Reservoir Engineering

Reservoir Engineering

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1 INTRODUCTION TO RESERVOIR ENGINEERING 2 RESERVOIR PRESSURES AND TEMPERATURES 3 RESERVOIR FLUIDS COMPOSITION 4 PHASE BEHAVIOUR OF HYDROCARBON SYSTEMS 5 BEHAVIOUR OF GASES 6 PROPERTIES OF RESERVOIR LIQUIDS 7 FUNDAMENTAL PROPERTIES OF RESERVOIR ROCKS 8 ROCK PROPERTIES MEASUREMENT 9 PERMEABILITY-ITS VARIATIONS 10 FLUID FLOW IN POROUS MEDIA 11 DRIVE MECHANISMS 12 VAPOUR LIQUID EQILIBRIA 13 EQUILIBRIUM RATIO PREDICTION AND CALCULATION 14 PVT ANALYSIS 15 MATERIAL BALANCE EQUATION 16 MATERIAL BALANCE EQUATION APPLICATION 17 WATER INFLUX 18 IMMISCIBLE DISPLACEMENT 19 EXAMINATION AND MODEL SOLUTIONS

Reservoir Engineering

Petroleum Engineering

RESERVOIR ENGINEERING

RE

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. 1.Fundamentals of Reservoir Engineering.

L.P.Dake. Elsevier. 1978 ISBN:0-444-41667-6

2.The Practise of Reservoir Engineering.

L.P.Dake. Elsevier. 1994. ISBN: 0-444-82094-9

3.Principles of Petroleum Reservoir Engineering.

G.H.Chierici. Springer-Verlag 1994. ISBN:3-540-56037-8

4.Fundamental Principles of Petroleum Reservoir Engineering

B.F. Towler. Society of Petroleum Engineers Inc ISBN:55563-092-8

5.Applied Reservoir Engineering

B.C.Craft & M.F.Hawkins. Prentice Hall. 1959.

6.The Properties of Petroleum Fluids 2nd Ed

W.D.McCain Pennwell Books . 1990 ISBN:0-87814-335-1

7.Petroleum Engineering Principles and Practise.

J.S.Archer & C.Wall.Graham & Trotman. 1986. ISBN:0-86910-715-9

8.Petroleum Reservoir Engineering.

J.W.Amyx,D.M.Bass & R.L.Whiting. McGraw-Hill. 1960. ISBN:07-001600-3

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.

Units Parameter Symbol To convert From Oilfield Units

to

Multiply by

Area

A

ft2

m cm in

9.29 x 10 9.29 x 10 1.44 x 10

Density

ρ

lb/gal (ppg)

kg/m33 g/cm3 lb/ft lb/bbl

1.198 x 10-12 1.198 x 10 7.48 42

Force

F

lb

N dyne

4.45 4.45 x 105

L, l D, d h

ft micron

m mile m in.

30.48 x 10 1.89 x 10 1.0 x 10-6-5 3.94 x 10

Mass

m

lb

kg tonne short ton

4.54 x 10-1-4 4.54 x 10-4 5.0 x 10

Power

HP

horsepower (HP)

kW

7.46 x 10

Pressure

P

lb/in2

Pascal bar dyne/cm3 atmosphere

6.89 x 10-23 6.89 x 10 4 6.89 x 10 6.89 x 10-2

Velocity

v

ft/sec

m/sec

30.48 x 10

Viscosity

µ

cp

pascal-sec

1.0 x 10

Volume

V

bbl

m cm ft gallon in3

1.59 x 10 1.59 x 10 5.615 42 9.70 x 103

Flowrate

Q

gpm

m /sec ft /min bbl/min. bbl/day

6.31 x 10 1.337 x 10 2.381 x 10 3.429 x 10

Length Depth Height

2 2 2

3 3 3

3 3





























-2 2 2

-2 -4

-1

-2

-3

-1 5

-5 -1 -2 1

Pounds mass conversion to pounds force. The relationship F=m.g may be used with non-metric units. If those units do not form a consistent set of units, the more general form F=k.m.g must be used, where the constant k is a conversion factor dependent upon the units used. For example, in imperial engineering units, F is in “pounds force” or “lbf”, m is in “pounds mass” or “lb”, and g is in feet per second squared. However, in this particular system, you need to use the more general form above, usually written F=m.g/gc with the constant normally used for this purpose gc = 32.174 lbm.ft/(lbf.s2) (this is the reciprocal of the k above). Thus 1lbf= 1lbm x g where g = 1in imperial units and is taken to be the ratio g/gc where g = acceleration due to gravity (ft/sec2) and gc = gravitational constant -32.17 ft-lbm/lbf-sec2 In SI units, Force in Newtons = Mass in kilograms* Acceleration due to gravity F=mg 1N =1kg * 9.81ms-2 and the constant gc in this case is 1 and is usually omitted because the units system is inconsistent 22/07/14

Preface

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Reservoir Engineering

Preface

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:

5 4 3 2 1

- Core material for examination purposes - Core material less analytical than 5 - examinable. - Between 4 & 2 - General awareness. Not so examinable with respect to analysis of detail. - Other information not examinable.

OM- Material covered in another module not for examination purposes in Reservoir Engineering. Equations – It is not necessary to memorise complicated equations. Equations unless asked to be derived will be given. Clearly some basic equations one should know and would not be given e.g.

Darcy’s Law, PV = nzRT STOOIP equation Equilibrium Ratio K=y/x

Mobility Ratio = M =

krw ′ kro′ / µw µo

Capillary pressure equation, Pc =

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2σ Cosθ = ρ gh r

Insitute of Petroleum Engineering, Heriot-Watt University

3

4

Chapter 2 Reservoir Pressures 1 5 2 5 3 5 4 4 5 3 6 3

Chapter 1 Introduction Section – grading 1.1 2/OM 1.2 2/OM 1.3 2/OM 2 4 3 – 3.1 4 3.2 4 3.3 4 3.4 3 4 4 5 OM 5.1 4 5.2 3 5.3 OM 5.4 OM 5.5 OM 5.6 OM 5.7 2 5.8 2 5.9 1 5.10 1 6 OM 7 3 8 3 8.3 3 9 3 10 3 Chapter 5 Gases 1.1 5 1.2 5 1.3 5 1.4 5 1.5 5 1.6 5 1.7 5 1.8 5 1.9 5 2.1 5 2.2 5 2.3 5 2.4 1 2.5 5 3 5 4 2 5 2 6 2

Chapter 4 Phase Behaviour All material 5

Chapter 3 Reservoir Composition 1 5 2 5 2.1 5 2.2 3 2.3 3 2.4 3 3 4 4 3 5.1 3 5.2 2 5.3 2

Chapter 7 Reservoir Rocks 1 3 2 3 3 3 4.1 5 4.2 4 4.3 3 4.4 3 4.5 4 4.6 4 4.7 3 4.8 5 4.9 2 4.10 2 5 3 6 5 7.1 5 7.2 5 7.3 5 8.1 5 8.2 5 8.3 5

Chapter 6 Liquids 1 5 2 5 3 5 4 5 5 5 6 5 7 3 8.1 5 8.2 5 9 5 10 3 11 1 12 5

Chapter 10 Fluid Flow 1 3 2 3 3.1 3 3.2 3 3.3.1 3 3.3.2 3 3.3.3 3 3.3.3.1 5 3.3.4 5 3.4 5 3.4.1 3 3.5 5 4 1 5 5 5.2 5 5.3 5 6 5

Chapter 9 - Permeability Variations 1 3 2 5 3 4

Chapter 8 Rock Measurement 1.1 2 2.1 2 2.2 2 3.1 2 3.2 2 4.1 3 4.2 3 4.3 3 4.4 3 5 2 6.1 5 6.2 3 6.3 5 6.4 3 7 2

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Vapour Liquid Equilibrium 2 2 Eq 11 – 5 4 5 5 5 3

Chapter 14 PVT 1 4 2 2 3.1 5 3.2 5 3.3 5 3.4 3 3.5 3 4 5 5 3 6 5 7 3 8.1 2 8.2 2 9 5 10 5 11 5 12 5 13 3 14 3 15 1

Chapter 13 Equilibrium Ratio 1 3 2 3 3 2 4 2

Chapter 12 1 2 3 4.1 4.2 4.3 5

Chapter 11 Drive Mechanisms All sections 5

Insitute of Petroleum Engineering, Heriot-Watt University

MB Application 5 5 5 5 5 (5.1.2.2 Eq46 -1 ) 4 4 2 5 5 2 2 1 3 1

Chapter 17 Water Influx 1 5 2.1 2 2.2 3 2.3 3 2.4 3 2.5 5 3 5 4 4 5 3 6 2 7 2

Chapter 16 1 2 3 4 5.1 5.2 5.3.1 5.3.2 5.3.3 5.3.4 5.3.5 5.4 5.5 5.6 6

Chapter 15 Material Balance 1 5 2 3 3 5 4 5 5 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 Chapter 18 Immiscible Displacement 1 5 2 5 3.1 3 3.2 5 3.3 3 ( Eqn 6 – 9 -should be expected to know ) 3.4 4 ( post equation 14 – 5) 4 4 5.1 2 5.2 5 ( from equation 72, – 2) 6.1 3 6.2 3 6.3 3 6.4 3 6.5 3 6.6 1 6.7 4 7 5 8 1

Preface

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Reservoir Engineering

Introduction To Reservoir Engineering O N E

47/2 47/3

Completed Producers

C Gw

C

C.I. = 50ft.

960

47/7 47/8

G

0

w

9 95500 00

Proposed Well Locations x Abandoned Wells

955000 8 95 50 8 94 00 94 50 93 00 8 8 8 93 8 B 250 9 A2 00 8 92 8 A

47/8-1x 8

A

8

8 A A5

A A4

A

Platform A 91001.0

0.9

50

91

A6

93 'Proven' 5

00 93 50 92 9200

Probability that the reserve is at least as large as indicated.

A3

0

x 47/8-2

0.5

0.1 0

'Proven + Probable'

'Proven + Proable + Possible' Recoverable Reserve

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Introduction To Reservoir Engineering O N E

C O N T E N T S 1 INTRODUCTION 1.1 Reserves Estimation 1.2 Development Planning 1.3 Production Operations Optimsation 2 RESERVOIR ENGINEERING TECHNIQUES 3 RESERVE ESTIMATING 3.1 Definitions 3.2 Proven Reserves 3.2.1 Exercises – Reserve Definitions 3.3 Unproved Reserves 3.3.1 Probable Reserves 3.3.2 Possible Reserves 3.4 Reserve Status Categories 3.4.1 Developed: 3.4.1.1 Producing 3.4.1.2 Non-producing: 3.4.2 Undeveloped Reserves:

6 OTHER APPRAISAL ROLES 7 DEVELOPMENT PLANNING 7.1 Reservoir Modelling 7.2 Technoconomics 7.3 Coping with Uncertainty 8 PRODUCTION OPERATIONS OPTIMISATION 8.1 Development Phase 8.2 History Matching 8.3 Phases of Development 9 THE UNIQUENESS OF THE RESERVOIR 10 CONCLUSIONS

4 PROBABILISTIC REPRESENTATION OF RESERVES 5 VOLUME IN – PLACE CALCULATIONS 5.1 Volume of Oil and Gas in-Place 5.2 Evolution of Reserve Estimate 5.3 Reservoir Area 5.4 Reservoir Thickness 5.5 Reservoir Porosity 5.6 Water Saturation 5.7 Formation Volume Factors 5.8 Recovery Factors 5.9 Production Capacity 5.10 Hydrocarbon Pore Volume Map

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LEARNING OBJECTIVES Having worked through this chapter the Student will be able to:

2

•

Show using a block diagram the integration of reservoir engineering with other petroleum engineering and other subjects.

•

Sketch a diagram showing the probability versus recoverable reserves indicating, proven, proven + probable and proven + probable + possible reserves.

•

Present a simple equation for volumes of oil and gas in-place.

•

Illustrate with a sketch the impact of different technical parameters on the associated uncertainties on a project.

•

Describe in general terms, in the context of production operations optimization, history matching.

•

Draw a sketch showing the various phases of production from build up to economic limit.

•

Draw a sketch illustrating the various recovery scenarios from primary to tertiary recovery.

Reservoir Engineering

Introduction To Reservoir Engineering O N E

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/ Geology

Petrophysics

Reservoir Engineering

Economics (Project viability)

Production Process Engineering General Engineering Platform Topsides Design

2

Figure 1 Reservoir Engineering in Relation to Other Activities (adapted Dake ).

Dake2 has usefully specified the distinct technical responsibilities of reservoir engineers as:

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•

Contributing, with the geologists and petrophysicists , to the estimation of hydrocarbons in place.

•

Determining the fraction of discovered hydrocarbons that can be recovered.

•

Attaching a time scale to the recovery.

•

Day-to-day operational reservoir engineering throughout the project lifetime.

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Petroleum Engineering

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 commences 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 categories: (i) Reserves Estimation (ii) Development Planning (iii) Production Operations Optimisation

1.1 Reserves Estimation

The underground reserves of oil and gas form the oil company’s main assets. Quantifying such reserves forms therefore a very important objective of the practising reservoir engineer but it is also a very complex problem, for the basic data is usually subject to widely varying interpretations and on top of that, reserves may be affected significantly by the field development plan and operating practice. It is an on-going activity during exploration development planning and during production. It is clearly a key task of the appraisal phase for it is at the heart of determining project viability. 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 hydrocarbon 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’, laboratory 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-wide3. These definitions have been used in the summary of reserve definitions which follow.

1.2 Development Planning

Oilfield development, particularly in the offshore environment, is a ‘front loaded’ investment. Finance has to be committed far in advance not only of income guaranteed by the investment, but frequently also of good definitive data on the character of the field. Much of the responsibility for this type of activity falls on the reservoir engineers because of their appreciation for the complex character of sub-surface fluid behaviour under various proposed development schemes. 4

Reservoir Engineering

Introduction To Reservoir Engineering O N E

1.3 Production Operations Optimisation

Producing fields will seldom behave as anticipated and, of course, by the very nature of this sort of activity, the balance of forces in the reservoir rock gets severely upset by oil and gas production. The reservoir engineer is frequently called upon to ‘explain’ a certain aspect of well performance, such as increasing gas-oil ratio, sand and/or water production and more importantly will be asked to propose a remedy. The actual performance of the reservoir as compared to the various model predictions is another ongoing perspective during this phase.

2

RESERVOIR ENGINEERING TECHNIQUES

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 of the estimate and the interpretation of these data. The relative degree of uncertainty may be conveyed by placing reserves into one of two principal classifications, either proved or unproved. 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.

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Estimation of reserves is carried out under conditions of uncertainty. The method of estimation is called deterministic if a single best estimate of reserves is made based on known geological, engineering, and economic data. The method of estimation is called probabilistic when the known geological, engineering, and economic data are used to generate a range of estimates and their associated probabilities. Identifying reserves as proved, probable, and possible has been the most frequent classification method and gives an indication of the probability of recovery. Because of potential differences in uncertainty, caution should be exercised when aggregating reserves of different classifications. Reserves estimates will generally be revised as additional geologic or engineering data becomes available or as economic conditions change. Reserves do not include quantities of petroleum being held in an inventory, and may be reduced for usage or processing losses if required for financial reporting. Reserves may be attributed to either natural energy or improved recovery methods. Improved recovery methods include all methods for supplementing natural energy or altering natural forces in the reservoir to increase ultimate recovery. Examples of such methods are pressure maintenance, gas cycling, waterflooding, thermal methods, chemical flooding, and the use of miscible and immiscible displacement fluids. Other improved recovery methods may be developed in the future as petroleum technology continues to evolve.

3.2 Proven Reserves

Proven reserves are those quantities of petroleum which, by analysis of geological and engineering data, can be estimated with reasonable certainty to be commercially recoverable, from a given date forward, from known reservoirs and under current economic conditions, operating methods, and government regulations. Proved reserves can be categorised as developed or undeveloped. 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 actually recovered will equal or exceed the estimate. Establishment of current economic conditions should include relevant historical petroleum prices and associated costs and may involve an averaging period that is consistent with the purpose of the reserve estimate, appropriate contract obligations, corporate procedures, and government regulations involved in reporting these reserves. In general, reserves are considered proved if the commercial producibility of the reservoir is supported by actual production or formation tests. In this context, 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.

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Reservoir Engineering

Introduction To Reservoir Engineering O N E

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.

3.2.1 Exercises – Reserve Definitions

The section on Reserve Definitions as put together by the SPE and the World Petroleum Congress, defines the various aspects of reserve definitions. These definitions, are important both to companies and countries, and they can have very significant commercial impact. The following tests are presented to help understand the working of these earlier definitions. Test 1 There are 950 MM stb (million stock tank barrels) of oil initially in place in a reservoir. 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. 950

500

400

MM stb

The Reserves are: 450

400

500

MM stb

STOIIP is:

Turn to page 9 for answers Test 2 Before starting production it was estimated that there was a 90% chance of producing 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.

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Institute of Petroleum Engineering, Heriot-Watt University

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Petroleum Engineering

Reservoir Engineering

Tick the correct answers. Proved reserves (MM stb): 200

400

500

600

700

200

400

500

600

700

Possible reserves 200 100

400

500

600

700

100

Probable reserves 100

Turn to page 9 for answers Test 3 What is wrong with the following definitions? 1. Reserves are those quantities of petroleum which are anticipated to be recovered from a petroleum accumulation. Test 4 1. We have a structure in our licence area which we intend to explore. We anticipate it to contain a STO IIP of 2000 MM stb, and recovery factor of 65% using primary methods (30%), secondary (25%) and tertiary (10%) recovery methods. What are the reserves? 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. STOIIP = 2000MM stb

Recovery factor = 35%

What are the reserves? Test 1 answer There are 950 MM stock tank barrels in place. It is estimated that 500 MM stb can be produced and 100 MM stb have been produced then 400 recoverable reserves remain. 950

√

500 X

400

X

MM stb

The Reserves are: 450

X

400 √

500

X

MM stb

STOIIP is:



Test 2 answer Proved: 100 MM stb Probable: 500 - 100 = 400 MM stb Possible: 700 - 500 = 200 MM stb Proved: 100 MM stb Proved & Possible: 500 MM stb Proved & Probable & Possible: 700 MM stb

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Introduction To Reservoir Engineering O N E

Test 3 answer Reserves are those quantities of petroleum which are anticipated to be commercially recovered from a petroleum accumulation. Clearly economics is a very important aspect of the definition. 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 1. Answer is zero by SPC/WPC definition. 2. Intentions and anticipations are not the basis for reserves. In this case no well has yet been drilled. Note: Some companies allocate potential reserves for internal use but these cannot be used for public and government Figures. Reserves are those quantities of petroleum which are anticipated to be commercially recovered from a known accumulation. Requirements for “Proved” include 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. Facilities An important perspective which might be forgotten by the reservoir engineer, is that for reserves to be classified as “proven”, all the necessary facilities for processing and the infrastructure for transport must either be in place or that such facilities will be installed in the future, as backed up by a formal commitment. Contribution to the Proved Reservoir Area 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 Ultimate recovery = 2 000 x 0.35 = 700 MM stb Minus production to date = 200 Reserves = 500 MM stb Reserves are those quantities of petroleum which are anticipated to be commercially recovered from known accumulations from a given date forward. i.e. Reserves refer to what can be produced in the future. Figure 2 gives a schematic of reserves showing the progression with time.

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Institute of Petroleum Engineering, Heriot-Watt University

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Petroleum Engineering

Potential

SPE / WPC Definitions

RESERVE CATEGORIES Probability Levels

P10

Reservoir Engineering

P50

P90

Possible

Possible

Probable

Probable

Provan

Provan

Cumulative Production

Time

Seismic Discovery of Start of Dev Start of Data Well Planning Production PERIOD TYPE OF DATA METHOD

Before Drilling Exploration Well Geophysical and Geological

Prior and During Delineation, Evaluation, Appraisal Development Geophysical, Geological, Petrophysical and Well Test Data

Mostly Probabilistic

Abandonment Production

Geophysical, Reservoir Performance Geological, and Production Data Petrophysical and Well Tests and Production Data

Deterministic and Probabilistic

Figure 2 Variations of Reserves During Field Life.

What are the amounts termed that are not recoverable? The quantity of hydrocarbons that remains in the reservoir are called remaining hydrocarbons in place, NOT remaining reserves! Reserves which are to be produced through the application of established improved recovery methods are included in the proved classification when : 1 Successful testing by a pilot project or favourable response of an installed program in the same or an analogous reservoir with similar rock and fluid properties provides support for the analysis on which the project was based, and, 2

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:

3

After a favourable production response from the subject reservoir from either;

4

10

a

A representative pilot or

b

An installed program where the response provides support for the analysis on which the project is based and It is reasonably certain the project will proceed.

Introduction To Reservoir Engineering O N E

3.3 Unproved Reserves

Unproved reserves are based on geologic and/or engineering data similar to that used in estimates of proved reserves; but technical, contractual, economic, or regulatory uncertainties preclude such reserves being classified as proved. Unproved reserves may be further classified as probable reserves and possible reserves. Unproved reserves may be estimated assuming future economic conditions different from those prevailing at the time of the estimate. The effect of possible future improvements in economic conditions and technological developments can be expressed by allocating appropriate quantities of reserves to the probable and possible classifications.

