Heriot-Watt University - Reservoir Engineering PDF
Heriot-Watt University - Reservoir Engineering PDF
Heriot-Watt University - Reservoir Engineering PDF
Reservoir Engineering notes cover an extensive amount of material. They are support
material for the examination in this topic but are also considered to be useful material
in subsequent career use. Not all the material in the text can be covered in a limited
time examination.
In the context of the examination a student should consider the learning objectives at
the front of each section which should help in the level of detail and analysis which
is required in relation to an examination covering the various topics.
Detailed below is a graded analysis of each section which should help the candidate
in examination preparation. These should be considered alongside the learning
objectives.
Grading structure:
OM- Material covered in another module not for examination purposes in Reservoir
Engineering.
Darcy’s Law,
PV = nzRT
STOOIP equation
Equilibrium Ratio K=y/x
5
1 INTRODUCTION TO RESERVOIR ENGINEERING
5 BEHAVIOUR OF GASES
9 PERMEABILITY-ITS VARIATIONS
11 DRIVE MECHANISMS
14 PVT ANALYSIS
17 WATER INFLUX
18 IMMISCIBLE DISPLACEMENT
This Reservoir Engineering module covers material presented in a range of reservoir engineering texts and a
number of the figures and examples are based on these texts and copyright is currently being sought. The student
may find the more detailed analysis in these texts supportive when going through these notes. The following
books are considered useful in building up a reservoir engineering library.
4.Fundamental Principles of Petroleum Reservoir B.F. Towler. Society of Petroleum Engineers Inc
Engineering ISBN:55563-092-8
7.Petroleum Engineering Principles and Practise. J.S.Archer & C.Wall.Graham & Trotman.
1986. ISBN:0-86910-715-9
9.PVT and Phase Behaviour of Petroleum Reservoirs A. Danesh. Elsevier. ISBN: 0-444-82196-1
Adrian C Todd
All rights reserved no part of this publication may be reproduced, stored in a retrieval system or
transmitted in any form or by any means, electronic, mechanical, photocopying, recording or
otherwise without the prior permission of the Copyright owner.
2
Reservoir Engineering notes cover an extensive amount of material. They are support
material for the examination in this topic but are also considered to be useful material
in subsequent career use. Not all the material in the text can be covered in a limited
time examination.
In the context of the examination a student should consider the learning objectives at
the front of each section which should help in the level of detail and analysis which
is required in relation to an examination covering the various topics.
Detailed below is a graded analysis of each section which should help the candidate
in examination preparation. These should be considered alongside the learning
objectives.
Grading structure:
OM- Material covered in another module not for examination purposes in Reservoir
Engineering.
Darcy’s Law,
PV = nzRT
STOOIP equation
Equilibrium Ratio K=y/x
5
Introduction To Reservoir Engineering
CONTENTS
4 PROBABILISTIC REPRESENTATION OF
RESERVES
7 DEVELOPMENT PLANNING
7.1 Reservoir Modelling
7.2 Technoconomics
7.3 Coping with Uncertainty
LEARNING OBJECTIVES
Having worked through this chapter the Student will be able to:
• Show using a block diagram the integration of reservoir engineering with other
petroleum engineering and other subjects.
• Calculate given the prerequisite data proved, probable and possible reserves.
• Describe briefly with the aid of a sketch the various maps used to represent
reservoir; area, thickness porosity, saturation.
• Describe briefly the use of the production (well0 test to determine reservoir
flowability and properties.
Introduction To Reservoir Engineering
1 INTRODUCTION
With the petroleum industry’s desire to conserve and produce oil and gas more efficiently
a field of specialisation has developed called Petroleum Reservoir Engineering. This
new science which can be traced back only to the mid 1930’s has been built up on a
wealth of scientific and practical experience from field and laboratory. In the 1959
text of Craft & Hawkins1 on Applied Reservoir Engineering it is commented that “as
early as 1928 petroleum engineers were giving serious consideration to gas-energy
relationships and recognised the need for more precise information concerning
physical conditions as they exist in wells and underground reservoirs. Early progress
in oil recovery methods made it obvious that computations made from wellhead or
surface data were generally misleading.” Dake2, in his text "The Practise of Reservoir
Engineering", comments that “Reservoir Engineering shares the distinction with
geology in being one of the ‘underground sciences’ of the oil industry, attempting
to describe what occurs in the wide open spaces of the reservoir between the sparse
points of observation - the wells”
The reservoir engineer in the multi-disciplinary perspective of modern oil and gas
field management is located at the heart of many of the activities acting as a central
co-ordinating role in relation to receiving information processing it and passing it on
to others. This perspective presented by Dake2 is shown in the figure below.
Exploration
Geophysics/ Petrophysics
Geology
Reservoir Engineering
Economics Production
(Project viability) Process Egineering
General Engineering
Platform Topsides Design
2
Figure 1 Reservoir Engineering in Relation to Other Activities (adapted Dake )
The responsibility of the first is shared with other disciplines whereas the second is
primarily the responsibility of the reservoir engineer. Attaching a time scale to recovery
is the development of a production profile and again is not an exclusive activity. The
day-to-day operational role is on going through the duration of the project.
A project can be conveniently divided into two stages and within these the above
activities take place, the appraisal stage and the development phase. The appraisal
phase is essentially a data collection and processing phase with the one objective of
determining the ‘viability’ of a project. The development phase covers the remaining
period if the project is considered viable from the time continuous production com-
mences to the time the field is abandoned. Reservoir engineering activity in various
forms takes place during both of these stages.
The activities of reservoir engineering fall into the following three general catego-
ries:
Before any production has been obtained, the so-called ‘volumetric estimate of
reserves’ is usually made. Geological and geophysical data are combined to obtain
a range of contour maps with the help of a planimeter and other tools the hydrocar-
bon bearing rock volumes can be estimated. From well log petrophysical analysis,
estimates of an average porosity and water saturation can be made and when applied
to the hydrocarbon rock volume yield an estimate of oil in place (STOIIP). Since
it is well known that only a fraction of this oil may in fact be ‘recoverable’, labora-
tory tests on cores may be carried out to estimate movable oil. The reserve estimate
finally arrived at is little more than an educated guess but a very important one for
it determines company policy.
In 1987 the Society of Petroleum Engineers in collaboration with the World Petroleum
Congress published definitions with respect to reserves and these are now accepted
world-wide 3. These definitions have been used in the summary of reserve defini-
tions which follow.
Introduction To Reservoir Engineering
In the past the traditionally available reservoir engineering tools were mainly
designed to give satisfactory results for a slide rule and graph paper approach. For
many problems encountered by reservoir engineers today this remains a perfectly
valid approach where the slide rule has been replaced by the calculator. Increasingly,
however, the advance of computing capability is enabling reservoir engineering
modelling methods (‘simulations’) to be carried out at the engineers desk, previously
considered impossible.
The basis of the development of the 'model' of the reservoir are the various data
sources. As the appraisal develops the uncertainty reduces in relation to the quality
of the forecasts predicted by the model. Building up this ‘geological’ model of the
reservoir progresses from the early interpretation of the geophysical surveys, through
various well derived data sets, which include drilling information, indirect wireline
measurements, recovered core data, recovered fluid analysis, pressure depth surveys,
to information generated during production.
3. RESERVE ESTIMATING
The Society of Petroleum Engineers SPE and World Petroleum Congress WPO1987
agreed classification of reserves3 provides a valuable standard by which to define
reserves, the section below is based on this classification document.
3.1 Definitions
Reserves are those quantities of petroleum which are anticipated to be commercially
recovered from known accumulations from a given date forward.
All reserve estimates involve some degree of uncertainty. The uncertainty depends
chiefly on the amount of reliable geologic and engineering data available at the time
Unproved reserves are less certain to be recovered than proved reserves and may
be further sub-classified as probable and possible reserves to denote progressively
increasing uncertainty in their recoverability.
If deterministic methods are used, the term reasonable certainty is intended to express
a high degree of confidence that the quantities will be recovered. If probabilistic
methods are used, there should be at least a 90% probability that the quantities actu-
ally recovered will equal or exceed the estimate.
Introduction To Reservoir Engineering
the term proved refers to the actual quantities of petroleum reserves and not just
the productivity of the well or reservoir. In certain cases, proved reserves may
be assigned on the basis of well logs and/or core analysis that indicate the subject
reservoir is hydrocarbon bearing and is analogous to reservoirs in the same area that
are producing or have demonstrated the ability to produce on formation tests.
The area of the reservoir considered as proved includes (1) the area delineated by
drilling and defined by fluid contacts, if any, and (2) the undrilled portions of the
reservoir that can reasonably be judged as commercially productive on the basis of
available geological and engineering data. In the absence of data on fluid contacts, the
lowest known occurrence of hydrocarbons controls the proved limit unless otherwise
indicated by definitive geological, engineering or performance data. Reserves may be
classified as proved if facilities to process and transport those reserves to market are
operational at the time of the estimate or there is a reasonable expectation that such
facilities will be installed. Reserves in undeveloped locations may be classified as
proved undeveloped provided (1) the locations are direct offsets to wells that have
indicated commercial production in the objective formation, (2) it is reasonably
certain such locations are within the known proved productive limits of the objective
formation, (3) the locations conform to existing well spacing regulations where
applicable, and (4) it is reasonably certain the locations will be developed. Reserves
from other locations are categorised as proved undeveloped only where interpretations
of geological and engineering data from wells indicate with reasonable certainty that
the objective formation is laterally continuous and contains commercially recoverable
petroleum at locations beyond direct offsets.
Before looking at further detail we will carry out some tests to help emphasise the
above definition.
Test 1
There are 950 MM stb ( million stock tank barrels) of oil initially in place in a res-
ervoir. It is estimated that 500 MM stb can be produced. Already 100 MM stb have
been produced. In the boxes below, identify the correct answer.
Before starting production it was estimated that there was a 90% chance of produc-
ing at least 100 MM stb, 50% chance of producing 500 MM stb and 10% chance of
producing 700MM stb. That is we are sure we can produce at least 100MM stb, and
we will probably produce as much as 500 MM stb, and we will possibly produce as
much as 700 MM stb.
Test 3
Test 4
Test 5
A reservoir has been discovered by drilling a successful exploration well, and drilling
a number of producing wells. We have even produced some 200 MM stb of oil.
Introduction To Reservoir Engineering
Test 1 answer
There are 950 MM stock tank boards in place. It is estimated that 500 MM stb can
be produced and 100 MM stb have been produced then 400 recoverable reserves
remain.
Test 2 answer
Test 3 answer
Economic Variables
What economic factors are used in the calculations? What oil and gas price do we
use for proved reserve estimates? Is inflation taken into account? Do we predict
future price trends? Do we apply discount factors to calculate present value of the
project? Are all these used in proved reserve calculations? The current economic
conditions are used for the calculations, with respect to prices, costs, contracts and
government regulations.
Test 4 answer
The following sources are required for proved reserves. Maps (from seismic and/
geological data). Petrophysical logs. Well test results and rock properties from core
analysis tests on recovered core.
This comes from drilled and produced hydrocarbons, the definition of the gas and oil
and water contacts or the highest and lowest observed level of hydrocarbons. Also
the undrilled area adjacent to the drilled can be used.
Test 5 answer
Possible Possible
Probability Levels
P50
Probable Probable
Time
Seismic Discovery of Start of Dev Start of Abandonment
Data Well Planning Production
What are the amounts termed that are not recoverable? The quantity of hydrocar-
bons that remains in the reservoir are called remaining hydrocarbons in place, NOT
remaining reserves!
10
Introduction To Reservoir Engineering
(ii) It is reasonably certain that the project will proceed. Reserves to be recovered
by improved recovery methods that have yet to be established through
commercially successful applications are included in the proved classification
only:
(i) After a favourable production response from the subject reservoir from either
(b) An installed program where the response provides support for the analysis
on which the project is based and
(3) Incremental reserves attributable to infill drilling that could have been classified
as proved if closer statutory spacing had been approved at the time of the
estimate,
(5) Reserves in an area of the formation that appears to be separated from the
proved area by faulting and the geologic interpretation indicates the subject
area is structurally higher than the proved area,
(2) reserves in formations that appear to be petroleum bearing based on log and
core analysis but may not be productive at commercial rates,
(3) incremental reserves attributed to infill drilling that are subject to technical
uncertainty,
12
Introduction To Reservoir Engineering
3.4.1. Developed:
Developed reserves are expected to be recovered from existing wells including reserves
behind pipe. Improved recovery reserves are considered developed only after the
necessary equipment has been installed, or when the costs to do so are relatively minor.
Developed reserves may be sub-categorised as producing or non-producing.
3.4.1.1 Producing:
Reserves subcategorised as producing are expected to be recovered from comple-
tion intervals which are open and producing at the time of the estimate. Improved
recovery reserves are considered producing only after the improved recovery project
is in operation.
3.4.1.2. Non-producing:
Reserves subcategorised as non-producing include shut-in and behind-pipe reserves.
Shut-in reserves are expected to be recovered from (1) completion intervals which
are open at the time of the estimate but which have not started producing, (2) wells
which were shut-in for market conditions or pipeline connections, or (3) wells not
capable of production for mechanical reasons. Behind-pipe reserves are expected to
be recovered from zones in existing wells, which will require additional completion
work or future recompletion prior to the start of production.
Whereas in the deterministic approach the volumes are determined by the calculation
of values determined for the various parameters, with the probalistic statistical analysis
is used, using tools like Monte Carlo methods. The curve as shown in the figure 3
below presents the probability that the reserves will have a volume greater or equal
to the chosen value.
1.0
0.9 'Proven'
as large as indicated.
'Proven + Proable
0.1 + Possible'
0
Recoverable Reserve
On this curve:
The proven reserves represent the reserves volume corresponding to 90% probability
on the distribution curve.
The probable reserves represent the reserves volume corresponding to the difference
between 50 and 90% probability on the distribution curve.
The possible reserves represent the reserves volume corresponding to the difference
between 10 and 50% probability on the distribution curve.
As with the deterministic approach there is also some measure of subjectivity in the
probalistic approach. For each of the elements in the following equation, there is a
probability function expression in low, medium and high probabilities for the particular
values. A schematic of a possible distribution scenario for each of the elements and
the final result is given below in the figure 4.
P = p50
p10
14
Introduction To Reservoir Engineering
Probable = 240 MM stb which together with the proven makes up the P50 figure.
of 740MMstb
Possible = 120 MM stb which together with the proven and probable makes up the
P10 value of 860MMstb
60
Proven 500 MMstb P50
50
Probable 240 M
40
30
P+P+P = 860 MMstb
20
120 P10
10
Proven Probable Possible
0
0 200 400 600 800 1000
Reserves / MMstb
As a field is developed and the fluids are produced the shape of the probability curve
changes. Probability figures for reserves are gradually converted into recovery leav-
ing less uncertainty with respect to the reserves. This is illustrated in figure 6.
80
70
Probability / %
60
50 P50
40
Proved ultimate recovery.
30
20
10 P10
Production Proved reserves
0
0 200 400 600 800 1000
Reserves / MMstb
5.1 The volume of oil and gas in-place depends on a number of parameters :
The aerial coverage of the reservoir. A
The thickness of the reservoir rock contributing to the hydrocarbon volume. hn
The pore volume, as expressed by the porosity ,φ , the reservoir quality rock.
The proportion of pore space occupied by the hydrocarbon ( the saturation ).
1-Sw
The simple equation used in calculation of the volume of fluids in the reservoir, V,
is
V=Ahnφ(1-Sw): (1)
where:
A= average area
hn = nett thickness. nett thickness = gross thickness x nett: gross ratio
φ = average porosity
Sw = average water saturation.
When expressed as stock tank or standard gas volumes, equation above is divided
by the formation volume factor Bo or Bg.
V = Ahnφ (1 − Sw ) / Bo (2)
16
Introduction To Reservoir Engineering
The recovery factor, RF, indicates the proportion of the in-place hydrocarbons ex-
pected to be recovered. To convert in place volumes to reserves we need to multiply
the STOIIP by the recovery factor so that:
The line over the various terms indicates the average value for these spatial
parameters.
The reservoir area A, will vary according to the category; proven, probable or pos-
sible, that is being used to define the reserves.
Oil
Using this data and possible suggested structure we can carry out some oil in place
calculations and estimate reserves. These figures however are not admissible in public
reserve estimates. They are useful inside the company to justify project expenditure!
The question is where do we locate the first exploration well and get involved in large
exploration expenditure costs. Figure 8 suggest three alternatives
Oil
In figure 9 an exploration well has been drilled and a core recovered and the struc-
ture of the field with respect to formations and contacts redefined. The redefined
structure can now be used to provide an estimate of reserves according to the
three, proven, probable and possible perspectives. Figure 10
Oil
18
Introduction To Reservoir Engineering
Proved
e
l
ab
ble
bl
si Oil ible
ba
ob
os s
Pos
o
Pr
P
Pr
Figure 10 After The Exploration Well Was Drilled.
Subsequent appraisal wells are now drilled to give better definition of the reserves
of the field. Well 2 aimed at defining the field to the left identifies some additional
isolated hydrocarbon structure with its own oil water contact. Figure 11. The well, as
well as increasing the proven reserves, further identifies previous unknown reserves.
The next appraisal well is aimed at defining the reserves in the other direction. Dur-
ing well testing on wells 1or 2 indications of faulting are also helping to define the
flowing nature of the accumulation. Figure 12 for the further appraisal well confirms
the accumulation to the right and also identifies the impact of the fault with a new
oil water contact. Subsequent appraisal wells and early development give greater
definition to the field description. Figure 13
ven
Pro Proven
Oil
Gas
ven
Pro Proven
Oil
Gas
ven
Pro Proven
Oil
From a deterministic perspective the various reserve estimates, that is, proven,
probable and possible can be further determined. The indication of the various
elements based on the top structure map are shown. Figure 14
20
Introduction To Reservoir Engineering
Probable
1
Proved
3
4
2
Possible
Fault B
ounda
ry
Porosity
Fluid Boundary
Contact
ry
Bo unda
Fault
7
Figure 15 Structure Contour Map.
se lu
St m
ru e
ctu
re
(units ss)
Hydrocarbon Water
Contact Elevation
o
Area Contained by Contour
7
Figure 16 Reservoir cross section.
Figures 17 & 18 show an example of a top structure map and cross section of the
Rough Gas field in the North Sea.
47/2 47/3
Completed Producers
Gw
x Abandoned Wells
9
95500
00 C.I. = 50ft.
C
w
G
0
955000
960
95 50 8
94 00 8
94 50 8
93 00 8 8
8 93 8
B 250 8 A
9 A2 00
A
8 92
47/7 47/8 8 A A4
47/8-1x A5
A
8 A Platform A
A3
9100 A6
50
91 93
9200
50
92
93 50
00
x 47/8-2
5
Figure 17 Top Sand Structure Map Rough Gas Field.
22
Introduction To Reservoir Engineering
Depth (ft) A3 A1 A4
subsea
9000 A2 A5
Unc
onf
9200 orm
Rot ity
lieg
Fault
end
Fault
Unc es
onfo
9400 rmit
y
Tentative
9600 hydrocarbon/
water contact
9800
Carboniferous
Sands
5
Figure 18 Schematic Cross Section of The Rough Field.
Isopach C I
25 Units
75
100
125
150
0
7
Figure 19 Net Pay Thickness Isopach.
130
C
Gw
120
0
11 0
10
A2
47/7 47/8 A4
47/8-1 x A5
80
A1
70
11
6
A3
11
100
0
90
A6
x 47/8-2
5
Figure 20 Rough Field Isopach.
The isopach map can also be used to calculate reservoir volume. For example in figure
21 the area under a plot of net pay thickness vs. area contained within the contour
provides a net pay volume. These plots can be generated for each section or rock
type. The thickness plots for each section are called isoliths.
40
Area Enclosed = Net Rock Volume
Net Pay Isopach Value
80
120
140
OWC
180
Area Contained by Contour
7
Figure 21 Hydrocarbon Volume From Net Pay Isopach.
24
Introduction To Reservoir Engineering
∑φ h k n, k
φw = k =1
hn (4)
where φk is the average porosity derived from the log over a small thickness hn,k
within the net pay thickness, hn.
These values of porosity can then be plotted to generate an isoporosity map as il-
lustrated in figure 22. The example of an isoporosity map for the Rough Field is
shown in figure 23.
Porosity C I
5%
25
20
15
5 10
7
Figure 22 Iso Porosity Map.
47/2 47/3
Gw
C
C
w
G
A2
x A5
A
A1
12%
%
10
A3
8%
A6
6%
47/8-2
x
7
Figure 23 Rough Field Iso Porosity Map.
∑S w, k φ k h n,k
Sw,w = k =1
φwh n (5)
The values of Sw,w can be plotted and contours of constant saturation (isosaturation)
presented. Figure 24.
Shale
15 20
25
30 35
40
WOC
4
Figure 24 Iso Saturation (sw) Map.
A more detailed description together with exercises are given in the mapping section
of the geology module.
26
Introduction To Reservoir Engineering
φ (1 − Sw )
0.1611 0.0979 0.1741
k pb
ER, o = 0.4185 (Sw ) 0.3722
Bob µob pa
(6)
φ (1 − Sw )
0.0422 0.0770
k µ wi p
ER, o = 0.54898 (Sw )− o.1903 pi − 0.2159
Boi µoi a
(7)
b refers to bubble point conditions, i is the initial condition and a, refers to abandonment
pressure.
RESIDUAL SATURATIONS
1.00 1.00
Reservoirs
0.50 0.50
0.05 0.05
−σ +σ
MEDIAN
0 0
2 5 10 20 30 40 50 60 70 80 95 98
Figure 25 Log - Probability Residual Oil Saturation For Water Drive and Solution Gas
6
Drive Reservoirs. (API )
2 5 10 20 30 40 50 60 70 80 95 98
OIL RECOVERY EFFICIENCY AT FIELD ABANDONMENT
RESIDUAL SATURATIONS
1.00 1.00
Water Drive
IN PERCENT OF OIL PLACE
0.05 0.05
−σ +σ
MEDIAN
0 0
2 5 10 20 30 40 50 60 70 80 95 98
6
Figure 26 Log - Probability of Oil Recovery For Various Drive Mechanisms. (API )
28
Introduction To Reservoir Engineering
4 5 4 3 2 1
0.5
123
0.25
7
Figure 27 Isocapacity Map.
47/2
Gw
C
C
w
G
120 A2
47/7 47/8 100 A4
80 A5
47/8-1 x
60 Platform B
40
A3
0
A6
x 47/8-2
Contour Intervals 20 millidarcies
5
Figure 28 Rough Field Permeability Map.
where:
_ _
Sh = 1 − Sw
Figure 29 gives an HPT map and the Rough Field HPT map is given in figure 30
12
15
11
0
14 10
14
13
13
12
11
10
9
0
7
Figure 29 Hydrocarbon Pore Thickness Map.
10
A2
9
8
A4
A5 7
A1 6
A3 5
A6
4
5
Figure 30 Rough Field Hydrocarbon Pore Thickness.
30
Introduction To Reservoir Engineering
A considerable amount of reservoir data can be obtained from these well tests
sometimes called DST’s ( drill stem tests). It has been the practise over recent years
for the produced fluids to be flared since there is unlikely to be an infrastructure to
collect these fluids. Now that companies are moving to a zero or reduced hydrocarbon
emission policy the nature and facilities required for these tests are changing. A
feature of the flaring approach is a public demonstration of the productivity of the
well being tested.
Production casing
Production tubing
Cement
Packer
Perforations
Down hole
pressure monitor
32
Introduction To Reservoir Engineering
q bbls / day
Flow 1 Flow 2 Well shut in
Pi
Pressure build up
Well test analysis is a powerful reservoir engineering tool and is treated in depth in
a subsequent module of the Petroleum Engineering course.
The nature of the fluids is key to reservoir behaviour and also subsequent processing
in any development. The collection and analysis of these fluids is an important
role and is at the focus of PVT analysis. This topic is covered in Chapter 14 PVT
Analysis. The pressure profile in a well is another important aspect of reservoir
characterisation and can be used to identify fluid contacts. When used during the
early stages of production it can be a powerful means of refining the structure and
hydrodynamic continuity characteristics of the reservoir. This is covered in the next
chapter. Like PVT analysis where the information is based on samples removed
from the reservoir, core analysis is based on recovered core from the formation.
Various tests on this material and its reaction to various fluids provides many of the
reservoir engineering parameters important in determining the viability of a project.
Core analysis also provides a cross check for indirect measurements made downhole.
These core analysis perspectives are covered in chapters 7 and 8.
In any project new data is always being generated. Indeed for a reservoir, its
characteristics are unlocked over the whole lifetime of the project. The duration of
the appraisal stage clearly is a techno economic decision related to the confidence
to go ahead based on a good foundation of quality data and forecasts. Fine tuning
can always be carried out but this is costly if this delays the development stage. It
is important to identify and fill the gaps for the largest uncertainties, and having
sufficient information to design a system which is safe and cost effective. The
difficulty is making the decision on the data under which a line is drawn which
defines the basis for field development design. In reservoir development the reservoir
is always revealing its properties, indeed it is in the production phase that the true
characteristics are revealed.
7 DEVELOPMENT PLANNING
The starting point will invariably be a reservoir map used to calculate reserves, but
in addition use will be made of the material balance equation (chapter 15), together
with some drive concepts (chapter 11), to predict reservoir behaviour. One of the
problems faced in making predictions is to adequately take into account knowledge
about geological trends and, although individual well models can be adjusted to reflect
local conditions, there is no practical ‘desk calculator’ technique for using say, the
material balance equation and well models to come up with a predictive reservoir
performance. Displacement models such as those derived by Buckley and Leverett
(chapter 18), mainly from observations in the laboratory, give some insight into
reservoir behaviour but again do not significantly assist in allowing the engineer to
study the effect of alternative development plans on a heterogeneous reservoir.
With insight and ingenuity, the reservoir can be divided into a number of simple
units that can be analysed by the traditionally available techniques but such an
approach remains unsatisfactory. Over recent years the integration of geological and
geophysical perspectives is contributing considerably to the ‘confidence’ in reservoir
modelling.
7.2 Technoeconomics
For hydrocarbon accumulations found on dry land the traditional reservoir engineering
techniques available for field development planning were, in fact, quite adequate. This
is mainly so because land development operations offer a high degree of planning
34
Introduction To Reservoir Engineering
flexibility to oil companies and hence allow them to make optimal use of the latest
information. In an offshore environment this is not the case; once platforms have
been ordered most development options are closed. It is with respect to offshore field
development planning that reservoir simulation models have found their greatest
application potential.
It is clear from what we have overviewed in this chapter and the topics which will be
covered in the subsequent chapters that there are many parameters which contribute
to the viability of the various aspects of successful oil and gas production. It is also
clear that the various forms of data required, the confidence in the absolute values
vary according to the type, and therefore the final impact on the final result will vary
according to the particular parameter.
The following list summarises some of the principal uncertainties associated with
the performance of the overall reservoir model. The type of data can for example
be subdivided into two aspects “static” and “dynamic” data .
Static Properties
• Reservoir structure
• Reservoir properties
• Reservoir sand connectivity
• Impact of faults
• “thief” sands
Dynamic Properties
• Relative permeability etc
• Fluid properties
• Aquifer behaviour
• Well productivity (fractures, welltype, condensate drop out etc.)
The impact of each of these parameters will vary according to the particular field but
it is important that the company is not ignorant of the magnitude of the contributing
uncertainties, so that resources can be directed at cost effectively reducing specific
uncertainties. Figure 33 illustrates an outcome which might arise from an analysis
of various uncertainties for a particular field. It demonstrates for this particular field
and at the time of analysis the impact of the various data has on the final project cost.
Clearly in this case the aquifer behaviour uncertainties has the least impact whereas
reservoir structure and well productivity uncertainties had the most significant. An-
other field would result in different impact perspectives, and therefore a different
strategy to reduce overall project uncertainty would be required.
Sand
conectives
Aquifer
behaviour
Fluid properties
Relative
permeabilities etc.
Thief zones
Faults
- Changes +
During the development phase Dake2 has identified a number of roles for the Reservoir
Engineering which are targeted at optimising production. It is an irony that some
of the best data is generated during the production phase. Through production the
reservoir unveils more of its secrets. Some of these may cause modifications to the
development, perhaps in defining new well locations. The nature of the hydrodynamic
continuity of the reservoir is mainly revealed through pressure surveys run after a period
of production. This may define zones not being drained and therefore modifications
to the well completions might result.
As production progresses fluid contacts rise and therefore these contacts need to
be monitored and the results used to decide, for example, to recomplete a well as a
result of, for example excessive water production. As is pointed out in the chapter on
reservoir pressure, development wells before they are completed provide a valuable
resource to the reservoir engineer to enable surveys of pressure to be run to provide
a dynamic pressure-depth profile.
36
Introduction To Reservoir Engineering
More simple approaches not requiring the resources of a complex simulator can also
be used to up date early predictions, for example material balance studies.
Once production has been obtained, the additional data becomes available and makes
an important contribution to the refining of the initial reserves estimates. Two tech-
niques historically used are decline curve analysis and material balance studies.
In material balance studies, the pressure-volume behaviour of the entire field is
studied assuming an infinite permeability for the reservoir. By assuming an initial
oil-in-place from volumetric calculations, the pressure is allowed to decline following
fluid withdrawal. This decline is matched against the observed pressure behaviour
and, if necessary, the original oil-in-place figure is modified until a match is obtained.
In the presence of a water drive, additional variables are included by allowing water
influx into the ‘tank’. Water influx is governed by mathematical relationships such
as van Everdingen and Hurst (These concepts are covered in Chapters 11, 12, and
13 MB/MB Applications and Water Influx).
Decline curves are plots of rate of withdrawal versus time or cumulative withdrawal
on a variety of co-ordinate scales. Usually a straight line is sought through these ob-
servations and extrapolated to give ultimate recovery and rates of recovery. Decline
curves only use rates of withdrawal and pay relatively little attention to the reservoir
and flowing pressures. A change in the mode of operation of the field could change the
slope of the decline curve; hence, this is one of the weaknesses of this technique.
A noteworthy feature of these two approaches is that the engineer in fact ‘fits’ a sim-
ple model to observe data and uses this model to predict the future by extrapolation.
As more data becomes available the model gets ‘updated’ and predicted results are
adjusted. Decline curve analysis has not been used to the same extent as in the 60’s
and 70’s. With the power of computing and the efforts made to integrate geological
understanding , the physics of the flow and behaviour of rock and fluid systems into
reservoir simulation, the ‘fitting” and the uncertainty of earlier methods are being
superseded by integrated reservoir simulation modelling.
The routine company function will generate the need for on going production pro-
file updates. The generation of these is generally the responsibility of the reservoir
engineer, who might chose simple analytical approaches to the more costly reservoir
simulation methods.
Plateau phase
Production rate
Artificial lift
Economic limit
Time - years
The challenge facing the industry is the issue of the proportion of hydrocarbons left
behind. The ability to extract a greater proportion of the in-place fluids is obviously
a target to be aimed at and over recent years recoveries have increased through the
application of innovative technology. Historically there have been three phases of
recovery considered. Primary recovery, which is that recovery obtained through the
natural energy of the reservoir.
Much effort was put into enhanced oil recovery (EOR) research up until the mid
seventies. Sometimes it is termed tertiary recovery. When the oil price has dropped
the economics of many of the proposed methods are not viable. Many are based on
38
Introduction To Reservoir Engineering
the injection of chemicals which are often oil based. The subject of EOR has not been
forgotten and innovative methods are being investigated within the more volatile
oil price arena. Figure 35 gives a schematic representation of the various phases of
development and includes the various improved recovery methods. More recently
a new term has been introduced called Improved Oil Recovery (IOR). IOR is more
loosely defined and covers all approaches which might be used to improve the recov-
ery of hydrocarbons in place. Clearly it is not as specific as EOR but provides more
of an achievable target than perhaps some of the more sophisticated EOR methods.
As we have entered into the next millennium it is interesting to note that a number
of major improved recovery initiatives are being considered particularly with respect
to gas injection. One perspective which make a project more viable is that of the
disposal of gas for example which is an environmental challenge in one field can
be the source of gas for another field requiring gas for a gas injection improved oil
recovery process.
Primary
Recovery
C
Natural Artifical Lift O
Pump gas lift etc. N
Flow V
E
N
Secondary T
Recovery I
O
N
A
Natural Pressure L
Flow Maintenance
Water, gas injection
Tertiary
Recovery
E
O
R Thermal Gas Chemical Microbial
As we have discussed the role of the reservoir engineer in combination with other
disciplines is to predict the behaviour of the reservoir. Whereas in the early years of
oil exploration little attention was paid to understanding the detailed characteristics
of the reservoir, it is now recognized that detailed reservoir properties associated with
often complex physical and chemical laws determine field behaviour. The unlocking
of these characteristics and understanding the laws enable engineering plans to be
put in place to ensure optimised developments are implemented. This is schemati-
cally illustrated in figure 36.
Development
Plan
Reservoir Description
Unique
Dynamic and Static
At one extreme for example in a blow - out situation, a reservoir produces in an un-
controlled manner only restricted by the size of the well through which is producing.
Optmised development however based on a thorough understanding of the reservoir
enables the reservoir to be produced in a controlled, optimised manner.
In many other industries the effort expended on one project can be utilised in engi-
neering a duplicate or a similar size unit elsewhere. Such opportunities are not pos-
sible in the engineering of a reservoir. Reservoirs are unique in many aspects. The
composition of the fluids are unique, the rock characteristics and related properties
are unique, the size and shape are unique and so on. From our perspective this reser-
voir description is dynamic as the reservoir over a period of time gives up its secrets.
From the reservoir’s perspective however the description is static, except with the
changes resulting from the impact of fluid production or injection. The challenge
to those involved is reducing the time it takes for our dynamic description to match,
our static description known only to the reservoir or whoever was responsible for
its formation! The answer perhaps is more of a philosophical nature. The reality is
shown in figure 37 where the top structure map for a North Sea gas field with a ten
year gap shows the impact of knowledge gained from a number of wells as against
that interpreted from the one well. Considerable faulting is shown not as a result of
major geological a activity over the ten years but knowledge gained from the data
associated with the new wells.
40
Introduction To Reservoir Engineering
21
00
0
0
200
100
53°10 53°10
SHELL/ESSO 49/26 AMOCO 49/27
20
2200
00
21
21
49/26.1 00
53°05 53°05
12
00
100
80 0
0
20
10
00
00
20 100
10
20
100
20
00 0
00
0
00
00
0
210
2000
2°00 2°20
Present interpretation of Leman Gas-field, showing contours on top of Rotliegendes in feet below sea-level
Figure 37 (a) The Leman Field as it Appeared to be When The Exploration Well Was
The Leman field as it appeared to be when the exploration well was drilled
Drilled.
Depth in feet
0 1 Miles
0 1 2 KMS
70
A permanent platform
63
00 63
00
53°05 53°05
620
0 6400
6300
69
00
610
6900 0
6300 6
90 0
690
0 64
69
0
00
0
6300
70 69 6400
00 00
53°00 53°00
Present interpretation of Leman Gas-field, showing contours on top of Rotliegendes in feet below sea level.
The coverage of the reservoir has also changed effecting the equity associated with
the blocks. This illustrates the early benefits to be gained from drilling a number of
exploration wells. These equity agreements, are called unitisation agreements and such
agreements are shortened when good quality and comprehensive reservoir descrip-
tion data is available. Clearly there can never be sufficient description, however the
10. CONCLUSION
It is also important that the Petroleum Reservoir Engineer has a thorough basic
understanding in general, historical and petroleum geology. The influence of geological
history on the structural conditions existing in a reservoir should be known and
considered in making a reservoir engineering study. Such a study may also help to
identify and characterise the reservoir as to its aerial extent, thickness and stratification
and the chemical composition, size distribution and texture of the rock materials.
In his latest text, Dake2 comments on some of the philosophy of approach to reser-
voir engineering, and identifies the importance of pinning down interpretation and
prediction of reservoir behaviour to well grounded laws of physics.
Reservoir forecasting has moved on considerably since wells were drilled with little
interest and concern into the production and forecasting of what was happening in
the reservoirs thousands of feet below. The approach to coping with uncertainty as
jokingly reflected in the cartoon below, (Figure 38) is no longer the case as sophisti-
cated computational tools enable predictions to be made with confidence and where
uncertainty exists the degree of uncertainty can be defined.
42
Introduction To Reservoir Engineering
"We feed the geological data for the area, the computer produces a schematic topological
overview designating high probability key points, then we stick the printout on the wall and
Lever throws darts at it."
REFERENCES
CONTENTS
1 INTRODUCTION
2 ABNORMAL PRESSURES
6. RESERVOIR TEMPERATURE
LEARNING OBJECTIVES
Having worked through this chapter the Student will be able to:
• Having worked through this chapter the student will be able to:
• Describe briefly and sketch the pressure gradients associated with overpressured
and underpressured reservoirs.
• Describe briefly , sketch and present equations for the pressures in a water
supported oil and gas bearing formation.
Reservoir Pressures and Temperatures
1. INTRODUCTION
Oil and gas accumulations are found at a range of sub-surface depths. At these depths
pressure exists as a result of the depositional process and from the fluids contained
within the prous media. These pressures are called lithostatic pressures and fluid
pressures. These pressures are illustrated in figure 1.
The lithostatic pressure is caused by the pressure of rock which is transmitted through
the sub-surface by grain-to grain contacts. This lithostatic or sometimes called geostatic
or overburden pressure is of the order of 1 psi/ft. The lithostatic pressure gradient
varies according to depth, the density of the overburden, and the extent to which the
rocks are supported by water pressure. If we use this geostatic pressure gradient of
1 psi/ft. then the geostatic pressure Pov, in psig at a depth of D feet is
The geostatic pressure is balanced in part by the pressure of the fluid within the pore
space, the pore pressure, and also by the grains of rock under compaction. In un-
consolidated sands, loose sands, the overburden pressure is totally supported by the
fluid and the fluid pressure Pf is equal to the overburden pressure Pov . In deposited
formations like reservoir rocks the fluid pressure is not supporting the rocks above
but arises from the continuity of the aqueous phase from the surface to the depth D in
the reservoir. This fluid pressure is called the hydrostatic pressure. The hydrostatic
pressure is imposed by a column of fluid at rest. Its value depends on the density of
the water ρw, which is affected by salinity. In a sedimentary basin, where sediment
has settled in a region of water and hydrocarbons have been generated and trapped,
we can expect a hydrostatic pressure. For a column of fresh water the hydrostatic
pressure is 0.433 psi/ft. For water with 55,000 ppm of dissolved salts the gradient is
0.45 psi/ft; for 88,000 ppm of dissolved salts the gradient is about 0.465 psi/ft.
Its variation with depth is given by the equation.
Pf = ρwDg (2)
There is another fluid pressure which arises as a result of fluid movement and that
is called the hydrodynamic pressure. This is the fluid potential pressure gradient
which is caused by fluid flow. This however does not contribute to in-situ pressures
at rest.
FP GP
Depth (Ft.)
Overpressure Overburden
Pressure (OP)
Underpressure Normal
Figure 1 Gives the relationship between the lithostatic pressure and the hydrostatic
1
pressure.
dP
Pw = x D + 14.7psia
dD water
(3)
This equation assumes continuity of water pressure from the surface and constant
salinity. In most cases even though the water bearing sands are divided between
impermeable shales, any break of such sealing systems will lead to hydrostatic pres-
sure continuity, but the salinity can vary with depth.
Reservoirs whose water pressure gradient when extrapolated to zero depth give an
absolute pressure equivalent to atmospheric pressure are called normal pressured
reservoirs.
EXERCISE 1
If the average pressure gradient in a region is 0.47 psi/ft, calculate the pore
pressure in a normally pressurised formation at 7400ft. Convert the pressure from
psi to KPa, then express the pressure in MPa. What is the pressure gradient in
KPa/m?
Reservoir Pressures and Temperatures
2. ABNORMAL PRESSURE
Under certain conditions, fluid pressures may depart substantially from the normal
pressure. Overpressured reservoirs are those where the hydrostatic pressure is greater
than the normal pressure and underpressured reservoirs are below normal pressure.
Figure 1. They are called abnormal pressured reservoirs and can be defined by the
equation:
dP
Pw = x D + 14.7 psia + C
dD water
(4)
where C is a constant, being positive for overpressured and negative for an under-
pressured system.
For abnormally pressured reservoirs, the sand is sealed off from the surrounding strata
so that there is not hydrostatic pressure continuity to the surface.
Conditions which cause abnormal fluid pressure in water bearing sands have been
identified by Bradley 2 and include (Figure 2):
FP-Too High
Upthrust
(a)
Original Deposition
North Sea
Glacier
(c)
Normal Surface Greenland 3 km thick
1300 psi/1000 m ice
• Osmosis between waters having different salinity, the sealing shale acting as a
semi-permeable membrane. If the water within the seal is more saline than the
surrounding water, the osmosis will cause a high pressure and vice versa.
Overpressured reservoirs are common in Tertiary deltaic deposits such as the North
Sea, Niger delta and the Gulf Coast of Texas. In the North Sea one mechanism for
overpressure is the inability to expel water from a system of rapidly compacted
shales.
With abnormally pressured reservoirs a permeability barrier must exist, which inhibit
pressure release. These may be lithological or structural. Common lithological
barriers are evaporates and shales. Less common are the impermeable carbonates
and sandstones. Structure permeability barriers may result from faults which, in
some cases, seal. The subject on of abnormal pressures is covered more fully in
the Geology Module
If reservoirs are all normal pressured systems then the pressure gradient for these
reservoirs would be virtually all the same, other than from the influence of salinity.
The figure below shows the water pressure gradients for a number of reservoirs in
the North Sea and indicates the significant overpressuring in this region. Often these
overpressuring show regional trends. For example the fields depicted in figure 3
show an increase in abnormal pressure in the south east direction. Clearly if all these
reservoirs were normally pressured then the pressure depths values would lie on the
same gradient line with a zero depth pressure value of atmospheric pressure.
Reservoir Pressures and Temperatures
8,000
Statfjord OWC
Brent OWC
9,000
Thistle OWC
Cormorant
OWC
4
10,000
1
2
Subsea Depth (Feet)
Ninian
OWC
11,000 Heather
OWC
3
Lyell
5
12,000
Alwyn
N.W. Alwyn
S.W> Ninian
Pressure, psig
3
Figure 3 Examples of overpressured reservoirs in the North Sea
Pressure gradients in hydrocarbon systems are different from those of water systems
and are determined by the oil and gas phase in-situ specific gravities, ρo and ρg of
each fluid.
The pressure gradients are a function of gas and oil composition but typically are:
Depth (Ft.)
13
8500
12
0.17 psi/ft
ρf = 0.39 gm/cc
11
10
Gas-Oil Contact 9
8
8600 7
6
Depth (Ft.)
0.29 psi/ft
ρf = 0.67 gm/cc
5 Oil-Water Contact
8700
4
0.47 psi/ft
ρf = 1.09 gm/cc
3
2
8800 1
Figure 4 Pressure distribution for an oil reservoir with a gas cap and an oil-water contact.
The nature of the pressure regime and the position and recognition of fluid contacts
are very important to the reservoir engineer in evaluating reserves, and determining
depletion policy.
EXERCISE 2
If the pressure in a reservoir at the OWC is 3625 psi, calculate the pressure at the top
if there is a 600ft continuous oil column. If a normal pressure gradient exists outwith
the reservoir, calculate the pressure differential at the top of the reservoir. Redo the
calculations for a similar field, but this time containing gas.
Reservoir Pressures and Temperatures
Water is always present in reservoir rocks and the pressure in the water phase Pw
and the pressure in the hyrocarbon phase Po are different . If P is the pressure at the
oil/water contact where the water saturation is 100%, then the pressure above this
contact for the hydrocarbon and water are :
Po = P - ρogh (8)
Pw = P - ρwgh (9)
The difference between these two pressures is the capillary pressure Pc: see Chapter
8. In a homogenous water-wet reservoir with an oil-water contact the variation of
saturation and phase pressure from the water zone through the capillary transition
zone into the oil is shown in Figure 5). In the transition zone the phase pressure
difference is given by the capillary pressure which is a function of the wetting phase
saturation. (Chapter 8).
Oil Zone
Vertical Oil Phase Pressure
Depth po = pFWL - ρogh
D
pc (Sw)
h=
∆ρg WOC Water Gradient
FWL
(pc = o)
0 Swc 1 pFWL
Water Saturation, Sw Pressure, P
Pc = Po - Pw (10)
at hydrostatic equilibrium
Pc(Sw) = ∆ρgh
∆ρ = ρw-ρo
h = height above free water level
The difference in depth between the oil-water contact and the free water level depends
on the capillary pressure which in turn is a function of permeability, grain size etc.
Providing the phase is continuous the pressures in the respective phases are:
On the depth-pressure diagram the intersection of the continuous phase pressure line
occurs at the free water level.
Earlier tests for vertical pressure logging have been replaced by open-hole testing
devices that measure the vertical pressure distribution in the well, and recover for-
mation samples.
One such device which was introduced in the mid seventies which has established
itself in reservoir evaluation is the repeat formation tester RFT (Schlumberger trade
name). It was initially developed as a device to take samples. Over the years however
its main application is to provide pressure -depth profiles over reservoir intervals. The
device places a probe through the well mud cake and allows small volumes of fluid
to be taken and pressure measurements to be made (Figure 6). It can only be operated
therefore in an open hole environment. The unit can be set at different locations in
the well and the pressure gradient thereby obtained. This device has been superseded
by different tools provided by a number of wireline service providers. The principle
is the same of measuring with a probe in open hole the pressure depth profile.
10
Reservoir Pressures and Temperatures
Packer
Filter
Flow Line
Piston Formation
Pressure Guage
Equalising Valve
(To Mud Column) Flow Line
Chamber 1
Probe Closed
Chamber 2
These open hole pressure measurements have proved valuable at both the appraisal
stage and can be used to establish fluid contacts. It has also proved particularly valu-
able during the development stage in accessing some of the dynamic characteristics
of the reservoir. The pressure changes in different reservoir layers resulting from
production reveal the amount of interlayer communication and these pressure meas-
urements can be a powerful tool in understanding the characteristics of the reservoir
formation.
In 1980 Amoco3 published a paper with respect to the Montrose Field in The North
Sea which illustrates the application of pressure-depth surveys. Figure 7 shows the
pressure depth survey in 1978 of a well after production since mid 1976. Only the
top 45ft of the 75ft oil column had been perforated. The initial pressure gradient in-
dicates the oil and water gradients at the condition of hydrostatic equilibrium. The
second survey shows a survey after a period of high production rate, and reveals the
reservoir behaviour under dynamic conditions. The various changes in slope in the
pressure profile reveal the partial restricted flow in certain layers. Similar surveys
in each new development wells (Figure 8) show the similar profiles and enable the
detailed layered structure of the reservoir to be characterised which is important for
reservoir simulation purposes.
Layer 3
8300
8500
2600
8600
2650 Layer 5
8700
8800
14 16 18 20 22 24 26
Reservoir pressure - MPa
2500 8200
True vertical subsea depth - feet
8300
2550
8400
8500
2600
8600
2650 8700
symbol ?Well number Date
22/17-A6 05/04/77
A8 27/01/78 8800
A11 20/12/77
2700 A15 15/08/78
A17 02/11/78 8900
A18 28/03/79
9000
18 20 22 24 26 28
Reservoir pressure - MPa
12
Reservoir Pressures and Temperatures
6. RESERVOIR TEMPERATURE
The temperature of the earth increases from the surface to centre. The heat flow out-
wards through the Earth’s crust generates a geothermal gradient, gc. This temperature
variation conforms to both a local and regional geothermal gradient, resulting from
the thermal characteristics of the lithology and more massive phenomenon associated
with the thickness of the earth’s crust along ridges, rifts and plate boundaries.
In most petroleum basins the geothermal gradient is of the order of 1.6˚F/100 ft.
(0.029 K/m) The thermal characteristics of the reservoir rock and overburden give
rise to large thermal capacity and with a large surface area in the porous reservoir
one can assume that flow processes in a reservoir occur at constant reservoir tem-
perature. The local geothermal gradient will be influenced by associated geological
features like volcanic intrusions etc. The local geothermal gradient can be deduced
from wellbore temperature surveys . However they have to be made under stabilised
conditions since they can be influenced by transient cooling effects of circulating
and injected fluids.
During drilling the local thermal gradient can be disturbed and by analysis of the
variation of temperature with time using a bottom hole temperature (BHT) gauge
the local undisturbed temperature can be obtained.
Solutions to Exercises
EXERCISE 1
If the average pressure gradient in a region is 0.47 psi/ft, calculate the pore pressure
in a normally pressurised formation at 7400ft. Convert the pressure from psi to KPa,
then express the pressure in MPa. What is the pressure gradient in KPa/m?
SOLUTION
Pressure in formation = 0.47 * 7400 = 3478 psi
SOLUTION
Typical pressure gradients are (psi/ft):
Water – 0.45
Oil – 0.35
Gas – 0.08
REFERENCES
14
Reservoir Fluids Composition
CONTENTS
1 INTRODUCTION
2 HYDROCARBONS
2.1 Chemistry of Hydrocarbons
2.2 Alkanes or Paraffinic Hydrocarbons
2.3 Isomerism
2.4 Unsaturated Hydrocarbons
2.5 Napthene Series
2.6 Aromatics
2.7 Asphalts
3 NON-HYDROCARBON COMPOUNDS
5 GENERAL ANALYSIS
5.1 Surface Condition Characterisation
5.2 Refractive Index
5.3 Fluorescence of Oil
LEARNING OBJECTIVES
Having worked through this chapter the Student will be able to:
• List the non- hydrocarbon compounds which might be present in small qualities
in reservoir fluids.
• Define the black oil model description of the composition of a reservoir fluid.
• Be aware of general analysis descriptors for petroleum fluids e.g. oAPI, refractive
index and flourescence.
• Calculate given the prerequisite data proved, probable and possible reserves.
Reservoir Fluids Composition
1 INTRODUCTION
Petroleum deposits occurring as a gaseous state are termed natural gas, in the liquid
state as petroleum oil or crude oil and in the solid state as tars, asphalts and waxes.
For a mixture with small molecules it will be a gas at normal temperature and pressure
(NTP). Mixtures containing larger molecules will be a liquid at NTP and larger
molecules as a solid state, for example, tars and asphalts.
The exact origin of these deposits is not clear but is considered to be from plant,
animal and marine life through thermal and bacterial breakdown.
The appearance varies from gases, through very clear liquids, yellow liquids to a
dark, often black, highly viscous material, the variety obviously being a function of
composition. Although the principal elements are carbon (84-87%), and hydrogen
(11-14%), crude oil can vary from a very light brown liquid with a viscosity similar
to water to a very viscous tar like material .
Water is always present in the pore space of a reservoir, since the original depositional
environment for the rocks was water. This water has subsequently been displaced
by the influx of hydrocarbons but not totally since surface tension forces acting in
the rock pore space cause some of the water to be retained.
The two compositional characterisation approaches used are the compositional model
and the black oil model. The basis of the compositional model is a multicomponent
description in terms of hydrocarbons and the black oil model is a two component
description in terms of produced oil, stock tank oil and produced gas, solution gas.
The compositional model is the topic covered in this chapter and the black oil model
is covered in the liquid properties chapter.
Institute of Petroleum Engineering, Heriot-Watt University
2 HYDROCARBONS
Hydrocarbons
Aliphatic Aromatics
(Paraffins) (Napthenes)
The hydrocarbons divide into two groupings with respect to the arrangement of the
carbon molecules and the bonds between the carbon molecules. The arrangement of
the molecules are open chain or cyclic and the bonds between the carbon are saturated
(single) bonds or unsaturated or (multiple) bonds.
The paraffin series begins with methane (CH4), and its basic formula is CnH2n+2.
Pentane to pentadecane are liquids and the chief constituents of uncracked gasoline.
Its higher members are waxy solids. In a given bore hole the wax may clog the pore
space next to the hole as gas expands and cools.
The paraffins are the largest constituent of crude oil and are characterised by their
chemical inertness. Clearly they would not have remained as they are if this were not
so.
2.3 Isomerism
From methane to propane there is only one way to arrange the branched chains however
above propane there are alternative arrangements and these are called isomers.
Reservoir Fluids Composition
Structural formulae do not represent the actual structure of the molecules. Isomers
are substances of the same composition that have different molecular structure and
therefore different properties, for example, normal butane and isobutane.
Pentane has three structures (isomers). Clearly the number of isomers increase as the
number of carbon atoms increases. Hexane has 5 isomers and heptane 9.
Table 1 below gives some of the basic physical properties of the more common
hydrocarbons of the paraffin series and Table 2 lists the state of the various pure
components demonstrating that components which might be solid on their own
contribute to liquid states when part of a mixture. Figure 2 gives some structural
formula for three paraffin compounds.
Density
Name Chemical Molecular Boiling Point Critical Gas Liquid
Formula Weight (°C) at normal Temp °C (air = 1) (water = 1)
conditions
sp.gr.
PARAFFINS H
H H C H H H H H H H H H
H H
H C H H C C C C C C C C H
H C C C H
H H H H H H H H H
H H H
Methane Iso-butane n-octane
Reservoir Fluids Composition
NAPHTHENES
H H H
H C H C
H H
H C C C
H H H H
C C H H
H H C C
H H H H
C C C
H H H H
Methyl
Cyclopentane Cyclohexane
2.6 Aromatics
The aromatic series (CnH2n-6) is an unsaturated closed-ring series, based on the benzene
compound and the compounds are characterised by a strong aromatic odour. Various
aromatic compounds are found in crude oils. The closed ring structure gives them
a greater stability than open compounds where double or triple bonds occur. Figure
4 gives the structural formula for two aromatic compounds.
AROMATICS
H H H
C C C
H C C H H C C C H
H C C H H C C C H
C C C
H H H
Benzene Naphthalene
The aromatic-napthene based crudes are usually associated with limestone and dolomite
reservoirs such as those found in Iran, the Arabian Gulf and Borneo.
Some crude oils used to be described, more from a refining perspective, according
to the relative amount of these non paraffin compounds. Crude oils would be called
paraffinic, napthenic or aromatic. It is not a classification of value in reservoir
engineering.
2.7 Asphalts
Asphalt is not a series by itself. Asphalts are highly viscous to semi-solid, brown-
black hydrocarbons of high molecular weight usually containing a lot of sulphur
and nitrogen, which are undesirable components, and oxygen. Asphalts are closely
related to the napthene series and because of their high nitrogen and oxygen content
they may be considered juvenile oil, not fully developed.
3 NON-HYDROCARBON COMPOUNDS
Although small in volume, generally less than 1%, non-hydrocarbon compounds have
a significant influence on the nature of the produced fluids with respect to processing
and the quality of the products.
The more common non-hydrocarbon constituents which may occur are:
sulphur, oxygen, nitrogen compounds, carbon dioxide and water.
Reservoir Fluids Composition
Oxygen compounds, up to 0.5% wt., are present in some crudes and decompose to
form napthenic acids on distillation, which may be very corrosive.
Nitrogen content is generally less than 0.1% wt., but can be as much as 2%. Nitrogen
compounds are complex . Gaseous nitrogen reduces the thermal quality of natural
gas and needs to be blended with high quality natural gas if present at the higher
levels.
Other compounds. Metals may be found in crude oils at low concentration and are
of little significance. Metals such as copper, iron, nickel, vanadium and zinc may be
present. Produced natural gas may contain helium, hydrogen and mercury.
Inorganic compounds The non-oil produced fluids like water will clearly contain
compounds arising from the minerals present in the rock, their concentration will
therefore vary according to the reservoir. Their composition however can have a
very significant effect on the reservoir behaviour with respect to their compatibility
with injected fluids. The precipitation of salts, scale, is a serious issue in reservoir
management.
Many of these salts need to be removed on refining as some generate HC1 when
heated with water.
For the oil refiner or chemical manufacturer the composition of the fluid is the key to
determine what chemical products can be extracted or processed from the material.
The petroleum engineer is not concerned with the fact that the oil might contain, albeit
in small concentrations, hundreds of different components. The petroleum engineer
wants as simple a description as possible which still enables the determination of the
physical properties and behaviour under different temperature and pressure conditions.
Two models are used in this industry to describe the composition for physical property
prediction purposes, the black-oil model and the compositional model.
The black-oil model is a 2 component description of the fluid where the two components
are, the fluids produced at surface, stock tank oil and solution gas. Associated with
this model are black-oil parameters like solution gas-oil ratio and the oil formation
volume factor. These parameters are discussed in the chapter on liquid properties.
Isomers, normal and iso are usually identified up to pentane. Non paraffinic
compounds are assigned to the next higher paraffin according to its volatility. The
material representing all compounds above the limiting carbon number are called the
C+ fraction , so C7+ for a limiting value of C6 and C10+ for a limiting value of C9.
The physical properties of paraffins up to the limiting C number are well known
and documented. The C+ component is however unique to the fluid and therefore
two properties are used to characterise it, apparent molecular weight and specific
gravity.
The behaviour of some fluids are complex and the paraffin based description may
have difficulty in predicting properties under certain conditions. Consideration may
be required to also identify napthenic and aromatic compounds, (PNA analysis),
which could be contributing to complex behaviour. This is particularly the case for
gas condensates existing at high pressures and high temperatures.
Figure 4 illustrates the compositional model and its application as reservoir fluids are
produced to surface. Although the individual components contribute to a single liquid
reservoir phase for an oil, when the fluids are produced to surface they produce a gas
phase, solution gas, and a liquid phase, stock tank oil. The distribution characteristics
of the individual components is complex and not just a function of temperature and
pressure. For reservoir fluids the composition is also an influence on the distribution.
This makes it a difficult task to predict this distribution perspective since reservoir
fluid compositions are unique. This topic is further dealt with in the chapter on
vapour liquid equilibrium. Improved methods of chemical analysis make it possible
to describe the oil up to a C value of C29. Although such definitions provide a very
accurate description, the associated computer effort in using such a comprehensive
description does lead to the use of pseudo components. Pseudo components are
obtained by grouping the various C number compositions, thereby reducing the
description to 4 or 5 "pseudo components". A number of methods exist to group the
various C values and other components.
10
Reservoir Fluids Composition
C1 C2 C3 C4 C5 C6 C7+
5. GENERAL ANALYSIS
The basis chosen is the fluids at surface conditions, the surface conditions being 14.7
psia or 101.3 kPa and 60oF or 298K. These conditions are called standard conditions.
For gas therefore this yields standard cubic feet SCF or standard cubic meters SCM.
It is useful to consider these expression not as volumes but as mass, the volume of
which will vary according to density. For liquids we express surface conditions as
stock tank volumes either stock tank barrels STB or stock tank cubic meters STM3.
The relative amount of gas to oil is expressed by the gas-oil ratio GOR SCF/STB.
Since there are so many types of oil, each with a wide range of specific gravity, an
arbitrary non-linear relationship was developed by the American Petroleum Institute
(API) to classify crude oils by weight on a linear-scaled hydrometer. The observed
readings are always corrected for temperature to 60oF, by using a prepared table of
standard values.
The API gravity of water is 10º. A light crude oil would have an API gravity of 40º,
while a heavy crude would have an API gravity of less than 20º. In the field, the API
gravity is readily measured using a calibrated hydrometer.
There are no definitions for categorising reservoir fluids, but the following table 5
indicates typical GOR, API and gas and oil gravities for the five main types. The
compositions show that the dry gases contain mostly paraffins, with the fraction
of longer chain components increasing as the GOR and API gravity of the fluids
decrease.
In chapter 4 we give a classification for the various reservoir fluid types in the context
of phase behaviour.
Type Dry Gas WetGas Gas Condensate Volatile Oil Black Oil
Composition (mol %)
C1 96.3 88.7 72.7 66.7 52.6
C2 3.0 6.0 10.0 9.0 5.0
C3 0.4 3.0 6.0 6.0 3.5
C4 0.17 1.3 2.5 3.3 1.8
C5 0.04 0.6 1.8 2.0 0.8
C6 0.02 0.2 2.0 2.0 0.9
C7+ 0.0 0.2 5.0 11.0 27.9
12
Reservoir Fluids Composition
When possible, pictures should be taken of the core showing the fluorescence. These
are very useful when accompanying reports to the head office which may be hundreds
if not a few thousand miles away.
The degree of fluorescence is indicated below for different compositions as reflected
in the API gravity.
It should be pointed out that most oils increase in API gravity with depth in a given
lithologic column with the reason being that younger juvenile oils, heavier with a
lower API gravity, have not yet been transformed from the initial formation conditions
to higher petroleum members. Two well-known exceptions to this pattern are found
in the Burgan sands of Kuwait and the shallow sands of the Bibi Eibat field in the
USSR where the high-gravity members are found higher up in the stratified column
than the low-gravity members.
Calculate the Specific Gravity (SG) of a 38o API oil. What is its density in lbs/cu.ft?
(62.32 lbs/cu.ft equals an SG of 1.0 and 43.28 API)
Now convert an oil with an SG of 0.744 to Degrees API.
EXERCISE 2
A reservoir oil is quoted as having a Gas Oil Ratio (GOR) of 604 scf/bbl. Convert
this to Standard Cubic Meters (SCM)gas per Stock Tank Cubic Meters (SM3)
1 Foot = 0.3048m
1 barrel = 5.615 cu ft.
1 barrel = 0.159 M3
EXERCISE 3
EXERCISE 4
14
Reservoir Fluids Composition
Solutions to Exercises
EXERCISE 1
Calculate the Specific Gravity (SG) of a 38o API oil. What is its density in lbs/
cu.ft?
(62.32 lbs/cu.ft equals an SG of 1.0 and 43.28 API)
Now convert an oil with an SG of 0.744 to Degrees API.
SOLUTION
SG = 0.835
EXERCISE 2
A reservoir oil is quoted as having a Gas Oil Ratio (GOR) of 604 scf/bbl. Convert
this to Standard Cubic Meters (SCM)gas per Stock Tank Cubic Meters (SM3)
1 Foot = 0.3048m
1 barrel = 5.615 cu ft.
1 barrel = 0.159 M3
SOLUTION
EXERCISE 3
SOLUTION
A reservoir with a GOR of 11,000 scf/bbl would be typically termed a ‘Gas Condensate
Reservoir’. The API gravity would probably be in the low 50’s.
SOLUTION
Compositional Model.
The compositional model is based on the paraffin series CnH2n+2. To keep the number
of components in the model manageable, long chain members are grouped together
and given an average property. These compounds are termed collectively as the ‘C+
fraction’. Typically this covers the hydrocarbons above Heptane and therefore is
called the C7+ fraction, which is characterised using the terms Apparent Molecular
Weight and Specific Gravity.
REFERENCES.
16
Phase Behaviour of Hydrocarbon Systems
CONTENTS
1 DEFINITIONS
4 MULTI-COMPONENT HYDROCARBON
4.1 Pressure Volume Diagram
4.2 Pressure Temperature Diagram
4.3 Critical Point
4.4 Retrograde Condensation
5 MULTI-COMPONENT HYDROCARBON
5.1 Oil Systems (Black Oils and Volatile Oils)
5.2 Retrograde Condensate Gas
5.3 Wet Gas
5.4 Dry Gas
Having worked through this chapter the Student will be able to:
General
• Define; system, components, phases, equilibrium, intensive and extensive
properties.
Pure Components
• Sketch a pressure-temperature (PT) diagram for a pure component and illustrate
on it; the vapour-pressure line, critical point, triple point, sublimation-pressure
line, the melting point line, the liquid, gas and solid phase zones.
• Define the critical pressure and critical temperature for a pure component.
• Describe briefly with the aid of a PT diagram the behavior of a pure component
system below( left|) and above ( right) of the critical point.
• Sketch the pressure- volume (PV) diagram for a pure component illustrating the
behavior above the bubble point, between the bubble and dewpoint and below
the dewpoint.
• Sketch a series of PV lines for a pure component with a temperature below, at
and above the critical temperature.
• Sketch the three dimensional phase diagram for pure component systems.
Two Components
• Plot a PV diagram for a 2 component system and identify key parameters.
• Plot a PV diagram for a 2 component system and identify key parameters and
the relationship to the vapour pressure lines for the two pure components.
• Sketch the critical point loci for a series of binary mixtures including methane
and indicate how a mixture a mixture of methane and another component can
exist as 2 phases at pressures much greater than the 2 phase limit for the two
contributing components.
• Draw a PT diagram for a two component system, to illustrate the cricondentherm,
cricondenbar and the region of retrograde condensation.
• Define the terms cricondentherm and cricindenbar.
• Explain briefly what retrograde condensation is.
Multicomponent Systems
• Sketch a PT and PV diagrams to illustrate the behaviour at constant temperature
for a fluid in a PVT cell. Identify key features.
• Draw a PT diagram for a heavy oil, volatile oil, retrograde condensate gas,
wet gas and dry gas. Illustrate and explain the behaviour of depletion from the
undersaturated condition to the condition within the phase diagram.
• Describe briefly with the aid of a sketch, the reasons for and the process of gas
cycling, for retrograde gas condensate reservoirs.
• Plot a PT diagram for a reservoir with a gas cap to illustrate the gas at dew point
and oil at bubble point.
Miscellaneous
• With the aid of sketch explain the process of critical point drying.
Phase Behaviour of Hydrocarbon Systems
Oil and gas reservoir fluids are mixtures of a large number of components which when
subjected to different pressure and temperatures environments may exist in different
forms, which we call phases. Phase behaviour is a key aspect in understanding the
nature and behaviour of these fluids both in relation to their state in the reservoir and
the changes which they experience during various aspects of the production process.
In this chapter we will review the qualitative aspects of the behaviour of reservoir
fluids when subjected to changes in pressure and temperature.
1 DEFINITIONS
• Components - those pure substances which produce the system under all
conditions.
For example, in the context of reservoir engineering, methane, ethane, carbon dioxide
and water are examples of pure components.
• Phases - This term describes separate, physically homogenous parts which are
separated by definite boundaries.1 Examples in the context of water are the three
phases, ice, liquid water and water vapour.
Intramolecular forces are the attractive and repulsive forces between molecules. They
are affected by the distance between the molecules. The attractive forces increases
as the distance between the molecules decreases until however the electronic field of
the molecules overlap and then further decrease in distance causes a repulsive force,
which increases as the molecules are forced closer together.
The molecules in gases are widely spaced and attractive forces exist between the
molecules whereas for liquids where the molecules are closer together there is a
repelling force which causes the liquid to resist further compression.
The hydrocarbon fluids of interest in reservoir systems are composed of many compo-
nents however in understanding the phase behaviour of these systems it is convenient
to reflect on the behaviour of single and two component systems.
Phase diagrams are useful ways of presenting the behaviour of systems. They are
generally plots of pressure versus temperature and show the phases that exist under
these varying conditions.
Phase Behaviour of Hydrocarbon Systems
1 2
C
Critical Point
oint
Solid Liquid
Melting P
Pressure
e
s ur 3
es
r Pr
u
po
Va
Vapour Gas
a tion
blim
Su
Triple Point
Temperature
• Define the black oil model description of the composition of a reservoir fluid.
Critical Point
The critical point C. is the limit of the vapour pressure line and defines the critical
temperature, Tc and critical pressure, Pc of the pure substance. For a pure substance
the critical temperature and critical pressure represents the limiting state for liquid and
gas to coexist. A more general definition of the critical point which is both applicable
to multi component as well as single component systems is; the critical point is the
point at which all the intensive properties of the gas and liquid are equal.
Triple Point
The triple point represents the pressure and temperature at which solid, liquid and
vapour co-exist under equilibrium conditions. Petroleum engineers seldom deal
with hydrocarbons in the solid state, however, more recently solid state issues are a
concern with respect to wax, asphaltenes and hydrates.
Sublimitation-Pressure Line
The extension of the vapour-pressure line below the triple point represents the con-
ditions which divides the area where solid exists from the area where vapour exists
and is also called the sublimation - pressure line.
P1 Pb P Pd P2
Liquid
Gas
Phase Behaviour of Hydrocarbon Systems
E F
Pc c
1
A G
B
oint Line
Solid Liquid
Pressure
P
Melting -
ne 4
e li
ssur
pre
u r-
po Gas
Va 2
Tc
Temperature
As the pressure is reduced, the pressure falls rapidly until a pressure is reached lying
on the vapour pressure line. A gas phase will begin to form and molecules leave the
liquid. At further attempts to reduce the pressure the volume of gas phase increases,
while liquid phase volume decreases but the pressure remains constant. Once the
liquid phase disappears further attempts to reduce pressure will be successful as the
gas expands.
Above the critical temperature, following the path 3 - 4, a decrease in pressure will
cause a steady change in the physical properties, for example a decrease in density but
there will not be an abrupt density change as the vapour pressure line is not crossed.
No phase change takes place.
Consider the behaviour of the system around the critical point. If we go from point
A to point B, by increasing the temperature, we go though a distinctive phase change
on the vapour pressure line where two phases, liquid and gas co-exist. If we now go
a different route to B, starting with the liquid state at ‘A’ increase the pressure iso-
thermally (constant temperature ) to a value greater than Pc at E. Then keeping the
pressure constant increase the temperature to a value greater than Tc at point F. Now
decrease the pressure to its original value at G. Finally, decrease the temperature
keeping the pressure constant until B is reached. The system is now in the vapour
state and this state has been achieved without an abrupt phase change. The vapour
states are only meaningful in the two phase regions. In areas far removed from the
two phase region particularly where pressure and temperature are above the critical
values, definition of the liquid or gaseous state is impossible and the system is best
described as in the fluid state.
c
700
Pressure - PSIA
Liquid
600
Vapor
500
400
40 60 80 100 120
Temperature - º F
Phase Behaviour of Hydrocarbon Systems
All Liquid
T > Tc SINGLE PHASE
1
Liquid state-rapid change of
pressure with small volume change
First Gas Bubble T < Tc Pressure remains constant while
both gas and liquid are present
Pressure
Dew Point
Last Drop of Liquid T2 > Tc
Gas
Bubble Point
2
TWO PHASE REGION
All Gas
Volume
For a pure substance vapour pressures at bubble point and dew point are equal to the
vapour pressure of the substance at that temperature. Above the critical point, ie 3
- 4 , the PV behaviour line shows no abrupt change and simply shows an expansion
of the substance and no phase change. This fluid is called a super critical fluid.
Critical Point
4
Pressure
De
Curve
w
Po
in
t
Pressure remains constant while Cur
Point
ve 2
both gas and liquid are present
Bubble
Volume
The pressure volume curve for pure component ethane is given in figure 7
The locus of the bubble points and dew points form a three-dimensional diagram
when projected in to a P-T diagram give the vapour pressure line (Figure 8).
900
800
C
700
Pressure - PSIA
11
0
ºF
600 90
ºF
400
0 0.05 0.10 0.15 0.20 0.25
10
Phase Behaviour of Hydrocarbon Systems
uid
Liq
Critical Point
Dew Point Line
id
qu
Pressure
Li
d
s
an
G
as
Ga
Vo
lu u id Critical Point
me
tur
e Liq
ra
pe
Vapor Pressure Curve
Te m
Pressure
s
Ga ure
rat
m pe
Te
Liquid
and
Gas
Bubble Point
Ga
s
Dew Point
Volume
The pressure-volume diagram for a specific n-pentane and n-heptane mixture is given
in Figure 10. Clearly a different composition of the two components would result
in a different shape of the diagram.
600
500
Critical point
45 45
0º 4º
F
400
Pressure - PSIA
425
º
400
300 º
300
º
100
0 0.1 0.2 0.3 0.4 0.5
If the vapour pressure lines for the pure components are drawn on the P-T diagram
then the two-phase region for the mixture lies between the vapour pressure lines.
In the figure 11 the critical temperature of the mixture TcAB lies between TcA and TcB
whereas the critical pressure PcAB lies above PcA and PcB. It is important to note that
the PcAB and TcAB of the mixture does not necessarily lie between the Pc & Tc of the
two pure components.
1
PCAB Critical Point
% Liq.
Liquid
PCA 100
CA
75
50
CB
PCB e
t Li n 25
P o in
- 0
b ble
Bu
2
Pressure
t Gas
Poin
Dew
A specific mixture composition will give a specific phase envelope lying between the
vapour pressure lines. A mixture with different proportions of the same components
will give a different phase diagram. The locus of the critical point of different mix-
ture compositions is shown in Figure 12 for the ethane and n-heptane system, and in
Figure 13 for a series of binary hydrocarbon mixtures. Figure 13 demonstrates that
for binary mixture e.g. Methane and n-decane two phases can coexist at conditions
of pressure considerably greater than the two phase limit, critical conditions for the
separate pure components. Methane is a significant component of reservoir fluids.
C2
Composition
No Wt % Ethane
1200 C 100.00
C1 90.22
C2 50.25
C3 9.78
C7 N-Heptane
1000
C1
Pressure, lbs./Sq. In. ABS
800
C
C3
A1
600
e
in
an e
L
i nt
Po
E th
A le
400 bb C7
Bu
e
li n
200 A2
i
an
nt
Po
e
pt
De
w He
N-
B2 B3
A3
B1 B
0
0 100 200 300 400 500 600
Temperature º F
14
Phase Behaviour of Hydrocarbon Systems
6000
Single Phase
5000
4000
2000
1000
ne
e e
ha
an ne ne an xane ptane
Eth pa ut a ent
et
Temperature º F
These limits are of particular significance in relation to the shape of the diagram in
figure 14.
Consider a single isotherm on Figure 14. For a pure substance a decrease in pressure
causes a change of phase from liquid to gas. For a two-component system below Tc
a decrease in pressure causes a change from liquid to gas.
We now consider the constant temperature decrease in pressure, 1-2-3 , in figure 14 at
a temperature between the critical temperature and the cricondentherm. As pressure
is decreased from 1 the dew point is reached and liquid forms, i.e., at 2 the system is
such that 5% liquid and 95% vapour exists, i.e. a decrease in pressure has caused a
change from gas to liquid, opposite to the behaviour one would expect. The phenom-
ena is termed Retrograde Condensation. From 2 - 3, the amount of liquid decreases
Cricondenbar 1
Liquid % Liq.
100
Pressure
75
ne
Li
2
t
in
50
Po
Cricondentherm
e l
bb
25
Bu
10
5 3
e
Dew Point Lin Gas
0
Temperature
Using two component systems we have examined various aspects of phase behaviour.
Reservoir fluids contain hundreds of components and therefore are multicomponent
systems. The phase behaviour of multicomponent hydrocarbon systems in the liq-
uid-vapour region however is very similar to that of binary systems however the
mathematical and experimental analysis of the phase behaviour is more complex.
Figure 15 gives a schematic PT & PV diagram for a reservoir fluid system. Systems
which include crude oils also contain appreciable amounts of relatively non-volatile
constituents such that dew points are practically unattainable.
16
Phase Behaviour of Hydrocarbon Systems
All Liquid
Liqu
id "a"
Critical Point
First Gas Bubble
Bubble Point
Bubble Point
uid
Liq
i ne
tL
Gas / 40% Liquid
%
in
80
Po
%
60
le
bb
%
Bu
40
Pressure
Pressure
% Lin
e
Last Drop of Liquid 20 int
Po Dew Point
w
De
Dew Point
All Gas
Temperature Volume
We will consider the behaviour of several examples of typical crude oils and natural
gases:
Figure 16 is a useful diagram to illustrate the behaviour of the respective fluid types
above. However it should be emphasised that for each fluid type there will be different
scales. The vertical lines help to distinguish the different reservoir fluid types.
Isothermal behaviour below the critical point designates the behaviour of oil systems
and the fluid is liquid in the reservoir, whereas behaviour to the right of the critical
point illustrates the behaviour of systems which are gas in the reservoir.
% Liquid TM
100
2
Pressure
T = Maximum temperature at
m
25 which two phases can coexist
20 e
15 t Li n C = Critical conditions
Poin
10 Dew X = Cricondentherm
5 Gas 5
0 Single Phase Region X5
Temperature
The two-phase region covers a wide range of pressure and temperature. Tc is higher
than the reservoir temperature. In figure 17 the line 1-2-3 represents the constant
reservoir temperature pressure reduction that occurs in the reservoir as crude oil is
produced for a black oil. These oils are a common oil type. The dotted line shows
the conditions encountered as the fluid leaves the reservoir and flows through the
tubing to the separator.
If the initial reservoir pressure and temperature are at 2, the oil is at its reservoir
bubble point and is said to be saturated, that is, the oil contains as much dissolved
gas as it can and a further reduction in pressure will cause formation of gas. If the
initial reservoir pressure and temperature are at 1, the oil is said to be undersaturated,
i.e. The pressure in the reservoir can be reduced to Pb before gas is released into the
formation. For an oil system the saturation pressure is the bubble point pressure.
18
Phase Behaviour of Hydrocarbon Systems
1 Undersaturated
Mole % Liq.
100
Liquid Critical Point
2 Saturated
Pb
Pressure
e
3
Lin
int
Po
75
le
bb
line
Sep.
Bu
int
Po
w
50 Gas
De
25
0
Temperature
As the pressure is dropped from the initial condition as a result of production of flu-
ids, the fluids remain in single phase in the reservoir until the bubble point pressure
corresponding to the reservoir temperature is reached. At this point the first bubbles
of gas are released and their composition will be different from the oil being more
concentrated in the lighter ( more volatile) components. When the fluids are brought
to the surface they come into the separator and as shown on the diagram, the separa-
tor conditions lie well within the two phase region and therefore the fluid presents
itself as both liquid and gas. The pressure and temperature conditions existing in the
separator indicate that around 85% liquid is produced, that is a high percentage and as
a result the volume of liquid at the surface has not reduced a great amount compared
to its volume at reservoir conditions. Hence the term low-shrinkage oil.
As the pressure is further reduced as oil is removed from the reservoir, point 3 will
be reached and 75% liquid and 25% gas will be existing in the reservoir. Strictly
speaking once the reservoir pressure has dropped to the bubble point, beyond that the
phase diagram does not truly represent the behaviour of the reservoir fluid. As we will
see in the chapter on drive mechanisms, below the bubble point gas produced flows
more readily than the associated oil and therefore the composition of the reservoir
fluid does not remain constant. The system is continually changing in the reservoir
and therefore the related phase diagram changes.
The summary characteristics for a black oil sometimes termed a heavy oil or low
shrinkage oil are as follows.
Broad-phase envelope
High percentage of liquid
High proportion of heavier hydrocarbons
GOR < 500 SCF/STB
Volatile oil contains a much higher proportion of lighter and intermediate hydocar-
bons than heavier black oil and therefore they liberate relatively large volumes of gas
leaving smaller amounts of liquid compared to black oils. For this reason they used
to be called high shrinkage oils. The diagram in figure 18 shows similar behaviour to
the black oil except that the lines of constant liquid to gas are more closely spaced.
Points 1 and 2 have the same meaning as for the black oil. As the pressure is reduced
below 2 a large amount of gas is produced such that at 3 the reservoir contains 40%
liquid and 60% gas.
At separator conditions 65% of the fluid is liquid, i.e. less than previous mixture.
The summary characteristics for a volatile sometimes termed a heavy oil or high
shrinkage oil when compared to black oils are as follows.
Liquid 2
Critical Point
Mole % Liq.
100
75
Pressure
3
e
lin
int
Sep.
po
ble
50 40
b
Bu
e
lin
Gas
t
in
po
w
25
De
Temperature
Clearly, for these fluids, it is the composition of the fluid that determines the nature
of the phase behaviour and the relative position of the saturation lines, (bubble point
and dew point lines), the lines of constant proportion of gas/liquid and the critical
point.
20
Phase Behaviour of Hydrocarbon Systems
For both of these fluids types one can prevent the reservoir fluid going two phase
by maintaining the reservoir pressure above its saturation pressure by injecting flu-
ids into the reservoir. The most common practise is the use of water as a pressure
maintenance fluid.
5.2 Retrograde Condensate Gas
If the reservoir temperature lies between the critical point and the cricondentherm a
retrograde gas condensate field exists and Figure 19 gives the PT diagram for such
a fluid. Above the phase envelope a single phase fluid exists. As the pressure de-
clines to 2 a dew point occurs and liquid begins to form in the reservoir. The liquid
is richer in heavier components than the associated gas. As the pressure reduces to
3 the amount of liquid increases. Further pressure reduction causes the reduction of
liquid in the reservoir by re-vaporisation. It is important to recognise that the phase
diagram below for a retrograde condensate fluid represents the diagram for a constant
composition system.
Before production the fluid in the reservoir exists as a single phase and is generally
called a gas. It is probably more accurate to call it a dense phase fluid. If the reservoir
drops below the saturation pressure the dew point, then retrograde condensation oc-
curs within the formation. The nature of this condensing fluid is only in recent years
being understood. It was previously considered that the condensing fluid would be
immobile since its maximum proportion was below the value for it to have mobil-
ity. It was considered therefore that such valuable condensed fluids would be lost to
production and the viability of the project would be that from the ‘wet’ gas.
Liquid
Critical Point
ine 2
tL
P oin
e
bl
ub
Pressure
3
B
Mole % Liq.
100
75
Sep.
50
25 e
t Lin Gas
oin
10
De wP
5
0
Temperature
One of the development options for such a field therefore is to set in place a pressure
maintenance procedure whereby the reservoir pressure does not fall below the
saturation pressure. Water could be used as for oils but gas might be trapped behind
the water as the water advances through the reservoir. Gas injection, called gas
Surface Separation
Condensate Sales
Injection Well
Production Well
Recent research has shown that the nature of oil forming in porous media by this ret-
rograde process may not be as first considered. The isolation of condensing liquids in
porous rock is dependant on the relative strength of the interfacial tension and viscous
forces working in the rock. If the relative magnitude of these is high then the fluid
will be trapped however if they are low as a result of low interfacial tension, which
is the case nearer the critical point, then the condensing liquids may be mobile and
move as a result of viscous and gravity forces. Condensate liquids have been able
to flow at saturations well below the previously considered irreducible saturation
proportion. Established relative permeability thinking is having to be reconsidered in
the context of gas condensates. The phenomena just described may give explanation
to the observation sometimes made of an oil rim below a gas condensate field.
Looking at the PT phase diagram one might consider that "blowing the reservoir down"
22
Phase Behaviour of Hydrocarbon Systems
quickly might be an option and as a result vaporise the condensed liquids in the for-
mation. This is not a serious option since once the reservoir pressure falls below the
dew point the impact of the increasing liquid proportion remaining in the reservoir
causes the phase diagram to move to the right relative to reservoir conditions, and any
vaporising will be of the lightest components which are likely to be in good supply
and therefore not of significant value.
The summary characteristics for a retrograde gas condensate fluid are as follows.
Contains more lighter HC’s and fewer heavier HC’s than high-shrinkage oil
API up to 60˚ API
GOR up to 70,000 SCF/STB
Stock tank oil is water-white or slightly coloured
5.3 Wet Gas
The phase diagram for a mixture containing smaller hydrocarbon molecules lies
well below the reservoir temperature. Figure 21. The reservoir conditions always
remain outside the two-phase envelope going from 1 to 2 and therefore the fluid ex-
ists as a gas throughout the reduction in reservoir pressure. For a wet gas system,
the separator conditions lie within the two-phase region, therefore at surface heavy
components present in the reservoir fluid condense under separator conditions and this
liquid is normally called condensate. These liquid condensates have a high propor-
tion of light ends and sell at a premium. The proportion of condensates depend on
the compositional mix of the reservoir fluid as represented by the iso-volume lines
on the PT diagram.
Liquid
1
Critical Point
Pressure
Mole % Liq.
