1 HW Solution - PEGN 423 – Petroleum Reservoir Engineering I – Fall 2016 PDF

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PEGN 423 – Petroleum Reservoir Engineering I – Fall 2016 HOMEWORK #1 Assignment: Tuesday, September 6, 2016 Due date: Thursday, September 15, 2016 NAME: SOLUTION Problem #1: Using dimensional analysis, present the Darcy’s Law equation in the following oil field units: (15 points)

1

Volumetric Permeability, Viscosity, Cross- Length, Pressure flow rate, k sectional L µ drop, Δ p q area, A ft3/s ft2 md cp ft psi

kA Δp µ L dimensional analysis: in 2 1cp ft 3 md ft 2 ft 2 ≡ ⋅144 2 ⋅ psi ⋅1.062315 ×10 −14 cp ft s md 2.088543 × 10 −5 lbf ⋅ s ft ft 2 q=

ft 3 ft 3 ≡ 7.3244 ×10 −8 s s

q = 7.3244 ×10−8

2

k A Δp µ L

Volumetric Permeability, Viscosity, Cross- Length, Pressure flow rate, k sectional L µ drop, Δ p q area, A ft3/day ft2 md cp ft psi

kA Δp µ L dimensional analysis: q=

s in 2 1cp ft 3 md ft 2 ft 2 ≡ ⋅144 2 ⋅ 86400 ⋅ psi ⋅1.062315 ×10 −14 cp ft d d ft md 2.088543 × 10 −5 lbf ⋅ s 2 ft ft 3 ft 3 ≡ 0.006328 d d

q = 0.006328

kA Δp µL

Volumetric Permeability, Viscosity, Cross- Length, Pressure flow rate, k sectional L µ drop, Δ p q area, A ft2 bbl/day md cp ft psi

3

kA Δp µ L dimensional analysis: bbl ft 3 1bbl ≡ 0.006328 ⋅ d d 5.6146 ft 3 bbl bbl ≡ 0.001127 d d q=

q = 0.001127

kA Δp µL

Useful conversion factors: -

For viscosity: To Obtain

Multiply

cp or mPa·s

lbf·s/ft2

psi·s

lbm/ft·s

cp or mPa·s lbf·s/ft2

1 4.788026E+04

2.088543E-05 1

1.450377E-07 6.944444E-03

6.719689E-04 3.217400E+01

psi·s lbm/ft·s

6.894757E+06 1.488164E+03

1.440000E+02 3.108100E-02

1 2.158403E-04

4.633056E+03 1

- For permeability: 1 µm2 = 1013.25 md

Problem #2: The oil-water contact of an undersaturated oil reservoir has been located from well logging interpretation at a depth of 4,975.2 ft. During the reservoir characterization, four types of rock where identified with different average porosity and permeability. Table 2.1 presents the porosity and permeability corresponding for each rock type. The four types of rock are distributed in five reservoir rock layers according to the depth intervals. Table 2.1 Data for problem 2 Porosity(and(Permeability(Data( k" (md)(

Ø(

Zone( 1(

575(

2( 3( 4(

111( 55( 10(

Zone( 1( 3( 2( 3( 4(

Capillary(Pressure(Data( Pc,(psi( Zone(2( Zone(3(

Sw(

Zone(1(

0.275(

1(

0.20(

0.38(

0.45(

0.80(

0.21( 0.2( 0.14(

0.95( 0.9( 0.85( 0.8( 0.75( 0.7( 0.65( 0.6( 0.55( 0.5( 0.45( 0.4( 0.35( 0.33( 0.3( 0.25( 0.24( 0.22( 0.21( 0.2( 0.19( 0.18( 0.17( 0.16(

0.20( 0.20( 0.20( 0.20( 0.20( 0.20( 0.24( 0.26( 0.29( 0.33( 0.37( 0.46( 0.58( 0.63( 0.75( 1.00( 1.10( 1.30( 1.40( 1.50( 1.70( 2.00( 2.50( 3.00(

0.38( 0.38( 0.38( 0.39( 0.40( 0.42( 0.45( 0.50( 0.55( 0.60( 0.70( 0.85( 1.10( 1.20( 1.40( 2.00( 2.30( 2.60( 2.80( 3.00( 3.50( 4.00( 5.00( 7.00(

