santa monica discovery well reservoir development plan · pdf file26.01.2010 ·...

Santa Monica Discovery Well Reservoir Development Plan

DONUT HOLE CONSULTANTS K. Chiu, J. Daniels, J. Dao, J. Davy, N. John, L. Hayum, S. Schoenberg, L. Zhu

GROUP 11

April 30th, 2014

INTRODUCTION

In this project, we will be looking at a discovery well drilled near the Santa Monica pier. The manager has requested a study for this reservoir by April 30, 2014, along with a development plan considering four scenarios to maximize the primary oil recovery:

• Scenario 0: No gas injection

• Scenario 1: Returning 50% of produced gas from beginning

• Scenario 2: Returning 50% of produced gas after reservoir pressure declines to 2000 psia

• Scenario 3: Returning all of produced gas into the reservoir

The above scenarios will be analyzed by comparing its production and economic data with that of

natural depletion. We made an optimal and comprehensive development plan of the reservoir.

ASSUMPTIONS

Several assumptions that we can make are as follows:

1) Gas-oil contact remains constant and the gas resulting from the 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.

7) No influence from any aquifers

VOLUMETRIC CALCULATIONS: Since there is no gas cap data given, we assume we do not have a gas cap. The equation for the original oil in place is:

𝑁 = 7758𝑉! ∗ ɸ ∗ (1 − 𝑆!)

𝐵!

where Vo is the rock volume in the oil zone Ø is the average porosity Sw is the average water saturation Boi is the formation volume factor of the oil

Our reservoir shape is assumed to be a rectangular-based pyramid with the top part cut off, as shown below.

Figure 1: Assumed Reservoir Shape

Since the assumption is that the trapping mechanism is anticline, we used the peak (or conical rule) to estimate this volume. To calculate the rock volume in the oil zone, we calculated the volume of our reservoir assuming the shape shown in Figure 1. We used the following equation:

𝑉! =ℎ3 𝐴! + 𝐴! + 𝐴!𝐴!

where Vo is the rock volume in the oil zone At is the area at the top of the reservoir (in pink) Ab is the area at the bottom of the reservoir (in yellow) h is the average thickness of the oil zone (height between the top and bottom)

At 1100 acres Ab 4500 acres H 30 ft Ø 25 % Sw 20 % Boi 1.225

Table 1: Reservoir Properties After determining the rock volume in the oil zone, we were able to find that our OOIP was 99,110,629.16 STB, which corresponds to the value of N used in the material balance calculations.

PRESSURE TESTING ANALYSIS – BUILD UP

The discovery well was put on production and after 74 hour was shut in for a build-up test. The well was produced at a constant allowable rate of 300 STB/day with a flowing pressure of 2400 psi. The total compressibility factor was considered to be ct = 16.26 * 10-6 (1/psi). The pressure was recorded and the Horner time was calculated in Table 2.

Table 2: Horner time

Then the pressure was plotted on a semi-log graph versus Horner time. This plot is shown below.

From

this graph we can calculate the slope m = -12.2, permeability of the reservoir can be computed through the equation:

𝑘 = −162.6𝑞𝜇!𝐵!𝑚ℎ

k = −162.6300×2.6513×1.216

−12.2×30 = 429.69 𝑚𝑑

deltat (t+deltat)/deltat p0.119 622.8487395 24750.183 405.3715847 24770.25 297 24790.55 135.5454545 24821.67 45.31137725 24883.19 24.19749216 24918.33 9.883553421 249616.65 5.444444444 249952.21 2.417352998 2503

2470

2475

2480

2485

2490

2495

2500

2505

1 10 100 1000

P, psi

(t+deltat)/deltat

Horner Plot

Then we found the degree of damage to the wellbore surroundings due to skin. Depending on the value of the skin factor, s, we may interpret how damaged the well is, and if a stimulation job should be performed on the well to increase production. P (∆t =1 hr) = 2508 psi P(∆t = 0 hr) = 2400 psi

𝑠 = 1.151𝑝!" 𝛥𝑡 = 0 − 𝑝!"!

