412 Final Report

PTE 412 Final Report 4/29/15

Jimmy Dao

Amey Dhaygude

Lin Du

Thomaz Lopes

Michael Onazi

Kevin Qi

Anuj Suhag

Team 11: Allied Engineering Co.

ALLIED ENGINEERING CO.

Pg 1

Table of Contents

Introduction………………………………………………………………………………………..2

Project Process. …………………………………………………………………………………...3

Volumetric Calculations…………………………………………………………………………...4

Pressure Testing Analysis……………………………………………………………………….…5

Permeability calculation…………………………………………………………………………...6

Fluid Properties……………………………………………………………………………………7

Forecasting Graphs………………………………………………………………………………..9

Economic Analysis………………………………………………………………………………16

Recommendations……………………………………………………………………………….19

Appendix…………………………………………………………………………………………20


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Introduction

The following reservoir study done by Allied Reservoir Engineering Co. guarantees the

optimum development plan for the new oil reservoir discovered in downtown Los Angeles. This

discovery well, drilled from the parking lot of the Convention Center intercepted oil and gas

bearing sand at a depth of 5570 ft. There is also an oil bearing formation at a similar depth of

5580 ft. that was found within 30 ft. of water, half a mile from Santa Monica pier. However,

there is no indication that the reservoirs are connected.

Our reservoir study and well development plan are essential documents prior to the

development of the fields. We have investigated the following pressure-maintenance scenarios to

maximize the primary oil recovery from the oil as recommended by the advisory board:

1. Returning 50% of produced gas from the beginning

2. Returning 50% of the produced gas after the reservoir pressure declined to 2000 psia

3. Returning all of produced gas into the reservoir

Using these scenarios, we can maximize the primary oil recovery and performed economic

analysis to consider the estimated profit from this discovery well.

Reservoir Properties

Area of the reservoir 4500 Acres

Area of Oil-Gas Contact 2250 Acres

Area of Top of Gas Cap 1125 Acres

Average Thickness 20 ft in the oil zone, 20 ft in the gas cap

Average Porosity 20% in the oil zone and the gas cap

Average Water Saturation 20% in the oil zone and in the gas cap

Initial Pressure @GOC 2500 psia

Economic Factors

Drilling and Completion Cost $2,325,000 per well

Stimulation Job (Hydraulic Fracturing) $500,000 per well

Maximum Safe Drawdown 200 psi

Allowable Rate 250 STB/Well

Economic Limit 5 STB/Well

Minimum flowing pressure 100 psig (with aid of pump)

Minimum Spacing 25 Acres

Oil Price $55/STB

Gas Price $ 2.5/MCF

Tax Rate 30%

Interest Rate 5%


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Assumptions

1. Gas-Oil Contact remains constant and the gas resulting from gas-cap expansion diffuses throughout the

oil column.

2. The wells can be completed for simultaneous gas injection and oil production.

3. The reservoir will be blown-down to recover the gas upon completion of the primary recovery.

4. The gas is separated from the oil at surface separator pressure of 400 psig.

5. The estimated cost of gas compression is $1/hp.

6. Non-Darcy effects for gas injection are negligible.

Project Process

Volumetric calculation

Well TestingFluid

Properties

Material balance

Calulate Np, Gp

four investigation

scenarios

Forecasting Graphs

optimum number of

well

Economic Analysis


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Volumetric Calculations Assumed that the reservoir is in the shape of a truncated cone.

Figure 1: Truncated Cone Structure

Using the following equation to calculate the volume:

V=1/3*(A1+A2+√𝐴1A2)

The original oil and gas in place can be found using the following two equations:

Calculating Oil and gas in Place by the Volumetric Method, Oil and gas in place by the

volumetric method is given by:


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From the equations above, results:

OOIP OGIP

74.57623498

MMSTB

1998.721374

MMSCF

Table 1: OOIP and OGIP

There is 74,576,235 STB of Oil in Place and 1,998,721,374 Scf of Gas in Place

Pressure Testing Analysis

A build-up test is made to analyze the wells.

Horner time can be calculated using the following equation:

Horner time= (t+Δ t)/Δ t

The calculated values are shown in the following table.

