term project 2 pte 508 numerical simulation of subsurface ... · pdf filepte 508 numerical...

TERM PROJECT 2

PTE 508 Numerical Simulation of Subsurface Flow and Transport Process

2-D 2-PHASE STUDY OF A RESERVOIR BY NUMERICAL SIMULATION AND

PARTIAL DIFFERENTIAL DISCRETIZATION

Thomas John, Nithin

USC ID: 5988 3597 37

PROJECT REVIEW

A 2-D 2-Phase 1-Layer reservoir was numerically simulated with partial differential

equations (PEDs) in MATLAB to study the performance of five wells (A, B, C, D, E) under

two given scenarios; primary recovery and secondary recovery.

The simulation was run through three primary tests to account for pressure and saturation

conditions. These included the LeakProofTest, SymmetryTest and MaterialBalanceTest.

In addition to the tests, primary recovery included producing the five wells at a constant

rate and bottom-hole pressure (BHP) until the field production went below 1.0 STB/day.

Secondary mechanism involved making a well; Well-C as an injector, and producing the

other wells at a BHP of 1000 psia. The mechanism was run for 10,000 days or when

water cut in Well A increased above 95%.

MATLAB Coding was extensively used for successful culmination of the project. Pertinent

plots and MATLAB codes are provided for delineation.

The reservoir description and filed contour map are given below:

With the given reservoir parameters and field location for each well, various tests were

performed prior to initiating the primary and secondary recovery mechanisms.

7. Calculation of initial reservoir parameters

A MATLAB code was generated to calculate initial reservoir parameters such as; Bo, Bw,

phi, krw, kro, So and Sw.

Both kro and krw were plotted against Sw in the same graph. Moreover, Bo and Bw were

also plotted with respect to pressure, P.

8. Leak-Proof Tests

A leak-proof test was conducted by setting the well rate to 0 for all the five wells and by

running a 100-day simulation with a time step of 10 days. Both pressure leak proof and

saturation leak proof tests were conducted and the results were plotted. In pressure leak

proof, the pressure was equal to the initial reservoir pressure. In saturation leak proof test,

both oil and water saturations were unchanged.

Figure: Pressure Leak Proof Test

Figure: Saturation Leak Proof Test

9. Symmetry Tests

A symmetric test was performed by setting bottom hole flowing pressure (BHFP) to 1000

psia and by running a 100 day simulation with a time step of 10 days. The pressure

distribution was made to be symmetric with the middle point of the grid system. The

pressure distributions were plotted for t = 0 days, 50 days and 100 days.

Figure: Symmetry Test for 0 days

Table: Symmetry Test Results for 0 days

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

1 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000

2 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000

3 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000

4 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000

5 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000

6 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000

7 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000

8 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000

9 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000

10 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000

11 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000 3000

10. Material Balance Test

The MBE Test is done to compare volumetric calculation of initial oil in place (OOIP) and

simulated OOIP. A tolerance level of 0.1% is accepted for accuracy.

The following equations were used to calculate Np and OOIP.

Tolerance Level (Error %) = ((Npn+1 + OIPn+1)/OOIP) – 1

A 100-day long simulation was run with time steps of 10 days and all the five wells were

made to produce at 1000 psia. For each step, the cumulative production and OIP were

calculated to account for accuracy.

Initial Oil in Place = OOIP = 1.9866E+06 STB

Time, t (days)

Original Oil in Place, OOIP (STB)

Cumulative Oil

Production, Np (STB)

Tolerance Level (%)

10 1986557.7 12728.34 0

20 1978251.4 6676.62 0.222596

30 1974425.2 5412.17 0.144087

40 1971280.2 4540.30 0.114831

50 1968603.9 3870.02 0.094558

60 1966295.2 3339.27 0.079307

70 1964280.8 2910.38 0.067381

80 1962505.9 2558.02 0.057807

90 1960928.4 2264.46 0.049963

100 1959515.6 2016.93 0.043429

11. i. Primary Recovery Study

A primary recovery study was initiated by calculating the initial oil in place (OIIP). The five

wells are made into producers by setting their BHFPs to 1000 psia. The reservoir is

produced until the production of Well A falls below 1.0 STB/d. Both oil production and

water production rates for the five wells are calculated and plotted at this time.