100
2
75 Sep.
50
25
5 Gas
0
Temperature
The reference wet gas, clearly does not refer to the system being wet due to the pres-
ence of water but due to the production condensate liquids.
1
Pressure
Critical Point
2
75
Liquid
50 Sep.
25
Gas
Temperature
Figure 16 gave a rather simplistic representation of the various types of fluids with
respect to the relative position of reservoir temperature with respect to the phase
diagram. In reality it is the phase diagram which changes according to composition
and the relative position of the reservoir temperature and separator conditions, and
these determine the character of the fluid behaviour. Figure 23 gives a better indica-
tion of the various reservoir types with respect to a specific pressure and temperature
24
Phase Behaviour of Hydrocarbon Systems
Pressure
Separator
Gas
Wet Gas Condensate Volatile Black
Dry Gas Oil Oil
Temperature (ºC)
Critical Point
CG
C
Reservoir
Liquid
Pd=Pb CL
Initial
Pressure
Reservoir
Pressure
Separator
Temperature
The diagram illustrates that at the gas-oil contact the gas is at its dew pressure, the oil
is at its bubble point pressure and the combination fluid lies on the constant propor-
tion quality line representing the ratio of the gas and oil as they exist in the reservoir
system. The gas cap may be dry, wet or condensate depending on the composition
and phase diagram of the gas.
8 CRITICAL POINT DRYING
Although not part of the topic of phase behaviour in the context of reservoir fluids it
is useful to illustrate the application in a very practical application in the context of
the evaluation of rock properties. Critical point drying has been used by a number
of sciences to prepare specimens of delicate materials for subsequent micro visual
analysis where conventional preparation techniques will destroy delicate fabric.
Critical point drying takes advantage of the behaviour of fluids around the critical
point where one can go from one phase type, like liquid to gas without a visually
observed phase change.
In the 1980’s it was observed in a UK offshore field that the interpreted permeability
for a well sand in the zone where water injection was proposed was different from
well injectivity tests when compared to the core analysis value where the value was
many times more. The extent of this difference was such that permeabilities from
the well test gave values which would prevent injection to take place whereas those
from the core tests would result in practical injectivities. Clearly the difference was
important.
26
Phase Behaviour of Hydrocarbon Systems
The company concerned embarked on a more sophisticated core recovery and analy-
sis process suspicious that perhaps the fabric of the rock was being affected by core
preparation methods. They resorted to critical point drying.
The core recovered from the water zone of the reservoir from a subsequent new well
was immersed and transferred to the test laboratory submerged in ‘formation water’.
At the laboratory a core plug sample was extracted, cut to size and loaded into a
core holder still submerged in the water. The core was then mounted in a flow rig
(figure 25) and an alcohol which is miscible with water displaced the water in the
core. Carbon dioxide at a pressure and temperature where it is in the liquid state was
then introduced which miscible displaced the alcohol. The temperature and pressure
was then adjusted taking them around the critical point rather than across the vapour
pressure line of the PT phase diagram (figure 26) ending up with a temperature and
pressure below the vapour pressure line with the fluid now in a gaseous state. After
this process the permeability was measured to be of the same order as that interpreted
from the well injectivity test.
The reason for this difference was subsequently demonstrated to be a very fragile
clay which during conventional core recovery and cleaning was damaged to an extent
that its pore blocking structure was destroyed.
T
P
Core In Holder
Critical Point
Pressure
LIQUID
Vapour
Pressure Line
GAS
Temperature
28
Behaviour of Gases
CONTENTS
1 IDEAL GASES
1.1 Boyle's Law
1.2 Charles' Law
1.3 Avogadro's Law
1.4 The Equation of State For an Ideal Gas
1.5 The Density of an Ideal Gas
1.6 Standard Conditions
1.7 Mixtures of Ideal Gases
1.7.1 Dalton's Law of Partial Pressures
1.7.2 Amagat's Law
1.8 Apparent Molecular Weight
1.9 Specific Gravity of a Gas
4 COEFFICIENT OF ISOTHERMAL
COMPRESSIBILITY OF GASES
5 VISCOSITY OF GASES
5.1 Viscosity
5.2 Viscosity of Mixtures
6 EQUATIONS OF STATE
6.1 Other Equations-of-State
6.2 Van de Waals Equation
6.3 Benedict - Webb - Rubin Equation (BWR)
6.4 Redlich - Kwong Equation
6.5 Soave, Redlich Kwong Equation
6.6 Peng Robinson Equation of State
6.7 Application to Mixtures
LEARNING OBJECTIVES
Having worked through this chapter the Student will be able to:
• Present the equation of state, EOS, for a ‘real gas’ and explain what ‘Z’ is,
PV=ZnRT.
• Define the gas formation volume factor and derive an equation fore it using the
EOS.
Behaviour of Gases
INTRODUCTION
A gas is a homogenous fluid that has no definite volume but fills completely the vessel
in which it is placed. The system behaviour of gases is vital to petroleum engineers
and the laws governing their behaviour should be understood. For simple gases these
laws are straightforward but the behaviour of actual hydrocarbon gases particularly
at the conditions occurring in the reservoir are more complicated.
We will review the laws that relate to the pressure, volume and temperatures of gases
and the associated equations. These relationships were previously termed gas laws;
it is now more common to describe them as equations of state.
1 IDEAL GASES
The laws relating to gases are straightforward in that the relationships of pressure,
temperature and pressure are covered by one equation. First consider an ideal gas.
An ideal gas is one where the following assumptions hold:
• Volume of the molecules i.e. insignificant with respect to the total volume of
the gas.
• There are no attractive or repulsive forces between molecules or between
molecules and container walls.
• There is no internal energy loss when molecules collide.
i.e.
1
V α or PV = constant, T is constant
P (1)
i.e.
V
V α T or = constant, P is constant
T (2)
One mole of a material is a quantity of that material whose mass in the unit system
selected is numerically equal to the molecular weight.
PV
= constant
T (3)
It is termed the Universal Gas Constant and has different values depending on the
unit system used, so that;
cu ft psia
10.732
R in oilfield units = lb mole R
Table 1 gives the values for different unit systems.
p V T n R
Behaviour of Gases
PV = nRT (4)
To find the volume occupied by a quantity of gas when the conditions of temperature
and pressure are changed from state 1 to state 2 we note that:
PV PV PV
n = is a constant so that 1 1 = 2 2
RT T1 T2
EXERCISE 1.
A gas cylinder contains methane at 1000 psia and 70°F. If the cylinder has a vol-
ume of 3 cu.ft assuming methane is an ideal gas calculate the mass of methane in
the cylinder.
m
ρg = weight / volume =
V
where ρg is the gas density
For 1 mole m = MW MW = Molecular weight
RT
V =
P
MW.P
∴ ρg =
RT (5)
EXERCISE 2.
ie
Pres Vres P V
= sc sc
Tres Tsc (6)
This relationship assumes that reservoir properties behave as ideal. This is NOT the
case as will be discussed later.
EXERCISE 3.
Assuming methane is at the conditions of exercise 1, calculate the volume the gas
would occupy at standard conditions.
Laws established over early years governing ideal gas mixtures include Dalton’s
Law and Amagat’s Law.
ie
P = PA + PB + PC + ..... (7)
therefore
Behaviour of Gases
RT RT RT
P = n A + n B + nC
V V V
RT
i.e. P = Σn j
V
Pj n
∴ = j = y j
P n
(8)
where yj = mole fraction of j component.
th
The pressure contribution of a component, its partial pressure, is the total pressure
times the mole fraction.
i.e.
V = VA + VB + VC (9)
RT RT RT
V = n A + n B + n C
P P P
RT
V = Σn j
P
Vj n
i.e. = j = y j
V n (10)
i.e, for an ideal gas the volume fraction is equal to the mole fraction.
EXERCISE 4.
(
AMW = Σ y j × MWj )
AMW for air = 28.97, a value of 29.0 is usually sufficiently accurate.
EXERCISE 5.
1.9 Specific Gravity of a Gas
The specific gravity of a gas, γg is the ratio of the density of the gas relative to that of
dry air at the same conditions.
ρg
γ g =
ρair (11)
MgP
M M
γ g = RT = g = g
M air P M air 29
RT
EXERCISE 6.
The equations so far listed apply basically to ideal systems. In reality, however,
particularly at high pressures and low temperatures the volume of the molecules are
no longer negligible and attractive forces on the molecules are significant.
Behaviour of Gases
The ideal gas law, therefore, is not too applicable to light hydrocarbons and their
associated fluids and it is necessary to use a more refined equation.
There are two general methods of correcting the ideal gas law equation:
PV = znRT (12)
where the factor ‘z’ is known as the compressibility factor and the equation is known
as the compressibility equation-of-state or the compressibility equation.
The compressibility factor is not a constant but varies with changes in gas composition,
temperature and pressure and must be determined experimentally (Figure 1).
P1V1 PV
= 2 2
z1T1 z 2 T2 (13)
z is an expression of the actual volume to what the ideal volume would be.
i.e.
Vactual
z =
Videal (14)
nt
ta
Compressibility factor, Z
ns
1.0
co
=
e
ur
r at
pe
m
Te
0.5
0
0 PRESSURE, P
T P
Tr = and Pr =
Tc Pc (15)
Where, Tc and Pc are the pure component critical temperature and pressure.
The compressibility factor ‘z’ follows this law. It is usually presented vs Tr and Pr.
Although in many cases pure gases follow the Law of Corresponding States, the gases
associated with hydrocarbon reservoirs do not. The Law has however been used to
apply to mixtures by defining parameters called pseudo critical temperature and
pseudocritical pressure .
For mixtures a pseudocritical temperature and pressure, Tpc and Ppc is used such
that:
where y is the mole fraction of component j and Tcj and Pcj are the critical temperature
and pressure of component j.
EXERCISE 7.
Calculate the pseudo critical temperature and pseudocritical pressure of the mixture
in exercise 4 .
For mixtures the compressibility factor (z) has been generated with respect to natural
gases 1, where ‘z’ is plotted as a function of pseudo reduced temperature, Tpr and
pseudo reduced pressure Ppr where
10
Behaviour of Gases
05
0.8
1.
1.7
1.5
1
1.
1.45
Compressibility Factor, z
2
1.
1.35
3
1.3
1.
0.6 4 1.5
1.25 1.
1.5
1.2
1.7
0.5 1.6
1.9 1.4
1.8
1.15
2.0 2.2
0.4 1.3
1.1 2.4
2.6
3.0
0.3 1.2
1.05
0.25 3.0
2.8
1.1 1.1
2.6 2.4
Compressibility of
Natural Gases
2.2
2.0 1.9 1.2 (Jan. 1, 1941)
1.0 1.0
1.8 1.1
1.7 1.05
1.6 1.4
1.3
0.9 0.9
7 8 9 10 11 12 13 14 15
Pseudo Reduced Pressure, Pr
Figure 2 Compressibility factors for natural gas1 (Standing & Katz, Trans AIME, 1942)
The use of this chart , figure 2 ,has become common practise to generate z values for
natural gases. Poettmann and Carpenter 2 have also converted the chart to a table.
Various equations have also been generated based on the tables.
EXERCISE 8.
Miscellaneous
Gases
Condens
650 ate Wel
l Fluid
s
600
550
500
s
Pseudocritical Temperature, R
se
s Ga
450 e ou
an
ell
sc
Mi
ids
400 Flu
W ell
e
sat
n den
Co
350
300
0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2
Gas Gravity (air = 1)
and
p pc Tpc′
p′pc =
Tpc + yH 2 S 1 − yH 2 S e ( ) (19)
T'pc and p'pc are used to calculate Tpr and ppr. The value for ε is obtained from
the figure 4 from the Wichert and Aziz paper
80
15
70
60
PER CENT C02
50
E
20
40
25
30
30
20
30
25
10
20
15
10
5 34.5
0
0 10 20 30 40 50 60 70 80
PER CENT H2S
Calculate the pseudo critical properties of the gas in exercise 4 if it also contained 3
lb of hydrogen sulphide, 10lb of carbon dioxide and 2.5lb of nitrogen
From Wichert & Azis chart for compositions of H2S and CO2 ε = 19
Tpc′ = Tpc - e = 371o R
p pc Tpc′
p′pc =
(
Tpc + yH 2 S 1 − yH 2 S e )
Ppc′ = 694.3
EXERCISE 10.
14
Behaviour of Gases
EXERCISE 11.
The petroleum industry expresses its reservoir quantities at a common basis of surface
conditions which for gases is standard cubic volumes. To convert reservoir volumes
to surface volumes the industry uses formation volume factors. For gases we have
Bg, the gas formation volume factor, which is the ratio of the volume occupied at
reservoir temperature and pressure by a certain weight of gas to the volume occupied
by the same weight of gas at standard conditions. The shape of Bg as a function of
pressure is shown in figure 5.
The gas formation volume factor can be obtained from PVT measurements on a gas
sample or it may be calculated from the equations-of-state discussed previously.
One definition of the gas formation volume factor is: it is the volume in barrels
that one standard cubic foot of gas will occupy as free gas in the reservoir at the
prevailing reservoir pressure and temperature.
Depending on the definition the units will change and the units will be; rb free
gas/scf gas or rm3 free gas/scm gas
.008
.006
Bg
rb/scf
.004
.002
V2 P Tz
Bg = = sc 2 2
Vsc P2 Tsc zsc (20)
volume at surface 1
= =E
volume in formation Bg
Usually the units of Bg are barrels of gas at reservoir conditions per standard cubic
foot of gas, ie bbl/SCF or cubic metres per standard cubic metre.
VR
Bg =
Vsc (21)
znRT
VR =
P (22)
zsc nRTsc
Vsc =
Psc (23)
T Psc cu. ft
∴ Bg = z . .
Tsc P SCF (24)
zT cu. ft
Bg = 0.0283
P SCF
zT cu. ft bbl
Bg = 0.0283 ×
P SCF 5.615 cu ft
or
zT res bbl
Bg = 0.00504
P SCF
16
Behaviour of Gases
zT cu. ft bbl
Bg = 0.0283 ×
P SCF 5.615 cu ft
or
zT res bbl
Bg = 0.00504
P SCF (25)
EXERCISE 12.
Calculate the gas formation factor for a gas with the composition of exercise 4
existing at the reservoir conditions given in exercise 8.
EXERCISE 13.
A reservoir exists at a temperature of 150°F (as for exercise 8) suitable for storing
gas. It has an areal size of 5 miles by 2 miles and is 200ft thick. The average porosity
is 20% and there is no water present. How much gas of the composition of exercise
4 can be stored at a pressure the same as in exercise 8 i.e. 3500 psia ? (1 mile=
5280 ft.)
4 Coefficient of Isothermal Compressibility of Gas-
es
The compressibility factor, z, must not be confused with the compressibility which is
defined as the change in volume per unit volume for a unit change in pressure, or
1 ∂V 1 ∂Vm
cg = − or = −
V ∂P Vm ∂P (26)
Vm is the specific volume or volume per mole.
cg is not the same as z, the compressibility factor.
PV = nRT or:
dV = − nRT
dP P2
1 − nRT 1
cg = =
V P 2 P (27)
znRT
V =
P
P nRT ∂z
cg = − 2
P − z
nRTz P ∂P
1 1 ∂z
cg = − .
P z ∂P (28)
P = Ppc Ppr
∂z ∂Ppr ∂z
=
∂P ∂P ∂Ppr
∂Ppr 1
=
∂P Ppc
∂z 1 ∂z
=
∂P Ppc ∂Ppr
1 1 ∂z
cg = −
Ppc Ppr zPpc ∂Ppr
Tpr
1 1 ∂z
c g Ppc = −
Ppr z ∂Ppr
Tpr
(29)
18
Behaviour of Gases
Since the pseudo reduced compressibility is a function of ‘z’ and pseudo reduced
pressure, the graph of Figure 2 can be used with Equation 29 to calculate values of
cpr.
5 VISCOSITY OF GASES
5.1 Viscosity
Viscosity is a measure of the resistance to flow. It is given in units of centipoise.
A centipoise is a gm/100 sec.cm. The viscosity term is called dynamic viscosity
whereas kinematic viscosity is the dynamic viscosity divided by the density.
dynamic viscosity
kinematic vis cos ity =
density
Kinematic viscosity has units of cm2/100 sec and the term is called centistoke.
1000
900
800 Viscosity of ethane
700
600
500 Pressure, psia
Viscosoty, micropoises
400 5000
4000
3000
300 2000
15000
200
750 1000
600
100 14.7
90
80
70
50 100 150 200 250 300 350 400
Temperature, deg F
The viscosity of gases at low pressures can be obtained from correlations presented
by different workers.
m
0.022 liu
He
en
rog
Nit
Air
0.020
e
x id
n Dio
0.018
rbo
Ca
id e
0.016 ulf
nS
ge
dro
Viscosity, cp
Hy
0.014 e
han
M et
e
ylen
Eth
0.012
ane
Eth
e
pan
0.010 pr o
ta e
n
ntan
e
i-Bu n-pe
t ane x ne
a
n-Bu n-He
tane ane
n-H ep n-Oct
0.008 N on ane
n-
ne
n-Deca
0.006
0.004
50 100 150 200 250 300 350 400
Temperature, ?ºF
Figure 7 and Figure 8 give the viscosities of individual components and paraffin
hydrocarbons at one atmosphere. For systems greater than 1 atmos the viscosities
can be obtained from the literature. Another way is by calculating the reduced
temperature and reduced pressure and use the chart developed by Carr6 which gives
a ratio of µ at reservoir conditions. This is given in Figure 9 in terms of pseudo
reduced conditions.
20
Behaviour of Gases
Correction added to
Correction added to
0.0015 1.0 0.0015
G = 20
Viscosity, c.p.
Viscosity, c.p.
1.5
0.014 0.0010 0.0010 G = 20 1.0
G = 06
Viscosity, at 1 atm, µ1, centipoise 0.013 0.0005 0.0005
G = 06
0.012 0
0 5 10 15
0
0 5 10 15
400
ºF Mole per cent N2 Mole per cent CO2
0.011
300
º F
0.010
0.009 200
º F
H2S
0.008
Correction added to
0.0015
100
Viscosity, c.p.
º F
0.007 0.0010
G = 20 1.5
1.0
0.006 0.0005
G = 06
0.005 0
0 5 10 15
Mole per cent H2S
0.004
10 20 30 40 50 60 70 80 90 100
Molecular Weight
6.0
3.5
3.0
Viscosity, µ / µA
20
ps
15 eu
do
2.5 red
uc
ed
10 pre
s su
8 re ,p
2.0 R
6
4
3
1.5
2
1.0
0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.4
Pseudoreduced Temperature, TR
j = 1, n
where:
n = number of components
The presence of other gases can also make a significant difference on the viscosity
(Figure 7).
EXERCISE 14.
Calculate the viscosity of the gas mixture in exercise 4 at 200°F and a pressure of
one atmosphere.
EXERCISE 15.
Use the gas gravity method to calculate the viscosity of the gas in exercise 4
EXERCISE 16.
Determine the viscosity of the gas in exercise 4 at 150°F and 3500 psia (ref ex 4, 7,
&8)
22
Behaviour of Gases
6 EQUATIONS OF STATE
PV = ZnRT
to describe the behaviour of gases is that the compressibility factor is not constant and
therefore mathematical manipulations cannot be made directly but must be carried out
through graphical or numerical techniques. Rather than use this modified equation
of state many have developed equations specifically to represent the behaviour of
real gases. It is an irony however that because of the long use of the equation above
incorporating z many of the real gas equation of states have been worked to calculate
z for use in the above equation.
The two corrective terms to overcome the limiting assumptions of the ideal gas
equation are:
(i) The internal pressure or cohesion term , which accounts for the cohesion forces,
is a/V2.
(ii) The co-volume b, which represents the volume occupied by one mole at infinite
pressure and results from the repulsion forces which occur when the molecules
move close together.
V3 - (+ b) V2 + (a/P)V - ab/P = 0
Z3 - Z2 (1 + B) + Z A - AB = 0 (32)
where
aP bP
A= 2
and B =
( RT ) RT (33)
Isotherm T1 is the single phase isotherm, Tc is the critical isotherm and T2 gives the
isotherm below the critical temperature.
T1>Tc
P
Tc
Psat
T2<Tc
At the critical point , for a pure substance , the equation of state should be such
that:
∂P ∂ 2P
= 2 =0
∂V T = Tc ∂V T = T
c
That is the critical isotherm exhibits a horizontal inflection point at the critical
point.
24
Behaviour of Gases
27 R2 Tc2 RT
a= and b =
64 Pc 8 Pc (34)
EXERCISE 17.
For the curve, T2<Tc, the pressure decreases rapidly in the liquid region with increasing
V; after crossing the liquid saturated line a minimum occurs, rises to a maximum
and then decreases at the saturated vapour line. Real behaviour does not follow this
behaviour. They contain a horizontal segment where saturated liquid and saturated
vapour coexist in varying proportions.
This equation is not able to represent gas properties over a wide rage of temperatures
and pressures and over subsequent years many equations have been developed. A
number are given including those which are finding favour in their application in
this industry.
PT Bo RT − Ao − Co / T 2 bRT − a
P= + + +
V V2 V3
aα C γ −γ
+ 3 o 2 1 + 2 exp 2
V
6
V T V V (35)
where a, b, c, Ao, Bo and Co are constants for a given gas.
These equations are derived for pure components for which the empirical parameters
need to be obtained. For mixtures mixing rules are required to obtain these
constants.
This modern development of cubic equations of state started in 1949 with the Redlich
and Kwong equation which involves only two empirical constants.
The term a(T) depends on the temperature and Redlich Kwong expressed this as a
function of the reduced temperature Tr using
ac
a(T ) =
TR
By applying the limiting condition at the critical points yields values of ac and b
related to critical constants. Such that ;
R2 Tc2 RT
ac = 0.42748 and b = 0.08664 c
Pc Pc (37)
RT acα
P= −
(V − b) [V (V + b)] (38)
where
α is a non dimensionless temperature dependent term which has a value of 1.0 at the
critical temperature.
α is obtained from
[ ( )]
2
α = 1 + m 1 − Tr
26
Behaviour of Gases
RT acα
P= −
V − b [V (V + b) + b(V − b)] (39)
R2 Tc2 RT
ac = 0.457235 and b = 0.0778 c
Pc Pc (40)
and
α is the same function as for the Soave equation except the ω function is
different;
These equations, in particular the SRK and PR equation are widely used in simulation
software used to predict behaviour in reservoirs, wells and processing. There are
other equations of state which are as competent at predicting physical properties
which have been developed mainly focusing on the need to improve the accuracy of
liquid volumes predictions. There is, however, great reluctance to change from those
presently used because of the investment in their associated parameters. An excellent
review of these equations and application is given by Danesh 9.
b = ∑ y j b j and a = ∑ ∑ yi y j ai aj 1 − kij ( )
j i j (41)
where the term kij is termed the binary interaction coefficients which are independent
of pressure and temperature. Values of binary interaction coefficients are obtained
by fitting equation of state (EOS) predictions to gas-liquid data for binary mixtures.
They have NO physical property significance. Each equation has its own binary
interaction coefficient.
Effort is underway and methods exist to not use binary interaction parameters but to
use physical property related parameters to enable good quality predictions.
A PVT cell contains 0.01 cu ft ( 300cc) of gas with at composition of ; methane 0.67
mol.frac, ethane 0.235 and n-butane 0.05. The temperature is increased to 300°C.
Use the SRK equation to calculate the pressure at this increased temperature. Use
binary interaction coefficients of C1-nC4 0.02, C2-nC4 0.01 and C1-C2 0.0
Solutions to Exercises
EXERCISE 1.
A gas cylinder contains methane at 1000 psia and 70 oF. If the cylinder has a volume
of 3 cu.ft assuming methane is an ideal gas calculate the mass of methane in the
cylinder.
SOLUTION
PV = nRT
n = m/M
where n = number of moles
m = mass
M = molecular weight
m = PMV/RT
lb
(1000 psia)16.04 (3cuft )
m= lbmole
10.73 psia.cuft 530 o R
lbmole.o R
( )
28
Behaviour of Gases
EXERCISE 2.
SOLUTION
MW.P
ρg =
RT
lb
(1000 psia)16.04
ρg = lbmole
10.73 psia.cuft 530 0 R
lbmole.oR
( )
lb
Density of gas, ρg = 2.82
cu. ft.
EXERCISE 3.
Assuming methane is at the conditions of exercise 1, calculate the volume the gas
would occupy at standard conditions.
SOLUTION
P1V1 PV P V
= 2 2 = sc sc
T1 T2 Tsc
P1 Tsc V
Vsc =
Psc T1
EXERCISE 4.
A gas is made up of the following components; 25lb of methane, 3 lb of ethane and 1.5
lb of propane. Express the composition of the gas in weight and mole fractions.
Gas A B C D
Components Weight Mol weight lb moles Mole fraction
1 Methane 25 16.04 1.559 0.921
2 Ethane 3 30.07 0.100 0.059
3 Propane 1.5 44.09 0.034 0.020
Totals 29.05 1
EXERCISE 5.
SOLUTION
Gas A B C
Components Mol weight Mol fraction
mw yi A*B
1 Methane 16.04 0.921 14.77
2 Ethane 30.07 0.059 1.77
3 Propane 44.09 0.020 0.89
1.000 17.43
Apparent Molecular weight= 17.43
EXERCISE 6.
SOLUTION
Mg M
γg = = g
Mair 29
Μg = AMW = 17.43
Gas gravity = 0.6
EXERCISE 7.
Calculate the pseudo critical temperature and pseudocritical pressure of the mixture
in exercise 4 .
30
Behaviour of Gases
SOLUTION
Gas A B C D.
Components Mol weight Mole fraction pc-psi Tc °R ppc Tpc
mw yi
EXERCISE 8.
For the gas of exercise 4 determine the compressibility factor at a temperature of 150
o
F and a pressure of 3500psia.
SOLUTION
EXERCISE 9.
Calculate the pseudo critical properties of the gas in exercise 4 if it also contained 3
lb of hydrogen sulphide, 10lb of carbon dioxide and 2.5lb of nitrogen
From Wichert & Azis chart for compositions of H2S and CO2 ε = 19
p pc Tpc′
p′pc =
( )
Tpc + yH 2 S 1 − yH 2 S e
Ppc′ = 694.3
EXERCISE 10.
SOLUTION
EXERCISE 11.
SOLUTION
Total mass of gas = 29.5 lb.
Apparent mol.wgt of gas exercise 5 = 17.43 lb./lb.mole
lb.moles of gas = 1.6924
Standard cubic feet of gas = 380.9 x 1.6924
= 644.68 scf
EXERCISE 12.
Calculate the gas formation factor for a gas with the composition of exercise 4 existing
at the reservoir conditions given in exercise 8.
SOLUTION
32
Behaviour of Gases
EXERCISE 13.
A reservoir exists at a temperature of 150oF (as for exercise 8) suitable for storing
gas. It has an areal size of 5 miles by 2 miles and is 200ft thick. The average porosity
is 20% and there is no water present. How much gas of the composition of exercise
4 can be stored at a pressure the same as in exercise 8 i.e. 3500 psia. ? (1 mile=
5280 ft.)
SOLUTION
EXERCISE 14.
Calculate the viscosity of the gas mixture in exercise 4 at 200°F and a pressure of
one atmosphere.
SOLUTION
Σµ j y j M j
µ mix =
Σy j M j
µmix = 0.0529/4.1451
µmix =0.01275 cp
EXERCISE 15.
Use the gas gravity method to calculate the viscosity of the gas in exercise 4
SOLUTION
EXERCISE 16.
Determine the viscosity of the gas in exercise 4 at 150oF and 3500 psia (ref ex 4, 7,
&8)
SOLUTION
From exercise 7
Ppc = 668.4
Tpc = 362.6
3500
Pr = P P = = 5.24
c 668.4
610
Tr = T T = = 1.68
c 362.6
EXERCISE 17.
SOLUTION
34
Behaviour of Gases
EXERCISE 18.
A PVT cell of volume 0.01 cu ft ( 300cc) contains 0.008 lb mole. of gas with
a composition of; methane 0.67 mol.frac, ethane 0.235 and n-butane 0.05. The
temperature is increased to 300°C. Use the SRK equation to calculate the pressure
at this increased temperature. Use binary interaction coefficients of C1-nC4 0.02,
C2-nC4 0.01 and C1-C2 0.0
SOLUTION
R2 Tc2 RT
ac = 0.42748 and b = 0.08664 c
Pc Pc
[ ( )]
2
α = 1 + m 1 − Tr
Methane 0.67 344 667 0.4759 8735 0.0104 0.49635 0.57546 5027
ethane 0.235 550 708 0.7223 21036 0.0979 0.63241 0.79033 16625
n-butane 0.05 766 551 1.2926 52429 0.1995 0.78701 1.00619 52753
b = ∑ y j b j and a = ∑ ∑ yi y j ai aj 1 − kij ( )
j i j
where a ij = (1- k ij )(a i a j )0.5
Methane 0.67 0.312 0.00 0.00 0.02 2123.7 1485.5 1037.29 4646.52
ethane 0.235 0.181 0.00 0.01 1485.5 1039.1 732.969 3257.56
n-butane 0.05 0.129 0.00 1037.3 732.97 527.535 2297.8
1 0.622 sum 10201.9
Vm = 1.25 cu ft / lb mole
b = 0.622 a cα = 10201.9
P = 8617.6 psia
REFERENCES
2. Poettmann FH and Carpenter PG The Multiphase Flow of Gas and Water through
Vertical Flow Strings with Application to the Design of Gas Lift Installations.
API Drilling and Production Practise. 1952, pp 279-91
4. Wichert, E and Aziz,K “ Calculate Z’s for sour gases” Hyd Proc.(May 1972)
51, 119-122
5. Katz, D.L., Handbook of Natural Gas Engineering, McGraw Hill, NY, 1959
6. Carr N et al. Viscosity of natural gases under pressure. Trans AIME 201, 264,
(1954)
7. Lee et al “The viscosity of natural gases.” Trans AIME 1966 237, 997-1000
36
Properties of Reservoir Liquids
CONTENTS
2 GAS SOLUBILITY, Rs
6 OIL COMPRESSIBILITY
8 FLUID DENSITY
8.1 Specific Gravity of a Liquid
8.2 Density Based on Ideal Solution Principles
10 VISCOSITY OF OIL
11 INTERFACIAL TENSION
Having worked through this chapter the Student will be able to:
• Define the oil formation volume factor Bo, and plot Bo vs. P for a reservoir
fluid.
• Define the Total Formation Volume factor Bt, and plot Bt vs. P alongside a Bo
vs. P plot.
• Use black oil correlations and their graphical form to calculate fluid
properties.
• Calculate the density of a reservoir fluid mixture, using ideal solution principles,
at reservoir pressure and temperature, using density correction chart for C1 &
C2 and other prerequisite data.
• Calculate the reserves and production of gas and condensate operating above
the dewpoint, given prerequisite data.
• List the comparisons of the black oil and compositional model in predicting
liquid properties
Properties of Reservoir Liquids
The black oil model considers the fluid being made up of two components - gas dissolved
in oil and stock tank oil. The compositional changes in the gas when changing pressure
and temperature are ignored. To those appreciating thermodynamics this simplistic
two component model is difficult to cope with. The Black Oil Model, illustrated in
Figure 1, is at the core of many petroleum engineering calculations, and associated
procedures and reports.
Associated with the black oil model are Black Oil model definitions in relation to
Gas Solubility and Formation Volume Factors.
Reservoir Fluid
Solution Gas
/ = Rs
/ = Bo
Although the gas associated with oil and the oil itself are multicomponent mixtures
it is convenient to refer to the solubility of gas in crude oil as if we were dealing with
a two-component system.
The amount of gas forming molecules in the liquid phase is limited only by the
reservoir conditions of temperature and pressure and the quantity of light components
present.
The solubility is referred to some basis and it is customary to use the stock tank
barrel.
For a fixed gas and crude, at constant T, the quantity of solution gas increases with
p, and at constant p, the quantity of solution gas decreases with T
Rather than determine the amount of gas which will dissolve in a certain amount of
oil it is customary to measure the amount of gas which will come out of solution as
the pressure decreases. Figure 2 illustrates the behaviour of an oil operating outside
the PT phase diagram in its single phase state when the reservoir pressure is above
its reservoir bubble point at 1. Fluid behaviour in the reservoir is single phase and
the oil is said to be undersaturated . In this case a slight reduction of pressure causes
the fluid to remain single phase. If the oil was on the boundary bubble point pressure
line at 2 then a further reduction in pressure would cause two phases to be produced,
gas and liquid. This saturated fluid is one that upon a slight reduction of pressure
some gas is released. The concept of gas being produced or coming out of solution
gives rise to this gas solubility perspective. Clearly when the fluids are produced to
the surface as shown by the undersaturated oil in figure 2 the surface conditions lie
within the two phase area and gas and oil are produced. The gas produced is termed
solution gas and the oil at surface conditions stock tank oil. These are the two com-
ponents making up the reservoir fluid, clearly a very simplistic concept.
The gas solubility Rs is defined as the number of cubic feet (cubic metre) of gas
measured at standard conditions, which will dissolve in one barrel (cubic metre)
of stock tank oil when subjected to reservoir pressure and temperature.
In metric units the volumes are expressed as cubic metre of gas at standard conditions
which will dissolve in one cubic metre of stock tank oil.
Properties of Reservoir Liquids
Solution Gas
Rsi scf/stb
Surface
Pi 1
2
+
Pressure
P
Stock Oil Tank
1 st b. oil
Temperature
Phase Diagram
Oil Reservoir
Figure 3 gives a typical shape of gas solubility as a function of pressure for a reser-
voir fluid at reservoir temperature. When the reservoir pressure is above the bubble
point pressure then the oil is undersaturated, i.e. capable of containing more gas. As
the reservoir pressure drops gas does not come out of solution until the bubble point
is reached, over this pressure range therefore the gas in solution is constant. At the
bubble point pressure, corresponding to the reservoir temperature, two phases are
produced, gas and oil. The gas remaining in solution therefore decreases.
The nature of the liberation of the gas is not straight forward. Within the reservoir
when gas is released then its transport and that of the liquid is influenced by the relative
permeability of the rock ( discussed in Chapter 10). The gas does not remain with its
associated oil i.e. the system changes. In the production tubing and in the separator
it is considered that the gas and associated liquid remain together i.e. the system is
constant. The amount of gas liberated from a sample of reservoir oil depends on the
conditions of the liberation. There are two basic liberation mechanisms:
400
200
Pb
The two methods of liberation give different results for Rs. This topic is covered in
more detail in the PVT analysis chapter.
Production of a crude oil at reservoir pressures below the bubble point pressure occurs
by a process which is neither flash or differential vaporisation. Once enough gas is
present for the gas to move toward the wellbore the gas tends to move faster than the
oil. The gas formed in a particular pore tends to leave the liquid from which it was
formed thus approximating differential vaporisation, however, the gas is in contact with
liquid throughout the path through the reservoir. The gas will also migrate vertically
as a result of its lower density than the oil and could form a secondary gas cap.
Fluid produced from reservoir to the surface is considered to undergo a flash process
where the system remains constant.
The volume occupied by the oil between surface conditions and reservoir or other
operating changes is that of the total system; the ‘stock tank oil’ plus its associated
or dissolved ‘solution gas’. The effect of pressure on the complex stock tank liquid
and the solution gas is to induce solution of the gas in the liquid until equilibrium is
reached. A unit volume of stock tank oil brought to equilibrium with its associated
Properties of Reservoir Liquids
gas at reservoir pressure and temperature will occupy a volume greater than unity
(unless the oil has very little dissolved gas at very high pressure).
The relationship between the volume of the oil and its dissolved gas at reservoir
condition to the volume at stock tank conditions is called the Oil Formation Volume
Factor Bo. The shape of the Bo vs. pressure curve is shown in Figure 4. It shows
that above the bubble point pressure the reduction in pressure from the initial pres-
sure causes the fluid to expand as a result of its compressibility. This relates to the
chapter on Phase Behaviour where for an oil the PV diagram shows a large decline
in pressure for a small increase in volume, being again an indication of the com-
pressibility of the liquid. Below the bubble point pressure this expansion due to
compressibility of the liquid is small compared to the ‘shrinkage’ of the oil as gas is
released from solution.
The oil formation volume factor, is the volume in barrels (cubic metres) occupied in
the reservoir, at the prevailing pressure and temperature, by one stock tank barrel
(one stock tank cubic metre) of oil plus its dissolved gas.
1.2
Bo rb./stb
1.1
Pb
1.0
1000 2000 3000
Pressure (psig)
These black oil parameters, Bo and Rs are illustrated in Figure 5 a,b,&c from Craft
and Hawkins 1 reservoir engineering text., where they present the Rs and Bo curve for
the Big Sandy field in the USA. The visual concept of the changes during pressure
and temperature decrease is also presented.
PB
P01 Free Gas Free Gas
(a) Free Gas 676 Cu. Ft. 567 Cu. Ft.
2.990 Cu. Ft.
1,310 BBL 1,333 BBL 1,210 BBL 1,040 BBL 1,000 BBL
P01 = 3500 PSIA PB = 2500 PSIA P = 1200 PSIA PA = 14.7 PSIA PA = 14.7 PSIA
T01 = 160º F T01 = 160º F T01 = 160º F T01 = 160º F T01 = 60º F
A B C D E
600
567SCF/STB
500
BUBBLE POINT PRESSURE
Solution Gas, SCF/STB
AT 1200 PSIA
INITIAL PRESSURE
400 RS = 337
300
(b)
200
100
0
0 500 1000 1500 2000 2500 3000 3500
Pressure, PSIA
Figure 5 Gas to oil ratio and oil formation volume factor for Big Sandy Field reservoir
oil 1.
1.40
2500 PSIA 3500 PSIA
Formation Volume Factor, BBL/STB
1.30
BUBBLE POINT PRESSURE
1200 PSIA
INITIAL PRESSURE
BO = 1.210
1.20
(b)
1.10 14.7 PSIA & 160º F
BO = 1.040
Figure 5b
Properties of Reservoir Liquids
The reciprocal of the oil formation volume factor is called the ‘shrinkage factor bo
1
bo =
Bo
The formation factor Bo may be multiplied by the volume of stock tank oil to find
the volume of reservoir required to produce that volume of stock tank oil. The
shrinkage factor can be multiplied by the volume of reservoir oil to find the stock
tank volume.
It is important to note that the method of processing the fluids will have an effect
on the amount of gas released and therefore both the values of the solution gas-oil
ratio and the formation volume factor. A reservoir fluid does not have single Bo or Rs
values. Bo & Rs are dependant on the surface processing conditions. This simplistic
reservoir model (Figure 6) demonstrates that the black oil model description of the
reservoir fluids is an after the event, processing, description in terms of the produced
fluids. This simplistic approach to modelling reservoir fluids becomes more difficult
to consider when one is involved in reservoirs which become part of a total reservoir
system (Figure 7).
Rs
BO
Rs 1
Rs 2
Bo 1
Bo 2
Rs 3
Rs 4
Bo 3
Bo 4
Figure 7 Integrated system of reservoir common pipeline and final collection system.
Bg (Rsb - Rs)
10
Properties of Reservoir Liquids
Gas
Bg(Rsb-Rs)
B0b Oil
Bt
Oil B0
Hg
Its application comes from the Material Balance equation (Chapter 15) where it is
sometimes used to express the volume of oil and associated gas as a function of pres-
sure. It is important to note that Bt does not have volume significance in reservoir
terms since the assumption in Bt is that the system remains constant. As mentioned
earlier if the pressure drops below the bubble point in the reservoir then the gas
coming out of solution moves away from its associated oil because of its favourable
relative permeability characteristics.
Figure 8b gives a comparison of the total formation-volume factor with the oil for-
mation-volume factor. Clearly above Pb the two values are identical since no free
gas is released. Below Pb the difference between the values represents the volume
occupied by free gas.
Bt Bo
Pressure Pb
The value of BT can be estimated by combining estimates of BO and calculation of
Bg and known solubility values for the pressures concerned.
Figure 9 depicts the behaviour below the bubble point when produced gas at the
surface comes from two sources, the solution gas associated with the oil entering the
wellbore plus free gas which has come out of solution in the reservoir and migrated to
the wellbore. The total producing gas to oil ratio is made up of the two components
solution gas Rs and the free gas which is the difference. The diagram illustrates the
volumes occupied by these two in the reservoir, the solution gas being part of Bo and
the free gas volume through Bg.
Free Gas
R= Rs + (R - Rs)
& Solution Gas
Pi
Surface +
Pressure
Temperature
Oil Reservoir
Reservoir
Gas Oil
Bo rb (oil and dissolved gas) /stb
6 OIL COMPRESSIBILITY
The volume changes of oil above the bubble point are very significant in the context
of recovery of undersaturated oil. The oil formation volume factor variations above
the bubble point reflect these changes but they are more fundamentally embodied
in the coefficient of compressibility of the oil, or oil compressibility.
1 ∂V
co = −
V ∂P T
in terms of formation volume factors this equation yields
12
Properties of Reservoir Liquids
1 ∂Bo
co = −
Bo ∂P T
Assuming that the compressibility does not change with pressure the above equation
can be integrated to yield ;
V2
co ( P2 − P1 ) = − ln
V1
where P1 & P2, and V1 & V2 represent the pressure and volume at conditions 1 & 2.
Over the years there have been many correlations generated based on the two com-
ponent based black oil model characterisation of oil. The correlations are based
on data measured on the oils of interest. These empirical correlations relate black
oil parameters, the variables of Bo and Rs to; reservoir temperature, and oil and gas
surface density. It is important to appreciate that these correlations are empirical and
are obtained by taking a group of data for a particular set of oils and finding a best fit
correlation. Using the correlation for fluids whose properties do not fall within those
for the correlation can result in significant errors. Danesh 2 has given an excellent
review of many of these correlations
A number of empirical correlations, based on largely US crude oils, and other loca-
tions across the world have been presented to estimate black oil parameters of gas
solubility and oil formation volume factor. The most commonly used is Standing’s
3
correlation. Other correlations include, Lasater 4, and recently Glaso 6
Pb = f (Rs, γg, po, T)
R 0.83
Pb = 18.2 10 (0.00091T − 0.0125( API )) − 1.4
s
γ g
(2)
His correlation for the oil formation volume factor is;
1.2
γ g 0.5
Bo = 0.9759 + 0.000120 Rs + 1.25T
ρo
(3)
Other correlations have been presented by Lasater 4 based on 137 Canadian ,USA
and South American crudes, Vasquez and Beggs5 using 6000 data points, Glaso 6 us-
ing 45 North Sea crude samples, and Mahoun 7 who used 69 Middle Eastern crudes.
Danesh2 gives a very useful table showing the ranges covered by the respective black
oil correlations
14
Properties of Reservoir Liquids
2000
Tank oil gravity, ºAPI
1500 50 30 10
1000
900
800
700
600
10
1.
500
bl
r b 400
pe
20
ft 1.90
1.
300
, cu
tio 1.80
ra
il
-o 200
30
1.70
as
1.
G
150 1.60
40
1.
50
100 1.50
1.
90
80
70
1.40
60
50
Temperature, ºF
40
100 1.30
120
30 140
160
180
200 il
20 220
ko
240
1.20
Gas gravity Air=1
260
1.00
an
0.90
ft
0.80
lo
0.70
bb
0.60
0.50
er
lp
bb
16
1.10 1.40 20
1.00 1.50
0.90
0.80
0.70 30
0.60
0.50 40
G 50
10
12 as 60
14
16 70
18
gr
20 av 80
22 90
24 ity
26 100
28
G
Ai
30
32
r=
as
-
34 1
36
oi
l
38 150
40 Ta
ra
42
ti
44 nk
200
o,
46
48
c
oi
50
u
52
lg
ft
54 ra
56 300
58
pe
vi
r
60 ty,
bb
ºA
l
PI 400
500
600
700
800
gravity
900
1000
200
1500
2000
300
Te
m
pe
400
ra
t
ur
Bubble-point
e
500
,º
F
600
Pr
e
700
ss
u
800
re
,
900
ps
1000
ia
Figure 11 Gas solubility as a function of pressure. Temperature, gas gravity and oil
1500
2000
3000
(STANDING)
4000
5000
6000
240 60
220 80
260 100
200 120
180 140
160
Properties of Reservoir Liquids
8 FLUID DENSITY
Liquids have a much greater density and viscosity than gases, and the density is affected
much less by changes in temperature and pressure. For petroleum engineers it is
important that they are able to estimate the density of a reservoir liquid at reservoir
conditions.
ρo
γo =
ρw (4)
The specific gravity of a liquid is the ratio of its density to that of water both at the
same T & P. It is sometimes given as 60˚/60˚, i.e. both liquid and water are measured
at 60˚ and 1 atmos.
The petroleum industry uses another term called ˚API gravity where
141.5
° API = − 131.5
γo (5)
There are several methods of estimating the density of a petroleum liquid at reservoir
conditions. The methods used depend on the availability and nature of the data of
data. When there is compositional information on the reservoir fluid then the density
can be determined using the ideal solution principle. When the information we have
is that of the produced oil and gas then empirical methods can be used to calculate
the density of the reservoir fluid.
Exercise 1.
Calculate the density at 14.7psia and 60 ºF of the hydrocarbon liquid mixture with
the composition given below:
Component Mol.
fract.
1b mol.
nC4 0.25
nC5 0.32
nC6 0.43
1.00
Solution Exercise 1
Solution
Component Mol. Mol. Weight Liquid Liquid
density
Liquids at their bubble point or saturation pressure contain large quantities of dis-
solved gas which at surface conditions are gases and therefore some consideration
for these must be given in the additive volume technique. This physical limitation
does not impair the mathematical use of a “pseudo liquid density “ for methane and
ethane since it is only a step in its application to determine a reservoir condition
density. This is achieved by obtaining apparent liquid densities for these gases and
determining a pseudoliquid density for the mixture at standard conditions which can
then be adjusted to reservoir conditions.
Standing & Katz 8 carried out experiments on mixtures containing methane plus other
compounds and ethane plus other compounds and from this were able to determine
a pseudo-liquid (fictitious) density for methane and ethane
18
Properties of Reservoir Liquids
0.5
0.4
0.3
0.4
Apparrent density of Methane, g/cc
0.3
0.2
Ethane - N - Butane
Ethane - Heptane
Ethane - Crystal oil
0.1 Methane - Cyclo Hexane
Methane - Benzene
Methane - Pentane
Methane - Hexane
Methane - Heptane
Methane - Propane
Methane - Crystal oil
Methane - Crude oil
Figure 12 Variation of apparent density of methane and ethane with density of the system 8.
To use the correlations a trial and error technique is required whereby the density of
the system is assumed and the apparent liquid densities can be determined. These
liquid densities are then used to compute the density of the mixture by additive vol-
umes and the value checked against the initial assumption. The procedure continues
until the two values are the same.
When non hydrocarbons are present, the procedure is to add the mole fractions of
the nitrogen to methane, the mole fraction of carbon dioxide to ethane and the mole
fraction of hydrogen sulphide to propane.
Methane 44.04
Ethane 4.32 Properties of
Propane 4.05 heptane +
Butane 2.84 API gravities = 34.2
Pentane 1.74 SG = 0.854
Hexane 2.9 Mol wt = 164
Heptane + 40.11
Solution Exercise 2
This trial and error method is very tedious so Standing and Katz devised a chart which
removes the trail and error required in the calculation. The densities have been con-
verted into the density of the heavier components, C3+, and the weight percent of the
two light components, methane and ethane in the C1+ and C2+ mixtures. Figure 13.
20
Properties of Reservoir Liquids
70
50 40 30 20 10 0
terial
60
a
lus m
ne p
50 70
etha
than
40 %e
Wt 60
10
em
st
sy
30 e 50
tir
en
in
ne 20
ha
et
m 40
t%
W
30
30
20
10
2. The gas solubility , the gas composition and the surface oil gravity is known
3. The gas solubility, and gas and liquid gravities are known.
Calculate the surface density of the mixture in exercise 2 using the chart of figure 13
The compressibility and thermal expansion effects have been expressed graphically
in Figures 14 and 15.
10
9
Density of pressure minus density at 60ºF β 14.7 psia lb/cu ft
15
8
,0
00
7
10
,00 8,000
0
6
6,0 5,0 4,0
5
00
Pr
es
su
00 00
4
re
, ps
ia
3,
3
000
2,
00
2 0
1,0
00
1
0
25 30 35 40 45 50 55 60 65
22
Properties of Reservoir Liquids
10
8
Density at 60ºF minus density at temperature, lb/cu ft
7
24
0
5
22
0
Te
mp
0 e ra
20
4 ture
18 ºF
0
3 16
0
140
2 120
100
1
80
60
0
25 30 35 40 45 50 55 60 65
Density at 60ºF and pressure P, lb/cu ft
Exercise 4.
Full compositional data may not always be available and the characterisation of the
produced fluids will vary from full compositional analysis to a description of the
fluids in terms of gas and oil gravity. The procedure just described is for the situa-
tion where the composition of the reservoir fluid is known. The procedures which
follow cover the situation where a less comprehensive analysis is available. These
methods make use of empirical correlations.
Exercise 5.
3. The Gas Solubility, and Gas and Liquid gravities are known.
Katz has produced a correlation (figure 16) to enable densities to be determined when
the only information on the gas is its solubility and its gravity. The figure gives ap-
parent liquid densities of gases against gravity for different API crudes
45
Apparent Liquid density of Dissolved Gas at
40
60 F and 14.7 psia, lb/cu. ft.
35 20 API Crude
30
30
40
50
25 60
20
15
0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4
Gas Gravity
24
Properties of Reservoir Liquids
Exercise 6.
Use the correlation of Katz to calculate the reservoir fluid density of a field with
a GOR of 500scf/STB with a gas gravity of 0.8 and a 35oAPI oil for reservoir
conditions of 4,000psia and a temperature of 180oF.
Katz method
The situation for a wet gas or gas condensate is different for a conventional oil when
one is considering the volume changes taking place upon release to surface condi-
tions. For a wet gas or condensate system liquid at surface is gas in the formation.
The comparison therefore with respect to conditions in the reservoir to those at the
surface is distinctly different from an oil system, where an oil in the reservoir produces
gas and liquids at the surface. For a wet gas or condensate, a gas in the reservoir
produces gas and liquids at the surface.
To calculate the properties of the reservoir fluid from the information on the produced
fluids requires a combination of the quantities and characteristics of these fluids. The
methods used depends on the level of detail of the characteristics of the produced
fluids. A number of methods are presented using examples which vary according
to the level of detail.
Exercise 7.
A gas condensate produces gas and liquids with the compositions detailed below,
with a producing GOR of 30,000 SCF/STB. Determine the composition of the
reservoir gas.
Component Composition
Gas Liquid
Methane 0.84
Ethane 0.08
Propane 0.04 0.15
Butane 0.03 0.36
Pentane 0.01 0.28
Hexane 0.12
Heptane + 0.09
1.00 1.00
Exercise 9.
Calculate the gas condensate formation factor for the example in exercise 8.
10 VISCOSITY OF OIL
The viscosity of oil at reservoir temperature and pressure is less than the viscosity
of the dead oil because of the dissolved gases and the higher temperature. Correla-
tions are available which enable the dissolved gas and pressure effect on the dead
oil viscosity to be determined. Danesh 2 has given a good review of many of the
empirical approaches. The favoured correlations are those of Beggs and Robinson
11
,
Egbogah and Ng 12 ,Vazquez and Beggs13 , and Labedi14 . Figure 17 gives plots,
presented by McCain 17 , of the correlation of dead oil viscosity from Egbogah and
Ng 12 , and figure 18s the impact of dissolved gas from the Beggs and Robinson 11
correlation.
1000
800
700
600
500
400
300
200 100º
100
80
70 150º
Viscosity of Gas-Free Oil, µoD, cp
60
50
40 200º
30 R
250º es
20 er
300º vo
irT
10
8
em
7 pe
6 ra
5 tur
e,
4
ºF
3
1
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
10 20 30 40 50
Stock - Tank Oil Gravity, ºAPI
200
0
100
80
70
60 0
50 10
40
io
0
at
10
50
il R
8
7
-O
6
as
5
00
G
4
10
n
io
3
0
t
150
lu
So
2 0
20 0
1
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.4 0.6 0.81 2 3 4 5 6 7 8 10 20 30 40 60 80100 200 300
Beggs and Robinson 11 examined 600 oil samples over a wide range of pressure and
temperature and came up with the following correlation.
µod = 10A - 1 (6)
Examination of these correlations has shown that they are not very reliable with
errors of the order of 25% (DeGetto15)
Beggs and Robinson 11 gave a correlation to give the impact of dissolved gas.
µob = CµodB
(7)
Vazquez and Beggs13 presented an equation to take into account pressure on viscosity
above the saturation pressure.
10,000
9,000
8,000
7,000
6,000
5,000
Bubble Point pressure, Pb, psia
4,000
3,000
Pre
2.000 ssu
50 re
00 60
00
40 psi
00 a
1.000
900 30
800 00
700
600
20
500
00
400
100
300
500
0
200 200
100
60 100
90
80
40 70
60
50
40
20
Viscosity of Oil Above Bubble Point, µo, cp
30
20
10
6 p
Point, c
10
Bubble
9
f Oil At
8
4 yo
Viscosit
7
6
5
4
2
3
2
1
0.6 1.0
0.9
0.8
0.4 0.7
0.6
0.5
0.2 0.4
0.3
0.2
0.1
0.1
Labedi (ref 14) also produced an empirical correlation to determine viscosity at pres-
sures above the bubble point
µo= µob + (P/Pb-1)(10-2.488µob0.9036 Pb0.6151 /100.0197oAPI ) (9)
Danesh 2 in his text compared the various correlations from a published experimental
viscosity value in a well known PVT report, using the following exercise.
28
Properties of Reservoir Liquids
Exercise. 10
µod = 10A -1
Log A = 3.0324 - 0.0202°API - 1.163 log Tx
µod = dead oil viscosity cp.
(Beggs 3.0324 0.0202 1.163)
(Egbogah 1.8653 0.025086 0.56441)
Beggs Egbolgah
API = 40.1
T = 220
Rs = 795
P = 5,000 psig
Pb = 2,635 psig
log A = -0.5031 -0.46
A = 0.3140 0.34
Viscosity
dead oil = 1.06 cp 1.21 cp
Measured value = 1.29 cp
11 INTERFACIAL TENSION
Interfacial tension decreases as temperature and pressure increases as shown for the
effect of temperature for pure components in figure 20 from McCain’s text 17 adapted
from Katz19 data.
35
30
Surface Tension, dynes per cm
25
20
Mol wt.
240
220
15 200
180
160
10 140
0
-200 -200 0 100 200 300 400 500 600
Temperature, ºF
Methane Ethane Propane n - Pentane n - Heptane
l - Butane n - Hexane
n - Octane
n - Butane
Figure 20 Interfacial tensions of hydrocarbons. (Adapted from Katz, et al., J. Pet. Tech.,
Sept. 1943.)
30
Properties of Reservoir Liquids
There are several methods for predicting IFT, and they require experimentally determined
parameters. Work on pure compounds have shown that IFT can be related to density,
compressibility and latent heat of vaporisation. The multicomponent perspective of
reservoir fluid properties has made use of the IFT/density relationships.
The Parachor method of McLeod 18 has gained acceptance where the IFT between
vapour and liquid is related to the density difference of the respective phases.
4
ρ − ρg
σ = Pσ L
M
(10)
where ρL and ρg are the density of the liquid and gas phases, and M is the molecular
weight. σ is the IFT . Pσ is called the parachor.
Katz19 has provided the parachors for pure components as shown in the table below and
they are also presented in the figure 21 prepared by MaCain using Katz’s19 data.
Component Parachor
Methane 77
Ethane 108
Propane 150.3
i-Butane 181.5
n-Butane 189.9
i-Pentane 225
n-Pentane 231.5
n-Hexane 271
n-Heptane 312.5
n-Octane 351.5
Hydrogen 34
Nitrogen 41
Carbon dioxide 78
Parachors have been shown to have a linear relationship with molecular weight ac-
cording to the relationship;
500
400
Parachor, P
300
i - C5
200
i - C4
100
Molecular Weight
Figure 21 Parachors for computing interfacial tension of normal paraffin hydrocarbons 19.
Pσ = 0.324Tc1/4vc7/8
To apply the parachor approach to mixtures the molar averaging approach of Weinaug
and Katz20 can be used.
4
ρ ρ
σ = ∑ Pσ x j L − yj g
j ML Mg
(12)
xj and yj are the mole fractions of the components in the liquid and gas phases.
32
Properties of Reservoir Liquids
1400
1200
1000
Parachor. P
800
600
400
200
Exercise 11.
Calculate the IFT of the following volatile oil mixture at 2315 psia and 190°F for
the oil with the following composition.
It is useful to summarise the differences between the Black Oil Model approach
compared to the Compositional Model in predicted fluid properties.
The suitability of the two approaches is largely related to the nature of the fluid. For
heavier oils where there are low GOR’s as compared to volatile oils with high GOR’s,
black oil models are likely to be suitable. For the more volatile systems where there
are more significant compositional variations in a reservoir as pressure is depleted,
compositional models are considered more capable of predicting fluid behaviour.
Compositional Models
• N components based on paraffin series
• Equation of state based calculations
• Feed forward calculation of fluid properties
In a subsequent chapter on vapour liquid equilibria we will describe how the volumes
and compositions of vapour and liquid equilibrium mixtures can be calculated when
a mixture is processed at a particular pressure and temperature. These calculations
although calculation intensive can be considered feed forward calculations and en-
able the effects of temperature and pressure changes to be determined on a particular
feed mixture.
The black oil approach which has been a major theme of this chapter uses the char-
acteristics of the produced fluids to determine the composition and properties of the
feed reservoir mixture, i.e. a back calculation. As will be seen in the section on PVT
reports, the quantities and characteristics of the produced fluids are dependant on the
pressure and temperature conditions used to separate the fluid.
At the back of this chapter are tables of physical properties which are useful in many
of the procedures described.
34
Properties of Reservoir Liquids
36
Properties of Reservoir Liquids
Solutions to Exercises
Exercise 1.
Calculate the density at 14.7psia and 60 ºF of the hydrocarbon liquid mixture with
the composition given below:
Component Mol.
fract.
1b mol.
nC4 0.25
nC5 0.32
nC6 0.43
1.00
Solution Exercise 1
Solution
Component Mol. Mol. Weight Liquid Liquid
density
Exercise 2:
Calculate the “surface pseudo liquid density” of the following reservoir
composition.
Methane 44.04
Ethane 4.32 Properties of
Propane 4.05 heptane +
Butane 2.84 API gravities = 34.2
Pentane 1.74 SG = 0.854
Hexane 2.9 Mol wt = 164
Heptane + 40.11
Exercise 3.
Calculate the surface density of the mixture in exercise 2 using the chart of figure
13
Solution Exercise 3
40
Properties of Reservoir Liquids
Exercise 4.
Solution Exercise 4
Density of following reservoir liquid at 6,000 psia and 180˚F.
Step 1
Pseudoliquid density at standard conditions
from exercise 3 ρo = 45 lb/cu ft
Step 2
Adjust to 60˚F and 5,500 psia
i.e. correction = +1.9 lb/cu ft (Figure 14)
i.e. ρo = 45 + 1.9 = 46.9 lb/cu ft at 60˚F 6,000 psi
Step 3
Adjust to 180˚F. (Figure 15)
i.e. thermal correction = -3.18 lb/cu ft
ρo = 46.9 - 3.18 = 42.32 lb/cu ft at 180˚ and 6,000 psia
ρo = 42.32 lb/cu ft @ 180˚F and 6,000 psia
Exercise 5.
A reservoir at a pressure of 4,000 psia and a temperature of 200oF has a producing gas
to oil ratio of 600 scf/STB. The oil produced has a gravity of 42 oAPI. Calculate the
density of the reservoir liquid. The produced gas has the following composition
Exercise 6.
Use the correlation of Katz to calculate the reservoir fluid density of a field with a
GOR of 500scf/STB with a gas gravity of 0.8 and a 35oAPI oil for reservoir condi-
tions of 4,000psia and a temperature of 180oF.
Katz method
Solution Exercise 6.
42
Properties of Reservoir Liquids
Pseudodensity
of reservoir fluid= 328.23 / 6.779 = 48.42
Exercise 7.
A gas condensate produces gas and liquids with the compositions detailed below,
with a producing GOR of 30,000 SCF/STB. Determine the composition of the
reservoir gas.
Component Composition
Gas Liquid
Methane 0.84
Ethane 0.08
Propane 0.04 0.15
Butane 0.03 0.36
Pentane 0.01 0.28
Hexane 0.12
Heptane + 0.09
1.00 1.00
Liquid
Component Mol. Fractn Mol.Wgt. Wgt. Liquid Liquid
lb mole lb/lb mol lb/lb mole density volume
lb/cu ft cu ft
C3 0.15 44.1 6.615 31.66 0.223
C4 0.36 58.1 20.916 35.78 0.585
C5 0.28 72.2 20.216 38.51 0.506
C6 0.12 86.2 10.344 41.3 0.25
C7+* 0.09 114.2 10.278 43.68 0.235
* C8 used for C7+ 68.369 1.799
Mol.Wgt. 68.369
liq.
Density of liquid= 38.00 lb/cu ft
GOR= 30000 scf/STB 213.39 lb/STB
= 79.16 lb mole gas/STB 3.12 lb mole /STB
Note: 1 lb mole = 379 SCF
GOR = 25.36 lb mole gas/lb mole liquid
Exercise 8.
The gas condensate reservoir above is contained in reservoir sands with an average
pay thickness of 100ft, with a porosity of 0.18 and a connate water saturation of 0.16.
The aerial extent of the field is 5 sq. miles. The initial reservoir pressure is 5,000 psia
and the reservoir temperature is 180 oF. Determine the initial reserves of the field in
terms of condensate and gas.
44
Properties of Reservoir Liquids
Solution Exercise 8
Exercise 9.
Calculate the gas condensate formation factor for the example in exercise 8.
Solution Exercise 9.
Bgc = bbls of gas in reservoir/STB condensate
Volume of gas in reservoir = 6.9696 x 1010 cu ft = 1.2412 x 1010 bbls
Condensate = 6.2935 x 106 STB
Bgc = 1972.2 bbls res gas/STB condensate
In some cases full compositional information may not be available but only black
oil descriptions of the oil and gas gravity for the gas. In this case correlations can be
used to provide the necessary data to calculate the same data as for exercise 8 & 9.
Calculate the viscosity of oil in the PVT report of chapter 12 at a pressure of 5,000psig
and 220°F. The °API of the oil is 40.1 and the GOR, Rs is 795 scf/ST
µod = 10A -1
Log A = 3.0324 - 0.0202°API - 1.163 log Tx
µod = dead oil viscosity cp.
(Beggs 3.0324 0.0202 1.163)
(Egbogah 1.8653 0.025086 0.56441)
Beggs Egbolgah
API = 40.1
T = 220
Rs = 795
P = 5,000 psig
Pb = 2,635 psig
log A = -0.5031 -0.46
A = 0.3140 0.34
Viscosity
dead oil = 1.06 cp 1.21 cp
Measured value = 1.29 cp
46
Properties of Reservoir Liquids
Exercise 11
Calculate the IFT of the following volatile oil mixture at 2315 psia and 190°F for
the oil with the following composition.
Solution Exercise 11
Component xj yi Pσ Equation 12
Co2 0.0159 0.0259 78.0 -0.0050
N2 0.0000 0.0022 41.0 -0.0006
C1 0.3428 0.8050 77.0 -0.2301
C2 0.0752 0.0910 108.0 -0.0108
C3 0.0564 0.0402 150.3 0.0161
i-C4 0.0097 0.0059 181.5 0.0046
n-C4 0.0249 0.0126 189.9 0.0154
i-C5 0.0110 0.0039 225.0 0.0105
i-C5 0.0141 0.0044 231.5 0.0147
C6 0.0197 0.0040 271.0 0.0278
C7+* 0.4303 0.0049 *586.6 1.6297
1.000 1.000 1.4723
from figure 23
1. Craft,BC & Hawkins, MF. Applied Reservoir Engineering” 1959 Prentice Hall,
NY
2. Danesh, A, PVT and Phase Behaviour of Petroleum Reservoir Fluids. 1998
Elsevier. pp 66-77
3. Standing MB “A pressure-Volume-Temperature Correlation for Mixtures of
Californian Oils and Gases”, Drill & Prod, Proc.275-287 (1947)
4. Lasater, J.A. “ Bubble Point Correlation “, Trans AIME, 213,379-381 (1958).
5. Vasquez,M and Beggs,HD “Correlations for Fluid Physical Property Prediction
“ JPT,968-970, (June 1980)
6. Glaso, O “Generalised Pressure Volume Temperature Correlations” JPT,785
795 (May 1980)
7. Marhoun,MA, “PVT Correlations for Middle East Crude Oils” JPT, 650-665
(May 1988)
8. Standing, M.B. and Katz,D.L. “ Density of Crude Oils Saturated with Natural
Gas” Trans AIME 146, 159 (1942)
9. Kessler, MG and Lee,BI,: “Improved Prediction of Enthalpy of Fractions,” Hyd
Proc.(Mar.1976) 55,153-158.
10. Standing,M “Volumetric and Phase Behaviour of Oil Field Hydrocarbon Systems”
SPE Dallas 1951
11. Beggs,HD. and Robinson,JR: Estimating the Viscosity of Crude Oil Systems”
JPT,27,1140-1141 (1975)
12. Egboghah,EO and Ng,JT: ‘An improved Temperature Viscosity Correlations
for Crude Oil Systems”, J.Pet Sci and Eng.,5,197-200 (1990)
13. Vasquez,M. and Beggs,HD :” Correlations for Fluid Physical Property Predictions”.
JPT,968 (June 1980)
14. Labedi,R: “Use of Production Data to Estimate Volume Factor, Density and
Compressibility of Reservoir Fluids”, J.of Pet.Sci and Eng. 4.375-90,(1990)
15. DeGhetto,G.,Paone,F. and Villa,M.: “Reliability Analysis of PVT Correlations
“,SPE 28904, Proc of Euro.Pet Conf. Lndn, 375-393 (Oct.,1994)
16. Danesh,A.,Krinis,D.,Henderson G.D., and Peden,J>M> “Visual Investigation
of Retrograde Phenomena and Gas Condensate Flow in Porous Media” 5th
European Symposium on Improved Oil Recovery ,Budapest (1988)
17. McCain,WD., “The Properties of Petroleum Fluids” Pennwell Books ,Tulsa,
Ok 1990. ISBN 0-87814-335-1
18. Macleod, DB., “On a Relation Between Surface Tension and Density,” Trans.,
Faraday Soc. (1923) 19,38-42.
19. Katz,DL.,”Handbook of Natural Gas Engineering”, McGraw Hill Book Co
Inc., New Yk,(1959)
20. Weinaug,KG and Katz,DL,: “Surface Tension of Methane-Propane Mixtures”.
I&EC,239-246 (1943)
21. Firoozabadi,A , Katz,D.L., Soroosh,H.M and Sajjadian,V.A.: “Surface Tension
of Reservoir Crude-Oil/Gas Systems Recognising the Asphalt in the Heavy
Fraction,” SPE Res Eng.(Feb) 1988,3,No 1, 265-272.
48
fundamental Properties of Reservoir Rocks
CONTENTS
INTRODUCTION
4. PERMEABILITY
4.1 Darcy's Law
4.2 Factors Affecting Permeability
4.3 Generalised Form of Darcy's Law
4.4 Dimensions of Permeability
4.5 Assumptions For Use of Darcy's Law
4.6 Applications of Darcy's Law
4.7 Field Units
4.8 Klinkenberg Effect
4.9 Reactive Fluids
4.10 Average Reservoir Permeability
6. POROSITY - PERMEABILITY
RELATIONSHIPS
LEARNING OBJECTIVES
Having worked through this chapter the Student will be able to:
• Define porosity and express it as an equation in terms of pore, bulk and grain
volume.
• Explain the difference between total and effective porosity.
• Define permeability and present an equation, Darcy’s Law, relating flow rate to
permeability in porous media.
• List the assumptions for the applicability of Darcy’s Law.
• Derive an equation based on Darcy’s Law relating flow of gas in a core plug
and the upstream and downstream pressures.
• Derive an equation relating flow rate to permeability for a radial incompressible
system.
• Comment on the difference between gas and liquid permeability (Klinkenberg
effect ).
• Sketch a figure relating liquid permeability to gas permeabilities plotted as a
function of reciprocal mean pressure.
• Briefly describe the impact of reservoir stresses on permeability and porosity
• Draw a sketch demonstrating the result of interfacial tension between oil, water
and a solid, and locate the contact angle and define its values for wetting and
non-wetting phases.
• Express the capillary pressure Pc as two equations, one in terms of interfacial
tension, contact angle and pore radius, and the other in terms of height and
density of fluids.
• Define the free water level.
• Draw the Pc or height vs. saturation capillary pressure curve and identify
significant features.
• Sketch and explain the impact of saturation, history, density difference and
interfacial tension in capillary pressure curves.
• Sketch the impact of capillary pressure effects on the saturation distribution of
stratified formations
• Define effective and relative permeability and plot typical shapes.
• Define imbibition and drainage in the context of capillary pressure and relative
permeability curves.
• Sketch the pore doublet model and use it to explain the retention of trapped oil
in large pores and briefly relate it to the principle behind some enhanced oil
recovery methods.
• Define mobility ratio.
• Sketch a shape for gas- oil relative permeability curves.
fundamental Properties of Reservoir Rocks
Introduction
The properties of reservoir rocks with respect to the fluids they contain and with respect
to the fluids which will be injected into them are important when characterising a
reservoir in terms of its reserves and the mobility of the fluids. This next section gives
a brief over view of these properties, and is followed by chapters on their measurement
and variation. In relation to the detailed description of rock characteristics the reader
is referred to the Geology module of this Petroleum Engineering course.
The reservoir engineer is concerned with the quantities of fluids contained within the
rocks, the transmissivity of fluids through the rock and other related properties.
The specifications of a reservoir rock are such that there must be a large enough
capacity to store economically viable amounts of hydrocarbon and the hydrocarbon
must flow at economical rates when penetrated by a well. The factors which may
affect the capacity and the flow properties are the porosity, permeability, capillary
pressure, compressibility and fluid saturation. In the case of a reservoir rock, these are
not standard characteristics determined before formation of the rock, but are closely
linked to the geological processes that brought the sediments together and deposited
them in the sequences and under the chemical and physical changes inherent in the
system.
In order to extract the fluids the rock must be permeable which requires that there be
sufficiently large, interconnecting pores.
Although a permeable rock must also be porous, a porous rock is not necessarily
permeable. Certain volcanic rocks are porous but not permeable because the voids are
not interconnecting; shale may be quite porous but impermeable because the pores are
extremely small, thereby preventing free movement of the fluids contained within.
Considering a common reservoir rock, sandstone, the grains making up this rock
are all irregular in shape. The degree of irregularity, or lack of roundness reflects
the source sediments and the physical and chemical processes to which they were
subsequently exposed. Violent crushing or grinding action between rocks causes
grains to be very irregular and sharp-edged. The tumbling action of grains along the
bottom of streams or seabeds smoothes sand grains. Wind-blown sand, as occurs in
moving dunes in deserts, results in sand grains that are even more rounded. Sand
grains that make up sandstone beds and fragments of carbonate materials that make
up limestone beds do not fit together congruently: the void space between the grains
forms the porosity.
For the reservoir engineer, porosity is one of the most important rock properties as
a measure of the space available for accumulation of hydrocarbon fluids.
3. Porosity
The first step in forming a sandstone, for example, is to have a source of material
which is eroded and transported to low lying depressions and basins such as would
be found off the coasts of a landmass. The material would consist of a mixture of
minerals, but for a sandstone, the majority would be made of quartz in the form of
grains. When these were deposited, they would be surrounded by sea water or brine,
and as the sediment thickness increased, the weight or the pressure produced by the
overlying sediments would force the grains together. Where they contacted each other
large stresses would be produced and a phenomenon called pressure solution would
occur which dissolved the quartz at the points of contact between the grains until the
stresses reduced to a level which was sustainable by the grains. The dissolved material
would be free to precipitate in other regions of the sediment. In this way the initially
loose material would be solidified with discrete connections between the grains.
Initially, if subsea, the pore spaces would be filled with brine, and as the lithification
process occurred, some pore spaces would be isolated with the brine trapped inside.
If the vast majority were interconnected then the initial pore fluid would be free to be
swept through the rock by other fluids such as hydrocarbons. In this way the geometry
of the grains produces an assembly of solids with voids in between them. The grains
vary in diameter but may be from a few microns to several hundred microns. The
geometry of the pore spaces is such that they have narrow entrances (pore throats)
where the edges of the grains touch each other and larger internal dimensions (between
the grains). The complicated nature of these interconnected pores is illustrated in
figure 1 which is a metal cast of the pores in a sandstone rock.
fundamental Properties of Reservoir Rocks
One method of classifying reservoir rocks, therefore, is based on whether pore spaces
( in which the oil and gas is found) originated when the formation was laid down
or whether they were formed through subsequent earth stresses or ground water
action.
The first type of porosity is termed original porosity and the latter, secondary or
induced porosity. This is illustrated in figure 2.
Cementing material
Sand grain
Isolated porosity 5%
The impact of isolated pore space clearly cannot contribute to recoverable reserves
of fluid nor contribute to permeable pore space as illustrated in figure 3.
dead End
Permeable Pore Space
Pore
Figure 3 Total, effective, isolated permeable and dead end pore space
Porosity is defined as the ratio of the void space or pore space (Vp) in a rock to the
bulk volume (Vb) of that rock and it is normally expressed as a percentage of total
rock volume. The porosity is usually given the symbol φ. The matrix volume is the
volume of the solid grains, Vm.
Void volume
Porosity = x 100
Bulk volume
pore volume
Porosity = × 100
void volume + grain volume
fundamental Properties of Reservoir Rocks
Total porosity is defined as the ratio of the volumes of all the pores to the bulk of a
material, regardless of whether or not all of the pores are interconnected. Effective
porosity is defined as the ratio of the interconnected pore volume of a material.
If the grains are represented by spheres stacked together as in figure 4, then the pore
space can be seen between the solid grains.
Induced or Secondary Porosity = porosity from fractures or vugs (large chambers formed
in certain carbonates and limestones caused by groundwater flow and dissolution).
The length of each side would be 2x radius, giving the bulk volume as:
Vb = (2r)3 = 8r3
The grain volume would be the equivalent of the volume of one sphere
4πr 3
Vm =
3
4πr 3
8r 3 −
φ=
Vb − Vm
= 3 = 1 - π = 0.476
3
Vb 8r 6
If the spheres fit in the cusps generated by the lower layer then a porosity of 26%
occurs. For a size distribution of spheres the ultimate minimum porosity would be
zero which would be the case if sufficient grains were available to completely fill the
pore spaces as shown in figure 6 for part filling of the void.
fundamental Properties of Reservoir Rocks
Several factors may combine to affect the porosity of a rock, but the main distinction
to be made is as follows based on the amount of connected pore volume, and whether
the pore space has been altered by dissolution or by fracturing after deposition and
lithification.
Pore Space
Limestone formations may have intergranular porosity. However, the pore openings
are more often inter-crystalline, that is spaces between microscopic crystals. They also
may take the form of pits or vugs caused by solution and weathering, or by shrinkage
of the matrix. These forms of porosity are called secondary porosity. Another type of
secondary porosity is that caused by fracturing and is very important in that it permits
many limestone rocks of otherwise low porosity to become excellent reservoirs.
Porosity may range from 50% to 1.5% and actual average values are listed below:
A point that needs to be emphasised is that the concept of ‘porosity’ is complex and
therefore difficult to define and determine. It may refer to spaces between sand grains
or it may refer to limestone caves: it may even exclude a fraction of the free water
(water not bound chemically) present in the rock. Sometimes good estimates, (i.e.
relevant to reservoir development problems) may be obtained from laboratory studies,
or core samples, and sometimes such measurements are irrelevant.
In general the downhole porosity may be related to the acoustic and radioactive
properties of the rock.
10
fundamental Properties of Reservoir Rocks
energy density is related to the number of atoms and therefore to the density of the
rock being bombarded. If the formation under test is known, for instance a sandstone,
then changes in the density measured within the sandstone result from a change in
the porosity of the formation rather than a change in the mineralogical nature of the
sandstone. This obviously relies on a good description of the geology of the formation.
In a porous formation, the pore fluid will also affect the response of the tool in that
the atoms of the fluid will also react to the bombardment and affect the energy
detected. With reference to calibration samples of different rock types, the effect of
both mineralogy and pore fluid content can be accounted for. Empirical relationships
have been developed to relate the porosity to the values of density which have been
logged. In the following relationship, the logged density, ρL, matrix density, ρm , and
the fluid density, ρf , are related to the porosity, φ
ρL = ρm (1- φ ) + ρf φ
ρ L − ρm
φ =
ρ f − ρm
The contribution of the matrix and the pore fluid are in relation to the relative amounts
of each, and these are related to the porosity. Typically, matrix densities and fresh
water density are as follows
∆TL − ∆Tm
φ =
∆Tf − ∆Tm
The contribution of the matrix and the pore fluid are in relation to the relative amounts
of each, and these are related to the porosity. Typically, matrix travel times and fresh
∑φ i
φa = i =1
n (1)
where φa is the mean porosity, φi is the porosity of the ith core measurement and n is
the number of measurements.
4 PERMEABILITY
The permeability is a term used to link the flowrate through and pressure difference
across a section of porous rock. The problem is complicated in that the number of
pore spaces, their size and the interconnections is not standard. Thus the application
of the general energy equation, for example as in the case of flow through pipes,
becomes very difficult for flow through porous media.
In petroleum engineering the unit of permeability is the Darcy, derived from the
empirical equation known as Darcy’s Law named after a French scientist who
investigated the flow of water through filter beds in 1856. His work provided the
basis of the study of fluid flow through porous rock.
12
fundamental Properties of Reservoir Rocks
k∆P. A
Q=
µL (2)
where:
Darcy’s law of fluid flow states that rate of flow through a given rock varies directly
with the pressure applied, the area open to flow and varies inversely with the viscosity
of the fluid flowing and the length of the porous rock. In terms of equating the
parameters, the constant of proportionality in the equation is termed the permeability.
The unit of permeability is the Darcy which is defined as the permeability which
will permit a fluid of one centipoise viscosity (= viscosity of water at 68°F) to flow
at a linear velocity of one centimetre per second under a pressure gradient of one
atmosphere per centimetre. Permeability has the units Darcys. Figure 8 illustrates
the concept and the units of permeability
∆p = 1 atmos
Q = 1 cm 3
sec 1cm 2
µ = 1 cp
L = 1 cm
k = 1 darcy
Sand
Length, L h1 h2
Flowrate, Q
His results showed that the flowrate was directly proportional to the area open to flow,
the difference in pressure and inversely proportionate to the length of the sandpack,
i.e.
1
Q ∝ A, ∆h,
L
or
A(h1 − h2 )
Q=k
L
where Q is the flow rate, A is the area of the end of the core, h1 and h2 are the static
heads of water at the inlet and outlet of the core (the equivalent of the static pressure),
L is the length of the core. K is the constant of proportionality. It is constant for a
particular sand pack. When other workers replicated the experiment, the results were
different to those of Darcy. This was accounted for by inclusion of the viscosity of
the flowing fluid and the equation becomes:
kA(h1 − h2 )
Q=
µL
where the original terms have the same meaning and µ is the viscosity of the fluid
in centipoise.
On a more theoretical basis, Poiseuille formulated the relationship between flow rate
and pressure drop for fluid flowing in a pipe. The form of the relationship is
πr 4 ∆P
Q= (3)
8µL
where Q is the flowrate, r is the radius of the tube, µ is the viscosity of the fluid and
L is the length of the tube. In this case the dependence of the flowrate / pressure drop
relationship can be seen to be dependent on the radius of the tube. In a similar manner,
the radius of the pores in a rock dictate the nature of the relationship, specifically, the
14
fundamental Properties of Reservoir Rocks
radius of the pore throats is of most significance, since these are the smallest radii
and therefore affect the flowrate/ pressure drop relationship most.
The practical unit is the millidarcy (mD) which is 10-3 Darcy. Formation permeabilities
vary from a fraction to more than 10000 milli-Darcies. At the low end of the range,
clays and shales have permeabilities of 10-2 to 10-6 mD. These very low permeabilities
make them act as seals between more permeable layers.
If sand grains of generally flat proportions are laid down with the flat sides non-
uniformly positioned and located in indiscriminate directions, both porosity and
permeability may be very high. To illustrate, if bricks are stacked properly, the space
between the bricks is very small; if the same bricks are simply dumped in a pile, the
space between the bricks might be quite large.
The shape and size of sand grains are important features that determine the size of the
openings between the sand grains. If the grains are elongated, large and uniformly
arranged with the longest dimension horizontal, permeability to fluid flow through the
pore channels will be quite large horizontally and medium-to-large vertically. If the
grains are more uniformly rounded, permeability will be quite large in both directions
and more nearly the same. Permeability is found generally to be lower with smaller
grain size if other factors (such as surface tension effects) are not influential. This
occurs because the pore channels become smaller as the size of the grains is reduced,
and it is more difficult for fluid to flow through the smaller channels.
-Z
s Vs
+x
0
+y
+Z
The x and y co-ordinates increase from zero to the left and out from the page; the
z co-ordinate increases downwards. The flow velocity in a particular direction can
be defined as the flowrate in that direction divided by the area open to flow. In any
direction, s, the flow velocity is termed Vs and is equated to the static pressure gradient
in that direction (i.e. the change in pressure, dP, over a small element of length, ds in
that particular direction) minus a contribution from the difference in head (because
of the difference in elevation) of the fluid across the section ds. Therefore,
k dp ρg dz
Vs = - ( − )
µ ds 1.0133 x10 ds
6
(4)
16
Q=k
L
kA(h1 − h 2 )
Q=
µL fundamental Properties of Reservoir Rocks
πr 4 ∆P
Q=
8µL
1 − h2 )
A(hand the change in elevation head is equal to the sine of the angle to the horizontal
k dp Q
ρg= k dzL
Vs = - ( − )
µ ds 1.0133x10 ds
6
= sine θ, where θ is in degrees.
kA(h1 − h 2 )
Q= The Darcy units are:
dz µL
= sin θ , where θ is in degrees.
ds
Vs = velocity along path s - cms-1
πr ∆P k 4
= permeability - Darcys
K dP Qρg= dz
µ = viscosity - centipoise
Vs = - ( − 8µL )
6
µ ds 1.0133x10 ds ρ = density of fluid - gcm-3
g = acceleration due to gravity - 980 cms-2
k dp ρg dz
L MVs = - µ ( M − 6
)
Vs = µ = ρ = ds3 1.0133x10 ds
L = 6pressure gradient along s−2- atm cm
-1
T LT
1.0133 x 10 converts from dynes cm to atmospheres
dz
M L dP = sinMθ , where θ is in degrees.