Time (t+Δ t)/Δ t Pressure

0.092 805.3478261 2474

0.127 583.6771654 2475

0.23 322.7391304 2478

0.46 161.8695652 2482

0.92 81.43478261 2485

2.3 33.17391304 2489

4.6 17.08695652 2493

9.2 9.043478261 2496

23 4.217391304 2500

Table 2: Calculated Pressure Values of Horner Time


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The Pressure is then plotted versus Horner Time on a semi-log plow.

Figure 2: Horner Plot

Permeability Calculation

The following equations are used to calculate permeability and skin factor:

k = −162.6qboμ

mh

s = 1.151 [pwf(Δt=0) − p1hr

m− log

k

ϕμctrw2

+ 3.23]

From Figure 2: Horner plot, the following data can be extracted:

Table 3: Slope, Permeability, and Skin Factor

2470

2475

2480

2485

2490

2495

2500

2505

1 10 100 1000

Pw

s

Horner Time

Pws Vs Horner Time


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Fluid Properties

Using the following data of gas saturation and relative permeability, a plot of the ratio of relative

permeabilites to gas and oil versus oil saturation can be plotted.

Sg% Kg/Ko Kro

3 0 0.85

7 0.001 0.4

13 0.0294 0.28

17.7 0.0535 0.22

21.5 0.084 0.17

24.8 0.125 0.13

28.8 0.198 0.09

32.2 0.29 0.06

35 0.405 0.04

37.4 0.54 0.02

38.2 0.9 0.01

40 - 0

Table 3: Gas Saturation and Relative Permeabilites


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Figure:3 Graphs of Viscosity & Oil formation volume factor Vs Pressure.

Figure 4: Ratio of Relative Permeabilties versus Oil Saturation

1.1

1.15

1.2

1.25

1.3

1.35

1.4

1.45

1.5

300 800 1300 1800 2300 2800

Bo

Pressure

Bo

Bo

0.4

0.6

0.8

300 800 1300 1800 2300 2800

Vis

cosi

ty

Pressure

μo

μo

y = 4E-08x6 - 2E-05x5 + 0.0024x4 - 0.1896x3 + 8.4297x2 - 198.5x + 1934.4

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 20 40 60 80 100

Krg

/K

ro

% So

Krg/Kro vs. So

krg/kro


Pg 9

Forecasting Graphs

Predicted Gp and Np

Figure 5: Relationship between GOR and G

p

The cumulative gas produced, Gp, and cumulative oil production, Np, are related to instantaneous

GOR by the following equation

𝐺𝑃 = ∫ (𝐺𝑂𝑅)𝑑𝑁𝑃

𝑁𝑃

0

According to the GOR and Pressure relationship we roughly guess a Rn at pressure Pn

Figure 6: Relationship between GOR and Pressure


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To calculate Np at pressure Pn:

N p n=N[Bon -Boi + (Rsi -Rsn )Bgn ]+G[Bgn -Bgi ]-Bgn[Gpn-1 -N pn(

GORn +GORn-1

2)

Bon -BgnRsn +Bgn (GORn +GORn-1

2)

Oil saturation: 𝑆𝑜 = (1 − 𝑆𝑤𝑖) (1 −𝑁𝑃

𝑁) (

𝐵𝑜

𝐵𝑜𝑖)

Gas saturation:

Sg = 1 − So − Swi

Rn = Rs + (krg

kro) (

μoBo

μgBg)

Then, we compare the Rn from the above equation. If the calculated Rn is equal to the Rn we

estimated then we can break and do the next calculation, if not, we need to do the iterations until

we get the same values.