3.3.1 Probable Reserves

Probable reserves are those unproved reserves which analysis of geological and engineering data suggests are more likely than not to be recoverable. In this context, when probabilistic methods are used, there should be at least a 50% probability that the quantities actually recovered will equal or exceed the sum of estimated proved plus probable reserves. In general, probable reserves may include : 1

Reserves anticipated to be proved by normal step-out drilling where subsurface control is inadequate to classify these reserves as proved,

2 Reserves in formations that appear to be productive based on well log characteristics but lack core data or definitive tests and which are not analogous to producing or proved reservoirs in the area, 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,

4

Reserves attributable to improved recovery methods that have been established by repeated commercially successful applications when; a

a project or pilot is planned but not in operation and

b

rock, fluid, and reservoir characteristics appear favourable for commercial application,

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,

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6

Reserves attributable to a future workover, treatment, re-treatment, change of equipment, or other mechanical procedures, where such procedure has not been proved successful in wells which exhibit similar behaviour in analogous reservoirs, and

7

Incremental reserves in proved reservoirs where an alternative interpretation of performance or volumetric data indicates more reserves than can be classified as proved.

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3.3.2 Possible Reserves

Possible reserves are those unproved reserves which analysis of geological and engineering data suggests are less likely to be recoverable than probable reserves. In this context, when probabilistic methods are used, there should be at least a 10% probability that the quantities actually recovered will equal or exceed the sum of estimated proved plus probable plus possible reserves. In general, possible reserves may include: 1

Reserves which, based on geological interpretations, could possibly exist beyond areas classified as probable,

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,

4

Reserves attributed to improved recovery methods when;

5

a

a project or pilot is planned but not in operation and

b

rock, fluid, and reservoir characteristics are such that a reasonable doubt exists that the project will be commercial, and

Reserves in an area of the formation that appears to be separated from the proved area by faulting and geological interpretation indicates the subject area is structurally lower than the proved area.

3.4 Reserve Status Categories

Reserve status categories define the development and producing status of wells and reservoirs.

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 completion 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

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Introduction To Reservoir Engineering O N E

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.

3.4.2 Undeveloped Reserves:

Undeveloped reserves are expected to be recovered: 1

From new wells on undrilled acreage,

2

From deepening existing wells to a different reservoir, or

3

Where a relatively large expenditure is required to; a

Recomplete an existing well or

b Install production or transportation facilities for primary or improved recovery projects.

4

PROBABILISTIC REPRESENTATION OF RESERVES

Probability that the reserve is at least as large as indicated.

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

0.5

'Proven'

'Proven + Probable'

'Proven + Proable + Possible'

0.1 0

Recoverable Reserve

Figure 3 Probabilistic Representation of Recoverable Reserves.

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Reservoir Engineering

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. Net rock volume.

Net rock average porosity

Connate Formation Estimated water volume recovery saturation factor factor

[ Vnr x φ x (1 - Swc) / B o] x RF Uniform

Triangular

P

Gaussian Uniform

= Reserves

=

p90

p50 p10

Figure 4 Probablistic Reserve Estimates.

The resulting calculations result in a probability function for a field as shown in the Figure 5 below, where the values for the three elements are shown Proven = 500 MM stb the P90 Figure. 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

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Introduction To Reservoir Engineering O N E

Reserves distribution for a new field.

100

P90

90

P10 = 860 MMstb P50 = 740 MMstb P90 = 500 MMstb

Probability / %

80 70 60

Proven 500 MMstb

50

Probable 240 M

40 30

P+P+P = 860 MMstb

20

120

10 0

P50

Proven 0

200

Probable 400 600 Reserves / MMstb

P10

Possible 800

1000

Figure 5 Reserves Cumulative Probability Distribution.

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 leaving less uncertainty with respect to the reserves. This is illustrated in Figure 6. 100

P90

90

Probability / %

80 70 60

P50

50 40

Proved ultimate recovery.

30 20 10 0

Production 0

200

P10

Proved reserves 400 600 Reserves / MMstb

800

1000

Figure 6 Ultimate Recovery and Reserves Distribution For a Mature Field.

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5

Reservoir Engineering

VOLUME IN-PLACE CALCULATIONS

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, f, 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=Ahnf(1-Sw) (1) where: A= average area hn = nett thickness. nett thickness = gross thickness x nett: gross ratio f = 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)

To convert volumes at reservoir conditions to stock tank conditions formation volume factors are required where Bo and Bg are the oil and gas formation volume factors. These are defined in subsequent chapters. The expression of original oil in place is termed the STOIIP. The recovery factor, RF indicates the proportion of the in-place hydrocarbons expected to be recovered. To convert in place volumes to reserves we need to multiply the STOIIP by the recovery factor so that:

Reserves = STOIIP x RF

(3)

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 possible, that is being used to define the reserves. Before examining the contributions of the various parameters it is worthwhile to give consideration of the evolution of the reserve estimate during the exploration and development stage.

5.2 Evolution of the Reserve Estimate

Figure 7 gives a cross section view of a reservoir structure as suggested from seismic and geological data.

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Introduction To Reservoir Engineering O N E

Oil

Suggested 0il and water contact Figure 7 Cross Section Interpretation From Seismic and Geological Data.

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 Suggest this location.

Oil

Suggested oil and water contact Figure 8 Alternative locations of Exploration Wells

In Figure 9 an exploration well has been drilled and a core recovered and the structure 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).

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Reservoir Engineering

Oil

Oil and water contact

Cored interval

Figure 9 Interpretation After Exploration Well Drilled and Cored.

ab ob

Pr

Oil

le

le ab

Pr

P

ob

ble

si os

Proved

e

sibl

Pos

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. During well testing on wells 1 or 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).

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Introduction To Reservoir Engineering O N E

Well 2.

ven

Pro

Well 1.

Proven

Proposed delineation well 3.

Oil

Initial appraisal stage. Figure 11 Further Delineation Well.

Well 2.

Well 1.

Well 3.

Gas

ven

Pro

Proven

Oil New oil water contact.

Figure 12 After Further Appraisal.

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Reservoir Engineering

Well 1. Well 4.

Well 2.

Well 3.

Gas

ven

Pro

Proven

Oil New oil water contact.

Figure 13 Final Appraisal Well.

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).

Probable

1

Proved 2

4

3

Possible Figure 14 Reserves Uncertainties by Deterministic Method.

5.3 Reservoir Area

The reservoir area can be obtained by separately evaluating the individual units making up the reservoir as obtained from various reservoir maps. These maps are derived from the evidence given from seismic and subsequent drilled wells. The maps generally indicate the upper and lower extent of the reservoir section or sections and the areal extent as defined by faults or hydrocarbon contacts. Figure 15 shows an areal section with the defined limits. The contour lines are lines of constant subsea depths. Figure 16 gives a cross section of a reservoir unit. The combination of the two representations of the unit(s) can be used to calculate the gross rock volume.

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Introduction To Reservoir Engineering O N E

Fault B

ounda

ry

Porosity Boundary

Fluid Contact

ary

ound

B Fault

Figure 15 Structure Contour Map.7

Heighest Elevation on Top Structure

To p

Contour Elevation (units ss)

Heighest Elevation on Base Structure

St R r Ro ese uctu re ck rvo Ba Vo ir se lu St m ru e c

tu

re

Hydrocarbon Water Contact Elevation

o

Area Contained by Contour

Figure 16 Reservoir cross section.

7

Figures 17 & 18 show an example of a top structure map and cross section of the Rough Gas field in the North Sea.

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Reservoir Engineering

47/2 47/3

Completed Producers

C Gw

Proposed Well Locations x Abandoned Wells

9 95500 00 0

G

w

C

C.I. = 50ft.

960

955000 8 95 50 8 94 00 4 9 50 93 00 8 8 8 93 8 B 250 9 A2 00 2 8 9 8 A

47/7 47/8

8 A

47/8-1x 8

8

A5

A

A A4

A

Platform A A3

9100

A6

93

00 93 50 92 9200

50

91

50

x 47/8-2

Figure 17 Top Sand Structure Map Rough Gas Field.5 Depth (ft) subsea 9000

A5

Unc

onf orm ity lieg end e s Unc onfo rmit y Rot

Fault

9400

A2

A4

Fault

9200

A1

A3

Tentative hydrocarbon/ water contact

9600 9800 Carboniferous Sands 5

Figure 18 Schematic Cross Section of The Rough Field.

5.4 Reservoir Thickness

Another representation of the reservoir formations is the reservoir thickness map. Where the areal contour maps show the thickness normal to the plane of the reservoir the contours are called isopachs. When the thickness is mapped as a vertical thickness then the contour is called an isochore. Not all the reservoir thickness will contribute to fluid recovery and will include non-productive strata. Those contours which include these non-productive material are called gross reservoir isopach and those where non-productive material is excluded are called net reservoir isopach maps. Those intervals contributing to flow are termed pay. The ratio of net to gross, hn/ht, is an important aspect in reservoir evaluation. Figure 19 shows a net pay thickness isopach and the isopach map for the Rough field is shown in Figure 20. 22

Introduction To Reservoir Engineering O N E

0

150

125

Isopach C I 25 Units

75

100

7

Figure 19 Net Pay Thickness Isopach. 47/2 47/3 GwC

140

130

C

Gw

120

47/7 47/8

0 11

0

10

A2

47/8-1 x

A5

A4 80

0 11

90

100

A3

A1

6 11

70

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.

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Reservoir Engineering

0

Net Pay Isopach Value

40

Area Enclosed = Net Rock Volume

80 120 140 OWC

180

Area Contained by Contour

7

Figure 21 Hydrocarbon Volume From Net Pay Isopach.

5.5 Reservoir Porosity

The variation of porosity can also be represented. The average porosity, φ, in a well can be calculated from the thickness-weighted mean of the porosities4. m



φw =

∑φ h k =1

k n, k

hn

(4)

where fk 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 illustrated in Figure 22. The example of an isoporosity map for the Rough Field is shown in Figure 23.

5

10

15

20

25

Figure 22 Iso Porosity Map.

24

Porosity C I 5%

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Introduction To Reservoir Engineering O N E

47/2 47/3

G w

C

C Gw

A

A1

6%

A3

A4

%

A5

8%

x

12%

47/8-1

10

47/7 47/8

14%

A2

A6

47/8-2

x

7

Figure 23 Rough Field Iso Porosity Map.

5.6 Water Saturation, Sw

The water saturation in a reservoir is influenced by the characteristics of the reservoir rock and the location with respect to the position above the free water level near the oil-water or gas-oil contact (see section Reservoir Rock Properties Chapter 7). The average water saturation Sw,w, can be calculated in a similar way to porosity by calculating the volume weighted mean across the producing elements of the formation, the pay. m



Sw , w =

∑S k =1

φh

w , k k n, k

φ w hn

(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. 22/07/14

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A more detailed description together with exercises are given in the mapping section of the geology module.

5.7 Formation Volume Factors Oil, Bo and Gas, Bg

These properties of the oil and gas which convert reservoir volumes to surface volumes, are generated from measurements made on fluid samples from the reservoir. They do not vary significantly across the reservoir when compared to the other rock related parameters. These parameters are covered in the gas properties and oil properties chapters. In some reservoirs where the formations are thick there is a compositional gradient over the depth. This variation in composition from heavier (less volatile components) to lighter components at the top results in a variation of the oil formation volume factor, Bo over the thickness. In such cases an average value based on values measured or calculated at depth would be a preferred value.

5.8 The Recovery Factor, ER

The proportion of hydrocarbons recovered is called the recovery factor. This factor is influenced by a whole range of factors including the rock and fluid properties and the drive mechanisms. The variability of the formation characteristics, the heterogeneity can have a large influence on recovery. The development process being implemented and the geometries and location of wells again will also have a large influence. Calculating recovery therefore in the early stages is not feasible and many assumptions have to be included in such calculations. It is in this area that reservoir simulation can give indications but the quality of the calculated Figure is limited by the sparse amount of quality data on which the simulation is based. The American Petroleum Institute6 has analysed the recoveries of different fields and correlations have been presented for different reservoir types and drive mechanisms. Figures 25 and 26 give the residual saturations and oil recovery efficiences for different drive mechanisms. The API also presents correlations for recoveries,ER, For sandstone and carbonate reservoirs with solution gas drive

ER, o

 φ (1 − Sw )  = 0.4185   Bob 

0.1611

 k     µob 

0.0979

(Sw )

0.3722

 pb     pa 

0.1741

(6)

For sandstone reservoirs with water drive

ER, o

 φ (1 − Sw )  = 0.54898   Boi 

0.0422

 k µ wi     µoi 

0.0770

p  a

(Sw )− o.1903  pi  − 0.2159 (7)

b refers to bubble point conditions, i is the initial condition and a, refers to abandonment pressure.

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Introduction To Reservoir Engineering O N E

2

5

10

20

30

40

50

60

70

80

95

98

RESIDUAL SATURATIONS 1.00

Sor In Water Drive Reservoirs

0.50

0.50

Sgr In Solution Gas Drive Reservoirs 0.10

0.10

0.05

0.05

−σ

0

+σ MEDIAN

Sor (OR Sgr) as Fraction of Total Pore Space

1.00

2

5

10

20

30

40

50

60

70

80

95

98

0

PERCENTAGE OF CASES LARGER THAN

Figure 25 Log – Probability Residual Oil Saturation For Water Drive 6 and Solution Gas Drive Reservoirs. (API )

5

OIL RECOVERY EFFICIENCY AT FIELD ABANDONMENT IN PERCENT OF OIL PLACE

2

10

20

30

40

50

60

70

80

95

98

RESIDUAL SATURATIONS

1.00

1.00

Water Drive Gas Cap Drive

0.50

0.50

Gas Cap Drive + Water Injection

0.10

0.10

Solution Gas Drive

0.05

0.05

0

+σ MEDIAN

−σ

2

5

10

20

30

40

50

60

70

80

95

98

0

PERCENTAGE OF CASES LARGER THAN

6

Figure 26 Log - Probability of Oil Recovery For Various Drive Mechanisms. (API ) 22/07/14

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Petroleum Engineering

Reservoir Engineering

5.9 Production Capability

Another concept, isocapacity, is used to signify production capability. Isocapacity denotes equal values of permeability-net thickness product. This product can be mapped instead of permeability. The Figure 27 shows an isocapacity map where the absolute permeability has been obtained as an arithmetic average in the zone.

5

4 0.5

4

3

1

2

123

0.25

7

Figure 27 Isocapacity Map.

The permeability map for the Rough Field is given in Figure 28

G w

C

C

Gw

47/2

47/7 47/8

120 A2 100 80

47/8-1 x

60

A4

A5 40

Platform B A3 0

A6 x 47/8-2

Contour Intervals 20 millidarcies

5

Figure 28 Rough Field Permeability Map.

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Introduction To Reservoir Engineering O N E

5.10 The Hydrocarbon Pore Volume Map

The hydrocarbon pore volume can be obtained by combining the net rock volume with a mean porosity and a mean hydrocarbon saturation. An alternative is the mapping of hydrocarbon thickness (HPT) at each well. HPT at a well in a given zone is:

_

_

HPT = φ .hn . Sh (8)

where:

Sh = 1 − Sw

Figure 29 gives an HPT map and the Rough Field HPT map is given in Figure 30

12 11 0 10 14

15

14 13

13

12 11

10 0

9

7

Figure 29 Hydrocarbon Pore Thickness Map.

0 10 A2

9

8

A5

A4

7

A1

6 A3

5 4

A6

5

Figure 30 Rough Field Hydrocarbon Pore Thickness.

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Petroleum Engineering

6. OTHER APPRAISAL ROLES In building up the ‘picture’ to enable the reserves estimates and recoveries to be determined the reservoir engineer will be involved in an number of aspects. One of the most powerful tools is the production test. In a well test an exploration or appraisal well is converted to a short term producing well, with all the associated facilities put in place to handle the produced fluids and monitor fluid rates. A downhole pressure monitoring device is also located in the well (Figure 31). The well is flowed at a constant rate, and sometimes two rates as illustrated in Figure 32a, a two rate test. The downhole pressure device responds to the production and pressure declines. After a short or longer time period depending on the nature of the test, the well is “shut in”, i.e. the flow is stopped. In the well the pressure builds up and eventually as monitored by the downhole pressure device, recovers to the original pressure, Figure 32b. It is in the analysis of the pressure draw down and build up curves and the rates that the reservoir engineer is able to determine the flowability of the reservoir. If the flowing interval thickness is known, the permeability can be calculated. The presence of faults can also be detected. 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.

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Surface casing

Production casing Production tubing Cement

Packer

Perforations Down hole pressure monitor

Figure 31 Production Test Assembly.

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31

q bbls / day

Petroleum Engineering

Flow 1

Reservoir Engineering

Flow 2

Well shut in

t

Pf. psig

Pi Pressure draw down Pressure build up

t

Figure 32 Production Test Analysis. Two Rate Test.

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. It is clear from what we have discussed that reservoir engineering is an important function in the appraisal of the reservoir. The focus for this appraisal so far has concentrated on determining the characteristics and potential flow behaviour of a reservoir under development. Clearly there could be a whole range of possibilities with respect to the plan that could be used to develop the field. This development planning perspective is an important part of the reservoir engineers role. Again it is

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a team effort involving the geological community who understand the ‘reservoir’ and the various engineers who have the responsibilities of designing and operating the hardware to enable production. An important part of any future development are the facilities that would be required for sustained production and it is therefore an important part of the appraisal stage to provide data for those who would have responsibility for good quality data predictions which will enable optimised facility design. 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

7.1 Reservoir Modelling

Given appraisal well data, and test results the reservoir engineer can consider some alternative development plans, relying heavily on experience and insight. Since the 80’s computer based reservoir simulation has played a major role. 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 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 22/07/14

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Petroleum Engineering

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.

7.3 Coping with Uncertainty

The challenge to the exploration & production business of the oil & gas industry is considerable. The looking for the “needle in the haystack” scenario is not too far from the truth, when compared to other industrial sectors. With the challenge of reserves being found in technically challenging areas and the oil price moving in response to political as well as demand scenarios, there is the need to define more accurately forecasts of production and recovery. Reducing uncertainty is the message of the current decade and not least in reservoir engineering and its related disciplines. 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. Another field would result in different impact perspectives, and therefore a different strategy to reduce overall project uncertainty would be required.

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Q

Reservoir area

P

Well production

Project Cost

Reservoir structure

Sand conectives Aquifer behaviour Fluid properties Relative permeabilities etc. Thief zones

Faults

-

Changes

+

Figure 33 Impact on a Project of Different Uncertainties.

8

PRODUCTION OPERATIONS OPTIMISATION

8.1 The Development Phase

The development phase covers the period from the time continuous production starts until the production from the field stops i.e. abandonment. The decision when to stop production clearly is a techno-economic decision based to a large extent on the costs of the development. Low volume producers can be allowed to continue in an onshore development where well operating costs might be low but the high costs associated with for example in an expensive offshore operation sets a much higher economic limit for the decision to abandon a field. During the development phase Dake2 has identified a number of roles for the Reservoir Engineer 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. 22/07/14

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Petroleum Engineering

8.2 History Matching

Throughout the production phase the comparison of the actual performance with that predicted during the appraisal stage and more recent predictions is made. It is during this stage that the quality of the reservoir simulation model comes under examination. The production pressure decline is compared to that predicted and the reservoir simulation model adjusted to match. This process is called history matching. Clearly if the simulation cannot ‘predict’ what has happened over the recent past it cannot be used with much confidence to forecast the future! 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 techniques 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-inplace 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 15, 16 and 17 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 observations 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 simple 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 profile 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.

36

Reservoir Engineering

Introduction To Reservoir Engineering O N E

8.3 Phases of Development

During the development there are a number of phases. Not all of these phases may be part of the plan. There is the initial production build up to the capacity of the facility as wells are brought on stream. There is the plateau phase where the reservoir is produced at a capacity limited by the associated production and processing facilities. Different companies work with different lengths of the plateau phase and each project will have its own duration. There comes a point when the reservoir is no longer able to deliver fluids at this capacity and the reservoir goes into the decline phase. The decline phase can be delayed by assisting the reservoir to produce the fluids by the use of for example ‘lifting’ techniques such as down-hole pumps and gas lift. The decline phase is often a difficult period to model and yet it can represent a significant amount of the reserves. These phases are illustrated in Figure 34

Production rate

Plateau phase Artificial lift Decline phase

Build up phase Economic limit Time - years

Figure 34 Phases of Production.

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. Secondary recovery is considered when the energy is supplemented by injection of fluids, for example gas or water, to maintain the pressure or partially maintain the pressure. The injected fluid also acts as a displacing fluid sweeping the oil to the producing wells. After sweeping the reservoir with water or gas there will still be remaining oil; oil at a high saturation where the water for a range of reasons, for example; well spacing, viscosity, reservoir characteristics to name just a few, has by-passed the oil. The oil which has been contacted by the injected fluid will not be completely displaced from the porous media. Because of characteristics of the rock and the fluids a residual saturation of fluid is held within the rock. Both of these unrecovered amounts, the by-passed oil and the residual oil are a target for enhanced recovery methods, EOR. 22/07/14

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Petroleum Engineering

Reservoir Engineering

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 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 recovery 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 Artifical Lift

Natural Flow

Pump gas lift etc.

Secondary Recovery Pressure Maintenance

Natural Flow

Water, gas injection

Tertiary Recovery E O R

Thermal

Gas

Steam In-situ combustion.