0.45( 0.45( 0.45( 0.48( 0.50( 0.55( 0.58( 0.62( 0.70( 0.80( 0.97( 1.15( 1.45( 1.60( 2.00( 2.70( 3.20( 3.60( 3.90( 4.20( 4.70( 5.60( 7.00( 9.00(

0.82( 0.85( 0.90( 0.95( 1.00( 1.04( 1.10( 1.20( 1.30( 1.45( 1.80( 2.20( 2.90( 3.20( 3.80( 5.40( 6.40( 7.00( 7.60( 8.20( 9.00( 11.00( 13.50( 15.00(

Zone-Depth(Data( Top( Bottom( (ft)( (ft)( ((((((4,877(( ((((((4,899(( ((((((4,899(( ((((((4,922(( ((((((4,922(( ((((((4,935(( ((((((4,935(( ((((((4,948(( ((((((4,948(( ((((((4,977((

Zone(4(

From a reservoir fluid analysis the water-oil interfacial tension is 55 dynes/cm, and the wetting angle is 30º.

2.1.- Determine the free-water level (depth at which Pc = 0 psi) if the oil density at reservoir conditions is 𝜌!" = 40 𝑙𝑏𝑚/𝑓𝑡^3 and 𝜌! = 64 𝑙𝑏𝑚/𝑓𝑡^3. (15 points) 2.2- Determine the initial water saturation distribution from the capillary pressure curves presented in Table 2.1. Plot depth (vertical axis) versus water saturation (horizontal axis). (20 points) First, calculate the heights for different capillary pressure values, using, P ⋅144 g c Δh = c → ( FWL = WOC + Δ h) ( ρ w − ρ o) g At Sw=100%, the height comes to 4.8 ft (Zone 4) and the Free Water Level: FWL=4.8 ft + 4975.20 ft=4980 ft Second, plot the changes in saturation as per the FWL as well as the WOC.

Depth&(ft)&

Initial&Fluid&Distribution-Problem& !5,200.00!! !5,195.00!! !5,190.00!! !5,185.00!! !5,180.00!! !5,175.00!! !5,170.00!! !5,165.00!! !5,160.00!! !5,155.00!! !5,150.00!! !5,145.00!! !5,140.00!! !5,135.00!! !5,130.00!! !5,125.00!! !5,120.00!! !5,115.00!! !5,110.00!! !5,105.00!! !5,100.00!! !5,095.00!! !5,090.00!! !5,085.00!! !5,080.00!! !5,075.00!! !5,070.00!! !5,065.00!! !5,060.00!! !5,055.00!! !5,050.00!! !5,045.00!! !5,040.00!! !5,035.00!! !5,030.00!! !5,025.00!! !5,020.00!! !5,015.00!! !5,010.00!! !5,005.00!! !5,000.00!! !4,995.00!! !4,990.00!! !4,985.00!! !4,980.00!! !4,975.00!!

Series1!

0!

0.2!

0.4!

0.6!

0.8!

1!

Water&Saturation&(Sw)&

Figure S2.1.- Plot of depth versus water saturation, representing the initial water saturation distribution in this reservoir.

Problem #3: Calculate the initial water saturation distribution in a heterogeneous reservoir, where the J-function relationship with water saturation is known as, (20 points)

J = 0.1+1.2 exp "#−6 (Sw − 0.15 )$%; for Sw > 0.15 Make a plot of Depth vs. water saturation (Sw). Table 3.1 presents the rock and fluid properties, and Table 3.2 presents the porosity and permeability at different depth intervals. Table 3.1 - Data of rock/fluid system 65 lbm/ft3 Water density, ρ w 42.53 lbm/ft3 Oil density, ρ o 74 dynes/cm Interfacial tension, σ 40º Contact angle, θ Free water level 4500ft

Table 3.2 - Porosity and permeability data! k((md)( φ Depth (ft) 4375 - 4400 0.1 35 4400 - 4415 0.15 145 4415 - 4425 0.12 68 4425 - 4450 0.18 388 4450 - 4465 0.08 10 4465 - 4480 0.14 112 4480 - 4500 0.08 10

First calculate Pc for different depths using,

Pc =

( ρw − ρo) 144

g Δh gc

Then calculate the J-function at the different depths using,

J=

0.217pc σ cosθ

k φ

Then calculate Sw using the given relationship for this reservoir, J = 0.1+1.2 exp "#−6 (Sw − 0.15 )$%; for Sw > 0.15