GORn = (Rs)𝑛 + (krg

kro)

𝑛

(μo

μg)

𝑛

(Bo

Bg)

𝑛

Gas production:

𝐺𝑝𝑛 = 𝐺𝑝(𝑛−1) + [𝑅𝑛 + 𝑅𝑛−1

2] (𝑁𝑝𝑛 − 𝑁𝑝(𝑛−1))

Using Matlab to calculate four investigation pressure-maintenance scenarios

For 0 injection

Returning 50% of produced gas from the beginning

Returning 50% of the produced gas after the reservoir pressure declined to 2000 psia

100% injection (Returning all of produced gas into the reservoir )


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1) For 0 injection

Table 4: 0 Injection Pressure-Maintenance Scenario

2) Returning 50% of produced gas from the beginning

Table 5: 50% of Produced Gas from Beginning Pressure-Maintenance Scenario


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3) Returning 50% of the produced gas after the reservoir pressure declined to 2000 psia

Table 6: 50% of Produced Gas after the Reservoir Pressure Declined to 2000 PSA Pressure-Maintenance

Scenario

4) 100% injection (Returning all of produced gas into the reservoir)

Time t No. of wells Np dNp P Q Rf %

1 24 1885349.1 1885349.1 2378.387934 215.2225 2.528082974

2 47 5577491.088 3692141.988 2165.287473 156.8827 7.478912133

3 47 8268813.806 2691322.719 2000.352056 123.1838 11.08773298

4 47 10382031.9 2113218.089 1879.398622 97.5693 13.92136771

5 47 12055833.24 1673801.342 1800.705556 97.5693 16.16578423

6 47 13729634.58 1673801.342 1749.946944 76.16395 18.41020075

7 47 15036227.14 1306592.562 1739.312522 76.16395 20.16222345

8 47 16342819.7 1306592.562 1763.240554 76.16395 21.91424614

9 47 17649412.26 1306592.562 1831.305815 76.16395 23.66626884

10 47 18956004.83 1306592.562 1954.216015 89.57301 25.41829153

Table 7: 100% Injection Pressure-Maintenance Scenario


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Forecasting Graphs

1. Cumulative Production Vs Time.

Figure 7: Np vs. t

2. Difference in Cumulative Production Vs Time.

Figure 8: dNp vs. t

0

5000000

10000000

15000000

20000000

25000000

30000000

0 2 4 6 8 10 12

Np

, ST

B

t, Years

Np Vs t

Np, 0 inj vs t

Np, 50 inj start vs t

Np, 50 inj after vs t

Np, 100% inj vs t

0

500000

1000000

1500000

2000000

2500000

3000000

3500000

4000000

0 2 4 6 8 10 12

dN

p, S

TB

t, Years

dNp Vs t

dNp 0 inj Vs t

dNp 50 start Inj

dNp 50 after Vs t

dNp 100% inj


Pg 14

3. Cumulative Gas Production vs Time

Figure 9: Gp vs. t

4. Difference in Cumulative Gas Production Vs time

Figure 10: dGp vs. t

0

2E+12

4E+12

6E+12

8E+12

1E+13

1.2E+13

1.4E+13

1.6E+13

1.8E+13

2E+13

0 2 4 6 8 10 12

Gp

, Scf

t, Years

Gp vs t

Gp 0 inj

Gp 50% start

Gp 50% after

0

2E+12

4E+12

6E+12

8E+12

1E+13

1.2E+13

1.4E+13

1.6E+13

1.8E+13

2E+13

0 2 4 6 8 10 12

dG

p,S

cf

t, Years

dGp Vs t

dGp 0 inj Vs t

dGp 50% start vs t

dGp 50% after vs t


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5. Flow Rate Vs Time

Figure 11: Q vs. t

6. Gas Oil Ratio Vs Time

Figure 12: GOR vs. t

0

50

100

150

200

250

0 2 4 6 8 10 12

Q, s

tb/d

ay

t,Years

Q vs t

Q, 0 inj vs t

Q, 50% start inj vs t

Q, 50% inj after

Q, 100% inj vs t

0

1000000

2000000

3000000

4000000

5000000

6000000

7000000

8000000

9000000

10000000

0 2 4 6 8 10 12

GO

R, S

cf/S

tb

t, Years

GOR Vs t

GOR 0 Inj Vs t

GOR 50 start vs t

GOR 50 after vs t


Pg 16

7. Gas Oil Ratio Vs Pressure

Figure 13: GOR vs. P

Economic Analysis

Reservoir Properties

Drilling and Completion: $2,325,000 per well

Stimulation Job (Hydraulic Fracturing) $500,000 per well

Economic limit 5 STb/ well

Oil price $55/STB

Gas Price $ 2.5/MCF

Tax Rate 30%

Interest Rate 5%

Estimated cost of gas compression is $1/hp

Estimation of the Optimum Number of Development Wells to Achieve Maximum

Economical Return

According to SPE 71431_ An Analytical Solution to Estimate the Optimum Number of

Development Wells to Achieve Maximum Economical Return

The optimum number of wells is given by the following equation (Wo):