𝑚− 𝑙𝑜𝑔

𝑘𝜙𝜇𝑐!𝑟!!

+ 3.23

s = 1.1512400− 2508

−12.2 − 𝑙𝑜𝑔429.69

0.25×2.6513×0.00001626×0.25! + 3.23

s = 3.7

The skin is less than 4 and the reservoir has high permeability values. Therefore, the reservoir need not be stimulated to increase production at this time. Any high skin factor calculated from future transient analyses would be a result of damage from drilling or production, not the integrity of the reservoir itself.

FLUID PROPERTIES

The given data on Reservoir Fluid Properties included Rs, bo, bg, oil viscosity, gas viscosity, oil density, and gas density, at declining values of Reservoir Pressure. Depending on the oil recovery scenario, the reservoir will have a certain gas saturation, Sg, at each value of declining reservoir pressure. The iterative process to forecast production versus time requires knowing the relative permeability ratio, krg/kro. To determine the relative permeability ratio, the data from the below table is used.

Sg krg/kro0 0

0.05 0.00050.1 0.001440.15 0.016140.2 0.043870.25 0.095380.3 0.1910.35 0.370080.4 0.710990.45 1.374250.5 2.697350.55 5.398940.6 10.95783

As long as gas saturation, Sg, is known, the relative permeability ratio can be interpolated from these data points. In our forecasting program we used a cubic spline interpolation, a form of interpolation that uses a piece-wise polynomial and is generally more accurate than linear interpolation.

TARNER METHOD CALCULATIONS

Table 5.1 Algorithm flow chart of material balance calculation

Using the N calculated prior, for each pressure step, Np was calculated using an estimate of Rn by

𝑁!" =𝑁 𝐵! − 𝐵!" + 𝑅!" − 𝑅! 𝐵! − 𝐵! 𝐺!(!!!) −

𝑅! + 𝑅!!!2 𝑁!

𝐵! − 𝐵!𝑅! +(𝑅! + 𝑅!!!)

2 𝐵!

After Np is calculated, the oil saturation, So, is calculated by

𝑆! = 1− 𝑆!" 1−𝑁!𝑁

𝐵!𝐵!"

With the oil saturation, we can get the gas saturation by:

S! = 1− S! − S!"

Using the available relative permeability data, determine the relative permeability ratio of !!"!!"

that corresponds to the gas saturation by interpolation method at Pn.

Obtain Np (using

material balance)

Iteration

Calculate So and Sg

Set initial conditions

Obtain krg/kro from the chart and interpolate

Calculate GOR

Compare GOR with the

estimation

Instantaneous Rn at pressure Pn was computed from the following equation:

R! = R! +k!"k!"

µμ!B!µμ!B!

With the Rn from the above equation, if the calculated Rn is equal to the estimated Rn then we move to the next step; if not we do the iterations until we get the same values. The amount of gas produced per iteration was computed from the following equation:

𝑅! − 𝑅!!!2 𝑁!" − 𝑁! !!!

The cumulative gas production was found as follows: For Scenario 0:

𝐺!" = 𝐺! !!! +𝑅! − 𝑅!!!

2 𝑁!" − 𝑁! !!!

For Scenario 1:

𝐺!" = 𝐺! !!! +𝑅! − 𝑅!!!

2 𝑁!" − 𝑁! !!!

2

For Scenario 2:

𝐺!" = 𝐺! !!! + !!!!!!!!

𝑁!" − 𝑁! !!! when Pressure > 2000 psia

𝐺!" = 𝐺! !!! +!!!!!!!

! !!"!!! !!!

! when Pressure < 2000 psia

For Scenario 3:

𝐺!" = 0

Based on the above equations, Np, Gp, ΔNp, ΔGp are calculated as programmed in MATLAB.

