Determination of Heterogeneity Index(HI) by Dykstra-Parsons Coefficient (Vk)
“Mohammed Redah Majid Mohammed”
“Basra University For Oil and Gas”
“2024-10-25”
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:/Users/mslme/Desktop/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
## Min. : 0.42 Length:819
## 1st Qu.: 657.33 Class :character
## Median : 1591.22 Mode :character
## Mean : 2251.91
## 3rd Qu.: 3046.82
## Max. :15600.00100 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$Freq
df2=data.frame(K.core,Frequency)
mo=head(df2,100)
mo1=head(df2,20)
mo1## K.core Frequency
## 1 15600 1
## 2 14225.3135 1
## 3 13544.9785 1
## 4 13033.5283 1
## 5 11841.7432 1
## 6 11117.4023 1
## 7 10860 1
## 8 10649.958 1
## 9 10540 1
## 10 9898.4785 1
## 11 9533.3623 1
## 12 9458.1729 1
## 13 9120 1
## 14 8820.1025 1
## 15 8820 2
## 16 8760 1
## 17 8742.7529 1
## 18 8689.8252 1
## 19 8390 1
## 20 8306.0459 1Then, the Number of samples with large Permeability will be found and the Cumulative Frequency Distribution.
df5=data.frame(mo)
m=seq(1,100)
df3=data.frame(m)
df3$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$rr,Cum_Freq=gg)
head(df6,20)## K.core Freq NO.samples Cum_Freq
## 1 15600 1 1 0.9803922
## 2 14225.3135 1 2 1.9607843
## 3 13544.9785 1 3 2.9411765
## 4 13033.5283 1 4 3.9215686
## 5 11841.7432 1 5 4.9019608
## 6 11117.4023 1 6 5.8823529
## 7 10860 1 7 6.8627451
## 8 10649.958 1 8 7.8431373
## 9 10540 1 9 8.8235294
## 10 9898.4785 1 10 9.8039216
## 11 9533.3623 1 11 10.7843137
## 12 9458.1729 1 12 11.7647059
## 13 9120 1 13 12.7450980
## 14 8820.1025 1 14 13.7254902
## 15 8820 2 15 14.7058824
## 16 8760 1 17 16.6666667
## 17 8742.7529 1 18 17.6470588
## 18 8689.8252 1 19 18.6274510
## 19 8390 1 20 19.6078431
## 20 8306.0459 1 21 20.5882353Now, the data is ready to be plotted.
df6$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 -0.009271Now, 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] 0.2710434According to Dykstra-Parsons method this value refer to Slightly Heterogeneity in the Reservoir.