Hydrocarbon miscible, CO2 N2 immiscible gas

Chemical

Microbial

Polymer surfactant/ polymer

Figure 35 Oil Recovery Mechanisms.

38

C O N V E N T I O N A L

Introduction To Reservoir Engineering O N E

9

THE UNIQUENESS OF THE RESERVOIR

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 schematically illustrated in Figure 36. Reservoir Behaviour Development Plan

Reservoir Description Unique Dynamic and Static

Figure 36 Relationship between Reservoir Description, and Reservoir Behaviour.

At one extreme for example in a blow – out situation, a reservoir produces in an uncontrolled 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 engineering a duplicate or a similar size unit elsewhere. Such opportunities are not possible 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 reservoir 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.

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2°00

Reservoir Engineering

2°10

2°20

100

0

00

200 0

21

53°10

SHELL/ESSO 49/26

53°10

AMOCO 49/27

2200

20

21

00

00

21

00

49/26.1

53°05

00

12

53°05

Gas /water contact Depths in metres scale 1 100,000

0 80

100

0

00

20 10

00

00

20

00

20

0 210

20 100 00 0

10

00

100

0

2000

2°00

2°20

Present interpretation of Leman Gas-field, showing contours on top of Rotliegendes in feet below sea-level

Figure 37a The Leman Field as it Appeared to be

The Leman field as itThe appeared to be when When Exploration Well the Wasexploration Drilled. well was drilled

2°10

2°00 53°10

2°20

2°30 53°10

SHELL/ESSO 49/26 AMOCO 49/27

Depth in feet Miles 0 1 0 1 2 km 70 00

Gas /water contact A permanent platform

00

00 63

63

53°05

53°05 6400 6300

620

0

69

00

610

0

6900

690

00

69

70

53°00

2°00

69

00

00

2°10

6300 6 900

0 64

6300 6400

00

53°00

2°20

2°30

Present interpretation of Leman Gas-field, showing contours on top of Rotliegendes in feet below sea level.

Figure 37b Leman Ten after Yearsdiscovery After Discovery. Leman field Field ten years

40

Introduction To Reservoir Engineering O N E

The coverage of the reservoir has also changed affecting 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 description data is available. Clearly there can never be sufficient description, however the economics of project management will determine when decisions have to be taken based on description to date. The value of extra information has to be balanced by the cost of delay in going ahead with a project.

10 CONCLUSION In order to accomplish these objectives the Petroleum Reservoir Engineer should have a broad fundamental background both theoretically and practically in the basic sciences and engineering. The basic areas are: (i) The properties of petroleum reservoir rocks (ii) The properties of petroleum reservoir fluids (iii) The flow of reservoir fluids through reservoir rock (iv) Petroleum reservoir drive mechanisms 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 areal 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 reservoir 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 sophisticated computational tools enable predictions to be made with confidence and where uncertainty exists the degree of uncertainty can be defined.

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"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."

Figure 38 A Past Approach to Uncertainty!

REFERENCES 1 2 3 4 5 6 7

42

Craft, B.C. and Hawkins, M.F. Applied Reservoir Engineering, Prentice-Hall Inc. 1959 Dake, L.P., The Practise of Reservoir Engineering. Elsevier. 1994 Society Of Petroleum Engineers. Reserves Definitions 1995. Chierici,G.L. Principles of Petroleum Reservoir Engineering. Vol 1 Springer Verlag 1994 Hollois, A.P. Some petroleum engineering considerations in the change over of the Rough Gas field to the storage mode. Paper EUR 295 Proc Europec. 1982, pg 175 API. A Statistical Study of the Recovery Efficiency. American Petroleum Institute. Bull D14, 1st Edition ,1967 Archer, J.S. and Wall,C.G. Petroleum Engineering Principles and Practise, Graham and Trotman ,1986.

Reservoir Engineering

Reservoir Pressures and Temperatures T W O

8,000

Statfjord OWC

Brent OWC

9,000

Thistle OWC

Cormorant OWC 4

11,000

1

2

Gr% Ninian Sw% 0 100OWC 100 0 θ% 0 50

Heather OWC

3

Reservoir pressure - psig 2500 Perforations

Layer 1

Lyell

2500

13,000

Note:

5000

True vertical subsea depth - metres

12,000

Layer 2

5

Alwyn Layer N.W. Alwyn

2550

3000

Top paleocene

3500

4000 8100

Original pressure gradient

8200

3

8300

S.W> Ninian

Water gradient lines drawn through known or projected oil/water contacts

6000 26007000

8400

Layer 4

8000

9000

8500

10,000

Pressure, psig

8600 Layer 5

2650

8700

True vertical subsea depth - feet

Subsea Depth (Feet)

10,000

8800 14

Reservoir Engineering 22/07/14

24 18 22 16 20 Reservoir pressure - MPa

26

Petroleum Engineering

Reservoir Pressures and Temperatures T W O

C O N T E N T S 1 INTRODUCTION 2 ABNORMAL PRESSURES 3 FLUID PRESSURES IN HYDROCARBON SYSTEMS 4 PRESSURE GRADIENTS AROUND WATEROIL CONTACT 5 TECHNIQUES FOR PRESSURE MEASUREMENT 6 RESERVOIR TEMPERATURE

Reservoir Engineering 22/07/14

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Petroleum Engineering

LEARNING OBJECTIVES Having worked through this chapter the Student will be able to:

2

•

Define the terms; lithostatic pressure, hydrostatic pressure and hydrodynamic pressure.

•

Draw the normal hydrostatic pressure gradient for water systems.

•

Define normal pressured reservoirs, overpressured reservoirs and underpressured reservoirs

•

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.

•

Illustrate how a downhole formation pressure device can be used to discriminate permeability layers after production has commenced.

Reservoir Engineering

Reservoir Pressures and Temperatures T W O

1. INTRODUCTION Determining the magnitude and variation of pressures in a reservoir is an important aspect in understanding various aspects of the reservoir, both during the exploration phase but also once production has commenced. 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 Pov = 1.0D

(1)

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 unconsolidated 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 rw, 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 = rwDg

(2)

where g is the acceleration due to gravity. 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.

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Petroleum Engineering

Depth (Ft.)

14.7 0

Reservoir Engineering

Pressure (psia)

FP

GP Overpressure

Underpressure

Overburden Pressure (OP)

Normal

(FP = Fluid Pressure, GP = Grain Pressure)

Figure 1 Gives the relationship between the lithostatic pressure and the hydrostatic 1 pressure.

Fluid pressure in hydrocarbon accumulations are dictated by the prevailing water pressure in the vicinity of the reservoir. In a normal situation the water pressure at any depth is:

dP Pw =   x D + 14.7psia  dD  water

(3)

where dP/dD is the hydrostatic pressure gradient 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 pressure 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?

4

Reservoir Pressures and Temperatures T W O

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 underpressured 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)

(b)

Original Deposition

Dense Shale Reservoir

Shale deposited too quickly to allow fluid equilbrium

North Sea (c)

Glacier

Normal Surface

Greenland 3 km thick 1300 psi/1000 m ice

Figure 2 Causes of overpressurring

•

22/07/14

Thermal effects, causing expansion or contraction of water which is unable to escape; an increase in temperature of 1˚F can cause an increase of 125 psi in a sealed fresh water system.

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Petroleum Engineering

•

Rapid burial of sediments consisting of layers of sand and clay. Speed of burial does not allow fluids to escape from pore space.

•

Geological changes such as uplifting of the reservoir, or surface erosion both of which result in the water pressure being too high for the depth of the burial. The opposite occurs in a down thrown reservoir.

•

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 inhibits 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.

6

Reservoir Engineering

Reservoir Pressures and Temperatures T W O

8,000

Statfjord OWC

Brent OWC

9,000

Thistle OWC

Cormorant OWC 4

Subsea Depth (Feet)

10,000

11,000

1

2

Heather OWC

Ninian OWC

3 Lyell 5

12,000

Alwyn N.W. Alwyn S.W> Ninian

13,000

Note:

5000

Water gradient lines drawn through known or projected oil/water contacts

6000

7000

8000

9000

10,000

Pressure, psig

3

Figure 3 Examples of overpressured reservoirs in the North Sea

3. FLUID PRESSURES IN HYDROCARBON SYSTEMS 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, ro and rg of each fluid. The pressure gradients are a function of gas and oil composition but typically are:

 dP  = (0.45 psi / ft)  dD  water dP

  = (0.35 psi / ft)  dD  oil

(6)

 dP  = (0.08 psi / ft)  dD  gas

(7)



22/07/14

(5)

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Petroleum Engineering

Reservoir Engineering

For a reservoir containing both oil and a free gas cap a pressure distribution results, as in the Figure 4. As can be seen, the composition of the respective fluids gives rise to different pressure gradients indicated above. These gradients will be determined by the density of the fluids which result from the specific composition of the fluids. Depth (Ft.) 13

8500

12

Depth (Ft.)

8600

11 10 9 Gas-Oil Contact

0.17 psi/ft rf = 0.39 gm/cc 8

7

6

0.29 psi/ft rf = 0.67 gm/cc 5

8700

Oil-Water Contact 4

0.47 psi/ft rf = 1.09 gm/cc

3

8800 4000

4050

4100

2

1 4150

Formation Pressure (PSI)

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. The data used for these fluid contacts comes from:

(i) (ii) (iii) (iv) (v) (vi) (vii)

Pressure surveys Equilibrium pressures from well tests Flow of fluid from particular minimum and maximum depth Fluid densities from reservoir samples Saturation data from wireline logs Capillary pressure data from cores Fluid saturation from cores

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.

8

Reservoir Pressures and Temperatures T W O

4. PRESSURE GRADIENTS AROUND THE WATER-OIL CONTACT 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 - rogh

(8)

Pw = P - rwgh

(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

Oil Phase Pressure po = pFWL - rogh Oil Gradient

Capilliary Transition Zone

Sw h=

Vertical Depth D

Water Phase Pressure pw = pFWL - rwgh

pc (Sw) rg

WOC

pc

Water Gradient

FWL

(pc = o) Water Zone 0

Swc

1 Water Saturation, Sw

pFWL Pressure, P

Figure 5 Pressure Gradients around the Water-Oil Contact

Pc = Po - Pw

(10)

at hydrostatic equilibrium Pc(Sw) = ∆rgh

∆r = rw-ro h = height above free water level

The free water level, FWL, is not coincident with the oil-water contact OWC. The water contact corresponds to the depth at which the oil saturation starts to increase from water zone. The free water level is the depth at which the capillary pressure is zero. 22/07/14

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Petroleum Engineering

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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: Po = PFWL - rogh

(11)

Pw = PFWL - rwgh

(12)

On the depth-pressure diagram the intersection of the continuous phase pressure line occurs at the free water level.

5. TECHNIQUES FOR PRESSURE MEASUREMENT 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 formation 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. Probe

Packer

Mud Cake Packer Filter

Flow Line Equalising Valve (To Mud Column)

Pressure Guage

Piston

Formation

Flow Line Chamber 1

Probe Closed

Chamber 2

Seal Valve to Upper Chamber

Seal Valve to Upper Chamber

Figure 6 Original Schematic of the RFT Tool

10

Probe Open and Sampling

Reservoir Pressures and Temperatures T W O

These open hole pressure measurements have proved valuable at the appraisal stage and can be used to establish fluid contacts. It has also proved particularly valuable 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 measurements can be a powerful tool in understanding the characteristics of the reservoir formation. By comparing current pressure information with those obtained prior to production, important reservoir description can be obtained which will aid reservoir depletion, completion decisions and reservoir simulation. 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 indicates 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. Sw% 100 0

θ% 0 50

Reservoir pressure - psig 2500

3000

Top paleocene Perforations

Layer 1 Layer 2

True vertical subsea depth - metres

2500

3500

4000 8100

Original pressure gradient

8200

Layer 3

8300

2550 8400

Layer 4

8500

2600

8600 Layer 5

2650

8700

True vertical subsea depth - feet

Gr% 0 100

8800 14

24 18 22 16 20 Reservoir pressure - MPa

26

Figure 7 RFT Pressure Survey in Development Well of Montrose Field 3. 22/07/14

Institute of Petroleum Engineering, Heriot-Watt University

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Petroleum Engineering

Reservoir pressure - psig 3000 3400 3200

True vertical subsea depth - metres

A15 A11 A17 A18

A6

A8

8000

Original pressure gradient

8100 8200 8300

2550

8400 8500

2600

8600

2650

symbol

2700

18

?Well number 22/17-A6 A8 A11 A15 A17 A18

20

Date 05/04/77 27/01/78 20/12/77 15/08/78 02/11/78 28/03/79

26 22 24 Reservoir pressure - MPa

8700 8800

True vertical subsea depth - feet

2450

2500

Reservoir Engineering

8900 28

9000

Figure 8 RFT Pressure Syrveys on a number of Montrose Wells3.

6. RESERVOIR TEMPERATURE

The temperature of the earth increases from the surface to centre. The heat flow outwards 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 temperature. 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. Without temperature surveys the temperature at a vertical depth can be estimated using a surface temperature of 15 oC (60 oF) at a depth D. T(D) = 288.2 + gcD (K)

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Reservoir Pressures and Temperatures T W O

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? Multiply kPa by 0.145 to get psi. 1 US foot = 0.3048m. SOLUTION Pressure in formation = 0.47 * 7400 = 3478 psig Converting to kPa

= 3478 / 0.145 = 23986 kPa

Converting to MPa

= 23986 / 1000 = 23.99 MPa

Pressure gradient

= 0.47 psi/ft = (0.47 / 0.145) kPa/ft = 3.2414 kPa/ft = (3.2414 /0.3048) kPa/m = 10.63 kPa/m

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. SOLUTION Typical pressure gradients are (psi/ft): Water – 0.45 Oil – 0.35 Gas – 0.08 Pressure at seal = 3625 - (600*0.35) = 3415 psi To calculate the pressure differential across seal, look at fluid gradient differential from OWC to seal 600ft above… Differential = (0.45-0.35) * 600 = 60 psi If the reservoir is gas then the differential becomes… (0.45 – 0.08) * 600 = 222 psi higher in the reservoir than surrounding area

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REFERENCES 1. Dake,L.P. Fundamentals of Reservoir Engineering. Elsevier 1986 2. Bradley,J.S. Abnormal Formation Pressure. The American Association of Petroleum Geologists Bulletin. Vol 59, No6, June 1975 3. Bishlawi,M and Moore,RL: Montrose Field Reservoir Management. SPE Europec Conference, London,(EUR166) Oct.1980

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Reservoir Fluids Composition T H R E E

Reservoir Fluid

Gas at Surface Conditions

Oil at Surface Conditions

C1

C2

C3

C4

C5

C6

C7+

The relative amounts of C1 - C7+ are a function of : Temperature, Pressure, Composition (particularly at high temperature)

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Reservoir Fluids Composition T H R E E

C O N T E N T S 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 4 COMPOSITIONAL DESCRIPTION FOR RESERVOIR ENGINEERING 4.1 Definitions of Composition in Reservoir Engineering 5 GENERAL ANALYSIS 5.1 Surface Condition Characterisation 5.2 Refractive Index 5.3 Fluorescence of Oil

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LEARNING OBJECTIVES Having worked through this chapter the Student will be able to:

2

•

Describe briefly the origin, nature and appearance of petroleum fluids.

•

Be aware that the principal components of petroleum fluids to be hydrocarbons.

•

Draw a diagram illustrating the classification of hydrocarbons and to identify; paraffin’s (alkanes), aromatics and cyclic aliphatics (napthas).

•

List the non-hydrocarbon compounds which might be present in small quantities in reservoir fluids.

•

Define the black oil model description of the composition of a reservoir fluid.

•

Define the compositional model description of a reservoir fluid.

•

Explain briefly what PNA analysis is and its application.

•

Describe briefly the concept of pseudo components in fluid composition characterization.

•

Be aware of general analysis descriptors for petroleum fluids e.g. °API, refractive index and fluorescence.

•

Be able to calculate the °API gravity given the specific gravity.

•

Be able to calculate the density of Stock Tank Oil given the API gravity and the appropriate equation.

Reservoir Engineering

Reservoir Fluids Composition T H R E E

1 INTRODUCTION Petroleum deposits vary widely in chemical composition and depending on location have entirely different physical and chemical properties. The very complex characteristics are evident from the many products which can be produced from oil and gas. What is petroleum? Petroleum is a mixture of naturally occurring hydrocarbons which may exist in the solid, liquid or gaseous states, depending on the conditions of temperature and pressure to which it is subjected.1 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 composition of crude oil consists mainly of organic compounds, principally hydrocarbons with small percentages of inorganic non-hydrocarbon compounds. Such as carbon dioxide, sulphur, nitrogen and metal compounds. The hydrocarbons may include the lightest (C1 methane ) to napthenes and polycyclics with high molecular weights. 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. For reservoir engineering purposes the description of the composition is an important characterisation parameter for the determination of a range of physical parameters important in various reservoir volumetric and flow calculations. It is not the concern of the reservoir engineer to determine the composition with respect to understanding the potential to separate the material to a range of saleable products. For this reason therefore simplistic characterisation approaches are used. 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. 22/07/14

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Reservoir Engineering

2 HYDROCARBONS 2.1 Chemistry of Hydrocarbons

The compositional model uses hydrocarbons as the descriptor since hydrocarbons represent the largest proportion in petroleum fluids. It is important to review briefly the chemistry of hydrocarbons. The hydrocarbon series is represented in figure 1 below Hydrocarbons Aliphatic Alkanes

Alkenes

Aromatics Alkynes

(Paraffins)

Cyclic Aliphatics (Napthenes)

Figure 1 Classification of Hydrocarbon.

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 (multiple) bonds.

2.2 Alkanes or Paraffinic Hydrocarbons

The largest series is the alkanes or paraffins which are open chain molecules with saturated bonds. Carbon has a valance of four and therefore the formula for these compounds is CnH2n+2. These saturated hydrocarbons include all the paraffins in which the valence of the carbon atoms is satisfied by single covalent bonds. This type of structure is very stable. Unsaturated hydrocarbons are those where the valence of some of the carbon atoms is not satisfied with single covalent bonds so they are connected by two or more bonds which make them less stable and more prone to chemical change. 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. 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. 4

Reservoir Fluids Composition T H R E E



normal butane CH3CH2CH2CH3

isobutane CH3CH CH3 CH3

-

B.Pt. 31.1˚F

-

B.Pt. 10.9˚F

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. Name

Methane Ethane Propane n-butane Isobutane n-pentane Isopentane n-hexane

Chemical Formula

Molecular Weight

Boiling Point (°C) at normal

Critical Temp °C

Density Gas Liquid (air = 1) (water = 1) conditions sp.gr.

16.04

-161.4

-82.4

0.554

0.415 (-614°)

C 2H 6 C4H10

CH4

30.07

-89.0

32.3

1.038

0.54 (-88°)

C 3H 8

44.09

-42.1

96.8

1.522

0.585 (-44.5°)

58.12

0.55

153.1

2.006

0.601 (0°)

C4H10

58.12

-11.72

134.0

2.006

0.557

C5H12

72.15

36.0

197.2

2.491

0.626

C5H12

72.15

27.89

187.8

2.491

0.6197

C6H14

86.17

60.30

228.0

2.975

0.6536

Table 1 Physical properties of common hydrocarbons.

ALKANES or PARAFFIN HYDROCARBONS Cn H 2n+2

No of carbon atoms 1 2 3 4 5 6 7 8 9 10 C5-C17 C18+

Name

State (ntp)

Methane Ethane Propane Butane Pentane Hexane Heptane Octane Nonane Decane

Gas Gas Gas Gas Liquid Liquid Liquid Liquid Liquid Liquid Liquid Solid

Table 2 Alkanes or Paraffin Hydrocarbons Cn H 2n + 2. 22/07/14

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H

PARAFFINS H H

C

H

H

H

Reservoir Engineering

Methane

H H

C

H H

C

C

C

H

H

H

H

H

H

H

H

H

H

H

H

H

C

C

C

C

C

C

C

C

H

H

H

H

H

H

H

H

Iso-butane

H

n-octane

Figure 2 Gives some standard formula for saturated hydrocarbons.

2.4 Unsaturated Hydrocarbons

These are hydrocarbons which have double or triple bonds between carbon atoms. They have the potential to add more hydrogen or other elements and are therefore termed unsaturated. There are termed the olefins, and there are two types, alkenes, for example ethylene, CH2=CH2, which have a carbon-carbon double bond and alkynes, for example acetylene,CH=CH which have a carbon carbon triple bond. Both compound types being unsaturated are generally very reactive and hence are not found in reservoir fluids.

2.5 Napthene Series

The napthene series (CnH2n) sometimes called cycloparaffins or alicyclic hydrocarbons are identified by having single covalent bonds but the carbon chain is closed and is saturated. They are very stable and are important constituents of crude oil. Their chemical properties are similar to those of the paraffins. A crude oil with a high napthene content is referred to as an napthenic based crude oil. An example is cyclohexane C6H12. Figure 3 gives the structural formula for two napthenic compounds. NAPHTHENES H

H H H H

H

C

H

C

H H

C

C

C

C

H H H

H H

Methyl Cyclopentane

H

H H

C

H

C

C

C

C C H

H H

H H

H

Cyclohexane

Figure 3 Structural formula for two naphenic compounds.

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.

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Reservoir Fluids Composition T H R E E

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 H Benzene

C

C

H

H

Naphthalene

Figure 4 Structural formula for two aromatic compounds.