The results for the four scenarios are tabulated as follows:

Scenario

Cumulative oil

production (STB)

Cumulative gas

production (SCF)

Instantaneous GOR, Rp

Recovery Factor

(%)

No injection 18854031.64 40563925896 4963.765573 19.023219 50% injection 22968548.68 38062160289 7503.208026 23.174657

50% injection at 2000 psi 22368753.63 38125686539 7074.909577 22.56948 100% injection 35869372.56 0 25046.35543 36.191247

FLOW RATE CALCULATIONS

In order to determine the IPR (Inflow Performance Relationship), Klins-Clark methodology was used and an inflow expression1 was proposed. For a saturated reservoir (Pr ≤ Pb);

(𝑄!)!"# =𝑄!

1− 0.295𝑃!"𝑃!

− 0.705(𝑃!"𝑃!

!

𝑑 = 0.28+ 0.72𝑃!𝑃!

(1.24+ 0.001𝑃!)

where; (Qo)max = 200 STB Pb = 2500 psi Pwf = 2400-300 psi The IPR curve is constructed by assuming various values of Pwf and then calculating the corresponding Qo. Furthermore, the productivity index, J, can then be calculated by the following way: We have a constant drawdown pressure (Pe – Pw).

𝐽 =𝑞

𝑃! − 𝑃!

Therefore, !!!= !

!!

This allows us to calculate !!!

. (We start with 𝐽𝐽𝑖= 1, because !

!!= 1.) So we can calculate J.

1 Ahmed, Tarek, Reservoir Engineering Handbook, Gulf Professional Publishing; 4 edition, January 26, 2010

Results (calculations by pressure step): No injection

Half Gas Injection from the beginning

Pressure Np (STB) deltaNp (STB) Gp (SCF) deltaGp (SCF) GOR J q(STB/Day/ Well) RF (%)2500 0.00 0.00 0.00 0.00 464.12 0.98 196.94 0.002400 1,697,742.58 1,697,742.58 823,766,613.64 823,766,613.64 506.31 0.87 174.69 1.712300 3,445,149.20 1,747,406.62 1,713,012,405.41 889,245,791.78 511.48 0.77 154.59 3.482200 5,382,023.99 1,936,874.79 2,626,940,424.46 913,928,019.05 432.23 0.68 136.48 5.432100 7,546,891.38 2,164,867.39 3,509,836,763.49 882,896,339.03 383.43 0.60 120.22 7.612000 9,709,284.09 2,162,392.71 4,670,803,039.40 1,160,966,275.90 690.35 0.53 105.67 9.801900 11,137,513.13 1,428,229.04 6,085,441,575.89 1,414,638,536.49 1,290.62 0.46 92.67 11.241800 12,230,135.37 1,092,622.25 7,863,858,216.60 1,778,416,640.72 1,964.70 0.41 81.11 12.341700 13,069,499.01 839,363.64 9,796,590,126.70 1,932,731,910.10 2,640.53 0.35 70.85 13.191600 13,770,146.83 700,647.82 11,875,668,549.60 2,079,078,422.90 3,294.20 0.31 61.76 13.891500 14,338,507.71 568,360.88 13,930,697,937.87 2,055,029,388.27 3,937.22 0.27 53.74 14.471400 14,861,678.51 523,170.80 16,142,631,725.83 2,211,933,787.96 4,518.65 0.23 46.66 15.001300 15,326,886.56 465,208.05 18,381,944,141.58 2,239,312,415.75 5,108.49 0.20 40.42 15.461200 15,745,225.35 418,338.79 20,630,310,136.61 2,248,365,995.04 5,640.53 0.17 34.92 15.891100 16,131,516.30 386,290.95 22,902,731,940.91 2,272,421,804.30 6,124.81 0.15 30.06 16.281000 16,492,573.84 361,057.54 25,181,981,078.30 2,279,249,137.39 6,500.60 0.13 25.76 16.64900 16,831,344.99 338,771.16 27,442,666,261.31 2,260,685,183.01 6,845.79 0.11 21.92 16.98800 17,163,639.04 332,294.05 29,737,297,917.77 2,294,631,656.46 6,965.06 0.09 18.47 17.32700 17,486,674.53 323,035.48 31,986,727,978.19 2,249,430,060.42 6,961.77 0.08 15.34 17.64600 17,804,182.45 317,507.93 34,186,188,965.43 2,199,460,987.25 6,892.76 0.06 12.45 17.96500 18,130,964.28 326,781.82 36,375,363,815.96 2,189,174,850.53 6,505.63 0.05 9.76 18.29400 18,473,647.20 342,682.92 38,499,426,720.23 2,124,062,904.27 5,891.04 0.04 7.22 18.64300 18,854,031.64 380,384.45 40,563,925,895.62 2,064,499,175.39 4,963.77 0.02 4.76 19.02