The cumulative reservoir production and recovery factor (RF) are calculated and plotted.

Finally, the reservoir pressure distribution for t = 0, 100, 200, 300, 400 and 500 days are

plotted.

OOIP =

It was found that the production rate of Well A falls below 1.0 STB/d after t = 530 days.

At this time, primary recovery is stopped and corresponding plots are generated from

MATLAB.

Figure: Oil Production Rate Vs Time_PrimaryRecovery

Figure: Water Production Rate Vs Time_PrimaryRecovery

Figure: Cumulative Production Rate Vs Time_PrimaryRecovery

Figure: Recovery Factor Vs Time_PrimaryRecovery

The cumulative oil production was NP = 6.984E+4 STB

Recovery Factor (RF) of the reservoir = RF = 3.46

Figure: Pressure Distribution Plots

ii. Secondary Recovery Study

A secondary recovery study was conducted on the reservoir. All the five wells were made

to produce until the oil production of Well A fell below 1.0 STB/d. Well C was made into

an injector by setting it’s BHFP to 2950 psia. The simulation is run until water cut is greater

than 95% or the simulation time reaches 10,000 days.

The oil saturation and pressure distribution of Well C was plotted. The cumulative

production and recovery factor were also plotted.

The secondary recovery was run for 10,000 days and the following pressure distribution

was observed. We also observe that water injection has begun at t = 550 days

The cumulative oil production was NP = 7.895E+05 STB

Recovery Factor (RF) of the reservoir = RF = 38.6160

CONCLUSIONS & SUGGESTIONS

From the reservoir simulation study conducted, we observe that secondary recovery

mechanism by water injection greatly increases the recovery factor and cumulative oil

production from the reservoir.

The RF increased from 3% to 35% from primary to secondary recovery. This was

achieved by making Well C as an injector to maintain the pressure inside the reservoir.

This was one of the most riveting and challenging term projects I have come across in

my academic study. It was interesting and greatly enhanced my coding skills in MATLAB.

I personally feel that a much more professional approach to MATLAB should be first

taught in class before moving on to the term project.

In sum, I’m satisfied with my work in the term project for PTE 508.

MATLAB CODES

P = 3000:-10:1000; %Pressure Range Swc = 0.2; %Connate Water Saturation So = 0.75:-0.01:0; %Oil Saturation Sw = 1-So; %Water Saturation Sn = (1-So-Swc)/(1-Swc); %Correy's Correlation

Cp = 2*10^-6; %Rock Compressibility Co = 10^-5; %Oil Compressibility Cw = 10^-6; %Water Compressibility

krw = Sn.^4; %Relative Permeability_water kro = ((1-Sn^2)).*(1-((Sn)))^2; %Relative Permeability_oil figure; plot(Sw,krw); hold on; plot(Sw,kro);

phi = 0.22*exp(Cp*(P-1000)); %Porosity

Bo = 1.25 * exp(-Co*(P-1000)); %Oil FVF Bw = 1.02 * exp(-Cw*(P-1000)); %Water FVF figure; plot(P,Bo); figure; plot(P,Bw); figure; plot(P,phi);

clc; clear all;

%Given Reservoir Data Lx = 1500; %length along x_axis in ft Ly = 1100; %length along y_axis in ft deltaz = 50; %thickness in ft k = 11; %reservoir permeability in mD rw = 0.25; %wellbore radius in ft muo = 5; muw = 1; %viscosity in cP; 'o' stands for oil and 'w' denotes water

nx = 15; %number of simulation rows ny = 11; %number of simulation columns nxy = nx*ny; Ones = ones(nxy,1); %initialization

dx = Lx/nx; dy = Ly/ny; deltat = 10; %time steps re = 0.14*(dx^2 + dy^2)^0.5; %drainage radius in ft

A = 34; B = 42; C = 83; D = 124; E = 132; %well locations

Soi(1:nxy) = 0.75; %initial oil saturation So(1:nxy) = 0.75; %oil saturation

Swi(1:nxy) = 0.25; %initial water saturation Sw(1:nxy) = 0.25; %water saturation

Sor = 0; %residual oil saturation Swc = 0.2; %connate water saturation Sn = (1-So-Swc)/(1-Swc); %Corey's correlation