P = 2 g = ds = 2 4.4 Dimensions Of Permeability
LT
2
T ds LT
2
k dp ρg dz
Vs = - ( − )
From Darcy’s equation, µ ds 1.0133 x10 ds the dimensions of each
6
K dP ρg dz
L kLT M ML ( −can be deduced
Vs = - term 6 )in terms of length, L, mass, M and time, T
= ( 2 2 − 3 2 ) µ ds 1.0133x10 ds
T M LT LT
L M M
L K Vs = µ = ρ = 3
= L MT M LT L
T LT Vs = µ = ρ = 3
K = L2 T LT L
M L dP M
P = 2
g = 2 = 2 2
M LT
L dP M T ds LT
K dP P ρg
= 2 dz
g = 2 ρg
= 2 2 dz
Vs = - ( − LT ), T ds L T6 = zero
µ ds 1.0113x106 Therefore,
ds 1.0113x10
the equationds in terms of the dimensions (and keeping permeability as k)
is
L kLT M ML
Q = ( 2 2 − 3 2)
Vs = Vx = T M L LT L kLT
T M ML
A = ( 2 2 − 3 2)
L K T M LT LT
kA dP =
Q = - T LT
µ dx
K = L2 L K
=
T LT
L
kA P2 ρg ρg
Q∫ dx = - ∫ dP V = - K ( dP − dz
),
dz
= zero
µ P1 s
µ ds 6 6
0 K =1.0113x10
L2 ds 1.0113x10
ds (5)
kA
Q(L - 0) = - (P − P )
µ 2 1 It can Q be seen that the dimensions reflect the nature of the constant of proportionality
Vs = Vx and = it should not be confused with, for example, the area open to flow, A, of the end
kA(P1 − P2 )
Q = of aAcore or a sand pack. In terms of metric units, since 1 atm = 14.73 psi = 1.013
(6)
µL kA dP
Q = - bar and 1 cp = 10 Pas it follows that
-3
µ dx
ρg 1Ddz = 9.87 xdz10 m2 ~ 1 x 10 m
-13 -12 2
dP ρg
Vs = Vx = - k( − 1mD
6 ), = 9.876 x 10=-16m
zero
2
~ 1 x 10-15m2
ds L1.0113x10kA P2ds 1.0113x10 ds
Q∫ dx = -
µ P∫1
Q dP
Vs = 0
A
kAof Petroleum Engineering, Heriot-Watt University
Institute 17
kA dP Q(L - 0) = - (P2 − P1 )
Q = - µ
µ dx
Other units of inches2 or cm2 could be used but they are all too large for porous media
and they would also require conversion to relate to permeabilities quoted in other
units. Darcys and milliDarcys are most commonly used.
2. Laminar flow, i.e. no turbulent flow. For most reservoir applications this is valid
however near to the well bore when velocities are high for example in gas production
turbulent flow occurs. Sometimes it is termed non- darcy flow. Figure 13
Laminar Flow
Turbulent Flow
Q
A
Q k . ∆P
=
A µ L
Q . L .
∴ K = µ
A ∆P
∆P
L
18
fundamental Properties of Reservoir Rocks
3. Rock 100% saturated with one fluid, i.e. only one fluid flowing.
In the laboratory this can be achieved by cleaning cores, however, there will be a
certain connate water saturation in the reservoir, and there may be gas, oil and mobile
water flowing through the same pore space. The concept of relative permeability can
be used to describe this more complex reservoir flow regime. Relative permeability
is discussed later.
4. Fluid does not react with the rock, i.e. it is inert and there is no change to the pore
structure through time.
There are cases when this may not happen, for example when a well is stimulated
during an hydraulic fracturing workover. The fluids used may react with the minerals
of the rock and reduce the permeability. In such cases, tests on the rock to determine
the compatibility of the treating fluids must be conducted before the workover.
5. Rock is homogeneous and isotropic, i.e. the pore structure and the material
properties should be the same in all directions and not vary. In reality, the layered
nature and large areal extent of a reservoir rock will produce variations in the vertical
and horizontal permeability.
Q
P2
L
P1
K dP ρg dz ρg dz
Vs = - ( − ), = zero
µ ds 1.0113 x10 ds
6 6
1.0113 x10 ds
Q
Vs = Vx =
A
kA dP
Q = -
µ dx
L P
kA 2
Q ∫ dx = -
µ P∫1
dP
0
kA
Q(L - 0) = - ( P2 − P1 )
µ
kA( P1 − P2 )
Q =
µL (6)
The final form is as formulated by Darcy and the permeability will have the units of
Darcys if the other units are:
The flow regime is the same as for the linear liquid system and from the basic Darcy
equation:
dP ρg dz ρg dz
Vs = Vx = - k ( − 6
), 6
= zero
ds 1.0113 x10 ds 1.0113 x10 ds
Q
Vs =
A
kA dP
Q = -
µ dx
In this case, the laboratory measurement of the gas flow would usually be conducted
downstream from the core at almost atmospheric conditions (i.e. there would not
be a large pressure drop across the flow meter). It is assumed that the gas used is
ideal, however, there needs to be a correction to the volumetric flow rate measured
to account for the higher pressure in the core. Figure 15.
20
fundamental Properties of Reservoir Rocks
P1 P2 Pb
P
L
A
Qb
Valve
Core Flow
measurement
QP = Q b Pb
Q b Pb
Q =
P
and substituting into the equation, separating the variables and integrating
produces
Q b Pb kA dP
=-
P µ dx
L P
kA 2
Q b Pb ∫ dx = -
µ P∫1
PdP
0
kA ( P22 − P12 )
Q b Pb (L - 0 = -
µ 2
2 µQb Pb L
k=
A( P12 − P2 2 ) (8)
Comparing the two expressions equations 6 and 7, it is seen that the gas flow rate is
proportional to the difference in the pressure squared, whereas the liquid flowrate
is proportional to the difference in the pressure. In well testing, the flow rates are
measured at the surface and for gas wells one of the diagnostic plots is the flowrate
versus difference in pressure squared plot. Neglecting the fact that the gas is real, it
gives an indication of the ability of the reservoir to produce gas.
P1 + P2
P=
2
and
Q = Volume flow rate at P
P Q = PbQb
and since
1 1 kA
(P1 + P2 )Q = (P1 − P2 )(P1 + P2 )
2 2 µL
kA( P1 − P2 )
Q=
µL (9)
The ideal gas permeability can be calculated from the liquid equation using mean
flowrate, Q measured at mean pressure.
Radial flow
re
Pe
rw
re
Pw h
rw
Well
Plan Elevation
Figure 16 Radial geometry with radial flow from the outer boundary to the wellbore
22
fundamental Properties of Reservoir Rocks
k dP ρg dz ρg dz
Vs = - ( − ), = zero
µ ds 1.0113 x10 ds 1.0113 x10 ds
6 6
Q
Substituting for flow velocity, Vs = Vr =
A
In this case the direction of flow is in the opposite sense to the co-ordinate system,
therefore
ds = -dr
A = 2πrh
Q k dP
= -
2πrh µ − dr (10)
Q k
(ln re − ln rw ) = ( Pe − Pw )
2πh µ
2πkh( Pe − Pw )
Q =
r
µ ln e
rw (11)
In this case the geometry is identical to that of the radial flow of incompressible
fluid with the modifications for the compressibility of a gas as per the linear gas
flow system.
Q k dP
= -
2πrh µ − dr
QP = Q b Pb
Qb Pb
Q =
P
Q b Pb k dP
= 2πrh
P µ dr
r 2πkh Pe 2 − Pw 2
Q bPb ln e =
rw µ 2
πkh
Qb =
r
(Pe 2 − Pw 2 )
µPb ln e
rw (10)
24
fundamental Properties of Reservoir Rocks
KA( P1 − P2 )
In order to convert the Darcy equation for liquid flow, Q =
µL
929cm 2 atm
( K )( Aft 2 )( 2
)( ∆Ppsia)( )
ft 14.696 psia
=
30.48cm
( µ )( Lft )( )
ft
bbl KA( P1 − P2 )
Q = 1.1271
day µL
and these produce the following version of Darcy’s equation in field units:
bbl KA( P1 − P2 )
Q = 1.1271
day µL (11)
Since the mean free path is a function of the size of the molecule, the permeability
is a function of the type of gas used in the permeability measurement. This gas
permeability is corrected for the Klinkenberg effect by plotting the gas permeability
at each reciprocal mean pressure. This is illustrated for hydrogen, nitrogen and carbon
dioxide in figure 17:
80
Gas Permeability: Millidarcies
60
40
Hydrogen
Liquid permeability
Nitrogen
20
Carbon Dioxide
0
0 1 2 3 4 5
Reciprocal Mean Pressure: (Atm.)
Pm is the mean pressure of the gas (the mean of the upstream and downstream pressures
either end of the core orp in figure 15). In effect, if the gas pressure is raised infinitely
high, the gas will perform as an incompressible liquid would, therefore if several
measurements of permeability are made at different mean pressures, the relationship
between mean pressure and permeability can be extrapolated to the equivalent
pressure conditions of a liquid. In reality, extrapolation to infinity is impossible, so
the reciprocal mean pressure is used and the results are extrapolated to zero reciprocal
mean pressure (i.e. 1/infinitely high mean pressure). This point corresponds to the
liquid permeability. The different gasses have different slopes, but they all extrapolate
to the same equivalent liquid permeability.
kG
kL =
b
l+
Pm (12)
where
kL = equivalent liquid permeability
kG = permeability to gas
Pm = mean flowing pressure
26
fundamental Properties of Reservoir Rocks
b = Klinkenberg constant for a particular gas and rock (slope of the gas permeability,
inverse mean pressure relationship).
The Klinkenberg effect is greatest for low permeability rocks and low mean
pressures.
It is worth considering the stresses associated with reservoir rock parameters. Figure
18 illustrates the likely configuration of a core extracted from a vertical well, and the
orientation of the core plug extracted for permeability and porosity measurements.
4 Inch
Formation
Within a reservoir the stresses in the formation can be expressed in three directions,
the major and two minor principal stresses. Figure 19a. The major principal stress
acting mainly in the vertical direction. Clearly the depositional environment and
formation structure will result in slight changes to these orientations.
28
fundamental Properties of Reservoir Rocks
Equal Stresses
Kh
(b)
(c)
Equal Stresses
In core analysis, service companies have been asked to measure porosity and
permeability under reservoir stress conditions. They have done this by applying
different stresses for the axial and radial stresses. As can be seen in Figure 19b for a
conventional plug the radial stress would be a combination of the major and a minor
principal stress. To enable the true stress field to be represented, a varying radial
stress distribution would be required. If a vertical plug was used, Figure 19c, then a
constant radial stress could be an acceptable value for the average minor stresses. In
this case, however, the permeability value would be Kv, the vertical permeability.
The effect of the overburden and the pore pressure on the matrix is to produce a net
force between the grains of the matrix (which, when the area over which the force
acts is accounted for produces a net stress). If the matrix is considered to be elastic,
that is, there is a unique relationship between the stress and the strain within the
matrix, then the matrix will strain as the stress is altered. If the stress increases, the
1.0
?Well Cemented
.8
Permeability: Fraction of Original
.6 ?Friable
?Unconsolidated
.4
.2
0
0 2000 4000 6000 8000 10000
Net Overburden Pressure: PSI
In general, the stress regime subsurface is considered to be hydrostatic (as in the case
of the pore fluid) and that the stresses can be resolved into one vertical stress, and
two horizontal stresses. For hydrostatic conditions, all of these are the same. In core
analysis, therefore, the porosity at equivalent subsurface conditions may be determined
by applying an external pressure to the core. This is usually done by inserting the core
into a cell rated for pressures up to 10000 psi (68.9MPa) and applying a stress to the
ends of the core and to the sides. The nature of these tests are such that usually the
stress applied to the sides of the core represents the horizontal stress and the stress
applied to the ends represents the vertical stress. Once trapped inside the cell, the
pore pressure may be increased to a representative level and measurements of pore
volume and permeability made under these stress conditions.
30
fundamental Properties of Reservoir Rocks
More recently, the effect of non-hydrostatic stress conditions has been shown to
be important in certain reservoir conditions, such as in tectonically active areas
(Columbia, South America where the formation of the Andes mountains is associated
with large horizontal stresses) or in areas associated with faults or very compressible
reservoir rocks such as some chalks. In this case the conventional test cells are not
appropriate and special true triaxial cells are required. In these cells the ends of the
core are subjected to the vertical stress as per the conventional cells, but the sides of
the core are wrapped in a cage of individual tubes which can be pressurised in banks
around the core to represent the different horizontal stresses.
In summary, when the properties of the cores are measured in the laboratory, they
can be subjected to
Hydrostatic stresses The effect of the magnitude of the stresses are measured
Real stress behaviour The effect of the magnitude and direction of the stresses
are measured
where Pcompacting is the grain-to-grain stress, Poverburden is the stress produced by the
weight of the overburden at a particular depth and Ppore pressure is the pressure of the
fluids in the pores. The expression shows the balance between the overburden and
the pore pressure in compacting the rock matrix: if the pore pressure declines, the
compacting stress increases and the pore volume declines. This assumes that the
overburden remains constant which is logical over the time period of a producing
reservoir. The balance can be represented by figure 21:
Po
Cap Rock Depth
Pf and Pc
Reservoir
Pc Pf
Pc
Grains
Pc
Pc
Pore space filled with fluid
Figure 21 The balance between overburden & rock stress and fluid pressure
Po = Pf + Pc Po = overburden pressure
Pf = fluid pressure
Pc = compacting stress
The effect of the change in the balance between the overburden stress and the pore
pressure is to change the compacting stress. If there is an increase in pore pressure,
then the pore volume will increase, however, this is rare and in the main, pore
pressure declines during production and the pore spaces compact under the increasing
compact stress. Two issues are significant: the initial porosity in the reservoir (i.e.
to correctly define the volume of oil in place) and the reduction in that porosity (or
pore volume) as the pressure declines (for material balance and simulation studies).
Figure 22 shows the relationship between porosity and depth (or stress). As the depth
(and stress) increases, the porosity declines. Care needs to be taken when assessing
porosity values: were they measured under overburden or at ambient conditions?
The shale sample shows a large change in porosity as the plate-like clay minerals
are compacted and fit together in a more congruent manner.
32
fundamental Properties of Reservoir Rocks
50
40
Sandstone
Porosity, φ
30
20
10 Shale
0
0 3000 6000
Depth of burial (ft) or stress (psi)
The rate of change of pore volume with pressure change can be represented by an
isothermal compressibility (assuming temperature is constant):
1 dv
Cf = -
v dP (15)
(i) Matrix volume compressibility - the change in volume of the rock grains. This is
very small and usually not of interest in sandstones since it is a purely mechanical
change in volume of the very stiff grains.
(ii) Bulk volume compressibility - the change in the unit volume of the rock. This
is of interest in reservoirs near the surface because of the problem of subsidence;
Changes in volume of the reservoir around faults which may cause the fault to slip
and alter the conductivity both through the fault and across it;
(iii) Pore volume compressibility - change in pore volume. This is of greatest interest
since the pore volume affects the porosity which affects reservoir performance.
Pipette
Sealed core
Pump
Pressure vessel
Figure 23 Measurement of the reduction inpore volume as the external stress (or compact-
ing pressure) is increased
The results show the change in pore volume relative to the original pore volume,
for a given change in the compacting pressure (this assumes that changes in the
compacting pressure have the same effects as changes in the pore pressure) which
can be substituted in to the isothermal compressibility as
1 dv p
Cp = -
v p dPc
where:
Cp = pore volume compressibility
vp = initial pore volume
dvp = change in pore volume (amount of fluid expelled)
dPc = change in compacting pressure
34
fundamental Properties of Reservoir Rocks
Typical values of pore compressibility are in the range 3 x10-6 psi-1 to 10 \x10-6 psi-1,
however, soft sediments can have compressibilities in the range 10 \x10-6 psi-1 to 20x10-6
psi-1 or 30 *10-6 psi-1. Figure 24 illustrates the values determined for some limestones
and sandstones.
10
3
0 10 20
Porosity %
100
Permeability (mD)
10
1
0 20 40 60 80
Hydrostatic stress (MPa)
Figure 25 The reduction in permeability for a range of sandstone samples (the porosity is
in the range 15% to 22%)
In true triaxial stress regimes, the stresses are not identical and the strain (and therefore
pore throat radii) may cause the sample to dilate in one direction and increase the
pore throat radii therefore enhancing the permeability. This can be illustrated better
by considering a fractured core (figure 26).
36
fundamental Properties of Reservoir Rocks
σ
v
Fracture
Permeability
σh maximum
σh maximum σh minimum
σv
Fracture
Permeability
σh maximum
Core Fracture opening under stress
σh minimum σ
h maximum
If the largest horizontal stress acts across the fracture (i.e. perpendicular to the faces of
the fracture) then it will be clamped shut; if the largest horizontal stress acts parallel
to the fracture, then it may split open. In this way the anisotropy (or difference in the
properties) may lead to different permeabilities and porosities from the same sample
if the stresses are applied in different ways around the core.
6. Porosity-Permeability Relationships
Whereas for porosity there are a number of downhole indirect measurement methods,
the same is not the case for permeability. The downhole determination of permeability
is more illusive. Down hole permeability is mainly obtained by flow and pressure
determination and requires other characteristics for example the flowing interval.
There has been a continued interest in porosity-permeability correlations, on the
basis if one has a good correlation of laboratory measured porosity and permeability
then down hole measurements of porosity could unlock permeability values for
those formations where recovered core has not been practical. Although porosity is
an absolute property and dimensionless, permeability is not and is an expression of
flow which is influenced by a range of properties of the porous media, including the
shape and dimensions of the grains and the porosity. Since porosity is an important
parameter in permeability it is not surprising for those rocks which have similar particle
characteristics that a relationship exists between porosity and permeability. Figure
27 below gives examples of permeability correlations for different rock types.
?Oolitic Limestone
Sucrosic Dolomite
Reef Limestone
Well Cemented
Hard Sand
100
Permeability: Millidarcies
Chalky
Limestone
10
Intercrystalline
Limestone and
Dolomite
Fine Grained
Friable Sand
1.0
0 5 10 15 20 25 30 35
Porosity: Percent
7 Surface Kinetics
If core for a particular section cannot be recovered, or for example is formed as a pile
of sand on the rig floor, then correlations like these in figure 27 are used. Porosity
measurements obtained indirectly from wireline methods can be used to obtain the
laboratory porosity vs down hole porosity cross plot. Using this laboration porosity
value the associated permeability value can be determined from an appropriate
correlation as in figure 27.
The simultaneous existence of two or more phases in a porous medium needs terms
such as the capillary pressure, relative permeability and wettability to be defined.
With one fluid only one set of forces needs to be considered: the attraction between
the fluid and the rock. When more than one fluid is present there are three sets of
active forces affecting capillary pressure and wettability.
Surface free energy exists on all surfaces between states of matter and between
immiscible liquids. This energy is the result of electrical forces. These forces cause
molecular attraction between molecules of the same substance (cohesion) and between
molecules of unlike substances (adhesion).
38
fundamental Properties of Reservoir Rocks
Surface tension (or interfacial tension) results from molecular forces that cause the
surface of a liquid to assume the smallest possible size and to act like a membrane
under tension.
W W W Different mass.
Different space
between molecules.
O O O
W: water molecule
O O: oil molecule
distance between molecules
Interfacial tension deforms the outer surface of immiscible liquids to produce droplets.
If the two liquids are present on a surface, the interfacial tension deforms the liquids
to produce a characteristic contact angle as shown in Figure 29.
A wetting phase is one which spreads over the solid surface and preferentially wets
the solid. The contact angle approaches zero (and will always be less than 90˚).
A non-wetting phase has little or no affinity for a solid and the contact angle will
be greater than 90˚
σwo
Contact angle, θ
Oil θ
σso
Water σsw
Solid
σ
sw Interfacial tension between the solid and water
σ Interfacial tension between the solid and oil
so
The contact angle describes the nature of the interaction of the fluids on the surface:
for the oil-water system shown above: an angle less than 90˚ indicates that the surface
is water wet. If the angle were greater than 90˚ then the surface would be oil wet.
The composition of the surface also affects the interfacial tension. Figure 30 shows the
effect of octane and napthenic acid on a water droplet on silica and calcite surfaces.
The water is not affected by the change in surface in the water/octane system, however,
the napthenic acid causes the water to wet the silica surface, but to be non-wetting
on the calcite surface.
30° 35°
Silica
Calcite
The Adhesion tension, At is defined as the difference between the solid water and
solid oil interfacial tension. This is equal to the interfacial tension between the water
and oil multiplied by the cosine of the contact angle,
40
fundamental Properties of Reservoir Rocks
If a container of oil and water is considered as in figure 31, the denser water lies
below the oil.
σcosθ σ
θ
OIL
h
radius, r
.c
Water
If a glass capillary tube of radius, r is inserted such that it pierces the interface between
the oil and water, the geometry of the tube and the imbalance in forces produced
between the glass, oil and water cause the interface to be pulled upwards into the tube.
If non wetting fluids were used, the interface in the tube may be pushed downwards.
Under equilibrium conditions, i.e. after the tube has pierced the original interface,
the adhesion tension around the periphery (2πr) of the tube can be summed to give
the total force upwards. Since the interface is static, this force must be balanced by
the forces in the column of water drawn up the tube and the equivalent column of oil
outside the tube, i.e. at point C, the force (or pressure) must be the same in the tube
as outside, therefore the excess force produced by the column of water is balanced
by the adhesion tension.
2σ wo Cosθ
gh( ρw − ρo ) =
r
It can be seen from the equations, capillary pressure can be defined both in terms of
curvature and in terms of interfacial tension, as expressed by the hydrostatic head.
2σCosθ
Pc = = gh( pw po )
rc (19)
where
Pc = capillary pressure
σ = surface tension
θ = contact angle
rc = radius of the tube
h = height of interface
ρw = the density of water
ρo = the density of oil.
For a distribution of capillaries, therefore, the capillary pressure will give rise to a
distribution of ingress of wetting fluid into the capillaries. The relative position of the
capillary rise is given with respect to the free water level, FWL, i.e. the point of zero
capillary pressure. Figure 32 illustrates the effect of three different capillary radii on
the rise of water. Figure 33 shows the behaviour for a full assembly of capillaries
and alongside the associated capillary pressure curve. In this figure it is important
to note five aspects.
The zone of varying water saturation with height above the 100% free water oil
contact is called the transition zone.
The formation containing irreducible water will produce only hydrocarbons whereas the
transition zone of varying water saturation will produce water and hydrocarbons.
The shape of the capillary pressure curves in the transition zone will depend on the
nature of the rock.
42
fundamental Properties of Reservoir Rocks
oil
oil
oil
θ oil
θ
h
FREE WATER
LEVEL
WATER WATER
Oil
Irreducible Water
Transition Zone
Pc
Oil water contact
Water
OWC
0
FWL 0% Sw 100%
Water 100% So 0%
Free water level
The height at which a wetting liquid will stand above a free level is directly proportional
to capillary pressure which is related to the size and size distribution of the pores.
It is also proportional to interfacial tension and the cosine of the contact angle and
Sw _ 2σCosθ
~ Pc =
re
Rock Property
(Permeability and Porosity)
Water wet, coarse grained sand and oolitic and vuggy carbonates with large pores
have low capillary pressure and low interstitial water contents. Silty, fine grained
sands have high capillary pressures and high water contents.
Figure 34 shows the capillary pressure curve for a reservoir where the water saturation
reduces above the aquifer. The 100% water saturation continues some distance above
the free water level corresponding to the largest pores of the rock, hD. Above this level
both the oil and water are present and the reservoir water saturation decreases with
increased height above the hydrocarbon water contact, since the larger pores can no
longer support the water by capillary action and the water saturation falls. Between
the 100% WOC and the irreducible saturation level is termed the transition zone.
44
fundamental Properties of Reservoir Rocks
Oil
Sand
Grain
Pc
h
Transition Zone
WOC
hp
FWL
0% Water Saturation 100%
Water
Consider the capillary pressure curves for the two rocks in figure 35. The first sample
(case 1) has a small range of connecting pore sizes. The second sample (case 2) has a
much larger range of connecting pore sizes, although the largest pores are of similar
size in both cases. Also, in case 2, the irreducible water saturation is reached at low
capillary pressure, but with the graded system, a much larger capillary pressure is
needed.
hI
Largest connecting pores
h about the same size.
Therefore simular hD
h
D
In addition to water transition zones, there can also be an oil/gas transition zone, but
this is usually less well defined.
Rock wettability influences the capillary pressure and hence the retentive properties
of the formation. Oil wet rocks have a reduced or negligible transition zone, and may
contain lower irreducible saturations. Low fluid interfacial tension reduces the transition
zone, while high interfacial tension extends it. Figure 36 illustrates this effect.
Height Above Water Level
A
Low Interfacial Tension
46
fundamental Properties of Reservoir Rocks
saturation results from the increase in the wetting phase (water) and the expulsion of
the hydrocarbons. In this case the saturation is determined by the large pore reducing
the capillary pressure effect and preventing water entering the larger pore. This is
the situation which occurs both when natural water drive imbibes into the formation
raising the water table level and in water injection processes. Clearly the two saturation
histories generate different saturation height profiles. Figure 37 shows the drainage
and imbibition effects on capillary rise.
Imbibition
0 100
Drainage Water Saturation: Percent Pore Space Imbibition
0 100
Fluid Density Difference Effect Water Saturation: Percent Pore Space
Water saturation
SHALE
Water saturation profile well C only
Height
profile
ES.
Well B only
NE R
m d
40
Transition K=
d
STO
m
19 0
zone 5
=1
0
K= d
D
5m
K=
SAN
25
0= 10 md
0= K=
200
30 SHALE
0=
1
2
3
4 100% Water Level
8 EFFECTIVE PERMEABILITY
8.1Definition
The idea of relative permeability provides an extension to Darcy’s Law to the
presence and flow of more than a single fluid within the pore space. When two or
more immiscible fluids are present in the pore space their flows interfere. Specific
or absolute permeability refers to permeability when one fluid is present at 100%
saturation. Effective permeability reflects the ability of a porous medium to permit
the passage of a fluid under a potential gradient when two or three fluids are present
in the pore space. The effective permeability for each fluid is less than the absolute
permeability. For a given rock the effective permeability is the conductivity of each
phase at a specific saturation. As well as the individual effective permeabilities being
less than the specific permeability, their sum is also lower.
48
fundamental Properties of Reservoir Rocks
Relative Permeability =
A typical set of effective permeability curves for an oil water system is shown in
figure 40 and for a gas oil system in figure 41.
1.0
0.9
0.8
Relative Permeability
0.7
k ro
0.6
k rw
0.5
0.4
0.3
0.2
0.1
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
S , Water Saturation, Fraction
W
0.7
k ro
0.6
0.5
0.4
0.3
0.2
0.1
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Liquid Saturation = SO + SWO, %
The introduction of a second phase decreases the relative permeability of the first
phase: for example, kor drops as Sw increases from zero. Secondly, at the point where
the relative permeability of a phase becomes zero there is still a considerable saturation
of the phase remaining in the rock. The value of So at kro = 0 is called the residual oil
saturation and the value of Sw at krw = 0 is called the irreducible water saturation.
The shapes of the relative permeability curves are also characteristic of the wetting
qualities of the two fluids (figure 42). When a water and oil are considered together,
water is almost always the wetting phase. This means that the water, or wetting phase,
would occupy the smallest pores while the non-wetting phase, or oil phase, would
occupy the largest pores. This causes the shape of the relative permeability curves
for the wetting and non-wetting phase to be different.
50
fundamental Properties of Reservoir Rocks
100
90 Water-Wet Drainage
(Decreasing S w )
80 Water-Wet Imbibition
(Increasing S w )
Oil-Wet Drainage
Relative Permeability, %
70 (Increasing S w )
K ro
60
50
40 Krw
30
20
10
0
0 10 20 30 40 50 60 70 80 90 100
Water Saturation, S W
Figure 42 Oil and Water Relative Permeability Curves for Water-Wet and Oil-Wet
Systems (Core Laboratories Inc)
This is illustrated by looking at the relative permeability to one phase at the irreducible
saturation of the other phase. The relative permeability to water at an irreducible oil
saturation of 10% (90% water) is about 0.6, figure 40, whereas the relative permeability
to the non-wetting phase, oil, at the irreducible water saturation of 0.3 approaches 1.0.
In this case it is 0.95. One practical effect of this observation is that it is normally
assumed that the effective permeability of the non-wetting phase in the presence of
an irreducible saturation of the wetting phase is equal to the absolute permeability.
Consequently, oil flowing in the presence of connate water or an irreducible water
saturation is assumed to have a permeability equal to the absolute permeability.
Similarly, gas flowing in a reservoir in the presence of irreducible water saturation
is assumed to have a permeability equal to the absolute permeability.
Relative permeability data when presented in graphical form are often referred to as
drainage or imbibition curves. (figure 42)
Water displacement of oil differs from gas displacement of oil since water normally
wets the rock and gas does not. The wetting difference results in different relative
permeability curves for the two displacements.
Water invasion results in water flow through both large and small pores as the water
saturation increases. Imbibition relative permeability characteristics influence the
displacement. Oil saturation decreases with a corresponding decrease in oil relative
permeability. Water relative permeability increases as water saturation increases.
Oil remaining after flood-out exists as trapped globules and is referred to as residual
oil. This residual oil is immobile and the relative permeability to oil is zero. Relative
permeability to water reaches a maximum value, but is less than the specific permeability
because the residual oil is in the centre of the pores and impedes water flow.
52
fundamental Properties of Reservoir Rocks
Water In Oil
Advancing water
Water In Oil
Water penetrating
smaller pores due to
capillary forces
Trapped oil
Water In Water
krw ©/µ w
M = mobility ratio =
kro ©/µ o (20)
Gas saturation less than the critical value is not mobile but it impedes the flow of
oil and reduces oil relative permeability. Successively smaller pore channels are
invaded by gas and joined to form other continuous channels. The preference of gas
for larger pores causes a more rapid decrease of oil relative permeability than when
water displaces oil from a water wet system. Figure 44 shows the alteration of relative
permeability as gas comes out of solution and flows at increasing saturation through
the oil reservoir. These gas/oil relative permeability curves are very significant in
relation to the drive mechanism of solution gas drive, which we will discuss in a
subsequent chapter.
54
Oil Water Gas
80 80 80
60 60 60
Krg
0 0 0
0 20 40 60 80 100 0 20 40 60 80 100 0 20 40 60 80 100
Gas Saturation: Percent Pore Space Gas Saturation: Percent Pore Space Gas Saturation: Percent Pore Space
Gas Saturation: 5% of Pore Space Gas Saturation: 20% of Pore Space Gas Saturation: 45% of Pore Space
Characteristic Sand During Oil Displacement Characteristic Sand During Oil Displacement Characteristic Sand During Oil Displacement
by Gas @ 5% Gas saturation by Gas @ 20% Gas saturation by Gas @ 45% Gas saturation
55
Rock Properties Measurement
CONTENTS
1. INTRODUCTION
1.1 Core Analysis
1.2 Core Definitions
2. SAMPLE PREPARATION
2.1 Whole Core Scanning
2.2 Core Cleaning
3. POROSITY MEASUREMENTS
3.1 Methods
3.2 Whole core versus conventional versus
sidewall samples
4. PERMEABILITY
4.1 Introduction
4.2 Impact of Stress
4.3 Steady State Permeability Methods
4.4 Unsteady State Permeability Measurements
5. FLUID SATURATION
5.1 Gas saturation
5.2 Oil saturation by retort
5.3 Water saturation
6. CAPILLARY PRESSURE
6.1 Introduction
6.2 Capillary Pressure Measurement Techniques
6.2.1 Porous Diaphragm (figure 22)
6.2.2 Centrifuge method ( Figure 23)
6.2.3 Dynamic method ( Figure 24)
6.2.4 Mercury Injection ( Figure 25)
6.3 Use of Laboratory Capillary Pressure Data
for Reservoir
6.4 Averaging capillary pressure data
7. EFFECTIVE PERMEABILITY
LEARNING OBJECTIVES
Having worked through this chapter the Student will be able to:
• Briefly describe the various stress conditions that can be imposed on a rock
sample.
Derive the Leverett J function and be aware of the major tortuosity related assumption
in its derivation.
Rock Properties Measurement
1. INTRODUCTION
Cores are recovered from the formation of interest using an annular shaped coring
bit. The integrity of the recovered core depends on the nature of the rock and can
vary from rock which is well formed to that which is friable in character or even is
so unconsolidated that it would form a pile of sand on the rig floor when recovered
from the core barrel. The core from the core barrel provides a record, over the well
section recovered, of the properties of the formation. Figure 1 illustrates the wide
range of measurements and procedures carried out on core samples 1.
Small Samples
• Grain Size Distribution Calbration of Wireline Logs
• Mineral Analysis
• X-Ray and SEM
Analysis
• Bio-Dating and
Association
Rock Properties Measurement
2. SAMPLE PREPARATION
The following analysis might be carried out on whole core. A core gamma log, an x-
ray analysis, a computer tomography CT scan and or an Nuclear Magnetic Resonance
NMR Scan.
Within a rock are naturally occurring gamma-ray emitters which can give a measurable
gamma-ray response that can be recorded with depth. If such a measurement can be
made on the whole core in the laboratory this whole core laboratory based measurement
can be used as depth check to relate to open hole measurements. Figure 3.
Recorder
Lead Shield
Conveyor Belt
Core
Sample for h
measurement
Io I
Rock Properties Measurement
Rotating energy
source and detector
Intensity profiles
Reconstruction algorithm
in computer.
The main benefit of Nuclear Magnetic Resonance, (NMR) imaging is that it is used
to provide a reconstruction of the fluids within a core, based on the frequency of the
excitation energy associated with a nudei. This excitation energy is supplied by an
oscillating magnetic field. The high energy attenuation associated with CT scanning
does not enable the distinctive density variations as possible with those from NMR
scanning.
These scans are able to identify localised variations in a core which if captured in
subsequent core analysis measurements could give rise to anomalous results.
Solvent extraction using centrifuge, Soxlet and Dean Stark refluxing solvent extractors
are commonly used to remove both oil and brine. No standard solvents are used and
organisations use their own preferences (figure 5).
Care needs to be taken to dry the samples particularly when hydrateable minerals are
present in the sample that break down at high temperatures. The drying procedure
is critical in that the interstitial water must be removed with no mineral alteration.
Humidity -controlled ovens are used when drying clay bearing samples to maintain
the proper state of hydration. Critical point can drying be used to clear core continuing
delicate clays like illite (see Phase Behaviour chapter - section 8.1).
3. POROSITY MEASUREMENTS
3.1 Methods
Figure 6 illustrates the methods used for routine determination of porosity.
Rock Properties Measurement
Valve
Vacuum Gauge
Displacement Pump
Pressure Gauge
Sight Glass
Core Sample
Sample in Place,
Mercury Micrometer Scale
Stopcock Open
Plunger
Sample Reference
Chamber Volume Pressure
Gauge
Valve Valve
Resaturation
In all porosity methods a bulk core sample volume has to be determined and this may
be carried out either by displacement of liquid or by callipering a shaped sample and
computation by the appropriate formula. Figure 7 shows the displacement method,
and figure 8 shows a mercury displacement pump.
Reference mark
Core plug
Mercury vessel
Single pan
Weighted
balance
base
_ 0.01 gm
+
Volume
Pressure read-out
read-out
Sample
chamber
Displacement
plunger
This method involves the independent determination of oil, gas and pure water
volumes of a fresh core sample. The oil and water can be obtained by retort ( Figure
9) and the gas by mercury injection. The pore volume is determined by summing the
three independent volumes.
10
Rock Properties Measurement
Thermocouple
Insulated Oven
Heating Elements
Sample Cup
Screen
Condensing Tube
Water Bath
(i) The Boyles Law based porosity determination method involves the compression
of a gas into the pore space or the expansion of gas from the pores of a prepared
sample. Depending on the instrumentation and the procedure, either pore volume
or grain volume can be determined. Figure 10 shows a typical set up for this and
is the most common method for measuring the grain volume. It involves setting up
a pressure in a known reference volume and then expanding the pressure into the
space containing the sample. With suitable calibration the grain volume is determined
using the ideal gas relation that PV=constant.
P Reference P1
Sample
chamber volume
Gas in
Pressure
regulator
The pores of a prepared sample are filled with a liquid of a known density. The
increase in weight of the sample divided by the fluid density is a measure of the
pore volume.
Total porosity is determined by this method as compared with effective porosity. The
sample is reduced to grain size after the dry weight and bulk volume are determined.
Grain volume is determined and subtracted from the bulk volume to yield the total
pore volume.
Whole core porosities tend to be slightly lower than small plug samples in certain
rock types. The whole core is likely to include tighter material than would be included
in a more carefully sampled plug.
For samples with medium to high porosity, sidewall and conventional samples agree
within one or two percent. During sidewall sampling low porosity highly cemented
materials tend to shatter and yield values greater than the true porosity.
4. PERMEABILITY
4.1 Introduction
The API recommended practice for the determination of permeability is also detailed
in API RP 40 which is a considerable improvement on API RP27.
There are essentially two approaches to measuring the permeability, the steady state
method where the pressure drop for a fixed flow rate is measured, generally a gas, or
the unsteady state method where the flow in the transient regeme is measured.
In the latter there are two types of test , the ‘pulse-decay’ method where two pressures
are set up and downstream of the contained sample. A slight increase in the upstream
pressure is imposed and the decay of this pressure through the sample is monitored.
The advent of very high speed data acquisition systems and accurate pressure
12
Rock Properties Measurement
transducers has made it possible to monitor these transient flow conditions. The other
approach is the pressure fall off method where a relatively low upstream pressure is
set and the decay of this pressure is monitored as it is released through the core to
the downstream open to atmosphere.
Core plug
Major for horizontal
principal stress k measurement
Formation
Figure 11 (a) Core recovered from vertical well and stress orientation in the reservoir.
If a core plug is recovered from a whole core recovered from a vertical well then the
stress orientations in a permeability test would be as shown in the sketch below. Figure
11b and 11c. These figures demonstrate that for a cylindrical horizontal core plug
it is difficult to impose a distinctive major principal stress on the core plug different
from one of the minor principal stresses whereas for a vertical orientated core plug
such distinctive stresses can be applied.
Isostatic Stress. Figure 12a. Under isostatic stress loading, equal stress is applied
to the sample in all directions, and sample strain can occur on all axes. Excessive
porosity reduction typically occurs when the imposed isostatic stress is equal to the
vertical reservoir stress ( i.e., the overburden stress).
14
Rock Properties Measurement
A
σ1 ∆D
∆L
Sample
σ1 σ1
Isostatic stress L
σ1
D
Triaxial Stress. Figure 12b. Under the true triaxial stress conditions, unequal stress
is applied to the three major axes of the sample. In the general case, strains will be
different on each axis. Typically a cube or rectangular prism -shaped sample will
be used.
σ1
∆L1
Triaxial stress
σ2 σ3
∆L ∆L
Biaxial Stress. Figure 12c. Biaxial stress loading conditions are a special case of triaxial
stress loading. In the biaxial stress loading of a cylinder , the stress parallel to the
cylinder’s axis is different from the stress applied around the cylinder’s circumference.
Strains can occur parallel to both the axis and diameter of the cylinder.
C
σ1 ∆D
∆L
Sample
σ σ
Biaxial stress L
σ1
D
σ1
∆L
Sample
Uniaxial stress σ σ L
σ1
D
Pressure
∆p
transducer Differential
Pressure. _ P D
P1 +
qr @ Pr, Tr Pa
L
Pressure Flow meter
regulator
Sample holder
The confining of the core in this case shows a Hassler type core holder where the radial
stress is low and is applied to ensure that flow of gas does not by-pass the core.
16
Rock Properties Measurement
Figure 14 shows a high pressure core holder designed to impose reservoir stresses.
The slideable inlet tube enables the strain of the stress core to be taken up. The stress
loading for this arrangement is isostatic.
Figure 15 shows a sophisticated core holder where a different axial stress can be
applied compared to the radial stress. In this arrangement the end faces of the core
plug need to be machined accurately to ensure that the loading of the axial stress
is distributed over the whole face. If not the core is liable to fragment. The stress
loading for this core plug is biaxial.
Reach Rod, X
Outlet
Ports
Cavity for
High Pressure
Nitrogen for
Axial Stress
N
Using a core plug removed from a horizontal well core it is possible using biaxial
stress loading to somewhat simulate the stress conditions, by considering the two minor
principal stresses as equal. However using biaxial stress conditions for a conventional
plug from a vertical well recovered core, then the stress conditions imposed do not
reflect those in the formation. The radial stress is a combination of the major principal
stress and one of the minor principal stresses and in the equipment these are equal.
If however, one is interested in measuring the vertical permeability from a sample
extracted from the whole core then biaxial stress conditions will reflect more readily
the reservoir stress condition.
A recent innovation has been the true triaxial cell 2 (Figure 16). In this arrangement
a series of axial tubes are hydraulically pressured between the confining rubber sleeve
of the core and the core holder body. This enables a stress pattern to be established
to represent a more realistic stress condition reservoir stress conditions.
18
Rock Properties Measurement
Platen
Threaded
end cap
Trapped tube
A A
Core
Rubber
sleeve Aluminium
cell body
Hydraulically
Maximum principal stress pressured tubes
σ
σ
1
1
Face of
core plus
1
1
σ σ
Section AA
The samples for analysis may be either the consolidated piece used for the porosity
determination or another sample but clearly it must be extracted and cleaned to
ensure that no water or oil are present. If interstitial water is very saline then it may
be necessary to remove salt.
Another recent innovation has been the probe permeameter. These devices were
initially invented to meet the need for a device to give indications of permeability of
an outcrop. The application of rock outcrops as analogues of subsurface formations
has been very valuable in developing geological/ reservoir modelling procedures. The
Flow
meter
Pressure
Pressure transducer
regulators
ri
Rock being ro
examined
The API RP40 document also describes a radial steady-state apparatus, figure 18,
where flow is from the outer to the inner radius. In this set up the preparation is not
easy and axial stresses are not balanced by radial stresses.
20
Rock Properties Measurement
Rubber Gaskets
Calibrated
Gas Burette
Springs
Mercury Manometer
rw re
P1 P
L
Pivot Ball
Piston
Regulators
Air Supply
Fill Vent
P1
Pc
VP Hydrostatic
confining
pressure
Fill/vac. Valve
∆p
+_
P
V1 VP V
Valve 1
Pc
5. FLUID SATURATION
Core analysis is sometimes used to measure the fluid saturations associated with
the core. Because of the large pressure variations between the reservoir and the
surface these saturations are not too representative of the values that would exist
in the formation, unless precautions have been taken to prevent evaporation during
pressure decline. Such precautions could be the application of pressure coring where
the down hole pressure is held in the core barrel as it is recovered to surface. At the
22
Rock Properties Measurement
surface prior to releasing the pressure the core in its container is frozen. It is then
slipped and stored in a frozen state. During controlled thawing of the core the fluids
produced and retained enable downhole saturation to be obtained.
Measurement of
collected water
Core plug
6.1 Introduction
The general laboratory procedure for capillary pressures to saturate a core sample
with a wetting phase and measure how much wetting measurement phase is displaced
from the sample when it is subjected to some given pressure of non-wetting phase.
Displacement takes place when the oil or non-wetting phase just exceeds the capillary
pressure corresponding to the largest pore. In other words the capillary force will
hold the water in the largest pore until the oil pressure is larger than the capillary
pressure of the largest pore.
The volume of the fluid displaced at a particular pressure also represents the pore
volume of all pores of that particular size. Once this pore volume has been displaced at
a particular pressure the pressure is increased and the new pore volume measured.
A plot of water volume displaced versus the displacement pressure will represent
a plot of the capillary pressure versus the percentage of the pores with a capillary
pressure greater than the subject capillary pressure.
Clearly a rock which contains a variety of pore sizes will have a capillary pressure
curve which is not discontinuous but is a smooth curve.
Since capillary pressure,
2σCosθ
Pc =
r
the curve can be calibrated to represent pore size versus percentage of pores less than
the subject pore size.
This procedure is closest to the actual saturation in the reservoir but the method is
time consuming varying from 10 to 40 days for a single sample.
24
Rock Properties Measurement
Nitrogen Pressure
Saran Tube
Crude Oil
Neoprene Stopper
Scale of
Squared Paper
Nickel-Plated
Spring Seal of
Core Red Oil
Kleenex Paper
Ultra-Fine
Fritted Glass
Disk
Brine
When the sample is rotated at various constant speeds a complete capillary pressure
curve can be obtained. The advantage of the method is the increased speed of obtaining
the data in that the complete curve can be established in a few hours.
Core
Oil inlet
To atmosphere
Porcelain
plate
Oil burette
The main advantages are that the test takes considerably less than the diaphragm
method, a matter of one or two hours. The disadvantages are the difference in wetting
properties and permanent loss of the core sample. Also there is concern on the
pore size to pressure relationship since the desaturation of some large pores may be
determined by access via smaller pores.
26
Rock Properties Measurement
Regulating Valve
Lucite Window To
Atmosphere
Cylinder
U-Tube
Manometer
Lucite Window
Pc
h=
( ρw − ρo )g (1)
The interfacial tension and contact angle values will depend on the characteristics of
the fluids. The relationship between Pc mercury/air and Pc oil/water is often taken
as 10:1 but these interfacial tension and contact angle values should be checked
before converting data.
The equations below give the procedure for generating a height saturation profile for
the reservoir from a laboratory based Pc vs saturation capillary pressure data.
Pc L (σCosθ ) R
(σCosθ ) L Pc R
h= =
( ρw − ρh )g ( ρw − ρh )g (5)
where:
h = height in feet above the free water level corresponding to zero capillary
pressure
PcR = capillary pressure at initial reservoir conditions (psi)
PcL = capillary pressure in the laboratory (psi)
(σCosθ)R = interfacial tension cosine of the contact angle (initial reservoir
conditions)
(σCosθ)L = interfacial tension cosine of the contact angle (laboratory conditions)
ρw = density of water at initial reservoir conditions
ρh = density of hydrocarbon at initial reservoir conditions
28
Rock Properties Measurement
The purpose of this exercise is to show that in a well, the water saturation not only
varies with the height above the free water level, but also due to variations in rock
properties.
A well penetrates a reservoir which from cuttings is known to consist of rock types
A and B from which a set of air-mercury measured capillary pressure curves are
available, taken in a nearby well. Figure E1. During logging the lowest 100% Sw
was found at the bottom of the well in rock type B as indicated in the figure E2.
The porosity at this level is 15%.
Specific gravities of the water and oil are 1.03 and 0.80 respectively at reservoir
conditions. The density of water is 62.4 lbm/ft3.
Questions
3. Estimate permeabilities
4. Which intervals would you recommend for completion based on the criteria
Sw<50% and k<0.1mD.
type B rock
00
10
100
0
0
0 0 100%
Pore space unoccupied by mercury
30
Rock Properties Measurement
Saturations
h Porosity 0 Oil 100%
Rock (ft) 1% 10
type (1 cm for 100 Water 0 k
10 ft) Unit No. (mD)
%
B 1%
10%
A
1%
B
%
%
A
%
%
B 1%
10%
A
1%
B
10%
100 Sw 1%
in B
type rock
found at
this level
Figure E2 Opposite
Institute of Petroleum Engineering, Heriot-Watt University 31
6.4 Averaging capillary pressure data
Capillary pressure measurements are not part of routine core analysis and a
comprehensive set of capillary pressure data is not always available. Leverett4 in 1941
generated a function which related capillary pressure to porosity and permeability,
which is commonly termed the Leverett J Function. The application of this function
was to be able to generate capillary pressure information when laboratory data was
not available. Capillary pressure data are obtained from core samples which represent
an extremely small part of the reservoir. The ‘J’ function is used to combine all the
capillary data to classify a particular reservoir.
The theory behind the J Function is outlined below and is based on figure 26 considering
flow through a core, which is assumed to be a bundle of capillary tubes.
Lcap
Lcore
πr 4 ∆P
q=
8µL cap (6)
For n tubes
nπr 4 ∆P
qn =
8µL cap (7)
nπr 2
φ=
A (8)
qµL core
k=
A∆P (9)
32
Rock Properties Measurement
8K L cap
r2 =
φ L core (10)
L cap
L core is the tortuosity of the bundle of tubes.
On the assumption that the reservoir rock has the same tortuosity at all points,
then
1
K 2
r = constant
φ (11)
2σCosθ
Pc = 1
K 2
constant
φ (12)
or
1
K 2
Pc
1 φ
= =J
constant σCosθ (13)
A set of capillary pressure data from a set of 9 core plugs taken from different depths
in a well is shown in figure 27 and shows the wide variation in shape of these curves
reflecting the different pore characteristics as given in the table below.
1
1
1
1
1
11
1
10
Pc (PSIG)
0
10 0 0 0 0 0 0 0 0 100
Sw %
A plot of the J function for a set of capillary pressure curves is given in figure 28 and
shows the impact of bringing together different rocks under one curve
34
Rock Properties Measurement
100
1100
1000
00
00
00
_
1
k )
_
Pc( ϕ
00
00
00
00
00
100
10
10 0 0 0 0 0 0 0 0 100
Sw %
The data for figure 27 however would not generate such a good function. The big
assumption in Leverett's model is that of constant tortuosity. Clearly different
rock types will have different tortuosities as a result of the pore characteristics and
composition of the rock. However within a rock type the J function could be a useful
route to obtain capillary pressure data if porosity, permeability and saturation data
is available.
Examination of field data has shown that by plotting J versus a better correlation
(Sw − Swc )
(1 − Swc ) is obtained suggesting that the S reflects the tortuosity variations within
wc
the various rocks. Figure 29
K
φ
Dimensionless Capillary Pressure Pc
σ
0
1
1
1
1
10
0
0 0.1 0. 0. 0. 0. 0. 0. 0. 0. 1.0
(
Normalised Wetting Phase Saturation Sw* = Sw-Swc
1-Swc (
7. EFFECTIVE PERMEABILITY
It is not the intention of these notes to review in detail the various approaches
to measuring effective permeabilities to multiphase systems. There has been
considerable activity in this area for gas - oil, oil - water, and three phase gas - oil
- water systems.
There are two approaches to measuring relative permeability, using an unsteady state
method or a steady state method.
36
Rock Properties Measurement
In the unsteady state method, a displacement process is set up where one fluid displaces
another and the flow rates and pressure drops are monitored as a function of time
for a fixed rate process. The saturations are obtained by calculation the remaining
volumes of the respective fluids. It is more difficult to generate relative permeabilities
as a function of saturation in this way and some would consider the method is more
suited to generate end-point effective permeability values.
In the steady state method a range of constant rate tests are set up and the pressure
drop noted when equilibrium has been achieved. Figure 30 gives a sketch of a typical
steady state set up.
Differential
pressure
transducer
∆P
Differential
pressure
transducer
∆P
Oil
Oil - water
separator and
production monitor
Brine recycle
system
Pressure
control
system
The focus is again on three phase relative permeability which has been the subject
of many papers and correlations. It is however of great interest now that large WAG,
water - alternating gas injection processes are being used to improve recovery.
The purpose of this exercise is to show that in a well, the water saturation not only
varies with the height above the free water level, but also due to variations in rod
properties.
A well penetrates a reservoir which from cuttings is known to consist of rock types
A and B from which a set of air-mercury measured capillary pressure curves are
available, taken in a nearby well. Figure E1. During logging the lowest 100% Sw
was found at the bottom of the well in rock type B as indicated in the figure E2. The
porosity at this level is 15%.
Specific gravities of the water and oil are 1.03 and 0.80 respectively at reservoir
conditions. The density of water is 62.4 lbm/ft3.
QUESTIONS
3. Estimate permeabilities
4. Which intervals would you recommend for completion based on the criteria Sw<50%
and k<0.1mD.
SOLUTION
1. The first step is to convert the air-mercury capillary pressure data to oil-water.
Conversion values:
lb f
Pcair/hg = 10 Pc water oil -
in 2
lb f 144in 2 lbm
Pc 2 2
= h ftx (1.03 − 0.8) x 62.4 3 xg
in ft ft
lb f 144in 2 lbm
Pc oil / water 2 = h( ft ) x (1.03 − 0.8) 62.4 3 xg
n ft
2
ft
38
Rock Properties Measurement
1 lbf = 1 lbm xg
This occurs in rock type B. φ = 15%. From capillary pressure curve 100% water
saturation at 15 psi i.e. 15 ft.
The free water level now provides the basis for the water saturation profile
determination.
The water saturation value is determined at each level where the rock properties
change but noting where the 100% water saturation value occurs for each rock type.
At the first change, the height is 20ft from rock type B, 15% φ to type B 10% φ
From the capillary pressure curves the respective saturations are 75% and 100%
Figure E4. For rock type B 10%, the 100% water saturation level is at 27ft when the
saturation decreases. The next rock change is at 41ft above the Free Water Level,
from rock type B 10% to type B 14% with a water saturation value of 73% and
44%. The 44% is based on an estimate of the capillary pressure curve for a value
of porosity of 14% between the 15% and 10% curves. This process is continued
through all the depths of the rock property changes and the total saturation profile
generated.
4. The estimates of permeability are based on porosity permeability trends from the
limited data given for the various rock types of the capillary pressure curves. In
unit 1 rock type B 15% the permeability is 35mD Unit 2, B 10% the permeability
is 15mD Unit 3 B 14%, interpolation suggests a value around 32mD and so on
through the units.
type B rock
00' 00
10' 10
100' 100
0' 0
0'
0' 0'
1 psi
0' 0
0 0 100%
Water saturation
40
Rock Properties Measurement
Saturations
Porosity
0 Oil 100%
Rock 1% 10
h
type 100 Water 0 k
(ft) Unit No. (mD)
0
1 0.0
A
1 0.0
B 00 1
1
1 1
A 1 0.
1
B 10
11 1
10 0.1
A 0.0
0.0
100
10
B
1
A 0.
0
B
10 mm
1
100% WL
1' 1
FWL
Figure E4
1. Archer. S., Wall. C., Petroleum Engineering Principles and Practice, Graham
and Trotman 1986
2. Recommended Practices for Core Analysis. American Petroleum Institute.
Recommended Practise 40. Second Edition , Feb 1998.
3. Smart. B,
4. Leverett. M,C., Capillary Behaviour in Porous Solids. Trans AIME 1941
5. Amyx et al Petroleum Reservoir Engineering McCranhill 1960
42
Permeability - Its Variations
CONTENTS
1 INTRODUCTION
2 AVERAGE PERMEABILITIES FOR SEVERAL
LAYERS
2.1 Beds in Parallel
2.2 Layers in Series - Linear Flow
Having worked through this chapter the Student will be able to:
Permeability - Its Variations
1 INTRODUCTION
Even on a local scale the value of permeability is not necessarily the same in all direc-
tions. Permeability is an anisotropic property, (Figure 1) i.e. its value is dependent
on direction. Porosity is an isotropic property however.
Ky
Kx Kz
The sedimentary nature of rocks is such that vertical permeability is less than hori-
zontal permeability and horizontal permeabilities in the principal directions will also
be different.
On a reservoir scale, thin streaks of very low permeability material can reduce the
effective vertical permeability to a value lower than the actual rock values would
indicate. Whereas core analysis represents microscale observations, data obtained
from well tests represent microscale behaviour.
1.0 1.0
a
12.0 17.5
b
2.5 >0.5
c
"Sample plug" 15.0
36.5 Poros 17.5%
Perm. 19 mD d
1.0
e
<0.5
38.5
21.0
f
>0.5
Location of mini-
g
permeability measurement 20.0
Permeability in mD h
Although the conditions for use of Darcy's Law state that the rock should be homo-
geneous and isotropic, in reality reservoirs do not conform to this restraint. If one
examined the variation say in a core as illustrated in figure 3, very large variations
in permeability occur and vary according to the scale of measurement. Conventional
core analysis takes a sample at around 1 per foot, probe permeability is able to sample
at much closer intervals. A well test result can reflect the permeability over tens of
feet. There is considerable effort taking place now in developing up-scaling methods
for representation of permeability for different applications.
Permeability - Its Variations
K Log Permeability K
300
180
25
160
200
Waren and Price2 demonstrated that the most probable behaviour of a heterogeneous
systems tends towards that of the geometric mean.
kG = (k1 x k2 x k3 ...kn )
1/ n
(1)
Values for average permeability can be generated by considering the formation being
made up as a composite with different layers. There are two main types of layering
to be considered: linear and radial.
Linear Flow
Consider the simple linear beds in parallel. Figure 4.
K3 h3
The average permeability can be developed using the Darcy flow equation:
QT = Q1 + Q2 + Q3 (2)
k1 A1 ( P1 − P2 ) k2 A2 ( P1 − P2 )
Q1 = , ...etc. (3)
µL µL
QT = k ∑ Ai ( P1 − P2 ) / µL (4)
k1 A1 ( P1 − P2 ) k2 A2 ( P1 − P2 ) k3 A3 ( P1 − P2 )
= + +
µL µL µL (5)
k=
∑k A i i
∑A i (6)
If all the beds have the same width the A ∝ h so k is the arithmetic average:
kA =
∑k h i i
∑ h i (7)
Permeability - Its Variations
pe
re
rw
pw
h1 Q1 K1
hT
h2 Q2 K2
h3 Q3 K3
2πhi ki ( Pe − Pw )
Qi = (8)
r
µ ln e
rw
2πhT k ( Pe − Pw ) 2π ( Pe − Pw )
Q = ∑ Qi = = ( k1 + k2 + k3 + .......) (9)
r r
µ ln e µ ln e
rw rw
k=
∑h k i i
hT (10)
This value can be compared with that obtained through well flow tests or pressure
build-up tests.
K1 K2 K3
L1 L2 L3
The average permeability of linear beds in series is obtained by adding the pressure
drop across each bed.
L
k=
L
∑ ki
i (13)
Permeability - Its Variations
re r1
rw
pw
P1 Pe
k1 k2
2πk 1 h( P1 − Pw (14)
Q1 =
r
µ ln 1
rw
2πk2 h( Pe − P1 )
Q2 =
r
µ ln e
rw (15)
Total flow:
2πkavg .h( Pe − Pw )
QT =
r
µ ln e
rw (16)
i.e.
( Pe − Pw ) = ( Pe − P1 ) + ( P1 − Pw ) (17)
re r r
QT µ ln Q2 µ ln e Qµ ln 1
rw r1 rw
= +
2πkavg.h 2πk2 h 2πk1h (18)
At steady-state flow:
QT = Q1 = Q2
(19)
ln re / rw
kavg. =
ln r1 / rw ln re / r1
+
k1 k2 (21)
Permeability variations can often be traced from well to well throughout the reservoir,
there by enabling a layered reservoir system to be developed. Many fields demonstrate
this layering phenomenon, leading to very large variations in permeability.
For example Figure 8 shows the considerable variation in permeability for the various
sand units making up the Brent sands in the North Viking Graben area of the North sea.
Using an average value for the permeability can lead to large errors and misleading
results in reservoir modelling.
Lower
ness
11
6
Permeability (Darcies)
In some cases it is not possible to correlate permeabilities from well to well, and it is
more difficult to put together a reservoir model to be used to examine flow behaviour
with such a reservoir.
Modelling reservoirs with average reservoir properties can only be valid if:
• reservoir sands are homogeneous
• random variations in reservoir properties occur across the sand
10
Permeability - Its Variations
• ordered distributions of the properties observed in one well do not correlate with
other wells.
Figure 9 shows the permeability variation in a reservoir when plotted on a linear scale
as against the log scale, for a section of a reservoir. It would be easy to wrongly estimate
an average permeability looking at the log permeability presentation, which from the
linear permeability presentation such as average clearly cannot be interpreted.
Permeability Distributions
10 20 30 40
Log (Permeability mD)
20
40
80
Thickness (ft)
20
40
60 Linear scale
80
Thickness (ft)
The difference in, for example, the behaviour of a waterflood when modelled as a
homogeneous system as against a layered system is illustrated this process is shown
schematically by figure 10.
The example demonstrates that for these heterogeneous systems with large perme-
abilities contrasts, that in water flooding water flows preferentially down the high
permeability section.
Gas
Limited flaring
Re-inject
qo sales Separation
Injection
Pump
qwi qo + qwp
qwp Sea Level
Purify /
Dump /
inject
Seawater
for injection
Sea Bed
Reservoir
Figure 11 gives the one dimensional reservoir simulation process and figure 12 the
predicted outcome of oil production, water-cut and oil recovery. As is shown the
piston like displacement arrives at the production well around 2500 days after the
start of injection at which the oil recovery is around 48%. A very good project if this
is an accurate simulation of the process.
12
Permeability - Its Variations
_
k
average Water Oil
WATER CUT
1.0
0.8
0.6
1D
0.4
0.2
0 TIME (days)
2000 4000 6000
µw = 0.5 cp
OIL RATE
1.0 kv = kh
2D = { kv = 0.1kh
1D 1D =
0.5
0 TIME (days)
2000 4000 6000
RECOVERY FACTOR
0.5 49.7%
0.4
1D
0.3
0.2
0.1
0 TIME (days)
2000 4000 6000
0
50
20
70
40 600 Days
50
60
µw = 0.5 cp
1000 2000 3000
kv = kh
0
50
20
70
1200 Days
40
50
60
0
50%
20
70%
40 2400 Days
60
It is also interesting to note an increase in the water saturation along the base of the
reservoir figure 13c and 13d. This is due to the impact of gravity, the density difference
of the denser water causing the water to move towards the base of the reservoir. After
14
Permeability - Its Variations
4800 days this gravity segregation perspective is clearly seen as the low permeability
zone is being swept by this gravity impact. At the top of the reservoir no such benefit
is generated and oil is unswept. This unswept oil is sometimes termed attic oil.
Figure 14 gives the oil rate, water cut and recovery for the simulation where the
various permeability layers have been identified in the 2-D simulation. They show
that the project is not as attractive as that forecast by the previous 1-D simulation.
The two lines in the 2 D case show the impact of vertical permeability, a significant
perspective in relation to gravity flow.
WATER CUT
1.0
0.8
0.6
1D
0.4
2D
0.2
0 TIME (days)
2000 4000 6000
µw = 0.5 cp
OIL RATE
1.0 kv = kh
2D = { kv = 0.1kh
1D 1D =
0.5 2D
0 TIME (days)
2000 4000 6000
RECOVERY FACTOR
0.5 49.7% 48.8%
0.4 39.0%
1D
2D
0.3
0.2
0.1
0 TIME (days)
2000 4000 6000
The example clearly demonstrates that for these heterogeneous systems with large
permeability contrasts, that in water flooding water flows preferentially down the
high permeability section.
With such a result the reaction is what can be done to improve the process? The im-
mediate reaction is not to complete the high permeability layers, forcing the fluids
through the more restricted lower permeabilities. This suggestion is illustrated in figure
15, where the simulation clearly shows that there is little impact. Near the injection
point there is evidence of some displacement into the lower permeability zone, but
once into the formation the water finds the easiest route the fluids move through the
central high permeability zone. The only way to impact the displacement would be
to reduce the permeability of the high value layer some distance into the formation.
The use of time setting polymers could provide such a fluid diversion. Clearly there
are technical risks associated with such a process.
20 50%
70%
40
50%
60
Figure 15 Water Injection Profile When High Permeability Layer Not Completed.
The recovery of the attic oil is a challenge. Gravity segregation based methods have
been suggested where the injection of a light fluid, for example nitrogen, would have
a similar impact on the unswept oil in the upper layers as the water has on the base
layers. In the example presented it has been assumed that there is strong pressure and
flow communication across the layers if this were not the case then the flow profiles
would be significantly different.
Although the example of water flooding has been used, the phenomena will also
occur in gas injection schemes, being therefore very relevant to the development of
gas condensate reservoirs by gas cycling.
The example has illustrated the importance of permeability contrast in a formation. The
topic is further covered in some depth in the chapter on immiscible displacement.
16
Permeability - Its Variations
Figure 16 a and b illustrate the geological process which result in the permeability
decreasing with depth (16a) and the permeability increasing with depth (16b) 3.
Considerable activity in the nineties was focused on the impact of a range of geologi-
cal scenarios on permeability and the development of realistic reservoir simulation
models. The subject of upscaling, unheard of in the seventies is now an integral part
of building realistic reservoir flow models.
Coarser Sediment in
Shallow Turbulent Water
Sea Level
Coarser Sediment in
Shallow Turbulent Water
Sea Level
Wave Energy
Increasing
e
Siz
rain itio
n
gG os
Wave Energy
in ep ize
Increasing
s
rea D
us in S Advance
Inc eo ra ion
an G sit
Fine Sediment in ult sing e po of Bar
m
Si crea
Deeper Quiet Water sD
In ou Advance
ne
Fine Sediment in u lta of Bar
Deeper Quiet Water Sim
GR GR
Profile (A) Profile (B)
GR GR
Profile (A) Profile (B)
Injection
Permeability
Injection
Permeability Oil
Water Oil
Depth
Water Production
Depth
Production
(a) Favorable
(a) Favorable
Levee
e Levee
Weak Fin
Er o si o
Current
on Growthinofe
Weak siti F tion )
Channel po
Enrof
e
Current
D Point Bar cre e
Migration Strong nd n Ac e Lin
a i o
siBoan
S it Growth l of (Tim
Channel Current o s ra
tion )
ep ate aBar
LPoint ce cre e
dD
nokf B
La face
nk
a rse Su
r
GR
Co
Permeability GR
Injection
Permeability
Injection
Oil
Depth
Oil
Depth
Production
(a) Unfavorable
Water
Production
(a) Unfavorable
Water
Figure 16(b) Effect of Unfavourable Permeability in Waterflooding 3.
1. Van Veen, F.R., "Geology of the Leman Gas Field Petroleum and the Continental
Shelf of N.W Europe." (Woodland, A.W., ed.) Supplied Science Pub. Barkins
1975. 223
3. Archer, J. S., Wall, C. C., Petroleum Engineering. Graham and Trotman 1988
London
18
Fluid Flow In Porous Media
10
CONTENTS
Having worked through this chapter the student will be able to:
• Understand the nature of fluid flow in a porous medium and the relation between
time, position and saturation
• Understand the use of the line source solution in radial systems to determine the
pressure at any point in a reservoir under transient flow conditions
• Understand the application of line source solution to multiple well/ multiple rate
histories in a transient flow reservoir
• Understand the basis of well test analysis, and use of the line source solution to
determine the reservoir permeability and skin factor
Fluid Flow In Porous Media
10
1 Introduction
The ability to determine the productivity of a reservoir and the optimum strategy to
maximise the recovery relies on an understanding of the flow characteristics of the
reservoir and the fluid it contains. The physical means by which fluid diffuses through
a rock (or any other porous medium) depends on the interaction between the fluid (and
its properties) and the rock (and its properties). In terms of energy, the process may
at first sight appear to be similar in concept to the application of the general energy
equation to flow through pipes, although in this case the container through which the
fluid flows is made of very small tubes. It is precisely because of the geometry and
dimensions of the tubes that the application of the general energy equation would be
impossible: the description of a real pore network in a whole reservoir would be too
complex. Coupled with this is the interaction between the material of the tubes (or
pores) and the fluids. Surface chemistry effects start to dominate the flow when very
small tubes are considered and when multiphase flow occurs in them. Thus, complex
force fields are produced from not only the viscous pressure drop but also the effects
of surface tension and capillary pressure.
The combination of these factors dictates the nature of the fluid flow and one of the
initially unusual aspects is the time taken for pressure to change in the reservoir or for
fluid to migrate from one location to another. For instance, if a large body of water,
such as a swimming pool were drained, for all intents and purposes, the level of water
in the swimming pool would be the same as the water drained out. It would take an
appreciable amount of time for the water to drain (i.e. it would not be instantaneous),
but the pressure or level of the water in the pool would be the same at all locations of
the pool. The pressure in the pool would equilibrate almost immediately. Contrast this
with, for example, a water saturated reservoir rock in which the water could flow, but
where the permeability of the reservoir and the compressibility and viscosity of the
water dictated that the transfer of the water through the reservoir was not instantaneous
(as in a swimming pool), but took an appreciable time. In this case pressure changes
in one part of the reservoir may take days, even years to manifest themselves in other
parts of the reservoir. In this case, the flow regime would not be steady state while
the pressure was finding its equilibrium and a major problem, therefore, would be
that Darcy’s Law could not be applied until the flow regime became steady state. In
some way, the diffusion through the reservoir needs to be examined: Darcy’s Law
is one expression of that diffusion process, but time dependent scenarios must also
be examined.
To illustrate this, consider the following model of a linear reservoir with a well at
the left side (Figure 1).
Outlet
(constant flowrate)
Profile
after
time t
-50
height of water in tubes
-100 t=0
t=1
t=2
t=3
t=4
-150 t=5
t=6
t=7
-200
Figure 1 Model of a linear reservoir and the pressure response measured after different
times
Each tube contains water, the height of which represents the pressure at that part of
the reservoir. The tubes are connected to each other at the base by a small diameter
tube which restricts the flow. Under initial conditions, the height of the fluid is
identical in each of the tubes (assuming the model is level). The outlet at one end is
at a lower level than the model and when it is opened the fluid immediately drains
from the model and the level of the water in the tubes decreases. The energy to
drive this system is the potential energy stored in the height of the water columns:
there is no high pressure inlet to the model. As is shown in figure 1, to reduce the
pressure in the model, the fluid needs to be expelled, but because of the permeability
of the rock (the restrictions in the bottoms of the tubes) it takes time for the fluid in
the tubes nearest the outlet to move (or expand in the case of pressurised fluid in a
reservoir) and therefore it takes time for the pressure to change. When the flow is
started from the outlet, there is an immediate reduction in the pressure in tube 1 and
this pressure perturbation moves through the rest of the fluid at a rate dictated by
the rock permeability and fluid properties. This produces a variation in the pressure
along the model. The pressure profile takes time to develop from the outlet (at tube
1) to the tube farthest from the outlet (tube 10) and at time, t=1, the pressure in tube
10 is still equal to the pressure at the initial time, t=0. This is termed a transient flow
condition as the fluid is trying to reach pressure equilibrium. When the fluid in tube
10 starts to expand and flow, all of the fluid in the whole model is now expanding
and flowing to the outlet. Tube 10 represents the limit of the fluid volume: there are
no more tubes behind to supply fluid at the initial pressure. Therefore, as the pressure
Fluid Flow In Porous Media
10
perturbation moves through the model from tube 1 to tube 10, the rate of pressure
change in the fluid is not limited by the volume of the fluid: it is as if the volume of
fluid was infinite in extent. During the transient period, the reservoir is often referred
to as infinite acting.
On inspection, a profile has been developing across the tubes during the transient
period. At the end of the transient period, the fluid in all of tubes is expanding producing
a decline in the pressure in all of the tubes. The shape of the pressure profile across
all of the tubes remains essentially constant and as time continues, the profile sinks
through the model until the water in the tube nearest the outlet empties. During this
time, the water in the model has not been replaced so steady state conditions have not
been achieved, however, since the gradient between the pressures in each adjacent tube
is not changing, the system can be considered to be in pseudo-steady state or semi-
steady state: the pressure gradient is constant but the absolute pressure is declining.
This mimics the situation in a real reservoir where the pressure is perturbed around
a well and the pressure disturbance moves out into the rest of the reservoir until it
reaches the outer boundary. If this is sealing and no flow occurs across the boundary,
then the reservoir pressure will decline (neglecting any injection into the reservoir)
in a pseudo-steady state manner. If the boundary is nonsealing (i.e. it is the water
oil contact and the aquifer water is mobile) then the aquifer water will flow into the
reservoir and a steady state will be achieved if the flowrates match.
The flow described in this model is trivial, but it illustrates the problem of applying
Darcy’s Law to real reservoirs: the effect of time on flow may be considerable and
if only steady state flow relationships were available then either permeability of the
reservoir would remain unknown or unrealistic flow periods would be required to
measure an essentially simple rock property.
The actual flow patterns in producing reservoirs are usually complex due mainly to
the following factors:
(i) The shapes of oil bearing formations and aquifers are quite irregular
(ii) Most oil-bearing and water bearing formations are highly hetereogenous with
respect to permeability, porosity and connate water saturation. The saturations
of the hydrocarbon phases can vary throughout the reservoir leading to different
relative permeabilities and therefore flow patterns
(iii)The wellbore usually deviates resulting in an irregular well pattern through the
pay zone
(iv)The production rates usually differ from well to well. In general, a high rate well
drains a larger radius than a lower rate well
(v) Many wells do not fully penetrate the pay zone or are not fully perforated
(i) The drainage area of the well, reservoir or aquifer is modelled fairly closely
by subdividing the formation into small blocks. This results in a complex series
of equations describing the fluid flow which are solved by numerical or semi-
numerical methods.
(ii) The drained area is modelled by a single block to preserve the global features and
inhomogeneities in the rock and fluid properties are averaged out or substituted
by a simple relationship or pattern of features (such as a fracture set, for example).
The simplifications allow the equations of flow to be solved analytically.
In the following sections, dealing mainly with oil, the compressibility of the flowing
fluid may depend on the pressure. It will always be assumed that the product of
compressibility and pressure, cP, is smaller than one, i.e. cP<<1. If it is not (as in the
case of a gas) then the pressure dependence of compressibility must be taken into
account.
Y U
Ux Uz
Uy
Fluid Flow In Porous Media
10
The components of the flow velocity vector, U are:
Ux = -(kx/µ)(δP/δx)
Uy = -(ky/µ)(δP/δy)
Uz = -(kz/µ)(δP/δz+ρg) (2.1)
where
k = permeability (m2) in the direction of X, Y, Z. The Z direction has an elevation
term, ρg, included to account for the change in head.
P = pressure (Pa)
µ = viscosity (Pas)
ρ = density (kg/m3)
g = acceleration due to gravity (m/s2)
U = flow velocity (m/s) = (m3/s/m2)
These components are similar to Darcy’s law in each of the three directions.
2.2.1 Linear Horizontal Model of a Single Phase Fluid
In this geometry, the flow is considered to be along the axis (in the x direction) of
a cuboid of porous rock. The total length of the cuboid is L and fluid flows into the
rock at the left end (x=0) and exits at the right end (x=L). There is no flow in the
other directions at any time i.e. Uy = Uz =0 for all values of x, y, z and time, t (in a real
reservoir, there may be flows in different directions in different parts of the reservoir
and there may be cross flows from different layers within the reservoir). The rock is
100% saturated with the fluid.
k ∂P
U x = −
µ ∂x (2.2a)
δ (Uρ ) δρ
= −φ ;0 ≤ x ≤ L
δx δt (2.‑2b)
where
k = permeability (in the X direction), (mD)
ρ = density, (kg/m3)
U = flow velocity (m/s)
t = time (s)
φ = porosity
µ = viscosity, Pas
P = pressure, Pa
x = distance, (m)
The latter equation is obtained from a mass balance as follows (figure 3):
dx
area, A
x=L
x+dx
porosity, φ X axis
x
flowrate, qin
x=0
isometric view
dx
X axis
plan view
In Figure 3, fluid flows into the end of the cuboid at position x=0, through the rock
only in the X direction and out of the cuboid at x=L. In the middle of the cuboid,
an element from position x to position x+dx is examined. The bulk volume of the
element is the product of the area, A and the length, dx, i.e. the bulk volume = A*dx.
The pore volume of the element is therefore the product of the bulk volume and the
porosity, φ, i.e. the pore volume = A*dx*φ. If the flow was steady state then the
flowrates into and out of the volume (qin and qout) would be identical and Darcy’s Law
would apply. If the flow rates vary from the inlet of the volume to the outlet, i.e. qin
≠ qout then either the fluid is accumulating in the element and qin > qout or the fluid is
depleting from the element qout > qin (which is possible in a pressurised system since
the pressure of the fluid in the element may reduce causing it to expand and produce
a higher flow rate out of the element). Therefore, there is a relationship between the
change in mass, m, along the cuboid and the change in density, ρ, over time as the
mass accumulates or depletes from any element. In terms of mass flowrate,
Fluid Flow In Porous Media
10
Mass flow rate through the area, A = qρ ((m3/s)*(kg/m3) = kg/s)
Mass flow rate through the area, A at position x = (qρ)x
Mass flow rate through the area, A at position x+dx = (qρ)x+dx
Mass flowrate into a volume element at x minus mass flowrate out of element at x +
dx
=(qρ)x - (qρ)x+ dx
The mass flow rate out of the element is also equal to the rate of change of mass
flow in the element,
i.e.
δ ( qρ )
(qρ ) x + dx = (qρ ) x + * dx
δx
δ ( qρ )
Therefore the change in mass flow rate = − * dx
δx
i.e. if the change in mass flowrate is positive it means the element is accumulating
mass; if the change is negative it is depleting mass.
This must equal the rate of change of mass in the element with a volume = A*dx*φ
δρ
The rate of change of mass is equal to Aφdx
δt
δ ( qρ ) 1 δρ
hence − =φ
∂x A δt
δ (Uρ ) δρ
− =φ
∂x δt
or
δ (Uρ ) δρ
= −φ
∂x δt (2.2b)
Substituting the parameters of equation 2.2a in 2.2b gives
δ kρ δP δρ
= −φ
δx µ ∂x δt (2.3)
Equation 2.3 shows the areal change of pressure is linked to the change in density
over time. Realistically, it is pressure and time that can be measured successfully in
a laboratory or a reservoir, therefore a more useful relationship would be between
the change in pressure areally with the change in pressure through time. The density
can be related to the pressure by the isothermal compressibility, c, defined as:
Since
δρ δρ δP δP
= = cρ
δt δP δt δt (from above)
then
(2.5)
This is the partial differential equation for the linear flow of any single phase fluid
in a porous medium which relates the spatial variation in pressure to the temporal
variation in pressure. If it were applied to a laboratory core flood, it could describe
the pressure variation throughout the core from the initial start of the flood when the
flowrate was increased from zero to a steady rate (the transient period) as well as the
steady state condition when the flow into the core was balanced by the flow out of
the core. Inspection of the equation shows that it is non-linear because of the pressure
dependence of the density, compressibility and viscosity appearing in the coefficients
kρ
µ and φcρ. The pressure dependence of the coefficients must be removed before
simple solutions can be found, i.e. the equation must be linearised. A simple form of
linearisation applicable to the flow of liquids such as undersaturated oil is to assume
their compressibility is small and constant. More complex solutions are required for
more compressible fluids and gasses.
δ δP φµcρ δP
ρ =
δx δx k δt
(2.6)
10
Fluid Flow In Porous Media
10
δρ δP δ 2P
+ ρ 2
δx δx δx
δρ δρ δP
=
δx δP δx
cρ(δP/δx)2 + ρ(δ2P/δx2).
δ 2 P φµc δP
=
δx
2 k δt (2.7)
The assumption is made that the compressibility is small and constant, therefore
φµc
the coefficients k are constant and the equation is linearised. In equation (2.7)
k
φµc is termed the diffusivity constant. For liquid flow, the above assumptions are
reasonable and have been applied frequently, but can be applied only when the
product of the compressibility and pressure is much less than 1, i.e. cP <<1.0. Thus
the requirement for small and constant compressibility. The compressibility in this
case is the saturation weighted compressibility, i.e. the effect of the oil, water and
formation compressibilities:
where
c is the saturation weighted compressibility
co is the compressibility of oil
cw is the compressibility of the connate water
cf is the compressibility of the formation (pore volume)
So is the oil saturation
Swc is the connate water saturation
Solutions of the linear diffusivity equation are needed when dealing with linear flow
from aquifers. For solutions dealing with well problems a radial model is required.
Y
h
rw X
re
radial element
h
wellbore
qρr qρr+dr
dr
r
section in the XZ plane
U = q/2πrh (2.9)
12
Fluid Flow In Porous Media
10
From Darcy’s Law (taking account of the flow direction and the co-ordinate
direction):
k δP
U=
µ δr (2.10)
δ ( qρ ) δρ
= 2πrhφ
δr δt (2.11)
Eliminating U and q through equations 2.9 to 2.11 gives the non-linear equation:
1δ k δP δP
ρr = φcρ
r δr µ δr δt (2.12)
(2.13)
(a) In the case of water encroachment from an aquifer into a reservoir, the inner
boundary corresponds to the mean radius of the reservoir, the outer boundary to the
mean radius of the aquifer.
(b) In the case of the pressure regime around a wellbore, the inner boundary corre-
sponds to the wellbore radius, rw, the outer boundary to the boundary of the drain-
age area. In general the wellbore radius, rw is a mathematical concept, however, the
following are widely treated as valid:
The solution of the equation requires the initial conditions and the boundary
conditions.
Outer Boundary
(a) If there is no flow across the outer boundary it is a closed system and the flow
velocity, U will equal zero. The pressure gradient, δP/δr will also be zero
(b) If there is flow across the outer boundary, the reservoir pressure will
be maintained at a constant value equal to the initial reserv oir pressure, Pi.
Inner Boundary
There are two main cases for the inner boundary which represent either maintaining
a constant pressure or a constant flow rate. These are representative of possible flow
regimes in the reservoir during either water flooding or production from a well.
Steady-state refers to the situation in which the pressure and the rate distribution
in the reservoir remain constant with time. Unsteady state is the situation in which
the pressure and/or the flow rate vary with time. Semi-steady is a special case of
unsteady state that resembles steady-state flow. These differences in the flow regimes
have ramifications in practical reservoir engineering since working solutions to the
diffusivity equation are usually limited to a particular flow regime. For instance, in
a pressure build up test in a well, the determination of an accurate average reservoir
pressure will depend strongly on the flow regime the well is in and therefore which
working solution is used.
14
Fluid Flow In Porous Media
10
In this flow regime, one of the conditions for solution of the diffusivity equation is that
the flow rate is constant. This can be applied to the flow of oil towards a full length
perforated well, and to the flow of water to a producing reservoir from an aquifer.
The flow can be described approximately as the radial flow of a single phase from the
outer radius ‘b’ of a right hollow cylinder towards its inner radius, ‘a’. It is assumed
that the cylinder consists of a homogeneous porous medium.
In the case of drainage by a well, ‘a’ is the radius of the well, rw and ‘b’ is the radius
of the external boundary, re. The flow rate, q at radius, r = rw is the production rate of
the well. In the case of natural water influx into a reservoir, ‘a’ is the mean reservoir
radius, ‘b’ is the mean aquifer radius, and q is the volume flow rate of water across
the initial oil-water contact.
The radial constant terminal rate case is determined by the following system of
equations:
1 δ δP φµc δP
(r ) = ;a ≤ r ≤ b
r δr δr k δt (3.1)
2πrkh δP
q= ;r = a
µ δr (3.2)
with the initial condition that the pressure at all points is constant
a ≤ r ≤ b, t = 0; P = Pi = constant (3.3)
and the boundary conditions that at the wellbore the flowrate is constant after the
production starts
r = a, t ≥ 0 : q = constant (3.4)
and at the outer boundary, the pressure is either a constant (and equal to the initial
pressure) in the case of pressure maintenance
r = b, t ≥ 0 : P = Pi = constant (3.5a)
or there is a sealing boundary with no flow across it in which case the pressure
gradient at the boundary is zero
δP
r = b, t ≥ 0: = 0
δr (3.5b)
The solution of the equations 3.1 to 3.4 and 3.5a & equations 3.1 to 3.4 and 3.5b are
well known and can be referenced in “Pressure buildup and flow tests in wells” by
δP
i.e. flowrate, q = constant and the pressure gradient, = 0 for all values of radius,
r and time, t δt
δP dP dr 2πkh
therefore, = and the flow equation becomes q = dP
δr dr r µ
integrating between the limits rw and r gives:
qµ r
P − Pw = ln
2πkh rw
(3.6)
qµ re
Pe − Pw = ln
2πkh rw
(3.7)
which is identical to the relationship described for a radial system by Darcy’s Law.
In this case, the pressure at the external radius of the reservoir is required and the
only way to measure it in the reservoir would be to drill a well at the external radius.