Pressure Np (STB) deltaNp (STB) Gp (SCF) deltaGp (SCF) GOR Jq(STB/Day/

Well)RF (%)

2500 0.00 0.00 0.00 0.00 0.98 196.94 464.12 0.002400 1,697,742.58 1,697,742.58 411,883,306.82 823,766,613.64 0.87 174.69 506.31 1.712300 3,838,227.72 2,140,485.14 955,470,257.96 1,087,173,902.29 0.77 154.59 509.51 3.872200 6,373,612.23 2,535,384.51 1,534,098,531.29 1,157,256,546.65 0.68 136.48 403.37 6.432100 9,070,232.76 2,696,620.54 2,143,566,445.44 1,218,935,828.30 0.60 120.22 500.68 9.152000 11,280,536.81 2,210,304.05 3,039,602,647.04 1,792,072,403.21 0.53 105.67 1,120.89 11.381900 12,778,810.24 1,498,273.43 4,194,614,634.67 2,310,023,975.25 0.46 92.67 1,962.70 12.891800 13,995,897.07 1,217,086.82 5,655,178,775.52 2,921,128,281.70 0.41 81.11 2,837.50 14.121700 15,021,439.73 1,025,542.66 7,335,854,467.52 3,361,351,384.01 0.35 70.85 3,717.76 15.161600 15,920,685.54 899,245.81 9,204,327,921.21 3,736,946,907.37 0.31 61.76 4,593.53 16.061500 16,692,574.83 771,889.30 11,152,699,631.66 3,896,743,420.90 0.27 53.74 5,503.11 16.841400 17,394,262.31 701,687.47 13,231,566,178.79 4,157,733,094.27 0.23 46.66 6,347.56 17.551300 18,032,691.84 638,429.53 15,396,600,577.89 4,330,068,798.19 0.20 40.42 7,217.19 18.191200 18,613,739.28 581,047.44 17,611,037,300.81 4,428,873,445.85 0.17 34.92 8,027.26 18.781100 19,150,992.01 537,252.73 19,867,961,495.67 4,513,848,389.71 0.15 30.06 8,776.19 19.321000 19,654,751.77 503,759.76 22,153,995,781.01 4,572,068,570.68 0.13 25.76 9,375.59 19.83900 20,129,925.20 475,173.43 24,448,138,431.51 4,588,285,301.01 0.11 21.92 9,936.45 20.31800 20,591,328.30 461,403.10 26,768,112,415.04 4,639,947,967.05 0.09 18.47 10,175.89 20.78700 21,044,340.05 453,011.75 29,080,083,642.37 4,623,942,454.66 0.08 15.34 10,238.34 21.23600 21,491,941.10 447,601.05 31,366,400,006.61 4,572,632,728.47 0.06 12.45 10,193.39 21.68500 21,949,227.76 457,286.66 33,638,329,369.33 4,543,858,725.45 0.05 9.76 9,679.74 22.15400 22,432,234.41 483,006.65 35,872,850,008.16 4,469,041,277.65 0.04 7.22 8,825.35 22.63300 22,968,548.68 536,314.27 38,062,160,288.51 4,378,620,560.70 0.02 4.76 7,503.21 23.17

50% Injection After 2000 psi

Full injection

Based upon the above results, the 100% injection rate from the start of production seems to be optimal, as evident from its higher recovery factor.