Pwf = 1000; %BHFP in psia Pwfinj = 1000; %Injection Pressure in psia constant = 1;

P(1:nxy) = 3000; %reservoir pressure in psia Po(1:nxy) = 3000; %reservoir pressure in psia Pr(1:nxy) = 3000; %reservoir pressure in psia

Co = 10^-5; %oil compressibility in psia^-1 Cw = 10^-6; %water compressibility in psia^-1 Cphi = 2*10^-6 ; %rock compressibility in psia^-1

Boi(1:nxy)=1.25*exp(-Co*(P(1:nxy)-1000)); %oil FVF in RB/STB Bwi(1:nxy)=1.02*exp(-Cw*(P(1:nxy)-1000)); %water FVF in RB/STB

Phi_ini(1:nxy) = 0.22*exp(-Cphi*(P(1:nxy)-1000)); %porosity as a function of

pressure

VB = (Lx*Lx*deltaz)/5.615; %bulk volume in cu.ft VPi=VB*Phi_ini(1); %pore volume in cu.ft OIIP = VPi*Soi(1)/Boi(1); %oil in place in BBL

%matrix cell allocation Sm = Ones; Nm = Ones; Wm = Ones; Em = Ones; Cm = Ones; Rm = Ones; Smo = Ones; Nmo = Ones; Wmo = Ones; Cmo = Ones; Emo = Ones; O_ON=1; W_ON=1;

for t = 1:10 for a = 1:50

for i = 1:nx; for j = 1:ny; m = (j-1)*nx+i; if t>50 constant = 0; Pwfinj = 2950; end if t == 1 if a == 1 Bo(t,m) = 1.25*exp(-Co*(Po(m)-1000)); Bw(t,m) = 1.02*exp(-Cw*(Po(m)-1000));

So(t,m) = So(m); Sw(t,m) = 1-So(m); phi(t,m) = 0.2*exp(-Cphi*(Po(m)-1000));

elseif a>1 Bo(t,m) = 1.25*exp(-Co*(P(t,m)-1000)); Bw(t,m) = 1.02*exp(-Cw*(P(t,m)-1000)); So(t,m) = So(t,m); Sw(t,m) = 1-So(t,m); phi(t,m) = 0.2*exp(-Cphi*(P(t,m)-1000));

end

else if a == 1 Bo(t,m) = 1.25*exp(-Co*(P(t-1,m)-1000)); Bw(t,m) = 1.02*exp(-Cw*(P(t-1,m)-1000)); So(t,m) = So(t-1,m); Sw(t,m) = 1-So(t-1,m); phi(t,m) = 0.2*exp(-Cphi*(P(t-1,m)-1000));

elseif a>1 Bo(t,m) = 1.25*exp(-Co*(P(t,m)-1000)); Bw(t,m) = 1.02*exp(-Cw*(P(t,m)-1000)); So(t,m) = So(t,m); Sw(t,m) = 1-So(t,m); phi(t,m) = 0.2*exp(-Cphi*(P(t,m)-1000));

end end end end

for i = 1:nx for j = 1:ny m = (j-1)*(nx)+i; Sn(t,m)=(1-So(t,m)-Swc)/(1-Swc);

if Sw(t,m)>Swc; krw(t,m) = Sn(t,m)^4; %relative permeability of water

else krw(t,m) = 0; end

if So(t,m)>Sor; kro(t,m) = (1-Sn(t,m)^2)*(1-Sn(t,m))^2; %oil relative

permeability else kro(t,m)=0; end

lmbdo(t,m)=k*kro(t,m)/(muo); %mobility of oil lmbdw(t,m)=k*krw(t,m)/(muw); %mobility of water

Ce(m)=(Cphi+(1-So(t,m))*Cw+So(t,m)*Co); Jo(m)=O_ON*(7.08E-

3*k*kro(t,m)*deltaz)/(muo*Bo(t,m)*2.302585*log(re/rw)); %PI of oil Jw(m)=W_ON*(7.08E-