This is uneconomic, therefore a mean reservoir pressure,P , is used. It is found from
routine bottom hole pressure measurements and well tests conducted on the wells
in a reservoir, it includes the effect of the area of influence of each well. In simple
terms, the volume drained by each well is used to weight the bottom hole pressure
measurements made in the well; all of the weighted pressures of all of the wells in
the reservoir are then averaged. Figure 5 shows a well in a reservoir and its area of
influence. Volumetrically, this volume is drained by the well and the mean reservoir
pressure,P , is related to the pressure, P of elements of volume, dV being drained.
The total volume is V.
16
Fluid Flow In Porous Media
10
initial pressure profile
wellbore
Pi
rw
re
re
1
P= ∫ PdV
V rw
(3.8a)
qµ r
P = Pw + ln
2πkh rw
2 re qµ r
P= 2 ∫ w
P + ln rdr
r e rw 2πkh rw
2 qµ re r
P - Pw = ∫ ln rdr
r 2e 2πkh rw rw
re
2 qµ 1 2 r re 1 r 2
P - Pw = 2 r ln − ∫ dr
r e 2πkh 2 rw rw rw r 2
2 qµ r 2e re r 2w rw r 2e r 2w
P - Pw = ln − ln − −
r 2e 2πkh 2 rw 2 rw 4 4
r 2w
assuming is negligible
4
qµ re 1
P - Pw = ln −
2πkh rw 2
(3.10)
EXERCISE 1
A well produces oil at a constant flowrate of 15 stock tank cubic metres per day (stm3/d).
Use the following data to calculate the permeability in milliDarcys (mD).
Data
porosity, φ 19%
formation volume factor for oil, Bo 1.3rm 3/stm 3 (reservoir cubic metres per
stock tank cubic metre)
net thickness of formation, h, 40m
viscosity of reservoir oil, µ 22x10-3 Pas
wellbore radius, rw 0.15m
external radius, re 350m
initial reservoir pressure, Pi 98.0bar
bottomhole flowing pressure, Pwf 93.5bar
qreservoir = qstock tank x Bo
1bar = 105 Pa
EXERCISE 2
A well produces oil from a reservoir with an average reservoir pressure of 132.6bar.
The flowrate is 13stm3/day. Use the following data to calculate the permeability.
Data
porosity, φ, 23%
formation volume factor for oil, Bo 1.36rm3/stm3
net thickness of formation, h 23m
viscosity of reservoir oil, µ 14x10-3 Pas
wellbore radius, rw 0.15m
external radius, re 210m
average reservoir pressure, 132.6bar
bottomhole flowing pressure, Pwf 125.0bar
18
Fluid Flow In Porous Media
10
EXERCISE 3
A reservoir is expected to produce at a stabilised bottomhole flowing pressure of 75.0
bar. Use the following reservoir data to calculate the flowrate in stock tank m3/day.
Data
porosity, φ 28%
formation volume factor for oil, Bo 1.41rm3/stm3
net thickness of formation, h 15m
viscosity of reservoir oil, µ 21x10-3 Pas
wellbore radius, rw 0.15m
external radius, re 250m
average reservoir pressure, 83.0bar
bottomhole flowing pressure, Pwf 75.0bar
permeability, k 125mD
r
rD =
dimensionless time, rD : rw
kt
dimensionless time, tD : tD =
φµcrw2
2π kh
dimensionless pressure, PD : PD ( rD ,t D ) = ( Pi − Pr,t )
(at a dimensionless radius qµ
and at a dimensionless time)
where
r = radius in question
rw = wellbore radius
k = permeability
t = time in question
φ = porosity
µ = viscosity
1 δ δPD δPD
rD =
rD δrD δrD δt D (3.11)
zero flowrate
flowrate, q
time
(a)
bottomhole flowing pressure, Pwf
Pi transient
late transient
time
(b)
Figures 6a and 6b show the response of a reservoir at a wellbore when a flow rate, q, is
suddenly applied. The pressure of the flowing fluid in the wellbore, Pwf falls from the
initially constant value, Pi (static equilibrium) through time and the constant terminal
rate (CTR) solution of the diffusivity equation describes this change as a function
of time. The CTR solution is therefore the equation of Pwf versus t for a constant
production rate for any value of the flowing time. The pressure decline, Figure 6(b),
20
Fluid Flow In Porous Media
10
can normally be divided into three sections depending on the value of the flowing
time and the geometry of the reservoir or part of the reservoir being drained by the
well. This figure represents the pressure change at the wellbore through time which
is equivalent to the pressure change (or change in the height of water) in the cylinder
nearest the outlet in the model represented in Figure 1.
Initially, the pressure response can be described using a transient solution which
assumes that the pressure response at the wellbore during this period is not affected
by the drainage boundary of the well and vice versa. This is referred to as the infinite
reservoir case, since during the transient flow period, the reservoir appears to be infinite
in extent with no limits to the fluid available to expand and drive the system.
The transient period is followed by the late-transient when the boundaries start to
affect the pressure response. This is analogous to the pressure disturbance having
moved along the line of tubes in the model in figure 1. The nature of the boundaries
affects the type of solution used to describe the pressure change since a well may drain
an irregularly shaped area where the boundaries are not symmetrical or equidistant
from the well.
The next phase in the pressure decline is termed semi-steady state or pseudo steady
state where the shape of the pressure profile in the reservoir is not changing through
time and the wellbore pressure is declining at a constant rate. It is analogous to the
model depicted in figure 1 where the level of water in all of the tubes is falling and
no additional water is being added to tube 10 to maintain absolute pressure profile. If
the pressure profile developed in the reservoir around the well had remained constant,
true steady state conditions would have occurred and the steady state solutions as
mentioned in the previous section would have applied.
The solution describes pressure drop as a function of time and radius for fixed values
of external radius, re, and wellbore radius, rw, rock and fluid properties. It is expressed
in terms of dimensionless variables and parameters as:
PD = f(tD,rD,reD) (3.12)
where
tD = dimensionless time
rD = dimensionless radius
reD = re/rw = dimensionless external radius.
e α m D J1 (α m reD )
2
∞ − t 2
2t D 3
i.e. PD ( t D ) = 2 + lnreD − + 2 ∑ 2 2
reD 4 (
m =1 α m J1 (α m reD ) − J1 (α m )
2
) (3.14)
where
αm are the roots of J1 (α m reD )Y1 (α m ) − Ji (α m )Y1 (α m reD ) = 0
This series has been evaluated for several values of dimensionless external radius,
reD, over a wide range of values of dimensionless time, tD. The results are presented
in the form of tables (from Chatas, AT3, “A Practical Treatment of non-steady state
Flow Problems in Reservoir Systems,” Pet. Eng. August 1953) in “Well Testing” by
J Lee, SPE Textbook series, Vol 1. A summary of the use of the tables for constant
terminal rate problems is as follows in Table 1. It reports the dimensionless pressure
at some dimensionless time for various configurations of reservoir. It is the solution
to equation 3.14.
22
Fluid Flow In Porous Media
10
Table Presents Valid for
3 i PD as a function of tD for 1.5< reD2 <10 (from table) finite reservoirs, but if the
value of tD is smaller
than that listed for a given
value of reD then the
reservoir is infinite acting
and therefore table 2 is used.
EXERCISE 4
A reservoir at an initial pressure, Pi of 83.0bar produces to a well 15cm in diameter.
The reservoir external radius is 150m. Use the following data to calculate the pressure
at the wellbore after 0.01 hour, 0.1 hour, 1 hour, 10 hours and 100hours of production
at 23stm3/d
Data
porosity, φ 21%
formation volume factor for oil, Bo 1.13rm3/stm3
net thickness of formation, h 53m
viscosity of reservoir oil, µ 10x10-3 Pas
wellbore radius, rw 0.15m
external radius, re 150m
initial reservoir pressure, Pi 83.0bar
permeability, k 140mD
compressibility, c 0.2x10-7Pa-1
fluid production
flow to the
wellbore
Data
porosity, φ 25%
net thickness of formation, h 0.2m
viscosity of fluid, µ 2x10-3 Pas
wellbore radius, rw 0.2m
external radius, re 2m
initial reservoir pressure, Pi 2bar
permeability, k 1200mD
compressibility, c 0.15x10-7Pa-1
EXERCISE 6
A discovery well is put on test and flows at 2.9stm3/d. Using the following data. calculate
the bottomhole flowing pressure after 5 minutes production.
Data
porosity, φ 17%
net thickness of formation, h 40m
viscosity of reservoir oil, µ 14x10-3 Pas
formation volume factor of oil, Bo 1.27rm3/stm3
wellbore radius, rw 0.15m
external radius, re 900m
initial reservoir pressure, Pi 200bar
permeability, k 150mD
compressibility, c 0.9x10-9Pa-1
24
Fluid Flow In Porous Media
10
3.3.3 The Line Source Solution
This solution assumes that the radius of the wellbore is vanishingly small relative to
the mean radius of the reservoir. It allows the calculation of the pressure at any point
in an unbounded reservoir using the flowrate at the well. The benefits are clear in
that no flow rates other than those measured in the producing well are required and
from which the pressure at any location can be calculated. The disadvantage is that
the solution works for infinite acting reservoirs only and if barriers are met, then the
solution fails to represent the true flow regime. The technique of superposition can
be used to combine the effect of more than one well in an infinite acting reservoir
and this technique can represent the effect of a barrier. The barrier is equivalent to the
pressure disturbance produced by a second, imaginary well producing at the same rate
and having the same production history as the real well with both these wells in an
infinite acting reservoir. This solution has found a lot of use in well test analysis.
In constant terminal rate problems, the flowrate at the well was given by
2πrhk δP
q=
µ δr r = r w
(3.15)
and for a line source, the following boundary condition must hold:
lim δp qµ
2y =
y → 0 δy 2πkh for time, t > 0.
φµcr 2
y=
4kt and substituting into the diffusivity equation
1 δ δP φµc δP
(r ) =
r δr δr k δt
gives
d 2 p dp
y + (1 + y) = 0
dy 2 dy
p → pi as y → ∝
lim δp qµ
2y =
y → 0 δy 2πkh
dp
If p′ =
dy then
dp C1 − y
i.e. p′ = = e (3.16)
dy y
where C and C1 are constants of integration. Since
lim δp qµ lim
2y = = 2C1e − y
y → 0 δy 2πkh y → 0
qµ
then C1 = and equation 3.16 becomes
4πkh
dp qµ e − y
=
dy 4πkh y which is integrated to give
y
qµ e − y
4πkh ∫∞ y
p= dy + C 2
or
∞
qµ e − y
4πkh ∫y y
p=− dy + C 2
qµ
p= Ei(-y) + C 2
4πkh
qµ φµcr 2
p i − p(r,t) = − Ei(- )
4πkh 4kt (3.17)
The term Ei(-y) is the exponential integral of y (the Ei function) which is expressed
as
∞
e−y
Ei( − y) = − ∫ dy
y
y
26
Fluid Flow In Porous Media
10
It can be calculated from the series
yn
Ei( − y) = γ + lny −
n!n
Solutions to the exponential integral can be coded into a spreadsheet and used with
the line source solution. Practically, the exponential integral can be replaced by a
simpler logarithm function as long as it is representative of the pressure decline. The
25φµcr 2
t>
limitation that y<0.01 corresponds to time, t, from the start of production k
. The equation can be applied anywhere in the reservoir, but is of significance at the
wellbore (i.e. for well test analysis) where typical values of wellbore radius, rw, and
reservoir fluid and rock parameters usually means that y<0.01 very shortly after
production starts. Therefore the line source solution can be approximated by
qµ γφµcr 2
P = Pi + ln
4πkh 4kt
qµ 4kt
P = Pi − ln
4πkh γφµcr 2 (3.18)
and if the pressure in the wellbore is of interest,
qµ 4kt
Pwf = Pi − ln
4πkh γφµcrw2 (3.19)
The values of exponential integral have been calculated and presented in Matthews
and Russel’s Monograph and are produced in Table 4. The table presents negative
values, i.e. -Ei(-y). For values of y<0.01, the ln approximation can be used. For values
>10.9, the decline in pressure calculated is negligible.
where rw is the wellbore radius. The value of 100 has been derived form the analysis
of the responses of real reservoirs; it can be varied according to the nature of a specific
well and reservoir. The time involved here is not the same as the dimensionless time,
tD calculated for other models of fluid flow in a reservoir (e.g. the input parameters for
the Hurst and van Everdingen solutions require the dimensionless time at the radius
where the dimensionless pressure drop is required - this may be the wellbore and rw
would be used or it may be some other radius).
where re is the external radius. The reservoir boundaries begin to effect the pressure
distribution in the reservoir after this time, the infinite acting period ends and the line
source solution does not represent the fluid flow.
EXERCISE 7
A well and reservoir are described by the following data:
Data
porosity, φ 19%
formation volume factor for oil, Bo 1.4rm3/stm3
net thickness of formation, h 100m
viscosity of reservoir oil, µ 1.4x10-3 Pas
compressibility, c 2.2 x10-9Pa-1
permeability, k 100mD
wellbore radius, rw 0.15m
external radius, re 900m
159
initial reservoir pressure, Pi 400bar
well flowrate (constant) 159stm3/day = 24x3600 stm3/second
skin factor 0
(3) the pressure in the reservoir at a radius of 50m after 4 hours production
(4) the pressure in the reservoir at a radius of 50m after 50 hours production
28
Fluid Flow In Porous Media
10
EXERCISE 8
A well flows at a constant rate of 20stm3/day. Calculate the bottomhole flowing pressure
at 8 hours after the start of production.
Data
porosity, φ 25%
formation volume factor for oil, Bo 1.32rm3/stm3
net thickness of formation, h 33m
viscosity of reservoir oil, µ 22.0x10-3 Pas
compressibility, c 0.6x10-9Pa-1
permeability, k 340mD
wellbore radius, rw 0.15m
external radius, re 650m
initial reservoir pressure, Pi 270bar
well flowrate (constant) 20stm3/day
skin factor 0
EXERCISE 9
Two wells are drilled into a reservoir. Well 1 is put on production at 20stm3 /day. Well
2 is kept shut in. Using the data given, calculate how long it will take for the pressure
in well 2 to drop by 0.5bar caused by the production in well 1. Well 2 is 50m from well 1.
Data
porosity, φ 18%
formation volume factor for oil, Bo 1.21rm3/stm3
net thickness of formation, h 20m
viscosity of reservoir oil, µ 0.8x10-3 Pas
compressibility, c 43x10-9Pa-1
permeability, k 85mD
wellbore radius, rw 0.15m
external radius, re 1950m
initial reservoir pressure, Pi 210bar
well flowrate (constant) 20stm3/day
skin factor 0
Distance well 1 to well 2 50m
Data
porosity, φ 16%
formation volume factor for oil, Bo 1.13rm3/stm3
net thickness of formation, h 10m
viscosity of reservoir oil, µ 5x10-3 Pas
compressibility, c 14x10-9Pa-1
permeability, k 10mD
wellbore radius, rw 0.15m
external radius, re 780m
initial reservoir pressure, Pi 86bar
well flowrate (constant) 2stm3/day
skin factor 0
EXERCISE 11
A well is put on production at 15stm3/day. The following well and reservoir data are
relevant.
Data
porosity, φ 21%
formation volume factor for oil, Bo 1.2rm3/stm3
net thickness of formation, h 23m
viscosity of reservoir oil, µ 5x10-3 Pas
compressibility, c 22 x10-9Pa-1
permeability, k 130mD
wellbore radius, rw 0.15m
external radius, re 800m
initial reservoir pressure, Pi 120bar
well flowrate (constant) 15stm3/day
skin factor 0
30
Fluid Flow In Porous Media
10
and completion (where the wellbore fluids alter the wettability of the near wellbore
formation as fluid leaks off into it, or solids suspended in the drilling fluids are
deposited in the pore spaces and become trapped thereby physically hindering the
flow of fluid and reducing the permeability) or during production (where sand or
precipitates from the hydrocarbon fluids or from formation brines can alter wettability
and plug pore spaces). Alternatively, wellbores intersecting fractures may exhibit
enhanced permeabilities as the fractures offer much greater conductive paths to the
fluids around the wellbore, thus enhancing the permeability. This situation may also
be required as part of the reservoir management: hydraulic fractures or acidising
workovers are performed on wells to bypass zones of reduced permeability which
have developed during production.
In these cases, the Ei equation fails to model the pressure drop in these wells properly
since it uses the assumption of uniform permeability throughout the drainage area
of the well up to the wellbore. Figure 7 shows the effect of a reduction in permeability
around a wellbore. The skin zone does not affect the pressures in the rest of the
formation remote from the wellbore, i.e. it is a local effect on the pressure drop at
the wellbore.
bottomhole following pressure, Pwf
Pwf(no skin)
actual pressure profile through skin zone
∆P skin
Pwf(skin)
skin zone
permeability, Ks permeability, K
rw rs
radius, r
Figure 7 Variation of the permeability around the wellbore changes the local pressure
profile
It can be shown that if the skin zone is considered equivalent to an altered zone of
uniform permeability, ks, with an outer radius, rs, the additional drop across this zone
(∆Ps) can be modelled by the steady-state radial flow equation. It is assumed that
after the pressure perturbation caused by the start of production has moved off into
the rest of the formation, the skin zone can be thought of as being in a steady state
flow regime. The pressure drop associated with the presence of a skin is therefore
the difference in the bottomhole flowing pressures at the well when skin is present
and when skin is not present, i.e.
Equation 22 simply states that the pressure drop in the altered zone is inversely
proportional to the permeability, ks rather than to the permeability, k of the rest of the
reservoir and that a correction to the pressure drop in this region must be made.
When this is included in the line source solution it gives the total pressure drop at
the wellbore:
qµ qµ k rs
Pi − Pwf = − Ei( − y) + ∆Ps = − Ei( − y) − 2 − 1 ln
4πkh 4πkh k s rw
(3.23)
If at the wellbore the logarithm approximation can be substituted for the Ei function,
then:
qµ γφµcrw2 k r
Pi − Pwf = − ln( ) − 2 − 1 ln s
4πkh 4kt k s rw
(3.24)
k r
s = − 1 ln s
k s rw (3.25)
qµ γφµcrw2
Pi − Pwf = − ln( ) − 2s
4πkh 4kt (3.26)
Equation 3.26 shows that a positive value of skin factor will indicate that the permeability
around the well has been reduced (by some form of formation damage). The absolute
value reflects the contrast between the skin zone permeability and the unaltered zone
permeability and the depth to which the damage extends into the formation. Part of the
essential information from a well test is the degree of formation damage (skin factor)
around a well caused by the drilling and completion activities. Alternatively, a well
may have a negative skin factor, i.e. the permeability of the skin zone may be higher
than that of the unaltered zone, caused by the creation of highly conductive fractures
or channels in the rock. The extent of the damage zone cannot be predicted accurately
and there may be variations vertically in the extent of the damage zone therefore this
simple model may not characterise the near wellbore permeability exactly.
An altered zone near a particular well affects only the pressure near that well, i.e. the
pressure in the unaltered formation away from the well is not affected by the existence
of the altered zone around the well.
32
Fluid Flow In Porous Media
10
EXERCISE 12.
A discovery well is put on well test and flows at 286stm3/day. After 6 minutes production,
the well pressure has declined from an initial value of 227bar to 192bar. Given the
following data, calculate the pressure drop due to the skin, ∆Pskin , and the mechanical
skin factor.
Data
porosity, φ, 28%
formation volume factor for oil, Bo 1.39rm3/stm3
net thickness of formation, h, 8.5m
viscosity of reservoir oil, µ 0.8x10-3 Pas
compressibility, c 2.3 x10-9Pa-1
permeability, k 100mD
wellbore radius, rw 0.15m
external radius, re 6100m
initial reservoir pressure, Pi 227bar
bottomhole flowing pressure
after 6 minutes 192bar
well flowrate (constant) 286stm3/day
EXERCISE 13
A reservoir and well are detailed in the following data. Use this data to calculate the
skin factor around the well after producing for 1.5 hours.
Data
porosity, φ 23%
formation volume factor for oil, Bo 1.36rm3/stm3
net thickness of formation, h 63m
viscosity of reservoir oil, µ 1.6x10-3 Pas
compressibility, c 17 x10-9Pa-1
permeability, k 243mD
wellbore radius, rw 0.15m
external radius, re 4000m
initial reservoir pressure, Pi 263.0bar
bottomhole flowing pressure
after 6 minutes 260.5bar
well flowrate (constant) 120stm3/day
δP dP
= = constant
δt dt
and where there is no flow across the outer boundary at r = re of the drainage zone,
i.e.
δP
= 0 at r = re
δr
In a similar manner to the steady state flow regime, the pressure difference between
the wellbore and, say, the external radius, or the pressure difference between the
wellbore pressure and the initial pressure, or the pressure difference between the
wellbore pressure and the average reservoir pressure can be calculated depending on
the physical measurements which have been taken. Usually, an average pressure is
known in a reservoir and this is used to determine the pressure drop. Figure 8 shows
the pressure profile in the reservoir and the values which may be relevant.
well with constant flow rate, q
initial pressure
Pi
pressure profile in reservoir calculated average pressure
height of formation
flowing pressure, P
Pe
Pwf
rw re
radius, r
Figure 8 Pressure profile in a reservoir under semi steady state flow conditions
Under semi steady state conditions, the pressure profile can be averaged over the
volume of the reservoir. This gives the average reservoir pressure at a particular time
in the stage of depletion of the reservoir. If there are several wells in a reservoir, each
well drains a portion of the total volume. For stabilised conditions, the volume drained
by each well is stable and in effect the whole reservoir can be subdivided into several
portions or cells. The average pressure in each cell can also be calculated from the
stabilised pressure profile. The calculation of the average pressure is determined from
the material balance of the initial pressure and volume of fluid and its isothermal
compressibility. The expansion of the fluid in each cell manifests itself as a volume,
or flow rate, at the well, i.e.
34
Fluid Flow In Porous Media
10
cV ( Pi − P ) = qt
(3.27)
where V = pore volume of the radial cell; q = constant production rate; t = total
flowing time, c = isothermal compressibility.
dV
q=
dt
dV qdt dt
= =q
dP dP dP
1 dV
since c = −
V dP T
dP
q = − cV
dt
dP q
=−
dt cV (3.28)
dP q
=−
dt cπre2 hφ (3.29)
1 δ δP φµc δP
(r ) =
r δr δr k δt
gives
1 δ δP φµc q
(r ) = −
r δr δr k cπre2 hφ
which is
1 δ δP qµ
(r ) = − 2
r δr δr πre hk
Integration gives
dP qµ 1 r
= − 2
dr 2πkh r re (3.31)
or
qµ r2 rw2
Pr − Pwf = lnr − − lnr −
2πkh 2re2 2re2
w
qµ r r2
Pr − Pwf = ln −
2πkh rw 2re2 (3.32)
rw2
2
The term 2 re is considered negligible, and in the case where the pressure at the
external radius, re is considered (including the skin factor, s, around the well),
qµ re 1
Pe − Pwf = ln − + s
2πkh rw 2 (3.33)
If the average pressure is used, then the volume weighted average pressure of the
drainage cell is calculated as previously in the steady state flow regime, i.e.
re
2
P=
r2e ∫ Prdr
rw (3.9)
where rw and re are the wellbore and external radii as before, and P is the pressure in
each radial element, dr at a distance r from the centre of the wellbore. In this case,
r
2 qµ e r r2
re 2πkh r∫w rw 2re2
P − Pwf = 2 r ln − dr
36
Fluid Flow In Porous Media
10
and integrating gives
(i)
re re r
r r2 r e
1 r2
∫r rw
r ln dr =
ln
2 r
w
− ∫r r 2 dr
w rw w
2 qµ e r r2
re r
re
r2 r rer 2
r wf re2 2πkh ∫ rw 2re2
= ln 2 qµ− r P − P 2 = r ln − dr
P − Pwf2 = r2w r ∫4rln − dr rw
re 2πkhw
rw
rw
rw 2re2
r2 r r2
≈ e ln e − e re
r r2
r
e re 1 r2
2 r 4 r
r ∫ 1 rr2
re
r
w
r2
re r re ln dr = ln − ∫ dr
2 r r 2
(ii)
∫r r ln rw dr = 2 ln rw r rw− r∫ r 2 dr w w rw rw
re
w w w
r 2 r r
r2 e
re r 2 r
re
r 2 re = ln −
re
r3 r4 2
re= ln − 2 rw rw 4 rw
∫ 2r 2 dr = 8r 2
e rw 8
≈ 2 rw rw 4 rw
re2 re re2
rw e 2 2 ≈ ln −
r r r 2 rw 4
≈ e ln e − e
and substitution into equation 3.32 2 with rw inclusion
4r e 3 of the skin factor
r gives
r r 4 e re2
r 4 e re2 ∫ 2 dr = 2 ≈
re r
r3 2re 8re r w 8
32 ≈
P − Pwf = ∫ 2re2 ln e −
q µ dr =r rw
+ s 8
rw 4e r w
8r
2rπw kh
(3.34)
qµ re 3
− wf
- Pwf), (Per- Pwf),3( P -P
The pressure differences (Prqµ Pwf) do=not change
ln with−time,
+ swhereas
P − P = ln e
− + s 2πkh rw 4
Pr, Pe, Pw and do change.
wf
2πkh rw 4
q
q P = Po + (t o − t )
EXERCISE 14P = Po + (t o − t ) cV
cV in
A well has been on production a reservoir which is in asemi-steady
qt state flow regime.
qt the bottomhole
For the following data, calculate Pi − pressure,
P =flowing P
cV
P = Pi −
wf
cV
Data
formation volume factor for oil, Bo 1.62rm3/stm3
net thickness of formation, h 72m P - P = qµ ln re − 3 + 2kt
3 -3 2kt 2πkh r
viscosity of reservoir oil, µ qµ 1.2x10r 4 φµcre 2
i wf
Pi - Pwf = ln e − +Pas w
permeability, k 2πkh
r123mD
w 4 φµcre 2
wellbore radius, rw 0.15m
external radius, re 560m
average reservoir pressure, 263.0bar qµ re 3
P −3 Pwf = ln −
qµ
well flowrate (constant) 216stm re 3 /day 2πkh rw 4
P − Pwf = ln −
skin factor 2πkh rw 0 4
re 3
re 3 ln −
3.4.1 Using The − Reservoir Pressure,
ln Initial rw Pi 4
r 4
If the pressure dropwfrom initial pressure conditions is required then equation
2 3.27
12 re 3 1 2 re 3 1 re
may be written as:
1 re 3 1 re 2 3 r 1 2 re2 r 2 2 2 ln r
2ln − = ln 3 − =
2ln − = ln − =w ln − lnw e w
2 rw 2 2 rw 2 2 rw
r 2
Institute of Petroleum Engineering, Heriot-Watt University r 2 37 e
e 1 rw
1 rw = ln 3
2
q
P = Po + ( t o − t )
cV (3.35)
qt
P = Pi −
cV (3.36)
where q is the volume flow rate, c is the isothermal compressibility, V is the original
volume to is a reference time after which flow starts, t is the flowing time, Po is the
pressure at the reference time and P is the pressure at time t after the flow starts. P
is the average reservoir pressure after time, t. Subtracting equation 36 from equation
34 gives
qµ re 3 2 kt
Pi - Pwf = ln − +
2πkh rw 4 φµcre 2
(3.37)
Using the average reservoir pressure and assuming no skin factor, the pressure drop
is described by equation 34 as
qµ re 3
P − Pwf = ln −
2πkh rw 4 (3.34)
re 3
ln −
Expressing the terms
rw 4 as
38
Fluid Flow In Porous Media
10
re 3 1 re 3 1 re 23
2 2
1
2ln − = ln
− = ln − ln e
2 rw 2 2 rw 2 2 rw
r 2
e
1 r
= ln w3
2
e2
1 (πre )
2
= ln
2 2 23
πrw e
The area drained (for a radial geometry) is πre2 therefore the logarithm term
becomes
( 4πre ) =
2
(4A)
2 2
3
(1.781 x 31.6 x rw2 )
4πrw e
where A is the area drained, and Dietz shape factor, CA (for a well in a
radial drainage area) =31.6.
The final form of the generalised semi steady state inflow equation for an average
reservoir pressure is
qµ 1 4A
P − Pwf = ln + s
2πkh 2 γC A rw
2
(3.38)
For the pressure drop between initial reservoir pressure conditions and some bottom
hole flowing pressure during semi steady state flow, equation 3.37 can be expressed
as
qµ 1 4A 2πkt
Pi − Pwf = ( ln + )
2πkh 2 γC A rw φµcA
2
(3.39)
or
qµ 1 4A 2πkt
Pwf = Pi − ( ln + )
2πkh 2 γC A rw φµcA
2
(3.40)
2πkh 1 4A kt rw2
(P - Pwf ) = ln + 2π
qµ 2 γC A rw2 φµcrw2 A
or
1 4A rw2
PD t D = ln + 2πt D
2 γC A rw2 A (3.41)
The term involving the wellbore radius can be accommodated by using the following
modified dimensionless time
rw2
t DA = t D
A
in which case
1 4A
PD t D = ln + 2πt DA
2 γC A rw2
The calculation of the Dietz shape factors and their limitations in use is presented
in Lee and reproduced in Table 5. There are a series of common simple shapes with
wells located close to certain barriers and the shape factors associated with them.
There are also values of tDA which indicate the use of the shape factors.
(i) The infinite system solution with less than 1% error for tDA < X in this case, X
is the value of the maximum elapsed time during which a reservoir is infinite acting
and the Ei function can be used. The time, t is calculated by
φµcA
t < t DA
k
This time is different to that quoted earlier in the section on the line source solution
and reflects the subjective decision as to the acceptable accuracy of the solution using
the Ei function.
(ii) The solution with less than 1% error for tDA > X in this case, the semi steady
state solution can be used with the results having an error less than 1% for an elapsed
time, t
φµcA
t > t DA
k
(iii) The solution which is exact for tDA > X in this case, the semi steady state
solution can be used with the results being exact for an elapsed time, t
φµcA
t > t DA
k
40
Fluid Flow In Porous Media
10
For a real reservoir under semi steady state conditions, the volume of reservoir drained
by a well can be determined from its flow rate, and this volume correlated to the
structural map of the reservoir to determine the shape. The values of shape factor can
then be used to locate the position of the well relative to the boundaries of the area
being drained. This is not an exact procedure and variations in the heterogeneity of
the reservoir can alter the pressure responses, however, it is an analytical step in the
characterisation of the reservoir.
EXERCISE 15
For each of the following geometries, calculate the time in hours for which the reservoir
is infinite acting
Geometry
1. Circle
2. Square
3. Quadrant of a square
Data
Area of reservoir, A 1618370m2
viscosity of reservoir oil, µ 1.0x10-3 Pas
permeability, k 100mD
porosity, φ, 20%
compressibility, c 1.45 x10-9Pa-1
The times are calculated by the dimensionless time, diffusivity of the reservoir and the
area of the reservoir. The dimensionless time accounting for the reservoir drainage
area is found for the conditions in Table 5.
In reality, only flow rates and pressures at wells can be measured directly, and the most
important unknown factor in the diffusivity equation is the permeability. Therefore,
rather than calculate a pressure drop for a given set of conditions, the pressure drop
can be continuously measured and the permeability calculated.
This is part of the objectives of well testing and for illustration, the following example
calculates the permeability and skin factor for a well in a reservoir. It is important to
note that these examples all assume that an initially undisturbed reservoir is brought
on production, i.e. that there has been no previous production in the reservoir therefore
the pressure is at its initial value. In well test analysis, the previous history of a well
must be accounted for. The section on superposition will introduce the concepts of
a multi-rate history for a well.
Data
porosity, φ 18%
formation volume factor for oil, Bo 1.2rm3/stm3
net thickness of formation, h 6.1m
viscosity of reservoir oil, µ 1x10-3 Pas
compressibility, c 2.18 x10-9Pa-1
wellbore radius, rw 0.1m
initial reservoir pressure, Pi 241.3bar
well flowrate (constant) 238stm3/day
0.0 241.3
1.0 201.1
2.0 199.8
3.0 199.1
4.0 198.5
5.0 197.8
7.5 196.5
10.0 195.3
15.0 192.8
30.0 185.2
40.0 180.2
50.0 176.7
60.0 173.2
70.0 169.7
80.0 166.2
90.0 162.7
100.0 159.2
2. Make an estimate of the area being drained by the well and the Dietz shape
factor.
42
Fluid Flow In Porous Media
10
EXERCISE 17
An appraisal well is tested by producing at a constant rate of 200stm3/day for 107
hours. The following table of flowing bottomhole pressures and time were recorded
during the test. Using the data,
Data
porosity, φ 22%
formation volume factor for oil, Bo 1.3rm3/stm3
net thickness of formation, h 21m
viscosity of reservoir oil, µ 1.9x10-3 Pas
compressibility, c 4.3 x10-9Pa-1
wellbore radius, rw 0.15m
initial reservoir pressure, Pi 378.7bar
well flowrate (constant) 200stm3/day
0.0 378.7
1.1 326.41
2.1 324.7
3.2 323.8
4.3 323.1
5.4 322.1
8.0 320.5
10.7 318.8
16.1 315.5
21.4 312.2
32.1 305.6
42.8 300.8
53.5 296.0
64.2 291.2
74.9 286.3
85.6 281.5
96.3 276.7
107.0 271.9
In the constant terminal rate solution of the diffusivity equation, the rate is known to
be constant at some part of the reservoir and the pressures are calculated throughout
the reservoir. Conversely, in the constant terminal pressure solution, the pressure is
known to be constant at some point in the reservoir, and the cumulative flow at any
particular radius can be calculated. The constant terminal pressure solution is not as
confusing as the constant terminal rate solution simply because less is known about
it. Only one constant terminal pressure solution is available, so there is no decision
to be made over which to use as in the case of the constant terminal rate solutions.
Hurst and Van Everdingen produced the solutions for cases of an infinite radial system
with a constant pressure at the inner boundary and for constant pressure at the inner
boundary and no flow across the outer boundary. These can model, for example, a
wellbore whose bottomhole flowing pressure is held constant whilst flow occurs
in the reservoir, or they can model a reservoir surrounded by an aquifer. The same
geometrical and property conditions apply as for the constant terminal rate solutions:
a radial geometry of constant thickness with a well in the centre, and with fixed rock
and fluid properties throughout, however, in this case there is a pressure drop from
an initial pressure to some constant value. In the case of aquifer encroachment, the
radius of the “well” is the radius of the initial oil water contact. The constant terminal
pressure solution is most widely used for calculating the water-encroachment (natural
water influx) into the original oil and gas zone due to water drive in a reservoir. This
topic is covered in the chapter on water influx.
5. Superposition
In the analyses so far, the well flow rate has been instantly altered from zero to
some constant value. In reality, the well flowrates may vary widely during normal
production operations and of course the wells may be shut in for testing or some other
operational reason. The reservoir may also have more than a single well draining it
and consideration must be taken of this fact. In short, there may be some combination
of several wells in a reservoir and/or several flowrates at which each produce. The
calculation of reservoir pressures can still be done using the previous simple analytical
techniques if the solutions for each rate change, for example, are superposed on each
other. In other words, the total pressure drop at a wellbore can be calculated as the
sum of the effects of several flowrate changes within the well, or it may be the sum
of the effects caused by production from nearby wells.
There is also the possibility of using infinite acting solutions to mimic the effects
of barriers in the reservoir by using imaginary or image wells to produce a pressure
response similar to that caused by the barrier.
44
Fluid Flow In Porous Media
10
These two properties form the basis for generating the constant terminal rate and
constant terminal pressure cases. The solutions may be added together to determine the
total effect on pressure, for example, from several applications of the equation. This is
illustrated if a typical problem is considered: that of multiple wells in a reservoir.
(Pi-Pwf)Total at Well Y
+ (Pi-P)Due to well Z
Assuming unsteady state flow conditions, the line source solution can be used to
determine the pressure in well Y. It is assumed here that the logarithm function can
be used for well Y itself and that there will be a skin around the well. The effects of
wells X and Z can be described by the Ei function. There is no skin factor associated
with the calculation of pressure drop caused by these wells, since the pressure drop
of interest is at well Y (i.e. even if wells X and Z have non-zero skin factors, their
skin factors affect the pressure drop only around wells X and Z). The total pressure
drop is then:
where
qY is the flowrate from well Y
qX is the flowrate from well X
qZ is the flowrate from well Z
rwY is the radius of well Y
rXY is the distance of well Y from the X well
rZY is the distance of well Z from the X well
the rest of the symbols have their usual meaning
This technique can be used to examine the effects of any number of wells in an infinite
acting reservoir. This could be to predict possible flowing well pressures amongst a
group of wells, or to deliberately use the interaction between wells to check reservoir
continuity. These interference tests and other extended well tests are designed to
characterise the reservoir areally rather than to determine only the permeability and
skin factor around individual wells.
EXERCISE 18
Two wells, well 1 and well 2, are drilled in an undeveloped reservoir. Well 1 is completed
and brought on production at 500stm3/day and produces for 40 days at which time Well
2 is completed and brought on production at 150stm3/day. Using the data provided,
calculate the pressure in Well 2 after it has produced for 10 days (and assuming Well
1 continues to produce at its flowrate). Therefore, Well 1 produces for 50days when
its pressure influence is calculated; Well 2 produces for 10 days when its pressure
influence is calculated.
The wells are 400m apart and the nearest boundary is 4000m from each well.
Data
porosity, φ, 21%
formation volume factor for oil, Bo 1.4rm3/stm3
net thickness of formation, h, 36m
viscosity of reservoir oil, µ 0.7x10-3 Pas
compressibility, c 8.7 x10-9Pa-1
permeability, k 80mD
wellbore radius, rw (both wells) 0.15m
initial reservoir pressure, Pi 180.0bar
Well 1 flowrate (constant) 500stm3/day
Well 2 flowrate (constant) 150stm3/day
skin factor around both wells 0
46
Fluid Flow In Porous Media
10
5.2 Principle of Superposition and Approximation of Variable - Rate
Pressure Histories
The previous section illustrated the effect of the production from several wells in a
reservoir on the bottomhole flowing pressure of a particular well. Of equal interest
is the effect of several rate changes on the bottomhole pressure within a particular
well. This is a more realistic situation compared to those illustrated previously where
a well is simply brought on production at a constant flowrate for a specific period
of time. For instance, a newly completed well may have several rate changes during
initial cleanup after completion, then during production testing then finally during
production as rates are altered to match reservoir management requirements (for
example limiting the producing gas oil ratio during production). A simple pressure
and flowrate plot versus time would resemble Figure 10.
q2
q1 (q2 - q1)
flowrate, q
t1 time, t
t1 time, t
The well has been brought onto production at an initial flowrate, q1. The bottomhole
flowing pressure has dropped through time (as described by the appropriate boundary
conditions and the flow regime) until at time t1, the flowrate has been increased to q2
and this change from q1 to q2 has altered the bottomhole flowing pressure (again as
described by the boundary conditions and the flow regime). The total (i.e. the real
bottomhole flowing pressure) is calculated by summing the pressure drops caused by
the flowrate q1 bringing the well on production, plus the pressure drop created by the
flowrate change q2 - q1 for any time after t1. During the first period (q1) the pressure
drop at a time, t, is described by
qiµ
∆P( t ) = Pi - Pwf = ∆PD ( t )
2πkh (5.2)
where ∆PD(t) is the dimensionless pressure drop at the well for the applicable boundary
condition.
qiµ (q − q)
∆P( t ) = ∆PD ( t ) + 2 µ ∆PD ( t - t1 )
2πkh 2πkh (5.3)
In this case, the pressure drop is that caused by the rate q1 over the duration t, plus
the pressure drop caused by the flowrate change q2 - q1 over the duration t - t1. In fact,
the pressure perturbation caused by q1 still exists in the reservoir and is still causing
an effect at the wellbore. On top of that, the next perturbation caused by flowrate
change q2 - q1 is added or superposed to give the total pressure drop ( at the wellbore
in this case).
In mathematical terms:
qiµ
0 ≤ t ≤ t1: ∆P(t) = ∆PD (t)
2πkh (5.4)
qiµ q − q1
t > t1 : ∆P(t) = ∆PD (t) + 2 µ ∆PD (t - t1 )
2πkh 2πkh (5.5)
In this 2nd equation, the first term is ∆P from flow at q1 : 2nd term is the incremental
term ∆P caused by increasing rate by an increment (q2-q1). These expressions are
valid regardless of whether q2 is larger or smaller than q1 so that even if the well is
shut in, the effects of the previous flowrate history are still valid.
The dimensionless pressure drop function depends as mentioned on the flow regime
and boundaries. If unsteady state is assumed and the line source solution applied,
then
Pi − Pwf 1 −φµcr 2 w
∆PD = = − Ei ( )
qµ / 2πkh 2 4 kt (5.6)
and the equation for time, t less than or equal to t1 would be as expected
q1µ −φµcr 2 w
∆P(t) = - Ei ( )
4πkh 4 kt (5.7)
For times greater than t1 the additional pressure drop is added to give
This approach can be extended to many flowrate changes as illustrated in figure 11.
48
Fluid Flow In Porous Media
10
different flow rates
q4
q3
flowrate, q
q2
q1
time, t
q1µ (q − q1 ) µ (q − q 2 ) µ
∆P(t) = ∆PD (t) + 2 ∆PD (t − t1 ) + 3 ∆PD (t − t 2 ) + ...
2πkh 2πkh 2πkh
(q − q n −1 ) µ
+ n ∆PD (t − t n −1 )
2πkh
(5.9)
or
q1µ n
q i − q i −1
∆P(t) =
2πkh
∆PD (t) + ∑ q1
∆PD (t − t i −1 )
i=2 (5.10)
This is the general form of the principle of superposition for multi rate history wells.
For the specific case where the well is shut in and the pressure builds up, an additional
term is added to reflect this. Assuming that the well was shut in during the nth flowrate
period, the pressure builds during the shut in time, ∆t (i.e. ∆t starts from the instant
the well is shut in) back up towards the initial reservoir pressure according to
q1µ n
q i − q i −1 q µ
Pi − Pws =
2πkh
∆PD (t) + ∑ ∆PD (t n-1 − t i −1 + ∆t) − n-1 ∆PD ( ∆t)
1= 2 q1 2πkh
(5.11)
It is the bottomhole flowing pressure, Pwf, that is of interest, and it can be calculated
using the line source solution. There is the possibility of a skin zone around the well,
so this must be accounted for. If no other flowrate change occurred, then eventually
unsteady state would give way to either semi steady state or steady state conditions
and the bottomhole flowing pressure would either decline at a steady rate or (if steady
state) would remain constant at some level. Assuming that this did not occur and that
unsteady state conditions still existed when the flowrate was changed to q2 then the
change q2 - q1 would cause a second pressure perturbation that would move out into
the reservoir, following the first one created when the well was put on production. The
reservoir is still in unsteady state conditions i.e. the first pressure perturbation has not
met any barriers so the reservoir fluid still reacts as if it were an infinite volume and
this behaviour is still causing a decline in the pressure at the wellbore even though a
second pressure perturbation has been created and is moving out into the reservoir.
The pressure drop due to this flowrate change can be calculated by the line source
solution and added to that produced by bringing the well onto production.
Eventually at time t2, the flowrate is changed again. This time, the pressure perturbation
caused by q3 -q2 follows the first and second perturbations into the reservoir, and
again, as long as the reservoir fluid still behaves as if it were infinite in volume, the
pressure drop created by this flowrate change can be added to the changes produced
by the others to give the total pressure drop.
50
Fluid Flow In Porous Media
10
real well flowrate history
q3
q1
flowrate, q
q2
time, t
t1 t2
time, t
flowrate, q
q2 - q1
t1
time, t
q3 - q2
time, t
t2
The pressure drop produced by bringing the well onto production is calculated by the
logarithmic approximation of the Ei function (it is assumed that the checks have been
made to the applicability of the Ei function and its logarithmic approximation).
− q1µ γφµcrw2
∆P1 = ( Pi − Pwf )1 = ln − 2s
4πkh 4kt
The next pressure drop is that produced by the flowrate change q2 - q1 at time, t1. It
is still the bottomhole flowing pressure that is to be determined, therefore any skin
zone will still exist and still need to be accounted for. The second pressure drop is:
−(q 3 - q 2 ) µ γφµcrw2
∆P3 = ( Pi − Pwf )3 = ln − 2s
4πkh 4k(t - t 2 )
The total pressure drop at the wellbore caused by all of the flowrate changes is
EXERCISE 19
Two wells are brought on production in an undeveloped reservoir. Using the data below,
calculate the bottomhole flowing pressure in each well. Well 1 produces at 110stm3/day
for 27 days at which time Well 2 starts production at 180stm3/day and both produce
at their respective rates for a further 13 days when the bottomhole flowing pressures
are calculated. Therefore Well 1 produces for 40 days when its pressure influence
is calculated; Well 2 produces for 13 days when its pressure influence is calculated.
Data
porosity, φ, 19%
formation volume factor for oil, Bo 1.2rm3/stm3
net thickness of formation, h, 36m
viscosity of reservoir oil, µ 1x10-3 Pas
compressibility, c 10 x10-9Pa-1
permeability, k 110mD
wellbore radius, rw (both wells) 0.15m
external radius, re 7000m
initial reservoir pressure, Pi 250.0bar
Well 1 flowrate (constant) 110stm3/day
Well 2 flowrate (constant) 180stm3/day
skin factor around both wells 0
52
Fluid Flow In Porous Media
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EXERCISE 20
A well is completed in an undeveloped reservoir described by the data below. The well
flows for 6 days at 60 stm3/day and is then shut in for a day. Calculate the pressure
in an observation well 100m from the flowing well.
Data
porosity, φ, 19%
formation volume factor for oil, Bo 1.3rm3/stm3
net thickness of formation, h, 23m
viscosity of reservoir oil, µ 0.4x10-3 Pas
compressibility, c 3 x10-9Pa-1
permeability, k 50mD
wellbore radius, rw (both wells) 0.15m
external radius, re 6000m
initial reservoir pressure, Pi 180.0bar
flowrate (constant) 60stm3/day
skin factor around well 0
EXERCISE 21
A well in a reservoir is brought on production at a flowrate of 25stm3/day for 6 days.
The production rate is then increased to 75stm3/day for a further 4 days. Calculate,
using the data given, the bottomhole flowing pressure at the end of this period, i.e.
10 days.
Data
porosity, φ, 21%
formation volume factor for oil, Bo 1.31rm3/stm3
net thickness of formation, h, 20m
viscosity of reservoir oil, µ 0.6x10-3 Pas
compressibility, c 8 x10-9Pa-1
permeability, k 75mD
wellbore radius, rw (both wells) 0.15m
external radius, re 5000m
initial reservoir pressure, Pi 200.0bar
1st flowrate (constant) 25stm3/day
1st flowrate period 6days
2nd flowrate (constant) 75stm3/day
2nd flow period 4days
skin factor around well 0
Therefore if an imaginary well is placed at a distance from the real well equal to
twice the distance to the boundary, and the flowrate histories are identical, then the
principle of superposition can be used to couple the effect of the imaginary well
to the real well in order to calculate the real well’s bottomhole flowing pressure.
Figure 13 illustrates the problem and the effect of superposition. Figure 14 shows a
simplification of the model.
Figure 13 The pressure effect of the barrier in the real reservoir can be represented by an
imaginary well
54
Fluid Flow In Porous Media
10
Can be modelled as
drainage boundary
between wells
L reservoir L L
boundary
actual well image well
qµ γφµcrw2 qµ −φµc(2L)2
Pi − Pwf = − ln( ) − 2s − Ei
4πkh 4kt 4πkh 4kt
where the symbols have their usual meaning, and L is the distance from the real well
to the fault. The skin factor is used in the actual well, but not in the other (image)
well since it is the influence of this image well at a distance 2L from it that is of
interest.
Data
porosity, φ, 19%
formation volume factor for oil, Bo 1.4rm3/stm3
net thickness of formation, h, 20m
viscosity of reservoir oil, µ 1x10-3 Pas
compressibility, c 9 x10-9Pa-1
permeability, k 120mD
wellbore radius, rw 0.15m
external radius, re 4000m
initial reservoir pressure, Pi 300.0bar
flowrate (constant) 120stm3/day
flowrate period, t 50days
distance to fault, L 300m
skin factor around well 0
There are other examples of the use of image wells to mimic the effect of boundaries
on flow. The larger networks require computer solution to relieve the tedium. To
complicate the simple fault boundary described earlier, consider the effect of a well
near the corner of a rectangular boundary. In this case, there are more image wells
required to balance the flow from the real well. Figure 15 shows the boundary and
the image wells.
Figure 17
image well 1 image well 3
L2 L2 Representation of a well
surrounded by boundaries
L1 L1
R3
L1 L1
L2 L2
Four pressure drop terms are required to determine the pressure at the actual well.
The total pressure drop then is the sum of the pressure drops caused by all of the
wells at the actual well.
56
Fluid Flow In Porous Media
10
Pi - Pwf = (∆P)rw + (∆P)2L1 + (∆P)2L2 + (∆P)r3
(Pi-Pwf)Total at the actual well = (Pi -P)at the actual wellbore radius, rw
+ (Pi-P)Due to image well 1 at distance 2L1
+ (Pi-P)Due to image well 2 at distance 2L2
+ (Pi-P)Due to image well 3 at distance R3
actual well
image wells
i7 i6 i3 i2 i1 i4 i5
parallel equidistant
boundaries
Even more complex patterns can be devised. Mathews, Brons and Hazebroek
(Matthews, CS, Brons, F and Hazebroek, P, A Method for the Determination of Average
Pressure in a Bounded reservoir. Trans. AIME.201) studied the pressure behaviour of
wells completely surrounded by boundaries in rectangular shaped reservoirs. Figure
17 shows the network of wells set up to mimic the effect of the boundaries.
EXERCISE 23
A well in a reservoir is producing close to two intersecting faults as shown below.
Using the data given, calculate the bottomhole flowing pressure after 32 days and
indicate the effect of the faults on the bottomhole flowing pressure. The production
rate is constant at 100stm3 /day
fault
L1 70m
fault
L2
120m
well
Data
porosity, φ, 22%
formation volume factor for oil, Bo 1.5rm3/stm3
net thickness of formation, h, 36m
viscosity of reservoir oil, µ 1x10-3 Pas
compressibility, c 9 x10-9Pa-1
permeability, k 89mD
wellbore radius, rw 0.15m
external radius, re 6000m
initial reservoir pressure, Pi 240.0bar
flowrate (constant) 100stm3/day
flowrate period, t 32days
distance to fault, L1 70m
distance to fault, L2 120m
skin factor around well 0
58
Fluid Flow In Porous Media
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EXERCISE 24
A well is 80m due west of a north-south fault. From well tests, the skin factor is 5.0.
Calculate the pressure in the well after flowing at 80stm3/day for 10 days.
Data
porosity, φ, 25%
formation volume factor for oil, Bo 1.13rm3/stm3
net thickness of formation, h, 23m
viscosity of reservoir oil, µ 1.1x10-3 Pas
compressibility, c 10.1 x10-9Pa-1
permeability, k 125mD
wellbore radius, rw 0.15m
external radius, re 6000m
initial reservoir pressure, Pi 210.0bar
flowrate (constant) 80stm3/day
flowrate period, t 32days
distance to fault, L 80m
skin factor around well 5.0
6. Summary
The basic partial differential equation expressing the nature of fluid flow in a porous
rock has been illustrated in the context of petroleum reservoirs. Only oil and water
have been used as the simplifications for solving the diffusivity equation have
required the compressibility of the fluid to be small and constant. This is the reason
that the compressibility of the fluid in the examples has not changed with pressure
as would be expected. So, for instance, the same value of compressibility is used for
the fluid at the wellbore which may be under a lower pressure than the same fluid at,
for example, the external radius of the reservoir.
In gasses, the same diffusion process occurs, but the pressure dependence of the
gas is accommodated by various mathematical devices which again lead to simple
working solutions.
The assumptions made concerning the geological structure and the petrophysical
properties of the rock may appear radical: to assume a reservoir is circular, horizontal
and has identical permeability in all directions is a great simplification of the problem.
Yet these simple analytical solutions allow an appreciation of the role of the fluids
and the rock in a producing reservoir. For more realistic treatments of real reservoirs,
approximations to the diffusivity equation are made from which simple algebraic
relationships can be formed. This process is encapsulated in reservoir simulation
where the reservoir (with its properties) is subdivided into small blocks within which
the flow equations have been approximated by simple relationships. These can then
be solved by a process of iteration to achieve an acceptable result. The great potential
of this process is the ability to represent the shape of the reservoir and the changing
properties, vertically and horizontally, throughout the reservoir.
The subject of Well Testing is considerable and is covered in the separate module
with that title.
Summary of the application of analytical solutions of the Diffusivity equation in this chapter
traditional assumption specific reservoir short time application short time application
for most reservoirs aquifer influxes formation testing devices formation testing devices
thin layers thin layers
60
Fluid Flow In Porous Media
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Solutions to Exercises
EXERCISE 1
A well produces oil at a constant flowrate of 15 stock tank cubic metres per day
(stm3/d). Use the following data to calculate the permeability in milliDarcys (mD).
Data
porosity, φ 19%
formation volume factor for oil, Bo 1.3rm3/stm3 (reservoir cubic metres per
stock tank cubic metre)
net thickness of formation, h, 40m
viscosity of reservoir oil, µ 22x10-3 Pas
wellbore radius, rw 0.15m
external radius, re 350m
initial reservoir pressure, Pi 98.0bar
bottomhole flowing pressure, Pwf 93.5bar
qreservoir = qstock tank x Bo
1bar = 105 Pa
Solution EXERCISE 1
the steady state inflow equation (accounting for fluid flowrate at reservoir conditions
in m3/s and pressure in Pa) is
qµBo re
Pe − Pwf = ln
2πkh rw
qµBo r
k= ln e
2π (Pe − Pwf )h rw
15x22x10 −3 x1.3 350.00
k= ln
24x3600x2πx(98.0 − 93.5)x10 x40 0.15
5
= 341x10 −15 m 2
= 341mD (3.7)
Data
porosity, φ, 23%
formation volume factor for oil, Bo 1.36rm /stm3
3
Solution EXERCISE 2
the steady state inflow equation (accounting for fluid flowrate at reservoir conditions
in m3/s and pressure in Pa) is
qµBo re 1
P − Pwf = ln −
2πkh rw 2
qµBo re 1
k= ln −
( )
2π P − Pwf h rw 2
13 x14 x10 −3 x1.36 ln 210.00 − 1
k=
24 x 3600 x 2π (132.6 − 125.0) x10 x 23
5
0.15 2
k = 176 x10 −15 m 2
k = 176 mD
62
Fluid Flow In Porous Media
10
EXERCISE 3
A reservoir is expected to produce at a stabilised bottomhole flowing pressure of 75.0
bar. Use the following reservoir data to calculate the flowrate in stock tank m3/day.
Data
porosity, φ 28%
formation volume factor for oil, Bo 1.41rm3/stm3
net thickness of formation, h 15m
viscosity of reservoir oil, µ 21x10-3 Pas
wellbore radius, rw 0.15m
external radius, re 250m
average reservoir pressure, 83.0bar
bottomhole flowing pressure, Pwf 75.0bar
permeability, k 125mD
Solution EXERCISE 3
the steady state inflow equation (accounting for fluid flowrate at reservoir conditions
in m3/s and pressure in Pa) is
qµBo re 1
P − Pwf = ln −
2πkh rw 2
q=
( P − P )2πkh
wf
r 1
µBo ln e −
rw 2
(83.0 − 75.0) x10 5 x 2π 125 x10 −15 x15
q=
250.00 1
21x10 −3 x1.41x ln −
0.15 2
q = 46 x10 −6 stm 3 / s
Data
porosity, φ 21%
formation volume factor for oil, Bo 1.13rm3/stm3
net thickness of formation, h 53m
viscosity of reservoir oil, µ 10x10-3 Pas
wellbore radius, rw 0.15m
external radius, re 150m
initial reservoir pressure, Pi 83.0bar
permeability, k 140mD
compressibility, c 0.2x10-7Pa-1
Solution EXERCISE 4
Using Hurst and Van Everdingen’s solution for Constant Terminal Rate, the
dimensionless external radius and the dimensionless time are calculated and used
with the appropriate solution to determine the dimensionless pressure drop. The
dimensionless pressure drop is then turned into the real pressure drop from which
the bottomhole flowing pressure is calculated.
re 150.00
reD = = = 1000
rw 0.15
kt 140x10 -15 xt
tD = = = 0.148t
φµcrw2 0.21x10x10 -3 x0.2x10 −7 x0.152
the bottomhole flowing pressure, Pwf is found from re-arrangement of the dimensionless
2πkh
PD = ( pi − pwf )
pressure q µ . Accounting for the oil formation volume factor,
Bo, the bottomhole flowing pressure, pwf, is:
64
Fluid Flow In Porous Media
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qµBo
Pwf = Pi − PD
2πkh
23x10x10 −3 x1.13
Pwf at 0.01hour = 83.0x10 5 − −15
x1.3846 = 82.1x10 5 Pa
24x3600x2π 140x10 x53
i.e. Pwf at 0.01 hour =82.1bar
time PD Pwf
(hour) (bar)
0.00 0 83.0
0.01 1.3846 82.1
0.10 2.4146 81.4
1.00 3.5473 80.7
10.00 4.6949 80.0
100.00 5.8462 79.2
fluid production
flow to the
wellbore
Data
porosity, φ 25%
net thickness of formation, h 0.2m
viscosity of fluid, µ 2x10-3 Pas
wellbore radius, rw 0.2m
external radius, re 2m
initial reservoir pressure, Pi 2bar
permeability, k 1200mD
compressibility, c 0.15x10-7Pa-1
Solution EXERCISE 5
Using Hurst and Van Everdingen’s solution for CTR, the dimensionless external
radius and the dimensionless time are calculated and used with the appropriate
solution to determine the dimensionless pressure drop. The dimensionless pressure
drop is then turned into the real pressure drop from which the bottomhole flowing
pressure is calculated.
re 2.0
reD = = = 10
rw 0.2
66
Fluid Flow In Porous Media
10
time time tD PD expression
(hour) (second) (4t)
0.001 3.6 14.4 1.808 table 2
0.005 18.0 72.0 3.048 table 3 since tD
is not less than
2
0.25reD i.e. not
infinite acting
2t D 3
PD = + lnreD −
0.100 360.0 1440.0 30.35 2
reD 4
qµ
Pwf = Pi − PD
2πkh
0.1x 2 x10 −3
Pwf at 0.001 hour = 2 x10 5 − −15
x1.814 = 1.97 x10 5 Pa
24 x 3600 x 2π 1200 x10 x 0.2
time PD Pwf
(hour) (bar)
0 0 2.00
0.001 1.808 1.97
0.005 3.048 1.95
0.100 30.35 1.53
Data
porosity, φ 17%
net thickness of formation, h 40m
viscosity of reservoir oil, µ 14x10-3 Pas
formation volume factor of oil, Bo 1.27rm3/stm3
wellbore radius, rw 0.15m
external radius, re 900m
initial reservoir pressure, Pi 200bar
permeability, k 150mD
compressibility, c 0.9x10-9Pa-1
Solution EXERCISE 6
Using Hurst and Van Everdingen’s solution for CTR, the dimensionless external
radius and the dimensionless time are calculated and used with the appropriate
solution to determine the dimensionless pressure drop. The dimensionless pressure
drop is then turned into the real pressure drop from which the bottomhole flowing
pressure is calculated.
re 900
reD = = = 6000
rw 0.15
qµ
Pwf = Pi − PD
2πkh
68
Fluid Flow In Porous Media
10
EXERCISE 7
A well and reservoir are described by the following data:
Data
porosity, φ 19%
formation volume factor for oil, Bo 1.4rm3/stm3
net thickness of formation, h 100m
viscosity of reservoir oil, µ 1.4x10-3 Pas
compressibility, c 2.2 x10-9Pa-1
permeability, k 100mD
wellbore radius, rw 0.15m
external radius, re 900m
initial reservoir pressure, Pi 400bar
159
well flowrate (constant) 159stm3/day = stm3/second
skin factor 0 24x3600
(3) the pressure in the reservoir at a radius of 50m after 4 hours production
(4) the pressure in the reservoir at a radius of 50m after 50 hours production
Solution EXERCISE 7
The line source solution is used to determine the pressures required at the specified
radii and at the specified times (i.e. using the flowrate measured at the wellbore, the
pressures at the other radii and times are calculated by the line source solution). SI
units will be used so time will be converted to seconds. Checks are made to ensure
that:
(i) there has been adequate time since the start of production to allow the line
source solution to be accurate
A. Check Ei applicability
t >13.2s
φµcre2
t<
the reservoir is infinite acting if the time, 4k (3.21)
i.e.
t<
4x100x10 -15
t < 1185030s
(1) the bottomhole flowing pressure after 4 hours production, Pwf at 4 hours
25φµcrw2
t>
the ln approximation is valid if the time, k
2
25x0.19x1.4x10 −3 x2.2x10 −9 x0.15
t>
100x10 -15
t > 3.3s
qµBo γφµcrw2
Pwf = Pi + ln
(ii) 4πkh 4kt (taking account of the conversion from stock
tank to reservoir conditions via the formation volume factor for oil, Bo, flow rates
in reservoir m3/s and pressures in Pascal).
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Fluid Flow In Porous Media
10
qµBo 159x1.4x10 −3 x1.4
= = 28703
4πkh 24x3600x4πx100x10 −15 x100
(2) the pressure after 4 hours production at a radius of 9m from the wellbore
25φµcr 2
t>
the ln approximation is valid if the time, k
t > 3.3hours
qµBo γφµcr
2
P = Pi + ln
(ii)
4πkh 4kt (taking account of the conversion from stock tank
to reservoir conditions via the formation volume factor for oil, Bo and also the fact
that the radius, r, is now at 9m from the wellbore).
25φµcr 2
t>
the ln approximation is valid if the time, k
qµBo φµcr
2
P = Pi + Ei −
(ii) 4πkh 4kt (taking account of the conversion from stock tank
to reservoir conditions via the formation volume factor for oil, Bo and also the fact
that the radius, r, is now at 50m from the wellbore).
P = 400x105 + 28703xEi(-0.254)
P = 400x105 +28703x-1.032
= 400x105 -29622
= 39970378Pa
= 399.7bar
(4) the pressure after 50 hours production at a radius of 50m from the wellbore
25φµcr 2
t>
the ln approximation is valid if the time, k
t > 365750s
72
Fluid Flow In Porous Media
10
t > 101.6 hours
qµBo φµcr
2
P = Pi + Ei −
(ii)
4πkh 4kt (taking account of the conversion from stock
tank to reservoir conditions via the formation volume factor for oil, Bo and also the
fact that the radius, r, is now at 50m from the wellbore and the time is now 50hours
after start of production).
P = 400x105 + 28703xEi(-0.020)
Ei(-0.020) = -3.355
P = 400x105 +28703x-3.355
= 400x105 -96300
= 39903700Pa
= 399.0bar
Summary
0 all 400.0
4 0.15 396.4
4 9.00 398.8
4 50.00 399.7
50 50.00 399.0
Data
porosity, φ 25%
formation volume factor for oil, Bo 1.32rm3/stm3
net thickness of formation, h 33m
viscosity of reservoir oil, µ 22.0x10-3 Pas
compressibility, c 0.6x10-9Pa-1
permeability, k 340mD
wellbore radius, rw 0.15m
external radius, re 650m
initial reservoir pressure, Pi 270bar
well flowrate (constant) 20stm3/day
skin factor 0
Solution EXERCISE 8
The line source solution is used to determine the pressures required at the specified
radius and at the specified time. Checks are made to ensure that:
(i) there has been adequate time since the start of production to allow the line source
solution to be accurate
(ii) the reservoir is infinite acting.
A Check Ei applicability
100φµcrw2
t>
k
t > 21.8s
φµcre2
the reservoir is infinite acting if the time, t <
4k
i.e.
74
Fluid Flow In Porous Media
10
0.25 x 22 x10 −3 x 0.6 x10 −9 x 650 2
t<
4 x 340 x10 −15
t < 1025184s
25φµcr 2
the ln approximation is valid if the time, t >
k
25 x 0.25 x 22 x10 −3 x 0.6 x10 −9 x 0.152
t>
340 x10 −15
t > 5.5s
qµBo γφµcrw2
Pwf = Pi + ln
(ii) 4πkh 4 kt (taking account of the conversion from stock
tank to reservoir conditions via the formation volume factor for oil).
Data
porosity, φ 18%
formation volume factor for oil, Bo 1.21rm3/stm3
net thickness of formation, h 20m
viscosity of reservoir oil, µ 0.8x10-3 Pas
compressibility, c 43x10-9Pa-1
permeability, k 85mD
wellbore radius, rw 0.15m
external radius, re 1950m
initial reservoir pressure, Pi 210bar
well flowrate (constant) 20stm3/day
skin factor 0
Distance well 1 to well 2 50m
Solution EXERCISE 9
The line source solution is used to determine the time equivalent to the specified
pressure drop at well 2. Checks are made to ensure that:
(i) there has been adequate time since the start of production to allow the line source
solution to be accurate
(ii) the reservoir is infinite acting.
A Check Ei applicability
100φµcrw2
t>
k
t > 164 s
it is expected that the time will be in excess of 164 seconds therefore the line source
solution is acceptable
76
Fluid Flow In Porous Media
10
B Check reservoir is infinite acting
φµcre2
t<
the reservoir is infinite acting if the time, 4k
i.e.
t < 69250235s
t < 19236 hours
25φµcr 2
t>
the ln approximation is valid if the time, k
t > 41s
Now,
qµBo γφµcr502 m
Pi − Pat 50 m from well 1 =− ln
4πkh 4 kt
P − Pat 50 m from = γφµcr50 m
2
e i well
qµBo 4 kt
−
4πkh
27.57 x10 −9
t=
3.4 x10 −13 xe −4.77
t = 9561863s
t = 2656hours
t = 111 days
This time is within the limits for the use of the ln approximation to the Ei function and
within the limits to the reservoir being infinite acting therefore the result is correct.
78
Fluid Flow In Porous Media
10
EXERCISE 10
A well in a reservoir has a very low production rate of 2stm3/day. Calculate the flowing
bottomhole pressure after 2 years production.
Data
porosity, φ 16%
formation volume factor for oil, Bo 1.13rm3/stm3
net thickness of formation, h 10m
viscosity of reservoir oil, µ 5x10-3 Pas
compressibility, c 14x10-9Pa-1
permeability, k 10mD
wellbore radius, rw 0.15m
external radius, re 780m
initial reservoir pressure, Pi 86bar
well flowrate (constant) 2stm3/day
skin factor 0
Solution EXERCISE 10
The line source solution is used to determine the pressures required at the wellbore
after 2 years production. Checks are made to ensure that:
(i) there has been adequate time since the start of production to allow the line source
solution to be accurate
(ii) the reservoir is infinite acting.
A Check Ei applicability
100φµcrw2
t>
k
t > 2520s
t > 0.7 hours
φµcre2
t<
the reservoir is infinite acting if the time, 4k
t < 170352000s
t < 5.4 years
25φµcr 2
t>
the ln approximation is valid if the time, k
t > 630s
qµBo γφµcrw2
Pwf = Pi + ln
(ii) 4πkh 4kt (taking account of the conversion from stock
tank to reservoir conditions via the formation volume factor for oil).
80
Fluid Flow In Porous Media
10
EXERCISE 11
A well is put on production at 15stm3/day. The following well and reservoir data are
relevant.
Data
porosity, φ 21%
formation volume factor for oil, Bo 1.2rm3/stm3
net thickness of formation, h 23m
viscosity of reservoir oil, µ 5x10-3 Pas
compressibility, c 22 x10-9Pa-1
permeability, k 130mD
wellbore radius, rw 0.15m
external radius, re 800m
initial reservoir pressure, Pi 120bar
well flowrate (constant) 15stm3/day
skin factor 0
SOLUTION EXERCISE 11
The line source solution is used to determine the pressures required at the specified
radii and at the specified time. Checks are made to ensure that:
(i) there has been adequate time since the start of production to allow the line source
solution to be accurate
(ii) the reservoir is infinite acting.
A Check Ei applicability
100φµcrw2
t>
k
t > 400s
φµcre2
t<
the reservoir is infinite acting if the time, 4k
i.e.
t < 28430769s
t < 7897 hours
(1) the bottomhole flowing pressure after 2 hours production, Pwf at 2 hours
25φµcrw2
t>
the ln approximation is valid if the time, k
qµBo γφµcrw2
Pwf = Pi + ln
ii) 4πkh 4kt (taking account of the conversion from stock tank
to reservoir conditions via the formation volume factor for oil, Bo).
(2) the presure after 2 hours production at a radius of 10m from the wellbore
82
Fluid Flow In Porous Media
10
25φµcr 2
t>
the ln approximation is valid if the time, k
t > 444231s
t > 123hours
qµBo φµcr
2
P = Pi + Ei −
(ii)
4πkh 4kt (taking account of the conversion from stock
tank to reservoir conditions via the formation volume factor for oil, Bo and also the
fact that the radius, r, is now at 10m from the wellbore).
P = 120x105 + 27724xEi(-0.62)
Ei(-0.62) = -0.437
P = 120x105 +27724x-0.437
= 120x105 -12115
= 11987885Pa
= 119.88bar
(3) the pressure after 2 hours production at a radius of 20m from the wellbore
25φµcr 2
t>
the ln approximation is valid if the time, k
P = Pi + Ei −
(ii)
4πkh 4kt (taking account of the conversion from stock tank
to reservoir conditions via the formation volume factor for oil, Bo and also the fact
that the radius, r, is now at 20m from the wellbore).
P = 120x105 + 27724xEi(-2.48)
Ei(-2.48) = -0.026 (by linear interpolation between adjacent values in the tables)
(4) the pressure after 2 hours production at a radius of 50m from the wellbore
25φµcr 2
t>
the ln approximation is valid if the time, k
t > 11105769s
t > 3085hours
qµBo φµcr
2
P = Pi + Ei −
(ii)
4πkh 4kt (taking account of the conversion from stock tank
to reservoir conditions via the formation volume factor for oil, Bo and also the fact
that the radius, r, is now at 50m from the wellbore).
84
Fluid Flow In Porous Media
10
qµBo 15x5x10 −3 x1.2
= = 27724
4πkh 24x3600x4πx130x10 −15 x23
P = 120x105 + 27724xEi(-15.5)
P = 120x105 +27724x0
= 120x105 -0
= 12000000Pa
= 120.00bar
The following figure illustrates the nature of the infinite acting reservoir in that the
pressure at 50m after 2 hours production is still the initial pressure of 120bar.
Pressure v Distance
120.0
119.5
Pressure (bar)
119.0
118.5
118.0
117.5
117.0
0 10 20 30 40 50 60
Distance from centre of well (m)
Data
porosity, φ, 28%
formation volume factor for oil, Bo 1.39rm3/stm3
net thickness of formation, h, 8.5m
viscosity of reservoir oil, µ 0.8x10-3 Pas
compressibility, c 2.3 x10-9Pa-1
permeability, k 100mD
wellbore radius, rw 0.15m
external radius, re 6100m
initial reservoir pressure, Pi 227bar
bottomhole flowing pressure
after 6 minutes 192bar
well flowrate (constant) 286stm3/day
SOLUTION EXERCISE 12
The line source solution is used to determine the skin factor at the wellbore after 6
minutes production. Checks are made to ensure that:
(i) there has been adequate time since the start of production to allow the line
source solution to be accurate
(ii) the reservoir is infinite acting.
A Check Ei applicability
100φµcrw2
t>
k
t > 11.6s
86
Fluid Flow In Porous Media
10
φµcre2
t>
the reservoir is infinite acting if the time, 4k
i.e.
t < 47926480s
t < 555 days
25φµcrw2
t>
the ln approximation is valid if the time, k
t > 2.9s
qµBo γφµcrw2
Pi - Pwf = − ln − 2s
(ii)
4πkh 4kt (taking account of the conversion from
stock tank to reservoir conditions via the formation volume factor for oil).
−3 −9 2
γφµcrw2 1.781x0.28x0.8x10 x2.3x10 x0.15
= -15
= 143371x10 −9
4kt 4x100x10 x6x60
Pi - Pwf γφµcrw2
2s = + ln
qµBo 4kt
4πkh
qµB
∆Ps = 2s
4πkh
∆Ps = 2x0.65x344610 = 447993Pa = 4.5bar
88
Fluid Flow In Porous Media
10
EXERCISE 13
A reservoir and well are detailed in the following data. Use this data to calculate the
skin factor around the well after producing for 1.5 hours.
Data
porosity, φ 23%
formation volume factor for oil, Bo 1.36rm3/stm3
net thickness of formation, h 63m
viscosity of reservoir oil, µ 1.6x10-3 Pas
compressibility, c 17 x10-9Pa-1
permeability, k 243mD
wellbore radius, rw 0.15m
external radius, re 4000m
initial reservoir pressure, Pi 263.0bar
bottomhole flowing pressure
after 6 minutes 260.5bar
well flowrate (constant) 120stm3/day
SOLUTION EXERCISE 13
The line source solution is used to determine the skin factor at the wellbore after 1.5
hours production. Checks are made to ensure that:
(i) there has been adequate time since the start of production to allow the line source
solution to be accurate
(ii) the reservoir is infinite acting.
A Check Ei applicability
100φµcrw2
t>
k
100x0.23x1.6x10 -3 x17x10 −9 x0.152
t>
243x10 -15
t > 58s
φµcre2
t<
the reservoir is infinite acting if the time, 4k
t < 102979424s
t < 1192 days
25φµcr 2
t>
the ln approximation is valid if the time, k
t > 14.5s
qµBo γφµcrw2
Pi - Pwf = − ln − 2s
(ii)
4πkh 4kt (taking account of the conversion from
stock tank to reservoir conditions via the formation volume factor for oil).
−3 −9 2
γφµcrw2 1.781x0.23x1.6x10 x17x10 x0.15
= -15
= 47762 x10 −9
4kt 4x243x10 x1.5x3600
Pi - Pwf γφµcrw2
2s = + ln
qµBo 4kt
4πkh
(263.0 - 260.5)x10 5
2s = + ln( 47762x10 -9 )
15710
2s = 15.9 − 10.0
s = 2.95
90
Fluid Flow In Porous Media
10
EXERCISE 14
A well has been on production in a reservoir which is in a semi-steady state flow
regime. For the following data, calculate the bottomhole flowing pressure, Pwf
Data
formation volume factor for oil, Bo 1.62rm3/stm3
net thickness of formation, h 72m
viscosity of reservoir oil, µ 1.2x10-3 Pas
permeability, k 123mD
wellbore radius, rw 0.15m
external radius, re 560m
average reservoir pressure, 263.0bar
well flowrate (constant) 216stm3/day
skin factor 0
SOLUTION EXERCISE 14
Substitute the values into the semi-steady state flow equation
qµBo re 3
P − Pwf = ln − + s
2πkh rw 4
qµBo re 3
Pwf = P - ln − + s
2πkh rw 4
Pwf = 25647120Pa
Pwf = 256.5bar
Geometry
1. Circle
2. Square
3. Quadrant of a square
Data
Area of reservoir, A 1618370m2
viscosity of reservoir oil, µ 1.0x10-3 Pas
permeability, k 100mD
porosity, φ, 20%
compressibility, c 1.45 x10-9Pa-1
The times are calculated by the dimensionless time, diffusivity of the reservoir and
the area of the reservoir. The dimensionless time accounting for the reservoir drainage
area is found for the conditions in Table 5.
SOLUTION EXERCISE 15
1. Circle
φµcA
t < t DA
k
t < 469327s
t < 130hours
2. Square
φµcA
t < t DA
k
t < 422395s
t < 117hours
92
10
Fluid Flow In Porous Media
4. Quadrant of a square
φµcA
t < t DA
k
t < 117332s
t < 33hours
Data
porosity, φ 18%
formation volume factor for oil, Bo 1.2rm3/stm3
net thickness of formation, h 6.1m
viscosity of reservoir oil, µ 1x10-3 Pas
compressibility, c 2.18 x10-9Pa-1
wellbore radius, rw 0.1m
initial reservoir pressure, Pi 241.3bar
well flowrate (constant) 238stm3/day
0.0 241.3
1.0 201.1
2.0 199.8
3.0 199.1
4.0 198.5
5.0 197.8
7.5 196.5
10.0 195.3
15.0 192.8
30.0 185.2
40.0 180.2
50.0 176.7
60.0 173.2
70.0 169.7
80.0 166.2
90.0 162.7
100.0 159.2
2. Make an estimate of the area being drained by the well and the Dietz shape fac-
tor.
SOLUTION EXERCISE 16
The description of the test is such that this is the first time the well has been put on
production and the reservoir pressure will decline at a rate dictated by the solutions
of the diffusivity equation. The pressure decline has been recorded at the wellbore (as
in the table of data) and it is expected that there will be an unsteady state (transient)
period initially followed by a semi steady state or steady state flow period. It is
thought to be an isolated block therefore there would be a depletion of the reservoir
pressure under semi steady state conditions expected. The initial unsteady state or
transient flow period can be used to determine the permeability and skin factor of
the well, and the subsequent semi steady state flow period can be used to detect the
94
Fluid Flow In Porous Media
10
reservoir limits. SI units will be used at reservoir conditions, therefore flowrates are
in m3/s and the formation volume factor for oil is used to convert from stock tank to
reservoir volumes. The pressure related items are in Pascal.
1. The permeability and skin factor can be determined from the initial transient
period using the line source solution:
qµ 4kt
Pwf = Pi − ln 2
+ 2s
4πkh γφµcrw
or
qµ
m=
4πkh
From this, the unknown value, i.e. the permeability, k, can be calculated. Once the
permeability is known, the equation 3.26 can be rearranged to determine the other
unknown, the skin factor, as:
Pi − Pwf 4kt
2s = − ln
m γφµcrw2
Any coherent set of data points can be used to determine the permeability and skin,
however, it is not clear when the data represent the line source solution. Therefore all
of the pressure data are plotted and a linear fit attached to those data which show the
linear relationship between the bottom hole flowing pressure, Pwf and the logarithm
of time, lnt. Table 7 and figure 9 illustrates this.
0.0 241.3
1.0 201.1 0.0
2.0 199.8 0.7
3.0 199.1 1.1
4.0 198.5 1.4
5.0 197.8 1.6
7.5 196.5 2.0
10.0 195.3 2.3
15.0 192.8 2.7
30.0 185.2 3.4
40.0 180.2 3.7
50.0 176.7 3.9
60.0 173.2 4.1
70.0 169.7 4.2
80.0 166.2 4.4
90.0 162.7 4.5
100.0 159.2 4.6
190
180 P
170
160
150
0 1 2 3 4 5
In flowing time, t (hours)
The plots of bottomhole flowing pressure show that the transient period (for which
the logarithm approximation is valid) lasts for approximately 4 hours and from the
plot, the slope, m, can be determined to be 1.98bar/log cycle. Substituting this into
the equation gives:
96
Fluid Flow In Porous Media
10
(converting from stock tank cubic metres/day to reservoir cubic metres/second and
from bar to Pascal producing a permeability in terms of m2 which is then converted
to mD).
To determine the skin factor, the slope, m, of the line is theoretically extrapolated
to a convenient time. This is usually a time of 1 hour. The bottomhole pressure
associated with this time is calculated and this is used to determine a pressure drop
(Pi - Pwf ) during the time (t1 hour - t 0). This is then equal to the pressure drop calculated
from the ln function plus an excess caused by the skin. In this case, a real pressure
measurement was recorded at time 1 hour. This is not necessarily the same number
as calculated from the extrapolation of the linear section of the relationship since the
real pressure recorded at time 1 hour may not be valid for use with the Ei function.
Although it was recorded. It may have been too early for the Ei function to accurately
approximate the reservoir flow regime.
In this case P1 hour =201.2bar and therefore (by rearranging equation 3.26)
2s=20.25-13.02 = 7.23
s=3.6
2. To determine the area drained and the shape factor, the data from the semi steady
state flow regime are required. From equation 3.29, there will be a linear relation-
ship between bottomhole flowing pressure and time. This is related to the area of
the drained volume and the shape factor.
To determine the gradient of the pressure decline, the bottomhole flowing pressure
and time are plotted using Cartesian co-ordinates as in figure 10:
200
190
170
160
150
0 20 40 60 80 100 120
Flowing time, t (hours)
dP q
=−
dt cAhφ
where q is the flowrate, c is the compressibility, A is the area of the reservoir, h is the
thickness and φ is the porosity. Taking account of the formation volume factor, Bo,
qBo
A=−
dP
chφ
dt
238 x 1.2
A=−
24 x 3600 x 2.18x10 -9 x 6.1 x 0.18 x - 9.72
A = 142076m2
qµ 1 4A 2πkt
Pwf = Pi − ( ln + + s)
2πkh 2 γC A rw φµcA
2
The linear extrapolation of this line to small values of t gives the specific value of
Pwf of 194.2 bar at t=0. In reality, at t=0, the flowrate has not started, so this will be
named P0. Inserting this value in equation 3.39 at t=0, converting bar to Pascal and
including the skin factor gives:
qµ 4A
Pi − P0 = ln 2 − lnC A + 2s
4πkh γrw
i.e.
4 x 142076
(241.3 − 194.2) x10 5 = 1.98 x10 5 ln 2
− lnC A + 2 x 3.62
1.781x0.1
98
Fluid Flow In Porous Media
10
EXERCISE 17
An appraisal well is tested by producing at a constant rate of 200stm3/day for 107
hours. The following table of flowing bottomhole pressures and time were recorded
during the test. Using the data,
Data
porosity, φ 22%
formation volume factor for oil, Bo 1.3rm3/stm3
net thickness of formation, h 21m
viscosity of reservoir oil, µ 1.9x10-3 Pas
compressibility, c 4.3 x10-9Pa-1
wellbore radius, rw 0.15m
initial reservoir pressure, Pi 378.7bar
well flowrate (constant) 200stm3/day
0.0 378.7
1.1 326.41
2.1 324.7
3.2 323.8
4.3 323.1
5.4 322.1
8.0 320.5
10.7 318.8
16.1 315.5
21.4 312.2
32.1 305.6
42.8 300.8
53.5 296.0
64.2 291.2
74.9 286.3
85.6 281.5
96.3 276.7
107.0 271.9
SOLUTION EXERCISE 17
(1) The permeability and skin factor can be calculated from the transient flow period
using the line source solution (if the reservoir is in transient flow) since
qµBo 4kt
Pwf = Pi − ln 2
+ 2s
4πkh γφµcrw
y = c1 + m X + c2
y = mx + c
Time Bottomhole
In time
flowing
pressure
(hours) (bar)
0.0 378.7
1.1 0.1 326.4
2.1 0.8 324.7
3.2 1.2 323.8
4.3 1.5 323.1
5.4 1.7 322.1
8.0 2.1 320.5
10.7 2.4 318.8
16.1 2.8 315.5
21.4 3.1 312.2
32.1 3.5 305.6
42.8 3.8 300.8
53.5 4.0 296.0
64.2 4.2 291.2
74.9 4.3 286.3
85.6 4.4 281.5
96.3 4.6 276.7
107.0 4.7 271.9
It can be seen that the slope changes after about 5 hours, therefore the data until 5
hours is used to determine a straight line fit giving the figure below.
330.0
320.0
bottomhole flowing
pressure (bar)
310.0
270.0
0.0 1.0 2.0 3.0 4.0 5.0
ln time
(i) Permeability
100
Fluid Flow In Porous Media
10
qµBo
m=
4πkh
k = 90mD
Extrapolation of the line to a time of 1 hour gives the pressure, P1 hour as 326.6bar.