Pressure Np (STB) deltaNp(STB)

Gp (SCF) deltaGp (SCF) GOR J q(STB/Day/ Well)

RF (%)

2500 0.00 0.00 0.00 0.00 464.12 0.98 196.94 0.002400 1,697,742.58 1,697,742.58 823,766,613.64 823,766,613.64 506.31 0.87 174.69 1.712300 3,445,149.20 1,747,406.62 1,713,012,405.41 889,245,791.78 511.48 0.77 154.59 3.482200 5,382,023.99 1,936,874.79 2,626,940,424.46 913,928,019.05 432.23 0.68 136.48 5.432100 7,546,891.38 2,164,867.39 3,509,836,763.49 882,896,339.03 383.43 0.60 120.22 7.612000 9,709,284.09 2,162,392.71 4,670,803,039.40 1,160,966,275.90 690.35 0.53 105.67 9.801900 11,137,513.13 1,428,229.04 5,378,122,307.64 1,414,638,536.49 1,290.62 0.46 92.67 11.241800 12,528,314.06 1,390,800.94 6,557,824,319.32 2,359,404,023.36 2,102.25 0.41 81.11 12.641700 13,704,198.17 1,175,884.10 8,050,383,847.72 2,985,119,056.80 2,974.98 0.35 70.85 13.831600 14,722,184.83 1,017,986.66 9,785,312,581.12 3,469,857,466.78 3,842.12 0.31 61.76 14.851500 15,586,936.01 864,751.18 11,638,018,041.05 3,705,410,919.87 4,727.77 0.27 53.74 15.731400 16,366,894.64 779,958.63 13,643,020,258.57 4,010,004,435.05 5,554.84 0.23 46.66 16.511300 17,070,844.31 703,949.67 15,748,127,524.43 4,210,214,531.72 6,406.85 0.20 40.42 17.221200 17,707,352.09 636,507.78 17,913,956,017.19 4,331,656,985.52 7,203.84 0.17 34.92 17.871100 18,292,477.29 585,125.20 20,130,001,718.67 4,432,091,402.96 7,945.37 0.15 30.06 18.461000 18,838,260.50 545,783.21 22,380,803,756.75 4,501,604,076.15 8,550.58 0.13 25.76 19.01900 19,350,652.17 512,391.67 24,644,273,663.47 4,526,939,813.45 9,119.27 0.11 21.92 19.52800 19,846,076.65 495,424.48 26,936,756,143.16 4,584,964,959.38 9,389.97 0.09 18.47 20.02700 20,330,561.32 484,484.67 29,224,060,498.82 4,574,608,711.32 9,494.46 0.08 15.34 20.51600 20,807,370.26 476,808.93 31,487,903,577.11 4,527,686,156.57 9,497.16 0.06 12.45 20.99500 21,292,688.23 485,317.97 33,739,067,911.02 4,502,328,667.82 9,056.98 0.05 9.76 21.48400 21,803,495.83 510,807.60 35,954,354,231.95 4,430,572,641.86 8,290.34 0.04 7.22 22.00300 22,368,753.63 565,257.80 38,125,686,538.87 4,342,664,613.86 7,074.91 0.02 4.76 22.57

Pressure Np (STB)deltaNp(STB)

Gp (SCF)deltaGp(SCF)

GOR Jq(STB/Day/

Well)RF (%)