Wo= Np

{ln(1+i)C−[365QVCln(1+i)]0.5}

−365QC

0

10000

20000

30000

40000

50000

60000

70000

0 500 1000 1500 2000 2500 3000

GO

R, S

cf/S

tb

P, Psia

GOR Vs P

GOR, 0 inj Vs P

GOR, 50% start Vs P

GOR, 50% after vs P

GOR, 100% inj vs P


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Initial daily oil production rate, Q

Oil price at well after income tax, V

PV of capital investment after income tax, C

Interest rate, i

The results of optimum number of wells are given in the following table:

0 Injection 50% starting 50% after 100% injection

Time t No. of wells No. of wells No. of wells No. of wells

1 24 24 30 24

2 47 47 50 47

3 47 47 70 47

4 47 47 70 47

5 47 47 70 47

6 47 47 70 47

7 47 47 70 47

8 47 47 70 47

9 47 47 70 47

10 47 47 70 47

Table 8: Optimum Number of Wells for each Scenario

The net present value of an oil field development project can be expressed approximately

by the equation,

NPV(W)=dfNp V-CW-Z

By replacing W by Wo in equation:

NPV(Wo)=365 WoQiV/[(365 WoQi/ Np) + ln(1+i)] - C Wo - Z


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NPV break even curves are shown blew:

Figure 14: NPV vs. time

-2E+08

0

200000000

400000000

600000000

800000000

1E+09

1.2E+09

1.4E+09

1.6E+09

1.8E+09

1 2 3 4 5 6 7 8 9 10 11

NP

V

t (years)

NPV x time

No injection

50% gas injected

50% gas injected,after 2000 psi100% gas injected


Pg 19

Recommendations

The most Economical plan is as follows:

Have the optimal number of 70 wells

The best scenario occurs when returning 50% of the produced gas after the reservoir

pressure declined to 2000 psia

NPV is 1541 MM$, return rate is 117.17%.

The recovery factor is 34.65%

No stimulation or fracturing required

Table 9: Recovery Factor, Numbers of Wells, NPV, and Return Rate for Scenarios


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Appendix Matlab Code

%% Tarner's Method to determine NP, GP, dNp, and dGp. % Author - Team_11 % Date -04/29/2015

% Text about file input/output

clear; clc;

%% PVT Data, Volumetrics PVT=[2500 1.498 721 0.001048 0.48 0.0148 2300 1.463 669 0.001155 0.49 0.0146 2100 1.429 617 0.00128 0.508 0.0143 1900 1.395 565 0.00144 0.527 0.014 1700 1.361 513 0.001634 0.544 0.0137 1500 1.327 461 0.001884 0.654 0.0134 1300 1.292 409 0.002206 0.578 0.0132 1100 1.258 357 0.002654 0.609 0.0129 900 1.224 305 0.0033 0.633 0.0126 700 1.19 253 0.004315 0.661 0.0124 500 1.156 201 0.006163 0.692 0.0121 300 1.121 149 0.010469 0.729 0.0119]; %% Initializing Vectors P = PVT(:,1); bo = PVT(:,2); Rs = PVT(:,3); bg = PVT(:,4); mu_o= PVT(:,5); mu_g= PVT(:,6); %% Some constants N = 74576235; %possibly change value of So because it's above bbl pt

pressure 74576235,1998721374 G = 1998721374; %% Step Sizes changed from 200 to 100 PSIA P1 = 2500:-100:300; P1 = P1(:); %% Saturation So = [73 67 62.3 58.5 55.2 51.2 47.8 45 42.6 41.8 40]; So = So/100; so = linspace(77,40,23); so = so(:); Sg = [0.03; 0.07; 0.13; 0.177; 0.215; 0.248; 0.288; 0.322; 0.35; 0.374;