Pi - Pwf(1hour) 4kt
2s = − ln
m γφµcrw2
2s = 21.5 - ln(17993.4)
This is obtained from the semi-steady state part of the flow. A plot of linear pressure
decline with time indicates this flow regime (i.e. an expansion of a fixed volume of
fluid) and this is shown in the figure below.
310.0
290.0
270.0
0.0 50.0 100.0 150.0
time (hours)
Since the pressure decline rate is related to the volume, the area, A, of the drainage
cell can be calculated assuming a constant thickness, h, and a constant porosity.
dP qBo
=−
dt cAhφ
qBo
A=−
dP
chφ
dt
dP -0.45x1x10 5
= −0.45bar / hour = = −12.5Pa / s
dt 1x3600
200x1.3
A=− -9
= 12118m 2
24x3600x4.3x10 x21x 0.22x -12.5
qµ 1 4A 2πkt
Pwf = Pi − ln + + s
2πkh 2 γC A rw φµcA
2
and extrapolation of the line to small values of time gives a pressure, Po of 320.8bar.
Insertion of these values at time = 0 gives
qµ 4A
Pi − P0 = ln 2 − lnC A + 2s
4πkh γrw
i.e.
4 x 12118
(378.7 − 320.05) x10 5 = 2.42 x10 5 ln 2
− lnC A + 2 x 5.9
1.781x0.15
58.7x105 = 2.42x105 (14.01- lnCA +11.8)
lnCA =14.01+11.8-24.27 = 1.54
102
Fluid Flow In Porous Media
10
EXERCISE 18
Two wells, well 1 and well 2, are drilled in an undeveloped reservoir. Well 1 is
completed and brought on production at 500stm3/day and produces for 40 days at
which time Well 2 is completed and brought on production at 150stm3/day. Using the
data provided, calculate the pressure in Well 2 after it has produced for 10 days (and
assuming Well 1 continues to produce at its flowrate). Therefore, Well 1 produces for
50days when its pressure influence is calculated; Well 2 produces for 10 days when
its pressure influence is calculated.
The wells are 400m apart and the nearest boundary is 4000m from each well.
Data
porosity, φ, 21%
formation volume factor for oil, Bo 1.4rm3/stm3
net thickness of formation, h, 36m
viscosity of reservoir oil, µ 0.7x10-3 Pas
compressibility, c 8.7 x10-9Pa-1
permeability, k 80mD
wellbore radius, rw (both wells) 0.15m
initial reservoir pressure, Pi 180.0bar
Well 1 flowrate (constant) 500stm3/day
Well 2 flowrate (constant) 150stm3/day
skin factor around both wells 0
SOLUTION EXERCISE 18
The line source solution is used to determine the bottomhole flowing pressure at Well
2 after 10 days production, accounting for the effect of 50days production from Well
1. Checks are made to ensure that:
(i) there has been adequate time since the start of production to allow the line source
solution to be accurate
A Check Ei applicability
100φµcrw2
t>
k (3.10)
t >36s
φµcre2
t<
the reservoir is infinite acting if the time, 4k
i.e.
t < 63945000
t < 740 days
The bottomhole flowing pressure at Well 2 is the sum of the pressure drops caused
by its production and by the pressure drop generated by the production of Well 1.
Pwf at Well 2 = Pi -∆Pwell2 flowing for 10 days - ∆Pwell1 flowing for 40+10 days 400m away
25φµcr 2
t>
the ln approximation is valid if the time, k
t > 9s
qµBo γφµcrw2
Pwf = Pi + ln
4πkh 4kt
qµBo γφµcrw2
Pi - Pwf = − ln
4πkh 4kt
−3 −9 2
γφµcrw2 1.781x0.21x0.7x10 x8.7x10 x0.15
= -15
= 185 x10 −9
4kt 4x80x10 x10x24x3600
104
Fluid Flow In Porous Media
10
Pi - Pwf = -47011x ln(185x10-9)
Pi - Pwf = -47011x -15.5
Pi - Pwf =728671Pa
(B) At 10 days production from well 2, well 1 has been producing for 50 days and
its contribution to pressure drop at Well 2 is calculated as follows.
25φµcr 2
t>
check ln approximation to Ei function k
t > 63945000s
t > 740 days
qµBo φµcr1-2
2
Pi - Pwf at Well2 caused by Well 1 = − Ei −
4πkh 4kt
φµcr1-2
2
0.21x0.7x10 −3 x8.7x10 −9 x 400 2
= = 0.148˚˚˚
4kt 4x80x10 -15 x50x24x3600
Ei(-0.148) = -1.476
Data
porosity, φ, 19%
formation volume factor for oil, Bo 1.2rm3/stm3
net thickness of formation, h, 36m
viscosity of reservoir oil, µ 1x10-3 Pas
compressibility, c 10 x10-9Pa-1
permeability, k 110mD
wellbore radius, rw (both wells) 0.15m
external radius, re 7000m
initial reservoir pressure, Pi 250.0bar
Well 1 flowrate (constant) 110stm3/day
Well 2 flowrate (constant) 180stm3/day
skin factor around both wells 0
SOLUTION EXERCISE 19
The line source solution is used to determine:
the bottomhole flowing pressure at well 2 flowing for 13 days plus the pressure
influence on it of well 1 flowing for 40 days
the bottomhole flowing pressure at well 1 flowing for 40 days plus the pressure
influence on it of well 2 flowing for 13 days
(i) there has been adequate time since the start of production to allow the line source
solution to be accurate
A Check Ei applicability
100φµcrw2
t>
k
100x0.19x1x10 -3 x10x10 −9 x0.152
t> ˚
110x10 -15
t > 39s
106
Fluid Flow In Porous Media
10
time is 13 days, therefore line source is applicable.
φµcre2
t<
the reservoir is infinite acting if the time, 4k
i.e.
t < 211590909s
t < 2449 days
The bottomhole flowing pressure at Well 2 is the sum of the pressure drops caused
by its production and by the pressure drop generated by the production of Well 1.
Pwf at Well 2 = Pi -∆Pwell2 flowing for 13 days - ∆Pwell1 flowing for 27+13 days 350m away
25φµcr 2
t>
the ln approximation is valid if the time, k
t > 10s
−3 −9 2
γφµcrw2 1.781x0.19x1x10 x10x10 x0.15
= -15
= 154 x10 −9
4kt 4x110x10 x13x24x3600
(B) At 13 days, contribution to pressure drop at Well 2 from production from Well
1
25φµcr 2
t>
the ln approximation is valid if the time, k
t > 52897727s
t > 612 days
qµBo φµcr1-2
2
Pi - Pwf at Well2 caused by Well 1 = − Ei −
4πkh 4kt
qµBo 110x1x10 −3 x1.2
− =−
4πkh 24x3600x4πx110x10 −15 x36
φµcr1-2
2
0.19x1x10 −3 x10x10 −9 x350 2
=
4kt 4x110x10 -15 x40x24x3600
Ei(-0.153) = -1.447
108
Fluid Flow In Porous Media
10
Pwf Well2 = 250.0 - 7.9 - 0.4bar
Pwf Well2 = 241.7bar
The bottomhole flowing pressure at Well 1 is the sum of the pressure drops caused
by its production and by the pressure drop generated by the production of Well 2.
Pwf at Well 1 = Pi -∆Pwell1 flowing for 40 days - ∆Pwell2 flowing for 13 days 350m away
25φµcr 2
t>
the ln approximation is valid if the time, k
t > 10s
qµBo γφµcrw2
Pwf = Pi + ln
4πkh 4kt
qµBo γφµcrw2
Pi - Pwf = − ln
4πkh 4kt
−3 −9 2
γφµcrw2 1.781x0.19x1x10 x10x10 x0.15
= -15
= 50.1x10 −9
4kt 4x110x10 x40x24x3600
(B) At 40 days, contribution to pressure drop at Well 1 from production from Well
2
t > 52897727s
t > 612 days
qµBo φµcr1-2
2
Pi - Pwf at Well1 caused by Well 2 =− Ei −
4πkh 4kt
φµcr1-2
2
0.19x1x10 −3 x10x10 −9 x350 2
= = 0.471
4kt 4x110x10 -15 x13x24x3600
Ei(-0.471) = -0.597
110
Fluid Flow In Porous Media
10
EXERCISE 20
A well is completed in an undeveloped reservoir described by the data below. The well
flows for 6 days at 60 stm3/day and is then shut in for a day. Calculate the pressure
in an observation well 100m from the flowing well.
Data
porosity, φ, 19%
formation volume factor for oil, Bo 1.3rm3/stm3
net thickness of formation, h, 23m
viscosity of reservoir oil, µ 0.4x10-3 Pas
compressibility, c 3 x10-9Pa-1
permeability, k 50mD
wellbore radius, rw (both wells) 0.15m
external radius, re 6000m
initial reservoir pressure, Pi 180.0bar
flowrate (constant) 60stm3/day
skin factor around well 0
SOLUTION EXERCISE 20
The line source solution is used to determine the pressure in the observation well
after 6 days production from the flowing well then 1 day shut in at the flowing well.
Checks are made to ensure that:
(i) there has been adequate time since the start of production to allow the line source
solution to be accurate
A Check Ei applicability
100φµcrw2
t>
k
t>10.3s
i.e.
t < 41040000s
t <475 days
25φµcr 2
t>
the ln approximation is valid if the time, k
t > 1140000s
t > 13 days
φµcr1-2
2
0.19x0.4x10 −3 x3x10 −9 x100 2
= = 0.019
4kt 4x50x10 -15 x7x24x3600
φµcr1-2
2
0.19x0.4x10 −3 x3x10 −9 x100 2
= -15
= 0.132
4k(t - t 1 ) 4x50x10 x(7 - 6)x24x3600
Ei(-0.019) = -3.405
Ei(-0.132) = -1.576
112
Fluid Flow In Porous Media
10
60 0 - 60
Pi − Pobs well = −35982857 x − 3.405 + x − 1.576
24x3600 24x3600
[
Pi − Pobs well = −35982857 −2.36x10 −3 + 1.09x10 −3 ]
Pi - Pobs well = 45698Pa = 0.5bar
Pobs well = 180.0 - 0.5 = 179.5bar
Data
porosity, φ, 21%
formation volume factor for oil, Bo 1.31rm3/stm3
net thickness of formation, h, 20m
viscosity of reservoir oil, µ 0.6x10-3 Pas
compressibility, c 8 x10-9Pa-1
permeability, k 75mD
wellbore radius, rw (both wells) 0.15m
external radius, re 5000m
initial reservoir pressure, Pi 200.0bar
1st flowrate (constant) 25stm3/day
1st flowrate period 6days
2nd flowrate (constant) 75stm3/day
2nd flow period 4days
skin factor around well 0
SOLUTION EXERCISE 21
The line source solution will be used to assess the effects of variables rates on the
bottomhole flowing pressure. Checks are made to ensure that:
(i) there has been adequate time since the start of production to allow the line source
solution to be accurate
(ii) the reservoir is infinite acting.
A Check Ei applicability
100φµcrw2
t>
k
φµcre2
t<
the reservoir is infinite acting if the time, 4k
i.e.
114
Fluid Flow In Porous Media
10
0.21x0.6x10 −3 x8x10 −9 x5000 2
t<
4x75x10 -15
t < 84000000s
25φµcr 2
t>
the ln approximation is valid if the time, k
−3 −9 2
γφµcrw2 1.781x0.21x0.6x10 x8x10 x 0.15
= -15
= 155.8 x10 −9
4kt 4x75x10 x10x24x3600
25 (75 - 25)
Pi − Pwf = −41698595 xln(155.8x10 -9 ) + ln(389.6x10 -9 )
24x3600 24x3600
Data
porosity, φ, 19%
formation volume factor for oil, Bo 1.4rm3/stm3
net thickness of formation, h, 20m
viscosity of reservoir oil, µ 1x10-3 Pas
compressibility, c 9 x10-9Pa-1
permeability, k 120mD
wellbore radius, rw 0.15m
external radius, re 4000m
initial reservoir pressure, Pi 300.0bar
flowrate (constant) 120stm3/day
flowrate period, t 50days
distance to fault, L 300m
skin factor around well 0
SOLUTION EXERCISE 22
The line source solution will be used to assess the effects of the rate and the boundary
on the bottomhole flowing pressure. Using an image well 600m from the real well (i.e.
2x distance to the fault) with identical pressure and rate history as the real well, the
effect of the boundary on the infinite acting reservoir can be overcome. The bottomhole
flowing pressure in the real well will be the pressure drop caused by the production
from the real well plus a pressure drop from the image well 600m away.
The line source solution will be used. Checks are made to ensure that:
(i) there has been adequate time since the start of production to allow the line source
solution to be accurate
A Check Ei applicability
100φµcrw2
t>
k
100x0.19x1x10 -3 x9x10 −9 x0.152
t>
120x10 -15
t >32s
116
Fluid Flow In Porous Media
10
B Check reservoir is infinite acting
φµcre2
t<
the reservoir is infinite acting if the time, 4k
t < 57000000s
t <660 days
Checking for the validity of the ln approximation, for the real well
25φµcr 2
t>
the ln approximation is valid if the time, k
Checking for the validity of the ln approximation, for the image well
25φµc(2L)2
t>
the ln approximation is valid if the time, k
t > 128250000s
t> 1484 days
For this case, then, the ln approximation will predict the bottomhole flowing pressure
around the real well, but the effect of the image well 600m away will need to be
predicted by the Ei function.
Pi − Pwf = − ln − Ei −
4πkh 4kt 4πkh 4kt
Ei(-0.297) = -0.914
The fault 300m away pulled the bottomhole flowing pressure down by an extra
58928Pa or 0.6bar.
118
Fluid Flow In Porous Media
10
EXERCISE 23
A well in a reservoir is producing close to two intersecting faults as shown below.
Using the data given, calculate the bottomhole flowing pressure after 32 days and
indicate the effect of the faults on the bottomhole flowing pressure. The production
rate is constant at 100stm3 /day
fault
L1 70m
fault
L2
120m
well
Data
porosity, φ, 22%
formation volume factor for oil, Bo 1.5rm3/stm3
net thickness of formation, h, 36m
viscosity of reservoir oil, µ 1x10-3 Pas
compressibility, c 9 x10-9Pa-1
permeability, k 89mD
wellbore radius, rw 0.15m
external radius, re 6000m
initial reservoir pressure, Pi 240.0bar
flowrate (constant) 100stm3/day
flowrate period, t 32days
distance to fault, L1 70m
distance to fault, L2 120m
skin factor around well 0
SOLUTION EXERCISE 23
The line source solution will be used to assess the effects of the rate and the boundary
on the bottomhole flowing pressure. Three image wells with identical pressure and
rate histories as the real well will be used as shown below.
image well 1
image well 3
L1
fault r3
L1 70m
fault
L2 L2
120m image well 2
well
The line source solution will be used. Checks are made to ensure that:
(i) there has been adequate time since the start of production to allow the line source
solution to be accurate
(ii) the reservoir is infinite acting.
A Check Ei applicability
100φµcrw2
t>
k
φµcre2
t<
the reservoir is infinite acting if the time, 4k
t < 200224719s
t < 2317 days
Checking for the validity of the ln approximation, for the real well
25φµcr 2
t>
the ln approximation is valid if the time, k
120
Fluid Flow In Porous Media
10
therefore ln approximation is valid.
Checking for the validity of the ln approximation, for the image well 1
25φµc(2L1)2
t>
the ln approximation is valid if the time, k
therefore the ln approximation is not valid and the Ei function is used. The distances
to image wells 2 and 3 are greater, therefore they must also need to use the Ei
function.
The distance r3 is
r3 = (2L1)2 + (2L2)2
r3 = (140)2 + (240)2
r3 = 277.8m
qµBo γφµcrw2
Pi − Pwf = − ln
4πkh 4kt
qµBo φµc(2L1)
2
− Ei −
4πkh 4kt
qµBo φµc(2L2)
2
− Ei −
4πkh 4kt
qµBo φµcr32
− Ei −
4πkh 4kt
−3 −9 2
γφµcrw2 1.781x0.22x1x10 x9x10 x0.15
= -15
= 80.6 x10 −9
4kt 4x89x10 x32x24x3600
ln(80.6x10-9) = -16.3
Ei(-0.039) = -2.706
Ei(-0.116) = -1.689
−3 −9 2
φµcr32 0.22x1x10 x9x10 x277.8
= = 0.155
4kt 4x89x10 -15 x32x24x3600
Ei(-0.155) = -1.436
The effect of the boundary is to pull the bottomhole flowing pressure down by an
extra 2.5bar.
122
Fluid Flow In Porous Media
10
EXERCISE 24
A well is 80m due west of a north-south fault. From well tests, the skin factor is 5.0.
Calculate the pressure in the well after flowing at 80stm3/day for 10 days.
Data
porosity, φ, 25%
formation volume factor for oil, Bo 1.13rm3/stm3
net thickness of formation, h, 23m
viscosity of reservoir oil, µ 1.1x10-3 Pas
compressibility, c 10.1 x10-9Pa-1
permeability, k 125mD
wellbore radius, rw 0.15m
external radius, re 6000m
initial reservoir pressure, Pi 210.0bar
flowrate (constant) 80stm3/day
flowrate period, t 32days
distance to fault, L 80m
skin factor around well 5.0
SOLUTION EXERCISE 24
The fault can be represented by an image well twice the distance from the real well
as the fault is. The pressure effect this image well has on the real well augments the
pressure drop in the well caused by the production, however, there is an additional
pressure drop over the skin zone around the real well which must be taken into
account.
The line source solution will be used. Checks are made to ensure that:
(i) there has been adequate time since the start of production to allow the line source
solution to be accurate
(ii) the reservoir is infinite acting.
A Check Ei applicability
100φµcrw2
t>
k
100x0.25x1.1x10 -3 x10.1x10 −9 x0.152
t>
125x10 -15
t > 50s
φµcre2
t<
the reservoir is infinite acting if the time, 4k
t < 199980000s
t < 2315 days
Checking for the validity of the ln approximation, for the real well
25φµcr 2
t>
the ln approximation is valid if the time, k
−3 −9 2
γφµcrw2 1.781x0.25x1.1x10 x10.1x10 x 0.15
= = 257.6 x10 −9
4kt 4x125x10 -15 x10x24x3600
[ ]
Pi − Pwf = −31857 ln(257.6x10 −9 ) − 2x5.0 − 31857xEi( −0.165)
124
Fluid Flow In Porous Media
10
Ei(-0.165) = -1.383
tD pD tD pD tD pD
0.06 0.251 0.22 0.443 0.40 0.565 0.52 0.627 1.0 0.802 1.5 0.927
0.08 0.288 0.24 0.459 0.42 0.576 0.54 0.636 1.1 0.830 1.6 0.948
0.10 0.322 0.26 0.476 0.44 0.587 0.56 0.645 1.2 0.857 1.7 0.968
0.12 0.355 0.28 0.492 0.46 0.598 0.60 0.662 1.3 0.882 1.8 0.988
0.14 0.387 0.30 0.507 0.48 0.608 0.65 0.683 1.4 0.906 1.9 1.007
0.16 0.420 0.32 0.522 0.50 0.618 0.70 0.703 1.5 0.929 2.0 1.025
0.18 0.452 0.34 0.536 0.52 0.628 0.75 0.721 1.6 0.951 2.2 1.059
0.20 0.484 0.36 0.551 0.54 0.638 0.80 0.740 1.7 0.973 2.4 1.092
0.22 0.516 0.38 0.565 0.56 0.647 0.85 0.758 1.8 0.994 2.6 1.123
0.24 0.548 0.40 0.579 0.58 0.657 0.90 0.776 1.9 1.014 2.8 1.154
0.26 0.580 0.42 0.593 0.60 0.666 0.95 0.791 2.0 1.034 3.0 1.184
0.28 0.612 0.44 0.607 0.65 0.688 1.0 0.806 2.25 1.083 3.5 1.255
0.30 0.644 0.46 0.621 0.70 0.710 1.2 0.865 2.50 1.130 4.0 1.324
0.35 0.724 0.48 0.634 0.75 0.731 1.4 0.920 2.75 1.176 4.5 1.392
0.40 0.804 0.50 0.648 0.80 0.752 1.6 0.973 3.0 1.221 5.0 1.460
0.45 0.884 0.60 0.715 0.85 0.772 2.0 1.076 4.0 1.401 5.5 1.527
0.50 0.964 0.70 0.782 0.90 0.792 3.0 1.328 5.0 1.579 6.0 1.594
0.55 1.044 0.80 0.849 0.95 0.812 4.0 1.578 6.0 1.757 6.5 1.660
0.60 1.124 0.90 0.915 1.00 0.832 5.0 1.828 7.0 1.727
0.65 1.204 1.0 0.982 2.0 1.215 8.0 1.861
0.70 1.284 2.0 1.649 3.0 1.506 9.0 1.994
0.75 1.364 3.0 2.316 4.0 1.977 10.0 2.127
0.80 1.444 5.0 3.649 5.0 2.398
Table 3 pD vs. tD - Finite radial system with closed exterior boundary, constant rate at inner
boundary
126
Fluid Flow In Porous Media
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reD = 4.5 reD = 5.0 reD = 6.0 reD = 7.0 reD = 8.0 reD = 9.0 reD = 10.0
tD pD tD pD tD pD tD pD tD pD tD pD tD pD
2.0 1.023 3.0 1.167 4.0 1.275 6.0 1.436 8.0 1.556 10.0 1.651 12.0 1.732
2.1 1.040 3.1 1.180 4.5 1.322 6.5 1.470 8.5 1.582 10.5 1.673 12.5 1.750
2.2 1.056 3.2 1.192 5.0 1.364 7.0 1.501 9.0 1.607 11.0 1.693 13.0 1.768
2.3 1.702 3.3 1.204 5.5 1.404 7.5 1.531 9.5 1.631 11.5 1.713 13.5 1.784
2.4 1.087 3.4 1.215 6.0 1.441 8.0 1.559 10.0 1.653 12.0 1.732 14.0 1.801
2.5 1.102 3.5 1.227 6.5 1.477 8.5 1.586 10.5 1.675 12.5 1.750 14.5 1.817
2.6 1.116 3.6 1.238 7.0 1.511 9.0 1.613 11.0 1.697 13.0 1.768 15.0 1.832
2.7 1.130 3.7 1.249 7.5 1.544 9.5 1.638 11.5 1.717 13.5 1.786 15.5 1.847
2.8 1.144 3.8 1.259 8.0 1.576 10.0 1.663 12.0 1.737 14.0 1.803 16.0 1.862
2.9 1.158 3.9 1.270 8.5 1.607 11.0 1.711 12.5 1.757 14.5 1.819 17.0 1.890
3.0 1.171 4.0 1.281 9.0 1.638 12.0 1.757 13.0 1.776 15.0 1.835 18.0 1.917
3.2 1.197 4.2 1.301 9.5 1.668 13.0 1.810 13.5 1.795 15.5 1.851 19.0 1.943
3.4 1.222 4.4 1.321 10.0 1.698 14.0 1.845 14.0 1.813 16.0 1.867 20.0 1.968
3.6 1.246 4.6 1.340 11.0 1.757 15.0 1.888 14.5 1.831 17.0 1.897 22.0 2.017
3.8 1.269 4.8 1.360 12.0 1.815 16.0 1.931 15.0 1.849 18.0 1.926 24.0 2.063
4.0 1.292 5.0 1.378 13.0 1.873 17.0 1.974 17.0 1.919 19.0 1.955 26.0 2.108
4.5 1.349 5.5 1.424 14.0 1.931 18.0 2.016 19.0 1.986 20.0 1.983 28.0 2.151
5.0 1.403 6.0 1.469 15.0 1.988 19.0 2.058 21.0 2.051 22.0 2.037 30.0 2.194
5.5 1.457 6.5 1.513 16.0 2.045 20.0 2.100 23.0 2.116 24.0 2.096 32.0 2.236
6.0 1.510 7.0 1.556 17.0 2.103 22.0 2.184 25.0 2.180 26.0 2.142 34.0 2.278
7.0 1.615 7.5 1.598 18.0 2.160 24.0 2.267 30.0 2.340 28.0 2.193 36.0 2.319
8.0 1.719 8.0 1.641 19.0 2.217 26.0 2.351 35.0 2.449 30.0 2.244 38.0 2.360
9.0 1.823 9.0 1.725 20.0 2.274 28.0 2.434 40.0 2.658 34.0 2.345 40.0 2.401
10.0 1.927 10.0 1.808 25.0 2.560 30.0 2.517 45.0 2.817 38.0 2.446 50.0 2.604
11.0 2.031 11.0 1.892 30.0 2.846 40.0 2.496 60.0 2.806
12.0 2.135 12.0 1.975 45.0 2.621 70.0 3.008
13.0 2.239 13.0 2.059 50.0 2.746 80.0 3.210
14.0 2.343 14.0 2.142 60.0 2.996 90.0 3.412
15.0 2.447 15.0 2.225 70.0 3.246 100.0 3.614
Table 3 (continued)
-Ei(-y),0.000<2.09,interval=0.01
0.0 + ∞ 4.038 3.335 2.959 2.681 2.468 2.295 2.151 2.027 1.919
0.1 1.823 1.737 1.660 1.589 1.524 1.464 1.409 1.358 1.309 1.265
0.2 1.223 1.183 1.145 1.110 1.076 1.044 1.014 0.985 0.957 0.931
0.3 0.906 0.882 0.858 0.836 0.815 0.794 0.774 0.755 0.737 0.719
0.4 0.702 0.686 0.670 0.655 0.640 0.625 0.611 0.598 0.585 0.572
0.5 0.560 0.548 0.536 0.525 0.514 0.503 0.493 0.483 0.473 0.464
0.6 0.454 0.445 0.437 0.428 0.420 0.412 0.404 0.396 0.388 0.381
0.7 0.374 0.367 0.360 0.353 0.347 0.340 0.334 0.328 0.322 0.316
0.8 0.311 0.305 0.300 0.295 0.289 0.284 0.279 0.274 0.269 0.265
0.9 0.260 0.256 0.251 0.247 0.243 0.239 0.235 0.231 0.227 0.223
1.0 0.219 0.216 0.212 0.209 0.205 0.202 0.198 0.195 0.192 0.189
1.1 0.186 0.183 0.180 0.177 0.174 0.172 0.169 0.166 0.164 0.161
1.2 0.158 0.156 0.153 0.151 0.149 0.146 0.144 0.142 0.140 0.138
1.3 0.135 0.133 0.131 0.129 0.127 0.125 0.124 0.122 0.120 0.118
1.4 0.116 0.114 0.113 0.111 0.109 0.108 0.106 0.105 0.103 0.102
1.5 0.100 0.099 0.097 0.096 0.094 0.093 0.092 0.090 0.089 0.088
1.6 0.086 0.085 0.084 0.083 0.081 0.080 0.079 0.078 0.077 0.076
1.7 0.075 0.074 0.073 0.072 0.071 0.070 0.069 0.068 0.067 0.066
1.8 0.065 0.064 0.063 0.062 0.061 0.060 0.060 0.059 0.058 0.057
1.9 0.056 0.055 0.055 0.054 0.053 0.052 0.052 0.051 0.050 0.050
2.0 0.049 0.048 0.048 0.047 0.046 0.046 0.045 0.044 0.044 0.043
2.0<y<10.9,interval=0.1
y 0 1 2 3 4 5 6 7 8 9
2 4.89x10-2 4.26x10-2 3.72x10-2 3.25x10-2 2.84x10-2 2.49x10-2 2.19x10-2 1.92x10-2 1.69x10-2 1.48x10-2
3 1.3x10-2 1.15x10-2 1.01x10-2 8.94x10-3 7.89x10-3 6.87x10-3 6.16x10-3 5.45x10-3 4.82x10-3 4.27x10-2
4 3.78x10-3 3.35x10-3 2.97x10-3 2.64x10-3 2.34x10-3 2.07x10-3 1.84x10-3 1.64x10-3 1.45x10-3 1.29x10-3
5 1.15x10-3 1.02x10-3 9.08x10-4 8.09x10-4 7.19x10-4 6.41x10-4 5.71x10-4 5.09x10-4 4.53x10-4 4.04x10-4
6 3.60x10-4 3.21x10-4 2.86x10-4 2.55x10-4 2.28x10-4 2.03x10-4 1.82x10-4 1.62x10-4 1.45x10-4 1.29x10-4
7 1.15x10-4 1.03x10-4 9.22x10-5 8.24x10-5 7.36x10-5 6.58x10-5 5.89x10-5 5.26x10-5 4.71x10-5 4.21x10-5
8 3.77x10-5 3.37x10-5 3.02x10-5 2.70x10-5 2.42x10-5 2.16x10-5 1.94x10-5 1.73x10-5 1.55x10-5 1.39x10-5
9 1.24x10-5 1.11x10-5 9.99x10-6 8.95x10-6 8.02x10-6 7.18x10-6 6.44x10-6 5.77x10-6 5.17x10-6 4.64x10-6
10 4.15x10-6 3.73x10-6 3.34x10-6 3.00x10-6 2.68x10-6 2.41x10-6 2.16x10-6 1.94x10-6 1.74x10-6 1.56x10-6
128
Fluid Flow In Porous Media
10
Table 4 (opposite) Values of the exponential integral, -Ei(y)
60º
27.1 0.2 0.07 0.09
Table 5 (continued)
130
11
Drive Mechanisms
CONTENTS
1 DEFINITION
4 SUMMARY
4.1 Pressure and Recovery
4.2 Gas / Oil Ratio
LEARNING OBJECTIVES
Having worked through this chapter the Student will be able to:
• Describe briefly with the aid of sketches solution gas drive distinguishing
behaviour both above and below the bubble point.
• Describe briefly with the aid of sketches the reservoir performance characteristics
of water drive reservoir.
• Describe briefly with the aid of sketches the rate sensitivity aspect of water
drive reservoir.
• Summarise the characteristics of solution gas drive, gas cap drive and water
drive reservoirs.
Drive Mechanisms
11
RESERVOIR DRIVE MECHANISMS
In the previous chapters we have considered the physical properties of the porous
media, the rock, within which the reservoir fluids are contained and the properties and
behaviour of the fluids. In this chapter we shall examine the various methods used to
calculate the performance of different reservoir types, we will introduce the various
drive mechanisms responsible for production of fluids from a hydrocarbon reservoir.
In this qualitative description of the way in which reservoirs produce their fluids we will
see how the various basic concepts come together to give understanding to the various
driving forces responsible for fluid production. One of the main preoccupation’s of
reservoir engineers is to determine the predominant drive mechanism, for dependant
on the drive mechanism different recoveries of oil can be achieved.
As well as presenting natural drive mechanisms we will also review various artificial
drive mechanisms.
1 DEFINITION
A reservoir drive mechanism is a source of energy for driving the fluids out through
the wellbore. It is not necessarily the energy lifting the fluids to the surface, although
in many cases, the same energy is capable of lifting the fluids to the surface.
There are a number of drive mechanisms, but the two main drive mechanisms are
depletion drive and water drive. Other drive mechanisms to be considered are
compaction drive and gravity drive. These drive mechanisms are natural drive
energies and are not to be confused with artificial drive energies such as gas injection
and water injection.
2.1 Depletion Drive Reservoirs
A depletion type reservoir is a reservoir in which the hydrocarbons contained are
NOT in contact with a large body of permeable water bearing sand. In a depletion
type reservoir the reservoir is virtually totally enclosed by porous media and the only
energy comes from the reservoir system itself. Figures 1 and 2 illustrate the types of
accumulations which can give rise to depletion drive characteristics.
In figure 1 the hydrocarbons are enclosed in isolated sand lenses which have been
generated by a particular depositional environment. Over geological time the
hydrocarbons have found their way into the porous media. The surrounding rocks may
have permeability but it is so low as to prevent energy transfer from other sources.
In figure 2 is illustrated another depletion type reservoir where a mature reservoir has
been subjected to faulting, resulting in the isolation of a part of the reservoir from
the rest of the accumulation. In a total field system, such a situation can give rise to
parts of the reservoir having different drive mechanism characteristics.
Gas
Oil
Water
Drive Mechanisms
11
2.2 Water Drive
Gas
Oil
Water
A water drive reservoir is one in which the hydrocarbons are in contact with a large
volume of water bearing sand. There are two types of water drive reservoirs. There
are those where the driving energy comes primarily from the expansion of water as
the reservoir is produced, as shown in figure 3 The key issue here is the relative
size and mobility of the water of the supporting aquifer relative to the size of the
hydrocarbon accumulation.
Water drive may also be a result of artesian flow from an outcrop of the reservoir
formation, figure 4. In this situation either surface water or seawater feeds into the
outcrop and replenishes the water as it moves into the reservoir to replace the oil. The
key issues here are the mobility of the water in the aquifer and barriers to flow from
the outcrop to the reservoir. It is not often encountered, and the water drive arising
from the compressibility of an aquifer, figure 3, is the more common.
Outcrop
of sand
Oil well
Water flow
Old land
surface
New land
surface
Oil
Drive Mechanisms
11
dominant energy in earlier years. Gravity drainage can be significant and effective
in steeply dipping reservoirs which are fractured.
Of the drive mechanisms mentioned the major drive mechanisms are depletion drive,
which are further classified into solution gas drive and gas cap drive and water drive.
Gravity Drive typically is active during the final stages of a depletion reservoir.
Closed in
1000
Initial
Z GOC
Present
GOC
ΟWC
Gas
Oil
Water
Inactive aquifer
Gravity drive typically is active during the final stage
of a depletion reservoir.
Drive Mechanisms
11
Gas cap present initially
Oil at interface is at Pb
Gas cap
Oil
Edge water
Water coning
Gas Cap
Oil zone
Water Water
Original condition
Gas Cap
Oil zone
Water Water
50 % Depleted
Sometimes it may be only water drive in the above situations. If the hydrocarbons
are taken out at a rate such that for every volume of oil removed water readily moves
in to replace the oil, then the reservoir is driven completely by water. On the other
10
Drive Mechanisms
11
hand there may be only depletion drive. If the water does not move in to replace the
oil, then only the gas cap would expand to provide the drive.
Having considered the basic aspects of the drive types we will now examine their
respective characteristics in relation to production, recovery and pressure decline issues.
Plateau phase
Production
Decline phase
Production
build up
0 Abandonment
Time
The next stage represents the period when the productivity of the production facility
is at its design capacity and the wells are throttled back to limit their productivity.
This period is called the plateau phase when production is maintained at the design
capacity of the facilities. Typical production rates for the plateau period cannot be
presented since it depends on the techno-economics of the field. Clearly for a field
with a very large front loaded capital investment there is an incentive to have a high
production rate during the plateau phase , say 20% of the STOIIP, whereas for a
lower cost onshore field 5% might be acceptable. Governments will also impose
their considerations on this aspect as well.
A time will come when the reservoir is no longer able to deliver fluids to match the
facilities capacity and the field goes into the decline phase. This phase can be delayed
by methods to increase production. Such methods could include artificial lift, where
the effort required to lift the fluids from the reservoir is carried out by a downhole
pump or by using gas lift to reduce the density of the fluid system in the well.
There comes a time when the productivity of the reservoir is no longer able to
generate revenues to cover the costs of running the field, This abandonment time
again is influenced by the size and nature of the operation. Clearly a single, stripper
well, carrying very little operational costs, can be allowed to produce down to very
low rates. A well, as part of a very high cost offshore environment however, could
be abandoned at a relatively high rate when perhaps the water proportion becomes
too high or the productivity in relation to all production is not sufficient to meet the
associated well and production costs.
We will now review the performance characteristics of the various mechanisms in
light of the forgoing production phases.
12
Drive Mechanisms
11
3.1.1 Solution Gas Drive, Oil Production ( Figure 14 )
After a well is drilled and production starts for a solution gas drive reservoir, the
pressure drops in the vicinity of the well. The initially pressure drop is rapid as flow
results from the low compressibility of the system above the bubble point. Pressure
continues to decline and solution gas drive becomes effective as gas comes out of
solution. Mobility of gas occurs and the reduced mobility to oil and resulting decreasing
oil relative permeability further causes the pressure to decline and productivity to oil
flow decrease. Initially when all wells are on stream the oil production is high but
the production rapidly declines and there is a short plateau and decline phase until
an economic limit is reached.
Oil Reservoir
G.O.R Prod Pressure
Oil
Prod
G.O.R
Reservoir
Pressure
Time-Year
A good analogy for this type of reservoir is the champagne bottle opened by a champion
to spray the contents over enthusiastic supporters - a short lived high production
scenario followed by rapid decline!
When the bubble point is reached in the near well vicinity, the initial gas which
comes out of solution is immobile and therefore oil entering the wellbore is short of
the previous level of solution gas. Theoretically at the surface the producing GOR
level is less than the original GOR 2-3 in figure 15.
As the pressure continues to decline the productivity of the well continues to decline
from the combined impact of reducing relative permeability and drop in bottom hole
pressure. The production GOR goes though a maximum as oil eventually is produced
into the well bore with a low solution GOR and the associated gas which has come
out of solution has progressed much faster to the well and contributed to earlier gas
production 4-5 in figure 15.
GOR constant
above bubble 4
point pressure
Producing GOR.
Rsi
1 2 5
3
Pb
Pressure
When the pressure drops below the bubble point throughout the reservoir a secondary
gas cap may be produced and some wells have the potential of becoming gas
producers.
3.1.3 Pressure
At first the pressure is high but as production continues the pressure makes a rapid
decline.
3.1.4 Water Production, Well Behaviour, Expected Oil Recovery and Well
Location
Since by definition there is little water present in the reservoir there should be no
water production to speak of. Because of the rapid pressure drop artificial lift will
be required at an early stage in the life of the reservoir. The expected oil recovery
from these types of reservoirs is low and could be between 5 and 30% of the original
oil-in-place. Abandonment of the reservoir will depend on the level of the GOR
and the lack of reservoir pressure to enable production. Well locations for this drive
mechanism are chosen to encourage vertical migration of the gas, therefore the wells
producing zones are located structurally low, but not too close to any water contact
which might generate water through water coning. Figure 16.
14
Drive Mechanisms
11
Secondary
gas cap
Gas Cap
Oil zone
Water Original condition Water
Gas Cap
Oil zone
Water Water
50 % Depleted
Pressure
G.O.R
500 10 5000
Oil
5 Prod
250 2500
Rate
G.O.R BSW %
20
10
0 0
0 1 2 3 4 5 6 7
Time-Year
16
Drive Mechanisms
11
3.2.4 Water Production, Well Behaviour, Expected Oil Recovery and Well
Locations
Like solution gas drive there should be negligible water production. The life of the
reservoir is largely a function of the size of gas cap but it is likely to be a long flowing
life. The expected oil recovery for such a system is of the order of 20 to 40% of the
original oil-in-place. The well locations, similar to solution gas drive, are such that
the production interval for the wells should be situated away from the gas oil contact
but not too close to the water oil contact to risk water coning.
3.3 Water Drive
The majority of water drive reservoirs predominantly get their drive energy from the
compressibility of the aquifer system. The effectiveness of water drive depends on
the ability of the aquifer to replace the volume of the produced oil. The key issues
with a water drive reservoir are therefore the size of the aquifer and permeability.
This is because the only way for a low compressibility system to be effective is
for its relative size to the oil accumulation to be large, and the permeability of the
aquifer to water to enable flow though the aquifer and into the oil zone. These key
issues set a considerable challenge to the reservoir engineer since to predict water
drive behaviour, requires such information, which in pre production periods can
only be obtained from exploration activity to determine the extent and properties of
the aquifer. It is difficult to obtain justification to expend such exploration costs in
determining the size of a water accumulation!
Three sketches below illustrate the various types of production profiles for different
aquifer types and the influence of rate sensitivity. In figure 19 we have the artesian
type aquifer where there is communication to surface water though an outcrop. In
this case if oil is produced at a rate less than the aquifer can move water into the oil
zone, then the reservoir pressure, as measured at the original oil water contact, remains
constant. The producing gas-oil ratio also remains constant since the reservoir is
undersaturated. These reservoirs will enable a plateau phase, however as in all water
drive reservoirs the decline of the reservoirs is not due to productivity loss through
pressure decline but the production of water. The encroaching aquifer with perhaps
its favourable mobility will preferentially move through the oil zone and if there
are high permeability layers will move through these. Eventually the water-cut, the
proportion of water to total production becomes too high and the well is abandoned
to oil production.
Oil well
Water flow
Reservoir pressure
Pi
Production
Rsi GOR
Water production
Time
Figure 20 illustrates a more typical water drive reservoir where the drive energy comes
from the compressibility of the aquifer system. In this case if the oil withdrawal
rate is less then the rate of water encroachment from the aquifer then the reservoir
pressure will slowly decline, reflecting the decompression of the total system , the
oil reservoir and the aquifer. Clearly this pressure decline is related to the size of
the aquifer. The larger the aquifer the slower the pressure decline. As with all water
drive reservoirs productivity of the wells remains high resulting from the maintained
pressure, however the productivity of the well to oil reduces as water breakthrough
occurs. So a characteristic of water drive reservoirs is the increasing water production
alongside decreasing oil production.
18
Drive Mechanisms
11
Pi
Oil production rate
Reservoir
Production pressure
Rsi GOR
Water production
Time
Figure 21 illustrates the rate sensitive aspect of water drive reservoirs. If the oil
withdrawal rate is higher than the water influx rate from the aquifer then the oil
reservoir pressure will drop at a rate greater than would be the case with aquifer
support alone, as the compressibility of the oil reservoirs supports the flow. If this
pressure drops below the bubble point then solution gas drive will occur, as evidenced
by an increase in the gas-oil ratio. Cutting back oil production to a rate to less than
the water encroachment rate restores the system to water drive, with the gas-oil ratio
going back to its undersaturated level.
When two drive mechanisms function as above then we have what is termed
combination drive ( water drive and solution gas drive).
Water drive reservoirs have good pressure support. The decline in oil production is
related to increasing water production as against pressure decline.
Reservoir pressure
Ps
ratio
B/d
psi
PROD
50 250
1000 5000 Oil production
Water production
GOR
25
Water
Bsw
0
0 0 69 0
70 71 72 73 74 75
Because water drive, through pressure maintenance provides the most optimistic
recoveries, artificial water drive is often part of the development strategy because of
the uncertainties of the pressure support from the associated aquifer. In the North Sea
for example many reservoirs have associated aquifers. The risk of not knowing either
20
Drive Mechanisms
11
the extent or activity of the aquifers is such that many operators are using artificial
water drive systems to maintain pressure so that solution gas drive does not occur
with the consequent loss of oil production.
4 SUMMARY
The following summaries and tables give the main features associated with the
various drive mechanisms.
Water-drive -pressure declines slowly and abandonment occurs when the water cut
is too-high at around 50% of recovery, but depends on local factors.
Gas-cap drive - the pressure shows a marked decline and economic pressures
are reached around 20% of the original pressure when about 30% of the oil is
recovered.
Solution- gas drive - the pressure drops more sharply and at 10% of the pressure
reaches, an uneconomical level of recovery at about 10% of the oil-in-place.
Water drive - the curve for a water drive system shows a gas/oil ratio that remains
constant. Variations from this indicate support from solution gas drive or other
drive mechanisms
Gas-cap drive - for this drive the gas/oil ratio increases slowly and continuously.
Solution- gas drive - the curve for a solution gas drive reservoir shows that the
gas/oil ratio increases sharply at first then later declines.
Characteristics Trend
1. Reservoir Pressure Declines rapidly and continuously
2. Gas/Oil Ratio First low then rises to a maximum and
then drops
3. Production Rate First high, then decreases rapidly and
continues to decline
4. Water Production None
5. Well Behaviour Requires artificial lift at early stages
6. Expected Oil Recovery 5-30% of original oil-in-place
Characteristics Trend
1. Reservoir Pressure Falls slowly and continuously
2. Gas/oil ratio Rises continuously
3. Production Rate First high, then declines gradually
4. Water Production Absent or negligible
5. Well Behaviour Cap Long flowing life depending on size of gas cap
6. Expected Oil Recovery 20 to 40% of original oil-in-place
Characteristics Trend
1. Reservoir Pressure Remains high
2. Gas/Oil Ratio Remains steady
3. Water Production Starts early and increases to appreciable
amounts
4. Well Behaviour Flow until water production gets excessive
5. Expected Oil Recovery up to 60% original oil-in-place.
Figures 22 and 23 give the pressure and gas-oil ratio trends for various drive
mechanism types
22
Drive Mechanisms
11
Reservoir pressure trends for reservoirs under various drives.
100
60
20
Dissolved
gas drive
0
0 20 40 60 80 100
Oil produced - percent of original oil in place
Figure 22
Reservoir gas - oil ratio trends for reservoirs under various drives.
5
1
Water drive
0
0 20 40 60 80 100
Oil produced - percent of original oil in place
Figure 23
CONTENTS
2 IDEAL SOLUTIONS
2.1 Raoult's Law
2.2 Dalton's Law
2.3 Ideal Equilibrium Ratio
5 SEPARATOR PROBLEMS
5.1 Gas / Oil Ratio
5.2 Oil Formation Volume Factor
5.3 Optimum Pressure of Separator System
5.4 Example of Separator Problem
LEARNING OBJECTIVES
Having worked through this chapter the Student will be able to:
• Derive and explain the use of equations to determine the dew point pressure
and bubble point pressure of a fluid mixture.
• Describe in general terms the impact of separator conditions the gas-oil ratio
and oil formation volume factor.
Vapour Liquid Equlibria
12
1 INTRODUCTION - THE IMPORTANCE OF VAPOUR-LIQUID
EQUILIBRIUM
Predicting the relative amount of the phases and their respective physical properties
is an essential element in all of the above operations. The topic of vapour-liquid
equilibria is also at the heart of many subsequent and other process operations and
therefore there have been a range of approaches into the solution of the problem.
Figure 1, and 2 illustrate the complex nature of oil and gas production where,
particularly in a major offshore province, as well as onshore, a number of reservoirs
produce into a common transport line and associated treatment facilities (Figure 1).
Each of the fields with their unique composition clearly contribute to a compositional
blend entering a treatment facility (Figure 2), where further separation occurs.
CUSTOMERS
ONSHORE FACILITIES
WELLS WELLS
RESERVOIR RESERVOIR
TERMINAL
Composition = (TA*CA)*(TB*CB)+(TC*CC)
TA + TB + TC
Throughput = TA + TB + TC
FIELD A FIELD C
Composition CA Composition CC
Throughput TA Throughput TC
FIELD B
Composition CB
Throughput TB
The allocation of revenue based on quality of product and oil injected into a common
pipeline provides a considerable challenge to metering and compositional analysis.
To the reservoir engineer, the main issues are the multiphase behaviour in the formation
and the relationship between the fluids in the reservoir and those produced at surface
conditions.
The critical element in reservoir simulation is the grid block where the saturations
and flow behaviour of the respective fluids; gas, oil and water are required. The
grid block therefore can be considered as a ‘separator’ and vapour-equilibrium
calculations are required out to determine the relative amounts of the phases which
lead to saturation values and relative permeability of the phases, and the composition
of these phases which lead to important physical property values of density, viscosity and
interfacial tension.
Vapour Liquid Equlibria
12
In the previous chapter the considerations of the relative amounts of gas and liquid
were considered in the simplistic two component black oil approach. In this chapter
we will consider approaches to vapour-liquid equilibrium from a compositional model
consideration both from an ideal behaviour perspective and then the consideration
of real systems.
On a pressure temperature phase diagram of a multi-component mixture, the area
bounded by the bubble point and dew-point curves defines the conditions for gas and
liquid to exist in equilibrium. It is an over-simplification to describe the system as
involatile oil with associated solution gas. The behaviour of the individual components
and their influence on the composition of the mixture need to be considered.
If a sample of bubble point fluid is brought to surface to separator conditions, the fluid
enters the two phase region at a temperature and pressure much lower than reservoir
conditions. In the separator the liquid and gas phases, in equilibrium, are withdrawn
separately. Large volumes of gas are formed at these separator conditions, and the
liquid volume shrinks substantially because of decreased temperature and conversion
of some of the fluid into the gas phase. The separator liquid is collected in the stock
tank, at which additional temperature and pressure drop may occur, more gas may
be released depending on the separator conditions to stock tank conditions.
If Vo is the volume of liquid at reservoir conditions and Vst is the volume of stock
tank oil. The oil formation volume factor Bo is :
Vo
Bo =
Vst
If Vg and Vst are the total volume of gas and oil collected from the separator and stock
tank. The solution gas to oil ratio is :
Vg
Rs =
Vst
The volume factors can be determined directly in the laboratory or from equilibrium
calculations.
• Reservoir calculations
• Process calculations
2 IDEAL SOLUTIONS
Before we consider the behaviour of real systems we will first examine the behaviour
of an ideal solution, where no chemical interaction occurs and where no inter-molecular
forces occur when mixing components.
These ideal solutions result in no heating effects when ideal solutions are mixed and
the volume of the mixture equals the sum of the volumes the pure components would
occupy at the same pressure and temperature.
pj = xjpvj (1)
pj = yjp (2)
where yj is the composition of the vapour and p is the pressure of the system
Vapour Liquid Equlibria
12
y j pvj
=
xj p (4)
i.e. the ratio of the component in the vapour and liquid phases is given by the ratio of
the vapour pressure of the pure component to the total pressure of the system. This
ratio is termed the Equilibrium ratio, Kj .
If n is the total number of moles of mixture and zj is the mole fraction of component
j in the mixture.
where nL and ng are the moles of liquid and gas such that nL + ng = n
From equation 4.
pvj
z j n = x j nL + x j ng
p
zjn
∴ xj =
p
nL + vj ng
p (6)
Similarly:
c c
zjn
∑y = ∑ j
j =1 n +
p
= 1.0
j =1
g .n L
pvj
(8)
Using these equations in a trial and error method the compositions of vapour and
liquid streams in a flash separation can be determined.
The equilibrium ratio Kj is defined as the ratio of the composition of j in the vapour
to liquid phase, i.e.
yj
Kj =
xj (11)
Other names for Equilibrium ratio, include K-factors, K-values, equilibrium vapour
-liquid distribution ratios.
Fugacity
Lewis2 introduced the concept of fugacity, for use in equilibrium calculations, to
extrapolate or correct vapour pressures. This is required since a pure component only
has a vapour pressure up to its critical point. The fugacity is a thermodynamic quantity
defined in terms of the change in free energy in passing from one state to another.
For an ideal gas , the fugacity is equal to its pressure, and the fugacity of each
component is equal to its partial pressure. The ratio of fugacity to pressure is termed
the fugacity coefficient, φ. For a multicomponet system,
fi
φ=
Pzi (12)
All systems behave as ideal gases at very low pressures, therefore φ > 1 when P >
0
fg = fL. (13)
Vapour Liquid Equlibria
12
The fugacity coefficient, φ of a pure component can be calculated from the following
general equation (Danesh).
p v
z − 1 1 RT
ln φ = ∫ dp = ( z − 1) − ln z + ∫ − P dv
p RT v
o ∞ (14)
The ratio of the fugacity at the state of interest to that at a reference state is called
the activity εi = fi/fio
The ratio of activity to concentration is called the activity coefficient Θi, where
Θi = εi/xi
Therefore fi=Θixifio
(15)
Different methods have been developed for treating vapour-liquid equilibrium for
non ideal systems.
The previous K value is based on both ideal and ideal solutions laws. To extend
the principle of equilibrium ratio to multicomponent hydrocarbon mixtures to the
pressures and temperatures relevant to petroleum engineers, methods of treating non
ideal systems need to be established.
The subject of non ideal equilibrium ratios are treated later in the text. We assume
in this section that K values are available either from whatever source, experimental,
NGPSA data charts, or from equations of state and other predictive methods.
The equilibrium equations which are used for a process separator are the same as
those within a grid block or element of a reservoir simulator.
yj
F T&P
zj
xj
It is common to express the feed F as 1.0 or 100 moles and express L and V as
fractions or percentages of F.
i.e. F = 1 = L + V (16)
For component j
yj
Kj =
xj (18)
By definition:
m m m
∑x = ∑y = ∑z
j j j =1
j =1 j =1 j =1 (19)
10
Vapour Liquid Equlibria
12
where m is the number of components.
zj = xjL + xjKjV
zj = xj (L + KjV)
zj
xj =
L + K jV (20)
and:
m m
zj
∑ x = ∑ L + K V = 1.0
j
j =1 j =1 j (21)
similarly:
m m
zj
∑ yj = ∑ j =1 V + L K j
= 1.0
j =1 (21a)
These equations are the key equations in vapour-liquid equilibrium calculations and
their use is the same whether in those calculations to determine phase behaviour in
a separator or those which take place within the numerous grid blocks of a reservoir
simulator. Clearly in the latter the amount of calculations is considerable since each
grid block can be considered a separator. In a large compositional based simulation
a study thousands of grid blocks will be used.
The method of calculation is therefore as follows for each separation element:
(1) Select Kj for each component at the temperature and pressure of the system;
(For the determination of K see the later section.)
(3) Calculate either V, L, ∑xj or ∑yj from equation 21, 21a, 22 and 22a;
(4) Either:
(i) check V&L calculated against assumed V or L;
(ii) determine if ∑xj or ∑yj = 1.0;
(5) Repeat the calculation until assumed value is calculated value or until ∑xj
and ∑yj = 1.0.
This phase equilibrium perspective can also be used to calculate the reservoir saturation
pressures for a particular temperature, ie. the dewpoint and bubble point pressures.
Vapour
yj = Zj
Liquid Xj
Vapour
Liquid
zj = yj
(22)
or: zj = xjKj
12
Vapour Liquid Equlibria
12
4.3 Bubble Point Calculation
The bubble point is when the first bubble of gas appears. At this point the composition
of this bubble of gas is higher in lighter hydrocarbons whereas the composition of
the liquid is essentially the composition of the system. Figure 5.
Vapour = Yj
Liquid
Xj = Zj Vapour
Liquid
zj = xj
(24)
or:
yj
zj =
Kj
The mixture at the bubble point is therefore in equilibrium with a quantity of liquid
having a composition defined by the above equation.
Also:
m m
∑y = ∑z K
j j j = 1.0
j =1 j =1 (25)
The dew-point and bubble point when either temperature or pressure are known are
determined by trial and error techniques until the above relationships are satisfied.
The dew-point pressure or bubble point pressure are estimated, K values obtained
and equations 23 or 25 checked. If the summation ≠ 1, different pressure values are
tried until convergence is reached. When convergence is reached the respective dew
point or bubble point pressure has been obtained.
5 SEPARATOR PROBLEMS
Feed
To pipeline / tanker
Liquid
Separator Stocktank
at P sep and T sep at P st and T st
Two-stage separation
Feed
To pipeline / tanker
Liquid Liquid
Three-stage separation
14
12
Vapour Liquid Equlibria
nST is the moles of liquid in stock tank for n1 moles into first separator:
m
∴ nST = n1 ∏ Li
i =1 (26)
m = number of stages
Li = mole fraction of liquid off ith stage
n1 = moles of feed to first stage
∴
If n1 = 1
then:
m
n ST = ∏ Li
i =1 (27)
m
n ST = ∏ Li
i =1 = mole fraction of STO in the feed.
∴ If nj = 1
m i −1
n gT = ∑ Vi ∏ L j
i =1 j =1 (28)
n ST MST
(VST )m = ρST
(29)
n gT Vm n gT Vm ρST
RT = =
(VST )m n ST MST (30)
If the feed to the first stage is a single-phase liquid into its point of entry into the
production stream then Bo can be calculated.
Bo =
(Vres ) = Mres ρST
VST ρres MST nST (31)
where
lb.res. fluid
Mres = molecular weight of reservoir fluid =
lb.mole. fluid
and
lb.mol. stock tan k fluid
nST =
lb.mol.res. fluid
16
Vapour Liquid Equlibria
12
It is the equilibrium characteristics of the individual components as a function
of temperature, pressure and composition which influence these total separation
characteristics for the mixtures at each separation stage.
33.2
tio
Total gas-oil ra
r
500 cto 1.26
32.6 e fa
olum
Formation v
Figure 7 Effect of separator pressure in a two-stage separation process (Amyx, Bass &
Whiting)
Equilibrium ‘flash’ calculations, which the above are called, are used in many other
applications. In reservoir engineering, flash calculations are at the core of compositional
simulation.
Step 1: Calculate the composition and quantities of separator gas and liquid using
equation 21.
zj
∑ x = ∑ L + K V = 1.0
j
j
Component xj = zj
L+KjV yj=kjxj
C1 0.0356 0.7399
C2 0.0299 0.1219
C3 0.0919 0.1066
i-C4 0.0170 0.0084
n-C4 0.0416 0.0143
i-C5 0.0198 0.0028
n-C5 0.0313 0.0034
C6 0.0511 0.0017
C7+ 0.6818 0.0010
1.0000 1.0000
Step 2: Calculate the compositions and quantities of stock tank and liquid using equation
21, noting that the composition of the feed to the stock tank is the composition of
the liquid from the separator.
18
Vapour Liquid Equlibria
12
Component Component, Equilibrium xj= zj
of feed to mole ratio at V=0.13 L+KjV V=0.1351
separator fraction zj 14.7 psia V=0.14
and 76°F, Kj
C1 0.0356 161 0.00163 0.00152 0.00157
C2 0.0299 30.7 0.00615 0.00580 0.00597
C3 0.0919 8.15 0.04763 0.04593 0.04674
i-C4 0.0170 3.27 0.01313 0.01290 0.01301
n-C4 0.0416 2.30 0.03559 0.03519 0.03538
i-C5 0.0198 0.90 0.02006 0.02008 0.02007
n-C5 0.0313 0.675 0.03286 0.03279 0.03274
C6 0.0511 0.20 0.05703 0.05755 0.05729
C7+* 0.6818 0.0089* 0.78264 0.79164 0.78720
1.0000 0.99654 1.00340 1.00000
* Used K for C9
The summation equals 1.0 so VST = 0.1351 and LST = 0.8649 and the compositions
of the stock tank gas and liquid are:
xj yj = Kjxj
0.016 0.2534
0.0060 0.1831
0.0467 0.3810
0.0130 0.0425
0.0354 0.0814
0.0201 0.0181
0.0327 0.0221
0.0573 0.0115
0.7872 0.0070
1.0000 1.0001
Step 3: Calculate the density and molecular weight of the stock tank oil.
0.181
= 0.001
Weight fraction ethane in ethane plus = 220.837
0.026
= 0.0001
Weight fraction methane in STO = 220.863
lb
ρSTO = 53.73
cu.ft. From chapter 6, figure 13.
53.73
γ STO = = 0.861
62.4
141.5
o
API = − 131.5 = 32.8
0.861
V1ρSTO
RSP = 2130
L1 LST MSTO
Similarly:
2130V2 ρSTO
RST =
L2 MSTO
RT = RST + RSP
(2130)(0.4291)(53.73) SCF
RSP = = 450
(0.5709)(0.8649)(220.9) STB
(2130)(0.1351)(53.73) SCF
RST = = 81
(0.8649)(220.9) STB
SCF
RT = 531
STB
Step 5: Calculate the density and molecular weight of the reservoir liquid at reservoir
conditions.
20
Vapour Liquid Equlibria
12
Component Component, Molecular Weight xjMj Liquid Liquid
of reservoir mole weight Mj density at volume at
liquid fraction xj 60°F and 60°F and
14.7 psia 14.7 psia
ροj xjMj / ροj
C1 0.3378 16.0 5.405
C2 0.0694 30.1 2.089
C3 0.0982 44.1 4.331 31.66 0.1368
i-C4 0.0133 58.1 0.773 35.12 0.0220
n-C4 0.0299 58.1 1.737 36.45 0.0477
i-C5 0.0125 72.2 0.902 38.96 0.0232
n-C5 0.0193 72.2 1.393 39.35 0.0354
C6 0.0299 86.2 2.577 41.30 0.0624
C7+ 0.3897 263 102.491 55.25 1.8550
1.0000 Mor =121.699 2.1825 cu.ft.C3+
lb / lb mole res oil lb.mole res oil
∑C2+ = 116.294
∑C3+ = 114.205
114.205 lb
= 52.33
Density of propane plus = 2.1825 cu ft
2.089
= 0.018
Weight fraction ethane in ethane plus = 116.294
5.405
= 0.044
Weight fraction methane in reservoir oil = 121.699
ρ
po = 49.1 lb.cu ft. From figure 13, Chapter 6.
(121.7)(53.73)
Bo =
(46.04)(220.9)(0.5709)(0.8649)
res bbl
Bo = 1.302
STB
Determination of K valves
In the procedures developed, it has been assumed that values for K are available. In
the next chapter we will examine the procedures for determining K.
22
Vapour Liquid Equlibria
12
REFERENCES
(1) Danesh, A, "PVT and Phase Behaviour of Petroleum Reservoir Fluids." 1998
Elsevier. pp 105-206
(5) GPSA: Engineering Dat Book, Gas Processors Association. Tulsa Oklahoma,
Gn Education 1972
CONTENTS
1 INTRODUCTION
1.1 Importance of the Critical Point
1.2 Equilibrium Ratio
Having worked through this chapter the Student will be able to:
• Comment on the evolution of EOS in VLE application . (The student would not
be expected to reproduce the equations.)
Equilibrium Ratio Prediction and Calculation
1 INTRODUCTION
In the previous chapter we introduced the concept of vapour-liquid equilibrium
and its role in attempting to predict the behaviour of multi-component hydrocarbon
mixtures in the two phase region, in particular the ability to predict phase densities
and composition as a function of pressure and temperature. In the chapter we also
reviewed some of the basic equations and procedures which are at the foundation of
these multicomponent calculations
There are two distinctive features of vapour equilibrium calculations in the context
of petroleum engineering compared to other sectors like process engineering. In
process engineering the calculations are mainly related to either a single process unit
or a few in series, whereas in reservoir engineering the calculations are often part of
multi grid block reservoir simulation where the numbers of “process units” can be
considered in hundreds or even thousands. In process engineering a full compositional
description is available for the prime objective is to determine compositional split
to obtain specific compositional objectives. In petroleum engineering however, the
objective is not to determine full compositional description of produced phases, but
to determine phase volumes and phase properties. The challenge is to carry this out
with a minimal composition description without compromising the quality of physical
property predictions.
In many ways the characteristics of the critical state that make it theoretically and
practically significant are also the characteristics that make it one of the more difficult
conditions to measure experimentally. The very fact that density differences between
phases vanish, that the rate of volume change with respect to pressure approaches
infinity, or that infinitesimal temperature gradients can be responsible for a transition
from 100% liquid to 100% vapour all make the critical condition one of the more
difficult to measure or observe accurately. For obvious economic reasons, it is a
condition that cannot be obtained by experiment in any practical way for the many
systems for which it is required. Consequently, many attempts have been made to
develop methods for predicting the critical properties based on generalised empirical
or semi-empirical procedures.
The literature abounds with methods and equations associated with phase equilibrium
prediction. The methods can be broadly classified as:
There are a range of approaches to expressing the vapour liquid distribution behaviour
of reservoir fluid hydrocarbon systems, as outlined below.
Equilibrium Ratio Prediction and Calculation
separation calculations is not straightforward, since no one before would have been
able to carry out tests on the fluid!!
An approach is required that the distribution of each component of the mixture between
the liquid and vapour phases be experimentally determined for a range of temperatures
and pressures. The resultant measurements are expressed as an equilibrium ratio Kj,
for component ‘j’ defined as:
yj
Kj =
where:
xj (1)
Kj = equilibrium ratio
yj = mole fraction of component ‘j’ in the vapour phase
xj = mole fraction of component ‘j’ in the liquid phase
This approach is really only useful for light hydrocarbons. The data is usually expressed
graphically as a plot of Kj versus pressure for a constant temperature.
The oil industry has relied on experimentally determined equilibrium ratios although
increasingly over recent years the move has been towards Equation of State derived K
values. Clearly the empirical K values are obtained from known mixtures, the challenge
is to determine the applicability of these empirical values to new mixtures.
At high pressures the equilibrium ratio is a function of temperature, pressure and also
the composition of the mixture. This compositional influence is of great significance.
Hence, pure component K-values measured for one mixture cannot be accurately
applied to another mixture. Figure 1 and 2 present the K values for a heavy oil and at
the other extreme a condensate at a temperature of 200˚F. The K values converge to
unity at 5,000 psia and 4,000 psia respectively. This point is called the convergence
pressure. Close examination of these two sets of K values demonstrate that these
distribution coefficients do not have the same values at particular pressures and
temperatures, confirming this compositional influence on K values.
If the temperature at which the equilibrium ratios are presented is the critical temperature
then the convergence pressure would be the critical pressure.
At a given temperature, as the system pressure increases, the K-values of all components
of the system converge to unity when the system pressure reaches the convergence
pressure. In other words, it is the pressure for a system at a given temperature when
vapour-liquid separation is no longer possible. Naturally, it is equally impossible to
have a vapour-liquid separation at a given temperature in which the system pressure
is greater than the convergence pressure.
Institute of Petroleum Engineering, Heriot-Watt University
100
et
M
10 ha
ne
Et
ha
ne
Pr
op
Bu
an
e
t an
Pe es
K=1 n ta
He ne
s
xa
ne
s
pt
H
an
e
0.1 es
an
d
He
av
ie r
0.01
10 100 1,000 10,000
Pressure, psia
Figure 1 Equilibrium Ratios at 200˚F for Low-Shrinkage Oil(Amyx Bass & Whiting)1
40
20
10 M
8 et
ha
6 ne
4 Et
ha
ne
2 Pr
Equilibrium Ratio, K
o pa
ne
1 Bu
0.8 tan
0.6 es
0.4 Pe
nta
ne
s
0.2 He
xa
nes
0.1
0.08
0.06
0.04 He
pta
ne
sa
0.02 nd
Heavier
0.01
100 1,000 10,000
Pressure, psia
Figure 2 Equilibrium Ratios at 200˚F for a Condensate Fluid (Amyx Bass & Whiting)1
Equilibrium Ratio Prediction and Calculation
CONSTANT
TEMPERATURE
Component 1
In Kj
Kj - 1.0
Component 2
In Pc
In P
Figure 3 Variation of K Values with pressure for a typical Binary Hydrocarbon System
Variations of Equilibrium
Constant With Pressure
at 200º F
M
10.0 et
ha
ne
Et
an
h
Pr e
op
Bu an
ta e
ne
Pe
Equilibrium Constant, K
s
H nta
1.0 ex ne
an s
es
H
ep
ta
ne
0.1 s
an
d
He
av
ie
r
0.01
10 100 1,000 10,000
Pressure, psia
Figure 4 Variation of K Values with pressure for a crude oil containing Light
Hydrocarbons
The procedure is to convert our mixture into a two component mixture based on
methane then seek to identify a compound which has the same physical properties
as our heavier C2+ component
The chart, Figure 5 from the NGPSA Manual, can be used for this2. The method is
to take as the lightest component present in any significant quantity (1.0% or greater
in a raw mixture; it is usually methane) as the light component of a two component
system. The heavy component is estimated from the composition of the remaining
components. A visual estimate is usually good enough. Join the heavy and light
component together as shown on the Figure 5 and read off PK against the operating
temperature.
Step 1: Assume the liquid phase composition or make an approximation. (If there
is no guide, use the total feed composition).
Step 2: Identify the lightest hydrocarbon component which is present at least 0.1
mole % in the liquid phase.
Step 3: Calculate the weight average critical temperature and critical pressure for
the remaining heavier components to form a pseudo binary system. (A shortcut
approach good for most hydrocarbon systems is to calculate the weight average Tc
only).
Step 4: Trace the critical locus of the binary consisting of the light component and
pseudo-heavy component. When the averaged pseudo heavy component is between
two real hydrocarbons, interpolation of the two critical loci must be made.
Convergence Pressures for Hydrocarbons
(Critical Locus)
20,000
C -Kens
1 ol
10,000
H2
9,000
8,000
-C 3
H2
H2
7,000
-C 2
-C 2
6,000
H 2-
C -nC
5,000 1 10
nC 6
C1-nC9
4,000
400
nC7-nC19
e
ne
en
nC13
ne
ne
n
ne
ne
yle
ta Kensol
ne
e
nC15
h an
ne
ta
ta
nC12
yle
xa
uta
200 Bu
ta
ne
e
an
Eth
en
en
a
nC14 nC18
op
Nitrog
ne
ep
He
r
an
Met
P
I-B
-
op
-I P
H
P
ct
ca
N-
Eth
Pr
N-
nC17
-O
De
e
N
an
N-
ec
xad
100 He
n-
Equilibrium Ratio Prediction and Calculation
-300 -200 -100 0 100 200 300 400 500 600 700 800 900
Temperature, ºF
20,000
C thru' Mid-Continent
10,000 1 Crude
9,000
8,000 Absorber O
7,000 verhead
6,000
5,000
C 1 thru' Distillate Crude
4,000
C thru' Gaso
1 lin e
3,000
Convergence Pressure, PSIA
2,000
1,000
900
1
,C
800
ne
700
tha
600
Absorbers, Crude Flashing Towers
Me
500
e
in
de
ol
ru
as
400
C
G
t
en
te
t
gh
tin
illa
300
Li
on
st
Di
-C
id
e
ud
M
61
200
Cr
.1
wt.
ol
M
il
O
an
Le
100
-100 0 100 200 300 400 500 600 700 800 900
Temperature, ºF
Step 6: Using the convergence pressure determined at Step 5, together with the
system, obtain K-values for the components from the appropriate convergence-
pressure K-charts.
Step 7: Make a flash calculation with the feed composition and the K-values from
Step 6.
Step 8: Repeat Steps 2 through 7 until the assumed and calculated convergence
pressures check within an acceptable tolerance, or until the two successive calcula-
tions give the same light and pseudo heavy components check within an acceptable
tolerance.
When the convergence pressure so determined is between the values for which charts
are provided, interpolation between charts may be necessary depending on how close
the operating pressure is to the convergence pressure.
If K-values do not change rapidly with PK(PK>>P) then the set of charts nearest to
calculated Pic may be used. Otherwise, a crossplot of K values versus PK all at constant
temperature and pressure, must be constructed for interpolation.
For those components characterised as a C7+ fraction Katz has suggests using a K
value of 15% of that of C7+. Danesh3 makes reference to other correlations to estimate
the critical properties of C7+ fractions. McCain6 has also presented pseudo critical
properties of C7+ as a function of molecular weight and specific gravity, his correlation
figures are shown below in figure 7.
10
Equilibrium Ratio Prediction and Calculation
1700
1600
Pseudocritical temperature, °R
1500
Specific gravity of heptanes plus = 1.0
.95
1400
.90
.85
1300
.80
1200 .75
.70
1100
1000
900
100 150 200 250 300
Molecular weight of heptanes plus
500
450
400
Pseudocritical pressure, °R
200 .70
150
100
100 150 200 250 300
Molecular weight of heptanes plus
5
The following example has been presented by McCain in his text . and is helpful as
a worked example in determining the appropriate convergence pressure charts
Methane 0.0356
Ethane 0.0299
Propane 0.0919
i-butane 0.0170
n-butane 0.0416
i-pentane 0.0198
n-pentane 0.0313
Hexane 0.0511
Heavier* 0.6818
1.0000
Solution
First, the composition of the liquid must be expressed as weight fraction.
* Molecular weight and gravity of C7+ assumed to be 263 and 0.886 respectively
12
Equilibrium Ratio Prediction and Calculation
The calculated values of weight averaged Tc and Pc are close to Kensol and the mid
continent crude point of figures 5 and 6. The location of the temperature 75°F with
the methane - component (Kensol or mid - continental crude) is at a convergence
pressure of around 10,000 psia, a value much greater than the assumed value of 2,000.
The calculator needs to be repeated using the higher convergence pressure related K
value data. The calculation procedure will converge when the estimated convergence
pressure is the same as the calculated convergence pressure.
f (P,V,T) = 0
where the pressure P, temperature T and molar volume V are related by a mathematical
function. Most equations of state have been developed by fitting an analytical expression
to pure component PVT data. To extend the application of the developed equation
to mixtures, the parameters of the equation must take into account the composition
of the mixture and for simplicity require only the insertion of pure component data.
Since most equations are effectively only mathematical models, the equations tend
in general to be more complex, that is contain a large number of parameters, as the
required level of accuracy increases. The basic assumption in the development of an
equation of state is that at a critical point:
∂P ∂2 P
= 2 = 0
∂V T ∂V T
Equations of state can be used for the following purposes:
3 prediction of gas phase properties of pure fluids and their mixtures using a
minimum amount of experimental data.
In the gas property chapter we reviewed the topic of equations of state. Currently
although a number of different equations could be used, the industry favours two, the
Peng Robinson equation of state and the Soave modification of the Redlich Kwong
equation of state.
where fiv and fiL are the fugacities of component in the vapour and liquid phases
respectively.
The fugacities can be expressed for a vapour and liquid therefore by:
fiL = φi xip
and:
fiv = φi yip
Therefore:
fiL
φ xp y
Ki = iL = i = i
φig fig ni ( )
y i p
The fugacity coefficients for the liquid and gas can be calculated using the following
equation
14
Equilibrium Ratio Prediction and Calculation
1 RT ∂p
V
RT ∫∞ V ∂ni T , V , n
ln φi = − dV − ln z
i (4)
McCain5 and also Ahmed6 give good descriptions of the application of EOS in
equilibrium calculations.
At the present time the preferred EOS are the 3 parameter Peng-Robinson (PR) 8,9
and the Soave-Redlich Kwong (SRK)10. Other equations exist and are more accurate
in predicting some properties. The considerable investment in binary interaction
parameters for the preferred equations is such that there is a reluctance to use some
recently developed EOS’s. Danesh's3 text gives a good review of the EOS.
Since the applications of the equations are applied to mixtures, mixing rules are required
to determine the values for the parameters in the particular EOS being used.
We will use the Peng Robinson equation as our basis but others could be used.
RT aT
p= −
Vm − b Vm (Vm + b) = b(Vm − b) (5)
The equation is set up for both liquid and gas using the following mixing rules to
calculate b, and aT . The rules are presented in the context of gas ie. y, clearly for
liquids, x values are used.
b = ∑ yi bi
and i (6)
where kij are the binary interaction coefficients, and kii = kjj = o and kij = kji.
The value of bi and aTi for the individual components are calculated as follows
RTci
bi = 0.07780
Pci (8)
R2 T 2 ci
ac i = 0.45724
Pci (10)
α i 0.5 = 1 + m(1.0 − Tri0.5 ) (11)
where
The Peng Robinson equation can be written as a cubic equation in terms of z, the
compressibility factor,
where
aP bP
A= B=
R2 T 2 RT (14)
The solution to the above equation gives three roots for z. The highest value is the
liquid phase z factor and when the equation is solved using vapour compositions then
the lowest root is the z factor for the vapour.
If the Peng Robinson equation is combined with the fugacity coefficient equation the
following equation results for the fugacity coefficient of each component.
A z + (21.5 + 1) B
ln φi = − ln( z − B) + ( z − 1) Bi′ − 1.5 ( Ai′ − Bi′) ln 1.5
2 z − (2 + 1) B (15)
and
Ai′ =
1
aT
0.5 0.5
(
2 aTi ∑ yi aTi 1 − kij )
1 (17)
Following all these steps independently for the liquid and gas phases the fugacities
of the gas and liquid phases can be calculated.
Having presented these equations we will now describe the process to calculate K
values , vapour liquid equilibrium ratios and compositions given a system composition,
temperature and pressure.
16
Equilibrium Ratio Prediction and Calculation
Step 1: Estimate the K values of the system. The Wilson11 equation is good for
this purpose.
p T
Ki = ci exp 5.37(1 + ω i )1 − ci
p T
(18)
Step 2: Carry out vapour equilibrium calculations using estimated K values using
the iterative procedure outlined previously. That is estimate the V/L ratio and iter-
ate until convergence is obtained, that is when compositions sum to unity. We now
have liquid and vapour compositions to use in equation of state calculations.
Step 3: Using liquid compositions, calculate the A & B values for the EOS and then
solve the z-value form of the EOS, to determine z. The lowest root (value) is the
z value for the liquid.
Step 4: Calculate the compositional coefficients A'i and B'i for the liquid components
and calculate the fugacity coefficients of the components of the liquid
Step 6: Calculate fgi and fLi fgi=yipφi and fLi= xipφI . Check if fgi= fLi.
If this value is greater than 10-12 then the whole process has to be repeated from step
1, except that the K values used are the calculated K values arising from step 5 i.e.
εi = fLi - fgi
based on K values.
A value of ε of 0.001 can be used for the sum of the errors.
φ Li
Ki =
φ gi (19)
( KiE − KiC )
2
εi =
KiE KiC (20)
The iteration is complete when these tolerance limits are met and the compositions
of the respective phases are those which have been been determined at the last
iteration. Calculations can then proceed to provide volumetric and density data for
the respective phases.
Danesh3 has given a flow diagram for the above flash calculation.
Input zi, P, T,
Component Properties
Calculate ZL Calculate ZV
Adjust Ki =
Ki old (fiL/fiV)
NO
NO
Is
Σ(1- fiL/fiV)2<10-12
?
YES
YES
Print xi,yi,vL,vV,nL,nV
End
EXERCISE 1.
Calculate the liquid and vapour phase composition when the mixture with the compo-
sition given in Table 1 is flashed to 2000 psia and 100°C. Use the Peng- Robinson
equation of state with binary interaction parameters given in Table 2.
Table 1: Multicomponent system.
Component Composition
mole fraction
Methane 0.55100
n-Pentane 0.11400
n-Decane 0.14600
n-Hexadecane 0.18900
18
Equilibrium Ratio Prediction and Calculation
No. Component 1 2 3 4
1 Methane 0.0000
2 n-Pentane 0.0236 0.0000
3 n-Decane 0.0489 0.0000 0.0000
4 n-Hexadecane 0.0600 0.0070 0.0000 0.0000
Ideally what is required is a full analysis of a crude oil in which all components
are quantified and identified so that the effects on the mixture behaviour due to
their structural and chemical properties can be taken into consideration. The phase
equilibrium prediction method which is employed, must match the quality and extent
of the experimental data available, for example, the use of a highly accurate method
might be invalidated if the composition analysis has been carried out on a basis of
generalised crude fractions such as boiling range. However, this is where flexibility
must be built into the program.
At present there is a tendency to increase the extent to which a crude oil composition
is analysed and this can only serve to make phase equilibrium prediction more
accurate. Condensate well fluids, however, are still generally reported on a basis of
a heptanes plus weight fraction, that is the collective weight fraction of heptane and
heavier components, because these heavier components are generally only present
in minute quantities.
It is thus desirable that any proposed phase equilibrium method should be able to
utilise detailed experimental data based on identifying the individual components
of a crude oil as well as data derived from a more generalised approach as discussed
above.
20
Equilibrium Ratio Prediction and Calculation
Pressure, PSIA
10 2 30 4 50 6 7 8 9 100 2 300 4 500 6 7 8 91,000 2 3,000 4 6 7 8 9 10,000
1,000 1,000
9 9
8 Plotted from 1947 tabulation 8
7 of G. G. Brown, University of 7
6 Michigan. Extrapolated and 6
3 3
2 2
100 100
9 9
8 8
7 7
6 6
5 5
4 4
3 3
Temperature °F
2 2
50
40 0
0
30
25
10 10
0
0
9 9
20
8 8
0
K= y/x K= y/x
7 7
6 6
10
5 0 5
80
4 4
60
3 40 3
20
2 2
0
-2
0
1.0 -4 1.0
9
0 9
8
-6 8
7 0 7
6 6
5 5
4 4
3 3
2 2
0.1 0.1
9 9
8 8
7 7
6 6
5 5
4 4
3 3
2 2
3 3
2 2
Temperature oF
100 50 100
9 0 9
8 8
7 45 7
6 0 6
5 40 5
0
4 38 4
0
36
3 0 3
34
0
2 32 2
0
30
0
28
10 0 10
9 26 9
8 0 8
24
K= y/x K= y/x
7 7
6 0 6
5 22 5
0
4 4
200
3 3
18
0
2 2
16
0
14
1.0 0 1.0
9 9
8
12 8
7 0 7
6 6
5 100 5
4 4
3 80 3
2 2
60
0.1 40 0.1
9 9
8 8
7 7
6 20 6
5 5
4 4
3 0 3
2 2
-20
22
Equilibrium Ratio Prediction and Calculation
Solution to Exercise
EXERCISE
Calculate the liquid and vapour phase composition when the mixture with the com-
position given in Table 1 is flashed to 2000 psia and 100°C. Use the Peng- Robinson
equation of state with binary interaction parameters given in Table 2.
Table 1: Multicomponent system.
Component Composition
mole fraction
Methane 0.55100
n-Pentane 0.11400
n-Decane 0.14600
n-Hexadecane 0.18900
No. Component 1 2 3 4
1 Methane 0.0000
2 n-Pentane 0.0236 0.0000
3 n-Decane 0.0489 0.0000 0.0000
4 n-Hexadecane 0.0600 0.0070 0.0000 0.0000
SOLUTION
Step 0: Calculate the coefficients of components in the mixture, using the properties
(critical temperature Tc and critical pressure Pc) of pure compounds:
R2 T 2 ci (10)
aci = 0.45724
Pci
RT
bi = 0.07780 ci
Pci (8)
( ( )
2
α i = 1.0 + m 1.0 − T r )
(11)
ai = ci αi (9)
Iteration One
Initial K-values can be calculated from Wilson Equation:
Pci T
Ki = Exp (5.37)(1.0 + ω i )1.0 − ci ))
P T (15)
Step 2: Determine the trial value of phase fraction (vapour fraction here) by solving
the vapour equilibrium equation:
N
Zi
∑L =V
i=l + KL
V (5)
Use the determined vapour fraction (Vf) to calculate the liquid and the vapour phase
composition from material balance:
Liquid:
N N
a = ∑ ∑ xi x j ai a j ( ) (1. − k )
0.5
(7)
ij
i=l J =l
N
b = ∑ xi bi (8)
i=l
N N
a = ∑ ∑ yi y j ai a j ( ) (1. − k )
0.5
ij
Vapour: i=l J =l (7)
24
Equilibrium Ratio Prediction and Calculation
N
b = ∑ yi bi (8)
i=l
aP bP
A= 2 2 B=
RT RT (14)
Phase a b A B
Liquid 74.88118 0.18470 10.86872 0.82089
Vapour 1.83396 0.02737 0.26619 0.12165
Calculate Z - factors of the liquid and vapour phases (Peng - Robinson EOS):
bi A 1 0.5 N bi
ln φi = − ln( Z − B) + ( Z − l ) − 2 ai ∑ y j a j l − kij − x
b 2 2B a
0.5
( )
j =l b
Z + ( 2 + 1) B
ln
Z − ( 2 − 1) B
(15)
Fugacity of components in the vapour phase:
fi v = yi Pφi v
Fugacity of components in the Liquid phase:
fi l = xi Pφi l
Liquid Vapour
Component φil fil φiv fiv
Methane .23811E+01 .67429E+02 .90058E+00 .12150E+03
n-Pentane .66312E-01 .17708E+01 .36516E+00 .41838E+00
n-Decane .19023E-02 .67189E-01 .14686E+00 .41455E-02
n-Hexadecane .43075E-04 .19707E-02 .51625E-01 .41965E-04
Final Iteration
Calculate composition dependent coefficient for z - factor calculations for both liquid
and vapour:
N N
a = ∑ ∑ xi x j ai a j ( ) (1. − k ) (7)
0.5
Liquid:
ij
i=l J =l
N
b = ∑ xi bi
i=l (8)
N N
a = ∑ ∑ yi y j ai a j ( ) (1. − k )
0.5
Vapour: (4)
ij
i=l J =l
N
b = ∑ yi bi
i=l (8)
aP bP
A= B=
R2 T 2 RT (14)
Phase a b A B
Liquid 48.03656 0.14799 6.97233 0.6577
Vapour 2.11326 0.02929 0.30673 0.1301
Calculate z - factors of Liquid and Vapour and Vapour Phase (Peng - Robinson
EOS):
Calculate the fugacity of each component in the liquid and vapour phase:
bi A 1 0.5 N bi
ln φi = − ln( Z − B) + ( Z − l ) − 2 ai ∑ y j a j l − kij − x
b 2 2B a
0.5
( )
j =l b
Z + ( 2 + 1) B
ln
Z − ( 2 − 1) B
Liquid Vapour
Component φil fil φiv fiv
Methane .22032E+01 .11881E+03 .90269E+00 .11882E+03
n-Pentane .66155E-01 .13105E+01 .33093E+00 .13106E+01
n-Decane .19873E-02 .53832E-01 .11847E+00 .53831E-01
n-Hexadecane .45319E-04 .15983E-02 .36226E-01 .1598E-02
2 RT
Vm 2
Molecular volume P
PM
P2
2 RT
1. Amyx, J.W, Bass, D.M, Whiting, RL, Petroleum Reservoir Engineering, McGraw
Hill. New York 1960
5. McCain, W.D. The Properties of Petroleum Fluids Penwell. 1st. Edition 1973
Starling, K.E., A.I.Ch.E.Jnl, 1972, 18, 6, 1184-1189.
