Dykstra-parsons Coefficient (Vk)

This method used Log-Normal distribution of permeability to define the Coefficient of permeability Variation, Vk.

Let’s upload some Real Data to work on it.

data<-read.csv("C:/karpur.csv",header=T)
summary(data)
##      depth         caliper         ind.deep          ind.med       
##  Min.   :5667   Min.   :8.487   Min.   :  6.532   Min.   :  9.386  
##  1st Qu.:5769   1st Qu.:8.556   1st Qu.: 28.799   1st Qu.: 27.892  
##  Median :5872   Median :8.588   Median :217.849   Median :254.383  
##  Mean   :5873   Mean   :8.622   Mean   :275.357   Mean   :273.357  
##  3rd Qu.:5977   3rd Qu.:8.686   3rd Qu.:566.793   3rd Qu.:544.232  
##  Max.   :6083   Max.   :8.886   Max.   :769.484   Max.   :746.028  
##      gamma            phi.N            R.deep            R.med        
##  Min.   : 16.74   Min.   :0.0150   Min.   :  1.300   Min.   :  1.340  
##  1st Qu.: 40.89   1st Qu.:0.2030   1st Qu.:  1.764   1st Qu.:  1.837  
##  Median : 51.37   Median :0.2450   Median :  4.590   Median :  3.931  
##  Mean   : 53.42   Mean   :0.2213   Mean   : 24.501   Mean   : 21.196  
##  3rd Qu.: 62.37   3rd Qu.:0.2640   3rd Qu.: 34.724   3rd Qu.: 35.853  
##  Max.   :112.40   Max.   :0.4100   Max.   :153.085   Max.   :106.542  
##        SP          density.corr          density         phi.core    
##  Min.   :-73.95   Min.   :-0.067000   Min.   :1.758   Min.   :15.70  
##  1st Qu.:-42.01   1st Qu.:-0.016000   1st Qu.:2.023   1st Qu.:23.90  
##  Median :-32.25   Median :-0.007000   Median :2.099   Median :27.60  
##  Mean   :-30.98   Mean   :-0.008883   Mean   :2.102   Mean   :26.93  
##  3rd Qu.:-19.48   3rd Qu.: 0.002000   3rd Qu.:2.181   3rd Qu.:30.70  
##  Max.   : 25.13   Max.   : 0.089000   Max.   :2.387   Max.   :36.30  
##      k.core            Facies           phi.cor.frac   
##  Min.   :    0.42   Length:819         Min.   :0.1570  
##  1st Qu.:  657.33   Class :character   1st Qu.:0.2390  
##  Median : 1591.22   Mode  :character   Median :0.2760  
##  Mean   : 2251.91                      Mean   :0.2693  
##  3rd Qu.: 3046.82                      3rd Qu.:0.3070  
##  Max.   :15600.00                      Max.   :0.3630

100 samples of K.core data will be worked on.

First , the Frequency should be found for K.core.

# Create dataframe
b<-table(data$k.core)
df=data.frame(b)
t=t(df)
t=as.data.frame(t)
r=rev(t)
r=t(r)
r=as.data.frame(r)
K.core=r$Var1
Frequency=r$Freq
df2=data.frame(K.core,Frequency)
mo=head(df2,100)
mo1=head(df2,20)
mo1
       K.core Frequency
1       15600         1
2  14225.3135         1
3  13544.9785         1
4  13033.5283         1
5  11841.7432         1
6  11117.4023         1
7       10860         1
8   10649.958         1
9       10540         1
10  9898.4785         1
11  9533.3623         1
12  9458.1729         1
13       9120         1
14  8820.1025         1
15       8820         2
16       8760         1
17  8742.7529         1
18  8689.8252         1
19       8390         1
20  8306.0459         1

Then, the Number of samples with large Permeability will be found and the Cumulative Frequency Distribution.

df5=data.frame(mo)
m=seq(1,100)
df3=data.frame(m)
df3$m=as.numeric(df3$m)
df5$Frequency=as.numeric(df5$Frequency)
f=df3$m+df5$Frequency
s=table(f)
df4=data.frame(s)
no=rbind(data.frame(f = 1, Freq = 1), df4)
rr=no$f
dd=data.frame( rr)
bb=as.numeric(rr)
gg=(bb/102)*100
df6=data.frame(K.core=df5$K.core,Freq=df5$Frequency,NO.samples=dd$rr,Cum_Freq=gg)
head(df6,20)
       K.core Freq NO.samples   Cum_Freq
1       15600    1          1  0.9803922
2  14225.3135    1          2  1.9607843
3  13544.9785    1          3  2.9411765
4  13033.5283    1          4  3.9215686
5  11841.7432    1          5  4.9019608
6  11117.4023    1          6  5.8823529
7       10860    1          7  6.8627451
8   10649.958    1          8  7.8431373
9       10540    1          9  8.8235294
10  9898.4785    1         10  9.8039216
11  9533.3623    1         11 10.7843137
12  9458.1729    1         12 11.7647059
13       9120    1         13 12.7450980
14  8820.1025    1         14 13.7254902
15       8820    2         15 14.7058824
16       8760    1         17 16.6666667
17  8742.7529    1         18 17.6470588
18  8689.8252    1         19 18.6274510
19       8390    1         20 19.6078431
20  8306.0459    1         21 20.5882353

Now, the data is ready to be plotted.

df6$K.core=as.numeric(df6$K.core)
df6$K.core=log(df6$K.core)
slop=gg
model1=lm(df6$K.core ~ slop)
plot(gg,df6$K.core,xlab ='Percent Samples with larger Permeability',ylab='log Permeability, md',axes = F)
axis(2,col = "darkgreen",col.axis="red")
axis(1,col = "darkgreen",col.axis="blue")
abline(model1, lwd=3, col='red')

#find the intercept and slop
model1

Call:
lm(formula = df6$K.core ~ slop)

Coefficients:
(Intercept)         slop  
   9.272671    -0.009271

Now, Heterogeneity Index Can be found as flow.

#find K50
K50=9.272671-0.009271*50
K50=exp(K50)
#find K84.1
K84.1=9.272671-0.009271*84.1
K84.1=exp(K84.1)
#find Heterogeneity Index
HI=(K50-K84.1)/K50
HI
[1] 0.2710434

According to Dykstra-Parsons method this value refer to Slightly Heterogeneity in the Reservoir.

Lorenz Coefficient (Lk)

Since 1950, the Lorenz Coefficient has provided a practical way to quantify layered permeability heterogeneity. Simple to calculate from a set of permeability and porosity measurements and naturally bounded between 0 – 1.

First , Thickness will calculate by using Depth.

#Calculate thickness from depth
thickness = c(0.5)
for (i in 2:819) {
  h = data$depth[i] - data$depth[i-1]
  thickness = append(thickness, h)
}

data = data[order(data$k.core, decreasing = TRUE), ]

Now, the Graph is ready to be plotted.

plot(FTV,FTF,xlab="Fraction of Total Volume",ylab="Fraction of Total Flow Capacity", pch = 10, cex = 0.2, col = 'red', text(0.4, 0.8, "B"))
text(0,0, "A", pos = 2)
text(1,1, "C", pos = 1)
text(1,0, "D", pos = 2)
abline(0,1, col = 'darkblue', lwd = 4)

``` According to Lorenz Coefficient (Lk) method this value refer to heterogeneous Reservoir.