A1 - Problems

Problems 1

NExT April 2000

Introduction to Well Testing

Problems 2

NExT April 2000

Exercise 1


List 4 Objectives of Well Testing

List 4 objectives of well testing. List as many as possible without referring to the notes.

1.

2.

3.

4.

Problems 3

NExT April 2000

Exercise 2


Define Variables Used In Well Testing

Define, give the units for, and name a common source for each ofthe following variables used in well testing. Complete as much of this exercise as possible before referring to the notes.

1. Porosity

2. Water saturation

3. Total compressibility

4. Oil compressibility

5. Formation volume factor

6. Viscosity

7. Wellbore radius

8. Net pay thickness

9. Permeability

Problems 4

NExT April 2000

Exercise 3


Calculate Compressibility for Undersaturated Oil Reservoir

Calculate total compressibility for the following situation. Assume solution gas/oil ratios do not include stock tank vent gas.

Undersaturated oil reservoir (above the bubblepoint)

Sw = 17%, TDS = 18 wt %, oil gravity = 27API,

Rso = 530 scf/STB, gas gravity = 0.85, Tf = 185F,

p = 3500 psi, cf = 3.610-6 psi-1

Tsep = 75F, p sep = 115 psia

From fluid properties correlations,

pb = 2803 psi

co = 1.158 x 10-5 psi-1

cw = 2.277 x 10-6 psi-1

Problems 5

NExT April 2000

Exercise 4


Calculate Compressibility for Saturated Oil Reservoir

Calculate total compressibility for the following situation. Assume solution gas/oil ratios do not include stock tank vent gas.

Saturated oil reservoir (below the original bubblepoint)

Sw = 17%, Sg = 5%, TDS = 18 wt %, oil gravity = 27API,

Rso = 530 scf/STB, gas gravity = 0.85, Tf = 185F,

p = 2000 psi, cf = 3.610-6 psi-1

Tsep = 75F, p sep = 115 psia


pb = 2803 psi

co = 1.429 x 10-4 psi-1

cg = 5.251 x 10-4 psi-1

cw = 4.995 x 10-6 psi-1

Problems 6

NExT April 2000

Exercise 5


Calculate Compressibility for Low-Pressure, High-Permeability Gas Reservoir

Calculate total compressibility for the following situation. Assume a dry gas.

Low-pressure, high-permeability gas reservoir

Sw = 20%, gas gravity = 0.74, Tf = 125F, p = 125 psi,

cf = 3.610-6 psi-1, cw = 4 x 10-6 psi [Tf is outside range of correlations]


cg = 8.144 x 10-3 psi-1

cw = 4x10-6 psi-1

Problems 7

NExT April 2000

Exercise 6

Introduction to Well TestingCalculate Compressibility for

High-Pressure, Low-Permeability Gas Reservoir

Calculate total compressibility for the following situation. Assume a dry gas.

High pressure, low permeability gas reservoir

Sw = 35%, TDS = 22 wt %, gas gravity = 0.67, Tf = 270F,

p = 5,000 psi, cf = 2010-6 psi-1


cg = 1.447 x 10-4 psi-1

cw = 3.512 x10-6 psi-1

Problems 8

NExT April 2000

Radial Flow and Radius of Investigation

Problems 9

NExT April 2000

Exercise 1


Factors That Affect Radius of Investigation

Without looking at the notes, choose the correct response to complete each statement. Check your answers by referring to the radius of investigation equation.

A) increases

B) decreases

C) does not affect

1. Increasing viscosity __________________ the radius of investigation.

2. Increasing permeability __________________ the radius of investigation.

3. Increasing formation volume factor __________________ the radius ofinvestigation.

4. Increasing test time __________________ the radius of investigation.

5. Increasing production rate __________________ the radius ofinvestigation.

6. Increasing net pay thickness __________________ the radius ofinvestigation.

7. Increasing porosity __________________ the radius of investigation.

8. Increasing total compressibility __________________ the radius ofinvestigation.

Problems 10

NExT April 2000

Exercise 2


Calculate Radius of Investigation for an Undersaturated Oil Reservoir

Calculate the time required to reach a radius of investigation of 745 feet for the following situation. Use the data and results from Exercise 3 in the section Introduction to Well Testing, with the following additional information.