The aromatic-napthene based crudes are usually associated with limestone and dolomite reservoirs such as those found in Iran, the Persian 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. Physical Properties of some Common Petroleum Reservoir Fluid Constituents Component Paraffins Methane Ethane Propane n-Butane Iso-Butane n-Pentane n-Hexane Iso-octane n-Decane Naphthenes Cyclopentane Methyl cyclo-pentane Cyclohexane Aromatics Benzene Toluene Xylene Naphthalene

Formula

Melting Point (˚C)

Normal Boiling Point (˚C)

Density (g/cm3) at 1 atm and 15˚C

CH4 C2H6 C3H8 C4H10 C4H10 C5H12 C6H14 C8H18 C10H22

-184 -172 -189.9 -135 -145 -131.5 -94.3 -107.4 030

-161.5 -88.3 -42.2 -0.6 -10.2 36.2 69.0 99.3 174.0

0.626 0.659 0.692 0.730

C5H10 C6H12 C6H12

-93.3 -142.4 6.5

49.5 71.8 81.4

0.745 0.754 0.779

C6H6 C7H8 C8H10 C10H8

5.51 -95 -29 80.2

80.1 110.6 144.4 217.9

0.885 0.867 0.880 0.971

Table 3 Physical properties of some common petroleum reservoir fluid constituents.

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2.7 Asphalts

Asphalt is not a series by itself. Asphalts are highly viscous to semi-solid, brownblack 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. Sulphur and its associated compounds represent 0.04% - 5% by weight. These corrosive compounds include sulphur, hydrogen sulphide (H2S ),which is very toxic, and mercaptans of low molecular weight ( these are produced during distillation and require special metals to avoid corrosion). Non-corrosive sulphur materials include sulphides. Sulphur compounds have a bad smell and both the corrosive and noncorrosive forms are undesirable. On combustion these products produce S02 and S03 which are undesirable from an environmental perspective. 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. Carbon Dioxide is a very common constituent of reservoir fluids, especially in gases and gas condensates. Like oxygen it is a source of corrosion. It reacts with water to form carbonic acid and iron to form iron carbonate. Carbon dioxide like methane has a significant impact on the physical properties of the reservoir fluids. 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. 8

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Reservoir Fluids Composition T H R E E

4. COMPOSITIONAL DESCRIPTION FOR RESERVOIR ENGINEERING 4.1 Definitions of Composition in Reservoir Engineering

In petroleum engineering, and specifically in reservoir engineering, the main issue is one of the physical behaviour and characteristics of the petroleum fluids. The composition of the fluid clearly has a significant impact on the behaviour and properties. In petroleum engineering therefore the description of the composition is a key to determine the physical properties and behaviour. 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. The compositional model is a compositional description based on the paraffin series CnH2n+2. The fluid is described with individual compositions of normal paraffins up to a limiting C number. Historically C6, more common now to go up to C9, or even higher. Components greater than the limiting C number are lumped together and defined as a C+ component. 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. 22/07/14

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Reservoir Engineering

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. Reservoir Fluid

Gas at Surface Conditions

Oil at Surface Conditions

C1

C2

C3

C4

C5

C6

C7+

The relative amounts of C1 - C7+ are a function of : Temperature, Pressure, Composition (particularly at high temperature)

Figure 5 Compositional Model.

5. GENERAL ANALYSIS 5.1 Surface condition characterisation

Reservoirs as well as having unique compositions also exist at specific pressures and temperatures. It is important therefore to provide a common basis for describing the quantities of fluids in the reservoir and throughout the production process. 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. It is useful to consider these expression not as volumes but as mass, the volume of which 10

Reservoir Fluids Composition T H R E E

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.

Degrees API =



141.5 -131.5 Sp.Gr.at 60ºF

(1)

Sp.Gr = specific gravity relative to water at 60oF. 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

Appearance Colourless at surface Gas

Initial GOR (scf/stb) ºAPI Gas S.G. (air=1)

WetGas

Gas Condensate

Volatile Oil

Black Oil

Colourless Gas + clear liquid

Colourless + significant clear/straw Colour

Brown liquid Some Red/Green Liquid

Black Viscous Liquid

No Liquids

>15000

3000-15000

2500-3000

100-2500

-

60-70

50-70

40-50

<40

0.60-0.65

0.65-0.85

0.65-0.85

0.65-0.85

0.65-0.85

88.7

72.7

66.7

52.6

Composition (mol %) C1

96.3

C3

C2

3.0

6.0

10.0

9.0

5.0

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

Table 5 Typical values for different reservoir fluids.

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5.2 Refractive index

The refractive index provides another indicator of the density of produced oils. The general refractive index range for oil is 1.39 to 1.49. The heavier the crude, the higher the refractive index and the lower the API gravity. This can be measured with a refractometer or by the same methods used in optical mineralogy with reference gravity oils.

5.3 Fluorescence of oil

The fluorescence of oil which is measured by its colour under ultraviolet light provides another indicator, and is often used by those analysing the cuttings as the well is drilled. The rock sample should be placed as quickly as possible under ultraviolet light since fluorescence of oil subsides with evaporation and the activity of ‘live’ oil decreases. If whole core is being examined then the whole core should be passed under UV light to determine the fluorescent colour and the pattern of oil-in-place in the cored interval. 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.



2˚ - 10˚ 10˚ - 18˚ 18˚ - 45˚ 45˚ - above

API API API API

non-fluorescent to dull brown yellow brown to gold gold to pale yellow blue-white to white

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.

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Reservoir Fluids Composition T H R E E

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.

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 (STM3) 1 Foot 1 barrel 1 barrel

= = =

0.3048m 5.615 cu ft. 0.159 m3

EXERCISE 3 A reservoir is said to contain an ‘initial GOR’ of 11,000scf/bbl. What type of reservoir is described, and what API oil could be typically expected from such a field?

EXERCISE 4 Define the ‘Black Oil Model’ and the ‘Compositional Model’

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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 Convert using the equation 1: API = (141.5 / SG) -131.5 38

= (141.5 / SG) -131.5

SG= 141.5 / (131.5 + 38)

SG = 0.835

Density = 0.8 x 62.32 1

= 49.86 lb/ft3

Similarly, to convert SG into API: API = (141.5 / 0.744) -131.5 API = 58.7o 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 604 scf/bbl = 604 * 0.3048 STM3/bbl = 17.09 SCM/bbl= 107.48 SCM/STM3 EXERCISE 3 A reservoir is said to contain an ‘initial GOR’ of 11,000scf/bbl. What type of reservoir is described, and what API oil could be typically expected from such a field?

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Reservoir Fluids Composition T H R E E

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. EXERCISE 4 Define the ‘Black Oil Model’ and the ‘Compositional Model’ SOLUTION Black Oil Model. Two component description of the reservoir fluid consisting of stock tank oil and solution gas. Compositional changes with varying pressure and temperature are ignored. Terms such as ‘Gas Oil Ratio’ and ‘Formation Volume Factor’ are black oil model terms. 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.

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REFERENCES. 1. Amyx, J.W., Bass, D.M., and Whiting, R.L."Petroleum Reservoir Engineering", McGraw-Hill Book Company, New York 1960

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Phase Behaviour of Hydrocarbon Systems F O U R

6000

Single Phase

5000

Pressure Lbs. (psia)

4000

Two Phases

3000

2000

Region of retrograde condensation et ha ne

1000

Eth

M

0 0

-100

e an

0

pa Pro

ne

100

e an xane ptane ne ut a ent ane N-B N- P N-He N-He N-Dec

200

Cricondenbar 300 400 500

600

1

700

Temperature º F % Liq. Liquid

Pressure

100

10 5 0

2

e Dew Point Lin

3 Gas

Cricondentherm

25

Po

e

50

Bu bb l

in t

Li

ne

75

Temperature

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Phase Behaviour of Hydrocarbon Systems F O U R

C O N T E N T S 1 DEFINITIONS 2 PHASE BEHAVIOUR OF PURE SUBSTANCES 2.1 The Phase Diagram 3 USE OF PHASE DIAGRAMS 3.1 Pressure - Temperature Diagrams (PT) 3.2 Pressure Volume Diagram (PV) 4 TWO COMPONENT SYSTEMS 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 6 COMPARISON OF THE PHASE DIAGRAMS OF RESERVOIR FLUIDS 7 RESERVOIRS WITH A GAS CAP 8 CRITICAL POINT DRYING

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LEARNING OBJECTIVES 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 . • 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.

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Phase Behaviour of Hydrocarbon Systems F O U R

Oil and gas reservoir fluids are mixtures of a large number of components which when subjected to different pressure and temperature 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 Before we consider the effect of temperature and pressure on hydrocarbon systems we will define some terms: •

System - amount of substances within given boundaries under specific conditions composed of a number of components. Everything within these boundaries are part of the system and that existing outside of the boundaries are not part of the system. If anything moves across these boundaries then the system will have changed.

• 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.

•

Equilibrium - When a system is in equilibrium then no changes take place with respect to time in the measurable physical properties of the separate phases.

•

Intensive and extensive properties - physical properties are termed either intensive or extensive. Intensive properties are independent of the quantity of material present. For example density, specific volume and compressibility factor are intensive properties whereas properties such as volume and mass are termed extensive properties; their values being determined by the total quantity of matter present.

The physical behaviour of hydrocarbons when pressure and temperature changes can be explained in relation to the behaviour of the individual molecules making up the system. Temperature, pressure and intermolecular forces are important aspects of physical behaviour. The temperature is an indication of the kinetic energy of the molecules. It is a physical measure of the average kinetic energy of the molecules. The kinetic energy increases as heat is added. This increase in kinetic energy causes an increase in the motion of the molecules which also results in the molecules moving further apart. 22/07/14

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The pressure reflects the frequency of the collision of the molecules on the walls of its container. As more molecules are forced closer together the pressure increases. Intermolecular 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 components however in understanding the phase behaviour of these systems it is convenient to reflect on the behaviour of single and two component systems.

2 PHASE BEHAVIOUR OF PURE SUBSTANCES 2.1 The Phase Diagram

It is beneficial to study the behaviour of a pure hydrocarbon under varying pressure and temperature to gain an insight into the behaviour of more complex hydrocarbon 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. Figure 1 gives a pressure - temperature phase diagram for a single-component system on a pressure temperature diagram and the following points are to be noted.

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Reservoir Engineering

Phase Behaviour of Hydrocarbon Systems F O U R

1

2

Melting P

Pressure

Solid

oint

C

Liquid

u po Va

blim

Su

a

Critical Point

r

s es Pr

e ur

3

Vapour

Gas

tion

Triple Point Temperature

Figure 1 Pressure temperature diagram for a single component system

Vapour Pressure Line The vapour pressure line divides regions where the substance is a liquid, 2, from regions where it is a gas, 3. Above the line indicates conditions for which a substance is a liquid, whereas below the line represent conditions under which it is a gas. Conditions on the line indicate where both liquid and gas phases coexist. 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 conditions which divides the area where solid exists from the area where vapour exists and is also called the sublimation - pressure line. Melting Point Line The melting line divides solid from liquid. For pure hydrocarbons the melting point generally increases with pressure so the slope of the line is positive. (Water is exceptional in that its melting point decreases with pressure). 22/07/14

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3 USE OF PHASE DIAGRAMS 3.1 Pressure -Temperature Diagrams (PT)

Consider the behaviour of a cell charged with a pure substance and the volume varied by the frictionless displacement of a piston as shown in Figure 2, below. P1

Pb

P

Pd

P2

Liquid Gas

Figure 2 Phase Changes With Pressure at Constant Temperature

For example, following the path 1 - 2 in Figure 3 on the pressure-temperature diagram, ie holding temperature constant and varying pressure by expansion of the cylinder.

3 E Pc

c A

B

Liquid

Melting -

Pressure

Solid

Point Lin e

1

F

ur

po Va

-

ure

ss pre

line 2

G

4 Gas

T

Temperature

Tc

Figure 3 Pressure-Temperature Diagram for a Single-Component System

6

Phase Behaviour of Hydrocarbon Systems F O U R

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 isothermally (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. The pressure-temperature diagram for ethane is given in Figure 4.

800 c

Pressure - PSIA

700 Liquid 600

Vapor 500

400 40

60

80

100

120

Temperature - º F

Figure 4 Pressure-Temperature diagram of Ethane

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3.2 Pressure Volume Diagram (PV)

The process just described in 3.1 can also be represented on a pressure-volume diagram at constant temperature (Figure 5). As the pressure is reduced from 1, a large change in pressure occurs with small change in volume due to the relatively low compressibility of the liquid. When the vapour pressure is reached gas begins to form. This point is called the bubble point, ie the point at which the first few molecules leave the liquid and form small bubbles of gas. As the system expands more liquid is vaporised at constant pressure. The point at which only a minute drop of liquid remains is called the dew point. Sharp breaks in the line denote the bubble point and dew point. PVT CELL

PV DIAGRAM 3

All Liquid

T > Tc

SINGLE PHASE

1

First Gas Bubble

T < Tc

Pressure

Last Drop of Liquid

Liquid state-rapid change of pressure with small volume change Pressure remains constant while both gas and liquid are present

Dew Point Gas

Bubble Point

T2 > Tc 4

2

TWO PHASE REGION All Gas

Volume

Figure 5 Pressure-Volume diagram for a Single-Component System

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. A series of expansions can be performed at various constant temperatures and a pressure volume diagram built up and the locus of the bubble point and dew point values gives the bubble point and dew point lines which meet at the critical point. Conditions under the bubble point and dew point lines represent the conditions where two phases coexist whereas those above these curves represent the conditions where only one phase exists. At the critical temperature the P,T curve goes through the critical point, Figure 6

8

Phase Behaviour of Hydrocarbon Systems F O U R

3

T = Tc

Liquid state rapid change of temperature with small volume change

T > Tc

1

SINGLE PHASE T < Tc 4

De

Curve

Bubble

Point

Pressure

Critical Point

w

Po in t Pressure remains constant while Cur ve both gas and liquid are present

2

TWO PHASE REGION Volume

Figure 6 Series of PV lines for a pure component

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

Pressure - PSIA

800 C

700

90

600

400 0

ºF

Two Phase Region

Liquid 500

11 0

A

B

D

0.05

0.10

0.15

ºF

Vapor 60 º F

0.20

0.25

Specific Volume - Cu. Ft. per lb.

Figure 7 Pressure-Volume Diagram of Ethane

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Bubble Point Line

uid

Liq

Critical Point

Vo lu

s

Ga

me

e

tur

ra pe

Tem

id

Critical Point

u Liq

Pressure

G as

an d

Pressure

Li qu id

Dew Point Line

Vapor Pressure Curve

s Ga ure rat pe m Te

Figure 8 Three Dimensional Phase Diagram for a Pure Component System

4 TWO COMPONENT SYSTEMS

Reservoir fluids contain many components but we will first consider a system containing two components, such a system is called a binary.

4.1 Pressure Volume Diagram

The behaviour of a mixture of two components is not as simple as for a pure substance. Figure 9 shows the P-V diagram of a two-component mixture for a constant temperature system.

Pressure

Liquid Bubble Point

Liquid

and

Gas

Dew Point

Ga

s

Volume

Figure 9 Pressure-Volume Line for a Two-Component System at Constant Temperature

10

Phase Behaviour of Hydrocarbon Systems F O U R

The isotherm is very similar to the pure component but the pressure increases as the system passes from the dew point to the bubble point. This is because the composition of the liquid and vapour changes as it passes through the two-phase region. At the bubble point the composition of the liquid is essentially equal to the composition of the mixture but the infinitesimal amount of gas is richer in the more volatile component. At the dew point the composition of vapour is essentially the mixture composition whereas the infinitesimal amount of liquid is richer in the less volatile component. Breaks in the line are not as sharp as for pure substances. 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

400

300

200

100 0

0º

Bubble Point Line

Pressure - PSIA

45

Critical point 45 4º F

425

º

400

º

350

º

300

º

0.1

Dew Point Line

0.2

0.3

0.4

0.5

Specific Volume - Cu. Ft. per lb.

Figure 10 Pressure-Volume Diagram for N-Pentane and N-Heptane (52.4 mole % Heptane) ref. 4

4.2 Pressure Temperature Diagram

Compared to the single line representing the vapour pressure curve for pure substances there is a broad region in which the two phases co-exist. The two-phase region of the diagram is bounded by the bubble point line and the dew point line, and the two lines meet at the critical point. Points within a loop represent two-phase systems (Figure 11). Consider the constant temperature expansion of a particular mixture composition. At 1 the substance is liquid and as pressure is reduced liquid expands until the bubble point is reached. The pressure at which the first bubbles of gas appear is termed the bubble point pressure. As pressure is decreased liquid and gas co-exist until a minute amount of liquid remains at the dew point pressure. Further reduction of pressure causes expansion of the gas. 22/07/14

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By carrying out a series of constant temperature expansions the phase envelope is defined and within the envelope contours of liquid to gas ratios obtained. These are called quality lines and describe the pressure and temperature conditions for equal volumes of liquid. The quality lines converge at the critical point.

4.3 Critical Point

In the same way as pure components, when more than one component is present liquid and gases cannot coexist, at pressures and temperatures higher than the critical point. The critical point for a more than one component mixture is defined as a point at which the bubble point line and dew point line join, ie. it is also the point at which all the intensive properties of the liquid are identical. This aspect is a very severe test for physical property prediction methods. 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 Critical Point

PCAB

PCA

% Liq.

Liquid

CA

100 75 50

PCB

Pressure

b Bu

ble

P

ine tL o in

Dew

Temperature

t Poin

TCA

CB

25 0 2

Gas

TCAB

TCB

Figure 11 Pressure-Temperature Diagram for a Two Component System

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 mixture 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. 12

Phase Behaviour of Hydrocarbon Systems F O U R

1400 C2

Composition No Wt % Ethane C 100.00 C1 90.22 C2 50.25 C3 9.78 C7 N-Heptane

1200

C1 800 C

C3 A1

an e

600

A

le

bb

Bu

i Po

nt

ne

Li

C7

A3 0

i

B2

B1 100

200

B3

De

w

Po

n

e

nt

A2

200

0

li n

e

400

E th

Pressure, lbs./Sq. In. ABS

1000

H N-

ep

ta

B 300

400

500

600

Temperature º F

Figure 12 Pressure-Temperature Diagram for the Ethane-Heptane System 2

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6000

Single Phase

5000

Pressure Lbs. (psia)

4000

Two Phases

3000

2000

et ha ne

1000

M

0 0

-100

Eth

e an

0

pa Pro

ne

100

e an xane ptane ne ut a ent ane N-B N- P N-He N-He N-Dec

200

300

400

500

600

700

Temperature º F

Figure 13 Critical Point Loci for a Series of Binary Hydrocarbon Mixtures 2

4.4 Retrograde Condensation

Within the two phase region our two component system there can be temperatures and pressures higher than the critical temperature where two phases exist and similarly pressures. These limiting temperatures and pressures are the cricondentherm and cricondenbar . The cricondentherm can be defined as the temperature above which liquid cannot be formed regardless of pressure, or expressed differently, as the maximum temperature at which two phases can exist in equilibrium. The cricondenbar can be defined as the pressure above which no gas can be formed regardless of temperature or as the maximum pressure at which two phases can exist in equilibrium. (Figure 14). 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 phenomena is termed Retrograde Condensation. From 2 - 3, the amount of

14

Phase Behaviour of Hydrocarbon Systems F O U R

liquid decreases and vaporisation occurs and the dew point is again reached where the system is gas. Retrograde condensation occurs at temperatures between the critical temperature and cricondentherm. The retrograde region is shown shaded in the Figure. Region of retrograde condensation

Cricondenbar Liquid

1

% Liq.

Pressure

100

10 5 0

2

e Dew Point Lin

3 Gas

Cricondentherm

25

Po

e

50

Bu bb l

in t

Li

ne

75

Temperature

Figure 14 Phase Diagram Showing Conditions for Retrograde Considerations

5. MULTI-COMPONENT HYDROCARBON 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 liquid-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.

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PVT CELL

Reservoir Engineering

PHASE DIAGRAM

All Liquid

Liqu id

"a"

Critical Point

First Gas Bubble

Bubble Point uid Liq

% % 40

%

20

%

w De

int Po

Lin

Pressure

60

e

Bu bb le

Pressure

Last Drop of Liquid

Po in

Gas / 40% Liquid

80

tL i ne

Bubble Point

Dew Point

Dew Point

All Gas

Temperature

Volume

Figure 15 Phase Diagrams for Multicomponent Systems

We will consider the behaviour of several examples of typical crude oils and natural gases: Low-shrinkage oil (heavy oil - black oil) High-shrinkage oil (volatile oil) Retrograde condensate gas Wet gas Dry Gas 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.

16

Phase Behaviour of Hydrocarbon Systems F O U R

Single Phase Region (Gas)

Single Phase Region (Liquid) Black Oil

P

Pressure

Bu

% Liquid 100 75

b

b

e Lin

Gas

m

P

int Po ble

Volatile Gas Oil Condensate CP

2

Two Phase Region

TM Where: P

b

50 25 20 15 10 5 0

Dew

Poin

e t Li n

Single Phase Region

= Bubble point pressure at indicated temperature

P

m

= Maximum pressure at which two phases can coexist

T

m

= Maximum temperature at which two phases can coexist

C = Critical conditions

Gas

X5

X

5

= Cricondentherm

Temperature

Figure 16 Phase diagram for reservoir fluids

5.1 Oil Systems ( Black Oils and Volatile Oils)

Figures 17&18 illustrate the PT phase diagrams for black and volatile oils. 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.