0.382; .4]; sg = -4.894e-08*P1.^2 + -3.451e-05*P1 + 0.4164; %% Rel Perm kro = [0.85 0.4 0.28 0.22 0.17 0.13 0.09 0.06 0.04 0.02 0.01 0]; krgkro = 0:0.03913043478:0.9; Krgro = [0; 0.001; 0.0294; 0.0535; 0.084; 0.125; 0.198; 0.29; 0.405; 0.54;

0.9; 0]; sg = sg(:); swi = 0.2; Ko =zeros(23,1); %% Vectorized Dummy Variables, Initial Conditions


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Np = zeros(23,1); Gp = zeros(23,1); Gp(1)=0; Np(1)=0; Npg=Np; Npg=zeros(23,1); Rcalc = zeros(23,1); R1 = 721; Rguess=zeros(23,1); Rguess=linspace(721,2000,23); Rguess=Rguess(:); q = zeros(23,1); q(1) = 250; %check to see if using 300 or 250 STB/day

%% Interpolations bo = interp1(P,bo,P1); bg = interp1(P,bg,P1); Rs = spline(P,Rs,P1); Rs = Rs(:); mu_g = spline(P,mu_g,P1); mu_o = spline(P,mu_o,P1); tol=100; m=(G*bg(1))/(N*bo(1)); %% Volume Compressibility Factor (Lumped together as C total) CT=[0 -0.0024 -0.0049 -0.0073 -0.0097 -0.0122 -0.0146 -0.0171 -0.0195 -0.0219 -0.0244 -0.0268 -0.0292 -0.0317 -0.0341 -0.0365 -0.0390 -0.0414 -0.0438 -0.0463 -0.0487 -0.0512 -0.0536];

%% Input/Output n = input('enter an injection scenario (Percent Injection) (0, .5, 1) : ');

switch n %Used to select different scenarios for injection

case 0 %No injection for i = 2:23


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Npg(2)=.0000001*N; Npg(i+1)=Npg(i)+100; krgkro(i) = (2264*sg(i)^5 - 1753.5*sg(i)^4 + 548.1*sg(i)^3 ... - 75.416*sg(i)^2 + 4.42*sg(i) - .0804);

Rcalc(i) =(Rs(i)+krgkro(i)*(mu_o(i)/mu_g(i))*(bo(i)/bg(i))); Rguess(i)= Rcalc(i); while abs(Rguess(i)-Rcalc(i))>1 Rguess(i)=Rcalc(i); end Np(i) = abs(((N*((Rs(1)-Rs(i))*bg(i)-((bo(i)-bo(1))/bg(i))... -(Gp(i)*bg(i)))))/(bo(i)+(bg(i)*((Rguess(i)... +Rguess(i-1))/2)-Rs(i)))); Gp(i) = abs(Gp(i)+((Rguess(i)+Rguess(i-1))/2)*(Npg(i)-Np(i-1)));

while (Npg(i)-Np(i))>1 Npg(i)=Npg(i)-10; end

end

case 1 %100 percent injection for i = 2:23 Npg(2)=.0000001*N; Npg(i+1)=Npg(i)+100; krgkro(i) = abs(-2264*sg(i)^5 + 1753.5*sg(i)^4 - 548.1*sg(i)^3 ... + 75.416*sg(i)^2 - 4.42*sg(i) + .0804);

Rcalc(i) =(Rs(i)+krgkro(i)*(mu_o(i)/mu_g(i))*(bo(i)/bg(i))); Rguess(i)= Rcalc(i); while abs(Rguess(i)-Rcalc(i))>1 Rguess(i)=Rcalc(i); end

Gp = zeros(23,1); Np(i) = abs(((N*((Rs(1)-Rs(i))*bg(i)-((bo(i)-bo(1))/bg(i))... +(Gp(i)*bg(i))))+(Gp(i))*bg(i))/(bo(i)+(bg(i)*((Rguess(i)... +Rguess(i-1))/2)-Rs(i)))); while (Npg(i)-Np(i))>1 Npg(i)=Npg(i)-10; end end

%% For 50% injection scenarios: Before 2000 Psia, and after 2000 Psia case .5 prompt=input('50% injection before 2000 psia, or after? (1=before,

0=after):'); %prompts user to select injection conditions

if prompt==1 %User selected to inject before 2000 Psia for i = 2:5 Npg(2)=.0000001*N; Npg(i+1)=Npg(i)+100; krgkro(i) = abs(-2264*sg(i)^5 + 1753.5*sg(i)^4 -

548.1*sg(i)^3 ...