Undersaturated oil reservoir (above the bubblepoint)

= 0.17

= 1.06 cp

ct = 1.36x10-5 psi-1

ko = 250 md

Problems 11

NExT April 2000

Exercise 3


Calculate Radius of Investigation for a Saturated Oil Reservoir

Calculate the time required to reach a radius of investigation of 745 feet for the following situation. Use the data and results from Exercise 4 in the section Introduction to Well Testing, with the following additional information.

Saturated oil reservoir (below the original bubblepoint)

= 0.17

= 1.185 cp

ct = 1.42 x 10-4 psi-1

kro = 0.8

k = 250 md (absolute permeability)

Problems 12

NExT April 2000

Exercise 4


Calculate Radius of Investigation for a Low-Pressure, High-Permeability

Gas Reservoir

Calculate the time required to reach a radius of investigation of 745 feet for the following situation. Use the data and results from Exercise 5 in the section Introduction to Well testing, with the following additional information.

Low-pressure, high-permeability gas reservoir

= 0.12

= 0.01151 cp

ct = 6.52 x 10-3 psi-1

k = 100 md

Problems 13

NExT April 2000

Exercise 5


Calculate Radius of Investigation for a High-Pressure, Low-Permeability

Gas Reservoir

Calculate the time required to reach a radius of investigation of 745 feet for the following situation. Use the data and results fromExercise 5 in the previous section, with the following additional information.

High-pressure, low-permeability gas reservoir

= 0.04

= 0.02514 cp

ct = 1.151 x 10-4 psi-1

k = 0.08

Problems 14

NExT April 2000

CharacterizingDamage and Stimulation

Problems 15

NExT April 2000

Exercise 1

Damage and Skin Factor Calculations

1. Calculate the additional pressure drop due to skin for a well producing at 2,000 STB/D. Oil formation volume factor is 1.07 RB/STB, viscosity is 19 cp, permeability is 5400 md, net pay thickness is 175 ft, skin factor is 11, and porosity is 1.2%.

2. Calculate the flow efficiency for the well in Problem 1, if the average reservoir pressure is 1,800 psi and the flowing bottomhole pressure is 1,600 psi.

3. Calculate the apparent wellbore radius for the well in Problem 1, if the bit diameter is 8 in.

4. Calculate the new skin factor if we create a 100-ft fracture in the reservoir in Problem 1.

Problems 16

NExT April 2000

Semilog Analysis for Oil Wells

Problems 17

NExT April 2000

Exercise 1

Determining permeability and skin factor from a constant-rate flow test

The data summarized below were recorded during a pressure drawdown test from an oil well. Estimate the effective permeability to oil and the skin factor using the graphical analysis technique for a constant-rate flow test.

q = 250 STB/D pi = 4,412 psia

h = 46 ft = 12%rw = 0.365 ft B = 1.136 RB/STB

ct = 17 x 10-6 psi-1 m = 0.8 cp

Pressure Drawdown Test Data for Exercise 1

t pwf t pwf2 3510.3 18 3414.53 3492.7 24 3402.04 3480.1 30 3392.36 3462.4 36 3384.38 3449.9 48 3371.810 3440.2 60 3362.112 3432.2 72 3354.115 3422.5

Problems 18

NExT April 2000

Exercise 1

3300

3350

3400

3450

3500

3550

3600

1 10 100

Time, hrs

Pre

ssu

re, p

si

Problems 19

NExT April 2000

Exercise 2

Determining permeability and reservoir pressure from buildup tests

A pressure buildup test was conducted on a well early in the life of an oil reservoir having the properties summarized below. Thewell was produced at a constant rate of 80 STB/D for 999 hours prior to being shut in. Determine the effective permeability to oil, the original reservoir pressure, and skin factor.

m = 2.95 cp ct = 15 x 10-6 psi-1

rw = 0.25 ft h = 32 ft

= 15% B = 1.25 RB/STBq = 80 STB/D tp = 999 hrs

pwf = 1847.8 psia

Pressure Buildup Test Datat HTR pws t HTR pws2 2615.1 18 2662.53 2623.9 24 2668.64 2630.1 30 2673.3