2500 0.00 0.00 0 0 464.12 0.98 196.94 0.002400 1,697,742.58 1,697,742.58 0 0 506.31 0.87 174.69 1.712300 4,234,819.78 2,537,077.20 0 0 505.24 0.77 154.59 4.272200 7,564,613.04 3,329,793.26 0 0 392.77 0.68 136.48 7.632100 10,748,035.88 3,183,422.84 0 0 813.16 0.60 120.22 10.842000 13,095,969.96 2,347,934.08 0 0 1,815.73 0.53 105.67 13.211900 14,908,914.31 1,812,944.34 0 0 3,032.38 0.46 92.67 15.041800 16,548,995.55 1,640,081.24 0 0 4,325.85 0.41 81.11 16.701700 18,089,439.40 1,540,443.85 0 0 5,788.71 0.35 70.85 18.251600 19,558,643.49 1,469,204.09 0 0 7,435.64 0.31 61.76 19.731500 20,941,644.87 1,383,001.38 0 0 9,344.68 0.27 53.74 21.131400 22,266,741.75 1,325,096.88 0 0 11,347.17 0.23 46.66 22.471300 23,542,006.41 1,275,264.66 0 0 13,575.34 0.20 40.42 23.751200 24,769,716.39 1,227,709.97 0 0 15,907.61 0.17 34.92 24.991100 25,960,555.93 1,190,839.54 0 0 18,304.68 0.15 30.06 26.191000 27,127,520.27 1,166,964.34 0 0 20,576.83 0.13 25.76 27.37900 28,276,791.47 1,149,271.20 0 0 22,950.69 0.11 21.92 28.53800 29,425,704.26 1,148,912.79 0 0 24,772.41 0.09 18.47 29.69700 30,592,477.23 1,166,772.97 0 0 26,301.90 0.08 15.34 30.87600 31,784,439.75 1,191,962.52 0 0 27,665.91 0.06 12.45 32.07500 33,027,955.13 1,243,515.38 0 0 27,919.66 0.05 9.76 33.32400 34,366,984.19 1,339,029.06 0 0 27,232.78 0.04 7.22 34.68300 35,869,372.56 1,502,388.37 0 0 25,046.36 0.02 4.76 36.19

ECONOMIC ANALYSIS: Cost Breakdown:

Drilling and Completion Cost $2,225,000 / well

Maximum Safe Drawdown 200 psi

Allowable Rate 200 STB/Well

Economic Limit 5 STB/Well

Minimum Flowing Pressure 100 psig (with aid of pump)

Minimum Spacing 25 acres

Oil Price $95/STB

Gas Price $3.5/MSCF

Tax Rate 50%

Interest Rate 6%

Limit of 10 wells per year for both production and injection wells.

The following calculation is used to determine Net Present Value:

𝑁𝑃𝑉 =net income !

(1+ 𝑖)!

!

!!!

Optimum Number of Wells We presume that the optimum number can be found through an expression of the maximum economic return over the life of the reservoir as a function of the number of development wells as proposed by Muskat’s economic method. The equation below expresses this relationship:

𝑁𝑃𝑉 𝑊 = 𝑑𝑓 ∗ 𝑁!𝑉 − 𝐶𝑊 − 𝑍 Whereby df is the discount factor associated with the cost per well, Np is the total amount of oil produced, V is the price of oil after tax, C is the cost of drilling, and Z is the present value of investment capital sans income tax. From this, we make several assumptions as follows:

• The total cumulative oil production is constant • Present net value is after income tax • All investments are incurred at t=0 • All wells have the same initial oil production and decline at the same rate over

time • Oil amount is calculated at the wellhead.

The simplified “Muskat” equation is as follows:

𝑁𝑃𝑉 𝑊 =365𝑊𝑄!𝑉

365𝑊𝑄!𝑁!

+ ln 1+ 𝑖− 𝐶𝑊 − 𝑍

Differentiating this with respect to W and solving via quadratic equation gives us an equation for Wo the optimum number of wells to maximize NPV(W).

𝑊! =𝑁! ln 1+ 𝑖 ∗ 𝐶 − 365𝑄!𝐶 ∗ ln 1+ 𝑖 !.!

