Problems 1

Introduction to
Well Testing

NExT April 2000

Problems 2

Exercise 1
Introduction to Well Testing
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.

NExT April 2000

Problems 3

Exercise 2
Introduction to Well Testing
Define Variables Used In Well Testing

Define, give the units for, and name a common source for each of
the 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

NExT April 2000

Problems 4

Exercise 3
Introduction to Well Testing
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 = 27°API,

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

p = 3500 psi, cf = 3.6×10-6 psi-1

Tsep = 75°F, 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

NExT April 2000

Problems 5

Exercise 4
Introduction to Well Testing
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 = 27°API,

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

p = 2000 psi, cf = 3.6×10-6 psi-1

Tsep = 75°F, p sep = 115 psia

From fluid properties correlations,

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

NExT April 2000

Problems 6

Exercise 5
Introduction to Well Testing
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 = 125°F, p = 125 psi,

cf = 3.6×10-6 psi-1, cw = 4 x 10-6 psi [Tf is outside range of

correlations]

From fluid properties correlations,

cg = 8.144 x 10-3 psi-1

cw = 4x10-6 psi-1

NExT April 2000

Problems 7

Exercise 6
Introduction to Well Testing
Calculate 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 = 270°F,

p = 5,000 psi, cf = 20×10-6 psi-1

From fluid properties correlations,

cg = 1.447 x 10-4 psi-1

cw = 3.512 x10-6 psi-1

NExT April 2000

Problems 8

Radial Flow and

Radius of Investigation

NExT April 2000

Problems 9

Exercise 1
Radial Flow and Radius of Investigation
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 of

investigation.

4. Increasing test time __________________ the radius of investigation.

5. Increasing production rate __________________ the radius of

investigation.

6. Increasing net pay thickness __________________ the radius of

investigation.

7. Increasing porosity __________________ the radius of investigation.

8. Increasing total compressibility __________________ the radius of

investigation.

NExT April 2000

Problems 10

Exercise 2
Radial Flow and Radius of Investigation
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)

φ = 0.17

µ = 1.06 cp

ct = 1.36x10-5 psi-1

ko = 250 md

NExT April 2000

Problems 11

Exercise 3
Radial Flow and Radius of Investigation
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)

φ = 0.17

µ = 1.185 cp

ct = 1.42 x 10-4 psi-1

kro = 0.8

k = 250 md (absolute permeability)

NExT April 2000

Problems 12

Exercise 4
Radial Flow and Radius of Investigation
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

φ = 0.12

µ = 0.01151 cp

ct = 6.52 x 10-3 psi-1

k = 100 md

NExT April 2000

Problems 13

Exercise 5
Radial Flow and Radius of Investigation
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 from
Exercise 5 in the previous section, with the following additional
information.

High-pressure, low-permeability gas reservoir

φ = 0.04

µ = 0.02514 cp

ct = 1.151 x 10-4 psi-1

k = 0.08

NExT April 2000

Problems 14

Characterizing
Damage and Stimulation

NExT April 2000

Problems 15

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.

NExT April 2000

Problems 16

Semilog Analysis for

Oil Wells

NExT April 2000

Problems 17

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 φ = 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 pwf
2 3510.3 18 3414.5
3 3492.7 24 3402.0
4 3480.1 30 3392.3
6 3462.4 36 3384.3
8 3449.9 48 3371.8
10 3440.2 60 3362.1
12 3432.2 72 3354.1
15 3422.5

NExT April 2000

Problems 18

Exercise 1
3600

3550

3500
Pressure, psi

3450

3400

3350

3300
1 10 100
Time, hrs

NExT April 2000

Problems 19

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. The
well 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
φ = 15% B = 1.25 RB/STB
q = 80 STB/D tp = 999 hrs
pwf = 1847.8 psia

Pressure Buildup Test Data

t HTR pws t HTR pws
2 2615.1 18 2662.5
3 2623.9 24 2668.6
4 2630.1 30 2673.3
6 2638.9 36 2677.1
8 2645.1 48 2683.1
10 2649.9 60 2687.7
12 2653.8 72 2691.4
15 2658.6

NExT April 2000

Problems 20

Exercise 2
2800

2750

2700
Pressure, psi

2650

2600

2550

2500
1000 100 10 1
Horner Time Ratio

NExT April 2000

Problems 21

Wellbore Storage

NExT April 2000

Problems 22

Exercise 1
Calculate WBS Coefficient For Single-Phase Liquid

Calculate the wellbore volume and WBS coefficient for a wellbore

filled 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?

NExT April 2000

Problems 23

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?

NExT April 2000

Problems 24

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 155°F,
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?

NExT April 2000

Problems 25

Manual Log-log Analysis

NExT April 2000

Problems 26

Exercise 1
1000

Pressure change, derivative, psi

100

10

1
0.001 0.01 0.1 1 10 100

Equivalent time, hr

Given the following data, analyze the data in the

log-log graph above.

q = 50 STB/D
h = 25 ft φ = 27.6 % rw = 0.36 ft
Bo = 1.099 RB/STB ct = 9.4 x 10−6 psi−1 µο = 5.28 cp

NExT April 2000

Problems 27

Exercise 2
10000
Adjusted pressure change, derivative, psi

1000

100

10
0.001 0.01 0.1 1 10 100

Adjusted equivalent time, hr

Given the following data, analyze the data in the

log-log graph above.
qg = 5108 Mscf/D
h = 4.4 ft φ = 10 % rw = 0.33 ft
Bgi = 0.781 RB/Mscf cti = 1.66 x 10−4 psi−1 µ gi = 0.0214 cp

NExT April 2000

Problems 28

Exercise 3
10000

Pressure change, derivative, psi

1000

100

10
0.001 0.01 0.1 1 10 100

Equivalent time, hr

Given the following data, analyze the data in the

log-log graph above.

q = 1200 STB/D
h = 26 ft φ = 21.6 % rw = 0.22 ft
Bo = 1.52 RB/Mscf ct = 16.6 x 10−6 psi−1 µο = 0.29 cp

NExT April 2000

Problems 29

Exercise 4
10000
Adjusted pressure change, derivative, psi

1000

100

10
0.1 1 10 100 1000 10000

Adjusted equivalent time, hr

Given the following data, analyze the data in the

log-log graph above.

qg = 380 Mscf/D
h = 6 ft φ = 18 % rw = 0.3 ft
Bgi = 0.744 RB/Mscf cti = 1.24 x 10−4 psi−1 µgi = 0.024 cp

NExT April 2000

Problems 30

Flow Regimes and

the Diagnostic Plot

NExT April 2000

Problems 31

Exercise 1
Flow Regimes and the Diagnostic Plot
FLOWREGM.WTD (Diagnostic Plot)
1000

100
Adjusted pressure change, psi

10

0.1

0.01
0.0001 0.001 0.01 0.1 1 10 100 1000 10000

Radial equivalent adjusted time, hr

Identify as many flow regimes as possible.

NExT April 2000