6 2638.9 36 2677.18 2645.1 48 2683.1

10 2649.9 60 2687.712 2653.8 72 2691.415 2658.6

Problems 20

NExT April 2000

2500

2550

2600

2650

2700

2750

2800

1101001000

Horner Time Ratio

Pre

ssu

re, p

si

Exercise 2

Problems 21

NExT April 2000

Wellbore Storage

Problems 22

NExT April 2000

Exercise 1

Calculate WBS Coefficient For Single-Phase Liquid

Calculate the wellbore volume and WBS coefficient for a wellborefilled with a single phase liquid. The well is 2600 ft deep and has 6 5/8, 24 lb/ft casing (5.921 ID). The bottomhole pressure is 1,690 psi. If the well is filled with water (cw = 4 x 10-6 psi-1) what is the wellbore storage coefficient?

Problems 23

NExT April 2000

Exercise 2

Calculate WBS Coefficient For Rising Liquid Level

Calculate the cross-sectional area and wellbore storage coefficient for a wellbore with a rising liquid level. The well is 2600 ft deep and has 6 5/8, 24 lb/ft casing (5.921 ID). the bottomhole pressure is 750 psi. If the well has a column of water of density 1.04 g/cm3, in it, what is the wellbore storage coefficient?

Problems 24

NExT April 2000

Exercise 3

Calculate WBS Coefficient for Single-Phase Gas

A wellbore is filled with a single-phase gas. the well has 7200 ft of 2 7/8 tubing (2.441 ID) and 375 ft of 6 5/8, 24 lb/ft casing (5.921 ID). the average temperature in the wellbore is 155F, and the average pressure is 2,775 psia. If the wellbore is filled with gas having 0.77 gas gravity and 0.2% CO2, what is the WBS coefficient?

Problems 25

NExT April 2000

Manual Log-log Analysis

Problems 26

NExT April 2000

Exercise 1

1

10

100

1000

0.001 0.01 0.1 1 10 100

Equivalent time, hr

Pre

ssu

re c

han

ge,

der

ivat

ive,

psi

Given the following data, analyze the data in the log-log graph above.

q = 50 STB/Dh = 25 ft = 27.6 % rw = 0.36 ft

Bo = 1.099 RB/STB ct = 9.4 x 106 psi1 = 5.28 cp

Problems 27

NExT April 2000

Exercise 2

10

100

1000

10000

0.001 0.01 0.1 1 10 100

Adjusted equivalent time, hr

Ad

just

ed p

ress

ure

ch

ang

e, d

eriv

ativ

e, p

si


qg = 5108 Mscf/Dh = 4.4 ft = 10 % rw = 0.33 ft

Bgi = 0.781 RB/Mscf cti = 1.66 x 104 psi1 gi = 0.0214 cp

Problems 28

NExT April 2000

Exercise 3

10

100

1000

10000

0.001 0.01 0.1 1 10 100

Equivalent time, hr

Pre

ssu

re c

han

ge,

der

ivat

ive,

psi


q = 1200 STB/Dh = 26 ft = 21.6 % rw = 0.22 ft

Bo = 1.52 RB/Mscf ct = 16.6 x 106 psi1 = 0.29 cp

Problems 29

NExT April 2000

Exercise 4

10

100

1000

10000

0.1 1 10 100 1000 10000

Adjusted equivalent time, hr

Ad

just

ed p

ress

ure

ch

ang

e, d

eriv

ativ

e, p

si


qg = 380 Mscf/Dh = 6 ft = 18 % rw = 0.3 ft

Bgi = 0.744 RB/Mscf cti = 1.24 x 104 psi1 gi = 0.024 cp

Problems 30

NExT April 2000

Flow Regimes and the Diagnostic Plot

Problems 31

NExT April 2000

Exercise 1Flow Regimes and the Diagnostic Plot

FLOWREGM.WTD (Diagnostic Plot)

0.01

0.1

1

10

100

1000

0.0001 0.001 0.01 0.1 1 10 100 1000 10000

Adj

uste

d pr

essu

re c

hang

e, p

si

Radial equivalent adjusted time, hr

Identify as many flow regimes as possible.

A1 - Problems

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