3*k*krw(t,m)*deltaz)/(muw*Bw(t,m)*2.302585*log(re/rw)); %PI of water

end end

for i = 1:nx; for j = 1:ny; m = (j-1)*nx+i; if t == 1 if a == 1 P(t,m) = Po(m); elseif a>1 P(t,m) = P(t,m); end

else if a == 1 P(t,m) = P(t-1,m); elseif a>1 P(t,m) = P(t,m); end end

if i == nx; %boundary condition (east) lmdEo(m) = (k*kro(t,m)/(muo))/Bo(t,m); lmdEw(m)= (k*krw(t,m)/(muw))/Bw(t,m); else if P(t,m+1)>P(t,m); lmdEo(m)=lmbdo(t,m+1)/(0.5*(Bo(t,m)+Bo(t,m+1))); lmdEw(m)=lmbdw(t,m+1)/(0.5*(Bw(t,m)+Bw(t,m+1))); else lmdEo(m)=lmbdo(t,m)/(0.5*(Bo(t,m)+Bo(t,m+1))); lmdEw(m)=lmbdw(t,m)/(0.5*(Bw(t,m)+Bw(t,m+1))); end end

if i == 1; %boundary condition (west) lmdWo(m)=lmbdo(t,m)/Bo(t,m); lmdWw(m)=lmbdw(t,m)/Bw(t,m); else if P(t,m-1)>P(t,m) lmdWo(m)=lmbdo(t,m-1)/(0.5*(Bo(t,m)+Bo(t,m-1))); lmdWw(m)=lmbdw(t,m-1)/(0.5*(Bw(t,m)+Bw(t,m-1))); else lmdWo(m)=lmbdo(t,m)/(0.5*(Bo(t,m)+Bo(t,m-1))); lmdWw(m)=lmbdw(t,m)/(0.5*(Bw(t,m)+Bw(t,m-1))); end end

if j == 1; %boundary condition (south) lmdSo(m)=lmbdo(t,m)/Bo(t,m); lmdSw(m)=lmbdw(t,m)/Bw(t,m);

else if P(t,m-nx)>P(t,m) lmdSo(m)=lmbdo(t,m-nx)/(0.5*(Bo(t,m)+Bo(t,m-nx))); lmdSw(m)=lmbdw(t,m-nx)/(0.5*(Bw(t,m)+Bw(t,m-nx))); else lmdSo(m)=lmbdo(t,m)/(0.5*(Bo(t,m)+Bo(t,m-nx))); lmdSw(m)=lmbdw(t,m)/(0.5*(Bw(t,m)+Bw(t,m-nx))); end end

if j == ny; %boundary condition (north) lmdNo(m)=lmbdo(t,m)/Bo(t,m); lmdNw(m)=lmbdw(t,m)/Bw(t,m); else if P(t,m+nx)>P(t,m) lmdNo(m)=lmbdo(t,m+nx)/(0.5*(Bo(t,m)+Bo(t,m+nx))); lmdNw(m)=lmbdw(t,m+nx)/(0.5*(Bw(t,m)+Bw(t,m+nx))); else lmdNo(m)=lmbdo(t,m)/(0.5*(Bo(t,m)+Bo(t,m+nx))); lmdNw(m)=lmbdw(t,m)/(0.5*(Bw(t,m)+Bw(t,m+nx))); end end end end

for i = 1:nx; for j = 1:ny; m = (j-1)*nx+i; Cmo(m) = -Bo(t,m)*(lmdWo(m)+lmdEo(m)+lmdNo(m)+lmdSo(m)); Nmo(m) = Bo(t,m)*lmdNo(m); Smo(m) = Bo(t,m)*lmdSo(m); Emo(m) = Bo(t,m)*lmdEo(m); Wmo(m) = Bo(t,m)*lmdWo(m); end end i = nx; %boundary condition (east) for j = [1:ny]; m = (j-1)*nx+i; Cmo(m) = Cmo(m)+Emo(m); Emo(m) = 0; end

i=1; %boundary condition (west) for j = [1:ny]; m = (j-1)*nx+i; Cmo(m) = Cmo(m)+Wmo(m); Wmo(m) = 0; end