−365 ∗ 𝑄!𝐶

Using this formula among the 4 scenarios, using each respective forecast from material balance for Qi, Np,, we find the following:

Scenario #1 Scenario #2 Scenario #3 Scenario #4

Wo 63 Wo 77 Wo 75 Wo 120

Np 18854031.64 Np 22968548.68 Np 22368753.63 Np 35869372.56

Qi 196.94 Qi 196.94 Qi 196.94 Qi 196.94

V 47.5 V 47.5 V 47.5 V 47.5

C 2225000 C 2225000 C 2225000 C 2225000.00

I 0.06 i 0.06 i 0.06 i 0.06

Z 35000000 Z 35000000 Z 35000000 Z 35000000.00

NPV(63) 545548170.07 NPV(77) 672241901.5 NPV(75) 653773121.90 NPV(120) 1069480788

Using the formula for NPV(W), we can more clearly illustrate the optimum number of wells in plots of NPV vs Number of Wells at their entire life. This yields results as expected, with full injection being most profitable, but also requiring the greatest number of wells.

All of these figures presumed a value of $90 million capital investment on behalf of an investor or investment company, at a return rate of 6%. Addressing this, we came to this conclusion that an investor would only loan approximately this amount due to rough estimations of start-up costs, including the cost of 10 wells within the first year, and the production costs to produce barrels at $10/BBL. We also further discussed the nature of an investor wanting to invest in such a project, it is unlikely that a company would provide such an investment in the millions for only a 6% return. Some rough numbers to follow appear more realistic, with a company asking for an upper bound of 20% within the first year. Knowing this, adjustments had to be made, as i, the interest rate must be applied to the df, thus yielding a different value for optimum number of wells, as well as different values for NPV.

$-‐

$100,000,000

$200,000,000

$300,000,000

$400,000,000

$500,000,000

$600,000,000

0 50 100 150 200

NPV

Number of Wells

$-‐ $100,000,000 $200,000,000 $300,000,000 $400,000,000 $500,000,000 $600,000,000 $700,000,000

0 20 40 60 80 100 120 140 NPV

Number of Wells

$-‐

$200,000,000

$400,000,000

$600,000,000

$800,000,000

$1,000,000,000

$1,200,000,000

0 50 100 150 200

NPV

Number of Wells

$-‐

$200,000,000

$400,000,000

$600,000,000

$800,000,000

0 50 100 150 200

NPV

Number of Wells

Case 1: Optimal return at 63 wells Case 2: Optimal return at 77 wells

Case 3: Optimal return at 75 wells Case 4: Optimal return at 120 wells

Optimu

m # of

Wells:

91

Optimu

m # of

Wells:

111

Optimu

m # of

Wells:

108

Optimu

m # of

Wells:

173

Np 18854031.64 Np

22968548.6

8 Np 22368753.63 Np 35869372.56

Qi 196.94 Qi 196.94 Qi 196.94 Qi 196.94

V 47.5 V 47.50 V 47.5 V 47.50

C 2225000 C 2225000.00 C 2225000 C 2225000.00

i 0.20 i 0.20 i 0.20 i 0.20

NPV(91)

348519653.0

7

NPV(111

)

433512915.

4

NPV(108

)

421278792.3

0

NPV(173

)

696665379.7

5

FORECASTING GRAPHS Np vs Time Plots To comprehend the plot of Np vs time, we will be taking an example of 10 wells per year. From the plots, we infer that for no injection, the life time of the reservoir is the shortest, while it reaches maximum for 100% injection.

Np vs Time for 10 wells/year

100% injection

50% injection at t=0

50% injection at P=2000

psi

No injection

ΔNp vs Time Plot From the plots we infer that, for 100% injection, the ΔNp peaks at the maximum value. The plot follows a trend of peaking off to a maximum value and then decreases to attain an almost constant value.