6. McCain, W.D. The Properties of Petroleum Fluids. Pennwell, 2nd Ed., 1990,
p 414-436.
11. Wilson, G.M., "A Modified Redlich - Kwong EOS Application to General
Physical Data Calculations." Paper No 15c presented at the AIChE 65th National
Meeting (May 1968) AI ChE.
28
PVT Analysis
CONTENTS
6 PVT TESTS
6.1 Flash Vaporisation
6.2 Differential Vaporisation
6.3 Separator Tests
6.4 Viscosity
6.5 Hydrocarbon Analysis
9 RETROGRADE CONDENSATION
Having worked through this chapter the Student will be able to:
• Describe the impact of well flow interruption on the sampling of ‘wet’ and gas
condensate systems
• List and describe the five main PVT tests for oils systems and their
application.
• Explain the two basic PVT tests for gas condensates, constant mass and constant
volume depletion.
• The gas-oil ratio and oil formation volume factor at the bubble point
pressure.
• Determine the bubble point pressure from a set of P vs. V relative volume test
data
• Determine the total formation volume factors above and below the bubble
point.
• Determine the oil formation volume factors and gas-oil ratios for pressures
below the bubble point pressure.
PVT Analysis
1 SCOPE
Reservoir fluid analysis provides some of the key data for the petroleum engineer.
The quality of the testing, therefore, is important to ensure realistic physical property
values are used in the various design procedures. As important is the quality of
the samples collected to ensure that the fluids tested are representative of the field.
Clearly, any subsequent high quality testing is of little value if the sample is not
representative.
The physical properties measured depend on the nature of the fluid under evaluation.
For a dry gas, the key parameters are, the composition, the specific gravity, the gas
formation volume factor, compressibility factor, z , and viscosity as a function of
pressure. The isothermal gas compressibility can be determined from the z value
with pressure variation.
For a wet gas, then all of the above parameters for a gas are required. However there
are some variations because of the production of liquids with gas. The formation
volume factor used is the gas condensate formation volume factor, Bgc, which is
the reservoir volumes of gas required to produce one stock tank volume (barrel or
M3) of condensate. The composition, specific gravity and molecular weight of the
produced condensates and produced gas are required. The composition and apparent
specific gravity of the reservoir gas are obtained by recombining the values for the
gas and condensates.
For an oil system, the following information is required; the bubble point pressure
at reservoir temperature, the composition of the reservoir and produced fluids, the
formation volume factor, the solution gas to oil ratio, the total formation volume factor,
and viscosity, all as a function of pressure. The coefficient of isothermal compressibility
of oil. The impact of separation on the above properties. The impact of operating
below the bubble point on the formation volume factor and solution GOR.
For a gas condensate system, properties measured reflect those for both wet gas and
oil analysis. The related property to the saturation pressure for the oil, the bubble
point pressure, for a gas condensate is the dew point pressure. Above the dew point
the compressibility characteristics of the gas are required, and the impact of allowing
the reservoir to drop below the dew point is another important evaluation.
Subsurface samples can only be representative of the reservoir contents when the
pressure at the point of sampling is above or equal to the saturation pressure. If this
condition is not fulfilled, one should take a surface sample. Even at pressures close
to the saturation pressure there is a serious possibility of sample integrity being lost
as a result of the system going two phase during transfer to the sample chamber.
A simple test can be carried out in the field at the well site to find out whether a reliable
sample has been obtained. In this test the sample cylinder is pressurised either by using
a piston cell or using mercury as the displacing fluid. Whereas mercury was the most
common fluid to be used as a pressure transfer and volume change fluid, because of
toxic and other concerns its use is diminishing. From the relation between injection
pressure and volume of mercury injected, the following properties are derived:
(1) the pressure existing within the sampler when it is received at the surface;
(2) the compressibility of the material within the sampler;
(3) the bubble-point pressure of the contents of the sampler.
If two or three samples taken at short time intervals show the same measured properties,
it is highly probable that good samples have been obtained.
The contents of the sampler are transferred by means of a mercury pump into a high
pressure shipping container. Two fillings of a sampler of 600 ml capacity are usually
sufficient for a complete PVT analysis. In recent years mercury has been replaced
using the application of piston cells by some companies.
PVT Analysis
Ground level
Tubing
Packer
Subsurface sampling
In the case of two or three stage separation the samples are taken from the high
pressure separator.
Inlet
Sight
Liquid
glass
outlet
Momentum
absorber Guage
Gas
Inlet outlet
Sight Liquid
glass Horizontal outlet
Separator
Sample Vessel
PVT Analysis
Sample Vessel
3.1 Introduction
The use and value of any PVT study or other analysis of a reservoir fluid is dependant
on the quality of the sample collected from the reservoir. Many PVT reports show
a variation of results from fluids from the same well. Sampling wet gas and gas
condensate fluids can give rise to errors. During the sampling procedure it is often
possible to alter the conditions such that the fluids sample are not representative of
those within the reservoir, the characteristics of which are being assessed during the
PVT report. It is important, therefore, in sampling reservoir fluids to ensure that the
conditions during which the samples are being taken are not altered to give rise to
false samples.
t
in
po
le
bb
Dewpoint curve
Bu
uid
Liq
uid
uid
Pressure
0%
Liq
Liq
10
80%
uid
70
Liq
uid
%
ur
80
Liq
o
Vap
uid
%
40
Liq
0%
%
20
d
10
q
ui
Li
5%
Temperature
At a particular point within the envelope the composition of each component in each
phase is constant. The separation of oil and gas as predicted by the phase diagram
results in each phase itself having a phase diagram. These phase diagrams intersect at
the separation temperature and pressure, the oil will exist at it’s bubble point and the
gas at it’s dew-point. Therefore, for example, in the separator, gas will be separated
and produce gas at it’s dew-point and the oil separated will be at it’s bubble point.
Gas at dewpoint
Separator
For a given system, therefore, a change in the temperature or pressure within the
phase envelope will result in alteration of the system and therefore alteration in the
characteristics of the two phases produced. The behaviour just described, therefore,
will have implications on the way samples are taken and on the techniques used to
collect the sample, for example, from the separator.
PVT Analysis
4. Separator
3. Wellhead
2. Bottom hole
1. Reservoir
Merits / Disadvantages
Location For Against
1. Reservoir Ideal Impossible
2. Bottom - hole 1- phase Representative?
Technology
Cost
Handling
3. Wellhead Cost 2- phase?
Representative
4. Separator Cost Gas / liquid volumes
1- phase Separator conditions
Buffer Representative?
Sampling
Volume
The bottom hole sample where the fluid is in single phase, probably, would be an ideal
situation. Present technology, however, is such that it is often difficult to produce a
single phase sample representative of the fluids bottom hole. The necessary pressure
drop to get the fluid from the bottom hole into the sample container can often give
The well behaviour can significantly influence the nature and characteristics of the
fluids which eventually arrive from the separator, for example, figure 9 in a flowing
well gas condensate entering the wellbore as it travels to the surface will experience
a drop in pressure likely to give rise to retrograde liquid behaviour in the wellbore.
The flow must be sufficient to lift this uniform liquid and gas fluid to the surface. If
the flow is slow it is possible that some liquid may fall back therefore altering the
overall composition moving up the wellbore.
Gas
Condensate
Gas
Gas Gas
Condensate
10
PVT Analysis
If the reservoir is shut in after flow then considerable changes can rise in the composition
of the fluid in the wellbore. The reservoir gas flowing into the wellbore sets up a
new equilibrium with condensed retrograde liquid which has rained down within the
wellbore. This separation in the wellbore gives rise to a lean gas near the top of the
well with a more than rich mixture at the bottom of the well. Figure 10.
Gas
Condensate
Lean gas
Gas
Pressure build up
Condensate
When the well is flowing after a shut-in period, figure 11 and sampling takes place
there will be a variation in the compositions produced at the surface and therefore
unrepresentative samples collected from the separator. For example, in the early
period after shut-in the lean gas at the top of the well enters the separator producing
a fluid with a GOR higher than that representative of the reservoir. As the fluids at
the bottom of the well move to the surface much richer as a result of the liquid having
collected at the bottom of the well fluids they are produced with a GOR lower than
that of the representative reservoir fluid. It is important, therefore, for the well to be
flowed for a sufficient period for all the fluid within the well to have been displaced
and also that in the near wellbore region which also could have been influenced by
the pressure and compositional changes experienced during the shut-in period.
Condensate
Higher GOR
Lower GOR
In assessing whether good samples have been taken it is important to know how
long it will take for unrepresentative samples to be displaced from the separator, the
wellbore and the near wellbore reservoir zone. For example, for a 12ft x 5ft diameter
separator with a liquid flowrate of the order of 200 barrels per day, it could take 1
hour to displace an 8.4bbls. If a contaminant enters a separator then the time to reach
1% of the original concentration could be over 4.5 hours. For example, if the tubing
is 4 inches in diameter with a length of 9000ft and an average tubing pressure of
5000psi and a temperature of 170°F and a gas flowrate of 5MM/scf/day, the volume
in the tubing would be 0.23MM standard cubic feet and the time to displace this gas
would be just over 1 hour.
12
PVT Analysis
figure 12a the sampling valve might be located on the lower portion of the valve. It
is likely that entrained liquids in the gas stream could collect at this point giving rise
to a very rich gas composition if these entrained liquids were collected in the sample
containers. Similarly sampling valves in the liquid line figure 12b could be located
on the upper portion of the line any gas which is carried through the line again will
collect in the dead volume of the sampling valve such that when samples are taken
the gas will enter the gas bottle of the container giving rise to an unrepresentative
sample. An alternative would be to locate both of these sampling valves on the side
of the pipe rather than at the top or bottom of the line. Figure 12c.
(a)
Separator
Liquid
(b)
(c)
(a) apparatus for the transfer and the recombination of separator oil and gas
samples;
(b) apparatus for measuring gas-volumes and for performing separator tests;
(c) the PV cell and displacement pumps and dispensing cell;
(d) high pressure viscometer;
(e) gas chromatograph.
Diagrams of the layouts for different systems are given in Figures 13, 14 and 15.
Separator
Viscometer
Trap
Bleed
Oil sample To liquid
Vacuum gage
bottle analysis
To gas
analysis
Gas meter
Displacement
fluid Vacuum pump
PVT cell in
thermal bath
Bleed
Precision pressure
gage Dead weight
Displacement tester
pump
14
PVT Analysis
Analysis
Separator
Viscometer
Recombination Trap
cell Bleed
Vacuum gage
To liquid
Gas sample analysis
bottle
Liquid To gas
sample analysis
bottle Gas meter
Vacuum pump
Displacement
fluid
PVT cell in
thermal bath
Bleed
Precision pressure
gage Dead weight
Displacement tester
pump
Analysis
9 Trap
Vacuum pump
3
Gas condensate
cell in thermal bath
Bleed
Precision pressure 4
gage
The GOR given by the field refers to the separator tank gas-oil ratio. In order to
recombine the separator gas and the separator oil in the correct ratio, the volumetric
ratio between tank oil and separator oil is determined in the laboratory
85% of the cylinder containing separator oil is occupied by oil. There is a gas cap
on top. This precaution is taken in view of the long transport time and the risk of
great fluctuations in temperature involved.
After the gas cap has been dissolved by pressing water into the cylinder at a pressure
higher than the separator pressure a given quantity of oil is then flashed at pressure
and room temperature through a separator operating under the same conditions as
6 PVT TESTS
At this stage we should remind ourselves of the main applications of the PVT data.
The three main application areas are;
Over recent years, as the data is subsequently applied to computer based simulation
tools, the ability to handle more complex descriptions of the fluids has led to more
extended compositional analysis, beyond the C7+ limit which was the basis for many
years. It is common practice in some PVT laboratories to measure co-position to C28
and then define a C29+ component
In reservoir calculations the PVT tests and subsequent report provides the source
of the reservoir engineering properties necessary to describe the behaviour of the
reservoir over its development and production. The tests conducted therefore have to
take into consideration the processes going on both above and below the saturation
pressure.
There are four main PVT tests for oil systems plus associated compositional
analysis:
16
PVT Analysis
PVT Facility
Valve 1
V1 Ps1
P
L1 Ts1
T Valve 2
V2 Ps2
L2 Ts2
Displacement
fluid or piston
PVT Cell in
temperature
controlled bath
The flash vaporisation test gives the relationship between P & V of a reservoir liquid
at constant (reservoir) temperature. Liberated gas remains in equilibrium with the
oil.
Gas
Oil
P
Thermal expansion
Hg
V2 − V1
β=
V2 (T2 − T1 )
V1 = volume of oil at T1
V2 = volume of oil at T2
To carry out a relative volume test run, the PV cell is set up as in Figure 18.
Valve 1
closed
P
V P
Tres
Pb
Vb V
PVT Cell
Pump
The PVT cell is filled with a certain quantity of reservoir liquid at a pressure above the
estimated bubble point and room temperature. After the PV cell has been filled (about
90 ml), it is immersed in a temperature bath and heated to reservoir temperature.
V2 − V1
β=
V2 (T2 − T1 )
where:
The thermal expansion factor is expressed, for example, in ˚C-1. The pump reading
taken at the moment when the pressure became constant is the first reading for the
PV curve. The pressure is now reduced by gradually withdrawing small quantities of
transfer fluid from the PVT cell and after each withdrawal equilibrium is established
by shaking the cell. After each equilibration the pressure and the volume are read.
By plotting the pressures against the volumes a curve is obtained showing a break
at the bubble point (Pb). Figure 18.
The saturation pressure or bubble point pressure is that pressure below which gas
is liberated. Hence, a two-phase system is formed, whereas above the bubble point
pressure a one phase system is present (undersaturated liquid).
18
PVT Analysis
The compressibility of the oil phase above the bubble point can now be calculated
from the graph.
V2 − V1
c=
V2 ( P2 − P1 )
where:
V2 = volume at pressure P2
V1 = volume at pressure P1
The main objectives of the flash vaporisation test are to provide the reservoir bubble
point pressure and together with the information from the separator test, the formation
volume factors above the bubble point.
The differential vaporisation differs from the flash in that the liberated gas is removed
from the cell stepwise. At each step below the bubble point the quantity of gas, oil
volume, density, gas expansion and gas compressibility are determined. The objectives
of the differential test therefore are to generate PVT data for conditions below the
bubble point. Figure 19 below indicates the differential process.
Gas
Oil P = Pb P1 < Pb P = P1 P < P1
Gas
At each stage - volume of oil and gas, density and Oil
composition of gas measured. Mercury /
Remaining oil is called RESIDUAL OIL displacing fluid
The bubble point Pb is the starting pressure for the differential test. The next step
is to reduce the pressure in the PV cell by expansion of the PV cell volume. The
reduction of pressure causes the system to go two phase. All the gas phase is removed
at constant pressure by reducing the cell volume as gas is withdrawn. The volume
of the remaining oil is then determined. The cell pressure is then again dropped by
expansion of the PV cell and the above process repeated until the cell pressure has
been dropped to atmospheric pressure. The pressure steps for the tests cover a range
of around 8-10 steps. All the above steps have taken place at reservoir temperature.
The final stage is to reduce the cell to 60°F keeping the pressure at atmospheric
pressure. The final oil volume is measured. This remaining all is termed residual
oil to distinguish it from stock tank oil which although at the same pressure and
temperature conditions has got there by a different process.
The cumulative weight of the amounts of gas withdrawn are used in the calculation of
the densities of the oil phase in the differential vaporisation process. These densities
can also be determined directly if a pressure pycnometer is available.
Flash liberation is considered to take place between the reservoir and through the
separator.
Differential liberation tests are carried out therefore to obtain oil formation volume
factors and GOR’s that can be used to predict the behaviour of a reservoir when the
pressure has dropped below the bubble point pressure.
20
PVT Analysis
With the equipment available, a single test or a multiple separation test can be carried
out.
Valve 1
V1 Ps1
Pb
L1 Ts1
Valve 2
V2 Ps2
Vres
Tres L2 Ts2
GOR = (V1+V2) / L2
PVT Cell
Bo = Vres / L2
Pump
The procedure for the separation test is as follows. The starting point is oil in the PVT
cell at its reservoir bubble point, ie. the same starting condition as the differential test.
Fluid is displaced from the PVT cell ensuring that the PVT cell contents remain at
bubble point pressure. The gas and liquids are collected from the separation stage(s)
and their respective properties measured. The final stage is at stock tank conditions.
The resulting fluid is termed stock tank oil.
A single separator test is carried out by flashing reservoir liquid at bubble point pressure
and reservoir temperature through the separator operating at the average annual
temperature and at pressures which may be expected in the field. The difference in
results when using a single or a double separator is that in the former case the total
gas/oil ratio is higher, the shrinkage is greater and the density of the tank oil is higher
than in the latter case.
The main objectives of the separator test are in combination with the flash vaporisation
and differential tests to provide formation volume factor and solution gas-oil ratios
over a full pressure range above and below the bubble point. In quoting these values
it is important to recognise that the values are separator condition specific.
The solution to gas-oil ratio, Rsb, is equal to (V1 + V2)/L2 Standard cubic volume
(SCF or SM3)/Stock Tank volumes (STB or STM3).
6.4 Viscosity
The viscosity is measured at reservoir temperature and at different pressures both
above and below saturation pressure. It is important for viscosity measurements below
the bubble point to generate the fluid for study by a differential mode to simulate the
nature of the fluid that would exist at these conditions.
Viscosity measurements were largely carried out with a “rolling ball” high pressure
viscometer consisting of a highly polished-steel capillary of 1/4" dia. which can be
closed at the top by means of a plunger and is provided at the bottom with a contact
which is connected with an amplifier. A steel or platinum ball rolls in the capillary, its
diameter hence slightly smaller than that of the capillary. When the ball reaches the
bottom, it makes contact between the wall of the capillary and the point of contact,
as a result of which a circuit is closed and a whistling sound is heard. The time of
rolling is a measure of the viscosity.
In recent years the pressure drop along a capillary tube of known length and internal
dimensions has been used. The viscosity being calculated using the Poiselle equation,
the laminar flow pressure drop equation for a pipe of a particular diameter and length.
Although being used it is clearly restricted by operating under a fixed flow regime,
laminar and velocities to ensure that there is a sufficient pressure drop to measure
and not too large to influence physical properties.
In some cases the fluids produced in the final separation stage are identified to a
higher C number together with aromatic and napha components.
Compositions are also made of the produced gases from the various tests.
22
PVT Analysis
The formation of solid deposits during oil production is a concern. Some heavy
hydrocarbons as mentioned above at low temperatures can form solid phases waxes and
in transfer lines and process facilities. The wax formation temperature is therefore an
important measurement. Ashphaltene, is another solid phase of concern. Asphaltenes
are large molecules largely of hydrogen and carbon with sulphur, oxygen or nitrogen
atoms. Asphaltenes do not dissolve in oil but are dispersed as colloids in the fluid.
From this rough indication constant temperature flows are taken at temperatures
just above and below the indicated WTC to give a more precise value. A plot of
differential pressure vs. flow indicates a more precise value of the WTC. Figures 21
& 22 below from a PVT report provided by Core Laboratories (UK) Ltd gives the
plots used to determine the WTC for a separator oil sample.
50
40
Differential Pressure, psig
30
10
0
75 85 95 105 115 125 135 145
Temperature, °F
24
PVT Analysis
45
40
35
20
15
10
0
0 5 10 15 20 25 30
Cumulative Volume Flow, cc
1 dV
c=
V dP (specify unit of pressure change)
1 dV
β=−
V dT (specify unit of temperature change)
(7) Gas formation or gas expansion factor: Bg or E, i.e. volume occupied under sc
(standard conditions) by that amount of gas which occupies unit volume at reservoir
temperature and pressure.
(12) Oil Formation Volume Factor or Shrinkage factor C, where C = ratio of the
volume of tank oil produced under specified separator conditions and then measured
under sc to the volume occupied by that quantity of oil and its dissolved gas under
reservoir conditions. depends on separator conditions.
(13) Solution gas to Oil Ratio or Gas-solubility factor D = that volume of gas liber-
ated under separator conditions and measured under sc which is held in solution at
reservoir temperature and any particular pressure by that quantity of oil which will
occupy unit volume when produced; depends on separator conditions.
Figure 23 below illustrates the volume relationship of fluids in an oil PVT tests to the
black oil description of volumes in a reservoir. In PVT analysis the basis of reference
is the bubble point, figure 23a whereas for the black oil system the reference state
is surface or stock tank conditions, figure 23c. The relationships between the two
are given in figure 23b.
V V V
Vt Vt B1
Cb
Vo l Vt Vi
Cb Vo I Vt Vi l Bo Bb Bt Bt
I Cb Cb Cb Cb
P Pb P Pi P Pb P Pi P Pb P Pi
(a) (b) (c)
26
PVT Analysis
Research has demonstrated that the IFT between hydrocarbon liquid and vapour phases
has a significant impact on residual condensate saturation and relative permeability.
Although not yet part of conventional PVT analysis the author considers that such
information will in future be required for PVT reports for more fluids where the
conditions are close to critical conditions and when dealing with gas condensates
fluids.
Tube
Tip
ds
Gas
de
Oil
The dimensions of the suspended droplet are controlled by the balance between the
surface and gravity forces. The equation relating the various parameters from this
method is:
When the IFT is very low, the droplet size is very small which requires an extremely
fine diameter tube to suspend it. A thin wire may be required for such situations. When
conditions are close to the critical point the pendant drop method may not work.
For very low IFT values other methods have been used. A light scattering technique
where the propagation of thermally stimulated waves in a vapour-liquid system has
been used3. A very successful and low cost method was developed at Heriot-Watt,
based on the curvature of the vapour-liquid interface on the viewing window of a
PVT cell4. Figure 25.
Enlarged veiw
Viewing window
of PVT device
Vapour
Vapour
h
Liquid q Liquid film
Liquid
Cross section
of window
The curvature is seen as a dark band of a specific thickness, due to the scattering
transmitted light across the viewing window of the PVT cell. The thickness, of the band
is directly proportional to the IFT. The equation relating the various measurements
and properties is as follows:
gde2 L
s= (ρ − ρ V ) )
l
where q = contact angle (assumed zero at low IFT's)
Then s = (rL - rV)gh2/2
It has been applied successfully to IFT values as low as 10-3 mN/m, where it is difficult
to distinguish the vapour-liquid phases.
28
PVT Analysis
9 RETROGRADE CONDENSATION
When studying the PVT behaviour of gas condensates, use is often made of a PVT
cell with a glass window to observe the dew point condition. Over recent years there
has been considerable developments in the types of PVT cells used in gas condensate
PVT measurements. It is not the intention of these notes to describe the different
cells. The key difference between an oil PVT cell compared to a condensate PV cell
is that the latter requires a window to observe the dew point condition.
(b) the liquid phase constitutes only a small part of the total volume in the PV-cell
A windowed cell makes accurate measurement of small liquid volumes possible.
In gas condensates there are large volumes of gas relative to liquid. The liquids
produced are generally small in volume. The relative small amounts of liquid can
yield errors in measurement.
Figure 26 below gives a schematic of the two main gas condensate tests.
P = Pd
Pressure is reduced
Gas Gas Gas
below dew point by
Vi V>Vi Vi expanding system.
Oil
Oil Volume is put back to
original volume by
Hg Hg Hg injection of mercury
at constant pressure
and removing some gas.
P = Pd P1 < Pd P = P1
Volume of condensate
measured.
Figure 26 Schematic of constant mass and constant volume depletion gas condersale tests.
There are two basic tests for gas condensates, Figure 26. The constant mass study,
the purpose for which is the z value determination above the dew point and the
determination visually of the dew point pressure. The constant volume depletion test
is carried out in at attempt to examine the potential liquid drop out in the reservoir
by retrograde condensation.
The constant mass study and the constant volume depletion study can be compared
to the flash vaporisation test (relative volume test) and the differential test for oils
respectively.
In the constant mass study, Figure 26, the system remains constant. No fluids are
removed. A portion of the reservoir fluid is charged to the gas condensate PVT cell.
The pressure is raised to that significantly above the anticipated dew point and then
the volume of the gas measured at reducing pressure steps. The dew point is observed
visually as liquid condenses on the PVT cell window. The retrograde liquid build
up is also measured at several pressures below the dew point, although the numbers
have little reservoir significance. Values of the compressibility factors z are obtained
above the dew point pressure.
30
PVT Analysis
The Constant Volume Depletion, CVD, test is carried out in an attempt to determine
the potential loss of liquids if the reservoir is depleted below the dew point. In the
test the same fluid at reservoir temperature as contained in the cell for the constant
mass test is used. The test consists of a series of pressure expansions and constant
pressure displacements to return the sample to a constant volume. This volume is
equal to the volume of the sample at the dew point pressure. This process is repeated
down to a low pressure.
Starting at the dew point the pressure is reduced by expansion of the cell. Liquids
condense and then gas is displaced at constant pressure until the volume returns to the
original volume. The quantity and compositions of the displaced fluids are determined
using gas chromatography methods. In one approach the produced gas from the cell
are pumped into a pre-weighed flask submerged in liquid nitrogen and condensed.
The condensed gas is then gradually allowed to return to ambient temperature. The
evolved gas and residual condensate are collected separately, weighed and analysed.
These compositions are recombined mathematically to the gas - oil ratio to determine
the produced gas composition. The remaining condensate phase is measured and
expressed as a percentage of the sample volume at the dew point pressure. . The
pressure is then reduced again and the process repeated. A series of relative liquid
amounts are obtained and a liquid dropout curve generated. Figure 27.
Retrograde
condensate
RLV %
Ps
Special tests include the measurement of interfacial tension, described earlier which
has come to be realised an important factor in understanding the behaviour of gas
condensates. Different methods have been proposed including pendant drop and
rising film techniques. The latter method can be carried out as a matter of routine
during the CVD test by observing the thickness of the rising film of the condensate
on the window of the cell. The thickness of this film is directly proportionate to the
IFT and the method can determine IFT values lower than conventional pendant
drop methods.
Having considered the various aspects of PVT analysis we will now consider the PVT
report and examine how we can generate the various reservoir engineering parameters
of interest. We will remind ourselves of the reason for the report and then using a
PVT report go through the main tests and interpret the detail.
Although the PVT report can be a source of information for a variety of applications
from reservoir, through well to surface facility calculations, the reservoir engineering
application has provided the main basis and structure of the report. The report is
structured to provide the much of the black oil model information together with
limited compositional data. The material balance equation which is covered in a latter
chapter also provides a basis for the PVT report. It is the PVT report which is the
source of much of the data embodied in the material balance equation. Thus, some
of the tabular information is set up to satisfy that need.
The PVT report can be used for a range of purposes; from its use in determining the
potential prospects of a hydrocarbon accumulation to history matching a reservoir
which has been on production for some time. The report should therefore cover all
past, present and future situations which may require calculations. To do this with a
minimum of tables and curves, the data are normalised to a reference state and only
data for the reference state given. In PVT data reporting as indicated in earlier the
reference state is the bubble point. The petroleum engineer must then “work back”
from the reference state to the particular situation.
As described in previously the laboratory tests are carried out in an attempt to simulate
the processes which take place in a reservoir and through the production system.
These will include the flash equilibrium separation of gas and oil in the surface traps
during production and for an oil below the bubble point the differential equilibrium
separation of gas and oil in the reservoir during pressure decline. In interpreting
the data the engineer needs to use both sets of data to provide the information for
reservoir calculations.
The PVT report is clearly specific to a particular fluid, collected from a specific well
under specific conditions. This sample may not be representative of the total field
system and therefore using subsequent reports it may be necessary to to adjust the
data for field application. In a PVT report therefore detail is given as to the manner
of obtaining the sample and the conditions that existed at the sampling time. Also,
the compositional analysis of the sample is given so that equilibrium calculations
can be made for conditions other than studied in the laboratory. In this area there has
been considerable progress in compositional analysis and although the report used in
this section only goes up to C7+ it is now common practise to characterise to higher
C numbers, even as light as e29.
We will now examine two PVT reports. The first is for an oil sample and is commonly
used in textbooks to illustrate the interpretation of a PVT report. The second is a PVT
study on a gas condensate fluid. It should be emphasised that reporting styles vary
from the different service providers. Both reports are provided by Core Laboratories.
Ltd.
32
PVT Analysis
The first report is for an undersaturated oil from Texas field attributed to the Good
Oil Company. The report is given at the end of this chapter. Although not covered
in this report there may be a number of data sheets reporting the validation of the
samples used and selected in the PVT report. These sheets would include the various
gas to oil ratios when taking the samples and other information as part of the sample
validation. A text description is usually given to the report describing the various tests
conducted and the principle observations. The source of these principle observations
and calculations will be covered as in the following sections. The pages of the report
often include processed data .
To determine black oil parameters of oil formation volume factor and gas to oil ratios
as a function of pressure, a combination of tables of the report are required.
We will first look at the separator tests the results of which provide the basis for the
Bo and GOR values.
12.1 Separator Tests of Reservoir Fluids
In the separator test, page 7 of 15 of the report, oil at reservoir temperature and the
bubble point pressure in a PVT cell has been carefully displaced from the cell through
a series of pressure and temperature steps. The tests show what quantity of surface
gases and stock tank oil results when one barrel of bubble point oil is flashed through
a certain surface trap sequence. The tabulation also gives the ˚API gravity of the
stock tank oil and, in some instances, the gravity of gas coming from the primary
trap. There are four separate tests reported. The first where the first separation is at
50 psig and 75°F and the tank separation at 0 psig and 75°F, the other tests where
the first stage conditions are at 100 psig, 200 psig and 300 psig. The temperatures
and conditions for the final stage are the same for each test.
Column 1 and 2 give the pressure-temperature conditions of the surface trap tests
that were investigated. These should be specified by the reservoir engineer at the
time the test is planned . A difficulty here is that the engineer specifying to the PVT
service company these separation conditions is unlikely to be involved in the ultimate
optimised surface separation conditions if the field ultimately is developed.
Exercise PVT 1.
What is the solution gas-oil ratio and formation volume factor resulting from the sepa-
rator test, at first stage of 300 psig and 2nd stage 0 psig and both at 75˚F ?
Solution.
This exercise illustrate the results using one of these tests, a two-stage separation;
a primary trap operating at 300 psig and 75˚F followed by a stock tank operating at
14.7 psia (0 psig) and 75˚F.
Columns 3 and 4 show the surface gas-oil ratio from the first stage and the tank.
The first stage ratio of 549 ft3/BSTO ( column 4) and the tank stage gas amount to
246 ft3/BSTO. It is important to read the footnotes of the report. Column 3 gives
the results in relation to the volumes at indicated P & T whereas column 4 gives the
volumes with respect to stock tank conditions of 14.65 psia and 60˚F. The solution
gas-oil ratio at bubble point conditions (2620 psig and 220˚F), is therefore Rsb is 549
+ 246 = 795 ft3/ BSTO when flashed through this surface trap arrangement.
If we compare these results for the 50psig, 0psig arrangement we obtain a Bob of
1.481 B/BSTO and a solution GOR of 778 ft3/ BSTO.
Clearly therefore Rsp, Bob, ˚API all vary with the separation pressure-temperature
situation. There is not one unique result. When reporting Bo and GOR data for a
reservoir therefore it is important to report that these are for a specific separation
route or averaged for a series of tests. The latter is not so useful since it is not so
straightforward to calculate the result for a different separation route using VLE
methods.
The results from the separation test are based on the bubble point condition and to
obtain volumetric information at other pressures we require the results from other
tests.
Remember in this test, the contents of the cell at reservoir temperature have been
expanded and the volumes measured. None of the contents has been removed, the
system has remained constant. In the style presented here the expansion of the fluid
as measured has been plotted and then the intersection of the two slopes of the liquid
phase expansion and the two phase, gas/liquid phases, has been interpreted as the bubble
point pressure and bubble point volume. All the volumes have then been normalised
to this bubble point condition and presented as a relative volume. (column 2).
The first and second columns of the Reservoir Pressure Volume Relations Data on
page 4 give the pressure volume relations of the original fluid at 220˚F. Note that the
data are presented in terms of a unit volume at the bubble point condition.
This flash vaporisation test gives us therefore the reservoir temperature bubble point
pressure, which in this case is 2620psig. i,e the point where the relative volume is
1.0.
34
PVT Analysis
Column 2 gives the volume of the system at pressure per unit system volume at 2620
psig and 220˚F. These are listed as relative volumes, relative to the bubble point.
Column 3 presents what is called the Y function, this function should provide a straight
line or a slight curve and can be used to pick out anomalous data.
We will now see how we can use the relative volume data to provide us with some
formation volume factors above the bubble point.
Exercise PVT 2.
What is the formation volume factor and the density of the oil at the last reservoir
pressure measured.
Solution.
The well characteristics give the last reservoir pressure as 3954 psig. @ 8500 ft: We
obtain the oil formation volume at 3954 psig by multiplying the formation volume
factor at the bubble point by the relative volume (to the bubble point). Why multiply?
Because:
vol reservoir oil vol bubble point oil vol reservoir oil
Bo = = ×
vol stock tank oil vol stock tank oil vol bubble point oil
and the reference bubble point oil volume cancels out. Therefore Boi, the initial
formation volume factor is 1.495 x 0.9778 = 1.4618 when the 300 psig primary trap
is involved. It is a different value if another separation pressure is used. The 0.9778
was obtained by interpolation between 3500 and 4000 psig in column 2.
Reservoir oil density at pressures greater than 2620 psig also make use of the relative
volume data of column 2, page 4. The added information we have is the density of
the bubble point oil. This is given in the summary data on page 3 of the report. We
see here that the specific volume at the bubble point, vb = 0.02441 ft3/lb. This comes
from direct weight-volume measurements on the sample in the PVT cell. We can
now calculate the density, roi, of the initial reservoir oil as:
1 1
ρoi = =
voi voi ⋅ vrel
1
ρoi = = 41.89lbs / ft 3
0.02441x 0.9778
The compressibility of the oil above the bubble point can also be obtained from the
relative volume test. The definition of compressibility is:
It makes no difference whether the volume units in the equation are relative volumes
to the bubble point, formation volumes, or specific volume values. To evaluate CO at
pressure p it is only necessary to graphically differentiate the p-vrel data in columns
∂v
1 and 2 to get ∂p at the pressure and divide by vrel. A less accurate value can be
obtained by the assumption:
1 ∆v
Co = −
vavg ∆p T
For example, to get Co at 4500 psig using relative volume values of 500 psi on each
side of 4500 psig:
1 (0.9639 − 0.9771)
Co = −
0.9639 + 0.9771 (5000 − 4000)
2
1 0.0219
Co = = 13.6(10 −6 )vol / vol / psi
0.9639 1000
The report also lists some compressibility numbers on page 3. These are not the
same as indicated above because they are changes in volume (in the pressure interval
indicated) per unit volume at the higher pressure. For example, the value of 13.48 (10-6)
for the 5000 and 4000 psi interval is obtained as:
1 (0.9771 − 0.9639)
−
0.9639 (5000 − 4000)
The compressibility data on page 2 are set up in this manner because of the way they
are used in one form of the material balance.
12.3 Total Formation Volume of Original Oil Below The Bubble Point
Pressure.
In the liquid properties chapter we introduced the total formation volume factor,
Bt. This factor is of little practical significance since it describes the volume of an
oil and its associated gas both above and below the bubble point, when the system
does not change. In reality below the saturation pressure the system changes as gas
and oil have different mobilities. In some forms of the material balance equation
Bt is used however to express oil volumes. We have just seen that to calculate the
formation volume factor of the oil above the bubble point we multiply the bubble
point formation volume by the relative volume given in column 2, page 4. If we
multiply Bob by vrel at pressures less than pb, we also get a formation volume factor,
36
PVT Analysis
the total formation volume Bt, of the original system. That is at p < pb we will have
two phases and the Bt is the volume in relation to both gas and liquid phases in
equilibrium at pressure p.
One form of the material balance equation makes use of the expansion of the original
oil between the initial system pressure and any subsequent pressure. This expansion
is given by the term:
Eo = N(Bt - Boi)
where N is the initial stock tank barrels in the reservoir and (Bt - Boi) is the expansion
per unit stock tank oil. Eo is, therefore, the expansion (bbl) of the original oil system.
Sometimes we see the expansion equation written:
Eo = N(Bt - Bti)
Figure 28 below illustrates the change of Bt and Bo with pressure over the total
pressure range.
Bt
Formation Volume
(Rsi-Rs)Bg Bti
5.61
Factor - B
Bo
Boi
p pb pi
Pressure p
Figure 28 Shape of Total Formation Volume Factor Bt and Oil Formation Volume Factor
Bo
(1) the reservoir pressure changes are not violent and large as are the pressure
changes in entering surface separators. The subsurface changes are more gradual
and might be considered to be a series of infinitesimal changes.
The differential data are reported on page 5. Note that the table is headed by the title
“Differential Liberation at 220˚F”. Probably the best way to understand these data
is to explain again the manner of obtaining the values.
To begin with, the laboratory starts with a known volume of the original system in
the PVT cell. This may be of the order of 100-200cm3. The volume at the bubble
point pressure (2620 psig in this instance) is determined accurately as it is a reference
for all subsequent measurements.
2620 psig
b'
Pressure
Resudual Oil
38
PVT Analysis
Referring to page 5, we see that the first pressure step was to 2350 psig. At this
pressure the original system will be in two phases. Its volume would be at b on the
adjoining sketch. Figure 29.
The first step in altering the overall system composition is made at 2350 psig by
removing the gas phase from the PVT cell while maintaining constant pressure.
The quantity of gas removed is determined by collecting it in a calibrated container.
The volume that the gas phase occupied in the cell is determined by the amount of
mercury or non contacting fluid injected during the removal process. Also, the gas
gravity is measured on the sample bled off. The volume of liquid remaining in the
cell is shown at b' in the sketch.
The above procedure is repeated by taking the 2350 psig saturated liquid to 2100
psig (point c) and removing a second batch of gas at that pressure. Again the volume
of the displaced gas in the cell at 2100 psig is determined as is the gravity of the
removed gas. The volume of liquid phase remaining after the second gas removal
step is illustrated by point c' in the sketch.
This process of removing batches of equilibrium gas continues until the cell pressure
at the last displacement is 0 psig. As indicated by the differential data on page 5,
there were ten equilibrium removals, all at 220˚F. The final volume of liquid phase
remaining in the cell at 0 psig and 220˚F is corrected by thermal expansion tables
(or by cooling the cell) to 0 psig and 60˚F. This 0 psig/60˚F liquid is called residual
oil. Note that residual oil and stock tank oil are not the same fluids. They are both
products of the original oil in the system but are generated by different pressure-
temperature routes.
Having now got to residual oil the data obtained are recalculated and presented on
the basis of a unit barrel of residual oil. By the time 0 psig and 220˚F had been
reached, the original system had liberated 854ft3/B residual oil. Column 2 expresses
the amount of gas in solution at the various pressures. This is the difference of the
854ft3 total liberated and the amount liberated between the original bubble point
pressure and that pressure.
It is important to understand why the solution gas-oil ratio determined from surface
flash by taking oil at its bubble point directly to separator and surface conditions
conditions compared to differential removal will be different, although the starting
and finishing conditions are the same. It is because the process of obtaining residual
oil and stock tank oil from bubble point oil are different. The first is a multiple series
of flashes at the elevated reservoir temperature ( the differential test); the second is
generally a one or two-stage flash at low pressure and low temperature (flash tests).
The quantity of gas released will be different and the quantity of final liquid will
be different because the changing composition of remaining liquid at each stage
will influence the distributions of components between the phases. Also, the quality
(gravity) of the products will be different (compare ˚API of residual oil vs ˚API of
stock tank oil). The only thing that will be the same for the two processes is the total
weight of end products.
Column 3 are the relative volumes of the liquid phase measured during the differential
liberation of gas. Note that (per the footnote) these are volumes at pressure p are
Above 2620 psig, the original bubble point, the system remained constant in
composition. Therefore, the relation of the relative oil volume at p to the bubble
point value, 1.600, must be the same as the relative volume in numbers in column
2, page 4 of the report.
The other data on page 5 are differential liberation that refer to the oil and gas phases
in the reservoir at 220˚F. Column 8 shows that the gravity of the gas liberated between
2620 psig and 2350 psig was 0.825. The next batch between 2350 psig and 2100 psig
was 0.818. The gas deviation (compressibility) factor of the first liberated gas was
0.846 at 2350 psig. The oil density at 2350 psig/220˚F was 0.6655 gm/cc.
Now we understand the basic difference between flash and differential data as given in
the standard PVT report, we can calculate flash solubilities and oil formation volume
factors below the bubble point from a combination of the differential and flash data.
It is important to appreciate that there are two separation stages in separating the oil
from its original solution gas when the fluid in the reservoir has dropped below the
bubble point. The drop of the reservoir pressure from the bubble point pressure to
a lower pressure is considered to be by a differential process. The separation of gas
from the reservoir pressure to the surface is then by the flash process.
(a) Differential solubility data at the bubble point state (2620 psig/220˚F) and
eleven pressures below the bubble point pressure gives a bubble point value at
854ft3/B residual oil. All fluids at pressures above pb have this amount of gas.
(b) The flash solubility of the bubble point oil for four different surface trap situ-
ations, where these vary from 778ft3/B stock tank oil to 795ft3/B stock tank oil for
a 300 psig primary trap-tank situation. These are shown on the sketch. Figure 30.
The 59 ft3/B difference in values is not experimental error but is a result of the total
differential process of the test. In reality there is only a small differential element
in the early stages of depletion.
The pressure depletion starts at the bubble point and the solution GOR is that from
the flash separator tests. We now need to develop the GOR curve below this value.
We will use the 300 psig primary - 0 psig tank situation and will examine the GOR
for the reservoir pressure of 1850 psig.
40
PVT Analysis
Rs ft3 /bbl
Two stage
surface flash
2620 psig
Pressure
Exercise PVT 3.
Calculate the solution GOR at 1850 psig using the 300/0psig separator data.
Solution.
Looking at the differential liberation data in column 2, page 5, we see that 242ft3 of
gas has come out of solution, per barrel of residual oil, when the pressure declined
from 2620 psig to 1850 psig. 854 - 684. In other words, we can say that the 1850 psig
saturated oil contains less gas by this amount. If this liquid were taken to the surface
and processed through the traps, it would also show somewhat less gas solubilities
than the 795ft3/B stock tank oil that the bubble point oil shows; but it would not be
242ft3 less because we now have a different oil base.
ft 3 B Re sidual Oil ft 3
⋅ =
B Re sidual Oil B Bubble Po int Oil B Bubble Po int Oil (2)
ft 3 B Bubble Po intoil ft 3
⋅ =
B Bubble Po int Oil B Stock Tank Oil B Stock Tank Oil (3)
ft 3
= ( ∆Rs ) flash
B Stock Tank Oil (4)
B Re sidual Oil 1
=
In equation (2) B Bubble Po int Oil 1.600
B Re sidual Oil
= 1.495
In equation (3) B Stock Tank Oil
1.495
(∆Rs ) flash = (∆Rs )diff ⋅ 1.600
and
1.495
( Rs )1850
flash
= 795 − ( ∆Rs ) flash = 795 − 242 ⋅
1.600
( R )1850
s flash = 795 − 2261 = 569 ft 3 / BTSO
Solution gas to oil ratio at 1850 psi = 569 scf/STB.
For those who prefer equations, this can be generalised as:
Bob
( Rs ) flash = ( Rsb ) flash − ( Rs )diff ⋅ V
b / VR
1.660
Differential 1.495
Volume Factor
@ 200º F
Two Stage
Surface Flash
2620 psig
1.0
Pressure
Figure 31 Comparison of differential data with flash for volume factor volumes.
The result from the separator test is the correct value, since it is based on stock tank
volumes. The differential data is used to calculate the change in this separator value
below the bubble point.
42
PVT Analysis
We can see that the relative oil volume and the formation volume factor at pressure
p can be related by transferring to the common point - the bubble point.
Let:
Then:
B Saturated Oil
=
B Bubble Po int Oil
B Bubble Po int
= Bo
B Stock Tank Oil
Therefore:
Bob
Bo = V VR .
Vb VR
Exercise PVT 4.
What is the oil formation volume factor at 1850 psig.
SOLUTION
At 1850 psig we would have: from page 5 relative volume of 1.479 B/B residual
oil
Bo
1850
= 1.3819 B / B stock tan k oil
Parts of a gas condensate report attributed to the Good Oil Exploration and
Production Company is given for a North Sea field. Not all the report is given,
information not included is the sample validation section. Table B1 is the summary
sheet giving information on the reservoir and the sampling details. Table C4 gives a
comprehensive compositional analysis of separator products and the calculated well
stream composition. The table C7 is the compositional analysis of the reservoir fluid
sample. This sample was generated by recombining the separator fluids and then
blowing down the recombined fluid to determine the fluid composition. This provides
a quality check against the calculated well stream composition obtained from the
composition of the separator fluids. Examination of the well stream and reservoir
fluid compositions show they are in close agreement.
The two key tests for gas condensates is the constant mass test or the constant
composition expansion test and the constant volume depletion test. Table D1 is the
constant composition expansion test. In the oil test the saturation pressure is obtained
by a distinctive change in slope of the pressure volume curves for the single phase
and two phase regions. For gas condensates no such distinct change in slope occurs
at the dew point. The dew point is obtained by observation of the condensation on
the window of the PVT cell. The distinctive nature of the condensation depends on
how near to the critical temperature the fluid is. If the fluid is near to the critical
temperature than the dew point is clearly observed, whereas for those fluids away
from the critical point the dew point is less distinct. Colour changes of the fluid also
occur around the dew point, getting darker as the dew point is approached.
For the constant composition expansion at 275 °F on page D1 relative volume results
are given over a pressure range from 8,000 psi g to 1198 psig. In the same way as
for the oil PVT analysis volumes are related to those at the saturation pressure, the
dew point. At each pressure step not only is the relative volume calculated but also
the compressibility factor, z. The dew point was observed to be at 5454 psig. The
table also records pressure steps below the dew point and down to 4,900 psig the
retrograde liquid volume is also reported. Clearly these values are not of great value
since in the reservoir constant composition does not occur below the dew point
because the system is considered to be changing as immobile condensate separates
from its associated gas.
The constant volume depletion test data is given on D8. In this test a series of
depletion steps have been carried out. Starting at the dew point pressure of 5454
psig the pressure was dropped to 4900psig. Between 5454 psig and 4900 the liquid
proportion was measured. At 4900psig gas was removed at constant pressure until
the volume at the dew point was obtained. A series of further depletion steps were
carried out at 4000, 3100,2200,1400 and 726 psig. At each stage gas was removed
to bring the contents back to the dew point condition volume.
44
PVT Analysis
The important liquid drop-out curve is presented in the curve on page D9. The dotted
line is the relative volume data from the constant mass test. The curve shows a very
high maximum liquid drop out of 44% at a pressure of 3900psig.
Over recent years as exploration activity has moved deeper into the sub-surface,
high pressure and high temperature hydrocarbon fluids have been found. A number
of these are now being produced providing a number of challenges in production
and design. Conventional PVT facilities do not cover the pressure and temperature
ranges covered by these fluids and special facilities are required. The temperature
and pressure ranges for such fluids are up to 250°C and 20,000 psi.
In order to handle these fluids and conditions and to enable visualisation of phenomena,
like for example IFT, very low volume apparatus are required. The move to Hp/HT
fluids has also put in question some of the physical property prediction methods which
have been based on pressure and temperature data not extending to HP/HT conditions.
An important feature of high pressure and high temperature is the role of water which
in conventional PVT practice is ignored. At these more extreme conditions recent
measurements are showing that the presence of water cannot be ignored because its
presence will contribute to the physical properties of the hydrocarbon fluids.
15 MERCURY
Historically the transfer fluid in PVT tests has been mercury. It has proved to be a
very effective fluid to generate variable volumes in PVT apparatus as well as being
non contamination with respect to the hydrocarbon fluids. Unfortunately health and
safety concerns with respect to personnel exposed to increased levels of mercury,
and it's incompatibility with certain materials e.g. Aluminium, are such that mercury
is being replaced by alternate systems. Such alternate systems are not as simple to
replace the "flexible metal" which mercury has proved to be. Although in these notes
we refer to mercury, the principals are the same, where for example the mercury is
replaced by a rigid piston driven by a safe fluid e.g. Water.
Exercise PVT 1.
What is the solution gas-oil ratio and formation volume factor resulting from the
separator test, at first stage of 300 psig and 2nd stage 0 psig and both at 75˚F ?
Solution.
This exercise illustrate the results using one of these tests, a two-stage separation;
a primary trap operating at 300 psig and 75˚F followed by a stock tank operating at
14.7 psia (0 psig) and 75˚F.
When one barrel of bubble point oil; defined as oil saturated at 2620 psig and 220˚F in
footnote 3, on page 7 of 15 is flashed (processed) through this separation arrangement,
the stock tank has a quality of 40.1˚API (column 5). The formation volume factor
of the bubble point oil, Bob = 1.495 B/BSTO (column 6). This would have been the
volume of oil displaced from the PVT cell divided by the volume collected at the final
stage and then corrected for the thermal reduction from 75˚F to 60˚F. The source of
this bubble point pressure value will be indicated later.
Columns 3 and 4 show the surface gas-oil ratio from the first stage and the tank.
The first stage ratio of 549 ft3/BSTO ( column 4) and the tank stage gas amount to
246 ft3/BSTO. It is important to read the footnotes of the report. Column 3 gives
the results in relation to the volumes at indicated P & T whereas column 4 gives the
volumes with respect to stock tank conditions of 14.65 psia and 60˚F. The solution
gas-oil ratio at bubble point conditions (2620 psig and 220˚F), is therefore Rsb is 549
+ 246 = 795 ft3/ BSTO when flashed through this surface trap arrangement.
If we compare these results for the 50psig, 0psig arrangement we obtain a Bob of
1.481 B/BSTO and a solution GOR of 778 ft3/ BSTO.
Clearly therefore Rsp, Bob, ˚API all vary with the separation pressure-temperature
situation. There is not one unique result. When reporting Bo and GOR data for a
reservoir therefore it is important to report that these are for a specific separation
route or averaged for a series of tests. The latter is not so useful since it is not so
straightforward to calculate the result for a different separation route using VLE
methods.
The results from the separation test are based on the bubble point condition and to
obtain volumetic information at other pressures we require the results from other
tests.
Exercise PVT 2.
What is the formation volume factor and the density of the oil at the last reservoir
pressure measured.
Solution.
The well characteristics give the last reservoir pressure as 3954 psig. @ 8500 ft: We
obtain the oil formation volume at 3954 psig by multiplying the formation volume
factor at the bubble point by the relative volume (to the bubble point). Why multiply?
Because:
70
PVT Analysis
vol reservoir oil vol bubble point oil vol reservoir oil
Bo = = ×
vol stock tank oil vol stock tank oil vol bubble point oil
and the reference bubble point oil volume cancels out. Therefore Boi, the initial
formation volume factor is 1.495 x 0.9778 = 1.4618 when the 300 psig primary trap
is involved. It is a different value if another separation pressure is used. The 0.9778
was obtained by interpolation bewteen 3500 and 4000 psig in column 2.
Reservoir oil density at pressures greater than 2620 psig also make use of the relative
volume data of column 2, page 4. The added information we have is the density of
the bubble point oil. This is given in the summary data on page 3 of the report. We
see here that the specific volume at the bubble point, vb = 0.02441 ft3/lb. This comes
from direct weight-volume measurements on the sample in the PVT cell. We can
now calculate the density, roi, of the initial reservoir oil as:
1 1
ρoi = =
voi vob ⋅ vrel
1
ρoi = = 41.89lbs / ft3
0.02441× 0.9778
The compressibility of the oil above the bubble point can also be obtained from the
relative volume test. The definition of compressibility is:
1 ∂v
Co = −
v ∂p T
It makes no difference whether the volume units in the equation are relative volumes
to the bubble point, formation volumes, or specific volume values. To evaluate COat
pressure p it is only necessary to graphically differentiate the p-vrel data in columns
∂v
1 and 2 to get ∂p at the pressure and divide by vrel. A less accurate value can be
obtained by the assumption:
1 ∆v
Co = −
vavg ∆p T
It makes no difference whether the volume units in the equation are relative volumes
to the bubble point, formation volumes, or specific volume values. To evaluate COat
pressure p it is only necessary to graphically differentiate the p-vrel data in columns
∂v
1 and 2 to get ∂p at the pressure and divide by vrel. A less accurate value can be
obtained by the assumption:
For example, to get Co at 4500 psig using relative volume values of 500 psi on each
side of 4500 psig:
1 (0.9639 − 0.9771)
C o = − 0.9639 + 0.9771
(5000 − 4000 )
2
1 0.0219
Co = = 13.6(10−6 )vol / vol / psi
0.9705 1000
The report also lists some compressibility numbers on page 3. These are not the
same as indicated above because they are changes in volume (in the pressure interval
indicated) per unit volume at the higher pressure. For example, the value of 13.48 (10-6)
for the 5000 and 4000 psi interval is obtained as:
1 ( 0.9771− 0.9639)
−
0.9639 (5000 − 4000)
The compressibility data on page 2 are set up in this manner because of the way they
are used in one form of the material balance.
Exercise PVT 3.
Calculate the solution GOR at 1850 psig using the 300/0psig separator data.
Solution.
Looking at the differential liberation data in column 2, page 5, we see that 242ft3 of
gas has come out of solution, per barrel of residual oil, when the pressure declined
from 2620 psig to 1850 psig. 854 - 684. In other words, we can say that the 1850 psig
saturated oil contains less gas by this amount. If this liquid were taken to the surface
and processed through the traps, it would also show somewhat less gas solubilities
than the 795ft3/B stock tank oil that the bubble point oil shows; but it would not be
242ft3 less because we now have a different oil base.
72
PVT Analysis
ft3
= (∆Rs ) flash
B Stock Tank Oil (4)
B Residual Oil 1
=
In equation (2) B Bubble Point Oil 1.600
Therefore:
(∆Rs ) flash = (∆Rs )diff ⋅ 1.495
1.600
and
1.495
( Rs )1850
flash
= 795 − ( ∆R s ) flash = 795 − 242 ⋅
1.600
( Rs )1850
flash
= 795 − 2261 = 569 ft 3 / BTSO
Solution gas to oil ratio at 1850 psi = 569 scf/STB.
For those who prefer equations, this can be generalised as:
B ob
( Rs ) flash = ( R sb) flash − (R s ) diff ⋅
Vb / VR
Exercise PVT 4.
What is the oil formation volume factor at 1850 psig.
SOLUTION
At 1850 psig we would have: from page 5 relative volume of 1.479 B/B residual
oil
1850
Bo = 1.3819 B / B stock tan k oil
CONTENTS
1. INTRODUCTION
2. LIST OF SYMBOLS
12. CONCLUSION
LEARNING OBJECTIVES
Having worked through this chapter the Student will be able to:
• Present a material balance (MB) equation for a dry gas reservoir with and without
water drive.
• Demonstrate the linear form of the MB equation for a gas reservoir with water
drive and comment on its application.
• Be able to derive the material balance equation including gas cap expansion,
water influx and core and water compressibility.
• Given the equation be able to identify the component parts of the MB equation,
eg. gas cap expansion etc.
Material Balance Equation
1. INTRODUCTION
In the chapter on Drive Mechanisms we reviewed qualitatively the various drive
energies responsible for hydrocarbon production from reservoirs. In this and sub-
sequent chapters we will introduce some reservoir engineering tools used in calculat-
ing reservoir behaviour. The petroleum engineer must be able to make dependable
estimates of the initial hydrocarbons in place in a reservoir and predict the future
reservoir performance and the ultimate hydrocarbon recovery from the reservoir. In
this chapter the material balance equation is presented.
The material balance equation is one of the basic tools in reservoir engineering.
Practically all reservoir engineering techniques involve some application of mate-
rial balance. Although the principle of conservation of mass underlies the material
balance equation, custom has established that the material balance be written on a
volumetric basis, because oilfield measurements are volumetric and significant fac-
tors can only be expressed volumetrically.
The equation was first presented by Schilthuis1 in 1936 and many reservoir engineering
methods involve the application of the material balance equation. Since the equation
is a volumetric balance, relating volumes to pressures, it is limited in its application
because of any time dependant terms. The equation provides a relationship with a
reservoir’s cumulative production and its average pressure. However when combined
with fluid flow terms, we have a basis to carry out predictive reservoir modelling, for
example to put a time scale to production figures.
Over recent years, as increasingly powerful computers have enabled the application of
large numerical reservoir simulators, some have looked down on the simple material
balance equation and the tank model of the reservoir which it represents.
Reservoir simulators however apply the material balance approach within each of
their multi-dimensional cells. The value of this classical tool is that it enables the
engineer to get a’feel’ of the reservoir and the contribution of the various processes
in fluid production. A danger of blind application of reservoir simulators is that the
awareness of the various components responsible for production might be lost to the
engineer using the simulation output in predictive forecasting.
When fluids (oil, gas, water) are produced from an oil reservoir, which may or may
not have a primary gas cap, the pressure in this reservoir will drop below the original
value. As a consequence of this pressure drop, a number of things will happen:
The question is now; if we start off with a “given” reservoir, and after some time we
have produced certain quantities of oil, gas and water, what can we say about the
average pressure in the reservoir, and what can we say about the average saturation
distribution? The answer to these questions can be obtained by considering our
reservoir at two stages:
(b) when we have produced certain amounts of oil, gas and water, by which time
the average pressure has declined to p (to be calculated).
Besides these natural phenomena the equation also has to be capable of handling
other factors affecting behaviour, for example injecting gas and or water.
There are a number of ways of developing the equation. We will look at two ap-
proaches, the first examining the equation as applied to specific reservoir types and
then a simple volumetric expansion approach.
NOTE:
2. LIST OF SYMBOLS
Material Balance Equation
Other subscripts
i at initial conditions
b at bubble point
The simplest material balance equation is that applied to gas reservoirs. The compress-
ibility of gas is a very significant drive mechanism in gas reservoirs. Its compressibility
compared to that of the reservoir pore volume is considerable. If there is no water
drive and change in pore volume with pressure is negligible (which is the case for a
gas reservoir), we can write an equation for the volume of gas in the reservoir which
remains constant as a function of the reservoir pressure p, the volume of gas produced
SCF, the original volume of gas, SCF, and the gas formation volume factor.
A representation of the equation for a gas drive reservoir with no water drive is
given below.
Bg - based on z, p, T
Gp
G Bgi = (G - Gp)Bg
Pi P
pV = znRT
If the gas reservoir is supported by water drive then as gas is produced water will
encroach into the gas pore space, and some of this water may be also be produced.
Figure 2 below illustrates the contact with a supporting aquifer. Because the mobil-
ity of gas is far greater than water, evidence in the form of produced water may be
delayed as the water keeps to the gas water contact. The support from the water would
be evidenced however by the pressure support given to the reservoir. In earlier years
this may not be so easy to detect.
Gp Wp
GBgi (G - Gp) Bg
= We - Wp
Water Water
EXERCISE 1
A gas reservoir without water drive contains 500 million standard cubic feet of gas
at an original pressure of 3,000psia. How much gas has been produced when the
reservoir pressure has declined to 2,900 psia. Use Bgi and Bg for the initial and
2,900psia pressure as 0.0010 and 0.0011 bbl/scf.
This simple example illustrates the significant amount of gas production associated
with a relatively small pressure decline.
Material Balance Equation
G. Bgi = (G − Gp ) Bg
0.00504 zi T
From equation 25 in Gas Pr operties chapter where Bg =
p
0.00504 zi T 0.00504 zT
G = (G − Gp )
pi p
zi z
G = (G − Gp )
pi p
p Gzi
Gp = G −
z pi
(3)
X
Gp
Cumulative gas
production X
O
p/z Pi/Zi
This procedure is often used in predicting gas reserves. Often the influence of water
drive is ignored resulting in a serious error in reserves.
EXAMPLE
A dry gas reservoir has produced as follows:
Data
Reservoir Temperature T= 100°F
Gas Gravity SG= 0.68
SOLUTION
To construct the graphical material balance plot we must first determine the P/Z
values.
Using figures 2 and 3 from the Gas Reservoir chapter for a gas gravity of:
SG= 0.68
Material Balance Equation
4200
P/Z (psia)
4100
4000
3900
3800
0 2,000 4,000 6,000 8,000 10,000 12,000 14,000
Cumulative gas production (MM scf)
Figure 4 (a)
3000
P/Z (psia)
2500
2000
1500
1000
500
0
0 2,000 4,000 6,000 8,000 10,000 12,000 14,000
Cumulative gas production (MM scf)
Figure 4 (b)
Initial Pressure
From figure 4, at Gp = 0:
From figure 5:
Zi= 0.81
Therefore:
Pi= 3,603
10
Material Balance Equation
1.05
1.75
1.10
1.2015
1.
1.65
1.25
1.30
1.35
Used to obtain p and z when
1.40
1.55
1.45
analyses result in p/z answers.
1 .5 0
1 .6 0
(pr/z) = (p/z)/pc
1.9.800
1.45
1 .7
z can be read from graph.
.0 0
2.2.20 0
p = (pr/z) z pc
2
40
1.35 2.60
2.80
3.00
1.25
1.15
Gas deviation factor, z
Reduced temperature, Tr
1.05 3.00
2.80 2.60
2.40 20
0.95 2.
2.00
1.90
1.80
0.85 1.70
1.60
1.50
0.75 1.45
1.40
0.65 1.35
1.30
1.25
0.55
1.20
0.45 1.15
1.10
0.35
1.05
0.25
0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0
(pr/z)
Great caution has to be taken when using this method. Water drive is considered to
be zero, that is the gas is being solely produced as a result of gas compressibility.
If water drive exists this will contribute to pressure support. If a plot of Gp vs p/z
deviates from linearity than that gives evidence of water drive support. Figure 6
from Dake illustrates this deviation. If a straight line is fitted to this data assuming
no pressure support from water then gas reserves are enhanced, beyond what they
are in actuality.
(a) (b)
3500 3500
P/Z P/Z
2700
O G
Gp G'>G Gp
applies.
Pi, Ti
Liquid
P
Sep. Dew point line
2 phase
Mixture
Gas
The equation already produced assumed that the formation of liquid condensate
causes insignificant error in the quality.
For condensate systems the Gp produced should include the produced condensate
and the produced water (originally dissolved in gas).
znRT
v=
P
12
Material Balance Equation
γ o SCF
v = 133,000
M o STB
GBgi = (G-Gpc) Bg
can be written to describe the change in gas cap volume due to oil production and
production of gas. In this case it is suggested that some gas has been produced from
the gas cap, Gpc.
Gpc
GBgi (G - Gpc) Bg
= Gas cap exp.
Oil Oil
where N is the original oil volume in the reservoir and the Np is the volume of oil
produced both expressed in stock tank barrels.
Clearly this is a poor assumption but is useful in illustrating the equation develop-
ment. Pore volume changes will be considered later in the context of pore space and
connate water.
As we observed in this mechanism, the produced fluids are now oil with its contained
solution gas and gas which has come out of solution from the oil. Not all of this re-
leased free gas will be produced to the surface, some will remain in the reservoir.
Free gas in reservoir = original gas in solution - remaining gas in solution - produced
gas(Gps).
The volume of free gas and the remaining oil can now be added to the original oil
volume.
The equation can be written in terms of the original stock-tank volume in the res-
ervoir.
N p Bo + Bg (G ps − N p R s )
N=
Bo − Boi + (R si − R s )Bg (9)
Np Gps
Free Gas
(NRsi-(N-Np)Rs-Gps)Bg
NBoi = Oil
(N-Np)Bo
14
Material Balance Equation
= (We-Wp) (11)
∴ Total change in volume = original oil volume - (oil volume + free solution gas)
Np Gpc Gps Wp
Gas cap expansion
Oil volume and free
NBoi =
solution gas
Net water encroachment
(
= NBoi - (N - Np)Bo + (NRsi-(N - Np)Rs - Gps)Bg )
(12)
The reduction in pore volume with decline in pressure is due to two factors..
Pore pressure
- sand grains
- pore space
When the pore pressure is reduced then the effective opposing pressure is increased
and the bulk volume is slightly reduced. At the same time since the rock grains are
compressible albeit only slightly the effect of a reduction in reservoir pressure will
be to expand the rock grains.
1 ∆Vpr
.
Compressibility of rock cf =
Vp ∆p
16
Material Balance Equation
1 ∆Vpw
cw =
Vpw ∆p (16)
Vpw = Vp x Swc
NBoi
Pore volume: Vp =
1 − Swc
NBoi
∆Vpw + pores = (cw Swc + c f )∆p
1 − Swc (19)
If the original volume of pores also includes that associated with the gas cap, then
the pore volume is equal to
NBoi
Vp = (1 + m)
1 − Swc
where m is the ratio of the original reservoir gas cap volume to the original reservoir
oil volume. Some choose not to add this m factor term since if free gas is present as
a gas cap then the compressibility associated with the gas is far greater than the pore
and water compressibility values.
(
= NBoi − ( N − N p ) Bo + ( NRsi − ( N − N p ) Rs − Gp ) Bg )
N p Bo + Bg (Gp − N p Rs ) − G( Bg − Bgi ) − (We − Wp )
N=
Boi
Bo − Boi + ( Rsi − Rs ) Bg + (C f + Cw Swc )∆p
1 − Swc
In some texts the pores connate water compressibility term includes a product with
(1+m). This includes pore volumes associated with a gas cap as mentioned above.
The equation can be rearranged for different applications. The following useful
rearrangement by Archer4 , Figure 11, helps to identify the constituent parts of the
equation.
[NRsi-(N-Np)Rs-Gps]Bg
In the literature sometimes the equation is presented using the total formation volume
factor Bt and the ratio of the initial reservoir free gas volume to the initial reservoir
oil volume m, Gp, is also expressed as a function of produced gas - oil ratio Gp.
18
Material Balance Equation
GBgi
• m=
NBoi
• Gp = NpRp
N p ( Bt + ( Rp − Rs ) Bg ) − (We − Wp )
N=
( Bg − Bgi )
Bt − Bti + (c f + cw Swc )∆ p Bti / (1 − Swc ) + mBti
Bgi (21)
All the terms of the general equation as just presented may not be significant all the
time.
For example above the bubble point a number of the parameters will be zero.
Above the bubble point the solution gas - oil ratio is constant and therefore
Gp-NpRs = 0 since only solution gas will be produced above the bubble point.
If we are operating above the bubble point then there will for the majority of reservoirs
be no gas cap, and therefore the gas in place term, G or m, will also be zero. (Some
reservoirs with a compositional gradient can have a gas cap and also at the lower
part of the formation a different com position with undersaturated fluid.) The term
in the denominator, Rsi - Rs, will also be zero.
The equation therefore above the bubble point reduces to a simple equation, associ-
ated with compressibility terms of the oil ( the formation volume factors ) and those
of the connate water and pore space. We will consider this later when we examine
some applications of the equation.
When the reservoir is below the bubble point then, the terms described above being
zero in the undersaturated condition, have significance and are not zero. However the
term in relation to the compressibility of the connate water and pore space although
contributing to the overall balance is very small when compared to that from free gas
compressibility. It could be argued that the absolute changes in the water and pore
compressibility term is less than the errors associated with the free gas terms, when
the system is below the bubble point
The equation also includes a term with respect to water drive, We. Other terms can also
be added to include artificial drive, for example gas injection, Gi and water injection
Wi. Clearly when any of these three drive supports, natural or otherwise, are not ac-
tive then clearly they are zero. Although there may not be any water drive, We or Wi
, there still could be water production as a result of mobilisation of connate water.
1 dv
c =−
v dP
The individual expressions of the compressibility of the oil, gas, water and rocks are,
co, cg, cw and cf. These compressibilities depend on the nature of the fluids and rocks
and between them have significant variations Gas is the most compressible down to
water and rock depending on its composition and nature.
Even these pore compressibilities are small but their significance should not be
neglected particularly above the bubble point.
20
Material Balance Equation
FLUID VOLUMES
Connate water
Water influx We
Reduction in total
pore volume down to
pressure P
Suppose that the pressure were reduced from pito p. Obviously this could not be
done without production, but let us see what effect such pressure reduction would
have. The volumes of the three phases will expand as shown in Figure 15. There
may also have been a water influx We. Also, the total available pore volume will
become smaller, through pore compressibility effects just described. Clearly, the new
fluid volumes, plus the water influx, do not fit any longer in the available pore space,
there is a shortage of space equivalent to the sum of the shaded areas in Figures 15.
Consequently, an equal volume of fluids can no longer be present in the formation,
and must therefore be the same as the reservoir volume at pressure p of the produced
fluids. With this in mind, we can state the material balance as follows:
N(Rsi-Rs)Bg.
mNBoi
Boi
mN scf
Bgi
22
Material Balance Equation
Bg
mNBoi
Bgi
B
mNBoi g − 1
Bgi (26)
1 dVw
cw =
Vw dP
dVw = cwVw∆p
Vw is the total volume of the water. This is a proportion Sw of the total pore volume.
The total pore volume is that associated with the oil and a gas cap. The pore volume
of the oil at a saturation of (1-Swc) or So is: NBoi, the pore volume including the water
is NBoi/(1-Swc).
For the gas cap, the hydrocarbon pore volume is mNBoi, and including water is
mNBoi/(1-Swc) .
The total pore volume associated with the gas cap and oil, including connate water,
is:
(1+m)NBoi.
(1 + m) NBoi Swc
(1 − Swc )
(1 + m) NBti Swc cw ∆p
(1 − Swc ) (27)
The change in volume of the pores associated with the total pore volume is there-
fore:
(1 + m) NBti Swc cw ∆p
(1 − Swc ) (28)
Water influx
If there is an aquifer, then as pressure is reduced water influxes into the reservoir
volume. This water influx is We (res.bbl.)
We can now add all these expansion terms and make then equivalent to the reservoir
volume of produced fluids. This gives the full material balance equation.
[ ] [
N p Bo + ( Rp − Rs ) Bg + Wp Bw = N ( Bo − Boi ) + ( Rsi − Rs ) Bg ]
Expansion of oil and dissolved gas (24)
[ ]
N p ( Bo + ( Rp − Rs ) Bg =
( B − Boi ) + ( Rsi − Rs ) Bg Bg c S + cf
NBoi o + m − 1 + (1 + m) w wc ∆p
Boi Bgi 1 − Swc
+(We − Wp Bw )
(28)
Injection terms
If there is water injection Wi and or gas injection Gi these can also be added, to the
equation either as added to the expansion terms or subtracted from the production
terms.
24
Material Balance Equation
The MB equation has some basic assumptions and limitations which can cause some
erros when applied to some reservoirs.
Pressure
The MB equation is a tank model treating the reservoir as a large tank at which the
pressure is constant throughout the reservoir at a particular time. It clearly ignores
pressure changes which may arise across the reservoir. In the radial flow section it
was clear that there are large pressure variations around the producing and injection
wells.
In order to apply the equation at a particular time an average pressure has to be se-
lected being representative of the reservoir pressure at the particular time. All fluid
properties are evaluated at this pressure. In the next chapter we will discuss this
topic further.
Temperature
Changes in a reservoir generally take place at isothermal, constant temperature, condi-
tions, unless major external temperatures are imposed thorough for example thermal
recovery processes and in some cases large cold water injection schemes.
Production Rate
When things happen is not part of the MB equation as there is no term present in-
cluding time, for this permeability would be required. Rate sensitivity is therefore
not part of the equation and for those situations, for example in water drive, which
are dependant on rate of production the material balance equation requires the ap-
plication of other equations.
• If production and pressure date are known as a function of time, and oil and gas
in place is available from a volumetric estimate, the water influx We can be
determined as function of time. Its magnitude has a direct bearing on secondary
recovery plans.
• If there is no evidence of a natural water drive (We=0) the oil in place can
be calculated from production and pressure data. This may have an influence
on the geological interpretation (volumetric estimate) and thus on the further
development of the reservoir.
• For a known oil in place, the pressure at future dates can be calculated for
a postulated production plan (making some assumptions regarding the future
water influx). The result of this calculation may help in:
Dake3 has also examined the status of the various parameters of the equation with
respect to the application of the equation. He divides the parameters into should be
known and potential unknown.
From this list there appears to be 6 knowns and 8 unknowns, demonstrating the
challenge facing reservoir engineering in needing sufficient independent equations
to solve to determine the number of unknowns. As Dake points out the situation in
reservoir simulation is even worse with more unknowns of reservoir geometry and
description in terms of porosity, and a variety of relative permeabilities.
In examining the ‘knowns’, he points out that although Np and Rp are generally the
best known, in old and remote fields good records may be such that oil, gas and water
production figures may not be so readily available. He points out that petrophysical
evaluation is always correct. So for example the connate water saturation Swc is obtained
by averaging its values over all intervals and wells associated with the analysis.
In relation to the unknowns, the material balance, once production and pressure
information is available, provides a useful route to upgrading the original estimate
of in place, STOIIP, N, which has previously been estimated from a combination of
26
Material Balance Equation
Waterdrive as was discussed in the drive mechanism chapter is a very effective drive
mechanism. In reservoir development it provides a major challenge in predicting its
role. To predict the influx of water from an aquifer requires a good characterisation
of the aquifer, its geometry and the important flow related properties. To determine
such for what is compared to the associated oil reservoir a very large system is very
costly and is difficult to justify, for something which only produces water! The nature
of waterdrive is best determined when its impact on actual reservoir performance
is observed. Clearly if water underlays the hydrocarbon formation as a bottom
water drive system, then the advancing water oil contact can be logged in the well.
However if edge water drive is occurring then actual well observations may not be
possible. Material balance provides an opportunity to determine the support from
water drive, but translating this information into specific aquifer characteristics is not
straightforward.
The size of gas cap although more accessible is not always easy to determine, since
it may be preferred for development reasons during drilling to drill through the gas
cap.
Up until recently, water and rock compressibility terms of cw and cf have largely been
assumed to be of little importance and their value if not readily available obtained
from text book type sources. Such assumptions can be very costly particularly for
those fields where compaction drive is very significant.
The material balance equation, a zero dimensional model, or tank model, requires
an average pressure and this average pressure is reflected implicitly in relation to
PVT parameters and explicitly in relation to compressibility of water and rock. This
average pressure determination may be obtained from a range of pressures from wells
within the drainage area. We will discuss this in the next chapter.
The material balance is also a backbone in all mathematical reservoir simulators,
where pressures in individual grid blocks are calculated (apart from production data)
on the basis of influxes from or effluxes to adjacent grid blocks. Over recent years
there has developed an perception by some that the ‘simple material balance’ approach
has been superseded by the more comprehensive reservoir numerical simulation, with
its potential of analysis at small dimension levels compared to the full field tank size
of the MB equation . Until his recent death, Dake and others have recognised the
value of the MB in 'feeling' the reservoir and also providing useful input to the many
uncertainties associated with implementing a full reservoir simulation study.
A range of sources provide the key data for the application of the MB equation. These
sources are also the source for other simulation tools.
Production Data
From well and reservoir records (data banks) or the subject of calculation.
Water Compressibility
cw : at oilfield temperatures and pressures: Should be determined.
Pore Compressibility
In the past has often been assumed from texts. Should be measured.
Reservoir Pressures
From pressure surveys in the field, or subject of the calculation. In the next chapter
we will see how an average pressure can be obtained from a reservoir where there
are different drainage zones.
Water Influx
The subject of water influx, We is covered in a subsequent chapter.
With the MB, the average saturation distribution (So, Sg, Sw with So + Sg + Sw = 1)
can be calculated. However, no conclusion may be inferred how a calculated gas
saturation is distributed, i.e. whether this free gas is spread more or less evenly over
the entire reservoir, or whether the gas is concentrated in some localised areas.
The most significant aspect of MB is that it does not contain time as a parameter. This
means that although an M.B. calculation may tell us what will happen, it cannot say
when it will happen. We can, for instance, calculate that the average pressure of a
given reservoir will drop by 1973 psi for an oil and gas production of 88 MM STB
and 59 MMM SCF respectively, but the material balance will not tell us whether this
28
Material Balance Equation
12. CONCLUSION
Solutions to Exercises
EXERCISE 1
A gas reservoir without water drive contains 500 million standard cubic feet of gas
at an original pressure of 3,000psia. How much gas has been produced when the
reservoir pressure has declined to 2,900 psia. Use Bgi and Bg for the initial and
2,900psia pressure as 0.0010 and 0.0011 bbl/scf.
SOLUTION 1
1. Schilthuis, R.J. Active Oil and Reservoir Energy, Trans AIME, 118:33-52,
1936.
30
Material Balance Equation Application
CONTENTS
1. INTRODUCTION
Having worked through this chapter the Student will be able to:
• Given the MB equation be able to present it in a short hand form as a basis for
use in linear forms.
• Using the various linear forms with sketches illustrate the MB equation for use
for:
• Reservoir with no water drive or gas cap.
• No water drive but with known gas cap.
• Comment with the aid of sketches the impact of water drive on the application
of MB equation in linear and other forms.
• Derive and use a simplified MB equation for application to an oil reservoir above the
bubble point, in terms of recovery, and oil, rock and water compressibility.
• Derive the instantaneous gas-oil ratio equation and use to explain the producing
GOR of a solution gas drive reservoir.
• Comment on how the instantaneous GOR equation can be used to history match
the ratio of the gas to oil relative permeabilities.
• Describe the procedure for the application of the Tarner or Tracy Tarner methods
in predicting solution gas drive performance.
• Be able to use both Tracy and Tracy Tarner methods for predicting future
performance given prerequisite equations and data.
• Comment how with well performance information a time line can be included
in material balance predictions.
Material Balance Equation Application
1. INTRODUCTION
In the previous chapter we developed the material balance, MB, equation identifying
the various elements. In this chapter we will examine the application of the equation
to different reservoir types and examine some techniques developed to predict
reservoir performance. There is no one universal solution to the MB equation and in
this chapter we will examine the equation and its application with respect to different
reservoir types.
In recent times the computing power behind other reservoir engineering tools like
numerical simulation, has cast a shadow of a lack of confidence in the old material
balance approach. Laurie Dake1 up until his sudden death in 1999, was a strong
proponent of the material balance equation and was sometimes interpreted as having a
negative perspective with respect to reservoir simulation. As demonstrated in his text,
"The Practise of Reservoir Engineering", such interpretation is far from the truth. To
quote Professor Dake in his defence of the use of the MB equation “It seems no longer
fashionable to apply the concept of material balance to oilfields, the belief that it is
now superseded by the application of modern numerical simulation. Acceptance
of this idea has been a tragedy and has robbed engineers of their most powerful
tool for investigating reservoirs and understanding their performance rather than
imposing their wills upon them, as is often the case when applying numerical
simulation directly in history matching. ........There should be no competition
between material balance and simulation instead they must be supportive of one
another: the former defining the system which is then used as input to the model.
Material balance is excellent at history matching production performance but has
considerable disadvantages when it comes to prediction, which is the domain of
numerical simulation modelling.”2
One reason for perhaps a lack of appreciation of the equation might be the immediate
impression of complexity through its many terms. A significant step forward in the
equation which had been originally presented by Schilthuis2 in 1936 was by Odeh
and Havlena3, who in 1963 examined the equation in its various linear forms.
In 1963 Havlena and Odeh3 presented a paper aimed at reducing the above problem.
Their method consists of re-arranging the material balance equation to result in an
equation of a straight line. The method requires the plotting of a variable group
versus another variable group with the variable group selection depending on the
drive mechanism.
Once linearity has been achieved, based on matching pressure and production data
then a mathematical model has been produced. This technique is referred to as history
matching, and the application of the model to the future enables predictions of the
reservoir’s future performance to be made.
[ ] [
N p Bo + ( Rp − Rs ) Bg + Wp Bw = N ( Bo − Boi ) + ( Rsi − Rs ) Bg ]
B (1 + m) NBoi (cw Sw + c f )∆p
+ mNBoi g − 1 + + We
Bgi (1 − Swc )
(1)
In some instances water formation volume factors are not included, i.e. Wp, We and
WiBw
Havlena and Odeh simplified the equation into a short hand form:
The left hand side of equation 2 represents the production terms in reservoir volumes
and are denoted by F, i.e.
(i) the expansion of the oil and its originally dissolved gas, Eo, where:
Eo = Bo - Boi + (Rsi -Rs) Bg ....bbl/STB (4)
(ii) the expansion of the pores and connate water Efw where:
Boi
E fw = (1 + m) (c f + Sw cw )∆p.....bbl / STB (5)
1 − Sw
B
Eg = Boi g − 1 ....bbl / STB
Bgi (6)
Material Balance Equation Application
With the above terms the material balance equation can be written:
Using this equation as a basis, Havelena and Odeh manipulated the equation making
different assumptions to produce a linear function.
F = NEo (8)
i.e. a plot of F vs Eo should produce a straight line through the origin Figure 1. This
is the simplest relation and is just a plot of observed production against determined
PVT parameters. The slope of the line gives the oil in place N.
N
F
O Eo
2.3 Gas Drive Reservoirs, No Water Drive and Known Gas Cap
Although We is zero, the gas cap has a volume as given by m, and the equation 7
becomes:
A plot of F vs (Eo + mEg) should produce a straight line through the origin with a
slope N. Figure 2. If m is not known then by making assumptions for m a number
of plots can be generated with the linear slope being the correct value for m.
m
large
m too
O Eo +mEg
Figure 2 F vs (Eo +mEg) Gas Drive, With Known Gas Cap, But No Water Drive.
2.4 Gas Drive Reservoirs with No Water Drive, N and G Are Unknown
If there is uncertainty in both the size of the oil and gas accumulation then Havlena
and Odeh suggest the following form of the material balance equation, by dividing
both sides by Eo.
F E (10)
= N +G g
Eo Eo
where:
Boi
G = Nm
Bgi
A plot of F/Eo vs Eg/Eo should be linear with an intercept of N and a slope of mN.