j = 1; %boundary condition (south) for i = [1:nx]; m = (j-1)*nx+i; Cmo(m) = Cmo(m)+Smo(m); Smo(m) = 0; end

j = ny; %boundary condition (north) for i = [1:nx]; m = (j-1)*nx+i; Cmo(m) = Cmo(m)+Nmo(m); Nmo(m) = 0; end

i = nx; %boundary condition (east) for j = [2:ny-1]; m = (j-1)*nx+i; Cmo(m) = Cmo(m)+Emo(m); Emo(m) = 0; end

i=1; %boundary condition (west) for j = [2:ny-1]; m =(j-1)*nx+i; Cmo(m) = Cmo(m)+Wmo(m); Wmo(m) = 0; end

for i=1:nx; for j=1:ny; m=(j-1)*nx+i; Em(m)=Emo(m)+Bw(t,m)*lmdEw(m); Wm(m)=Wmo(m)+Bw(t,m)*lmdWw(m); Sm(m)=Smo(m)+Bw(t,m)*lmdSw(m); Nm(m)=Nmo(m)+Bw(t,m)*lmdNw(m); Cm(m)=Cmo(m)-Bw(t,m)*(lmdWw(m)+lmdEw(m) + lmdNw(m) +

lmdSw(m))-(158*dx^2/deltat)*phi(t,m)*Ce(m); end end

i = nx; %boundary condition (east) for j = [1:ny]; m = (j-1)*nx+i; Cm(m) = Cm(m)+Em(m); Em(m) = 0; end

i = 1; %boundary condition (west) for j = [1:ny]; m = (j-1)*nx+i; Cm(m) = Cm(m)+Wm(m); Wm(m) = 0; end

j=1; %boundary condition (south) for i = [1:nx]; m = (j-1)*nx+i; Cm(m) = Cm(m)+Sm(m); Sm(m) = 0; end

j=ny; %boundary condition (north) for i = [1:nx]; m = (j-1)*nx+i; Cm(m) = Cm(m)+Nm(m); Nm(m) = 0; end

i = nx; %boundary condition (east) for j = [2:ny-1]; m = (j-1)*nx+i; Cm(m) = Cm(m)+Em(m); Em(m) = 0; end

i=1; %boundary condition (west) for j = [2:ny-1]; m = (j-1)*nx+i; Cm(m) = Cm(m)+Wm(m); Wm(m) = 0; end

for i = 1:nx; for j = 1:ny; m = (j-1)*nx+i; if m == A Cm(m) = Cm(m)-

887.53*dx*(Bo(t,m)*Jo(m)+Bw(t,m)*Jw(m))/(dy*deltaz); elseif m == B Cm(m) = Cm(m)-

887.53*dx*(Bo(t,m)*Jo(m)+Bw(t,m)*Jw(m))/(dy*deltaz); elseif m == C Cm(m) = Cm(m)-

887.53*dx*(constant*Bo(t,m)*Jo(m)+Bw(t,m)*Jw(m))/(dy*deltaz); elseif m == D Cm(m)=Cm(m)-

887.53*dx*(Bo(t,m)*Jo(m)+Bw(t,m)*Jw(m))/(dy*deltaz); elseif m == E Cm(m) = Cm(m)-

887.53*dx*(Bo(t,m)*Jo(m)+Bw(t,m)*Jw(m))/(dy*deltaz); else Cm(m) = Cm(m); end end end

for i = 1:nx; for j = 1:ny; m = (j-1)*nx+i; if t == 1 Pr(t,m)=Po(m); elseif t>1 Pr(t,m)=P(t-1,m); end

if m == A

Rm (m)= -158*(dx^2/deltat) * Ce(m)*phi(t,m) * Pr(t,m)-

887.53 * Pwf*dx * (Bo(t,m) * Jo(m) + Bw(t,m) * Jw(m))/(dy*deltaz); elseif m==B Rm (m)= -158*(dx^2/deltat)*Ce(m)*phi(t,m)*Pr(t,m)-