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1 Undersaturated Mole % Liq. 100

Lin e

2 Saturated

Critical Point

Pb

3

line

75

De

50

w

Po

Sep.

int

Po int

Bu bb le

Pressure

Liquid

Gas

25 0

Temperature

Figure 17 Phase Diagram for a Black Oil

As the pressure is dropped from the initial condition as a result of production of fluids, 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 separator 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.

18

Phase Behaviour of Hydrocarbon Systems F O U R



Broad-phase envelope High percentage of liquid High proportion of heavier hydrocarbons GOR < 500 SCF/STB Oil gravity 30˚ API or heavier Liquid - black or deep colour

Volatile oil contains a much higher proportion of lighter and intermediate hydocarbons 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 light oil or high shrinkage oil when compared to black oils are as follows.

Not so broad phase envelope as black oil Fewer heavier hydrocarbons Deep coloured API < 50˚ GOR < 8000 SCF/STB

1 2

Liquid

Critical Point

Mole % Liq. 100

3

50 40

e

Sep.

25

Gas

De w

po

in

tl

in

Bu bb le

po

int

lin

e

Pressure

75

0

Temperature

Figure 18 Phase Diagram for a Volatile Oil

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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. 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 fluids 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 declines 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 occurs 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 mobility. 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.

1

Mole % Liq.

e bl ub

P

tL oin

ine

Critical Point 2

3

B

Pressure

Liquid

100 75

Sep.

50 25 10 5 0

P Dew

t oin

e Lin

Gas

Temperature

Figure 19 Phase Diagram for a Retrograde Condensate Gas

20

Phase Behaviour of Hydrocarbon Systems F O U R

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 cycling ( Figure 20 ), is the preferred yet very expensive option. In this process the produced fluids are separated at the surface and the liquid condensates, high value product relative to heavy oil, are sent for export, in an offshore situation probably by tanker. The ‘dry’ gas is then compressed and reinjected into the reservoir to maintain the pressure above the dew point. Clearly with this process the pressure will still decline because the volume occupied by the gas volume of the exported liquid is not being replaced. Full pressure maintenance is obtained by importing dry gas equivalent to this exported volume from a nearby source. Eventually the injected dry gas displaces the ‘wet’ gas and then the field can be blown down as a conventional dry gas reservoir, if a suitable export route for the gas is then in place. The process described is very costly and carries with it a number of risks not least the possibility of early dry gas breakthrough. Imported Gas

Gas

Dry Gas Reinjection

Surface Separation Condensate Sales

Injection Well

Production Well Gas Water Contact

Figure 20 Gas cycling process

Recent research has shown that the nature of oil forming in porous media by this retrograde 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 22/07/14

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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" quickly might be an option and as a result vaporise the condensed liquids in the formation. 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 exists 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 proportion 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

Pressure

Critical Point

Mole % Liq. 100 75 50 25 5 0

2 Sep. Gas

Temperature

Figure 21 Phase Diagram for a Wet Gas

22

Phase Behaviour of Hydrocarbon Systems F O U R

The reference wet gas, clearly does not refer to the system being wet due to the presence of water but due to the production condensate liquids. In some locations where there are natural petroleum leakages at the surface, when condensates are produced they are sometimes called white oil. The summary characteristics for wet gas are as follows. GOR < 100,000 SCF/STB Condensate liquid > 50˚ API

5.5 Dry Gas

The phase envelope of the dry gas, which contains a smaller fraction of the C2-C6 components, is similar to the wet gas system but with the distinction that the separator also lies outside the envelope in the gas region (Figure 22). The term dry indicates therefore that the fluid does not contain enough heavier HC’s to form a liquid at surface conditions. The summary characteristics for a dry gas are as follows.

GOR > 100,000 SCF/STB

Pressure

1

Critical Point

Liquid

75 50 25

2 Sep. Gas

Temperature

Figure 22 Phase Diagram for a Dry Gas

6 COMPARISON OF THE PHASE DIAGRAMS OF RESERVOIR FLUIDS 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 22/07/14

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Pressure

these determine the character of the fluid behaviour. Figure 23 gives a better indication of the various reservoir types with respect to a specific pressure and temperature scales. As the proportion of heavier components in the respective fluids increases the phase envelope moves to the right.

Separator

Dry Gas

Gas Wet Gas Condensate

Volatile Oil

Black Oil

Temperature (ºC) Critical Point

Figure 23 Relative positions of phases envelopes

7 RESERVOIRS WITH A GAS CAP Figure 24 illustrates a simplification of the phase diagrams associated with an oil reservoir with a gas cap. The phase diagram for the gas cap fluid, the oil reservoir fluid and for a fluid representing the combination fluid of a mixture of gas and liquid in the same proportions as they exist in the reservoir are presented.

24

Phase Behaviour of Hydrocarbon Systems F O U R

Reservoir Temperature Reservoir Gas

Total Reservoir Fluid

CG

C Reservoir Liquid

Pd=Pb Pressure

Initial Reservoir Pressure

CL

Separator

Temperature

Figure 24 Phase Diagram for an Oil Reservoir with a Gas Cap

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 proportion 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.

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The company concerned embarked on a more sophisticated core recovery and analysis 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

Figure 25 Critical point drying system

Pressure

Critical Point Drying Route

Critical Point LIQUID

Vapour Pressure Line GAS Temperature

Figure 26 Critical point drying

26

Phase Behaviour of Hydrocarbon Systems F O U R

REFERENCES 1. Figure 1 Daniels, F Farrington: “Outlines of Physical Chemistry,” John Wiley & Sons,Inc New York, 1948 2. Figure 2 Brown,GG et al. “ Natural Gasoline and Volatile Hydrocarbons,” Natural Gasoline Association of America, Tulsa, Okl., 1948. Figure 10 Sage, S.G.,Lacy,W.N. Volumetric and Phase Behaviour of Hydrocarbons, Gulf Publishing Co.Houston 1949

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Behaviour of Gases F I V E

c

T1>Tc

P

Tc

Psat

T2
.008

Vsat (liq)

V

.006 Bg rb/scf Vsat (vap) .004 .002 1000

2000

3000

PRESSURE (psig)

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Behaviour of Gases F I V E

C O N T E N T S 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 2 BEHAVIOUR OF REAL GASES 2.1 Compressibility Factor For Natural Gases 2.2 Law of Corresponding States 2.3 Pseudocritical Properties of Natural Gases 2.4 Impact of Nonhydrocarbon Components on z Value 2.5 Standard Conditions For Real Reservoir Gases 3 GAS FORMATION VOLUME FACTOR 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

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LEARNING OBJECTIVES Having worked through this chapter the Student will be able to: • Present the ideal equation of state, PV=nRT. • Calculate the mass of an ideal gas given PV &T values. • Derive an equation to calculate the density of an ideal gas. • Convert a mixture composition between weight and mole fraction. • Present an equation and calculate the apparent molecular weight of a mixture. • Define and calculate the specific gravity of a gas. • Present the equation of state, EOS, for a ‘real gas’ and explain what ‘Z’ is, PV=ZnRT. •

Define the pseudocritical pressure and psuedocritical temperature and be able to use them to determine the ‘Z’ value for a gas mixture.

• Express and calculate reservoir gas volumes in terms of standard cubic volumes. • Define the gas formation volume factor and derive an equation fore it using the EOS. • Calculate the volume of gas in a reservoir in terms of standard cubic volumes given prerequisite data. • Be aware of the development of EOS’s to predict reservoir fluid properties.

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Behaviour of Gases F I V E

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 is 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.

Out of these assumptions come the following equations.

1.1 Boyle’s Law

At constant temperature the pressure of a given weight of a gas is inversely proportional to the volume of a gas. i.e.

1 or PV = constant, T is constant (1) P



Vα



P = pressure, V = volume, T = temperature.

1.2 Charles’ Law

At constant pressure, the volume of a given weight of gas varies directly with the temperature:

i.e.

V α T or

V = constant, P is constant (2) T

The pressure and temperature in both laws are in absolute units.

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1.3 Avogadro’s Law

Avogadro’s Law can be stated as: under the same conditions of temperature and pressure equal volumes of all ideal gases contain the same number of molecules. That is, one molecular weight of any ideal gas occupies the same volume as the molecular weight of another ideal gas at a given temperature and pressure. Specifically, these are: (i) 2.73 x 1026 molecules/lb mole of ideal gas. (ii) One molecular weight (in lbs) of any ideal gas at 60˚F and 14.7 psia occupies a volume of 379.4 cu ft.



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. eg.

one lb mole of methane CH4 = 16 lb one kg mole of methane CH4 = 16kg

1.4 The Equation of State for an Ideal Gas

By combining the above laws an equation of state relating pressure, temperature and volume of a gas is obtained.

PV = constant (3) T

R is the constant when the quantity of gas is equal to one mole. It is termed the Universal Gas Constant and has different values depending on the unit system used, so that;

cu ft psia

R in oilfield units = 10.732 lb mole R Table 1 gives the values for different unit systems. P psia atm atm atm atm mm Hg in.Hg

V

T

n

cu ft cu ft cc litre cu ft litre cu ft

R K K K R K R

lb - mole lb - mole gm - mole gm - mole lb - mole gm - mole lb - mole

R 10.73 1.3145 82.06 0.08206 0.730 62.37 21.85

Table 1 Values of R for different unit systems.

For n moles the equation becomes:

4

PV = nRT

(4)

Behaviour of Gases F I V E



T= absolute temperature oK or oR where ºK=273 +oC and oR=460 +oF

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:

n =

PV PV PV 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 volume of 3 cu.ft assuming methane is an ideal gas calculate the mass of methane in the cylinder.

1.5 The Density of an Ideal Gas

Since density is defined as the weight per unit volume, the ideal gas law can be used to calculate densities.

ρg = weight / volume =

where rg is the gas density For 1 mole m = MW

V =





m V MW = Molecular weight

RT P

∴ ρg =

MW.P (5) RT

EXERCISE 2. Calculate the density of the gas in the cylinder in exercise 1.

1.6 Standard Conditions

Oil and gas at reservoir conditions clearly occur under a whole range of temperatures and pressures. It is common practice to relate volumes to conditions at surface, ie 14.7 psia and 60˚F. ie 22/07/14

Pres Vres P V = sc sc (6) Tres Tsc

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sc - standard conditions

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res - reservoir conditions

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.

1.7 Mixtures of Ideal Gases

Petroleum engineering is concerned not with single component gases but mixtures of a number of gases. Laws established over early years governing ideal gas mixtures include Dalton’s Law and Amagat’s Law.

1.7.1 Dalton’s Law of Partial Pressures

The total pressure exerted by a mixture of gases is equal to the sum of the pressures exerted by its components. The partial pressure is the contribution to pressure of the individual component. Consider a gas made up of components A, B, C etc The total pressure of the system is the sum of the partial pressures ie

P = PA + PB + PC + ..... (7)

where A, B and C are components. therefore

P = nA i.e.

∴

RT RT RT + nB + nC V V V

P =

RT Σn j V

Pj n = j = yj P n



where yj = mole fraction of jth component. 6



(8)



Behaviour of Gases F I V E

The pressure contribution of a component, its partial pressure, is the total pressure times the mole fraction.

1.7.2 Amagat’s Law

Amagat’s Law states that the volume occupied by an ideal gas mixture is equal to the sum of the volumes that the pure components would occupy at the same temperature and pressure. Sometimes called the law of additive volumes. i.e.

V = VA + VB + VC (9) V = nA V =



i.e.

RT RT RT + nB + nC P P P

RT Σn j P

Vj n = j = yj V n

(10)

i.e, for an ideal gas the volume fraction is equal to the mole fraction. It is conventional to describe the compositions of hydrocarbon fluids in mole terms. This is because of the above laws. In some circumstances however weight compositions might be used as the basis and it is straight forward to convert between the two. 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.

1.8 Apparent Molecular Weight

A mixture does not have a molecular weight although it behaves as though it had a molecular weight. This is called the apparent molecular weight. AMW

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If yj represents the mole fraction of the jth component:

(

AMW = Σ y j × MWj

)

AMW for air = 28.97, a value of 29.0 is usually sufficiently accurate.

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EXERCISE 5. What is the apparent molecular weight of the gas in exercise 4



1.9 Specific Gravity of a Gas

The specific gravity of a gas, gg is the ratio of the density of the gas relative to that of dry air at the same conditions.

ρg (11) ρair



γg =



Assuming that the gases and air are ideal.

MgP

γg =

RT = M g = M g M air P M air 29 RT

Mg = AMW of mixture, Mair = AMW of air. EXERCISE 6. What is the gas gravity of the gas in exercise 4?

2 BEHAVIOUR OF REAL GASES 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. 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: (1) By using a correction factor in the equation PV = nRT (2) By using another equation-of-state

2.1 Compressibility Factor for Natural Gases

The correction factor ‘z’ which is a function of the gas composition, pressure and temperature is used to modify the ideal gas law to:

8

PV = znRT

(12)

Behaviour of Gases F I V E

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). To compare two states the law now takes the form:

P1V1 PV = 2 2 (13) z1T1 z 2 T2

z is an expression of the actual volume to what the ideal volume would be.

co ns ta nt

Vactual (14) Videal

Compressibility factor, Z

=

1.0

=

z

at ur e



pe r

i.e.

Te m



0.5

0

0

PRESSURE, P

Figure 1 Typical plot of the compressibility factor as a function of pressure at constant temperature.

Although all gases have similar shapes with respect to z the actual values are component specific. However through the law of corresponding states all pure gases are shown to have common values.

2.2 Law of Corresponding States

The law of corresponding states shows that the properties of many pure liquids and gases have the same value at the same reduced temperature (Tr) and pressure (Pr) where:

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Tr =

T P and Pr = (15) Tc Pc

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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:

Tpc = Σy jTcj and Ppc = Σ y j Pcj

(16)

where y is the mole fraction of component j and Tcj and Pcj are the critical temperature and pressure of component j. It should be emphasised that these pseudo critical temperature and pseudocritical pressures are not the same as the real critical temperature and pressure. By definition the pseudo values must lie between the extreme critical values of the pure components whereas the actual critical values for mixtures can be outside these limits, as was observed in the Phase Behaviour chapter. 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 F I V E

Compressibility Factors for Natural Gases as a Function of Pseudoreduced Pressure and Temperature.

1.1

Pseudo Reduced Pressure, Pr

1

0

2

3

4

5

6

3.0 2.8 2.6 2.4 2.2 2.0 1.9 1.8

0.9

5

1. 4

1.

1.5

05

1.6

1.

1.1 1.0 1.05 1.2 0.95

1.7

1

0.8

1.3 1.1

1.

1.7

1.45

0.7 0.6

1. 3

1.2

1.6 1.8

1.15

0.4

2.0

1.7 1.9

3.0

1.3

2.6

1.2

1.05

0.25

2.8

1.1

3.0

2.2 2.0

1.8 1.7 1.6

0.9 7

1.9

1.2 1.1 1.4 1.3

8

1.1

Compressibility of Natural Gases (Jan. 1, 1941)

2.6 2.4

1.4

2.2

2.4

1.1

0.3

1.5

4 1. 1.5

1.25

0.5

1.6

2

1.4 1.35 1.3

1.

Compressibility Factor, z

8

Pseudo Reduced Temperature

1.0

1.0

7

1.0

1.05

0.9 9

10

11

12

13

Pseudo Reduced Pressure, Pr

14

15

Figure 2 Compressibility factors for natural gas1 (Standing & Katz, Trans AIME, 1942).

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Tpr =

T P and Ppr = Tpc Ppc

Reservoir Engineering

(17)

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. For the gas of exercise 4 determine the compressibility factor at a temperature of 150°F and a pressure of 3500psia.

2.3 Pseudocritical Properties of Natural Gases

Pseudocritical Pressure, psia

The pseudocritical properties of gases can be computed from the basic composition but can also be estimated from the gas gravity using the correlation presented in Figure 3. Pseudocritical Properties of Natural Gases

700

Condens

650

Miscellaneou s

ate Wel l

Fluids

Gases

600

550

Pseudocritical Temperature, R

500

450

s

se

Ga

ids Flu ell W e sat de n n Co

400

350

300

e

an

ell

sc Mi

s ou

0.5

0.6

0.7

0.8

0.9

1.0

1.1

Gas Gravity (air = 1)

Figure 3 Pseudocritical properties of natural gases. 3

12

1.2

Behaviour of Gases F I V E

2.4 Impact of Nonhydrocarbon Components on z value.

Components like hydrogen sulphide, and carbon dioxide have a significant impact on the value of z. If the method previously applied is used large errors in z result. Wichert and Aziz 4 have produced an equation which enables the impact of these two gases to be calculated. T'pc = Tpc - e (18) and

p′pc =

p pcTpc′ Tpc + yH 2 S 1 − yH 2 S ε

(

(19)

)

T'pc and p'pc are used to calculate Tpr and Ppr. The value for e is obtained from the Figure 4 from the Wichert and Aziz paper

80

15

70

PER CENT C02

60

50

20

40

25

30

30

20 30

25

10 20 15

0

0

5

10

10

20

30

40

34.5

50

60

70

80

PER CENT H2S

Figure 4 Adjustment factors for pseudocritiacl properties for non hydrocarbon gases(Wichert & Aziz).

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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 Gas Components 1 2 3 4 5 6



Methane Ethane Propane Hydrogen sulphide Carbon Dioxide Nitrigen Total

Wgt Mol fraction weight

lb moles

Mole fraction

Pc-psi

Tc °R

Ppc psia

Tpc 255.70 26.17 10.81 28.25

25 3 1.5 3

0.56 0.07 0.03 0.07

16.04 30.07 44.09 34.08

0.035 0.002 0.001 0.002

0.743 0.048 0.016 0.042

667.00 708.00 616.00 1306

344 550 666 673

495.8 33.7 10.0 54.8

10

0.22

44.01

0.005

0.108

1071

548

116.1 59.38

2.5 45

0.06 1.00

28.02

0.002 0.0466

0.043 1.000

493

227

21.0 731

9.66 390

From Wichert & Azis chart for compositions of H2S and CO2 e = 19

Tpc′ = Tpc - ε = 371o R p′pc =



Weight

p pcTpc′ Tpc + yH 2 S 1 − yH 2 S ε

(

)

Ppc′ = 694.3

2.5 Standard Conditions for Real Reservoir Gases

As indicated in section 1.6 for ideal gases it is convenient to describe the quantity of gas to a common basis and this is termed the standard conditions, giving rise to the standard cubic foot and the standard cubic metre. The petroleum engineer is primarily interested in volume calculations for gaseous mixtures. Throughout the industry gas volumes are measured at a standard temperature of 60˚F (15.6˚C) and at a pressure of 14.7 psia (one atmosphere). These conditions are referred to as standard temperature and pressure STP. Standard Cubic Feet, the unit of volume measured under these conditions is sometimes abbreviated SCF or scf (SCM is Standard Cubic Metres). It is helpful to consider these expressions not as volumes but as an alternate expression of the quantity of material. For example a mass of gas can be expressed as so many standard cubic feet or metres. EXERCISE 10. Express the quantity of 1 lb mole of a gas as standard cubic feet.

14

Behaviour of Gases F I V E

EXERCISE 11. Express the mass of gas in exercise 4 as standard cubic feet.

3 GAS FORMATION VOLUME FACTOR 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.

Bg =

volume occupied at reservoir temperature and pressure volume occupied at STP

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 reservoir barrels free gas/scf gas or rm3 reservoir cubic meters free gas/scm gas

.008 .006 Bg rb/scf .004 .002 1000

2000

3000

PRESSURE (psig)

Figure 5 Gas Formation Volume Factor, Bg.

For example Bg for a reservoir at condition 2 is;

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V2 P Tz = sc 2 2 (20) Vsc P2 Tsc zsc



Bg =



‘sc’ refers to standard conditions. z at standard conditions is taken as 1.0

The reciprocal of Bg is often used to calculate volumes at surface so as to reduce the possibility of misplacing the decimal point associated with the values of Bg being less than 0.01, ie:



volume at surface 1 = =E volume in formation Bg



E is sometimes referred to as the expansion factor.

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 (21) Vsc



Bg =



Subscripts R and sc are reservoir and standard conditions respectively.



VR =



T and P at reservoir conditions:



Vsc =



z = 1 for standard conditions



∴ Bg = z



Since Tsc = 520˚Rm Psc = 14.7 psia for most cases



znRT (22) P zsc nRTsc (23) Psc T Psc cu. ft . . (24) Tsc P SCF

Bg = 0.0283 Bg = 0.0283 or

zT cu. ft P SCF zT cu. ft bbl × P SCF 5.615 cu ft

zT res bbl B = 0.00504 g

16

P SCF

(25)

Behaviour of Gases F I V E

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 GASES 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

cg = −

1  ∂V  1  ∂Vm    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. For an ideal gas:



PV = nRT or:

 dV  = − nRT  dP  P2



1 − nRT  1 cg =    =  V   P 2  P (27)

For real gases:



22/07/14

V =

znRT P

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Reservoir Engineering

dz P −z ∂ V   dP nRT =   ∂P  T P2 cg = −



cg =

P  nRT  ∂z  − z  P 2    nRTz  P  ∂P

1 1 ∂z − . P z ∂P

(28)

∂z/∂P can be obtained from the slope of the z vs P curve. The Law of Corresponding states can be used to express the above equation in another form

P = Ppc Ppr

∂z  ∂Ppr   ∂z  =  ∂P  ∂P   ∂Ppr  ∂Ppr 1 = ∂P Ppc



∂z  1   ∂z  = ∂P  Ppc   ∂Ppr 

Combining this equation with eqn 28 above yields

cg =



1 1  ∂z  − Ppc Ppr zPpc  ∂Ppr  Tpr

c g Ppc =

1 1  ∂z  − Ppr z  ∂Ppr  Tpr

(29)

Units of cg = P-1, and cgPc is dimensionless cgPpc is called pseudo reduced compressibility, cpr

18

Behaviour of Gases F I V E

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 density

kinematic viscosity =

Kinematic viscosity has units of cm2/100 sec and the term is called centistoke.