1 " J − 0.1 % Sw = 0.15 − ln $ ' 6 # 1.2 & if Sw < 0.15 then Sw = 0.15

Results shown in Table S3.1. The plot of depth versus water saturation is shown in Figure S3.1. Table S3.1. Calculated capillary pressure, J function and water saturation. Depth (ft)

Pc (psi)

J function

Sw

4375

19.50520833

1.396878597

0.15

4376

19.34916667

1.385703568

0.15

4377

19.193125

1.374528539

0.15

4378

19.03708333

1.36335351

0.15

4379

18.88104167

1.352178482

0.15

4380

18.725

1.341003453

0.15

4381

18.56895833

1.329828424

0.15

4382

18.41291667

1.318653395

0.15

4383

18.256875

1.307478367

0.15

4384

18.10083333

1.296303338

0.150514218

4385

17.94479167

1.285128309

0.152078418

1.27395328

0.153657439

4386

17.78875

4387

17.63270833

1.262778252

0.155251562

4388

17.47666667

1.251603223

0.15686108

4389

17.320625

1.240428194

0.158486293

4390

17.16458333

1.229253165

0.16012751

4391

17.00854167

1.218078136

0.161785049

4392

16.8525

1.206903108

0.163459239

4393

16.69645833

1.195728079

0.165150417

4394

16.54041667

1.18455305

0.166858932

4395

16.384375

1.173378021

0.168585142

4396

16.22833333

1.162202993

0.170329418

4397

16.07229167

1.151027964

0.172092143

1.139852935

0.17387371

4398

15.91625

4399

15.76020833

1.128677906

0.175674528

4400

15.60416667

1.85717573

0.15

4401

15.448125

1.838603972

0.15

4402

15.29208333

1.820032215

0.15

Depth (ft)

Pc (psi) 4403

J function

15.13604167

Sw

1.801460458

0.15

4404

14.98

1.7828887

0.15

4405

14.82395833

1.764316943

0.15

4406

14.66791667

1.745745186

0.15

4407

14.511875

1.727173429

0.15

4408

14.35583333

1.708601671

0.15

4409

14.19979167

1.690029914

0.15

4410

1.671458157

0.15

4411

13.88770833

1.652886399

0.15

4412

13.73166667

1.634314642

0.15

4413

13.575625

1.615742885

0.15

4414

13.41958333

1.597171127

0.15

4415

13.26354167

1.208641179

0.163197742

4416

13.1075

1.194421871

0.165349217

4417

12.95145833

1.180202563

0.167528829

4418

12.79541667

1.165983255

0.169737323

4419

12.639375

1.151763947

0.171975475

4420

12.48333333

1.137544639

0.174244093

4421

12.32729167

1.123325331

0.176544017

1.109106023

0.178876124

4422

14.04375

12.17125

4423

12.01520833

1.094886715

0.181241326

4424

11.85916667

1.080667407

0.183640578

4425

11.703125

2.079962483

0.15

4426

11.54708333

2.05222965

0.15

4427

11.39104167

2.024496817

0.15

4428

11.235

1.996763984

0.15

4429

11.07895833

1.969031151

0.15

4430

10.92291667

1.941298318

0.15

4431

10.766875

1.913565484

0.15

4432

10.61083333

1.885832651

0.15

4433

10.45479167

1.858099818

0.15

1.830366985

0.15

4434

10.29875

4435

10.14270833

1.802634152

0.15

4436

9.986666667

1.774901319

0.15

1.747168486

0.15

4437

9.830625

4438

9.674583333

1.719435653

0.15

4439

9.518541667

1.69170282

0.15

4440

9.3625

1.663969986

0.15

4441

9.206458333

1.636237153

0.15

4442

9.050416667

1.60850432

0.15

Depth (ft)

