Harvard University Statistical Simulation Technique Questions

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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 Qs   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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