Viscosoty, micropoises

Gas viscosity reduces as the pressure is decreased. At low pressures an increase in temperature increases gas viscosity whereas at high pressures gas viscosity decreases as the temperature increases. Figure 6 gives the values for pure component ethane.

1000 900 800 700 600 500

Viscosity of ethane Pressure, psia 5000

400

3000

300

4000

2000 15000

200 750

100 90 80 70 50

1000

600 14.7

100

150

200

250

300

350

400

Temperature, deg F

Figure 6 Viscosity of ethane.

The viscosity of gases at low pressures can be obtained from correlations presented by different workers.

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Petroleum Engineering

Reservoir Engineering

0.024

0.022

He

m liu

ro Nit

Air

0.020

0.018

Ca

Viscosity, cp

0.016

d Hy

en rog

0.014

n ge

nD rbo

lfid Su

iox

ide

e

e han Met e ylen Eth

0.012

ane Eth

e pan Pro n ta e i-Bu tane n-Bu

0.010

0.008

e ntan n-Pe ne a x n-He ane t p e n-H

ane n-Oct ane n o N nn n-Deca

e

0.006

0.004

50

100

150

200

250

300

350

400

Temperature, oF

Figure 7 Viscosity of paraffin hydrocarbon gases at one atmosphere.

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 (Lee). 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 F I V E

1.0

1.5

Gas Gravity (Air = 1) 2.0 2.5

0.015

Correction added to Viscosity, c.p.

0.010 0.009

0.006 0.005

0.0010

G = 20

ºF

300

ºF

200

ºF

100

ºF

0 0

CO2

0.0015

G = 06

0.0005

5 10 15 Mole per cent N2

0 0

1.5 1.0

G = 20

G = 06 5 10 15 Mole per cent CO2

1.5 1.0

0.0005

0.004 10

400

H 2S

0.0015

Correction added to Viscosity, c.p.

Viscosity, at 1 atm, m1, centipoise

0.011

0 0

3.5

0.0010

0.0005

0.012

0.007

1.5 1.0

G = 20

0.0010

0.013

0.008

N2

0.0015

0.014

3.0 Correction added to Viscosity, c.p.

0.016

0.5

G = 06 5 10 15 Mole per cent H2S

20

30

40

50 60 70 Molecular Weight

80

90

100

Figure 8 Viscosity of gases at atmospheric pressure.6

6.0 5.0

µ =

Viscosity at operating temperature and pressure, centipoises

µA =

Viscosity at 14.7 psia (1atm) and operating temperatures, centipoises

4.0

Viscosity, µ / µA

3.5 3.0

20 15

2.5

2.0

6 3

1.5

8

Ps eu do red

10

uc ed

Pre s

su

re, P

R

4

2 1

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

Figure 9 Viscosity ratio vs pseudo reduced temperature and pseudo reduced pressure.

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5.2 Viscosity of Mixtures

Another formula that is used for mixtures is:

mix

Σµ j y j M j Σy j M j

(30)

j=1, n where:

y j = mole fraction of jth component M j = molecular weight of component m j = the viscosity of jth component

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).

6

EQUATIONS OF STATE

6.1 Other Equations-of-State

As indicated at the start of section 2 the compressibility factor evolved out of the need to use an equation derived from ideal gas behaviour and incorporating a correction factor to suit real gas behaviour. One of the difficulties of the compressibility equation: 22



PV = znRT

Reservoir Engineering

Behaviour of Gases F I V E

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.

6.2 Van de Waals Equation 1873

The well known van der Waal’s equation was one of the earliest equations to represent the behaviour of real gases. This most basic EOS, corrects for the volume of the molecules and attractive and collision forces using empirical constraints a and b.

(P + a/V2) (V-b) = RT

(31)

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.

The equation can also be written as:

V3 - (+ b) V2 + (a/P)V - ab/P = 0 Such equations are therefore called cubic equations of state. The equation written to solve for z, the compressibility factor , becomes:

Z3 - Z2 (1 + B) + Z A - AB = 0



A=







(32)

where

aP bP and B = 2 ( RT ) RT

(33)

Values of a and b are positive constants for a particular fluid and when they are zero the ideal gas equation is recovered. One can calculate P as a function of V for various values of T. Figure 10 is a Figure of 3 isotherms. Also drawn is the curve for saturated liquid and saturated vapour. Isotherm T1 is the single phase isotherm, Tc is the critical isotherm and T2 gives the isotherm below the critical temperature.

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c

T1>Tc

P

Tc

Psat

T2
Vsat (liq)

V

Vsat (vap)

Figure 10 PV behaviours of pure components predicted by EOS.

At the critical point, for a pure substance , the equation of state should be such that:



 ∂ 2P  ∂P  =0 =    ∂V  T = Tc  ∂V 2  T = T c

That is the critical isotherm exhibits a horizontal inflection point at the critical point. The application of these conditions to the van de Waals equation yields:

a=

RT 27 R2 Tc2 and b = 64 Pc 8 Pc

EXERCISE 17. Calculate the critical constants for n- heptane.

24

(34)

Behaviour of Gases F I V E

For the curve, T2
6.3 Benedict-Webb - Rubin Equation (BWR) 1940

This equation developed for pure light hydrocarbons found considerable application in predicting thermodynamic properties of natural gases, since natural gases are essentially mixtures of light hydrocarbons and it can be written in a form similar to Van der Waals equation.

PT Bo RT − Ao − Co / T 2 bRT − a + + + V V2 V3 −γ aα C γ + 3 o 2 1 + 2  exp 2  6 V  V V T  V 

P=

(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.

6.4 Redlich-Kwong Equation 1949

Numerous equations were developed with increasing numbers of constants specific to pure components. More recently there has been a move back to the cubic equations like van der Waals. We will describe briefly those which have found favour in the oil and gas sector. This modern development of cubic equations of state started in 1949 with the Redlich and Kwong equation which involves only two empirical constants.

P =

RT a( T ) − V - b V (V + b )

(36)

where a and b are functions of temperature. The term a(T) depends on the temperature and Redlich Kwong expressed this as a function of the reduced temperature Tr using

a(T ) =

ac TR

By applying the limiting condition at the critical points yields values of ac and b related to critical constants. Such that ; 22/07/14

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Petroleum Engineering



ac = 0.42748

Reservoir Engineering

R2 Tc2 RT and b = 0.08664 c Pc Pc

(37)

6.5 Soave, Redlich Kwong equation

Soave, in 1972, modified the Redlick-Kwong (RK) equation and replaced the a/T0.5 term with a temperature dependent term aT where aT = aca. The Soave, Redlich-Kwong (SRK) equation is therefore:

P=

where

RT acα − (V − b) [V (V + b)]

(38)

a is a non dimensionless temperature dependent term which has a value of 1.0 at the critical temperature.

a is obtained from

[

(

α = 1 + m 1 − Tr

)]

2

where m = 0.480 + 1.574ω - 0.176ω 2 8 where ω is the Pitzer accentric factor .

6.6 Peng Robinson Equation of State 1975

Peng and Robinson modified previous equations in relation to the attractive term. They introduced it to improve the predictions of the Soave modification in particular for the calculation of liquid densities.

RT acα − V − b [V (V + b) + b(V − b)]



P=



ac = 0.457235

and

R2 Tc2 RT and b = 0.0778 c Pc Pc

(39)

(40)

a is the same function as for the Soave equation except the a function is different;

where m = 0.37464 + 1.54226w - 0.26992w2

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

26

Behaviour of Gases F I V E

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.

6.7 Application to Mixtures

When properties of mixtures are required mixing rules are required to combine the data from pure components. For both the SRK and PR equation

b=

j

y j b j and a =

i

j

(

)

yi yj ai aj 1 kij (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. EXERCISE 18. A PVT cell contains 0.01 cu ft ( 300cc) 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

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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 n where n m M

= nRT = m/M = number of moles = mass = molecular weight

m

= PMV/RT

m=

lb  (3cuft ) lbmole  10.73 psia.cuft  530 o R ( )  lbmole.o R 

(1000 psia)16.04





Mass of methane, m = 8.46 lb

EXERCISE 2. Calculate the density of the gas in the cylinder in exercise 1. SOLUTION

ρg =

ρg =



28

MW.P RT

(1000 psia)16.04

lb  lbmole 

10.73 psia.cuft  530 0 R ( )  lbmole.oR 

Density of gas, ρg = 2.82

lb cu. ft.

Behaviour of Gases F I V E

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



Vsc =

P1 Tsc V Psc T1

Vsc =

1000 psia 520 o R x3ft 3 o 14.7 psia 530 R

Vsc = 200.23 scf

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. SOLUTION Gas Components 1 Methane 2 Ethane 3 Propane Totals

A Weight 25 3 1.5 29.05

B Mol weight 16.04 30.07 44.09

C lb moles

D Mole fraction

1.559 0.100 0.034

0.921 0.059 0.020 1

EXERCISE 5. What is the apparent molecular weight of the gas in exercise 4

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SOLUTION Gas Components 1 2 3

A Mol weight mw 16.04 30.07 44.09

Methane Ethane Propane

B Mol fraction yi

0.921 0.059 0.020 1.000

C A*B

14.77 1.77 0.89 17.43

Apparent Molecular weight= 17.43

EXERCISE 6. What is the gas gravity of the gas in exercise 4 ? SOLUTION



γg =

Mg M = g Mair 29

Mg = AMW = 17.43 Gas gravity = 0.6

EXERCISE 7. Calculate the pseudo critical temperature and pseudocritical pressure of the mixture in exercise 4 . SOLUTION Gas Components 1 2 3



Methane Ethane Propane Total

A B Mol weight Mole fraction mw

16.04 30.07 44.09

yi

0.921 0.059 0.020 1.0

C Pc-psi 667.00 708.00 616.00

D. Tc °R 344 550 666

Ppc 614.3 41.7 12.4 668.4

Tpc 316.81 32.42 13.39 362.6

Pseudocritical pressure = 668.4 psia Pseudocritical temperature = 362 oR

EXERCISE 8. For the gas of exercise 4 determine the compressibility factor at a temperature of 150 o F and a pressure of 3500psia.

30

Behaviour of Gases F I V E

SOLUTION Ppr = P/Ppc, Tpr = T/Tpc

From exercise 7 Ppc = 668 psia, Tpc = 362.6°R



P = 3500 psia, and T = 150°F ie. 610°R



Ppr = 5.24, and Tpr = 1.68



From standing Katz chart, Figure 2



Compressibility factor, z = 0.88

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 Gas Components 1 2 3 4 5 6



Methane Ethane Propane Hydrogen sulphide Carbon Dioxide Nitrigen Total

Wgt Mol fraction weight

lb moles

Mole fraction

Pc-psi

Tc °R

Ppc psia

Tpc 255.70 26.17 10.81 28.25

25 3 1.5 3

0.56 0.07 0.03 0.07

16.04 30.07 44.09 34.08

0.035 0.002 0.001 0.002

0.743 0.048 0.016 0.042

667.00 708.00 616.00 1306

344 550 666 673

495.8 33.7 10.0 54.8

10

0.22

44.01

0.005

0.108

1071

548

116.1 59.38

2.5 45

0.06 1.00

28.02

0.002 0.0466

0.043 1.000

493

227

21.0 731

9.66 390

From Wichert & Azis chart for compositions of H2S and CO2 e = 19

Tpc′ = Tpc - ε = 371o R p′pc =



Weight

ppcTpc′ Tpc + yH 2 S 1 − yH 2 S ε

(

)

Ppc′ = 694.3

EXERCISE 10. Express the quantity of 1 lb mole of a gas as standard cubic feet. SOLUTION 22/07/14

Equation of state PV = RT for 1 mole R = 10.732 psia. cu.ft/lb.mole °R T = 60+460 = 520 °R, P = 14.65 psia or V for 1 lb.mole = RT/P = 380.9 scf/lb.mole.

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EXERCISE 11. Express the mass of gas in exercise 4 as standard cubic feet. 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 T = 150 oF ie 610 oR and P = 3500 psia Compressibility factor at these conditions from exercise 8 = 0.88 Bg using equation above = 0.0008 res bbl/scf 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

Volume of reservoir pore space = 5x2 x (5280)2 x 200 x 0.2 = 11,151,360,000 cu. ft. =1,985,994,657 bbls Bg , exercise 11 =0.00077299 res. bbls/SCF Volume of gas =2.56923E+12 scf

EXERCISE 14. Calculate the viscosity of the gas mixture in exercise 4 at 200°F and a pressure of one atmosphere.

32

Reservoir Engineering

Behaviour of Gases F I V E

SOLUTION Gas Components

Mol Weight Mole fraction yj 16.04 30.07 44.09

Methane Ethane Propane



µ mix =

0.921 0.059 0.020 1.000

Viscosity from fig 7 µj 0.013 0.0112 0.0098

√Mj

yj√Mj

µjyj√Mj

4.0050 5.4836 6.6400 SUM

3.6884 0.3233 0.1335 4.1451

0.0470 0.0036 0.0013 0.529

Σµ j y j M j Σy j M j

mmix = 0.0529/4.1451 mmix =0.01275 cp EXERCISE 15. Use the gas gravity method to calculate the viscosity of the gas in exercise 4 SOLUTION Gas Components Methane Ethane Propane



Mol Weight Mole fraction mw yj 16.040 30.070 44.090 0.000

0.921 0.059 0.020 1.000

14.7720 1.773 0.886 17.431

gg=AMW/Mair gg=AMW/29 Temperature = 150°F Mol weight air = 29.000 AMW of gas = 17.431 Gas Gravity = 0.601 mg = 0.01265 from Figure 8

EXERCISE 16. Determine the viscosity of the gas in exercise 4 at 150oF and 3500 psia (ref ex 4, 7, &8)

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Petroleum Engineering

SOLUTION From exercise 7

Ppc = 668.4

Tpc = 362.6



3500 = 5.24 Pr = P P = pc 668.4 610 Tr = T T = = 1.68 pc 362.6

From Lee correlation m / matmos = 1.75 Viscosity at atmospheric pressure From exercise 13 and 14 = 0.01275 cp Viscosity at conditions = 0.0223 cp EXERCISE 17. Calculate the critical constants for n- heptane. SOLUTION

R = 10.732. Tc for heptane = 973 oR and Pc = 397 psia Using equations above a = 115,872 cu ft 2/lb mole and b = 3.2878 cu ft./lb mole

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

34

Reservoir Engineering

Behaviour of Gases F I V E

SOLUTION Calculate the constants a and b for each component

ac = 0.42748

R2 Tc2 RT and b = 0.08664 c Pc Pc

where m = 0.480 + 1.574

- 0.176

2

[

(

Components

Y

Tc°R)

Methane

0.67

344

667

0.4759

8735

0.0104 0.49635 0.57546

ethane

0.235

550

708

0.7223

21036

0.0979 0.63241 0.79033 16625

(

)

= 1 + m 1 − Tr



0.05

n-butane

)]

2

Pc

766

551

bj

1.2926

ac

ω

52429

m

a

a=a*α 5027

0.1995 0.78701 1.00619 52753

Now calculate the mixture values.

b = ∑ y j b j and a = ∑ ∑ yi y j ai a j 1 − kij j



i

j

where a ij = (1-k ij )(a i a j )0.5 Components

Yi

b

Methane

0.67

0.312

ethane

0.235

0.181

n-butane

0.05 1

kij Methane 0.00

0.129

kij ethane

kij n-butane

aij Methane

aij ethane

aij n-butane

sum

0.00

0.02

2123.7

1485.5 1037.29 4646.52

0.00

0.01

1485.5

1039.1 732.969 3257.56

0.622

0.00

1037.3

732.97 527.535 2297.8 sum

10201.9

Now use SRK to calculate pressure.

P=

RT acα − (V − b) [V (V + b)]

Vm = 1.25 cu ft / lb mole b = 0.622

22/07/14

a cα = 10201.9

P = 8617.6 psia

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Petroleum Engineering

REFERENCES 1. Standing MB and Katz DL Density of Natural Gases. Trans AIME, 146(1942). p140 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 3. Brown GG et al. Natural Gasoline and Volatile Hydrocarbons” National Gasoline Assoc. of America, Tulsa, Okl. 1948 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 8. Pitzer K S et al The Volumetric and Thermodynamic Properties of Fluids II. Compressibility Factor, Vapour Pressure and Entropy of Vaporisation. J .Am. Chem. Soc. (1955) 77, No 13,3433-3440 9. Danesh, A PVT and Phase Behaviour of Petroleum Reservoir Fluids. 1998 Elsevier ISBN:0 444 82196 1 p129-162

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Reservoir Engineering

Properties of Reservoir Liquids S I X

30

20

0

10

ne p lus m a

70

30

t% W

ne (a) ha et m

in

e tir en

st sy

60 10

em

PB 20

1.310 BBL

1.333 BBL

P01 = 3500 PSIA T01 = 160º F

PB = 2500 PSIA T01 = 160º F

A

B

30

Solution Gas, SCF/STB

40

30

20

600

(b)

50

P01

500

10

PA

Free Gas 2.990 Cu. Ft.

1.040 BBL

1.000 BBL

PA = 14.7 PSIA T01 = 160º F

PA = 14.7 PSIA T01 = 60º F

C

D

E

567SCF/STB

200 100

0

500

1000

Free Gas 567 Cu. Ft.

1.210 BBL

300

0

Free Gas 676 Cu. Ft.

P = 1200 PSIA T01 = 160º F

AT 1200 PSIA RS = 337

400

PA

P

1500

2000

2500

INITIAL PRESSURE

0

BUBBLE POINT PRESSURE

%e than e in

etha

50

Density of system including methane and ethane, lb/cu ft

60

40

40

terial

50

Wt

Density of propane plus, lb/cu ft

70

3000

3500

Pressure, PSIA

Reservoir Engineering 22/07/14

Petroleum Engineering

Properties of Reservoir Liquids S I X

C O N T E N T S 1 COMPOSITION BLACK OIL MODELS 2 GAS SOLUBILITY, RS 3 OIL FORMATION VOLUME FACTOR, BO 4 TOTAL FORMATION VOLUME FACTOR, BT 5 BELOW THE BUBBLE POINT 6 OIL COMPRESSIBILITY 7 BLACK OIL CORRELATIONS 8 FLUID DENSITY 8.1 Specific Gravity of a Liquid 8.2 Density Based on Ideal Solution Principles 9 FORMATION VOLUME FACTOR OF GAS CONDENSATE, BGC 10 VISCOSITY OF OIL 11 INTERFACIAL TENSION 12 COMPARISON OF RESERVOIR FLUID MODELS

Reservoir Engineering 22/07/14

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Petroleum Engineering

LEARNING OBJECTIVES Having worked through this chapter the Student will be able to: • Define gas solubility, Rs and plot vs. P for a reservoir fluid. • Define undersaturated and saturated oil. • Explain briefly flash and differential liberation • 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. • Present an equation to express Bt in terms of Bo, Rs and Bg. • Express oil compressibility in terms of oil formation volume factor. • 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. • Define the formation volume factor of a gas condensate • 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

2

Reservoir Engineering

Properties of Reservoir Liquids S I X

1 COMPOSITION - BLACK OIL MODEL As introduced in the chapter on Composition, petroleum engineers require a compositional description tool to use as a basis for predicting reservoir and well fluid behaviour. The two approaches that are commonly used are the multicomponent compositional model described in the earlier chapter and the two component black oil model. The latter simplistic approach has been used for many years to describe the composition and behaviour of reservoir fluids. It is called the “Black Oil Model”. 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

Stock Tank Oil

Bo Bo = Oil Formation Volume Factor Rs = Solution Gas to Oil Ratio

Figure 1 "Black Oil Model"

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2 GAS SOLUBILITY 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. Solubility = f (pressure, temperature, composition of gas, composition of crude oil) 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 releases some gas. 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 components 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. At initial reservoir conditions, the subscript i is added and becomes Rsi. 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.

4

Reservoir Engineering

Properties of Reservoir Liquids S I X

Solution Gas

Surface

Rsi scf/stb

Pi 1

Pressure

2

P

+

Stock Oil Tank Temperature Phase Diagram

1 st b. oil Oil Reservoir

Bo rb.oil

Oil and Dissolved Gas

Figure 2 Production of reservoir hydrocarbons above bubble point

Figure 3 gives a typical shape of gas solubility as a function of pressure for a reservoir 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 7). 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:

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Petroleum Engineering

Reservoir Engineering

600

Rs scf/stb

Rsi 400

200 Pb 1000

2000

3000

Pressure (psig)

Figure 3 Solution Gas - Oil Ratio as a Function of Pressure.

Flash liberation

-

the gas is evolved during a definite reduction in pressure and the gas is kept in contact with the liquid until equilibrium has been established.