Figure 3.
mN
F/Eo
F = N + G Eg
Eo Eo
N
Eg /Eo
Material Balance Equation Application
Eo +Efw
F We
=N+ ....STB
Eo + E fw Eo + E fw (12)
Dake1 points out that the right hand side has two unknowns, N and We, and suggest
that the MB in this form is a powerful tool in assessing whether there is a supporting
aquifer or not. He suggest plotting F/(Eo+Efw) vs Np, or time or pressure drop,
∆p. The plot will take different shapes dependant on the energy support . Figure 4
illustrates this.
x C
x x x
x x
x x x
x
x + + + + +
x + + + + + B
Eo + Efw
x+
+
F
O Np, ∆p or time
The examples above in figure 4 give various scenarios as a result of plotting regular
production figures. Dake1 points out that for Curve A, the horizontal line, indicates
In the previous chapter we introduced the topic of the linear form of the MB equation for
a gas reservoir without water drive, the p/z plot. Craft and Hawkins4 in their text gave
a warning of the application of this approach both with respect to neglecting another
possible energy support such as water drive, and using Gp vs. p as against p/z.
In figure 5 below, the plots of Gp vs p or p/z illustrate the different values of gas in
place which can result from the linear extrapolation of the early production pressure
values. Using the Gp vs.p results in an under estimate of the gas in place whereas
neglecting water drive could result in a significant over estimate of gas in place.
5000
4000
Water drive
Pressure or P/Z
3000
Initial gas in place
Volumetric
Gp vs p/z
2000
Volumetric
Gp vs p
1000 Wrong
extrapolation
0
1 2 3 4 5 6 7
Cumulative Production, MMMSCF
Figure 5 Comparison of the Gp vs p/z and p for Volumetric and Water Drive Gas Reser-
voirs. (Craft and Hawkins4)
The material balance equation lends itself to gas field application in part because of
a more uniform pressure across the reservoir, because of the fluid diffusivity, k/φµc.
For gases because of low viscosity, this gives a diffusivity of the order of five times
that of oil. Because of such rapid equilibrium of pressure it is easy to neglect pressure
support from elsewhere. We will examine both the approach of Havlena and Odeh
and the long established p/z approach.
Material Balance Equation Application
Using the approach of Havlena and Odeh to gas reservoirs provides a history matching
approach to get a good estimate of gas initially in place to compare with previous
volumetric calculations and also determine if the reservoir is a volumetric depletion
reservoir or it is also supported with water drive.
Using the approach of fluid production being equal to the expansion of insitu gas +
water and pore expansion + water influx gives the following equation.
Fluid production = gas expansion + water expansion & pore compaction +water
influx
Gp Bg + Wp Bw = G( Bg − Bgi ) + GBgi
(c Sw wc + cf ) ∆p + W
e (13)
1 − Swc
F = Gp Bg + Wp Bw ...res.cu. ft
( )
Eg = Bg − Bgi .....rcf / scf
E fw = Bgi
(c S
w wc + cf ) ∆p.....rcf / scf
1 − Swc (14)
F = G (Eg + Efw) + We
When dealing with gas reservoirs the water and pore compressibility terms can
generally be neglected because of the large gas compressibility.
F = GEg + We (15)
F W
=G+ e
Eg Eg (16)
This provides a useful form to examine the production figures of a gas reservoir.
The figure 6 below illustrates the possible resulting plots from plotting F/Eg vs Gp,
time, or ∆p.
Volumetric depletion
Gp
The three plots demonstrate how that if there is pressure support from an aquifer the
curves deviate from the horizontal line applicable to the volumetric depletion. The
intercept also provides a figure for gas initially in place.
This can be also be a useful tool in gas field development since if water drive by
this means is detected, then it could be some time before there is evidence of water
production. The advancing water will only reveal itself when the gas-water contact
reaches the lower limit of the gas production wells which are generally set high in
the structure. Another consideration is the mobility ratio for water displacing gas.
krw ′ krg ′
M= /
µw µg
Where:
krw′ and krg′ are the end point relative permeabilities for water and gas.
M can be as low as 0.10 for this situation which interprets into the gas moving 100times
faster than the water. Clearly as the water table advances it will trap gas, in the form
of the residual gas saturation, Sgr. The higher the pressure of the reservoir the larger
the amount of gas which will be trapped. Dake1 developed an equation to determine
the volume of gas trapped behind an advancing water drive.
The pore volume, PV, of the reservoir, Vφ can be expressed in terms of the gas in
place.
1
PV = GBgi (18)
1 − Swc
10
Material Balance Equation Application
(1 − Sgr − Swc )
MGV = PV (1 − Sgr − Swc ) = GBgi
1 − Swc (19)
After an influx of WeBw into the gas reservoir, the proportion of this movable gas
volume swept by the water is:
α = We Bw / GBgi
(1 − S gr − Swc ) (20)
1 − Swc
This equation can be used to determine the amount of gas that is being ‘lost’ to
production through an advancing water drive. This can be reduced by depleting the
gas reservoir at a higher rate taking advantage of the higher diffusivity of the gas
relative to that of the water.
Gas produced from Reservoir = Gas initially in reservoir - Gas remaining in reservoir
(scf)
The above does not include the water production term, Wp. We is considered in the
above to be the net water influx.
As stated before the compressibility terms for water and pore are very small and can
be neglected which reduces the equation to:
Gp B WB
= 1 − gi 1 − e w
G Bg GBgi
(23)
From the equation of state and assuming constant temperature the gas formation
factors can be replaced with z/p, and the equation becomes:
Gp
1 −
p pi G
=
z zi We Bw / Bgi
1 −
G (24)
Institute of Petroleum Engineering, Heriot-Watt University 11
The term WeBw/GBgi is the proportion of the gas in place volume invaded by water,
and as can be seen in the equation the higher this proportion the higher the pressure
and vice versa to the extent that with depletion drive where there is no water drive
or compaction drive , the equation simplifies to:
p pi Gp
= 1 −
z zi G (25)
This is the well known linear equation of p/z vs. Gp which when plotted enables
the gas in place, G to be obtained when p/z =0. (Chapter 5.) If there is any other
pressure support the curve will deviate from linear as demonstrated in figure 7 below
by Dake1.
Material balance
pi/zi at abandonment
Strong waterdrive
Depletion
Moderate waterdrive
0
Gp Gp=6
The difficulty with the application of this approach is that in the early time periods,
the pressure support from the aquifer may not be felt and deciding what is the slope
of a linear line may take in points from pressures being supported by water drive,
leading to an over estimate of gas in place.
This danger has been known for some time, but the simplicity afforded by this p/z vs.
Gp plot can readily lead to erroneous interpretation. As pointed out by Dake1, warnings
by respected reservoir engineers of Craft & Hawkins 1959 ,Bruns & Fetkofitch 1965,
Agarwal, Al-Hussainy & Raimey1965, Dake 1978 were taken up by Cason in 1989 ,
when he said, “The theory showing that depletion drive gas reservoirs will exhibit
a straight p/z plot has been developed but the corollary; that a straight line p/z plot
proves the existence of depletion drive has not been proven”
As indicated above the high diffusivity of the gas in contact with an aquifer is such
that if gas is withdrawn at a rate greater than water can encroach into the gas reservoir
then the pressure will decline faster than if the gas production rate is slower enabling
the water drive to replace gas production . This rate effect also distorts the slope of
12
Material Balance Equation Application
the p/z plot for fields supported by water drive. Varying production rate is a common
characteristic in gas field management, where higher gas production in winter periods
will be followed by lower summer production. Clearly all these contributing factors
can offset other phenomena in p/z plots and lead to poor interpretation. The rate
effect is illustrated in figure 8 in the distortion generated with high winter and low
summer production.
p/z
0 Gp
Figure 8 Non Linear Impact of Gas Rate Production For Waterdrive Gas Fields.
Dake1 has also demonstrated how the p/z plot can be used to determine abandonment
conditions. When water drive reaches water break through, than gas will be remaining
in two forms. In the swept portion at a residual gas saturation and in portions by-
passed by the water at the original saturation (1-Swc).
S
Gp = G − αGBgi gr + (1 − α )GBgi / Bgab
1 − Swc (26)
α is the volumetric sweep efficiency at the abandonment, pressure Bgab, gas formation
volume factor at abandonment.
Gp
1 −
p pi G
=
zab zi Sgr 1−α
α +
1 − Swc α (27)
This is plotted on the p/z vs Gp plot of figure 7. It indicates the maximum gas recovery
points for the strong water drive and the moderate water drive. Dake explains further
application of this in his text .
As was discussed in the section on drive mechanisms, a depletion drive reservoir has
two phases of depletion. The first stage above the saturation pressure the undersaturated
state, and that below the saturation pressure from where the drive mechanism gains
its name, solution gas drive.
dV = C x V x ∆p
If we take the full MB equation developed in the previous chapter then the equation
simplifies above the bubble point to:
co =
( Bo − Boi )
Oil compressibility is
Boi ∆p (29)
N p Bo = NBoi co +
(
cw Swc + c f ) ∆p
1 − Ssw
(30)
14
Material Balance Equation Application
Above the bubble point only oil and connate water exists , therefore:
So + Swc = 1
c S + c S + cf
N p Bo = NBoi o o w wc ∆p (31)
1 − Swc
or:
N p Bo = NBoi ce ∆p
(32)
1
ce = (co So + cw Swc + c f )
1 − Swc (33)
N p Boi
= (ce ∆p)
N Bob (34)
The equations above have neglected water production. This may not be the case as
a reduction in pressure could mobilise connate water. The equation in such a case
would be:
During this phase pressure declines very rapidly and the recovery is low. There are
some exceptions to this where the compressibilities of the rock are exceptionally large,
such that the rock compressibility provides the major energy source. Such is the case
in some Venezuelan fields and in the carbonate field of Ekofisk in the North Sea.
As production wells are brought on stream their productivity can put a time line to
the MB calculations of production vs .pressure. The time for the reservoir pressure
to drop to a safe position above the bubble point, then is a basis for determining the
time availability for water injection installations to arrest the pressure decline.
The gas production can come from gas in solution in the reservoir and from production
free gas in the reservoir which has come out of solution.
qg
Free Gas =
Bg
Solution gas = QoRs
where:
qg = free gas flow rate, res.bbls/day
Bg = gas formation-volume factor, bbls/SCF
Qo = oil flow rate, STB/day
Rs = gas solubility SCF/STB
16
Material Balance Equation Application
qg
Qg = + Qo Rs
Bg (37)
where:
qo
Qo =
Bo (38)
where:
qg
+ Qo Rs
Bg
R=
qo / Bo (39)
since:
qo
Qo = (40)
Bo
qg
B
R = g + Rs
qo
Bo (41)
qg and qo are reservoir flow rate where for a radial system in terms of Darcy’s law
gives:
2πkeg h∆p
qg =
µ g ln re / rw
and:
2πkeo h∆p
qo =
µo ln re / rw
2πkeg h∆p
B µ lnr / k
R = g g e w + Rs
2πkeo h∆p
Bo µo lnre / kw
Bo keg µ o
R= + Rs
Bg keo µ g
(42)
When discussing solution gas drive (Chapter 10) we presented the typical shape for
the producing gas-oil ratio curve . The distinctive shape is again reproduced below
and shows the various phases.
Instantaneous
3
GOR
1
Rsi
2
Pb
PNp
Figure 9 Instantaneous Gas - Oil Ratio R For Solution Gas Drive Reservoir.
Examination of this plot in the context of the instantaneous GOR equation is consistent
with its shape. Above the bubble point at 1, there is no free gas, therefore keg is zero,
and R=Rs=Rsi. Once gas comes out of solution from the bubble point, theoretically
there will be a short time when critical gas saturation has not been reached, keg is still
zero but R = Rs < Rsi (point 2). In reality it is unlikely that this reduction in GOR
is seen since the whole reservoir is not at uniform saturation. From 2 to 3 gas has
reached the critical gas saturation, and keg increases with increasing gas saturation,
while correspondingly keo decreases. Gas is very mobile compared to the oil and the
reservoir gives up its free gas while the oil moving at an increasingly slower rate
is depleted of its solution gas. The curve goes through a maximum. To understand
this negative slope is to appreciate that the gas formation factor is increasing with
decreasing pressure. At high pressures the change in Bg is small but at lower pressures
the change is larger and greater than the increasing values of keg and µg.
This instantaneous GOR should not be confused with the cumulative producing GOR,
Rp. or the solution GOR, Rs. The instantaneous GOR, R, is the ratio of the total oil and
gas production rates at a particular moment in time. The cumulative GOR, Rp is the
18
Material Balance Equation Application
ratio of the cumulative gas and cumulative oil produced up to a particular moment
in time. These two GOR's are related to by the following two equations.
Np (42)
Rp = ∫ RdN p
o
ΣRi ∆N pi
Rp =
N p (43)
whereRi is the average instantaneous GOR over the period that ∆Npi of oil was
produced.
( N − N p ) Bo
So =
NBob / (1 − Swi ) (44)
where
Np Bo
So = (1 − ) (1− Swi )
N Bob
(45)
Equation 45 assumes that the oil saturation is uniform throughout the reservoir.
However due to vertical migration of free gas a secondary gas cap may be produced
and another equation may be required.
( N − N p ) Bo (1 − Swi )
So ′ =
NBob (1 − m©)
N p Bo
1 − (1 − Swi )
N Bob
So ′ =
1 − m© (46)
where:
So = oil saturation with oil zone
m = ratio of secondary gas cap to original oil zone size
The big assumption in equation. 46 is that the oil saturation in the secondary gas
cap is zero.
Predicting the future performance of a reservoir is difficult not the least because there
are many uncertainties associated with the reservoir. For example we may not know
the drive mechanism responsible for fluid production which would have an impact
on the parameters included in the MB equation. We may have doubts in relation to
the quality of the laboratory data, such as relative permeability.
The instantaneous gas-oil ratio equation when rearranged in another form can be
used for generating relative permeability data from production data or validating
laboratory generated relative permeability data.
Keg Bµ
= ( R − Rs ) o o
Keo Bg µo (47)
20
Material Balance Equation Application
Production data will provide data for R, and Np as a function of pressure. For each
pressure value, the PVT values would be available from the PVT report. The Np
value will enable the oil saturation to be calculated from the oil saturation equation,
and the corresponding gas/oil relative permeability ratio is calculated from the IGOR
equation (47).
There are a number of procedures which different authors have developed for predicting
performance, they are attributed to Schilthuis, the originator of the MB equation,
Tarner, Tracy& Tarner and Muskatt. The methods focus on the predictions below the
bubble because the predictions above the bubble point are straightforward.
In section 4.1.1. we derived the MB equation for application above the bubble point.
Such calculations need to done separately from those below the bubble point.
Np =
[ ]
N Bo − Bob + ( Rsi − Rs ) Bg + G( Bg − Bgi ) − Gp Bg
Bo − Bg Rs (48)
All the PVT related properties are determined from the PVT report using the reservoir
pressure.
At the start of the prediction the values of the parameters at the bubble point are
known, N,G,Bob,Rsi and Bgi.
Bo keg µo
R= + Rs
Bg keo µ g (42)
N B
So = 1 − p o (1 − Swi )
N Bob (45)
Compare the calculation of Gp with predicted Gp and repeat steps, 2-5 if they do not
agree. When the values agree go to step 1 and set a lower pressure and continue with
the Np and Gp vs pressure decline prediction.
N ( Bo + ( Rsi − Rs ) Bg − Bob ) − N p ( Bo − Rs Bg )
N p Rp = (50)
Bg
NpRp = cumulative gas produced SCF.
.
The procedure is of a trial and error approach using the two independent equations,
instantaneous GOR and MB:
22
Material Balance Equation Application
3. Using the assumed Np solve the oil saturation equation-equation 45 for So.
These enable keg/keo to be determined.
5. Calculate the gas produced during the pressure drop period, i.e.
( Ri + Ri +1 )
N p1 (51)
2
where Ri = instantaneous gas-oil ratio at start of period.
6. The total gas produced from the material balance equation and the instantaneous
GOR equation are compared, and the assumed value of Np adjusted and steps
2-6 repeated until the two values match to within an acceptable error. We then
go to the next step
2 Solve MB for Np2. This is the cumulative gas production at the end of the
second pressure.
N ( Bo + ( Rsi − Rs ) Bg − Bob ) − N p 2 ( Bo − Rs Bg )
G2 = N p 2 Rp 2 − N p1 Rp1 = − N p1 Rp1
Bg
(52)
3 Calculate gas produced during 2nd step by removing from cumulative gas,
gas produced during step 1, G1
4 With the assumed Np2 a value of So is obtained from the saturation equation.
(R i +1 + R i + 2 )
(N p2 − N p1 ) = G 2
2 (53)
7 If G2MB does not compare with G2GOR then a new value of Np2 is assumed, and
so on until convergence as before. By plotting Gmb and GGOR vs Np the point of
convergence can be determined.
The stepwise trial and error procedure is continued until the desired limit is
achieved.
Both of these procedures are not so well suited to computer application and Tracy
modified Tarners method to make it more suited to computer application.
N p (Bo − R s Bg ) + G p Bg − (We − Wp )
N=
Bo − Boi + (R si − R s )Bg + mBoi (Bg − Bgi ) / Bgi (54)
Bo − R s Bg
φn =
Bo − Boi + (R si − R s )Bg + mBoi (Bg − Bgi ) / Bgi (55)
Bg
φg =
Bo − Boi + (R si − R s )Bg + mBoi (Bg − Bgi ) / Bgi (56)
1
φw =
Bo − Boi + (R si − R s )Bg + mBoi (Bg − Bgi ) / Bgi (57)
For simplicity in presenting the equations if we assume no gas cap these become:
Bo − Rs Bg
φn =
Bo − Boi + ( Rsi − Rs ) Bg (58)
Bg
φg =
Bo − Boi + ( Rsi − Rs ) Bg (59)
1
φw =
Bo − Boi + ( Rsi − Rs ) Bg (60)
24
Material Balance Equation Application
These functions φn, φg and φw are only dependent on reservoir pressure and oil
properties, i.e. they can all be obtained from PVT data.
Using the above short-hand system the material balance equation can be written:
This is now a simplified form of the MB equation with the PVT related functions
conveniently grouped together.
Tracy considered two pressure conditions Pj and a lower pressure Pk and the oil
production ∆Np during this pressure interval.
Tracy differs from Tarner in estimating the producing GOR, Rk at the lower pressure
rather than the production ∆Np.
also:
and:
where R′ avg is the estimated average gas-oil ratio between pressure Pj and Pk.
R j + R′ K
′ =
Ravg
i.e. 2 (67)
In Equation 69, the only unknown is R′avg. All the other parameters are uniquely
determined from the PVT data or have been calculated during previous pressure
steps.
Rk can also be estimated if the liquid saturation is known using the instantaneous
GOR equation, equation 42.
Bo keg µo
Rk = + Rs
Bg keo µ g
N B
So = 1 − p o (1 − Sw )
N Boi
Tracy observed that φn, φg and φw are smooth curved and above Pb they have infinite
slope. An example of these curves is given in figure 10.
0.10 100
.08 80
.06 60
.04 40
.02 20
.010 10
.008 8
.006 6
φg φn
.004 4
.002 2
.001 1
0 5 10 15 20 25
Pressure, hundreds of psia
Using the above equations the step by step procedure of Tracy is as follows.
26
Material Balance Equation Application
Tracy’s Procedure
Set pressure step below Pb
(5) Using Σ∆Np. Determine So from saturation equation (equation 45) and thereby
keg/keo from So vs keg/keo data.
(7) Compare Rk with R´k. of step 1. If (Rk - R´k) is greater than tolerance, tolerance
limit set R´k = Rk and repeat steps 1 - 6.
(9) Estimate φg Gp + Np φn. This should be 1.00. If the error is greater than a preset
tolerance, calculations are repeated with an adjustment to Rk'.
Vp So
Nr = N − N p = (70)
Bo
Vp is the pore volume in rb (res bll). The change in this volume is;
dNr 1 dSo S dBo
= Vp − Vp 2o
dp Bo dp B o dp (71)
So Rs V
Gr = Vp + (1 − So − Swc )
Bo Bg (72)
∆G p ∆G r ∆G r / ∆p dG r / dp
R= = = =
∆N p ∆N r ∆N r / ∆p dN r / dp
The producing GOR can also be obtained through the instantaneous GOR equation
Bo keg µo
R= + Rs
Bg keo µ g (42)
Bg dRs
Χ( p) =
Bo dp (78)
1 dBo µo
Υ( p) =
Bo dp µ g (79)
1 dBg
Ζ( p) =
Bg dp (80)
28
Material Balance Equation Application
These are all pressure related functions and can be obtained from PVT data.
oil remaining ( N − N p ) Bo
So = = (1 − Swc )
one PV NBoi (81)
Which gives
Np So Boi
=1−
N 1 − Swc Bo (82)
All the procedures are similar , and are very dependant on reservoir, fluid and rock
data. The quality of a material balance study of a reservoir is related to the quality
of the data. There clearly should be sufficient data both with respect to quantity
and quality as in any simulation study the quality of the output is directly related to
the quality of the input. Another challenge is the definition of the average reservoir
pressure. We will briefly look at these two data perspectives.
Listed below are the data which are used in the various solution gas drive prediction
procedures. Some of them could be adjusted in history matching.
Tarner's method can be used for Gas Cap drive reservoir predictions. The equation
however may need alteration to account for gas coming out of solution migrating
into the gas cap.
B − Bgi
N ( Bo + ( Rsi − Rs ) Bg − Bob ) + mBob g − N p ( Bo − Rs Bg )
Bgi
N p Rp =
Bg
(83)
30
Material Balance Equation Application
In the figure 11 from Dake1 are presented the pressures for equilibrium conditions
and the well positions and boundaries for a non equilibrium condition.
Pressure
Time
pj,qj,Vj
Time
In figure 11 c the wells have their own pressure declines. Dake1 presents a volume
weighting for the pressures within each drainage area.
pj, Vj and qj, represent the pressure , volume and reservoir rate for the drainage area
j.
p = ∑ p j Vj / ∑ Vj
j j (84)
Dake suggests that determination of the volumes of each drainage zone is subjective
and suggests a production rate alternative, based on the time derivative of the
compressibility equation.
dV= cV∆p
dVj dp
= q j = cVj j = cVj p′ j
dt dt (85)
Vj ∝ q j / p′ j
(86)
∑ p q / p′
j
j j j
p=
∑ q / p′ j j
j (87)
Since the material balance is applied at regular periods, say six months , then the
change in fluid withdrawal (UWj) can be used over a pressure drop ∆pj. Then equation
87 can be expressed as:
32
Material Balance Equation Application
∑ p ∆UW / ∆p
j
j j j
p=
∑ UW / ∆p j j
j (88)
Dake comments that some might consider that this averaging approach somewhat
tedious and is better superseded by applying numerical simulation from the start. He
suggests that the two approaches are not exclusive and suggests that the MB approach is
a useful investigation tool prior to structuring a more complex simulation model.
None of the terms in the material balance equation have a time term and therefore,
the equation just provides a volume, pressure result. It does not indicate when, a
cumulative production is obtained at a particular pressure value. To do this we need
to incorporate another method where time is included. The productivity of the wells
could be used and has been suggested by Cole8.
Qo
J= STB / psi
pe − pw (89)
where:
Replacing Qo in equation 89 with the Darcy radial flow equation results in the
following equation:
7.07keo h( pe − pw )
J=
µo ln(re / rw ) Bo ( pe − pw ) (90)
For most reservoir producing conditions, the height and drainage area remains
constant and therefore:
7.07h
= Cons tan t = C
1nre / rw (91)
Combining equations 90 and 91, and eliminating the identical (pe - pw) terms gives:
or:
Jµo Bo
C=
Ko (93)
Once determined, i.e. at initial conditions, this value, c, can then be used for future
performance predictions.
An important aspect in the material balance equation is the water influx term. In the
next chapter we will examine in more detail this topic
Solution to Exercise
EXERCISE:
Data
Initial reservoir pressure pi = 3,000 psi
Reservoir temperature T = 220 ºF
Water compressibility cw = 3.0E-06 psi-1
Rock compressibility cf = 8.6E-06 psi-1
Water saturation Swc = 0.22
Separator pressure Psep = 300 psi
SOLUTION (A)
Separator
Pressure FVF
psig bblb/STB
300 1.495
34
Material Balance Equation Application
Oil Compressibility:
(co So + cw Swc + cf )
co =
(1 − Swc )
ce = 1.58E - 05 psi-1
Np
N
Boi
= ce ∆P
Bob
Np
N =0.0059 or 0.59% of the original oil in place
SOLUTION (B)
Np Bo − Boi
=
N Boi
Np 1.4950 − 1.4844
=
N 1.4844
Np
= 0.00714
N
Np
= 0.00714 or 0.71%
N
5. Tarner,J. “How Different Size Gas Caps and Pressure Maintenance Programs
Affect Amount of Recoverable Oil”, Oil Weekly, June 2,1944 No2 32-34
6. Tracy, G.W. “ Simplified Form of the Material Balance Equation “ Trans AIME
294,243, 1955
36
Water Influx
CONTENTS
Having worked through this chapter the Student will be able to:
• Calculate the total water influx resulting from a known aquifer volume in terms
of total aquifer compressibility and pressure drop over the aquifer.
• Sketch and describe the Schiltuis steady state model and the Van Everdingen
and Hurst Unsteady State Model for Water .
• Explain how a constant boundary pressure profile solution can be used for
declining pressure aquifer/ reservoir pressure.
• Calculate given prerequisite equations the water influx as a function of time for
a declining pressure profile.
• Describe and sketch the short hand linear forms of the MB equation for water
drive reservoirs for :
• Water drive and no gas cap
• Water drive and gas cap
• Describe the above for very small aquifers.
• Describe briefly the application of the Fetkovitch method for water influx
determination.
Water Influx
In the preceding chapters on drive mechanisms and material balance we identified the
positive characteristics of water drive. In this chapter we will examine the various
methods which can be used to predict the amount of natural water drive.
The driving force for water drive comes from the response to pressure being low-
ered as a result of oil production, and since the aquifer is part of this system it also
responds to this declining pressure. As pointed out in the material balance equation
chapter, fluid production is a response to the compressibility of the oil reservoir and
the same is true in most cases for aquifer water drives. The porous system represent-
ing the hydrocarbon reservoir and the aquifer are compressible. All its elements:
hydrocarbon, water, and rock expand as pressure declines. It is on the basis of this
compressibility that water encroachment is understood and calculated.
(a) Expansion of the water due to pressure drop within the aquifer
(d) Artesian flow, if any, where the outcrop is located structurally higher than the
hydrocarbon accumulation, and the water is replenished at the surface
The amount of water flowing into the hydrocarbon reservoir is also influenced by
other factors:
(a) The cross sectional area between the water zone and the hydrocarbon
accumulation
(b) The permeability of the rock in the aquifer
(c) The viscosity of the water
The decline in pressure resulting from oil or gas production moves with a finite velocity
( related to fluid flow) into and through the aquifer. The reduction in pressure causes
the aquifer, water and rock to expand. As long as this moving pressure disturbance
has not reached the external limits of the aquifer, the aquifer will continue to provide
expansion water to the hydrocarbon reservoir. In describing the size of aquifers we
refer to infinite and finite aquifers. Clearly there is not an aquifer which extends to
an infinite extent! The terminology indicates, where in the time considerations of the
analysis, the pressure disturbance has not reached the external limits of the aquifer.
Although natural water drive provides very effective recovery characteristics “there
are still more uncertainties attached to this subject in reservoir engineering, than to
any other. This is simply because one seldom drills wells into an aquifer to gener-
ate reservoir characteristic. Instead these properties have frequently to be inferred
from what has been observed in the reservoir. Even more uncertain is the geometry
and the areal continuity of the aquifer itself. The reservoir engineer should therefore
consult both production and exploration geologists. Due to these inherent uncertain-
ties the aquifer fit obtained from history matching is seldom unique and the aquifer
model may require frequent updating as more production and pressure data becomes
available.” Dake 1978.1
The compression of the void spaces in the reservoir and aquifer rock as a result of
pressure decline in the pore spaces can affect reservoir performance and contribute to
water influx from an aquifer. The compression of the void spaces results in a reduc-
tion in the pore volume of the reservoir as withdrawals continue.
Water Influx
From a practical standpoint it is usually difficult to separate the water expansion from
the rock compression. Therefore, these two effects, which are additive, are usually
combined into one term which, for convenience, is referred to as effective water
compressibility. The compressibility of water, as well as the compressibility of other
liquids, will vary slightly, according to the pressure and temperature imposed on the
water. Increasing the pressure will reduce the compressibility of water and increas-
ing the temperature will increase the compressibility of water. The compressibility
of fresh water at one atmosphere pressure and 60ºF is 3.3 x 10-6 bbl/bbl/psi.
Effective water compressibilities which have been used in reservoir engineering cal-
culations with good results vary from 1.0 x 10-6 bbl/bbl/psi to 1.0 x 10-4 bbl/bbl/psi.
We
OIL AQUIFIER
Wi
Oil/Water
Contact
Assuming no restrictions due to permeability etc. the maximum water influx associated
with an aquifer system, We, the water influx from the aquifer, can therefore be related to
the volume of the aquifer, its effective compressibility and the pressure drop over it.
We = cWi(pi-p) (1)
where:
The main problem facing the reservoir engineer is determining the characteristics of
the aquifer; its geometry, size and flow characteristics.
Exercise 1
(a) Calculate the volume of water an aquifer of 35,000 ft. radius can deliver to
a reservoir of 3,200 ft. radius rock and water compressibility under a 1,100
psi pressure drop throughout the aquifer. assume porosity = 0.22.
(b) Compare the available influx with the initial hydrocarbon volume of the
reservoir.
Data:
Aquifer radius rw = 35,000 ft.
Reservoir radius ro = 3,200 ft.
Water compressibility cw = 3.0E-06 psi-1
Rock compressibility cf = 5.0E - 06 psi-1
Reservoir thickness h = 45ft.
Pressure drop ∆P = 1,100 psi
Porosity φ = 0.22
Water saturation swc = 0.25
(i)
(ii)
Pressure
OIl
RESERVOIR AQUIFER
Wellbore
Reservoir outer boundary Aquifer Outer Boundary
Water Influx
Before examining the different models we will review the development of equations
which enable the pressure, time and distance solution to be obtained
+ dp
pp
q
Well
dq
+
q
r dr
The flow rate at any radius r + dr is q, Figure 3. The rate of flow at radius r will be
larger by the amount dq caused by:
(ii) Expansion of the fluid in the element due to pressure changing with time dp/
dt
Volume of element
V = 2πrh φ dr (2)
dV = - cV dp = - c 2πrh φ dr dp (3)
dv ∂p
dq = = 2πrh φcdr
dt ∂t (4)
∂q ∂p
= − 2πrh φc
∂r ∂t (5)
2πrkh ∂p
q=−
µ ∂r (6)
∂q −2πkh ∂ 2 p ∂p
= r ∂r 2 + ∂r
∂r µ (7)
dp −2πkh d 2 p dp
−2πrh φc = r dr 2 + dr
dt µ
Dividing by r gives
∂ 2 p 1 ∂p µcφ ∂p
+ =
∂r 2 r ∂r k ∂t
(8)
This is the diffusivity equation and describes the flow of a slightly compressible
fluid in porous media. The pressure with respect to distance and time is related by
the parameters φ,µ,c and k.
1 φµc
=
η k (9)
∂ 2 p 1 ∂p 1 ∂p
+ =
∂r
2
r ∂r η ∂t (10)
The name diffusivity equation comes from its application to the flow or diffusion of
heat. The equation is also applied in range of flow systems, heat, electricity as well
as flow in porous media as is the application in a reservoir situation.
All aquifers are finite in size, however there is a period of time when a pressure dis-
turbance created by production from a well has not travelled far enough and reached
the boundary of the aquifer. During this time the aquifer behaves as being infinite
and unsteady state flow applies. After the boundary influences the behaviour of the
system pseudosteady state flow starts.
The diffusivity equation demonstrates that the states of flow are influenced by the
initial conditions and the boundaries, the outer boundary having a significant influence.
In analysing behaviour, the two boundary conditions must be specified: the inner
boundary the oil-water interface, and the outer boundary the limit of the aquifer.
Conditions may be constant pressure, constant rate, closed boundary etc. The initial
condition describes the condition of the system at time, t=0, where a uniform pressure
distribution exists. To solve the equation for water encroachment we need to specify
the boundary and initial conditions.
In general for water influx calculations, the most common conditions are a closed
system, no flow at the outer boundary of the aquifer and constant rate or constant
pressure at the inner boundary. In general constant pressure is used in aquifer model-
ling, whereas in reservoir behaviour constant rate is assumed at the inner, well bore
boundary.
We will now consider the various aquifer models in light of the above discussion.
Pi p
Sand-filled pipe
Production
Aquifer
In this model, the aquifer tank pressure remains constant, and could represent an artesian
type aquifer recharged with water or an aquifer large compared to the reservoir.
The reservoir is considered to be relatively small in size with high permeability such
that a flat pressure profile exists. The relative sizes should be at least 10-20:1 .
C is the aquifer constant and contains the unchanging components of Darcy’s Law,
(units vol/time/pressure).
dWe
= C( pi − p)
dt (12)
(pi-p) is the boundary pressure drop .
Hurst(3) in 1943 proposed an equation which recognised that at least part of the aquifer
flow was transient.
dWe c1 ( pi − p)dt
=
dt log e at (13)
10
Water Influx
in water influx modelling. Before developing the equations we will consider the
hydraulic analogue of Craft & Hawkins2.Figure 5.
pi
p4 p3 p2
p1
p
Aquifer Reservoir
t
We = B∑ ∆pQ(t )
o
where B is the water influx constant in barrels per pounds per day per square inch,
∆p is the pressure decrement in pounds per square inch, and Q(t) is the dimension-
less water influx, which is a function of the dimensionless time. This equation will
be discussed later.
The unsteady-state hydraulic analog is shown above where the reservoir tank on
the right is connected to a series of tanks of increasing size which are connected by
sand-filled pipes of constant diameter and sand permeability, but of decreasing length
between the larger tanks. Initially all tanks are at a common level or pressure Pi rep-
resenting the original pressure across the system As production occurs, the pressure
in the reservoir tank drops causing water to flow from aquifer tank 1, and a resulting
lower pressure in tank 1. This pressure drop in tank 1 in turn generates flow from
tank 2, and so on. Clearly the pressure drops in the aquifer tanks are not uniform
but vary with time and production rate, and are progressive across the reservoir. An
illustration of these pressure profiles in a radial aquifer are shown in figures 6 and
7 for a constant rate of water influx and for a constant boundary pressure. Even if
there is an infinite number of aquifer tanks, it is evident that reservoir pressure can
never fully stabilise at constant production rate, because an ever-increasing portion
of the water influx must come from an ever-increasing distance.
10
T
=
T
00
=1
P1 T 00
10
T=
P10
P100
P1000
RW RE
Radius
Figure 6 Pressure Distributions in an aquifer at several time periods, for a constant rate of
water influx.2
T = 0 = Time
P1
Reservoir Pressure
10
1
=
T=
T 0
10
T=
1 000
T=
T = ∞∞
P
RW RE
Radius
The water analog can also be illustrated by the series of concentric circles in the figure
8 below . The figure represents the cylindrical elements in an aquifer surrounding a
circular reservoir. An analysis of the pressure in each element will enable the amount
of expansion of water each element can produce as a result of effective compress-
ibility in a pressure decline from pi to zero.
12
Water Influx
Reservoir 1 2 3 4
Aquifer
Reservoir 1 2 3 4
3. PERFORMANCE PREDICTION
Although a consideration of the nature of various aquifers lends support to the ana-
lytical expressions presented above, there is no certainty beforehand that any one of
the three will adequately represent the water influx into a particular reservoir, and
studies must be made to determine the most suitable expression.
For active water drive reservoirs, the use of steady state water influx equation will
not usually result in reliable predictions of reservoir performance. As the pressure
drop due to water expansion moves out further into the aquifer, the expanding water
will not move into the hydrocarbon reservoir at the same rate, because for a given
pressure drop the water has to move a greater distance in order to enter the oil or gas
zone. The favoured approach of analysis is the unsteady state model of Van Everdingen
& Hurst (4).
The diffusivity equation in radial form expresses the relation between pressure and
radius and time for a radial system such as drainage from an aquifer, where the driving
potential of the system is the water expandability and the rock compressibility :
This diffusivity equation is the same basic equation as has been used to calculate
heat flow and electrical flow, as well as fluid flow through porous media. The term
( is usually defined as the diffusivity constant (η) and will be essentially constant
for any given reservoir.
where:
k
η=
µφc
(a) The Pressure case, where the pressure at the inside boundary is known and
the outside boundary is closed, or the reservoir is infinite; and we want to calculate
the water influx.
(b) The Rate case, where the rate is known at the inside boundary. At the outside
boundary there is no flow or the pressure is constant or the reservoir is infinite, and
we want to calculate the total pressure drop.
To enable their analysis to be applicable for different reservoirs they produced a more
general solution of the diffusivity equation by generating dimensionless functions.
Dimensionless time, tD, in place of real rime, t, and dimensional radius, rD, which is
re/ro where re is the radius of the aquifer and ro is the radius of the oil reservoir.
1 δ δpD δpD
rD =
rD δrD δrD δrD (14)
where:
kt r 2πkh∆p
tD = , rD = e , pD =
µφcro2
ro qµ (15)
Since only basic equations have been utilised, the units for the above quantities
are:
tD = time, dimensionless
t = time, seconds
k = permeability, darcy
14
Water Influx
µ = viscosity, centipoise
φ = porosity, fraction
c = effective aquifer compressibility, vol/vol/atmosphere
ro = reservoir radius, centimetres
kt
t D = 6.323 x10 −3
µφcro2 (16)
The solution of equation 14 with constant terminal rate boundary conditions is used
in well testing. Hurst and Van Everdingen also derived the constant terminal pressure
solution which is used in water influx calculations.
qµ
q D (t D ) =
2πkh∆p (17)
It is the change in rate from zero to q due to a pressure drop ∆p applied at the outer
hydrocarbons reservoir boundary, ro , at time t=0.
t t
µ D
dt
2πkh ∆p ∫ qdt = ∫ q D (t D )
dt
dt D
o o D (18)
We µ φ µcro2
= Q( t )
2πkh∆p k (19)
Where:
We = cummulative water influx
Q(t) = dimensionless water influx
(20)
This equation gives the cumulative water influx for a fixed pressure drop ∆p. The
equation applies in Darcy units. However, when oilfield units are used the following
equation applies.
2πφcro2 h∆pQ( t )
We = = 1.119φcro2 h∆pQ( t )
5.615 (21)
where;
We = B x ∆pQ(t)
(22)
where:
B = 1.119φcr2oh (23)
Van Everdingen and Hursts’ paper presented the solution of equation 8 in the form
of dimensionless time, tD, and dimensionless water influx Q(t). Their solution of
the diffusivity equation can therefore be applied to any reservoir where the flow of
water into the reservoir is essentially radial in nature. They provided solutions for
external boundaries of an infinite extent and for those of limited extent. Tables 1 and
2 show the tabulated form of their solutions. In addition to being presented in tabular
form the dimensionless water influx as a function of dimensionless time, Dake1 also
reproduced Van Everdingen & Hurst data solutions in graphical form. The graphs
are presented in figures, 8a-e
16
Water Influx
Dimensionless water influx and dimensionless pressures for infinite radial aquifers (courtesy of SPE)3
tD Qt pD tD Qt tD Qt tD Qt
1.0 x 10-2 0.112 0.112 1.5 x 103 4.136 x 102 1.5 x 107 1.828 x 106 1.5 x 1011 1.17 x 1010
5.0 x 10-2 0.278 0.229 2.0 x 103 5.315 x 102 2.0 x 107 2.398 x 106 2.0 x 1011 1.55 x 1010
1.0 x 10-1 0.404 0.315 2.5 x 103 6.466 x 102 2.5 x 107 2.961 x 106 2.5 x 1011 1.92 x 1010
1.5 x 10-1 0.520 0.376 3.0 x 103 7.590 x 102 3.0 x 107 3.517 x 106 3 . 0 x 1 0 11 2.29 x 1010
2.0 x 10-1 0.606 0.424 4.0 x 103 9.757 x 102 4.0 x 107 4.610 x 106 4.0 x 1011 3.02 x 1010
2.5 x 10-1 0.689 0.469 5.0 x 103 11.88 x 102 5.0 x 107 5.689 x 106 5.0 x 1011 3.75 x 1010
3.0 x 10-1 0.758 0.503 6.0 x 103 13.95 x 103 6.0 x 107 6.758 x 106 6.0 x 1011 4.47 x 1010
4.0 x 10-1 0.898 0.564 7.0 x 103 15.99 x 103 7.0 x 107 7.816 x 106 7.0 x 1011 5.19 x 1010
5.0 x 10-1 1.020 0.616 8.0 x 103 18.00 x 103 8.0 x 107 8.866 x 106 8.0 x 1011 5.89 x 1010
6.0 x 10-1 1.140 0.659 9.0 x 103 19.99 x 103 9.0 x 107 9.911 x 106 9.0 x 1011 6.58 x 1010
7.0 x 10-1 1.251 0.702 1.0 x 104 21.96 x 102 1.0 x 108 10.95 x 106 1.0 x 1012 7.28 x 1010
8.0 x 10-1 1.359 0.735 1.5 x 104 3.146 x 103 1.5 x 108 1.604 x 107 1.5 x 1012 1.08 x 1011
9.0 x 10-1 1.469 0.772 2.0 x 104 4.679 x 103 2.0 x 108 2.108 x 107 2.0 x 1012 1.42 x 1011
1.0 1.570 0.802 2.5 x 104 4.991 x 103 2.5 x 108 2.607 x 107
1.5 2.032 0.927 3.0 x 104 5.891 x 103 3.0 x 108 3.100 x 107
2.0 2.442 1.020 4.0 x 104 7.634 x 103 4.0 x 108 4.071 x 107
2.5 2.838 1.101 5.0 x 104 9.342 x 103 5.0 x 108 5.032 x 107
3.0 3.209 1.169 6.0 x 104 11.03 x 104 6.0 x 108 5.984 x 107
4.0 3.897 1.275 7.0 x 104 12.69 x 104 7.0 x 108 6.928 x 107
5.0 4.541 1.362 8.0 x 104 14.33 x 104 8.0 x 108 7.865 x 107
6.0 5.148 1.436 9.0 x 104 15.95 x 104 9.0 x 108 8.797 x 107
7.0 5.749 1.500 1.0 x 105 17.56 x 104 1.0 x 109 9.725 x 107
8.0 6.314 1.556 1.5 x 105 2.538 x 104 1.5 x 109 1.429 x 108
9.0 6.861 1.604 2.0 x 105 3.308 x 104 2.0 x 109 1.880 x 108
1.0 x 101 7.417 1.651 2.5 x 105 4.066 x 104 2.5 x 109 2.328 x 108
1.5 x 101 9.965 1.829 3.0 x 105 4.817 x 104 3.0 x 109 2.771 x 108
2.0 x 101 1.229 x 101 1.960 4.0 x 105 6.267 x 104 4.0 x 109 3.645 x 108
2.5 x 101 1.455 x 101 2.067 5.0 x 105 7.699 x 104 5.0 x 104 4.510 x 108
3.0 x 101 1.681 x 101 2.147 6.0 x 105 9.113 x 104 6.0 x 109 5.368 x 108
4.0 x 101 2.088 x 101 2.282 7.0 x 105 10.51 x 105 7.0 x 109 6.220 x 108
5.0 x 101 2.482 x 101 2.388 8.0 x 105 11.89 x 105 8.0 x 109 7.066 x 108
6.0 x 101 2.860 x 101 2.476 9.0 x 105 13.26 x 105 9.0 x 109 7.909 x 108
7.0 x 101 3.228 x 101 2.550 1.0 x 106 14.62 x 105 1.0 x 1010 8.747 x 108
8.0 x 101 3.599 x 101 2.615 1.5 x 106 2.126 x 105 1.5 x 1010 1.288 x 109
9.0 x 101 3.942 x 101 2.672 2.0 x 106 2.781 x 105 2.0 x 1010 1.697 x 109
1.0 x 102 4.301 x 101 2.723 2.5 x 106 3.427 x 105 2.5 x 1010 2.103 x 109
1.5 x 102 5.980 x 101 2.921 3.0 x 106 4.064 x 105 3.0 x 1010 2.505 x 109
2.0 x 102 7.586 x 101 3.064 4.0 x 106 5.313 x 105 4.0 x 1010 3.299 x 109
2.5 x 102 9.120 x 101 3.173 5.0 x 106 6.544 x 105 5.0 x 1010 4.087 x 109
3.0 x 102 10.58 x 101 3.263 6.0 x 106 7.761 x 105 6.0 x 1010 4.868 x 109
4.0 x 102 13.48 x 101 3.406 7.0 x 106 8.965 x 105 7.0 x 1010 5.643 x 109
5.0 x 102 16.24 x 101 3.516 8.0 x 106 10.16 x 106 8.0 x 1010 6.414 x 109
6.0 x 102 18.97 x 101 3.608 9.0 x 106 11.34 x 106 9.0 x 1010 7.183 x 109
7.0 x 1022 21.60 x 101 3.684 1.0 x 107 12.52 x 106 1.0 x 1011 7.948 x 109
8 .0 x 1 0 24.23 x 101 3.750
9.0 x 102 26.77 x 101 3.809
1.0 x 103 29.31 x 101 3.860
1.0 x 10-1 0.395 1.75 x 10-1 0.553 3.5 x 10-1 0.829 8.0 x 10-1 1.363 2.00 2.427 3.00 3.170 5.0 4.454
1.1 x 10-1 0.414 2.00 x 10-1 0.597 4.0 x 10-1 0.897 9.0 x 10-1 1.465 2.20 2.574 3.25 3.334 5.5 4.727
1.2 x 10-1 0.431 2.25 x 10-1 0.638 4.5 x 10-1 0.962 1.00 1.563 2.40 2.715 3.50 3.493 6.0 4.986
1.3 x 10-1 0.446 2.50 x 10-1 0.678 5.0 x 10-1 1.024 1.25 1.791 2.60 2.849 3.75 3.645 6.5 5.231
1.4 x 10-1 0.461 2.75 x 10-1 0.715 5.5 x 10-1 1.083 1.50 1.997 2.80 2.976 4.00 3.792 7.0 5.464
1.5 x 10-1 0.474 3.00 x 10-1 0.751 6.0 x 10-1 1.140 1.75 2.184 3.00 3.098 4.25 3.932 7.5 5.684
1.6 x 10-1 0.486 3.25 x 10-1 0.785 6.5 x 10-1 1.195 2.00 2.353 3.25 3.242 4.50 4.068 8.0 5.892
1.7 x 10-1 0.497 3.50 x 10-1 0.817 7.0 x 10-1 1.248 2.25 2.507 3.50 3.379 4.75 4.198 8.5 6.089
1.8 x 10-1 0.507 3.75 x 10-1 0.848 7.5 x 10-1 1.229 2.50 2.646 3.75 3.507 5.00 4.323 9.0 6.276
1.9 x 10-1 0.517 4.00 x 10-1 0.877 8.0 x 10-1 1.348 2.75 2.772 4.00 3.628 5.50 4.560 9.5 6.453
2.0 x 10-1 0.525 4.25 x 10-1 0.905 8.5 x 10-1 1.395 3.00 2.886 4.25 3.742 6.00 4.779 10 6.621
2.1 x 10-1 0.533 4.50 x 10-1 0.932 9.0 x 10-1 1.440 3.25 2.990 4.50 3.850 6.50 4.982 11 6.930
2.2 x 10-1 0.541 4.75 x 10-1 0.958 9.5 x 10-1 1.484 3.50 3.084 4.75 3.951 7.00 5.169 12 7.208
2.3 x 10-1 0.548 5.00 x 10-1 0.982 1.0 1.526 3.75 3.170 5.00 4.047 7.50 5.343 13 7.457
2.4 x 10-1 0.554 5.50 x 10-1 1.028 1.1 1.605 4.00 3.247 5.50 4.222 8.00 5.504 14 7.680
2.5 x 10-1 0.559 6.00 x 10-1 1.070 1.2 1.679 4.25 3.317 6.00 4.378 8.50 5.653 15 7.880
2.6 x 10-1 0.565 6.50 x 10-1 1.108 1.3 1.747 4.50 3.381 6.50 4.516 9.00 5.790 16 8.060
2.8 x 10-1 0.574 7.00 x 10-1 1.143 1.4 1.811 4.75 3.439 7.00 4.639 9.50 5.917 18 8.365
3.0 x 10-1 0.582 7.50 x 10-1 1.174 1.5 1.870 5.00 3.491 7.50 4.749 10 6.035 20 8.611
3.2 x 10-1 0.588 8.00 x 10-1 1.203 1.6 1.924 5.50 3.581 8.00 4.846 11 6.246 22 8.809
3..4 x 10-1 0.594 9.00 x 10-1 1.253 1.7 1.975 6.00 3.656 8.50 4.932 12 6.425 24 8.968
3.6 x 10-1 0.599 1.00 1.295 1.8 2.022 6.50 3.717 9.00 5.009 13 6.580 26 9.097
3.8 x 10-1 0.603 1.1 1.330 2.0 2.106 7.00 3.767 9.50 5.078 14 6.712 28 9.200
4.0 x 10-1 0.606 1.2 1.358 2.2 2.178 7.50 3.809 10.00 5.138 15 6.825 30 9.283
4.5 x 10-1 0.613 1.3 1.382 2.4 2.241 8.00 3.843 11 5.241 16 6.922 34 9.404
5.0 x 10-1 0.617 1.4 1.402 2.6 2.294 9.00 3.894 12 5.321 17 7.004 38 9.481
6.0 x 10-1 0.621 1.6 1.432 2.8 2.340 10.00 3.928 13 5.385 18 7.076 42 9.532
7.0 x 10-1 0.623 1.7 1.444 3.0 2.380 11.00 3.951 14 5.435 20 7.189 46 9.565
8.0 x 10-1 0.624 1.8 1.453 3.4 2.444 12.00 3.967 15 5.476 22 7.272 50 9.586
2. 0 1.468 3.8 2.491 14.00 3.985 16 5.506 24 7.332 60 9.612
2.5 1.487 4.2 2.525 16.00 3.993 17 5.531 26 7.377 70 9.621
3.0 1.495 4.6 2.551 18.00 3.997 18 5.551 30 7.434 80 9.623
4.0 1.499 5.0 2.570 20.00 3.999 20 5.579 34 7.464 90 9.624
5.0 1.500 6.0 2.599 22.00 3.999 25 5.611 38 7.481 100 9.625
7.0 2.613 24.00 4.000 30 5.621 42 7.490
8.0 2.619 35 5.624 46 7.494
9.0 2.622 40 5.625 50 7.497
10.0 2.624
18
Water Influx
4.0
reD = 3.5
= 3.0
reD = 4.0 r eD
3.5 reD =∞
3.0
reD = 2.5
Qt 2.5
2.0
reD = 2.0
1.5
1.0
reD = 1.5
0.5
0
0 1.0 2.0 3.0 4.0 5.0 6.0 7.0
tD
Figure 9a Graphical form of Qt vs. tD for infinite and finite reservoirs. Dake1.
16
14
Qt reD = 5.0
12
reD = 4.5
10
8 reD = 4.0
6 reD = 3.5
reD = 3.0
4
0
0 10 20 30 40 50 60 70
tD
Figure 9b Graphical form of Qt Vs. to for infinite and finite reservoirs. Dake1.
20
Water Influx
110
100
∞
=
D
re
90
80
70
Qt
60
50 reD = 10.0
reD = 9.0
40
reD = 8.0
30
reD = 7.0
20 reD = 6.0
reD = 5.0
10
0
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300
tD
Figure 9c Graphical form of Qt Vs. to for infinite and finite reservoirs. Dake1
reD = 5.0
reD = 4.5
10 102
8 reD = 4.0
reD = 3.5
6
reD = 3.0
4
reD = 2.5
2
USE THIS SCALE FOR LINE reD = ∞ (CONT'D) ONLY
1 10
10 2 4 6 8 102 2 4 6 8 103 2 4 6 8 104
tD
103
102
10
6 8 104
reD = 10.0
reD = 9.0
reD = 8.0
reD = 7.0
reD = 6.0
reD = 5.0
reD = 4.5
reD = 4.0
reD = 3.5
reD = 3.0
reD = 2.5
4
)
reD = 15.0
t'd
on
(c
∞ =
2
r eD
6 8 103
tD
4
∞ =
r eD
2
6 8 102
4
2
10
102
10
1
8
6
4
8
6
4
2
Qt
Figure 8e Graphical form of Qt Vs. to for infinite and finite reservoirs. Dake1.
Although the solutions are for a radial system the solution can be applied where the
influx is not full radial but can be considered a segment of such. One of the simplest
modifications which can be made is to determine the fraction of a circular area through
which water is encroaching, and the equation is modified to:
B = 1.119φcr2ohf
(24)
where:
OWC
Reservoir
The graphical solutions demonstrate clearly the finite time it takes for a pressure
disturbance to reach the limit of the aquifer, when in the figures Q(t) becomes con-
stant. Dake' has indicated that this maximum value of Q(t) depends on the size of the
aquifer and is equal to:
It is significant to note that when these values for Q(t) are put in equation 20 for a full
radial system the following expression results.
r2 − r2
We = 2πφhcro2 ∆p0.5 e 2 o = π (re2 − ro2 )hφc∆p
ro (27)
Examination of this equation indicates that it is the total water influx resulting from
from the ∆p being instantly communicated throughout the aquifer.
Clearly for infinite acting radial aquifers there is no maximum Q(t) value since the
effect of the pressure drop is continually moving out into the aquifer. For an infinite
linear aquifer there is no plot of Q(t). The water influx can be directly calculated us-
ing the equation below: Dake1.
φkct
We = 2 hw × ∆p
πµ (ccs) (28)
24
Water Influx
φkct
We = 3.26 × 10 −3 hw × ∆p
π (bbls) (29)
Table 3 lists the summary of expressions for Hurst and Van Everdingen for both
radial and linear system1.
Radial Geometry
ro re
θ f= θ/390°
Oil reservoir
Aquifer
Linear Geometry
Aquifer
Oil reservoir
h
L
B = 2 π φ hcr 02 f constant
+ - hrs. = 0.000264
t - days = 0.00634
t - yrs. = 2.309
B = 1.119 fφhcr 20 (bbl/psi)
LINEAR SYSTEM kt kt
tD = t D = const ×
φµcL2 φµcr 20 → L2
w = width of aquifer
L = length of aquifer const. as for radial system
boundary to oil B = wLhφc B = 0.1781wLhφc
reservior boundary t in secs and B in cc/atm B-bbls/psi
Table 3 Summary of equations and constants for Van Everdingen and Hurst water influx
model
EXERCISE 2
Calculate the water influx at the end of 1,2 and 5 years into a circular reservoir with
an aquifer of infinite extent. Effective water permeability is 120md, water viscosity
is 0.8 cp, effective water compressibility is 1.0 x 10-6 bbl/bbl/psi, the radius of the
reservoir is 2,400 ft. reservoir thickness is 35ft, porosity is 22%, initial reservoir
pressure is 4,500 psig and present reservoir pressure is 4,490 psig.
Data Table 1
26
Water Influx
This example shows that for a given pressure drop, doubling the time interval will
not double the water influx. It also shows how to calculate the water influx as a result
of a single pressure drop.
3800
Pi 3780
P1
3750
3750
P2
3700
Boundary Pressure - Psia
3700
P3
3650
3620
P4
3600
3550
3500
0 3 6 9 12 15 18 21 24
Time - Months
Van Everdingen & Hurst proposed a method of calculating the results of a series
of successive pressure drops and adding the solutions together. By superimposing
the effects of a series of fixed pressure drops a steady declining pressure can be
simulated.
The method is illustrated in the Figure 12 where the progressive impact of a series
of fixed pressure drops is illustrated.
Pi Pi Pi
T=0
We0 = 0
owc owc
Pi P1 P1 P1 Pi
T=T
1
We1 = B∆P1QT1
owc owc
Pi P1 P2 P2 P2 P1 Pi
T=T
2
We1 = B∆P1QT2
We2 = B∆P2Q(T2-T1)
We = We1+We2
owc owc
Pi P1 P2 P3 P3 P3 P2 P1 Pi
T=T
3
We(1) = B∆P1Q(T3)
We(2) = B∆P2Q(T3-T1)
We(3) = B∆P3Q(T3-T2)
We = We1+We2+We3
In order to use the unsteady state method it is necessary to assume that the boundary
reservoir pressure declines in a series of steps. For example in figure 12 above it
is assumed that at the end of the first time period T1 the pressure at the reseservoir
aquifer boundary drops suddenly from pi to p1. It is further assumed that the pressure
stays constant for another time period, at the end of which it again drops suddenly
throughtout at the reservoir aquifer boundary to p2. These stepwise decreases in
reservoir pressure are continued for the length of time desired in the water influx
calculations.
If the boundary pressure in the reservoir is suddenly reduced from pi to p1, a pres-
sure drop, will be imposed across the aquifer. Water will continue to expand and
the new reduced pressure will continue to move outward into the aquifer. Given a
sufficient length of time the pressure at the outer edge of the aquifer will finally be
reduced to p1.
If some time after the boundary pressure has been reduced to p1 a second pressure
p2 is suddenly imposed at the boundary, a new pressure wave will begin moving
28
Water Influx
outward into the aquifer as a result of the decompression resulting from the second
pressure drop, decompressing further that decompressed from the first pressure drop.
This new pressure wave will also cause water expansion and therefore encroachment
into the reservoir. However, this new pressure drop will not be pi - p2 but will be
p1 - p2. This second pressure wave will be moving behind the first pressure wave.
Just ahead of the second pressure wave will be the pressure at the end of the first
pressure drop, p1.
Since these pressure waves are assumed to occur at different times, they are entirely
independent of each other. Thus, water expansion will continue to take place as a
result of the first pressure drop, even though additional water influx is also taking
place as a result of one or more later pressure drops. In order to determine the total
water influx into a reservoir at any given time, it is necessary to determine the water
influx as a result of each successive pressure drop which has been imposed on the
reservoir and aquifer.
The aquifer term, B is usually a constant for a given reservoir. Thus where the wa-
ter influx must be calculated for several different pressure drops, each of which has
been effective for varying lengths of time, instead of calculating the water influx for
each pressure step, the total water influx as a result of all the pressure steps can be
calculated as follows:
We = B(Σ∆p x Q(t))
(30)
P3
∆P4 = 1/2 (P2 - P4)
∆P4
∆P5
P4
0 1 2 3 4 5
TIME PERIODS
Rather than use the entire pressure drop for the first period a better approximation is
to consider that one half of the pressure drop, eg. 1/2 (pi - p1), is effective during the
entire first period. For the second period the effective pressure drop then is one-half
of the pressure drop during the first period, 1/2 (pi - p1) plus one-half of the pressure
drop during the second period, 1/2 (p1 - p2), which simplifies to:
1
/2 (pi - p1) + 1/2 (p1 - p2) = 1/2(pi - p2)
Similarly, the effective pressure drop for use in the calculations for the third period
would be one-half of the pressure drop during the second period, 1/2(p1 - p2) plus
one-half of the pressure drop during the third period, 1/2 (p2 - p3), which simplifies to
1
/2 (p1 - p3). The time intervals must all be equal in order to preserve the accuracy
of these modification.
EXERCISE 3
Data Table 1
Reservoir Radius ro = 2,500 ft
Effective water permeability kw = 105 mD
Effective water compressibility ce = 1.0E06 psi-1
Water viscosity µw = 0.8 cp
Reservoir thickness h = 30 ft
Porosity φ = 0.22
Initial reservoir pressure Pi = 3,800 psi
Water encroaching factor f = 1
30
Water Influx
0 3,800
1 6 3,780
2 12 3,750
3 18 3,700
4 24 3,620
i.e. as figure 11.
We have seen that water influx is due to expansion of the aquifer water and rock
as a result of a decline in pressure. Simply, a drop in reservoir pressure due to fluid
production is transmitted through the aquifer and the compressibility of the water
albeit small causes the water to expand and flow into the hydrocarbon reservoir.
As we have seen this equation is generally not sufficient to describe water influx be-
haviour, in particular for reasonably sized aquifers, because of the finite time required
for the pressure effect to be felt throughout the aquifer. In water influx calculations
therefore it is necessary to include this time dependency as a result of fluid flow. In
Havlena and Odeh’s paper they recognise this time dependant water influx perspec-
tive, where they use the dimensionless water influx term to express We, ie:
We = B∑∆pQt (30)
F
=N+B
∑ ∆pQt
Eo Eo (32)
∑ ∆pQt
A plot of F/Eo vs. Eo should give a straight line, as shown in figure 14.
∑∆pQt
correct
B
all
sm
incorrect
geometry
o
to
e
W
F/Eo
large
too
We
F ∑∆pQt
= N+B
Eo Eo
∑∆pQt
Eo
Eo
This line will be straight if the aquifer characteristics, B, and the radius of the aquifer
are correct. The intercept will be the oil in place, N, the slope B. Havalena and Odeh
suggest four other plots. Complete scatter, suggesting the calculations or basic data
is in error. A systematically upward or downward curve suggesting that ∑∆pQtDis
too small or to large. This means that re/ro and/or td is too small or too large . An S
shaped curve indicates that a better fit might be obtained by assuming linear water
influx.
Once the assumed values give realistic behaviour then the model, which has been
obtained by history matching, can be applied in predicting future reference reser-
32
Water Influx
voir performance. The assumption is taken in such a situation that the reservoir and
associated aquifer continue to behave as before. Because of this large assumption
and that no aquifer model is likely to be unique the validity of the model should be
updated as more pressure and production data becomes available.
This linearisation approach has been used as a means of determining the extent of
a supporting aquifer. In figure 15 below the curves show the results for a range of
dimesionless radius’s for a field in the North Sea6. The results show a straight line
fit with an infinite aquifer.
16
x o x
14 x o x
Re/Rw = 10 xx Re/Rw = 20 o x
o x
o x
12 x xx
oo x
NPBo + Wρ - Wi , MMMSTB
x o xx x
x xx x oxxx Re/Rw = 00
10
x x x
x xx
x xx
8 x
x x x
Eo
x x
x x
x x
6 x x
xx
x
x x
x
4 x
xx
x
x x
x
x
x
2 x
x x
6 MONTHS
0
∑∆PQ(Qt) , MMPSI
Eo
We = B′ ∆p′ (33)
where: ∆p′ = pi - p
and: B′ = Wicw
(34)
Wi is the water volume in the aquifer.
F ∆p′
= N + B′
Eo Eo (35)
B'
F/Eo
F ∆p'
= N + B'
Eo Eo
∆p'
Eo
4.2. Water Drive, Gas Cap of Known Size
Using a similar approach to the treatment of We as before and applying it to a gas
cap drive reservoir Havlena and Odehs’ equation yields:
F
=N+B
∑ ∆pQt
Eo + mEg Eo + mEg (36)
where We = B∑∆pQ
Again this gives a straight line function ,figure 16 to 17, if the geometry of the aquifer
and time are assumed correct. If the line is not straight then assumptions regarding
the aquifer need to be modified as for water drive systems without a gas cap.
34
Water Influx
F
Eo + mEg
F ∑∆pQt
=N+B
Eo + mEg Eo + mEg
N
∑∆pQt
Eo + mEg
Figure 18 Havalena and Odeh plot for water drive and known gas cap
F ∆p′
= N + B′
Eo + mEg Eo + mEg (37)
F
= N + B′
Eo + mEg Eo + mEg
N
Eo + mEg
Figure 19 Havlena and Odeh plot for small aquifers and known gas cap
With both the water influx equation and the material balance equation there are
two unknowns, We and pressure. It is therefore not straightforward to predict future
performance. However by combining the material balance equation with the unsteady
state equation we are able to predict reservoir performance, but it is a tedious trial
and error approach.
(1) Collect all the available reservoir production and subsurface sample data.
(2) Using the production data we can calculate a value for the aquifer constant, B,
in the unsteady state equation.
In order to determine the validity of the constant B it is important to carry this cal-
culation over a range of times. Values of B at these various times are calculated
using:
The Material Balance equation given in the Havlena and Odeh shorthand form:
36
Water Influx
W
e(MB) = F - N( Eo +mEg + Efw)
F are production terms.
Using the unsteady state equation and knowing the time and pressure drop,we can
calculate
∑∆pQ(t)
We ( mb )
B=
∑ ∆pQ(t ) (38)
The apparent value of B, is determined at several past production times and plotted
versus cumulative oil production. The best horizontal line is then drawn through the
various points. This is the value of B which can be used for all future calculations.
(3) Using the calculated aquifer characteristic, B, water influx over the past history
of the reservoir is calculated using both the unsteady state equation and the
material balance equation. Clearly these should reasonably agree since the past
production data has been used in calculating B.
(4) Using the trends of past production history of the reservoir the future production
trends of oil, gas and water production throughout the future prediction period can
be plotted and the following trial and error calculation steps carried out:
1. The first step is the estimation of the reservoir pressure at the end of the first
trial period (suggested 6 months). Gross water influx is calculated by
both equations, the MB and the Unsteady State Influx Equations. If the results
agree then the first estimate of pressure is correct; if not, then another pressure
must be selected and the procedure repeated until convergence achieved.
2. This procedure is carried out for each time period until the desired range of
production has been covered
Different combinations of production rates of oil, gas and water should be used and
a complete prediction of pressure decline made for each set of values.
There are a number of factors to consider with respect to the limitation of the
performance prediction, in particular, the extent of the aquifer, the size of the constant
B, and the time intervals of the calculation. All of these are significant factors in
the calculations. For example in later time periods the pressure influence could
have reached the aquifer limit and therefore we have a finite aquifer. Using the MB
equation to calculate water influx may not be so accurate as small changes in pressure
can cause significant changes in PVT terms. Clearly shorter time periods generates
more accurate predictions. Three to six month periods are suggested.
Although the Hurst and van Everdingen unsteady state influx theory provides a well
established method for predicting water influx its application is somewhat tedious
particularly as a result of the superposing solutions for each time step and the trial
and error approach for history match and predicting future performance. In 1971
Fetkovitch7 produced an approach for finite aquifers.
The method of Fetkovitch7 models the aquifer flow in the same way as oil flow from
a reservoir into a well, and the results of the model follow closely those of van Ever-
dingen and Hurst. Fetkovitch’s concept is that the productivity index approach can
be used to describe the water influx from an aquifer into a reservoir.
dWe
qw = = J ( pa − p)
dt (39)
where:
For a finite aquifer the total water influx arising from a total pressure drop is:
We = cWi ( pi − pa )
(40)
where:
therefore:
We
pa = pi 1 −
cWi pi (41)
or:
W
pa = pi 1 − e
Wei (42)
38
Water Influx
where:
Wei = cWipi is the maximum possible expansion of the aquifer arising from a pres-
sure drop of Pi.
dpa Jp
= − i dt
pa − p Wei (44)
Integrating this equation with initial conditions when t = 0(We=0,pa=pi) and a pressure
drop ∆p=pi-p is imposed at the hydrocarbon reservoir boundary and assuming boundary
pressure remains constant over the period of interest:
Jpi
ln( pa − p) = − dt
Wei (45)
therefore:
( pa − p) = ( pi − p)e
− Jpi t / Wei
(46)
dWe
= J ( pi − p)e
− Jpi t / Wei
dt (47)
We =
Wei
pi
( (
pi − p) 1 − e
− Jpi t / Wei
) (48)
Wei
We =
pi
( pi − p) = cWi ( pi − p)
(49)
which is the maximum amount of water influx that could occur once the pressure
drop has moved through the aquifer.
∆We1 =
Wei
pi
( (
pi − p1 ) 1 − e pi 1 ei
− J ∆t / W
)
(50)
where p1 is the average reservoir boundary pressure during the first time step.
∆We 2 =
Wei
pi
( (
pi − p2 ) 1 − e pi 2 ei )
− J ∆t / W
(51)
wherepa1 is the average aquifer pressure at the end of the first time interval.
∆We1
pa1 = pi 1 −
Wei (52)
∆Wen =
Wei
pi
( )(
− J ∆t / W
pan−1 − pn 1 − e pi n ei )
(53)
where:
n −1
∑ ∆Wej
pa( n−1) = pi 1 − 1
Wei
(54)
pn −1 + pn
pn =
2 (55)
40
Water Influx
Semi-steady State
2 πfkh
(drawndown expressed
asPa - P) re 3 khw
µ1n − 3 ...................(56)
ro 4 µl
The semi-steady state equations 56 are used in conjunction with Fetkovitch equations
53 and 54. The steady state equations, however, are applied differently. In the
steady state situation the assumption is made of the water influx from the aquifer
being replaced from another source thereby maintaining pressure Pi constant at the
external aquifer boundary. It is therefore unnecessary to use Fetkovitch’s method of
determining the average pressure. The aquifer productivity values are used directly
with the drawdown equation 47.
dWe
qw = = J ( pi − p)e − Jpi t / Wei
dt (47)
dWe
qw = = J ( pi − p)
dt (58)
i.e.
t
We = J ∫ ( pi − p)dt
o (59)
The expression of the above equation 57 is the same as the Schilthuis steady state
equation, equation 11.
It is not possible to apply the Fetkovitch method to very large aquifers, because of
transient flow effects. With large aquifers there is a finite time for the initial pressure
disturbance at the reservoir boundary to feel the effect of the aquifer boundary.
It is necessary for large aquifers to use the Hurst and Van Everdingen for the first time
intervals. Dake1 has demonstrated that a combination of the two procedures, using van
Everdingen and Hurst for the first time periods covering the more transient time with
Fetkovitch for the later time periods provides an effective approach to water influx
calculations. The reader is recommended to read Dake's text where some examples
are worked through, using both the van Everdingen, Hurst and Fetkovitch method.
In 1960 Carter and Tracy8 published a procedure for calculating water influx which
eliminates the superposition calculations described previously to accommodate a
steady declining pressure for a constant terminal pressure solution. The results show
close approximation to the procedure of van Everdingen and Hurst4. They used the
constant terminal rate solution of the diffusivity equation and their equation is;
B∆p − E p′
( t Dj ) e ( t Dj −1 ) ( t Dj )
We t = We t +
( Dj ) ( Dj −1 ) p − t p′ t
( t Dj ) Dj −1 ( )
Dj
(60)
The “j” refers to the present time step and the “j-1” refers to the previous time
step.
Rather than using the tables of PtD given by Van Everdingen and Hurst, Fanchi10 as
presented by Dake9 used a regression equation form of Van Everdingen and Hurst
functions as;
42
Water Influx
Carter-Tracy influence functions regression coefficients for the constant terminal rate case (17)
Regression coefficients
reD a0 a1 a2 a3
1.5 0.10371 1.66657 -0.04579 -0.01023
2.0 0.30210 0.68178 -0.01599 -0.01356
3.0 0.51243 0.29317 0.01534 -0.06732
4.0 0.63656 0.16101 0.15812 -0.09104
5.0 0.65106 0.10414 0.30953 -0.11258
6.0 0.63367 0.06940 0.41750 -0.11137
8.0 0.40132 0.04104 0.69592 -0.14350
10.0 0.14386 0.02649 0.89646 -0.15502
∞ 0.82092 -3.68 x 10-4 0.28908 0.02882
Table 5 Carter - Tracy influence functions regression coefficients for the constant terminal
rate case
The reader is referred to Dake's9 text to see worked examples on Dake's approach.
Solutions to Exercises
EXERCISE 1
(a) Calculate the volume of water an aquifer of 35,000 ft. radius can deliver to a res-
ervoir of 3,200 ft. radius rock and water compressibility under a 1,100 psi pressure
drop throughout the aquifer. assume porosity = 0.22.
(b) Compare the available influx with the initial hydrocarbon volume of the reser-
voir.
Data:
Aquifer radius rw = 35,000 ft.
Reservoir radius ro = 3,200 ft.
Water compressibility cw = 3.0E-06 psi-1
Rock compressibility cf = 5.0E - 06 psi-1
Reservoir thickness h = 45ft.
Pressure drop ∆P = 1,100 psi
Porosity φ = 0.22
Water saturation swc = 0.25
SOLUTION (A)
Wi = 3.78E+10 ft.3
We = 3.325E+08 ft.3
SOLUTION (B)
EXERCISE 2
Calculate the water influx at the end of 1,2 and 5 years into a circular reservoir with
an aquifer of infinite extent. Effective water permeability is 120md, water viscosity
is 0.8 cp, effective water compressibility is 1.0 x 10-6 bbl/bbl/psi, the radius of the
reservoir is 2,400 ft. reservoir thickness is 35ft, porosity is 22%, initial reservoir
pressure is 4,500 psig and present reservoir pressure is 4,490 psig.
Data Table 1
Reservoir Radius ro = 2,400 ft
Effective water permeability kw = 120 mD
Effective water compressibility ce = 1.0e-06 psi-1
Water viscosity µw = 0.8 cp
Reservoir thickness h = 35 ft
Porosity φ = 0.22
Initial reservoir pressure Pi = 4,500 psi
Current reservoir pressure P = 4,490 psi
Water encroaching factor f = 1
SOLUTION
B = 49.63 bbl/psi
k t
Dimensionless time td = 6.323x10-3 2
φ µ ce
ro (16)
After 1 year, i.e. t = 365 days
tD = 273 days
44
Water Influx
After :
1 Year => 1 365 273 98.0 4,490 10 48,626
2 Years => 2 730 546 175.3 4,490 10 86,988
3 Years => 5 1825 1,366 383.9 4,490 10 190,524
EXERCISE 3
Data Table 1
Reservoir Radius ro = 2,500 ft
Effective water permeability kw = 105 mD
Effective water compressibility ce = 1.0E06 psi-1
Water viscosity µw = 0.8 cp
Reservoir thickness h = 30 ft
Porosity φ = 0.22
Initial reservoir pressure Pi = 3,800 psi
Water encroaching factor f = 1
0 3,800
1 6 3,780
2 12 3,750
3 18 3,700
4 24 3,620
i.e. as figure 11.
B = 46.16 bbl/psi
k t
td = 6.323x10-3 φ µ ce ro
2
Dimensionless time (16)
Dimensionless influx rate Q(t); => is extracted from the table for infi-
nite aquifers
Water influx Σ
We = B*( ∆P*Q(t)) (30)
B = 46.15875 bbl/psi
t = 182.5 days
tD = 110.15 => Q(t) = 46.63 (By interpolation)
Period Time (t) (Duration) tD Pressure Effective ∆p Q(t) ∆P*Q(t) Water Influx
Months Days (psi) Pressure Drop (From Table (psi) We (bbl)
(psi) Infinite Aquifers)
Σ(∆P*Q(t)) = 466
21,522
* Where:
1
∆P1 = (P − P1 )
2 i
B = 46.15875 bbl/psi
t = 365 days
tD = 220.3 => Q(t) = 82.06 (By interpolation)
The first pressure drop, Pi - P1, has been effectivefor one year, but the second pres-
sure drop, Pi - P2, has been effective for only six months. Separate calculations must
be made for the two pressure drops because of this time difference and the results
added in order to determine the total water influx.
The equation:
Σ
We = B( ∆P Q(t))
46
Water Influx
Period Time (t) (Duration) tD Pressure Effective ∆p Q(t) ∆P*Q(t) Water Influx
Months Days (psi) Pressure Drop (From Table (psi) We (bbl)
(psi) Infinite Aquifers)
1 1
∆P2 = ( Pi − P1 ) + ( P1 − P2 )
2 2
1
∆P2 = ( Pi − P2 )
2
B = 46.15875 bbl/psi
t = 547.5 days
tD = 330.45 => Q(t) = 114.87 (By interpolation)
The first pressure drop, ∆P1, will have been effective the entire eighteen months, the
second pressure drop ∆P2, will have been effective for 12 months and the last pressure
drop, ∆P3, will have been effective for only 6 months. the table below summarises
the calculations:
Period Time (t) (Duration) tD Pressure Effective ∆p Q(t) ∆P*Q(t) Water Influx
Months Days (psi) Pressure Drop (From Table (psi) We (bbl)
(psi) Infinite Aquifers)
* Where:
1 1
∆P3 = ( P1 − P2 ) + ( P2 − P3 )
2 2
1
∆P3 = ( P1 − P3 )
2
B = 46.15875 bbl/psi
t = 730 days
tD = 440.60 => Q(t) = 146.23 (By interpolation)
The first pressure drop, ∆P1, has now been effective for the entire twenty four months,
the secondpressure drop, ∆P2, has been effectinve for eighteen months, the third pres-
sure drop, ∆P3, has been effective for twelve months and the fourth pressure drop,
∆P4, has been effective for only six months. the table below summarises the water
influx calculations:
Period Time (t) (Duration) tD Pressure Effective ∆p Q(t) ∆P*Q(t) Water Influx
Months Days (psi) Pressure Drop (From Table (psi) We (bbl)
(psi) Infinite Aquifers)
* Where:
1 1
∆P4 = ( P2 − P3 ) + ( P3 − P4 )
2 2
1
∆P4 = ( P2 − P4 )
2
48
Water Influx
REFERENCES
8. Carter,R.D. and Tracy,G.W. “An improved Method for Calculating Water Influx”,
Trans AIME. Vol 219, p 415-417., 1960.
CONTENTS
4. DISPLACEMENT THEORIES
4.1 Introduction
4.2 Buckley- Leverett Theory
4.3 Welge Analysis
4.4 Calculations for Oil Recovery
4.5 The Impact of Viscosity.
Having worked through this chapter the Student will be able to:
Immiscible Displacement
1. INTRODUCTION
In previous chapters we have examined the various fundamental properties associ-
ated with the behaviour of fluids when subjected to pressure and temperature changes
and the characteristics of reservoir porous media in relation to its pore volume and
transmission characteristics. At another extreme scale we have reviewed the various
drive mechanisms responsible for providing the energy to move hydrocarbons in a
reservoir. We have also examined the various volumetric methods used to relate the
volumes of fluids produced in relation to the overall pressure decline of the reservoir
and the original volumes in place and energy support provided by attached water
and gas.
It is the purpose of this next chapter to bring some of these topics together in the
context of those reservoirs where the principle drive mechanism is that associated
with the immiscible displacement of oil. The subject will be mainly presented in
the context of water displacing oil, and then later the application to gas displacing
oil will be covered.
The topic of water drive in the chapter on drive mechanisms showed that this drive
mechanism provided the highest recovery factor in relation to reservoir depletion.
For this reason therefore water drive provided by intervention, that is when water
is injected into the reservoir through injection wells, is common practise in oilfield
operation.
Most of the reservoir engineering texts cover this topic. The author considers that
Dake1,2 and the text of Chierici3 provide excellent detailed analysis of the topic
In the next sections; we will review some of the reasons for using water injection,
then review some of the basic properties used in prediction, derive the fractional
flow equation and then examine procedures used to determine the movement and
displacement of fluids within a reservoir.
Water injection is the main intervention method used in reservoir development, pri-
marily because of the associated recovery achieved and also the availability of the
injection fluid. Historically it is termed a secondary recovery process in recognition
It should be recognised that with water injection come many technical challenges
not the least the fact that injected fluids will eventually arrive at the producing wells
where they will present a disposal challenge. Historically, returning the fluids from
where they came was the straightforward answer. With increasing concern of environ-
mental contamination, disposal of produced oily water to the sea is being gradually
replaced by a recycle process where rather than a once through process, the water is
reinjected into the formation. This is termed produced water injection. Dake1 outlines
the following benefits of water injection:
Immiscible Displacement
Zone Isolation
OIL
AQUIFER OIL
AQUIFER
2.2 Permeability
A characteristic of a number of offshore oil producing regions, for example the North
Sea, is the moderate to high permeabilities, which enable production wells to be very
productive reducing the required number of wells. Since the major cost in offshore
production is the offshore structures then minimising well slots results in minimising
the number of platforms. Maintaining high productivity through pressure maintenance
can be obtained through water injection when good injectivities can be achieved.
′ kro′
krw
M= /
µ w µo (1)
In sectors like the North Sea, Dake1 points out that relatively low oil viscosities lead
to high flow rates and the favourable oil viscosity compared to water gives a mo-
bility ratio for some North Sea reservoirs of less than 1. This means that at least at
microscopic level the water cannot move faster than the oil and therefore displaces
the oil in a piston like manner. If M is greater than one, the case where oil viscosi-
ties are higher, then the higher velocity of the water causes an increasing instability
and water fingers through the oil and breaksthrough early compared to piston like
behaviour. The behaviour is illustrated in the sketch below, Figure 2. As pointed
out this behaviour only relates to the microscopic scale, and at reservoir scale the
various heterogeneities and the influence of gravity will have a big impact on the
reservoir flooding behaviour.
Injection Production
Water Oil
M<1
Stable Displacement
Injection Production
M>1
Unstable Displacement
Immiscible Displacement
Gr% Sw%
0 100
Reservoir pressure - psig
100 0
θ%
0 50 2500 3000 3500 4000
Top paleocene
Perforations Original 8100
Layer 1 pressure
gradient
Layer 2
2500 8200
True vertical subsea depth - metres
Layer 3
8300
8500
2600
8600
2650 Layer 5
8700
8800
14 16 18 20 22 24 26
Reservoir pressure - MPa
2500 8200
2550
8400
8500
2600
8600
2650 8700
symbol ?Well number Date
22/17-A6 05/04/77
A8 27/01/78 8800
A11 20/12/77
2700 A15 15/08/78
A17 02/11/78 8900
A18 28/03/79
9000
18 20 22 24 26 28
Reservoir pressure - MPa
Immiscible Displacement
Sea Level
Sea Bed
Crestal
Producer
OIL
WATER
Injector
Figure 5 Application of deviated wells from one structure to reach limits in the reservoir.
Thermal
Fracture
Producer
Produces
Premature
Water
Breakthrough
Desired Water
Injection Flood
Front
Injector
10
Immiscible Displacement
Injection
Pump q o Sales
Treat/
qw Dump/
Inject
Sea Level
q o + qw
qi
Seawater
Sea Bed
Reservoir
Dake2 points out an equation which links the reservoir engineer to the production
engineer which is:
If water breaksthrough prematurely then, since the water injection rate has to be
maintained to maintain the reservoir pressure, there is an inevitable reduction in oil
production.
qwp
fws =
qo + qwp where ‘s’ denotes surface conditions. (3)
Expressing the equation in terms of water production and substituting in (2) gives:
fws
qwp = qo
1 − fws (4)
and
B f
qwi = qo Bo + w ws
1 − fws (5)
In his text Dake2 gives an example of the use of these equations to highlight the
commercial impact underestimating water breakthrough and the serious impact of
not being able because of platform limitations to increase water handling facilities.
Behind these commercial calculations is the importance for the reservoir engineer to
predict the producing watercut as a function of oil recovery.
In this next section we will review the basic parameters which are used to predict the
displacement process and then present the basic theory of water drive presented by
Buckley and Leverett over 20 years ago. The theory is a combination of behaviour
at the microscopic scale and that at a macroscale and then is applied at a reservoir
scale.
3.1 Introduction
Before examining the various methods used in predicting the behaviour of reservoirs
under a constant injection process, such as water drive or gas injection, we will re-
view some of the important basic properties relevant to the application. The method
presented is applicable to both water injection and gas injection where an immiscible
displacement process occurs. An immiscible displacement process is where there is
no mixing of the respective injection and displaced phases at the pore level through
mass transfer of components. This is distinguished from a miscible displacement
process where the injected phase mixes with the displaced phase by mass transfer
of the components from the respective phases, for example in a CO2 enhanced oil
recovery process.
