data = read.csv(“karpur.csv”) head(data)
depth caliper ind.deep ind.med gamma phi.N R.deep R.med SP1 5667.0 8.685 618.005 569.781 98.823 0.410 1.618 1.755 -56.587 2 5667.5 8.686 497.547 419.494 90.640 0.307 2.010 2.384 -61.916 3 5668.0 8.686 384.935 300.155 78.087 0.203 2.598 3.332 -55.861 4 5668.5 8.686 278.324 205.224 66.232 0.119 3.593 4.873 -41.860 5 5669.0 8.686 183.743 131.155 59.807 0.069 5.442 7.625 -34.934 6 5669.5 8.686 109.512 75.633 57.109 0.048 9.131 13.222 -39.769 density.corr density phi.core k.core Facies 1 -0.033 2.205 33.9000 2442.590 F1 2 -0.067 2.040 33.4131 3006.989 F1 3 -0.064 1.888 33.1000 3370.000 F1 4 -0.053 1.794 34.9000 2270.000 F1 5 -0.054 1.758 35.0644 2530.758 F1 6 -0.058 1.759 35.3152 2928.314 F1 1- Dykstra-Parsons
This method required using only the K.core column from our data-set.
Here I am sorting the whole data-set descending based on K.core than spliting cored permeability as an individual vector:
#SORTING BASED ON K CORE
data = data[order(data\(k.core, decreasing = TRUE), ] K = data\)k.core
Next creating another vector containing the number of samples ≥ k, while the order of data is descending the 1st K will has only one permeability greater or equals to its value while the last K has 819 (data length) of permeability values greater or equals to its value.
sample = c(1:819)
The next parameter required is the percent of samples that have permeability greater or equal to the current K value, which represent the probability of current K value. note: the division on [length of the data-set +1] to neglect having probability of zero aNow, all parameters is ready to use. Let’s plot Y values (logarithmic permeability) and X values (probability):
xlab = “Portion of Total Sample Having Higher K” ylab = “Permeability (md)” plot(k_percent, K, log = ‘y’, xlab = xlab, ylab = ylab, pch = 10, cex = 0.5, col = “#001c49”) Now, let’s find the Linear Model that fits the data, lm() function used to build linear regression model:
log_k = log(K) model = lm(log_k ~ k_percent) plot(k_percent,log_k, xlab = xlab, ylab = ylab, pch = 10, cex = 0.5, col = “#001c49”) abline(model, col = ‘red’, lwd = 2) summary(model)
Call:
lm(formula = log_k ~ k_percent)
Residuals:
Min 1Q Median 3Q Max
-5.8697 -0.2047 0.1235 0.3150 0.4280
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 9.2584172 0.0377994 244.94 <2e-16 ***
k_percent -0.0426137 0.0006549 -65.07 <2e-16 ***
—
Signif. codes: 0 ‘’ 0.001 ’’ 0.01 ’’ 0.05 ‘.’ 0.1 ’ ’ 1
Residual standard error: 0.5404 on 817 degrees of freedom
Multiple R-squared: 0.8382, Adjusted R-squared: 0.838
F-statistic: 4234 on 1 and 817 DF, p-value: < 2.2e-16
Our model has good R2 value which is obvious from the plot. abline is just adding the fitted line.
Now let’s find K values at K50 & K84.1 , by define new data frame that contains the required values:
new_data = data.frame(k_percent = c(50, 84.1))
Now using predict () function to predict from given model:
predicted_values = predict(model, new_data)
so the logarithmic permeability of K50 = 7.13 & K84.1 = 5.67
to calculate heterogeneity index:
V=K50−K84.1K50
heterogenity_index = (predicted_values1 - predicted_values[2]) / predicted_values1
heterogenity_index
1
0.2038692
So, the heterogeneity index = 0.204 , which indicate slightly heterogeneous class of permeability distribution, the low value of heterogeneity allow to use homogeneous model with minimum error. 2- Lorenz Coefficient
This method is measuring the inequality of distribution of a property, in finance for example: Is the income distributed equally among the population? The same concept can be applied in reservoir engineering but between the flow capacity (KH) and storage volume (PHIK).
Using the same data:
data = read.csv(“karpur.csv”)
Then, we need to calculate the thickness of each permeability based on Depth[i] - Depth[i-1] , then update the data by sorting it based on Permeability.core:
#thickness calculation 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), ]
Then, calculate the flow capacity and storage capacity which is easily the multiplication between permeability and porosity with corresponding thickness. Then calculate the cumulative sum of each one using cumsum() function.
KH = thickness * data\(k.core PH = thickness * data\)phi.core cum_sum_kh = cumsum(KH) cum_sum_ph = cumsum(PH)
Finally we need to calculate the following terms:
∑cum.k.coretotal & ∑cum.phi.coretotal
frac_tot_vol = (cum_sum_ph/100) / max(cum_sum_ph/100) #to convert phi between (0,1) frac_tot_flow = cum_sum_kh / max(cum_sum_kh)
Now, plotting the percent of storage capacity on X-axis and the percent of flow capacity on Y-axis:
plot(frac_tot_vol,frac_tot_flow, pch = 10, cex = 0.2, col = ‘#001c49’, text(0.4, 0.8, “D”)) text(0,0, “A”, pos = 2) text(1,1, “B”, pos = 1) text(1,0, “C”, pos = 2) abline(0,1, col = ‘red’, lwd = 2) Now, We need to estimate the area under the curve (ADBCA).
using AUC() function that approximate the area under a curve, trapezoid the numerical technique that used:
require(“DescTools”)
Loading required package: DescTools
Warning: package ‘DescTools’ was built under R version 4.2.2
library(DescTools)
area = AUC(frac_tot_vol, frac_tot_flow, method=“trapezoid”)
The formula of Lorenz Coefficient:
LorenzCofficient=AreaADBAABCA
Area ADBA, is just area value - 0.5:
Lcoff = (area - 0.5) / 0.5
The heterogeneity Index:
Lcoff
1 0.4524741
is 0.45 , in Lorenz method if the value is closer to zero, it considers more heterogeneous. So the current value considers as heterogeneous or high heterogeneous reservoir.