Differential liberation

-

the gas being evolved is being continuously removed from contact with the liquid and the liquid is in equilibrium with the gas being evolved over a finite pressure range.

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 nor 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.

3 OIL FORMATION VOLUME FACTOR, B O 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 6

Properties of Reservoir Liquids S I X

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 pressure 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 compressibility 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. Units - rb (oil and dissolved gas)

Bo rb./stb

1.2

1.1

1.0

Pb 1000

2000

3000

Pressure (psig)

Figure 4 Oil formation volume factor

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.

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Reservoir Engineering

PA

PA

P PB

P01

(a)

Free Gas 430 Cu. Ft.

Free Gas 676 Cu. Ft.

1.310 BBL

1.333 BBL

1.210 BBL

1.040 BBL

1.000 BBL

P01 = 3500 PSIA T01 = 160º F

PB = 2500 PSIA T01 = 160º F

P = 1200 PSIA T01 = 160º F

PA = 14.7 PSIA T01 = 160º F

PA = 14.7 PSIA T01 = 60º F

A

B

C

D

E

567SCF/STB

500 AT 1200 PSIA RS = 337

400 300 200 100 0

0

500

1000

1500

2000

2500

INITIAL PRESSURE

BUBBLE POINT PRESSURE

Solution Gas, SCF/STB

600

(b)

Free Gas 567 Cu. Ft.

3000

3500

Pressure, PSIA

Figure 5 Gas to oil ratio and oil formation volume factor for Big Sandy Field reservoir oil 1.

2500 PSIA BOB = 1.333 1200 PSIA BO = 1.210

1.20

14.7 PSIA & 160º F BO = 1.040

1.10

1.00

14.7 PSIA & 60º F BO = 1.000

0

500

1000

1500 2000 Pressure, PSIA

Figure 5b

8

2500

INITIAL PRESSURE

1.30

3500 PSIA BOI = 1.310

BUBBLE POINT PRESSURE

(b)

Formation Volume Factor, BBL/STB

1.40

3000

3500

Properties of Reservoir Liquids S I X

The reciprocal of the oil formation volume factor is called the ‘shrinkage factor bo

bo =

1 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 on 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

Figure 6 Black oil description of reservoir fluid

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Reservoir Engineering

Multi Reservoir System Bo Rs

Rs 1 Bo 1

Rs 3 Bo 3

?

Rs 2 Bo 2

Rs 4 Bo 4

Figure 7 Integrated system of reservoir common pipeline and final collection system.

4 TOTAL FORMATION VOLUME FACTOR, BT In reservoir engineering it is sometimes convenient to know the volume occupied in the reservoir by one stock tank barrel of oil plus the free gas that was originally dissolved in it. A factor is used called the total formation-volume factor Bt, or the two-phase volume-factor and is defined as the volume in barrels that 1.0 STB and its initial complement of dissolved gas occupies at reservoir temperature and pressure, i.e. it includes the volume of the gas which has evolved from the liquid and is represented by: Bg (Rsb - Rs) i.e.

Bt = Bo + Bg (Rsb - Rs)

Rsb = the solution gas to oil ratio at the bubble point

10

(1)

Properties of Reservoir Liquids S I X

B0b

Oil

Gas

Bg(Rsb-Rs)

Bt Oil

Hg

B0

Hg

Figure 8a Total formation volume factor or two phase volume factor

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 pressure. 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 formation-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

Figure 8b Total and oil formation volume factor

The value of Bt can be estimated by combining estimates of Bo and calculation of Bg and known solubility values for the pressures concerned.

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Reservoir Engineering

5 BELOW THE BUBBLE POINT 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, R, is made up of the two components solution gas Rs and the free gas which is the difference (R-Rs). 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 & Solution Gas

Surface

R= Rs + (R - Rs)

Pi

Pressure

+ Stock Oil Tank

P

1 st b. oil

Temperature

Oil Reservoir

Gas

Oil

Reservoir Bo rb (oil and dissolved gas) /stb (R - Rs) Bg

rb (free gas) /stb

Figure 9 Production of reservoir hydrocarbons below bubble point

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. The equation for oil compressibility is

co = −

1  ∂V    V  ∂P  T

co = −

1  ∂Bo    Bo  ∂P  T

in terms of formation volume factors this equation yields

12

Properties of Reservoir Liquids S I X

Assuming that the compressibility does not change with pressure the above equation can be integrated to yield ;

co ( P2 − P1 ) = − ln

V2 V1

where P1 & P2, and V1 & V2 represent the pressure and volume at conditions 1 & 2.

7 BLACK OIL CORRELATIONS Over the years there have been many correlations generated based on the two component 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 locations 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’s3 correlation. Other correlations include, Lasater 4, and recently Glaso 6

Pb = f (Rs, gg, ro, T)

where Pb = bubble point pressure at ToF Rs = solution gas-oil ratio (cu ft/ bbl) gg= gravity of dissolved gas ro = density of stock-tank oil (specific gravity). Standing’s correlation for the calculation of Pb, bubble point pressure is:

Pb = 18. 2

Rs g

0.83

10

( 0. 00091 T − 0 .0125( API ))

-1.4

(2)

His correlation for the oil formation volume factor is; 1.2



22/07/14

  γ g  0.5  Bo = 0.9759 + 0.000120  Rs   + 1.25T    γ o  

Institute of Petroleum Engineering, Heriot-Watt University

(3)

13

Petroleum Engineering

Standing's correlations have been presented as nomographs enabling quick look predictions to be made. Figures 10 & 11 give the nomogram forms of these correlations for gas solubility and oil formation volume factor. Standing’s correlation is based on a set of 22 California crudes. 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 using 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

Reservoir Engineering

Properties of Reservoir Liquids S I X

Figure 10 Oil-formation volume factor as a function of gas solubility, temperature, gas gravity and oil gravity (Standing 10).

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Petroleum Engineering

Reservoir Engineering

Figure 11 Gas solubility as a function of pressure. Temperature, gas gravity and oil gravity (Standing 10).

16

Properties of Reservoir Liquids S I X

Correlation Ref Bubble - point pressure (psia) Temperature, °F Bo Gas - oil ratio (scf/stb) Oil Gravity, oAPI Gas Gravity Separator Pressure Searator Temperature °F

Standing 3 130-7000 100-258 1.024-2.15 20-1425 16.5-63.8 0.59-0.95 265-465 100

Lasater 4 45-5780 82-272

Vasquez-Beggs 5 15-6055 162-180 1.028-2.226 3-2905 0-2199 17.9-51.1 15.3-59.5 0.574-1.22 0.511-1.651 15-605 60-565 36-106 76-150

Glaso 6 165-7142 80-280 1.025-2.588 90-2637 22.3-48.1 0.65-1.276 415 125

Marhoun 7 130-3573 74-240 1.032-1.997 26-1602 19.4-44.6 0.752-1.367

Table 1 Black oil correlation and their ranges at application 2

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.

8.1 Specific Gravity of a Liquid

γo =

ρo 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

°API =

141.5 −131.5 γo

(5)

where go is specific gravity at 60˚/60˚. 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. 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.

8.2 Density based on Ideal Solution Principles

Mixtures of liquid hydrocarbons at atmospheric conditions behave as ideal solutions. An ideal solution is a hypothetical liquid where no change in the character of the liquids is caused by mixing and the properties of the mixture are strictly additive.

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Petroleum Engineering

Reservoir Engineering

Petroleum liquid mixtures are such that ideal-solution principles can be applied for the calculation of densities and this enables the volume of a mixture from the composition and the density of the individual components. The principle is illustrated using the following exercise. Data for the specific components are given in the tables at the end of the chapter EXERCISE 1 Calculate the density at 14.7psia and 60 °F of the hydrocarbon liquid mixture with the composition given below: Component

nC4 nC5 nC6

Mol. fract. lb mol. 0.25 0.32 0.43 1.00

Liquids at their bubble point or saturation pressure contain large quantities of dissolved 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 Correlations have been obtained by experiment giving apparent liquid densities of methane and ethane versus the pseudoliquid density (Figure 12).

18

Apparrent density of Methane, g/cc

Apparrent density of of Ethane, g/cc

Properties of Reservoir Liquids S I X

0.6

0.5

0.4

0.3

0.4

0.3

0.2

Ethane - N - Butane Ethane - Heptane Ethane - Crystal oil Methane - Cyclo Hexane Methane - Benzene Methane - Pentane Methane - Hexane Methane - Heptane Methane - Propane Methane - Crystal oil Methane - Crude oil

0.1

0.3

0.4

0.5

0.6

0.7

0.8

0.9

Density of system, 60ºF B atm. pressure

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 volumes 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. This trial and error method is very tedious so Standing and Katz devised a chart which removes the trial and error required in the calculation. The densities have been converted 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.

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Petroleum Engineering

30

20

0

10

ne p lus m a

70

%e than e in

etha

50

0

30

t% W

ne ha et m

in

e tir en

st sy

em

60 10 50 20 40

30

30

20

10

Density of system including methane and ethane, lb/cu ft

60

40

40

terial

50

Wt

Density of propane plus, lb/cu ft

70

Reservoir Engineering

Figure 13 Pseudo-liquid density of systems containing methane and ethane 10.

We shall examine through examples various ways of calculating downhole reservoir fluids densities dependant on the data available. The three considered are: 1. The composition of the reservoir fluid is known. 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. 1. The composition of the reservoir fluid is known. The procedure is illustrated using the following two exercises.

20

Properties of Reservoir Liquids S I X

EXERCISE 2 Calculate the "surface pseudo liquid density" of the following reservoir composition using the chart of Figure 13. Component Methane Ethane Propane Butane Pentane Hexane Heptane +

Mole percent 44.04 4.32 4.05 2.84 1.74 2.9 40.11

Properties of heptane + API gravities = 34.2 SG = 0.854 Mol wt = 164

The pseudodensity is converted to reservoir conditions firstly by taking the effect of pressure and secondly accounting for the effect of temperature. The variation of density with respect to pressure and temperature has been investigated and it has been demonstrated that thermal expansion is not affected by pressure. Standing & Katz took National Petroleum Standards data and with supplementary data produced correction factors for pressure and temperature to convert atmospheric density to reservoir density. The compressibility and thermal expansion effects have been expressed graphically in Figures 14 and 15.

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Reservoir Engineering

10

00

8

,0 15

7 6

4

1 0 25

ps ia

00

2

Pr es su re ,

0 3,

3

00 00 00 6,0 5,0 4,0

5

0 ,00 8,000 10

Density at pressure minus density at 60ºF & 14.7 psia lb/cu ft

9

2, 00 0 1,0

00

30

35

40

45

50

55

60

Density at 60ºF and 14.7 psia, lb/cu ft

Figure 14 Density correction for compressibility of liquids 10.

22

65

Properties of Reservoir Liquids S I X

10 9

Density at 60ºF minus density at temperature, lb/cu ft

8 7 6 24 0

0 22

5

20

0

4 3

16

18 0

Te mp era

tur e

ºF

0

14 0

120

2

100

1 0 25

80

60 30

35

40

45

50

55

60

65

Density at 60ºF and pressure P, lb/cu ft

Figure 15 Density correction for thermal expansion of liquids 10.

EXERCISE 3 Calculate the density of the reservoir liquid of exercise 2 at reservoir conditions of 5,500 psia and 180 oF

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 situation 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. 2. Reservoir Density when the Gas Solubility , the gas composition and the surface oil gravity are known 22/07/14

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Reservoir Engineering

By considering surface liquid as a single component and knowing the composition of the collected gas the techniques previously discussed can be used to determine reservoir liquid density. Again we will illustrate the procedure with an example EXERCISE 4 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 lb/cu.ft and its pressure gradient psi/ft. The produced gas has the following composition. The pore volume of the reservoirs is estimated at 1.2x109 cu.ft with an average water saturation of 20% estimate the stock tank barrels of oil in place.

Component Methane 0.71 Ethane 0.13 Propane 0.08 Butane 0.05 Pentane 0.02 Hexane 0.01

Mole Fraction

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 apparent liquid densities of gases against gravity for different API crudes Behaviour of Oil Field Hydrocarbon Systems

45

Apparent Density of Dissolved Gas at 60°F and 14.7 psia - lb per cu ft

40

35

20°

ude

I cr

AP

30°

30

40° 50° 60°

25

20

15

0.6

0.7

0.8

0.9

1.0

Gas Gravity

1.1

Air = 1

1.2

1.3

Figure 16 Apparent liquid densities of natural gases (Katz 194222)

24

1.4

Properties of Reservoir Liquids S I X

EXERCISE 5 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

9 FORMATION VOLUME FACTOR OF GAS CONDENSATE 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 conditions. 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. The formation-volume factor therefore for a condensate, Bgc is defined as the volume of gas in the reservoir required to produce 1.0 STB of condensate at the surface. The units are generally barrels of gas at res. conditions per barrel of stock tank oil. There are a number of methods of estimating Bgc. EXERCISE 6 A gas condensate produces gas and liquids with the compositions (mol fractions) detailed below, with a producing GOR of 30,000 SCF/STB. Determine the composition of the reservoir gas. Component Methane Ethane Propane Butane Pentane Hexane Heptane +

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Gas 0.84 0.08 0.04 0.03 0.01

Composition

1.00

Institute of Petroleum Engineering, Heriot-Watt University

Liquid 0.15 0.36 0.28 0.12 0.09 1.00

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Petroleum Engineering

EXERCISE 7 The gas condensate reservoir in exercise 6 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 areal 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 in terms of SCF gas and STB condensate. If the condensate reservoir has a dew point of 3,000 psia determine the amount of STB of condensate and SCF gas which has been produced when the reservoir pressure has dropped to 4,000 psia. EXERCISE 8 Calculate the gas condensate formation factor for the example in exercise 7.

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. Correlations 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 18 shows the impact of dissolved gas from the Beggs and Robinson 11 correlation.

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Reservoir Engineering

Properties of Reservoir Liquids S I X

1000 800 700 600 500 400 300 200

Viscosity of Gas-Free Oil, µoD, cp

100 80 70 60 50 40 30 20

100º 150º 200º 250º 300º

R es er

10

8 7 6 5 4 3

vo irT em

pe ra

tur e,

ºF

2

1 0.8

0.7 0.6 0.5 0.4 0.3 0.2

0.1 10

20

30

Stock - Tank Oil Gravity, ºAPI

40

50

Figure 17 Dead oil viscosities 17.

0

200

0

10

0

20 at io

20

0

50

G as -O

il R

10 8 7 6 5 4 3

tio

n

00 10 0 150 0 0 20

So lu

Viscosity of Gas-Saturated Oil, µo, cp

100 80 70 60 50 40 30

2

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

Viscosity of Gas-Free Oil, µoD, cp

Figure 18 Viscosities of saturated black oils 11.

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Petroleum Engineering

Reservoir Engineering

Beggs and Robinson 11 examined 600 oil samples over a wide range of pressure and temperature and came up with the following correlation. mod = 10A - 1





(6)

where, log A = 3.0324 - 0.0202oAPI -1.163 log T mod is the dead oil viscosity in cp and T is in oF.

Egbogah and Ng 12, had a different expression for A log A = 1.8653 - 0.025086oAPI -0.56441 log T 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.

mob = CmodB



where C = 10.715 (Rs + 100)-0.515 and B = 5.44 (Rs + 150)-0.338 mob is the saturated oil viscosity



(7)

Vazquez and Beggs13 presented an equation to take into account pressure on viscosity above the saturation pressure.



mo = mob (P/Pb)D where

(8) D

= 2.6 P1.187 e-11.513-8.98 x10-5P

This is presented in Figure 19 from McCain 17.

28

Properties of Reservoir Liquids S I X

10,000 9,000 8,000 7,000 6,000 4,000 3,000

Pre

2.000

1.000 900 800 700 600 500

60

00

psi

a

200

100 60 40

100 90 80 70 60 50 40

20

30 20

10 6

4

y of Viscosit

oint, cp

ubble P

Oil At B

2 1 0.6

0.4 0.2

0.1

10 9 8 7 6 5 4 3 2

Viscosity of Oil Above Bubble Point, µo, cp

200

00

00

re

00

300

30

40

ssu

00

20

400

50

0 100 500

Bubble Point pressure, Pb, psia

5,000

1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1

Figure 19 Viscosities of undersaturated black oils 17.

Labedi (ref 14) also produced an empirical correlation to determine viscosity at pressures above the bubble point

mo= mob + (P/Pb-1)(10-2.488mob0.9036 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.

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Petroleum Engineering

Reservoir Engineering

11 INTERFACIAL TENSION In recent years interfacial tension has become to be realised as an important physical property in the context of the recovery of reservoir held hydrocarbons, in particular for gas condensates. Interfacial tension arises from the imbalance of molecular forces at the interface between two phases. For many years it has been neglected but more recently it has been realised that in gas injection and condensation processes the magnitude of the various forces; surface, gravitational and viscous forces can have a significant impact on the mobility of the various phases. A major advance in knowledge has been that in the context of gas condensates where it was considered that in the tradition of relative permeability knowledge, liquid formation by retrograde condensation would be immobile. Recent research has shown that such fluids are mobile because of the associated low interfacial tension 16. Danesh 2 in his text covers the topic of interfacial tension extensively. Mentioned briefly below are some of the techniques which are currently used in predicting IFT for reservoir fluids. 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

Surface Tension, dynes per cm

30

25

20

Mol wt. 240 220 200 180 160

15

140

10

5

0 -200

-200

0

100

200

300

400

500

600

Temperature, ºF Methane

Ethane

Propane

n - Pentane

l - Butane n - Butane

n - Heptane

n - Hexane

n - Octane

Figure 20 Interfacial tensions of hydrocarbons. (Adapted from Katz, et al., J. Pet. Tech., Sept. 1943)

30

Properties of Reservoir Liquids S I X

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.

 ρ − ρg  σ =  Pσ L   M 

4

(10)



where rL and rg are the density of the liquid and gas phases, and M is the molecular weight. s is the IFT . Ps 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. Parachors, Ps, for IFT Component Methane Ethane Propane i-Butane n-Butane i-Pentane n-Pentane n-Hexane n-Heptane n-Octane Hydrogen Nitrogen Carbon dioxide

Parachor 77 108 150.3 181.5 189.9 225 231.5 271 312.5 351.5 34 41 78

Parachors have been shown to have a linear relationship with molecular weight according to the relationship; Ps = 21.99 + 2.892M

(11)

and also to the critical properties

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Petroleum Engineering

Reservoir Engineering

600

500

Parachor, P

400

300 i - C5

200

i - C4

100

0

50

100

150

200

Molecular Weight

Figure 21 Parachors for computing interfacial tension of normal paraffin hydrocarbons 19.

Ps = 0.324Tc1/4vc7/8 where Tc is in K and the critical volume vc is in cm3/gmol. To apply the parachor approach to mixtures the molar averaging approach of Weinaug and Katz20 can be used.

  ρ ρ  σ = ∑ Pσ  x j L − yj g   Mg    ML  j

4

(12)

xj and yj are the mole fractions of the components in the liquid and gas phases. Firoozabadi 21 has provided parachors to enable calculations to be made for heavy components using the following equation. Ps= -11.4 + 3.23M-0.0022M2 where M is the molecular weight of the heavy component. Figure 22.

32

(13)

Properties of Reservoir Liquids S I X

1400 1200

Parachor. P

1000 800 600 400 200 0

100

200 300 Molecular weight

400

500

Figure 22 Parachors of heavy fractions for computing interfacial tension of reservoir liquids. McCain17

12 COMPARISON OF RESERVOIR FLUID MODELS 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. The computational requirements of compositional models used to be a restriction when carrying out large reservoir simulation. The continued development of computing and associated equations of state modelling reduces these former restrictions. Companies are now focusing their attention on being capable of modelling the total process from fluid extraction from the reservoir, through well production and facility treatment. At present separate modelling occurs, and many of these models are not compatible. The black oil approach is certainly considered by many to be from a former era. The list below gives a summary comparison of the two approaches. Black Oil Models • 2 components - solution gas and stock tank oil • Bo, Rs, etc. • Empirical correlations • After the event description of fluid properties

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Petroleum Engineering

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 enable 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 characteristics 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

Reservoir Engineering

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MOL. WEIGHT 16.04 30.07 44.09 58.12 58.12 72.15 72.15 86.17 100.20 114.20 128.30 142.30 28.97 28.02 32.00 44.01 34.08 18.02

COMPOUND

METHANE C1 ETHANE C2 PROPANE C3 ISO-BUTANE C4 N-BUTANE C4 ISO_PENTANE C5 N-PENTANE C5 N-HEXANE C6 N-HEPTANE C7 N-OCTANE C8 N-NONANE C9 N-DECANE C10

AIR NITROGEN OXYGEN CARBON DIOXIDE HYDROGEN SULPHIDE WATER

Institute of Petroleum Engineering, Heriot-Watt University

-317.7 -320.4 -297.4 -109.3 -76.5 212.0

BOILING POINT at 14.7 psia °F -258.7 -127.5 -43.7 10.9 31.1 82.1 96.9 155.7 209.2 258.2 303.4 345.2 547 492 732 1072 1306 3206

239 227 278 548 673 1165

CRITICAL CONSTANTS PRESSUE TEMP psia °R 668 344 709 550 616 666 530 734 551 766 490 830 488 846 437 914 397 973 361 1025 332 1071 305 1112

0.9991

0.5072 0.5625 0.5836 0.6241 0.6305 0.6637 0.6875 0.7062 0.7211 0.7333

62.37

31.66 35.12 36.43 38.96 39.36 41.43 42.92 44.09 45.02 44.78

LIQUID DENSITY at 60°F, 14.7 psia gm/cc lb/cu.ft

Physical Properties of the Paraffin Hydrocarbons and Miscellaneous Compounds

Properties of Reservoir Liquids S I X

Table 1 Physical properties of the paraffin hydrocarbon and miscellaneous compounds.