887.53*Pwf*dx*(Bo(t,m)*Jo(m)+Bw(t,m)*Jw(m))/(dy*deltaz); elseif m==C Rm (m)= -158*(dx^2/deltat)*Ce(m)*phi(t,m)*Pr(t,m)-

887.53*Pwfinj*dx*(Bo(t,m)*constant*Jo(m)+Bw(t,m)*Jw(m))/(dy*deltaz); elseif m==D Rm (m)=-158*(dx^2/deltat)*Ce(m)*phi(t,m)*Pr(t,m)-

887.53*Pwf*dx*(Bo(t,m)*Jo(m)+Bw(t,m)*Jw(m))/(dy*deltaz); elseif m==E Rm (m)=-158*(dx^2/deltat)*Ce(m)*phi(t,m)*Pr(t,m)-

887.53*Pwf*dx*(Bo(t,m)*Jo(m)+Bw(t,m)*Jw(m))/(dy*deltaz); else Rm (m)=-158*(dx^2/deltat)*Ce(m)*phi(t,m)*Pr(t,m); end end end

for m = nxy-1:-1:1 Em(m+1) = Em(m); end for m = 2:nxy Wm(m-1) = Wm(m); end for m = 1:nxy-nx Sm(m) = Sm(m+nx); end for m = nxy-nx:-1:1 Nm(m+nx) = Nm(m); end

matrixsol = spdiags([Sm, Wm, Cm, Em, Nm],[-nx -1 0 +1 +nx],nxy,nxy); y = full(matrixsol);

Ptemp = matrixsol\Rm; %temporary pressure solution for m = 1:nxy Ptemp2(t,m) = Ptemp(m); %temporary pressure solution 2 end

if a>1 if max(abs(Ptemp2(t,1:nxy)-P(t,1:nxy)))<0.001; break

else for m = 1:nxy; P(t,m) = Ptemp2(t,m); end end else P(t,m)=Ptemp2(t,m); end

%%Rate Calculation for m = 1:nxy

if m == A qA_o(t,m)=Jo(m)*(P(t,m)-Pwf); qA_w(t,m)=Jw(m)*(P(t,m)-Pwf); elseif m == B qB_o(t,m)=Jo(m)*(P(t,m)-Pwf); qB_w(t,m)=Jw(m)*(P(t,m)-Pwf); elseif m == C qC_o(t,m)=constant*Jo(m)*(P(t,m)-Pwf); qC_w(t,m)=Jw(m)*(P(t,m)-Pwf); elseif m == D qD_o(t,m) = Jo(m)*(P(t,m)-Pwf); qD_w(t,m) = Jw(m)*(P(t,m)-Pwf); elseif m==E qE_o(t,m)=Jo(m)*(P(t,m)-Pwf); qE_w(t,m)=Jw(m)*(P(t,m)-Pwf); else qA_o(t,m)=0; qA_w(t,m)=0; qB_o(t,m)=0; qB_w(t,m)=0; qC_o(t,m)=0; qC_w(t,m)=0; qD_o(t,m)=0; qD_w(t,m)=0; qE_o(t,m)=0; qE_w(t,m)=0; end end

for m = 1:nxy if m == A && m==B && m==C && m==D && m==E

Cum_oil(t,m)=887.53*dx*(qA_o(t,m)+qB_o(t,m)+qC_o(t,m)+qD_o(t,m)+qE_o(t,m))/(d

y*deltaz); else Cum_oil(t,m)=0; end end

%RateFlux Calculations for i = 1:nx for j = 1:ny m = (j-1)*nx+i; if i == 1 if j == 1 OilFlux(t,m) =

Cmo(m)*P(t,m)+Emo(m)*P(t,m+1)+Nmo(m)*P(t,m+nx); elseif j == ny OilFlux(t,m) = Smo(m)*P(t,m-

nx)+Cmo(m)*P(t,m)+Emo(m)*P(t,m+1); else OilFlux(t,m) = Smo(m)*P(t,m-

nx)+Cmo(m)*P(t,m)+Emo(m)*P(t,m+1)+Nmo(m)*P(t,m+nx); end

elseif i == nx

if j == 1 OilFlux(t,m)=Wmo(m)*P(t,m-

1)+Cmo(m)*P(t,m)+Nmo(m)*P(t,m+nx); elseif j == ny OilFlux(t,m) = Smo(m)*P(t,m-nx)+Wmo(m)*P(t,m-