Pc (psi) 4443

J function

8.894375

Sw

1.580771487

0.15

4444

8.738333333

1.553038654

0.15

4445

8.582291667

1.525305821

0.15

4446

8.42625

1.497572988

0.15

4447

8.270208333

1.469840155

0.15

4448

8.114166667

1.442107322

0.15

1.414374488

0.15

4449

7.958125

4450

7.802083333

0.333917853

0.422517807

4451

7.646041667

0.327239495

0.427345388

4452

7.49

0.320561138

0.432316985

4453

7.333958333

0.313882781

0.437441454

4454

7.177916667

0.307204424

0.442728495

0.300526067

0.448188764

4455

7.021875

4456

6.865833333

0.29384771

0.453833997

4457

6.709791667

0.287169353

0.459677166

4458

6.55375

0.280490996

0.465732657

4459

6.397708333

0.273812639

0.472016484

4460

6.241666667

0.267134282

0.478546544

4461

0.260455925

0.485342924

4462

5.929583333

0.253777568

0.492428274

4463

5.773541667

0.247099211

0.499828262

4464

5.6175

0.240420854

0.507572137

4465

5.461458333

0.591326941

0.298827844

4466

5.305416667

0.574431885

0.304659796

0.55753683

0.310703242

4467

6.085625

5.149375

4468

4.993333333

0.540641774

0.316974099

4469

4.837291667

0.523746719

0.323490153

4470

4.68125

0.506851663

0.330271363

4471

4.525208333

0.489956608

0.337340227

4472

4.369166667

0.473061552

0.344722235

0.456166497

0.352446421

4473

4.213125

4474

4.057083333

0.439271442

0.360546056

4475

3.901041667

0.422376386

0.369059512

4476

3.745

0.405481331

0.378031362

4477

3.588958333

0.388586275

0.387513791

4478

3.432916667

4479

3.276875

0.37169122

0.397568439

0.354796164

0.408268828

4480

3.120833333

0.133567141

0.746088199

4481

2.964791667

0.126888784

0.783061265

4482

2.80875

0.120210427

0.830646363

Depth (ft)

Pc (psi)

J function

Sw

4483

2.652708333

0.11353207

0.897502404

4484

2.496666667

0.106853713

1

0.100175356

1

4485

2.340625

4486

2.184583333

0.093496999

1

4487

2.028541667

0.086818642

1

4488

1.8725

0.080140285

1

4489

1.716458333

0.073461928

1

4490

1.560416667

0.066783571

1

0.060105213

1

4491

1.404375

4492

1.248333333

0.053426856

1

4493

1.092291667

0.046748499

1

4494

0.93625

0.040070142

1

4495

0.780208333

0.033391785

1

4496

0.624166667

0.026713428

1

0.020035071

1

4497

0.468125

4498

0.312083333

0.013356714

1

4499

0.156041667

0.006678357

1

4500

0

0

1

Figure S3.1.- Plot of initial water saturation distribution with depth. 4360 4380

Depth (ft)

4400 4420 4440 4460 4480 4500 0

0.2

0.4

0.6

Water Saturation

0.8

1

Problem #4: Determine the fluid distribution in a reservoir using pressure gradient calculations. Figure 1 shows the diagram of a reservoir where the exploration well penetrates the reservoir near the top of the structure, observing only gas. There is no information about the presence of oil from this well. From a wireline formation test (Table 4.1) the reservoir pressure and temperature are measured, and a gas sample is obtained. Exploration well

Gas

?

! ! Figure 1. Diagram of reservoir with exploration well.

Table 4.1 - Wireline formation test info Depth 5100 ft Reservoir pressure 2377 psia Reservoir temperature 170 ºF 0.65 γ Gas specific gravity, g Gas z-factor, z 0.87

1.- Calculate the pressure at the top of the reservoir. From a geologic model the top of the reservoir is at 5000 ft. (15 points)

! 0.01877 ⋅ 0.65 ⋅ (−100 ) $ ! 0.01877γ g h$ && = 2371.71psia p2 = p1 exp # & = 2377exp ## 0.87⋅ 170 + 460 zT % " ( ) % "

2.- Assuming this is a gas reservoir at normal hydrostatic pressure, (dp/dz)w = 0.45 psi/ft, calculate the deepest possible gas-water contact. (15 points)

pw = pg Deepest Possible Gas-Water Contact (DPGWC)

$ !0.01877⋅ γ g ! dp $ DPGWC − D1 ) & # & ⋅ DPGWC +14.7 = p1 exp # ( " dh %w zT " % To solve this non-linear equation, a non-linear solver is needed. In this case we can use the Goal Seek or Solver from Excel. DPGWC = 5267.32 ft

3.- Plot the pressure distribution within the gas layer, with depth in the vertical axis and pressure in the horizontal axis. (10 points) Note: Consider the gas density as a function of pressure. The gas...