ΔNp vs Time for 10 wells/yr

Gp vs Time Plot From the plot of Gp vs Time, we infer that for 100% injection, the value of Gp = 0 and the reservoir lifetime is maximum. The cumulative gas produced tends to be maximum in the case of no injection of gas into the reservoir.

Gp vs Time for 10 wells/yr

No injection

100% injection


50% injection at

P=2000 psi

100% injection

50% injection at P=2000 psi


No injection

ΔGp vs Time Plot From the plot, we infer that at 100% gas injection, ΔGp = 0 for all time. The lifetime of a reservoir tends to be maximum in this case. Produced gas is low for the case of no-injection and maximum for the case of 50% injection at t=0.

ΔGp vs Time Plot for 10 wells/year

GOR vs Time Plot The Gas-Oil ratio was plotted against time and (GOR)100% inj was found to have the maximum peak, while the (GOR)no inj was found to be having the least peak. The average reservoir lifetime of the reservoir was found to be directly proportional to the GOR.

GOR vs time for 10 wells/year

100% injection

No injection

50% injection at

t=0 50% injection at P=2000 psi

100% injection

50% injection at P=2000 psi

No injection


Flow Rate vs Time Plot The daily production rate is declining from 200 STB/day. By the tenth year, it will have decreased below 50 STB/day. The full injection scenario, which is the optimal of the four scenarios, shows the trend that more gas injection ultimately decreases the flow rate slowest for each well.

ECONOMIC RESULTS Payback Period

From the charts shown above, we can see that scenario 4 has the shortest payback period, especially for the case that drills 10 wells per year.

0.00

50.00

100.00

150.00

200.00

250.00

0 2000 4000 6000 8000 10000 12000 14000

Flow Rate(STB/D/Well) vs. Time(Days)

No InjecSon 50% from beginning 50% from 2000psia 100% InjecSon

No Injection 2.450% Inj @ t=0 2.550% Inj @ P=2000 2.4100% Inj 2.3



Payback Period For 10 Wells Per Year (Years)



NPV vs Time Plot (10 wells per year)

From the above plot, we infer that for 100% injection, the NPV tends to be maximum and for the case of no injection, NPV tends to reach a minimum peak for the life of the field. NPV vs Time Plot (full injection)

-‐1000000000.00

0.00

1000000000.00

2000000000.00

3000000000.00

4000000000.00

5000000000.00

6000000000.00

7000000000.00

0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00

NPV

Time (Years)

NPV vs Time: 10 Wells per Year

No InjecSon 50% Inj @ t=0 50% Inj @ P=2000 100% Inj @ t=0

-‐1000000000.00

0.00

1000000000.00

2000000000.00

3000000000.00

4000000000.00

5000000000.00

6000000000.00

7000000000.00

0 5 10 15 20 25 30 35 40

NPV

Time (Years)

10/year 7/year 5/year

From the above plot of cumulative NPV vs time for 100% injection, 10 wells/year produced for a lower reservoir lifetime, but with a higher NPV. On the other hand, 5 wells/yr produced for a longer reservoir lifetime and lowest cumulative NPV. RECOMMENDATION AND CONCLUSION Based on our assumptions and calculations, we recommend the full gas injection. Drilling 10 wells from the beginning, and adding ten wells per year for this scenario, would give the highest recovery factor as well as the highest NPV, making this the most productive and profitable plan. However, our assumptions do assume a perfect, idealistic case that most likely does not exist in real life. Unlike above, it is most prudent to build 120 wells, yielding $6.25 billion over the lifetime of the well. But we still believe that this reservoir off the Santa Monica pier is a lucrative reservoir to produce, and returning the produced gas from the start of production would yield the greatest recovery as well as the most profit. REFERENCES R.D. Corrie, “An Analytical Solution to Estimate the Optimum Number of Development Wells to Achieve Maximum Economical Return”, SPE, Inmeaka S.A, Sept. 30, 2001 Ahmed, Tarek, Reservoir Engineering Handbook, Gulf Professional Publishing; 4 edition, January 26, 2010

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