layering.
Apply arithmetic average for porosity.
Add-up thicknesses for a set of layers into
new layer thickness.
11
Upgridding Optimization Example
Going from 246 layers to 46 layers can still match the
sweep efficiency reasonably well
12
Upgridding Optimization Example
13
Upgridding Optimization Example
14
Example: Layer Upgridding from 13 to 4 Layers
For this example: Use harmonic average for permeability
15
Example: Layer Upgridding from 13 to 6 Layers
For this example: Use harmonic average for permeability
16
Example: Layer Upgridding from 13 to 6 Layers
For this example: Use harmonic average for permeability
17
Example: Layer Upgridding from 13 to 7 Layers
For this example: Use harmonic average for permeability
18
Example: Layer Upgridding from 13 to 8 Layers
Use harmonic average for permeability
19
Example: Layer Upgridding from 13 to 9 Layers
Use harmonic average for permeability
20
Example: Layer Upgridding from 13 to 10 Layers
Use harmonic average for permeability
21
Summary: Comparing 3 levels of Vertical Upgridding with Lorenz Plots
22
23
24
UPSCALING
Objectives: Calculate for a coarser scale
grid the following:
✓
✓
✓
✓
✓
✓
Gross Thickness
Porosity
Permeability
Relative permeability
Capillary pressure
Water Saturation
25
26
SCALING UP
In general, the statistics of the effective properties
become narrower as the scale increases.
27
SCALING UP
✓
The standard deviation and range of the coarse
scale reservoir properties are smaller than those
for the fine scale.
✓
The distributions of the coarse scale
should preserve the overall character of
the fine-scale distributions.
28
Gross Thickness
29
Gross Thickness
30
Porosity
If, Ai is a constant, then
31
Porosity
If Ai is a constant.
32
Porosity
33
Water Saturation
34
35
Water Saturation
36
Numerical Example 1
Scaleup of Static Properties
This example illustrates the scaleup calculations
for gross thickness, porosity, and initial water
saturation.
The fine scale grid, consisting of 1X1 ft blocks,
and the reservoir properties for each block are
shown in the next Figure.
Calculate the effective properties of a coarse block
with the dimensions 5 X5 ft, which includes all
fine-scale blocks.
37
Numerical Example 1: Scaleup of Static
Properties of a fine-scale grid (1X1 ft)
38
Numerical Example 1: Scaleup of Static
Properties of a fine-scale grid (1X1 ft)
39
Numerical Example 1: Scaleup of Static
Properties of a fine-scale grid (1X1 ft)
40
41
Numerical Example 1: Scaleup of Static
Properties of a fine-scale grid (1X1 ft)
42
Scaleup for Dynamic Properties
Single Phase Flow: Permeability
Because permeability is a dynamic property,
scaleup depends on:
1. Relative special arrangement of fine-scale.
2. Boundary conditions
3. Anisotropy
43
Scaleup for Dynamic Properties
Single Phase Flow: Permeability
For some ideal special arrangements of
permeabilities and boundary conditions,
analytical equations for scaleup are available.
For more general cases, permeability scaleup is
based on numerical solutions of the partial
differential equations that govern single phase
flow.
44
Parallel Permeability Layers
45
Permeability Layers in Series
46
Permeability Layers in Series
47
Scaleup for Permeability
Permeability is completely uncorrelated
48
Scaleup for Permeability
The power average
49
Scaleup for Permeability Comparison
For the same fine-scale values:
✓ The arithmetic average gives the highest
✓ The harmonic average gives the lowest
✓ The geometric average gives a value in
between the above two approximations
50
Directional Upscaling of Permeability
k = k max k min
51
Directional Upscaling of Permeability
52
53
Example: Limitation of Directional Upscaling of Permeability:
Consider the 4 parallel layers between the no-flow boundary zones
54
How to get around this problem?: Use Flow-Based Upscaling:
Similar to Measuring K across a Plug
55
Scaleup for Permeability:
56
Scaleup Using Renormalization Method
57
58
Renormalization Method
59
Numerical Example 2: Scaleup of Permeability
of a fine-scale grid (5X5 ft)
60
Numerical Example 2: Scaleup of Permeability of a fine-scale grid (5X5 ft)
61
Numerical Example 2: Scaleup of Permeability of a fine-scale grid (5X5 ft)
62
63
Numerical Example 2: Scaleup of Permeability of a fine-scale grid (5X5 ft)
64
Numerical Example 2: Scaleup of Permeability of a fine-scale grid (5X5 ft)
65
Numerical Example 2: Scaleup of Permeability of a fine-scale grid (5X5 ft)
66
Numerical Example 2: Scaleup of Permeability of a fine-scale grid (5X5 ft)
67
2. Perform upgridding, and reduce the number of layers into 3 effective layers. Indicate layers
that are clustered together. Put your answer in a table format. Calculate the effective permeability
for each cluster of combined layers. (10 points)
PROBLEM # 4 (20 points)
The table below shows vertical permeability in mD for 10 layers of equal thickness (each layer is
10 ft thick). Each layer has a porosity value of 10%.
Layer Perm
5
2
15
3
1
9
4
5
6
7
12
175
350
0.8
0.5
1.5
0.1
8
9
10
1. Estimate the Lorenz coefficient for the tabulated data. Show your work and make necessary
plot(s). (10 points)
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