1)+Cmo(m)*P(t,m); else OilFlux(t,m) = Smo(m)*P(t,m-nx)+Wmo(m)*P(t,m-

1)+Cmo(m)*P(t,m)+Nmo(m)*P(t,m+nx); end

elseif j == 1 if i == 1 OilFlux(t,m) =

Cmo(m)*P(t,m)+Emo(m)*P(t,m+1)+Nmo(m)*P(t,m+nx); elseif i == nx OilFlux(t,m) = Wmo(m)*P(t,m-

1)+Cmo(m)*P(t,m)+Nmo(m)*P(t,m+nx); else OilFlux(t,m) = Wmo(m)*P(t,m-

1)+Cmo(m)*P(t,m)+Emo(m)*P(t,m+1)+Nmo(m)*P(t,m+nx); end

elseif j == ny if i == 1 OilFlux(t,m) = Smo(m)*P(t,m-

nx)+Cmo(m)*P(t,m)+Emo(m)*P(t,m+1); elseif i == nx OilFlux(t,m) = Smo(m)*P(t,m-nx)+Wmo(m)*P(t,m-

1)+Cmo(m)*P(t,m); else OilFlux(t,m) = Smo(m)*P(t,m-nx)+Wmo(m)*P(t,m-

1)+Cmo(m)*P(t,m)+Emo(m)*P(t,m+1); end else OilFlux(t,m) = Smo(m)*P(t,m-nx)+Wmo(m)*P(t,m-

1)+Cmo(m)*P(t,m)+Emo(m)*P(t,m+1)+Nmo(m)*P(t,m+nx); end end end

%% %Saturation Calculations for m = 1:nxy; X(t,m) = (OilFlux(t,m)-Cum_oil(t,m))/(158*dx^2/deltat); if t == 1 Y(t,m) = ((Phi_ini(m)*Soi(m))/(Boi(m))); else Y(t,m) = ((phi(t-1,m)*So(t-1,m))/(Bo(t-1,m))); end So(t,m) = (Bo(t,m)/phi(t,m))*(Y(t,m)+X(t,m)); Sw(t,m) = 1-So(t,m); end end

for m = 1:nxy;

OOIP_EntireGrid(t,m)=dx*dy*deltaz*phi(t,m)*So(t,m)/(5.6145835*Bo(t,m)); end

for m = 1:nxy; qTotal(t)=(qA_o(t,A)+qB_o(t,B)+qC_o(t,C)+qD_o(t,D)+qE_o(t,E));

if t == 1; CumA_o(t) = qA_o(t,A)*deltat; CumB_o(t) = qB_o(t,B)*deltat; CumC_o(t) = qC_o(t,C)*deltat; CumD_o(t) = qD_o(t,D)*deltat; CumE_o(t) = qE_o(t,E)*deltat; CumTotal(t) = CumA_o(t)+CumB_o(t)+CumC_o(t)+CumD_o(t)+CumE_o(t); elseif t >1; CumA_o(t) = qA_o(t,A)*deltat+CumA_o(t-1); CumB_o(t) = qB_o(t,B)*deltat+CumB_o(t-1); CumC_o(t) = qC_o(t,C)*deltat+CumC_o(t-1); CumD_o(t) = qD_o(t,D)*deltat+CumD_o(t-1); CumE_o(t) = qE_o(t,E)*deltat+CumE_o(t-1); CumTotal(t) = CumTotal(t-1)+qTotal(t)*deltat; end end

OOIP_Reservoir = sum(OOIP_EntireGrid,2); ProductionRate_Reservoir(t) =

(qA_o(t,A)+qB_o(t,B)+qC_o(t,C)+qD_o(t,D)+qE_o(t,E))* deltat; CumulativeOil_Reservoir(t,1) = sum(ProductionRate_Reservoir,2); MBE(t) = (OOIP_Reservoir(t,1)+CumTotal(t))/OIIP; RF(t) = CumulativeOil_Reservoir(t,1)/OIIP; if qTotal(t)<1; break end

end

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