Pg 23

+ 75.416*sg(i)^2 - 4.42*sg(i) + .0804);

Rcalc(i) =(Rs(i)+krgkro(i)*(mu_o(i)/mu_g(i))*(bo(i)/bg(i))); Rguess(i)= Rcalc(i);

Np(i) = N*((bo(i)-bo(1)+(Rs(1)-Rs(i))*bg(i)+m*bo(1)*... ((bg(1)/bg(i))-1)))-Gp(i)*bg(i)-0.5*Gp(i)/(bo(i)-

bg(i)*Rs(i));

Gp(i) = .5*(abs(Gp(i)+((Rguess(i)+Rguess(i-1))/2)*(Npg(i)-

Np(i-1))));

end

for k=6:23 Npg(6)=.00001*N;

Npg(k+1)=Npg(k)+100;

krgkro(k) = abs(-2264*sg(k)^5 + 1753.5*sg(k)^4 -

548.1*sg(k)^3 ... + 75.416*sg(k)^2 - 4.42*sg(k) + .0804);

Rcalc(k) =(Rs(k)+krgkro(k)*(mu_o(k)/mu_g(k))*(bo(k)/bg(k)));

Rguess(k)= Rcalc(k);

Np(k) = abs(((N*((Rs(1)-Rs(k))*bg(k)-((bo(k)-

bo(1))/bg(k))... -(Gp(k)*bg(k)))))/(bo(k)+(bg(k)*((Rguess(k)... +Rguess(k-1))/2)-Rs(k))));

Gp(k) = abs(Gp(k)+((Rguess(k)+Rguess(k-1))/2)*(Npg(k)-Np(k-

1))); while (Npg(k)-Np(k))>1 Npg(k)=Npg(k)-10; end end end

%% For 50% inject after 2000 PSIA

if prompt==0 %User selected injection after 2000 Psia for k=2:5 Npg(2)=.00001*N;

Npg(k+1)=Npg(k)+100;

krgkro(k) = abs(-2264*sg(k)^5 + 1753.5*sg(k)^4 -

548.1*sg(k)^3 ... + 75.416*sg(k)^2 - 4.42*sg(k) + .0804);


Pg 24

Rcalc(k) =(Rs(k)+krgkro(k)*(mu_o(k)/mu_g(k))*(bo(k)/bg(k)));

Rguess(k)= Rcalc(k);

Np(k) = abs(((N*((Rs(1)-Rs(k))*bg(k)-((bo(k)-

bo(1))/bg(k))... -(Gp(k)*bg(k)))))/(bo(k)+(bg(k)*((Rguess(k)... +Rguess(k-1))/2)-Rs(k))));

Gp(k) = abs(Gp(k)+((Rguess(k)+Rguess(k-1))/2)*(Npg(k)-Np(k-

1))); while (Npg(k)-Np(k))>124 Npg(k)=Npg(k)-10; end end

for i=6:23; Npg(2)=.0000001*N; Npg(i+1)=Npg(i)+100; krgkro(i) = abs(-2264*sg(i)^5 + 1753.5*sg(i)^4 -

548.1*sg(i)^3 + 75.416*sg(i)^2 - 4.42*sg(i) + .0804);

Rcalc(i) =(Rs(i)+krgkro(i)*(mu_o(i)/mu_g(i))*(bo(i)/bg(i))); Rguess(i)= Rcalc(i); Np(i) = N*((bo(i)-bo(1)+(Rs(1)-

Rs(i))*bg(i)+m*bo(1)*((bg(1)/bg(i))-1)))-Gp(i)*bg(i)-0.5*Gp(i)/(bo(i)-

bg(i)*Rs(i)); Gp(i) = .5*(abs(Gp(i)+((Rguess(i)+Rguess(i-1))/2)*(Npg(i)-

Np(i-1))));

while (Npg(i)-Np(i))>124 Npg(i)=Npg(i)-10; end

end end end

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