12
Immiscible Displacement
An illustration of these different perspectives is shown in figure 8, where the oil wa-
ter displacement process is illustrated at two, microscopic, and reservoir behaviour
scales. This scale up perspective is considerable and should not be forgotten, if not
‘giant leaps of faith” might be made using data beyond its range of applicability. The
engineering of sub surface behaviour such as a water injection process can be com-
pared to the engineering of an oil refining plant. In the later, the process takes place
in vessels and pipes of centimeters and metres size over an area of a some hectares.
In a reservoir, the pipes and vessels, ‘the pores” are of micron dimensions and are
considerable in number to cover depths of hundreds of metres with an area perhaps
of tens of square kilometres.
Over recent years, considerable effort has been put into scale-up considerations
in relation to reservoir simulation, where rock properties at microscopic level can
be combined with geological characteristics at various scales to provide greater
confidence in field scale predictions. This topic is covered in the Geology and
Reservoir Simulation modules .
Water
Oil
Rock Grains
Water
Injection Oil at Residual
Wells Oil Saturation
(b)
By-Passed
Oil at Original
Oil
Oil Saturation
Oil
Producers
Figure 8 (a) microscopic displacement (b) Residual oil remaining after a water flood.
Q k Dp
u= =
A µ Dl
in relation to figure 9.
14
Immiscible Displacement
DP
A Superficial
Q Fluid Velocity
distance
This equation is for single phase flow only and does not apply to flow resistance when
two phases ( for example oil and water) are present. For this purpose the concept of
relative permeability is used, which is a measure of the permeability of one of the
phases and is a function of the phase saturations.
kew
krw =
k (7)
where kew is the effective permeability to water calculated from Darcy’s law when
oil and water are present, and k is the absolute permeability (single phase).
Darcy’s Law in linear flow for the two fluids allowing for gravity effects in an inclined
configuration, figure 10, is
kkro A ∂po
qo = − + ro gSinθ
µo ∂x
kkrw A ∂pw
qw = − + rw gSinθ
µw ∂x (8)
qt
qi
qo
qw
y
x z
ø
O X
Production
Injection
O w (Plan view) X
O X
The relative permeabilities are a function of saturation and reflect the surface, and
wettability forces of the fluid-rock system. An example of relative permeabilitiy
curves for a water oil rock is given in figure 11.
k'ro
End point
relative
k'rw permability.
kr
16
Immiscible Displacement
Identified on the curves are the two conditions at the limiting saturation of the respec-
tive phases, the end point relative permeabilties for oil and water k'ro and k'rw.
k'ro – the relative permeability to oil in the presence of irreducible water saturation
and
k'rw– the relative permeability to water in the presence of residual oil saturation.
Dake2 reminds his readers that rock relative permeabilities are obtained from one
dimensional core flooding experiments, where often a cleaned core is flooded with
water and then the oil displaced with water. Two types of experiments are then
used. A viscous displacement of oil with water or a steady state experiment where
co-injection of water and oil at increasing ratios of water to oil.
Dake2 also notes that the relative permeability data, used in subsequent reservoir en-
gineering calculations are unlikely to be representative of field characteristics. They
have probably been carried out at flow rates orders of magnitude higher than in the
reservoir, often using a synthetic oil not necessarily representative of the reservoir
fluid, and with wetting characteristics probably different than in the reservoir.
In the viscous displacement experiment the injected water, starting at the irreducible
connate water level, Swc, where the water is immobile, generates increasing satura-
tions in the core as a result of displacing oil. This increases until the saturation in
the core, where there is no more oil mobile in the core and the saturation to water, is
1-Sor, where Sor is the residual oil saturation.
If we express the volume of the pores in the core plug as the pore volume, PV, then
the oil displaced from the core flood experiment, is the movable oil volume, MOV,
which is;
MOV = (1- Sor - Swc) PV
The importance of end point relative permeabilities was presented earlier in this
chapter in the context of mobility ratio, M, where ;
′ kro′
krw
M= /
µ w µo (1)
At the end point conditions this represents the maximum velocity of the water flow
compared to the maximum velocity of the oil.
qw
fw =
qw + qo (9)
qo = qt - qw
If the Darcy equations for water and oil are subtracted (using field units P in atmos.)
the equations become;
µ µ qµ ∂P Drg sin θ
qw w + o = t o + A c −
kkrw kkro kkro ∂x 1.0133 × 10 6 (11)
where;
If values for flow rates using Darcy’s Law are now substituted in fraction flow equation
(equation 9) it becomes;
1− G
fw =
µ k
1 + w ro
µo krw where G is a positive gravity number; (13)
The above term not only considers gravity effects but also includes a velocity term
,v, which is qt/A
The angle of dip. If water is being injected downdip than the gravity term, ∆ρgsinθ/
1,0133x106 will be positive, reducing the fractional flow of water and it would be
positive for gas being injected downdip in a gas displacing oil senario. The density
difference in gas displacing oil systems is larger and therefore the significance is
greater. If the dip angle is zero, ie. horizontal flow, then the gravity term is zero.
18
Immiscible Displacement
The impact of capillary pressure, is illustrated from the slope of the capillary pres-
sure and saturation with distance curves, figure 12 since;
1-Sor
Pc
- dPc Sw - dSw
Swf
+ dSw
+ dX
Swc
Swc 1-Sor
Sw X
i.e. the capillary pressure term is also positive increasing the fractional flow, for a
water displacing oil system as the two function gradients are negative. The capillary
pressure term is often neglected because the saturation with distance profile is unknown
being the objective of the displacement calculation, which we will consider later.
Velocity. This velocity is the superficial velocity, the rate divided by the cross sectional
area,A. The actual velocity is larger because of the impact of porosity. The impact
of velocity is small. Dake2 notes that the value for G for an edge water drive ,typical
of the North Sea, is 0.22kro and a comparative bottom water drive is 10.29kro. This
demonstrates the stability of the bottom water drive, where piston like displacement
will inevitably occur.
If both the angle of dip, and capillary pressure effects are neglected the fractional
flow equation becomes;
1
fw = µ k
1 + w ro
µo k rw
(15)
The fractional flow equation enables a fractional flow versus saturation curve to be
generated from relative permeability data. This curve is influenced by a number of
parameters not least the viscosity of the respective phases. Its shape varies but can
have a shape as given by figure 13 below.
ƒw
Swc 1-Sor
Sw
EXERCISE 1
Plot the water-oil relative permeability from the following data set. Indicate the end
point relative permeabilities.
Sw Krw Kro
0.20 0.0000 0.8800
0.25 0.0021 0.6710
0.30 0.0095 0.5170
0.35 0.0210 0.4070
0.40 0.0347 0.3135
0.45 0.0536 0.2420
0.50 0.0788 0.1793
0.55 0.1050 0.1320
0.60 0.1386 0.0891
0.65 0.1785 0.0550
0.70 0.2184 0.0297
0.75 0.2636 0.0110
0.80 0.3150 0.0000
20
Immiscible Displacement
A
Transition Zone
B
C Free Water
0 50 100 Level
k'rw
50
k'ro
kro krw
0 50 100
100
% Water Cut
50
0 50 100
% Water Saturation
Figure 14 The relationship between capillary pressure, relative permeability and frac-
tional flow in a reservoir4
The capillary pressure curves represents the transition zone saturation profile associated
with the advancing imbibition process as a result of water injection. If a well is located
at A, the well will only produce oil since although the water saturation is 10%, the
relative permeability to water is zero. At B, the 45 % saturation level the well will
produce both water and oil with a water cut of 50%. At location C, the advancing
water has isolated an irreducible oil saturation and the well produces only water.
The redistribution of the saturation profile giving rise to a height saturation function
is called vertical equilibrium, and depends on a number of factors, including; a large
vertical permeability, small reservoir thickness, a large density difference between
the injected and displaced fluids, high capillary forces, low fluid viscosities and
low injection rates. It is not the intention of this chapter to present the associated
procedures, which would be part of a full numerical simulation analysis.
1 35.0 0.5
2 4.5 0.5
3 0.4 1.0
EXERCISE 2
Water is to be injected into a horizontal core, with the relative permeability charac-
teristics of table 1, to displace oil. Determine the mobility ratios, and the fractional
flow curves for the following three cases.
4.DISPLACEMENT THEORIES
4.1 Introduction
To model the displacement process a number of theories have been successfully ap-
plied. These theories are aimed at providing the important predictions of reservoir
performance including the proportion of hydrocarbons recovered. In the methods
presented , there are a number of assumptions.
The displacement is incompressible, which implies that steady state conditions exist,
that is the pressures within the reservoir at any point remain constant.
qt=qo+qw=qi
where
Diffuse flow conditions exist. Diffuse flow means that the saturations at any point
in the direction of linear displacement are uniformly distributed over the thickness.
This diffuse flow assumption enables a one dimensional simple analysis to be used
for the displacement modelling. In a simple core flooding relative permeability test
such an assumption is not unreasonable. Diffuse flow can also be encountered in a
reservoir where the injection rates are high preventing the establishing of vertical
equilibrium and for low injection rates where the thickness of the reservoir is small
compared to the thickness of the transition zone.
22
Immiscible Displacement
dx
Water Water
qw rw x + dx + Oil
A qw rw x
Porosity L
Consider the linear system in which water is displacing oil. The systems has a porosity
of φ and we are considering the principle of conservation of mass around a volume
element of length, dx. Therefore;
Mass flow rate in –mass flow rate out =rate of increase of mass in the volume.
∂x
qw rw x − qw rw x + dx = Aφdx ( rw Sw )
∂t
(16)
or
∂ ∂
qw rw x − qw rw x + (qw rw )dx = Aφdx ( rw Sw )
∂x ∂t (1)
This becomes
∂ ∂
( qw rw ) = − Aφ ( rw Sw )
∂x ∂t (17)
∂qw ∂S
= − Aφ w
∂x t ∂t x (18)
Then
∂Sw ∂S dx
=− w
∂t x ∂x t dt sw (20)
Also
∂qw ∂q ∂S
= w ⋅ w
∂x t ∂Sw ∂x t (21)
∂qw dx
= Aφ
∂Sw t dt Sw (22)
For incompressible flow, the total injection rate, qt is constant, and the water flow
rate is the total rate times the fractional flow, qw=qt x fw. Rearranging equation 22
therefore gives:
dx q t ∂fw
v sw = =
dt s w aφ ∂sw sw
(23)
Clierici6 has presented a very thorough analysis of the displacement process for three
fractional flow curves.
Sw = Swi for 0 < x ≤ L,t = 0
24
Immiscible Displacement
dx q ∂fw
vSw = = t
dt Sw Aφ ∂Sw
Sw
and the equation is then integrated a general solution to the displacement process is
obtained which enables the calculation of Sw in terms of x and t.
qt dfw
[ x(S )]w
t
= x0 ( Sw ) + t
φA dSw sw (25)
This equation describes a series of straight lines, the characteristics, with an initial
qt dfw
ordinate value of x0(Sw) and a slope of
φA dSw Sw
1 x
A C qt dfw
Swi Αθ dSw Swi+ DSw
fw
w
�S
+
S iw
0
S1
B xo(Siw)
dfw S2
xo(S1)
dSw
xo(S2) 1 - S or
Sor
0
0 Swi S1 S2 1 0 t
Sw
The velocity of saturation is therefore maximum where Sw is just greater than Swi
and decreases to a minimum at Sw=1-Sor . Figure 16C, The progression of water
profiles are shown in figure 17 and shows the fraction of water at breakthrough at the
Water
Injector Producer
1
Sor
Sw
t=0 t1 t2 tBT
Swi
0
0 L
X
Figure 17 Progressive saturation profile for a concave downwards fractional flow curve6.
1 x
A C
fw
�Sw
S wi+
xo(Swi)
S1
0
S2 qt dfw
B xo(S1)
Αθ dSw Swi+ DSw
dfw
dSw xo(S2)
or
-S
1
Sor
0
0 Swi S1 S2 1 0 t1 t2 t3 t
Sw
Figure 18 Displacement of oil by water for a concave upwards fractional flow curve (light
oil displacement).6 A = concave upwards fractional flow curve. B = velocity of water satu-
ration. C = characteristics of water saturations Sw.
26
Immiscible Displacement
The implications of this are that the highest velocity is for the highest water saturation,
Sw=1-Sor and that saturations less than this cannot exist since they would be overtaken
by the Sw=1-Sor saturation. Figure 18C, There is therefore a quick build up of a shock
front with a saturation, Swf=1-Sor . The producing characteristics are shown in figure
19, where , until the shock front arrives water-free oil is produced and thereafter only
water is produced. The oil remaining in the reservoir with a saturation of Sor .
Water
Injector Producer
1
Sor
Sw t1 t2 t3 tBT
t=0
Siw
0
0 L
X
Figure 19 Progressive saturation profile for a concave upwards fractional flow curve.6
1 x
A C
fw
S wi+ �Sw
xo(Swi)
S1
xo(S1)
0
S wf
B S2
dfw xo(Sw,f)
dSw xo(S2)
1 - S or
Sor xo(1-Sor)
0
0 Swi S1 S2 t1 t2 t3
1 0 t
Sw
Figure 20 Displacement of oil by water for a rock with an S-shapeed fractional flow
curve (light oil displacement). A = S shaped fractional flow curve. B = velocity of water
saturation. C = characteristics of water saturations Sw.6
Water
Injector Producer
1
Sor
Sw t1 tBT
t=0 t2 t3
Siw Sw,f
0
0 L
X
The velocity of the stabilised shock front can be calculated from a material balance
across the front. Chierici4 explains this using figure 22. He designates R to represent
conditions ahead of the front and L those behind. Firstly for case two, piston like
displacement
q w,L q w,R
x
S w,L dxf S w,R
28
Immiscible Displacement
qw,Ldt-qw,Rdt = Af(Sw,L - Sw,R)dxf (26)
s
Since
qw= qtfw
dx f q ( fw , L − fw , R )
vf = = t
dt Aφ ( Sw , L − Sw , R ) (27)
Since qt=Aut
Then
dx f ut ( fw , L − fw , R )
vf = =
dt φ ( Sw, L − Sw, R ) (28)
This is the Rankine-Hugoniot condition for the frontal velocity of shock fronts for
physical systems.
Sw,L=1-Sor
fw,L=1
Sw,R=Swi
fw,R=0
Therefore:
(f w, L − fw , R )
=
1
= tan α
(S w, L − Sw , R ) 1 − Sor − Swi (29)
Where tan α is the angle on the Sw vs. fw curve joining (Swi,0) and (1-Sor,1)
fw
α
0
0 Swi 1-Sor 1
sw
dx u ∂f
vSw1− Sor = = t w
dt 1− Sor φ ∂Sw 1− S
or
dx u ∂f u ( fw , L − fw , R ) ut 1 u
vS1− Sor = = t w = t = = t tan α
dt 1− Sor φ ∂Sw 1− S φ ( Sw, L − Sw, R ) φ 1 − Sor − Swi φ
or
(30)
Sw,L=Swf
fw,L= fw(Swf)
Sw,R=Swi
fw,R=0
dx u df ut ( fw , L − fw , R ) ut fwf u
vSwf = = t w = = = t tan β
dt wf φ dSw φ ( Sw, L − Sw, R ) φ Swf − Swi φ
wf (31)
30
Immiscible Displacement
dfw fwf
tan β = =
dSw Swf − Swi
wf (32)
That is a tangent drawn to the fractional flow curve from the point (Swi,0) which meets
the curve at the conditions of the shock front. Figure 24 below.
fw(Swf) dfw
tan β = dS S
w w,f
fw
β
0
Swi Sw,t 1-Sor
Figure 24 Graphical Procedure for Determining the Conditions of the Shock Front
If we now consider the time, tbt it takes for this shock front to move though our lin-
ear system we generate a useful equation which we will use later in water injection
performance calculations.
L Lφ Swf − Swi
tbt = =
vf ut fw S
wf
(33)
1-Sor
Sw
Sw Swf
Swi
X1 X2
X
Before water arrives at the exit, the volume of oil produced is equivalent to the volume
water injected. Wi = qw x t. At breakthrough the volume of oil produced, Np is the
difference between the initial oil volume, (ALφ(1-Swi), less that remaining in terms
of an average saturation, Sw, at breakthrough, (ALφ(1-Sw))
Swf − Swi
N pbt = qw tbt = ALφ = ALφ ( Sw − Swi )
fwf (34)
Therefore:
Swf − Swi
Sw − Swi =
fwf (35)
from equation 32 this can be written as:
1
Sw − Swi =
dfw
dSw Swf
(36)
32
Immiscible Displacement
1 − fwf
Swbt − Swf +
dfw
dSw Swf
(38)
This is Welge's equation for average saturation and and combining with equation
16 gives:
dfw
=
(1 − f ) =
w S wf 1
dSw S wf
Sw − Swf Sw − Swi (39)
There is also a graphical significance in the above equation. The line of the tangent
drawn previously at the breakthrough point cuts the line fw=1 at an x-axis of
.
Sw
1.0
fwf
Swf, f
wf
fw
Data Summary
Oil formation volume factor Bo = 1.36 bbl/STB
Water formation volume factor Bw = 1.01 bbl/STB
Initial water saturation Swc = 0.20
Draw the fractional flow curves and calculate the cumulative oil recovery at break-
through for the following combinations.
1 35.0 0.5
2 4.5 0.5
3 0.4 1.0
PV = AfL. (40)
Before water breakthrough, only oil exits from the core at a rate equivalent to the
rate of water being injected, since it is an incompressible system. At breakthrough
therefore the pore volumes of fluids involved are;
( )
N pd bt = Wid bt = qid tbt = Sw bt − Swi (41)
Npdbt = pore volumes oil produced at water breakthrough. Widbt = pore volumes water
injected at water breakthrough
34
Immiscible Displacement
qi
qid =
AφL (42)
tbt is the time taken for the water to breakthrough, which is;
Wid bt
tbt =
qid (43)
S
w bt average water saturation at water breakthrough
At breakthrough , when x=L in the Buckley Leverett equation (equation 23), the
following convenient result exists;
Wi dfw
xSwbt = L =
Aφ dSw S wbt (44)
Therefore ;
1
Wi
Wid = = dfw
LAφ
dSw
S wbt
(45)
In equation 45 above therefore the oil recovery at breakthrough is also equal to the
inverse of the slope of the breakthrough characteristic, the slope of the line drawn from,
Swi, tangent to the fractional flow curve, figure 26. The oil recovery at breakthrough
is fully given by the equation below.
(
N pd dt = Wid bt = q id t bt = sw bt − swi = ) 1
dfw
dsw swbt
(46)
After breakthrough, the fractional flow saturation profile at the exit of the core in-
creases as shown in figure until the irreducible oil saturation, Sor, is reached where
fw=1, as the flood front moves through the core. Figure 27
Swbt Swe
Swbt= Swf
0 X L
From Welge’s equation 38 where the saturation value is now Swe, the average satura-
tion remaining is now
w S we (47)
This can also be expressed, in terms of injected pore volumes of water, using the
Buckley Leverett relationship, so that
Sw = Swe + 1 − fw( S we )W
id
(48)
The oil recovered associated with this average saturation is given by;
(49)
dfw
=
(1 − f )w S we
dSw S we (S w − Swe ) (50)
36
Immiscible Displacement
Sw
1-Sor
fw= 1
Sw,bt Sw
(1-fwe)
Swe,fwe
(Sw - Swe)
fw
dfw = 1-fwe
dSw Sw - Swe
Swe
Swbt
1. Generate a fractional flow vs. water saturation curve for the system to be studied,
using the appropriate relative permeability data.
2. Draw a tangent to the fractional flow curve from the initial Sw = Swi position at
fw=0.
At the point of tangency are the conditions of breakthrough ;
(i) fw=fwbt, Sw=Swbt and extrapolation of line to fw=1 gives the average water
Saturation value. Swbt
Also
dfw
( )
N pd bt = Wid bt = qid tbt = Sw bt − Swi = 1
dSw
(ii) S wbt
(41, 45)
Wid bt
tbt =
qid (43)
and
(
N pd = Sw − Swi = ( Swe − Swi ) + 1 − fw S we )W
id
(49)
Wi 1
Wid = =
LAφ dfwe
dSwe
(45)
Wide
(iii) Time te = (43)
qide
The results can then be plotted to produce recovery of oil as a function of time. As
shown in figure 29. Up to A, the breakthrough point, the recovery is linear as the oil
recovered is equal to the water injected however after this point the recovery follows
a shape determined by the fractional flow curve above the breahthrough saturation.
Breakthrough
Np Point
Time
38
Immiscible Displacement
1.0
µo
= 100
µw
A
µo
=1
fw µw
B
µo
= .01
µw
C
Swi 1-Sor
Sw
In this figure, Case A gives the shape for a system where the ratio of oil viscosity to
water viscosity is high, say around 100. This could be for a very dense, viscous oil
which gives rise to unstable displacement, with by-passed oil and premature water
breakthrough. To generate the oil production would require a considerable number of
pore volumes of injected water. To improve recovery for this system increasing the
temperature of the injected fluid can improve behaviour. Although the temperature
decreases the viscosity of both fluids, there is a greater impact on the oil. M>>1
In case B, the viscosity ratio is considerably lower giving rise to a more stable and
favourable displacement with a shock front developing. M = 1.
In case C, where the curvature of the fractional curve is opposite to that of case A,
the shape results from a low oil to water viscosity ratio. In this case which might
be representative of a light oil, the mobility ratio M is <<1 and leads to piston like
displacement. The three saturation profiles for these cases are illustrated in figure
31 below.
Sw A M around 1
M<1
M>>1 B
Swi
EXERCISE 4
Data Summary
Injection rate Qi = 1,200 bbl/d/well
Water viscosity µw = 0.5 cp
Oil viscosity µo = 4.5 cp
Initial water saturation Swc = 0.20
Residual oil saturation Sor = 0.20
Porosity φ = 0.22
Dip angle θ = 0°
Reservoir thickness h = 50 ft
Distance between injection wells w = 800 ft
Distance - injectors and producers L = 2,000 ft
Assuming that diffuse flow conditions prevail and that the injection project starts
simultaneously with oil production from the reservoir, determine:
40
Immiscible Displacement
The procedure we have just described was for diffuse flow conditions which is a one
dimensional problem. The uniform distribution of saturation over the thickness was
likened to the water flooding of a core plug. Clearly reservoirs are not so simple, the
next step therefore is to examine the scale-up of the one dimensional situation more
realistic reservoir situations, where the reservoir has thickness and vertical permeability
giving rise to, segregated flow and where the reservoir is not homogeneous but is
made up of different permeability layers.
5.1 Introduction
The analysis so far has been focused around the displacement in a core flood where
diffuse flow has been assumed. That is a uniform saturation distribution over the
thickness of the core. The challenge is now to consider how the application of this
one dimensional analysis so far can be applied to real field applications where at least
a two dimensional perspective is required.
The first real perspective is that of segregated flow where the relative density per-
spectives of the two fluids lead to a saturation distribution over the thickness of the
displacement path. This type of displacement was studied by Dietz 8. He considered
that “ water encroachment on a monoclinal flank can be studied in a representative
cross section of a field and the problem is therefore reduced to one or two dimensions.
A sharp interface, rather than a transition zone is assumed between the oil- bearing and
the flooded part of the formation. No pressure drop is assumed across the interface.”
The situation for segregated flow in two dimensions is illustrated in figure 32.
lug
Core P
Oil
Water
A useful indicator of the dominating forces is that in relation to the ratio of viscous
forces to capillary forces:
uµo
Nvc =
σCosθ (51)
uµo
Nvg =
k0 g( rw − ro ) (52)
Where Nvc and Nvg are the capillary and gravity numbers.
For vertical equilibrium , both of these are low, and indicate that as the displace-
ment front advances the saturation distribution is readjusted. This can be considered
therefore as a fixed saturation profile advancing through the oil.
If the velocity is low all the front will move at the same speed. This results in the
inclination of the front remaining constant and to the limit of zero velocity would
result in a horizontal interface. The gravity forces have enabled the front to stabilise
and for this reason, the term gravity stabilised is used. At high injection rates the
viscous forces dominate and cause a tongue of water to advance along the base of
the layer. If gas was being injected the tongue would advance across the top of the
layer, This displacement is termed non-stabilised.
42
Immiscible Displacement
t2
dx -dy t
a) β 1
t0
t2
b)
t1
t0
In figure a, we have stabilised flow with the angle of inclination of the front remaining
constant and in b, the non-stabilised front we have the advancing tongue.
Dietz(8) analysed this to determine the conditions for stabilised flow. Considering
figure 33a. for the incompressible displacement at the interface. At stable conditions
then on the interface the velocity of the oil and water are the same. Then applying
Darcy’s law
kkro′ δpo
uo = ut = − + ro g sin θ
µo δx
(53)
′ δpw
kkrw
uw = ut = − + rw g sin θ
µ w δx (54)
where uw and uo and ut are the oil water and total velocities at the interface. Combin-
ing these yields;
µ µ δ
ut o − w = − ( po − pw ) + Drg sin θ
kkro′ kkrw
′ δx (55)
µ µ dy
ut o − w = Drg cos θ + sin θ
kkro′ kkrw
′ dx (57)
′
kkrw
µw ′ ADrg sin θ 1 dy
kkrw
kkro′ − 1 = +1
µ w qt tan θ dx
µo (58)
′ kro′
krw kk ′ ADr sin θ
m= / , and G = rw , equation 1 and 14
µ w µo qt µo
dy 1
M − 1 = G + 1
dx tan θ (59)
The angle of the interface of the front can therefore be expressed as;
dy M − 1 − G
= − tan β = tan θ
dx G (60)
The condition for frontal stability is that dy/dx must be negative i.e.
G > M-1
The limiting case is when dy/dx =0 and the water underruns the oil. This will occur
when:
G = M-1
This provides a relationship for flow which gives the limiting flow rate
′ ADr sin θ
kkrw
qcrit =
µ w ( M − 1)) (61)
44
Immiscible Displacement
4.9 × 10 −4 kkrw
′ ADγg sin θ
qcrit =
µ w ( M − 1))
The various aspects of stability of the front can be summarised in the context of
equation 60.
M<1
Oil
β
Water
θ M=1
Oil
β
Water
θ
Oil M>1
β
Water
θ
1. M < 1 The front will always be stable regardless of the value of G. From Equation
60 above tanβ/tanθ <-1 Therefore β>θ. This is illustrated in figure Figure
34a.
3. M > 1. The front will only be stable if G > M-1. In this case β<θ . This is
illustrated in figure Figure 34c. The limiting condition will be at the maximum
velocity associated with the critical flow rate expressed in equation 61 above.
These three conditions can be summarised , such that when displacement of oil by
water takes place under vertical equilibrium conditions (segregated flow), then gravity
will always stabilise the flow when the mobility ratio is less than or equal to 1. For
h
Oil
Sw=Swi
So=1-Swi
1- hw
hw
X
Water
Sw=1-Sor
So=Sor
If we consider a point in the displacement path we see that for our homogeneous
formation experiencing strong gravity segregation. we have a sharp interface over the
formation thickness. At any point x along the injection path therefore, if hw represents
the fractional thickness below the water oil interface, Then at the position x,y on the
interface, hw =y/h where h is the formation thickness.
Sw − Swc
hw =
1 − Sor − Swc (64)
since the two irreducible water and oil saturations are constant. Then hw is directly
proportional to the average saturation Sw.
The average relative permeability over the thickness, the thickness averaged relative
permeability, can therefore be expressed as;
46
Immiscible Displacement
krw Sw
= hw krw ( Sw =1− Sor ) + (1 − hw )krw ( Sw = S wc ) (65)
Sw
krw = hw krw
′
Therefore
k′rw is the relative permeability to water in the presence of irreducible oil, the end
point relative permeability to water.
kro Sw
= hw kro ( Sw =1− Sor ) + (1 − hw )kro ( Sw = S wc ) (67)
Therefore kro Sw
= (1 − hw )kro′
k′ro is the relative permeability to oil in the presence of irreducible water, the end
point relative permeability to oil.
Sw − Swc
hw =
1 − Sor − Swc
(69)
we have
S − Swc
krw S = w ′
krw
w 1 − Sor − Swc (70)
and
1 − Sor − Sw
kro S = kro′
w 1 − Sor − Swc (71)
The equations above show that the relative permeability curves are linear functions
of the average water saturation. Figure 36 shows the presentation of these two
equations.
k′rw
kro Sw krw Sw
Swi Sw 1-Sor
Also plotted are the relative permeability curves for diffuse flow, the situation for
measurements made in small cores.
The linear thickness averaged relative permeabilities, sometimes termed the psuedo
relative permeabilties, can be used in combination with the one dimensional Buckley
Leverett theory to determine the breakthrough and recovery calculations for our two
dimensional segregated flow condition.
(i) Using the core determined relative permeability curves generate the linear
thickness averaged relative permeability lines. As shown previously only the
end point values are required for this purpose.
(ii) Calculate the fractional flow curve using the pseudo(thickness) average relative
permeabilties.
Dake1 amplifies further the topic with respect to recovery calculations for unstable
flow in a horizontal situation and generates the following relations.
Where volumes are expressed as moveable pore volumes MOV where MOV = PV
(pore volume) x (1 - Sor - Swc).
1
N pD =
M −1
( wiD M − wio − 1) (72)
48
Immiscible Displacement
1
N pDbt = (73)
M
In this equation therefore for stable displacement when M=1, piston like displacement,
then NpDbt=1. Also for other mobility ratios to recover one pore volume the pore
volumes of water injected are
WiD = M (74)
For the case of unstable displacement in a dipping reservoir where G<M-1, then
1
N pDbt =
M − G (75)
N pDbt =1 is when
and the maximum recovery of
M
Wid =
G + 1 (76)
EXERCISE 5
Additional Data
Absolute permeability K = 500 mD
Specific gravity of water in the reservoir γw = 1.02
Specific gravity of oil in the reservoir γo = 0.83
6.1 Introduction
So far we have considered one dimensional flow for diffuse flow conditions and two
dimensional conditions with respect to segregated flow in a homogeneous system.
One of the most significant perspectives in reservoirs by nature of the process of their
formation is that reservoirs are heterogeneous. This heterogeneity has a considerable
impact on the displacement process in a reservoir.
There are two distinct types of heterogeneity, vertical heterogeneity, and horizontal
heterogeneity. The impact of these changes in reservoir characteristics have an
Np
= EV × EA
N
(77)
where:
EA is the areal sweep efficiency, which is the fraction of the oil recovered in the areal
cross section by waterdrive, natural or injected and
EV is the vertical sweep efficiency, which is the fraction of the oil recovered in the
vertical cross section .
The proportion of the reservoir formation swept by the water leaves an oil satura-
tion at the irreducible saturation level and that unswept is at the original saturation
level. Figure 37.
Producer Producer
Injector
Producer Producer
Aerial Sweep Efficiency Vertical Sweep Efficiency
Residual Oil
50
Immiscible Displacement
In chapter 2, on reservoir pressure and in section 2.4 of this chapter, we discussed the
use of downhole pressure monitoring systems. In openhole use during the early pro-
duction of a field these pressure surveys are able to distinguish the various layers
in a formation and also their communication characteristics. These pressure sur-
veys have provided a very useful role in enabling waterdrive predictions to be made
which are representative of the formation. They also demonstrate the effectiveness
in allowing the producing reservoir to provide reservoir behaviour characteristics to
reduce uncertainties in future prediction senarios. Figure 38 shows a sketch from
Dake2 of the RFT surveys which would result from the two extremes of pressure
communication and where there is no communication.
a) Pressure
Original water
pressure gradient.
Depth
Pressure gradient
post production.
b) Pressure
Original water
pressure gradient.
Depth
Pressure gradient
post production.
Figure 38 Pressure profiles pre and post production (ref Dake) (a) layers in communication
(b) non communicating layers
Clearly in actual reservoirs there could be some layers in communication and others
not.
The procedure used for those systems where there is cross communication, cross flow
or no communication are very similar and follow the same pattern. We will follow
the general procedure and then look at the variations which are dependant on the
various communication and segregation perspectives.
(i) The first step is to identify the various layers in the reservoir and the respective
characteristics. Figure 39 shows the various layers as they exist in their natural order
in the reservoir. For each of the layers, the following characteristics are required;
thickness hi, permeability ki, porosity φi. Connate water saturation Swci, irreducible
oil saturation Sori, end point water relative permeability k'rwi and the oil end point
relative permeability k'roi, where i defines the layer.
(ii) Consider the vertical pressure communication between the layers and decide
if there is cross flow or no cross flow.
(iii) Decide the flooding order of the layers and generate the thickness average,
pseudo-relative permeabilities. As each layer floods out, then the following equa-
tions apply.
52
Immiscible Displacement
∑ h φ (1 − S )+ ∑ hφ S
n N
j j or j j j wc j
j =1 j = n +1
Sw = N
∑h φ
1
j j
(78)
Subscript j refers to the ordering of the flooding out of the layers.
∑ h k k′
j =1
j j rw j
krwn S wn
= N
∑h k1
j j
(79)
and
N
∑ h k k′ j j rw j
kron S wn
= n +1
N
∑h k 1
j j
(80)
(v) The problem can now be solved using the 1-D approach using the Welge pro-
cedure.
Clearly if there is a very strong vertical equilibrium with strong segregation then the
bottom layer will flood out first and therefore j=i ,in the description of the layers and
is illustrated in Figure 40.
′i
ki krw
vi ∞
φi 1− S − S
( ori wci ) (81)
j in the above equation then represents the layers in descending order of their
velocity.
For the case where there is no communication, Stiles10 in 1949 considered the flooding
out of the layers with respect to those reservoirs were the mobility ratio is close to
unity. His approach is illustrated in Figure 41 where the natural ordering of figure
39 is reordered according to the velocity in each layer.
4
1
6
Clearly the ordering of the layers impacts on the pseudo relative permeabilities which
results in a subsequent fractional flow curve. However once the fractional flow curve
has been generated the breakthrough and recovery analysis procedure is the same
as for the 1D case.
If neither of these extremes occur, for example where the thicknesses are similar,
the capillary pressure needs to be taken into account when generating the pseudo
54
Immiscible Displacement
Figure 42 from Dake1 gives a capillary pressure curve and relative permeability data
for a rock with a transition zone of around 30 feet in thickness.
Figure 42 illustrates the three conditions for a homogeneous reservoir for diffuse
flow,
segregated flow and when capillary pressure is significant, for the average relative
permeability curves (a) and the oil recovery as a function of time (b).
A
1.0
Diffuse Flow (Rock Curves)
krw kro
k'rw
0
0 1.0
Sw
B
.6
Diffuse Flow
Npd
Segregated Flow
(PV)
Intermediate (H ≈ h)
qi = 1000 rb/d
0
0 7 Wpd (PV)
0 35 Time (yrs)
Figure 42 (a and b) Pseudo relative permeability and oil recovery curves for diffuse flow,
segregated flow and capillary pressure influenced cases (Dake1)
.8 .8
Oil Oil
.6 .6
krw kro krw kro
.4 Water .4 Water
.2 .2
0 0
0 .2 .4 .6 .8 1.0 0 .2 .4 .6 .8 1.0
Sw Sw
Figure 43 Relative permeabilty curves for three layered system (a) high permeability at
top (b) high permeabilty at base
Dake1 illustrated the impact of the relative position of the high permeability layer
is illustrated in figure 43 the thickness average relative permeabilities. The impact
of water breakthrough is clear from the fractional flow curves figure 44. The top
curve gives almost immediate breakthough as the high permeability on the bottom
breaksthrough. Whereas with the high permeability on the top a better breakthrough
profile is obtained.
1.0
.8 High Permeability
at the Top
.6
High Permeability
fw at the Base
.4
.2
0
0 .2 .4 .6 .8 1.0
Sw
Figure 44 Fractional flow curve for layered systems high permeability at the top, high
permeability at the base
56
Immiscible Displacement
can result are shown in figure 45&46. This topic was also covered in the chapter on
permeabilty variations. The various saturation profiles resulting from four permeablity
distributions are shown in Figure 47.
Levee
e
Weak Fin
Eros io
Current
n
itio Growth of
Channel p os tion )
n of
De Point Bar cre e
Migration Strong nd Ac e L i n
Sa al (Ti
m
Ba
Current ter
La face
nk
e r
ars Su
Co
GR
Permeability
Injection
Oil
Depth
Production
(a) Unfavorable
Water
Coarser Sediment in
Shallow Turbulent Water
Sea Level
Wave Energy
Increasing
e
Siz
r ain ion
gG sit
s in e po
rea u sD
Inc eo
Advance
Fine Sediment in ltan of Bar
Deeper Quiet Water Si mu
GR GR
Profile (A) Profile (B)
Injection
Permeability
Oil
Water
Depth
Production
(a) Favorable
Super-Homogenous
D
Worst Case
D
D
Tunnel
Random
D
As we have seen the resulting fractional flow curve determines the quality of the water
flood as shown by the saturation at breakthrough and the subsequent oil production
with time, The shape of the fractional flow curve is influenced by a number of factors,
including the relative viscosities, end point permeabilties and permeability distribution.
Figure 48 from Dake illustrates the impact of mobility ratio and permeability on the
fractional flow curves for the two extreme conditions of permeability distributions.
Figure 48a for the situation where the permeability increases with depth and figure
48b with the permeability decreasing with depth.
58
Immiscible Displacement
A B
1.0 1.0
M=16.1
M=1 M=16.1
0.8 0.8 M=1
bt
0.6 0.6
fw fw
0.4 0.4
0.2 0.2
0 0
0 0.2 0.4 0.6 0.8 1.0 0 0.2 0.4 0.6 0.8 1.0
Sw Sw
Figure 48 Fractional flow curves for (a) coarsening downward permeability distribution
(b) coarsening upward (Dake)2
Considering figure 48a the curves for both extreme mobility ratios and unfavourable
high permeability at the base, both give a concave downwards shape. With a mo-
bility ratio of M=16 the breakthrough and displacement is very poor and although
the mobility ratio of M=1 is better the result leads to a long displacement process to
recover the oil. In figure 48b however with the high permeability contrast to the top,
the mobility ratio gives piston like displacement over the whole displacement and
even for a high unfavourable mobility ratio contrast of 16 there is a breakthrough
point better than the mobility ratio of 1 with the opposite permeability contrast.
The impact of these various permeability contrasts on the water production development
versus time is illustrated in figure 49. The early water breakthrough at a low water cut
requires a considerable time to displace the oil. In the other extreme case the piston
like displacement obtained with a favourable mobility ratio and a permeability profile
increasing to the top leads to a delayed waterbreakthrough then a rapid increase in
water cut over a much shorter period.
a
Co
Do
M = 16.1
ning
0.60
fw
se
M = 16.1
Coar
0.40
wn
Do
M=1
ing
0.20
n
ar
se
Co
Mov = 0.594 HCPV
It is not practical to include exercises on all the topics in this text. Dake1,2 and Chierici4
give a number of examples illustrating the various aspects.
So far we have considered the displacement process from a predictive perspective. The
displacement equations generated can also be used to examine actual field waterflood
performance and then use the generated information to determine future waterflood
strategy. This novel approach has been developed by Dake2 in his second text.
His method is based on the development of a reservoir fractional flow curve from
production data. Production data provides values of oil production, water injection,
exit fractional flows as a function of time, i.e. Np. Wi,and fws. When these are converted
to dimensionless pore volume based values they become Npd, Wid and fwe.
N pd = Sw − Swi = ( Swe − Swi ) + 1 − fw ( S we )W id
(49)
the unknown Swe can be obtained, thus enabling a reservoir fractional flow curve to
be generated. Figure50 gives the water cut developed for a reservoir and its eratic
nature and figure 51 gives the generated reservoir fractional flow curve for a fractured
limestone reservoir. The function in figure 51 is quite smooth and often coverts the
more erratic watercut versus time profile into a manageable curve.
60
Immiscible Displacement
0.9
0.8
0.7
Watercut 0.6
0.5
0.4
0.3
0.2
0.1
0
0 50 100 150 200 250 300 350
Cumlative Production (MMstb)
Core Data
at Sor
0.8
0.6
fwe
0.4
0.2
0.525 0.65
0
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
Swe (PV)
Extrapolation of the fractional flow curve for the last few data points to the fw=1 posi-
tion gives a value for the average recovery obtained at that point, which in the example
presented gives a value of Sw avg.=0.525. The core value of the water saturation at
irreducible oil saturation, suggests a prediction extension of the reservoir fractional
flow curve as shown. Analysis of this curve would provide the future oil recovery
against time for the development assuming injection rates remain the same.
It is interesting to note for the example cited, Dake2 indicates that the operator had
made provision in the original facility design for such upgrading of water injection
and handling facilities.
This option of the operator being able to progress round the extended reservoir gen-
erated fractional flow curve to generate better recovery is an interesting approach.
Dake2 suggests that this simple approach provides an effective means in relating
the ultimate recovery to the total water injected over the field life and is particularly
useful for the ’difficult’ field where the fractional flow curve has the characteristic
concave downward shape.
The discussion has so far has centered around immiscible water oil displacement.
The methods generated however can be used for gas displacing oil systems where
the process is immiscible. That is when there is no component transfer between the
gas and oil phases. Again for a full account of the topic the reader is referred to
Dake (2).
From the presentation focused on water oil displacement a number of key perspectives
have been identified, the mobility ratio, the permeability variation and the impact of
gravity. In the context of gas oil systems these topics are even more significant.
Consequently gas drive is very unstable since the gas is able to move considerably
faster than the oil under the same pressure differential.
This unfavourable mobility ratio results in a very unfavourable fractional flow curve
for horizontal flow.
1
fgh =
µ g kro
1+
µo krg
(83)
62
Immiscible Displacement
Typical one dimensional gas-oil relative permeability and fractional flow curves are
given in figure 52 and shows the very unfavourable form of the fractional flow curve.
In reality in field applications the full laboratory relative permeability curves are not
used, but only the ‘end point’ values, since in gas drive systems gravity dominates
leading to segregated flow with a sharp gas oil interface.
k'ro 1.0
k'rg fg
kr
The equations used for calculating recovery are the same as for the water –oil
displacement, that is one can use the Buckley- Leverett theory and the associated
Welge analysis. For the recovery of oil in gas drive therefore the recovery equation
becomes
Npd = Sge + (1-fge)Gid (84)
where Gid is the pore volumes of gas injected, and which is from the Buckley Leverett
equation equal to the reciprocal of the slope of the fractional flow curve at the exit
gas saturation and fractional flow.
arge gravity impact can cause the gas to override the oil. Its control therefore is
difficult
a)
Gas
Oil
Oil
Production
Gas
Injection
b)
Gas Oil
The prediction of performance can be carried out in the same way as for the layered
two dimensional systems for water, where thickness averaged pseudo relative per-
meabilities are generated and then the calculation reduced to the one dimensional
form.
The thickness average saturations and relative permeability equation are as before
but in the context of gas oil values. In this case of course the layers flood from the
top to the base.
∑ h φ (1 − S )
n
j j or j − Swc j
j =l
Sg = N
∑h φ j j
j =l (85)
n
∑ h k k′
j =l
j j rg j
krg = N
∑h k j j
j =l (86)
n
∑ h k k′
j =l
j j ro j
kro = N
∑h k
j =l
j j
(87)
64
Immiscible Displacement
From these equations the fractional flow curve is generated and it is important to
note in this case the gravity impact is involved since
1− G
fg =
µ k
1 + g ro
µo krg
(88)
q
v′ =
Aφ (1 − Sor − Swc ) (90)
Using the fractional flow curve the method of Welge (equation 49) can then be used
to generate oil recovery, in relation to gas injected in a similar way as for water
injection. kk ro aDrsinθ
G = 2.743 × 10−3
vµ
The following equations are used
o
in the analysis
where S'g and fg are the averaged saturations and fractional flows determined on
the fractional flow curve after breakthrough.
GiD is the gas injected in pore volumes and is the reciprocal of the slope of the frac-
tional flow curve.
fg
bt A: Horizontal (fg = fgh)
B: v = 1 ft/d
C: v = 0.25 ft/d
B C
Sg = 0 1-Sor-Swc
Sg
Figure 54 Influence of the velocity of gas flood on the stability of frontal advance (Dake2)
Dake2 develops the equation for the GOR in terms of the fractional flow value and
the respective formation volume factors.
Bo
5.615
Bg
GOR = + Rs
1
f − 1
g scf/stb (92)
fg is the pseudo fractional flow for gas in consideration of the gravity term for
inclined systems
where
1− G
fg =
µk
1 + g ro
µo krg (93)
66
Immiscible Displacement
Dake2 provides a field example of the application of these equations for immiscible
displacement in a heterogeneous reservoir.
Sea Level
Gas Imported
from another
Field Sea Bed
Wet Gas
Dry Gas
Production
Injection
Dry
Gas
Wet
Gas
In this process we have dry gas displacing wet gas. The process is very costly with
respect to the compression facilities required and therefore a range of techno-economic
studies would be required before such a scheme is implemented. Two key parameters
would be involved not the least the price of oil to justify the recovery of otherwise
lost liquids, and also the flooding behaviour. The ideal situation would be for all the
wet gas to have been displaced when the dry gas breaksthrough into the wells and
For a miscible dry gas displacing wet gas senario, the core relative permeabilties are
simple linear functions since no residual wet gas saturation remains after contact
with dry gas. Figure 56
Dry Wet kr
Gas Gas
1-Swc
Miscible Sgd
Front
Figure 56 Dry gas recycling flooding experiment and resulting one-dimensional relative
permeabilities
The other important parameter in the development of the fractional flow is the mobility
ratio, which using example fluid data gives;
′
krgd k′ 1 1
M= / rgw = / = 1.5
µ gd µ gw 0.02 0.03
The resulting ratio although greater than 1, provides a reasonable stable displacement
perspective.
The heterogeneity of the reservoir will also have an impact on the process. High
permeabilties at the top will favour earlier breakthrough since the drier, less dense
gas will tend to migrate to the upper part of the reservoir. Whereas high permeabilties
towards the base will counter balance the density impact. Figure 57. High permeability
layers within the formation like water flooding can be a concern providing an easy
path for injected gas leading to early breakthrough.
68
Immiscible Displacement
Production
Injection
Permeability
Dry Wet
Thickness
a)
Production
Injection
Permeability
Wet
Dry
Thickness
b)
Solutions to Exercises
EXERCISE 1
Plot the water-oil relative permeability from the following data set. Indicate the end
point relative permeabilities.
Sw Krw Kro
0.20 0.0000 0.8800
0.25 0.0021 0.6710
0.30 0.0095 0.5170
0.35 0.0210 0.4070
0.40 0.0347 0.3135
0.45 0.0536 0.2420
0.50 0.0788 0.1793
0.55 0.1050 0.1320
0.60 0.1386 0.0891
0.65 0.1785 0.0550
0.70 0.2184 0.0297
0.75 0.2636 0.0110
0.80 0.3150 0.0000
Table E1
Plot
1.00
0.90
Relative Permeability
Krw
0.80 Kro
0.70
0.60
0.50
0.40
0.30
0.20
0.10
0.00
0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00
Figure E1
EXERCISE 2
Water is to be injected into a horizontal core, with the relative permeability charac-
teristics of table 1, to displace oil. Determine the mobility ratios M, and the fractional
flow curves for the following three cases.
1 35.0 0.5
2 4.5 0.5
3 0.4 1.0
Table E2
SOLUTION
k©rw / µ w
Mobility ratio = (1)
k©ro / µo
1
fw for horizontal flow = fw = (15)
µ w kro
1+
krw µo
70
Immiscible Displacement
Case mw/mo M
1 0.01 25.06
2 0.11 3.22
3 2.50 0.14
Table E3
The Fractional flow in the reservoir for the three cases can be calculated as fol-
lows:
Table E4
1.00
0.90
0.80 M = 25
fw (rbbl/rbbl)
0.70
0.60
0.50 M = 3.22
0.40
Case 1
0.30 M = 00.4 Case 2
Case 3
0.20
0.10
0.00
0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00
Sw
Figure E2
Data Summary
Oil formation volume factor Bo = 1.36 bbl/STB
Water formation volume factor Bw = 1.01 bbl/STB
Initial water saturation Swc = 0.20
Compare the values of the producing watercut (at surface conditions) and the cumulative
oil recovery at breakthrough for the following combinations.
1 35.0 0.5
2 4.5 0.5
3 0.4 1.0
SOLUTION
1
fw =
µ k
1 + w ro
krw µo
qw
Bw
fws =
qw qo
Bw + Bo
Combining the above two equations leads to an expression for the surface watercut
as:
1
fws =
B 1
1 + w − 1
Bo fw
72
Immiscible Displacement
Fraction flow for case 1,2,3 calculated using fractional flow equation above and
tabulated in table E5.
Equation 41 in text.
Table E5
1.00
0.90
0.80
0.70
fw (rbbl/rbbl)
0.60
0.50
0.40
0.30 Case 1
0.20 Case 2
0.10 Case 3
0.00
0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00
Sw
Figure E3 Fractional flow plots for different oil - water viscosity ratios
EXERCISE 4
Data Summary
Injection rate Qi = 1,200 bbl/d/well
Water viscosity µw = 0.5 cp
Oil viscosity µo = 4.5 cp
Initial water saturation Swc = 0.20
Residual oil saturation Sor = 0.20
Porosity φ = 0.22
Dip angle θ = 0°
Reservoir thickness h = 50 ft
Distance between injection wells w= 800 ft
Distance - injectors and producers L= 2,000 ft
Assuming that diffuse flow conditions prevail and that the injection project starts
simultaneously with oil production from the reservoir, determine:
2) the cumulative oil production as a function of both the cumulative water injected
and the time.
SOLUTION
The fractional flow in the reservoir (for horizontal flow) is calculated from:
1
fw =
µ w kro
1+
krw µo
74
Immiscible Displacement
µw/µo = 0.11
Results of the fractional flow calculations for this case are summarised in table E.7
fractional
Sw Krw Kro Kro/Krw flow fw
Table E7
0.70 breakthrough Sw
0.60
0.50
0.40
0.30
0.20
0.10
0.00
0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00
Sw
Figure E4
Widbt * PV
t=
Qi * 5.615 * 365
PV = h*W*L*φ
PV = 1.76E+07 ft3
t = 2.66 years
or 972 days
The oil recovery after breakthrough, expressed in pore volumes, can be calculated
using the equation:
Where:
1
Wid =
dfw
dSw
Swe
Allowing Swe, the water saturation at the producing end of the block, to rise in in-
crements of 5% (for Swe >=Swbt) the corresponding values of Wid are calculated in
Table 2.2.
76
Immiscible Displacement
At breakthrough => 0.450 0.625 0.425 0.950 0.050 1.515 0.5008 10.84
0.665 0.372 0.3720 2.66
0.675 0.475 0.975 0.025 3.125 0.5531 22.36
0.475 0.275 0.735 0.265 0.373 0.3739 2.67
0.725 0.525 0.988 0.012 3.846 0.5712 27.52
0.525 0.325 0.841 0.159 0.658 0.4296 4.71
0.575 0.375 0.905 0.0905 0.877 0.4583 6.28
0.625 0.425 0.950 0.050 1.515 0.5008 10.84
0.675 0.475 0.975 0.025 3.125 0.5531 22.36
0.725 0.525 0.988 0.012 3.846 0.5712 27.52
Table E9
The values of fwe* in table E9, have been obtained from figure 2.1, for the corresponding
values of Swe*. The oil recovery, in reservoir pore volumes, is plotted as a function
of Wid and time in figure 2.2. The maximum possible recovery is one Movable Oil
Volume, i.e. (1-Swc - Sor) = 0.6 PV.
0.50
0.40
Npd (PV)
0.30
0.20
0.10
0.00
0 1 1 2 2 3 3 4 4 5
Wid (PV)
Figure E5
EXERCISE 5
Additional Data
Absolute permeability K = 500 mD
Specific gravity of water in the reservoir γw = 1.02
Specific gravity of oil in the reservoir γo = 0.83
SOLUTION
From the relative permeability relations given in figure E6, the end point relative
permeability are:
78
Immiscible Displacement
1.00
0.90 krw
k'rw (end points)
0.80
Relative Permeability
kro
0.70 k'ro (end points)
0.60
0.50
1.00
0.40
0.90 krw
0.30 k'rw (end points)
0.80
Relative Permeability
0.20 kro
0.70 k'ro (end points)
0.10
0.60
0.00
0.50
0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00
0.40
0.30 Water Saturation (Sw)
0.20
0.10
0.00 Sw K'rw K'ro
0.20 0.30 Figure
0.00 0.10 0.20 0.0000.50
0.40 E6 0.880
0.60 0.70 0.80 0.90 1.00
0.80 0.315 0.000
Water Saturation (Sw)
Sw K'rw K'ro
0.20 0.000 0.880
0.80 0.315 0.000
Table E10
′ / µw
krw
M=
kro′ / µo
M = 3.2216
d flow
low
ef
fus
egate
Dif
1.000
Segr
Segregated Flow
0.800
0.600
fw
Diffuse Flow
0.400
0.200
0.000
0.00 0.2 0.4 0.6 0.8 1.0
Sw
Figure E7
Npd = Wid
where: WiD is the cumulative water injection expressed in Pore Volume (PV):
N pd bt = Savg − Swi
Time
WiD NpD t
(PV) (PV) (Years)
80 0.50
0.40
Time
WiD NpD t
(PV) (PV) (Years)
0.50
0.40
Npd(PV) Diffuse Flow
Segregated Flow
0.30
0.20
0.10
0.00
0 1 2 3 4 5
Wid (PV)
Figure E8
In figure E8 plots of the oil recovery, assuming the segegated flow condition, are
compared with the results of exercise 2, assuming diffuse flow. The comparison shows
that, although the breakthrough occurs much earlier for segregated flow, the ultimate
recovery is obtained sooner and for a much smaller throughput of water.
82
HERIOT WATT UNIVERSITY
DEPARTMENT OF PETROLEUM ENGINEERING
Reservoir Engineering 1
Tuesday
09.30 - 13.30
8. State clearly any assumptions used and intermediate calculations made in numerical
questions. No marks can be given for an incorrect answer if the method of calculation is
not presented.
(b) the black oil model description for the characterisation of a reservoir fluid.
(2)
A2 Explain briefly the importance of characterising the permeability variations in a reservoir in relation
to the prediction of the behaviour of natural and injected water drive systems. The answer should
be limited to the behaviour in the vertical plane rather than the areal plane
(3)
A4 Derive two equations in terms of composition and equilibrium ratios to determine the dew point
and bubble point pressure of a reservoir fluid. Explain briefly how the equations are used, when
the temperature of the reservoir and composition of the fluid are known.
(3)
(3)
A6 Briefly explain the need for the development of transient flow solutions to the diffusivity equation
in reservoir engineering.
(3)
A7 Describe the method by which the line source solution may be adapted to accommodate a zone of
reduced permeability around a wellbore (a skin).
(3)
SECTION B
B1
(i) Draw a pressure temperature diagram for a retrograde-gas condensate fluid and indicate the
key features. What is gas cycling and why in some cases is it used?
(6 marks)
(ii) The dew point pressure of a condensate gas field is 6250 psia. The initial reservoir conditions
are 240oF and 8500 psia.
When the reservoir was initially tested, a condensate to gas ratio of 80 stock tank barrels per
million SCF of gas was obtained. The produced gas and condensate compositions were as follows:
Gas Condensate
C1 Methane 0.89 —
C3 Propane 0.07 0.21
C5 N-Pentane 0.04 0.61
C8 Octane — 0.18
The reservoir pore volume is considered to be 5 x 1011 cu ft with a connate water saturation of 0.17.
Calculate the condensate fluids produced (STB) and the gas produced (SCF) in producing the reservoir
down to a pressure of 6750 psia.
1 bbl = 5.615 cu ft
o
R = 460+oF
1 lb mole = 379.4 SCF
R = 10.73 psi cu ft/lb mole oR
B2
(i) Explain briefly the three following tests carried out on reservoir fluid samples in relation to a
PVT study, and comment on their application.
(6 marks)
(ii) Explain briefly the constant volume depletion test for gas condensate
(4 marks)
(iii) Table 1 gives the results for a volume/pressure investigation of a reservoir fluid at reservoir
temperature. The system composition remained constant throughout the test.
Table 1
(System constant)
In another test on the fluid a sample of oil at its bubble point pressure and reservoir temperature in a PVT
cell were passed through a two stage separator at 100 psig and 75oF and 0 psig and 60oF. 34 cc of oil were
displaced from the PVT cell and 27.4 cc of oil were collected from the last separator stage. 4976 cc of gas
were collected at standard conditions during the test.
In a further test the pressure in a PVT cell at reservoir temperature was reduced in stages and the gas
produced at each stage removed and the remaining oil volume measured. The total gas produced at
standard conditions was recorded and is presented in Table 2.
Table 2
Pressure in Cumulative Volume of Oil
PVT Cell Gas Produced in Cell
psig cc cc
(standard conditions)
(a) Determine the bubble point pressure of the reservoir fluid at reservoir temperature.
(c) The solution gas-oil ratio at 3650 psig and 2700 psig
(e) The total formation volume factors at 3650 psig and 1200 psig.
(10 marks)
B3
(i) In the context of capillary pressure define the free water level.
(3 marks)
(ii) Explain briefly the reason for significant oil saturation remaining in the water swept zones of
a reservoir after natural water drive or water injection.
(5 marks)
(iii) Core samples have been obtained from a well and air-mercury capillary pressure curves
generated for an oil reservoir system (see attachment Figure 1). The lowest limit of 100% Sw was
found at the bottom of the well in rock type A as shown in the attachment Figure 2.
(a) Determine the free water level and indicate it on the well diagram provided.
(c) Calculate the oil-in-place per unit cross-section over the thickness of the reservoir.
B4
(i) Water drive reservoirs are said to be ‘rate sensitive’. Explain briefly this statement with
respect to different aquifer characteristics
(4 marks)
(ii) Explain briefly how the constant terminal pressure solution of the Hurst and van Everdingen
unsteady state theory can be used to predict water influx into an oil reservoir with a declining
reservoir pressure.
(4 marks)
(iii) A water drive reservoir extends to a radius of 15,000 ft. Sealing faults restrict the shape of
the reservoir to form only a part of the full radial system. The supporting acquifer is considered to
extend to 90,000 ft. The reservoir shape is given below.
Aquifer
140º
Time (months) 0 6 12 18 24
Pressure (psia) 6700 6688 6642 6584 6508
After the first 6 months 80,000 bbls of water were calculated to have influxed from the acquifer.
K = 180 mD
µw = 0.6 cp
porosity = 0.19
water compressibility = 3 x 10-6 psi-1
pore/rock compressibility = 4 x 10-6 psi-1
The Hurst & van Everdinger equation for a full radial system is:
We = 1.119∅cRo2h∆pWD
where
Charts are supplied of dimensionless water influx WD versus dimensionless time tD (see 2 attachments)
where:
kt
tD = 2.309
µ w φcR 20
t = time (years)
k = permeability (millidarcies)
µw = viscosity (cp)
(12 marks)
B5
(i) A radial oil reservoir of constant thickness has a single vertical well situated at its centre,
perforated the full thickness of the reservoir. The pressure everywhere is the initial reservoir
pressure. The outer boundary of the reservoir is closed. Describe the development of the pressure
profile from the well to the outer boundary as production continues. Assume single phase flow
and that the whole oil reservoir can be produced with no technical or economic limitations.
(5)
(ii) A well flows at a constant rate of 200stm3/day. Calculate the bottomhole flowing pressure
at 8 hours after the start of production. The well is vertical, perforated along the full thickness of
the reservoir.
Data
porosity, _ 25%
formation volume factor for oil, Bo 1.30rm3/stm3
net thickness of formation, h 50m
viscosity of reservoir oil, _ 2.2x10-3 Pas
compressibility, c 0.8x10-9Pa-1
permeability, k 120mD
wellbore radius, rw 0.15m
external radius, re 650m
initial reservoir pressure, Pi 270bar
well flowrate (constant) 200stm3/day
skin factor 0
(15)
B6 A well has been on production in an oil reservoir. For the following data, calculate the bottomhole
flowing pressure, Pwf for
and briefly describe the main differences in the flow regimes (8)
Data
05
1.6
1.
0.8 1.7
1
1.5
1.
1.45
Compressibility Factor, z
0.7
2
1.4 1.6
1.
1.35
3
1.
1.3
0.6 4 1.5
1.25 1.
1.5
1.2
1.7
0.5 1.6
1.9 1.4
1.8
1.15
2.0 2.2
0.4 1.3
1.1 2.4
2.6
3.0
0.3 1.2
1.05
0.25 3.0
2.8
1.1 1.1
2.6 2.4
Compressibility of
Natural Gases
2.2
2.0 1.9 1.2 (Jan. 1, 1941)
1.0 1.0
1.8 1.1
1.7 1.05
1.6 1.4
1.3
0.9 0.9
7 8 9 10 11 12 13 14 15
Pseudo Reduced Pressure, Pr
Air/Mercury Capillary Pressure Curves
k(md) 55 25 5 0.05 0.01
h(ft) φ(%) 15 10 10 5 5
150
C D E
A B
Capillary Pressure (psi)
100
50
0 10 20 30 40 50 60 70 80 90 100
Pore Space Not Occupied by Mercury
h (ft) Scale 2cm = 10ft
Rock Type
Top of
Reservoir
Water Saturation
100 80 60 40 20 0
PHYSICAL PROPERTIES OF THE PARAFFIN HYDROCARBONS & MISCELLANEOUS COMPOUNDS
3.0
reD = 2.5
Qt 2.5
2.0
reD = 2.0
1.5
1.0
reD = 1.5
0.5
0
0 1.0 2.0 3.0 4.0 5.0 6.0 7.0
tD
20
reD = 7.0
reD = 8.0
18
reD = ∞ r eD =
6 .0
16
14
Qt reD = 5.0
12
reD = 4.5
10
8 reD = 4.0
6 reD = 3.5
reD = 3.0
4
0
0 10 20 30 40 50 60 70
tD
110
100
∞=
D
re
90
80
70
Qt
60
50 reD = 10.0
reD = 9.0
40
reD = 8.0
30
reD = 7.0
20 reD = 6.0
reD = 5.0
10
0
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300
tD
10
reD =∞
8 reD = 6.0
reD = 5.0
6 reD = 4.5
reD = 4.0
4 reD = 3.5
reD = 3.0
reD = 2.5
Qt 2
reD = 2.0
8
6 reD = 1.5
10-1
10-2 2 4 6 8 10-1 2 4 6 8 1.0 2 4 6 8 10
tD
reD = 15.0
102 103
'd)
= ∞ o nt
8 (c
r eD
6 =∞
r eD reD = 10.0
reD = 9.0
4 reD = 8.0
reD = 7.0
WD 2
∞ (CONT'D) ONLY
reD = 6.0
reD = 5.0
10 reD = 4.5
102
8 reD = 4.0
6 reD = 3.5
reD = 3.0
4
reD = 2.5
2
USE THIS SCALE FOR LINE reD =
1 10
10 2 4 6 8 102 2 4 6 8 103 2 4 6 8 104
tD
Compressibility Factors for Natural Gases as a
Function of Pseudoreduced Pressure and Temperature.
05
1.6
1.
0.8 1.7
1
1.5
1.
1.45
Compressibility Factor, z
0.7
2
1.4 1.6
1.
1.35
3
1.
1.3
0.6 4 1.5
1.25 1.
1.5
1.2
1.7
0.5 1.6
1.9 1.4
1.8
1.15
2.0 2.2
0.4 1.3
1.1 2.4
2.6
3.0
0.3 1.2
1.05
0.25 3.0
2.8
1.1 1.1
2.6 2.4
Compressibility of
Natural Gases
2.2
2.0 1.9 1.2 (Jan. 1, 1941)
1.0 1.0
1.8 1.1
1.7 1.05
1.6 1.4
1.3
0.9 0.9
7 8 9 10 11 12 13 14 15
Pseudo Reduced Pressure, Pr
Air/Mercury Capillary Pressure Curves
k(md) 55 25 5 0.05 0.01
h(ft) φ(%) 15 10 10 5 5
150
C D E
A B
Capillary Pressure (psi)
100
50
0 10 20 30 40 50 60 70 80 90 100
Pore Space Not Occupied by Mercury
h (ft) Scale 2cm = 10ft
Rock Type
Top of
Reservoir
Water Saturation
100 80 60 40 20 0
PHYSICAL PROPERTIES OF THE PARAFFIN HYDROCARBONS & MISCELLANEOUS COMPOUNDS
3.0
reD = 2.5
Qt 2.5
2.0
reD = 2.0
1.5
1.0
reD = 1.5
0.5
0
0 1.0 2.0 3.0 4.0 5.0 6.0 7.0
tD
20
reD = 7.0
reD = 8.0
18
reD = ∞ r eD =
6 .0
16
14
Qt reD = 5.0
12
reD = 4.5
10
8 reD = 4.0
6 reD = 3.5
reD = 3.0
4
0
0 10 20 30 40 50 60 70
tD
110
100
∞=
D
re
90
80
70
Qt
60
50 reD = 10.0
reD = 9.0
40
reD = 8.0
30
reD = 7.0
20 reD = 6.0
reD = 5.0
10
0
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300
tD
10
reD =∞
8 reD = 6.0
reD = 5.0
6 reD = 4.5
reD = 4.0
4 reD = 3.5
reD = 3.0
reD = 2.5
Qt 2
reD = 2.0
8
6 reD = 1.5
10-1
10-2 2 4 6 8 10-1 2 4 6 8 1.0 2 4 6 8 10
tD
reD = 15.0
102 103
'd)
= ∞ o nt
8 (c
r eD
6 =∞
r eD reD = 10.0
reD = 9.0
4 reD = 8.0
reD = 7.0
WD 2
∞ (CONT'D) ONLY
reD = 6.0
reD = 5.0
10 reD = 4.5
102
8 reD = 4.0
6 reD = 3.5
reD = 3.0
4
reD = 2.5
2
USE THIS SCALE FOR LINE reD =
1 10
10 2 4 6 8 102 2 4 6 8 103 2 4 6 8 104
tD
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Model Solutions to Examination
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Date:
Reservoir Engineering
Subject:
INSTRUCTIONS TO CANDIDATES
No. Mk.
1. Complete the sections above but do not seal until the examination is finished.
5. This book must be handed in entire with the top corner sealed.
6. Additional books must bear the name of the candidate, be sealed and be affixed to
the first book by means of a tag provided
1
2
Model Solutions to Examination
SECTION A
1. Compositional Model
Can also include non paraffin components in some cases PNA analysis
Black-oil model
of separators.
3
k rg Adp
qg =
µ g dx - D’arcy’s law
k ro Adp
qo =
µ o dx
Q o = q o Bo
Qg k Adp k ro Adp
= R s + rg
Qo µ g dxBg µ o dxBo
krg µ oB o
Instantaneous GOR = R s +
kroµ gB g
Therefore zj = kj x xj
ie zj = xj
Therefore zj = yj/kj
Composition of gas yj = zj kj
4
Model Solutions to Examination
b) Overpressure reservoirs
5
0 Pressure
Normal Pressured
Gas Gradient
GAS
Oil Gradient
Depth
OIL
Water Gradient
WATER
Over Pressured
long time to develop, or may never develop (i.e. the flow may never
tests were required and these required transient solutions which could
the wellbore is a line source (or sink). The calculation of the pressure
within the reservoir based on the well flow and drawdown can then be
made as long as the fluid is in transient flow. The effect of skin around
the well is to reduce the permeability such that for a given drawdown,
6
Model Solutions to Examination
the flow rate reduces. To maintain the original flow rate in the
skin will manifest itself after the pressure perturbation has moved off
into the reservoir, the flow across the skin zone can be assumed to be
undamaged region, a skin factor can be defined. Thus the skin factor is
Section B
Question 1
Critical Point
(i)
Cricondenbar Region of Retrograde Condensation
Gas Cycling: Gas cycling is the reinjection of separated gas back into
the reservoir - in some cases with gases from other near reservoirs.
7
It is used to keep the reservoir above the dew point line - preventing
(ii)
C1 16.04 0.89 - - - -
C3 44.09 0.07 0.21 31.66 9.2589 0.29
C5 72.15 0.04 0.61 39.36 44.0115 1.12
C8 114.2 - 0.18 44.09 20.5560 0.47
Σ 73.826 1.88
= 12500 SCF/STB
lb mole gas
GOR = 11.0127
lb mole oil
8
Model Solutions to Examination
z = 1.15
9
1010 × 1 moles liq
= 2.36 ×
12.0127 total moles
Gas produced
11.0127 SCF
= 2.36 × × 379.4
12.0127 total moles lb ml
QUESTION B2
PART 1
10
Model Solutions to Examination
pressure
P
Gas
P Oil
Hg
Pb
Vb V
b) Separator Test
• gas collected from each stage collected and final volume of oil at
conditions
11
c) Differential Test
PART 2
retrograde condensation
P Pd
12
Model Solutions to Examination
PART 3
Draw graph and determine intersection of oil and oil and gas slope
= 1.241 bbl/stb
= 1.02409 x 164.42/168.08
13
Same value since both above Bp.
SCF
= 177.99
bbl residual oil
= 167.33 SCF/STB
= 1019.72 – 167.33
14
Model Solutions to Examination
1800
1700
Pb = 1595 psi
1600
1500
1400
166 167 168 169 170 171 172 173 174
15
(e) BT @ 3650 = Bo @ 3650 = 1.2138
197.28 − 186.95 × 34
at 1200 = 186.95 +
1234 − 1113
= 186.95 + 2.9026
= 1.3713
QUESTION 10
The free water level is the position of zero capillary pressure and lies
below the last 100% water saturation level, the oil-water contact
(OWC).
16
Model Solutions to Examination
Pc
OWC
FWL 0
0 100%
Water Saturation
Water
narrow pores than in larger pores as above. The capillary forces then
isolate oil in larger pore as above which is then held by capillary forces
17
Pc = h∆pg
lb f lbm
pc 2
= h(ft )∆p 3 × g
1n ft
lbf = lbm x g
lb m × g 144in 2 lb
= Pc 2
× 2
= h ft × (1.03 − 0.8) × g × 62.4 m3
in ft ft
Pc × 144
h=
(1.03 − 0.8) × 62.4
2.30
=
0.23
h ft = Pc air/Hg psi
18
Model Solutions to Examination
h(ft) φ(%) 15 10 10 5 5
150
C D E
A B
Capillary Pressure (psi)
100
100
95.5
85.5 3
82'=18
5
69.5
49.5 50
36.75
23'
17.5
FWL
0 10 20 30 40 50 60 70 80 90 100
Pore Space Not Occupied by Mercury
19
h (ft) Scale 2cm = 10ft
Rock Type
Top of
Reservoir
103.2' 91
A (90)
95.5' 45
90
D (44)
85.5'
10 43
82'
(5)
E
69.5'
44
C
(35)
49.5' 83
A
(79)
36.75'
44 76
(22)
B
23' 23'
17.5' 46
A
(23)
6'
100% Sw in Rock Type FWL
A Found at This Level 0 20 40 60 80 100
Oil Saturation
Water Saturation
100 80 60 40 20 0
20
Model Solutions to Examination
Oil in Place: ∑h x So x φ
3.4924
21
QUESTION 11
(a) Artesian well type whose oil production is less than aquifer influx
(c) Sealed system where oil production is greater then aquifer system
Type b
P Type a
Type c
Pressure at OWC
Time
(ii) Hurst and Van Everdingen unsteady state solution is for a fixed
22
Model Solutions to Examination
cover the decline then the water flux is B∑∆PQt, where the effective
water influx for each ∆P as a function of time is calculated and added
Pi ∆P1=(Pi-P1)/2
∆P2=(Pi-P1)/2+(P1-P2)/2
=(Pi-P2)/2
P
∆P3=(P1-P2)/2+(P2-P3)/2
=(P1-P3)/2
1 2 3 4
T
We = 1.119φ cRo2h∆PQt
kt
t D = 2.309 ×
µ w φcR 2o
23
Time Yrs. 0 0.5 1.0 1.5 2.0
Pressure 6700 6688 6642 6584 6508
∆P - 6 29 52 67
tD 0 1.1574 2.3148 3.4722 4.6296
Qt (Chart) 0 1.88 2.74 3.65 4.33
6
We=BΣ∆pQt - 80,000 495,016 1,672,657 3.54x10
h = 50 ft.
B = 139.52 x h = 6976
h = thickness = 50 ft
24
Model Solutions to Examination
QUESTION 12
(i) The well is in the centre of a radial reservoir and the full thickness is
the flow. The well is assumed to have a constant flow rate from the
in terms of:
(a) an initial transient flow regime where the fluid acts as if it were
out into the reservoir. The transient nature of the flow regime
rate than the pressure farther into the formation. Initially, the
(b) a flow regime that mimics steady state flow, termed semi-steady
reservoir from the wellbore to the outer boundary does not change
reservoir remain the same (as in steady state flow) but the
pressure wave has developed across the reservoir and it slowly sinks
25
(ii) The line source solution is used to determine the pressures required
at the specified radius and at the specified time. Checks are made to
ensure that:
(i) there has been adequate time since the start of production to allow
A Check Ei applicability
100φµcrw2
t>
k
φµcre2
the reservoir is infinite acting if the time, t <
4k
26
Model Solutions to Examination
25φµcr 2
the ln approximation is valid if the time, t >
k
25x0.25x2.2x10 −3 x0.8x10 −9 x0.152
t>
120x10 -15
t > 2.1s
qµBo γφµcrw2
(ii) Pwf = Pi + ln (taking account of the conversion from
4πkh 4kt
stock tank to reservoir conditions via the formation volume factor for oil).
−3 −9 2
φµcrw2 0.25x2.2x10 x0.8x10 0.15
= = 716x10-9
4kt 4x120x10 -15 x8x3600
= 270x105 - 1191726
= 25808274Pa
= 258.1bar
27
QUESTION 13
qµBo re 1
P − Pwf = ln −
2πkh rw 2
qµBo re 1
Pwf = P − ln −
2πkh rw 2
Pwf = 260.4bar
qµBo re 3
P − Pwf = ln −
2πkh rw 4
qµBo re 3
Pwf = P − ln −
2πkh rw 4
Pwf = 260.7bar
28
Model Solutions to Examination
The main differences in the flow regimes is the effect of time. In the
steady state flow regime, there is the assumption that the original
through the reservoir to the oil water contact. There has been an
well, and the pressure profile that has developed in the reservoir is
state flow regime has also reached the outer boundary, but there is no
the constant pressure gradient in the steady state flow regime. In this
flow regime, the absolute pressure in the reseroir fluid declines
29
Course:- 28117
Class:- 289013
HERIOT-WATT UNIVERSITY
DEPARTMENT OF PETROLEUM ENGINEERING
Reservoir Engineering 1
8 State clearly any assumptions used and intermediate calculations made in numerical
questions. No marks can be given for an incorrect answer if the method of
calculation is not presented.
A1 Define:-
B8
(a) Methane is a significant component in reservoir fluids.
(b) A wet gas is producing at 40,000 SCF/STB, from a reservoir which has
been estimated from petrophysics to have a volume of 1.1 x 1010 cu ft.
and has a pressure of 3,000 psia and a temperature of 250°F
(ii) What are the reserves of gas and condensate at 3,000 psia?
(iii) How much gas will have been produced, when the pressure has
dropped to 2,000 psia?
[14]
(b) A laboratory cell, contained 290cc of reservoir liquid at its bubble point
of 2100 psia at 145°F. 21cc of mercury were removed from the cell and
the pressure dropped to 1700 psia. Mercury was then re-injected at
constant temperature and pressure and 0.153 SCF of gas was removed
leaving 270cc of liquid in the cell. The process was repeated reducing
the pressure to 14.7 psia and the temperature to 60°F. Then 0.45 SCF of
gas was removed and 207.5cc of liquid remained in the cell.
Determine:-
(b) The system below represents the common arrangement for measuring the
permeability of a core plug using a gas.
(d) Capillary pressure data are obtained from core samples which represent a
small part of the reservoir. Leverett derived a “J” Function using the
Poiseuille equation for laminar flow:-
πr 4 ∆P
q=
8µL
Derive the 'J' function and comment on one of its limiting assumptions.
[6]
B12
(a) Identify the various elements in the material balance equation below:-
( N − N p )Bo = NBoi − Bg [ NR si − ( N − N p )R s − G ps ] −
[(G − G pc )Bg − GBgi ] − ( We − Wp Bw ) −
c + Swc c w
NBoi (1 + m )∆p f − ( WinjBw + G injBg )
1 − Swc
(b) Simplify the material balance equation above so that it can be used for an
undersaturated reservoir without waterdrive and water production.
Derive the instantaneous gas-oil ratio equation and use the equation to
explain briefly the shape of the producing GOR of a depletion type
reservoir from a pressure above the bubble point to one significantly
below the bubble point pressure.
[6]
(d) Tarner’s method and Tracy's modification of Tarner's method use the
Material Balance equation, the Instantaneous GOR equation and the
Saturation equation to predict oil production as a function of pressure for
a solution gas drive reservoir from the bubble point pressure.
In a formation of 80ft thickness and 20% porosity, estimate how much the
well productivity index will be increased if 1000bbl of acid are injected
into the formation, assuming piston displacement of connate water and
oil to a residual saturation of 0.35.
End of Paper
Course:- 28117
Class:- 289013
HERIOT-WATT UNIVERSITY
DEPARTMENT OF PETROLEUM ENGINEERING
Reservoir Engineering 1
Friday 8 January 1999
09.30 - 13.30
A3 Derive the instantaneous gas-oil ratio equation and plot the shape of
the producing GOR versus pressure for a solution gas drive reservoir
[3]
(b) Explain briefly gas cycling with reference to gas condensate reservoirs.
[3]
(c) A wet gas reservoir is producing gas and condensate with the
compositions given below at a gas-oil ratio of 25,000 SCF/STB. The
average reservoir temperature is 260˚F and the initial reservoir pressure
is 8520 psia.
1bbl = 5.615 cu ft
0˚F = 460˚ R
1lb mole = 379.4 SCF
R = 10.73 psi cu ft/lb mole° R
Gas Condensate
C1 Methane 0.890 --
C3 Propane 0.075 0.215
C5 n-Pentane 0.035 0.620
C8 Octane -- 0.165
Attachments
B9
(a) After natural water drive or injected water drive residual oil is left
within that part of the rock contacted by the water. Comment briefly
with the aid of a sketch why this might be.
[5]
(b) Miscible gas injection can be used to recover the residual oil after a
water flood. Explain briefly what miscible gas injection is and why in
some cases gas injected is alternated with water injection in a WAG
(water alternating gas ) process.
[5]
(c) An edge water drive reservoir extends to a radius of 8,000 ft. Sealing
faults are such that the water influx only forms part of a full radial
system as the sketch below illustrates. The supporting aquifer extends
to a radius of 40,000ft. Over the first two years of production the
pressure decline is expected to be as follows:
Time( months) 0 6 12 18 24
Pressure(psia) 4640 4630 4612 4584 4448
After the first 6 month period 23,200 bbls. of water were estimated to
have influxed from the aquifer.
The properties common to the oil reservoir and the adjoining aquifer are
as follows:
The Hurst & van Everdingen equation for a full radial flow system is:
[10]
We = 1.119φcR o2 h∆pWD
where
where
kt
t D = 2.309
µ w φcR 2o
t = time ( years)
k = permeability (millidarcies)
µw = viscosity (cp)
(b) Explain briefly why surface samples from a wet gas or condensate
reservoir can be unrepresentative if collected too early after a shut
down or major well disturbance. What suggestions would you give to
get more representative samples.
[4]
(c) Table 1 gives the results for a constant mass study on an oil sample. In
the test the volume of live oil was measured as a function of pressure.
The temperature was maintained at the reservoir temperature of 220 ˚F.
In another test a sample of the same oil in a PVT cell,at its bubble point
and at 220 ˚F was passed through a two stage separator at 500 psig
and 160 ˚F and 0 psig and 60 ˚F. 38 cc of oil were removed from the
PVT cell and 28.34 cc of oil were collected from the final separator. A
total amount of 3492 cc of gas at 60˚F and 0 psig were collected from
both stages of the separation process.
(ii) The oil formation volume factor at 4400 psig and 3925 psig.
(iii) The solution gas to oil ratio at 4400 psig and 3925 psig.
(iv) The oil formation volume factor and solution gas to oil ratio you
would use for reservoir calculations at 1605 psig.
Pressure (psig) 5500 5000 4500 4350 4173 4000 3800 3675
Pressure (psig) 3643 3594 3493 3446 3293 3146 2954 2713
(b) A well penetrates an oil reservoir from which core samples have been
collected and capillary pressure curves generated using the air-mercury
method. The capillary pressure data for the specific rock types are
given in the attached figure. The lowest 100% Sw saturation level was
found at the bottom of the well in rock A as indicated.
(i) Determine the free water level and indicate it on the diagram.
[2]
(iii) Estimate the oil -in place per unit cross section area over the
reservoir thickness.
[5]
Data
The specific gravities of water and oil relative to water density at 60 ˚F are
1.03 & 0.8 respectively. The density of water at 60˚F is 62.4 lbm/cu.ft.
1 lbf = 1 lbm x g.
psi = lbf / sq inch
B12 You have been appointed as the engineering manager of a reservoir.
Given that you appreciate the importance of allowing for the activation
of rock mechanical phenomena ( stress-sensitivity) in the management
of the reservoir, explain how you would screen the reservoir for stress -
sensitivity at the reservoir scale by listing and describing:
C13 A “super well” has been devised for high permeability reservoirs in
which individual vertical wells are drilled in two concentric circles as
shown in the diagram. Each ring has the same number of wells but
staggered in position as shown. Yaxley has given an expression for the
Dietz shape factor for a well in a triangular drainage area, also shown as
a diagram. Sketch the approximate shape of the well drainage areas for
the “super well” configuration and show how Yaxley’s formula may
be used to obtain the productivity index of the wells in the inner and
outer rings.
4A
CA =
4 π re 3 θro
γ exp ln − + 2 ln
θ ro 4 πθ
2 π sin o
θ
θ θ
A= πre2 = re2
where 2π 2
or
Derive the Hawkins equation for the damage skin factor in an open-
hole completion. Supposing the skin factor, S , has been measured in a
transient well test what additional information could be used to
estimate the depth of damage, ra , and hence the damage zone
permeability, ka .
[20]
Closed Outer
Boundary
2∆p
qi = Cch At
ρ1 − σ2
Dt
σ = D
p
Dt = throat diameter
Dp = pipe diameter
[20]
End of Paper
q Balanced Production Using
Variable Chokes
Layer 1
S1 k1 h1 p
1
q
1
Layer 2
q S2 k2 h2 p2
2
q=q +q
1 2 Q14
BEHAVIOUR OF OIL FIELD HYDROCARBON SYSTEMS
05
1.6
1.
0.8 1.7
1
1.5
1.
1.45
Compressibility Factor, z
0.7
2
1.4 1.6
1.
1.35
3
1.
1.3
0.6 4 1.5
1.25 1.
15
.
1.2
1.7
0.5 1.6
1.9 1.4
1.8
1.15
2.0 2.2
0.4 1.3
1.1 2.4
2.6
3.0
0.3 1.2
1.05
0.25 3.0
2.8
1.1 1.1
2.6 2.4
Compressibility of
Natural Gases
2.2
2.0 1.9 1.2 (Jan. 1, 1941)
1.0 1.0
1.8 1.1
1.7 1.05
1.6 1.4
1.3
0.9 0.9
7 8 9 10 11 12 13 14 15
Pseudo Reduced Pressure, Pr
46
PVT Analysis