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Petroleum Engineering

Reservoir Engineering

SOLUTIONS TO EXERCISES Data from the physical properties Table 1. EXERCISE 1 Calculate the density at 14.7psia and 60 ºF of the hydrocarbon liquid mixture with the composition given below: Component

N-Butane C4 N-Pentane C5 N-Hexane C6

Mol. fraction lbmol

0.25 0.32 0.43 1.00

SOLUTION EXERCISE 1 Component

N-Butane C4 N-Pentane C5 N-Hexane C6

Mol Fraction

Molecular Weight

Weight

Density at 60F and 14.7 psi

Liquid Volume

lb mol

lb/lb mol

lb

lb/cu ft

cu ft

0.25 0.32 0.43

58.1 72.2 86.2

14.525 23.104 37.066

36.43 39.36 41.43

0.3987 0.5870 0.8947

sum 1 Density = 74.695 / 1.884 = 39.72 lb/cu ft

36

74.695

1.8804

Properties of Reservoir Liquids S I X

EXERCISE 2 Calculate the "surface pseudo liquid density" of the following reservoir composition using the chart of Figure 13. Component

Mole percent

Methane Ethane Propane Butane Pentane Hexane Heptane +

44.04 4.32 4.05 2.84 1.74 2.9 40.11

Properties of heptane + API gravities = 34.2 SG = 0.854 Mol wt = 164

SOLUTION EXERCISE 2 Component

Methane C1 Ethane C2 Propane C3 N-Butane C4 N-Pentane C5 N-Hexane C6 Heptane+ Sum For Heptane + Parameter Weight Specific Gravity Specific Gravity Liquid volume Liquid volume

Mole fraction

Mol. Weight Liquid Liquid Weight lb Density Volume lb/lb mol lb/cu ft at cu ft. 60F and 14.7psi

z 0.4404 0.0432 0.0405 0.0284 0.0174 0.0290 0.4011

M 16.04 30.07 44.09 58.12 72.15 86.17 164.00

1.0000

zM 7.06 1.30 1.79 1.65 1.26 2.50 65.78

ρo

zM/ρo

31.66 36.43 39.36 41.43 53.26

0.06 0.05 0.03 0.06 1.24

81.33

1.43

from this calculation Molecular weight * Mole fraction 141.5/(API+131.5) may be given weight/density SG * density water

Accounting for gasses Weight of propane + (lb) Liquid volume (cu ft) Density of propane + (lb/cu ft) Weight percent ethane in ethane + (%) Weight percent methane in methane + (%) Pseudo liquid density (Figure 13) (lb/cu ft)

72.97 1.43 51.06 1.75 8.69 46

141.5

* API equation - Degrees API = Sp.Gr.at 60ºF -131.5

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Petroleum Engineering

EXERCISE 3 Calculate the density of the reservoir liquid of exercise 3 at a reservoir conditions of 5,500 psia and 180 oF SOLUTION EXERCISE 3 Step 1 Pseudo liquid density at standard conditions from exercise 2. = 46 lb/cu ft Step 2 Correction for pressure at 5,500 psia From Figure 14 = 1.65 + 46 lb/cu.ft Density at pressure = 47.65 lb/cu.ft Step 3 Correction for temperature at 180°F From Figure 15 = -3.2 lb/cu.ft Density at 5,500 psia and 180° 47.65 - 3.2 = 44.45 lb/cu ft

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Reservoir Engineering

Properties of Reservoir Liquids S I X

EXERCISE 4 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 (lb/cu.ft) and its pressure gradient (psi/ft). The pore volume of the reservoir is estimated at 1.2x109 cu.ft with an average water saturation of 20 per cent. Estimate the stock tank barrels of oil in place. The produced gas has the following composition. Average values taken for normal and iso components.

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Component Methane C1 Ethane C2 Propane C3 Butane C4 (n&i) Pentane C5 (n&i) Hexane

Mole Fraction 0.71 0.13 0.08 0.05 0.02 0.01

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SOLUTION EXERCISE 4 Assumes that when a hydrocarbon gas disolves in a crude oil the combined volume of the system is equal to the volume of the oil plus that of the gas if it existed in the liquid state.

Step 1 Calculation of Pseudo Density For produced Gas: Component

Mole fraction

Molecular Weight

Volume of gas in 1 stb

lb moles of gas in 1 stb

weight of gas in 1 stb

liquid density*

Volume of gas in 1 stb

lb/lb mol

scf/stb

lbmole/stb

lb/stb

lb/cu ft

cu ft

M

zRs

zRs/379

MzRs/379

31.66 35.78 39.16 41.43

0.18 0.13 0.06 0.03

z Methane C1 Ethane C2 Propane C3 Butane C4 (n&i) Pentane C5 (n&i) Hexane

0.71 0.13 0.08 0.05 0.02 0.01

Sum

1.00

16.04 30.07 44.09 58.12 72.15 86.17

426 78 48 30 12 6

1.12 0.21 0.13 0.08 0.03 0.02

18.03 6.19 5.58 4.60 2.28 1.36

600

1.58

38.05

oil API SG SG = 141.5/(API+131.5) Density of water = 62.37lb/cu ft Density of oil = SG* density of water 42 0.816 Volume of 1stb oil = 5.615 cu ft Weight of 1stb oil = 5.615 cu ft = SG* density of water* 5.615 cu ft

water density lb/cu ft 62.37

0.40

*Average liquid densities (lb/cu ft) n-butane i-butane Average butane n-pentane i-pentane Average pentane

36.43 35.12 35.78 39.36 38.96 39.16

For produced oil

Calculate pseudo density at 14.7psia and 60F Weight of C3+in gas (lb) Weight of C3+ (cu ft)

13.83 0.40

Weight of oil and gas without C1 and C2

299.45

Volume of oil and gas without C1 and C2

6.01

Density of oil and gas without C1 and C2 (lb/cu ft) Weight % C2 = wt C2/(wt oil +C3+C2) Weight% C1 =wt C1/(wt oil +C3+ +C2+ C1) From Figure 13, pseudo liquid density at 14.7psia and 60F (lb/cu ft)

40

49.81 2.02 5.57 46.5

volume 1stb oil cu ft 5.615

weight of 1stb oil 285.62

Properties of Reservoir Liquids S I X

Step 2 Correction for reservoir temperature and pressure

Value at 14.7psia and 60F Figure 14 pressure Figure 15 temperature Density of reservoir fluid

correction at 4000psia Correction at 200F

Pseudo Density (lb/cu ft) 46.50

1.25

47.75

-3.50

44.25 44.25

Pressure gradient, psi/ft P = density*height*g In imperial units, 1 lb force = 1 lb mass*g, where g =1.00 Pressure/foot of depth (lb/sq ft)/ft = density = lb/(sq ft*1ft) 144sq in per 1 sq ft, pressure/foot of depth =psi/ft = density/144 psi/ft Pressure gradient = 0.31 psi/ft

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Estimate oil in place Vol of reservoir

1.20E+09 cu ft

Swc

20%

Vol oil in place

9.60E+08 cu ft

Mass of oil in place = vol* res fluid density

4.25E+10

lb

Mass of 1 stb oil =

285.62

lb

Mass of Rs gas

38.05

lb

Total

323.67

lb

Number STB =

1.31E+08

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Petroleum Engineering

Reservoir Engineering

EXERCISE 5 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 180°F. Katz method SOLUTION EXERCISE 5 Mass of gas per STB Gas gravity = 0.8 Mol. Weight of gas=mol weight air times gas gravity Mol. Weight air = 28.97 Mol. Weight of gas = mol weight air times gas gravity = 23.2 lb/lb mol GOR = 1 lb. mol = GOR= GOR=

500 379 1.319 30.58

* API=

35 Mass of 1stb = 298 Component

Wght/STB lb/STB

Gas Oil

30.58 298 328.38

scf/stb scf/lb mol lb.mol gas/stb lb.gas/stb lb Liquid Density from fig 16 lb/cu.ft 25

Pseudo density of reservoir fluid at surface Correction for pressure Figure 14 4000 psi Density at P & 60F Correction for temperature 180F Density at res P & T

Liquid volume cu.ft/STB 1.22 5.615 6.84 = 328.38/6.84 = 48.02 lb/cu ft = 1.12 49.14 -2.95 46.19 lb/cu.ft

141.5 * API equation - Degrees API = Sp.Gr.at 60ºF -131.5

42

Properties of Reservoir Liquids S I X

EXERCISE 6 Gas condensate 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 Methane C1 Ethane C2 Propane C3 Butane n&iso C4 Pentane n&iso C5 Hexane C6 Heptane + (use C8 Octane) Total

Gas 0.84 0.08 0.04 0.03 0.01

Composition Liquid

1.00

0.15 0.36 0.28 0.12 0.09 1.00

SOLUTION EXERCISE 6 Step 1. Determine lb moles gas produced per lb mols liquid Comp Mole frac. Mol.wgt Weight Liquid Density* Liquid volume Liquid lb/lb.mol lb/lb.mol lb/cu.ft cu.ft from table* of physical properties A B C=A*B D C/D normal iso Propane C3 0.15 44.09 6.614 31.66 0.209 density Butane n&isoC4 0.36 58.12 20.923 35.78* 0.585 36.43 35.12 Pentane n&iso C5 0.28 72.15 20.202 39.16* 0.516 39.36 38.96 Hexanes C6 0.12 86.17 10.340 41.43 0.250 Heptane+ ( use C8) 0.09 114.20 10.278 44.09 0.233 Total 1 68.3571 1.792 AMW= 68.357 * avg of n&iso Density of liquid= 38.14 lb/cu.ft 214.15 lb/STB 3.13 lb.mol/STB .=D/AMW GOR SCF/STB 30000 SCF/STB 1 lb.mole= 379 SCF GOR= 79.16 lb.mol gas/STB GOR= 25.27 lb.mol gas/lb.mol liquid Step 2. Recombination of Gas and liquid with the ratio 25.27 lb.mol gas/lb.mol liquid Component

Methane C1 Ethane C2 Propane C3 Butane n&isoC4 Pentane n&iso C5 Hexanes C6 Heptane+ ( use C8)

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Composition Gas lb.mol y 0.84 0.08 0.04 0.03 0.01 0 0 1

Liquid lb.mol x 0.15 0.36 0.28 0.12 0.09 1

lbmol.gas/ lb.mol oil 25.27 (25.27y) 21.22 2.02 1.01 0.76 0.25 0.00 0.00 25.27

Institute of Petroleum Engineering, Heriot-Watt University

lb.mol res gas/ lb.mol oil (25.27y+x) 21.22 2.02 1.16 1.12 0.53 0.12 0.09 26.27

Composition Res Fluid mol. Frac. 0.808 0.077 0.044 0.043 0.020 0.005 0.003 1.000

43

Petroleum Engineering

EXERCISE 7 The gas condensate reservoir in exercise 6 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 °F. Determine the initial reserves of the field in terms of condensate and gas in terms of SCF gas and STB condensate. If the condensate reservoir has a dew point of 3,000 psia determine the amount of STB of condensate and SCF gas which has been produced when the reservoir pressure has dropped to 4,000 psia.

44

Reservoir Engineering

Properties of Reservoir Liquids S I X

SOLUTION EXERCISE 7 Component

Mol Frac lb.mol yj

Critical Temperature °R °R Tcj yj*Tcj

Critical Pressure psia psia Pcj yjPcj 668 709 616 541 489 437 361

Methane C1 Ethane C2 Propane C3 Butane n&isoC4 Pentane n&iso C5 Hexanes C6 Heptane+ ( use C8) Total

0.808 0.077 0.044 0.043 0.020 0.005 0.003 1.000

344 550 666 750 838 914 1025

Reservoir pressure = Reservoir Temperature = Pseudo reduced pressure = Pseudo reduced temperature =

5000 180

psia °F = 640 °R 7.6 1.6

From Standing & Katz chart z = R = 10.73 Volume of reservoir

Volume of reservoir = 2.1076 x 109 PV = znRT V/n = zRT/P = Nu. of lb.moles in reservoir = GOR ( from ex.6) = In place gas = 1lb.mol gas = In place gas = Reserves of condensate = In place condensate =

278 42 29 32 17 4 4 406

540 55 27 23 10 2 1 658

iso temp

normal temp

iso pres

normal pres

734 830

766 846

530 490

551 488

0.98 cu.ft psi/lb.mol.°R Thickness Area Porosity gas sat cu.ft.

100 ft 5 sq miles 0.18 0.84

1 mile = 5280 ft 1.3939 x 108 sq.ft.

1.3460 cu ft/lb.mol = specific volume vol. of res/specific volume 1.5659 x 109 lb .mols in res 25.27 lb mols gas produced /lb mols. liquid 30,000 SCF/STB 26.27 lb.mols res fluid/lb moles liquid 1.5063 x 109 lb mols. gas 379 SCF 5.7087 x 1011 SCF Reserves of produced gas SCF/GOR, SCFper STB 1.9029 x 107 STB

Determine volumes of condensate and gas produced when the pressure has dropped to 4000 psia Determine value of z at 4000 psia Use previous values of Ppc and Tpc Reservoir pressure = Reservoir Temperature = Pseudo reduced pressure = Pseudo reduced temperature =

4000 psia 180 F = 640 R 6.1 1.6

From Standing & Katz chart z =

0.9

PV=znRT V/n=zRT/P = Nu. of lb.moles in reservoir at Number of moles produced down to GOR ( from ex.6)= Produced gas down to 4000 = 1lb.mol gas = Produced gas = Produced condensate = gas/GOR Produced condensate =

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1.5451 cu ft/lb.mol = specific volume 4000 psia vol. of res/specific volume 1.3640 x 109 lb .mols in res at 4000 psia 4000 psia = mols at initial pressure minus mols at 4000 psi pressure = 2.0182 x 108 lb. mols 25.27 lb mols gas produced /lb mols. liquid SCF/STB 26.27 lb.mols res fluid/lb moles liquid 1.9414 x 108 lb mols produced gas 379 SCF 7.3579 x 1010 SCF 2.4526 x 106 STB

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EXERCISE 8 Calculate the gas condensate formation factor for the example in exercise 7 at the initial reservoir pressure.

SOLUTION EXERCISE 8

Bgc= bbls of gas in reservoir /STB of produced condensate Volume of gas = 2.10761 x 109 cu ft 3.75353 x 108 bbls 7 Condensate = 1.9029 x 10 STB Bgc = 19.73 bbls of gas in reservoir /STB of produced condensate

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Properties of Reservoir Liquids S I X

REFERENCES 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.

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Fundamental Properties of Reservoir Rocks S E V E N

Core plug for horizontal k measurement Core plug for vertical k measurement

Whole core 100

Formation

Gas Permeability: Millidarcies

4 Inch 80

60

40

Hydrogen Liquid permeability

Nitrogen

20

Carbon Dioxide

0

0

1

2

3

4

5

Reciprocal Mean Pressure: (Atm.)

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Fundamental Properties of Reservoir Rocks S E V E N

C O N T E N T S INTRODUCTION 1 CHARACTERISTICS OF RESERVOIR ROCKS 2 PHYSICAL CHARACTERISTICS OF RESERVOIR ROCKS 3 POROSITY 3.1 Range of Values 3.2 Factors Which Affect Porosity 3.2.1 Packing and Size of Grains 3.2.2 Particle Size Distribution 3.2.3 Particle Shape 3.2.4 Cement Material 3.3 Subsurface Measurement of Porosity 3.3.1 Density Log 3.3.2 Sonic Log 3.3.3 Neutron Log 3.4 Average Porosity

6 POROSITY – PERMEABILITY RELATIONSHIPS 7 SURFACE KINETICS 7.1 Capillary Pressure Theory 7.2 Fluid Distribution in Reservoir Rocks 7.3 Impact of Layered Reservoirs 8 EFFECTIVE PERMEABILITY 8.1 Definition 8.2 Water Displacement of Oil 8.2.1 Water – Oil Relative Permeability 8.3 Gas Displacement of Oil and Gas – Oil Relative Permeability

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 5 STRESS EFFECTS ON CORE MEASUREMENTS 5.1 Stress Regimes 5.2 Compressibility of Poros Rock 5.3 Types of Compressiblilty 5.4 Measurements of Pore Volume Compressiblity 5.5 Effect of Stress on Permeability

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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 • 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.

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Fundamental Properties of Reservoir Rocks S E V E N

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.  

1. CHARACTERISTICS OF RESERVOIR ROCKS 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 contain enough oil or gas to make production economically viable, a reservoir rock must exceed: a minimum porosity, a minimum thickness, a minimum permeability, and a minimum area. 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.

2. PHYSICAL CHARACTERISTICS OF RESERVOIR ROCKS 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. The pore spaces (or interstices) in reservoir rock provide the container for the accumulation of oil and gas and these give the rock its characteristic ability to absorb 22/07/14

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and hold fluids. Most commercial reservoirs of oil and gas occur in sandstone, limestone or dolomite rocks, however, some reservoirs occur in fractured shale and even in basement rocks such as in Vietnam. Knowledge of the physical characteristics of the pore space and of the rock itself (which controls the characteristics of the pore space) is of vital importance in understanding the nature of a given reservoir. 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.

Figure 1 Metallic Cast of Pore Spaces in a Consolidated Sand 4

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Fundamental Properties of Reservoir Rocks S E V E N

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 Effective porosity 25%

Isolated porosity 5% Total porosity 30%

Figure 2 Effective, isolated and total porosity

Secondary porosity in limestone beds occurred as a result of fracturing, jointing, dissolution, recrystallisation or a combination of these processes. Where water is present in a carbonate formation, there is a continuous process of solution and deposition or recrystallization. If solution is greater than deposition in any zone, porosity will be developed between the crystal grains. An important type of porosity of this kind is found in dolomite zones which occur in conjunction with large limestone deposits. Dolomite may be deposited originally as a sedimentary rock, or it may be formed by replacing the calcium carbonate in limestone rock with magnesium. 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.

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Total Pore Space

Isolated Pore Space

Effective Pore Space

Permeable Pore Space

Dead End 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 f. The matrix volume is the volume of the solid grains, Vm.

Porosity =

Porosity =



Void volume x 100 Bulk volume

Bulk volume - Grain volume x 100 Bulk volume

Porosity =

pore volume × 100 void volume+grain volume

Bulk Volume Representation

Grain Volume Representation

Pore Volume Representation

Figure 4 Representation of bulk, grain and pore volumes

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Fundamental Properties of Reservoir Rocks S E V E N

These components are illustrated in Figure 4 for monosize spheres. 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.

Total Porosity =



Total Void Space Vb

Effective porosity =

Interconnected Void Space Vb

Induced or Secondary Porosity = porosity from fractures or vugs (large chambers formed in certain carbonates and limestones caused by groundwater flow and dissolution).

3.1 Range Of Values

The maximum porosity of porous media can be considered in relation to an assembly of spheres arranged as a cubic packing of spheres. If the sides of a cube are assumed to be formed by the lines drawn from the centre of each sphere to the adjacent spheres, the cube in Figure 5 would be produced.

Figure 5 Cube defined by the centres of each adjacent sphere

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 22/07/14

Vm =

4πr 3 3

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and the porosity (given the symbol f) would be



φ=

Vb − Vm = Vb

4πr 3 3 = 1 - π = 0.476 3 8r 6

8r 3 −

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.

Figure 6 Minimum porosity when all pore spaces are filled

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.

3.2 Factors Which Affect Porosity

The porosity (and permeability) of sandstone depend upon many factors, among which are the packing, size and shape of the grains, variations in size of grains, arrangement in which grains were laid down and compacted, and amount of clay and other materials which cement the sand grains together.

3.2.1 Packing And Size Of Grains

The absolute sizes of the sand grains which make up a rock do not influence the amount of porosity occurring in the rock. However variations in the range of sand grains sizes do influence considerably the porosity.

3.2.2 Particle Size Distribution

If spheres of varying sizes are packed together, porosity may be any amount from 48 per cent to a very small amount approaching 0 per cent as shown in Figure 7.

3.2.3 Particle Shape

If the sand grains are elongated or flat and are packed with their flat surfaces together, porosity and permeability may both be low we will discuss further in the context of permeability.

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Pore Space

Figure 7 Reduction in porosity due to a range of particle sizes

3.2.4 Cement Material

Sandstones are compacted and usually cemented together with clays and minerals. The porosity and permeability of a sandstone are both influenced to a marked degree by the amount of cementing material present in the pore space and the way this material occupies the pore space between the sand grains. The cementing material may be uniformly located along the pore channels to reduce both porosity and permeability or the cementing material may be located at the pore throats which reduces the ability of fluid to enter the pore, but may not reduce the overall porosity of the rock by a significant amount. 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: Recent sands (loosely packed) Sandstones (more consolidated) Tight/well cemented sandstones Limestones (e.g. Middle East) Dolomites (e.g. Middle East) Chalk (e.g. North Sea)



35 - 45% 20 - 35% 15 - 20% 5 - 20% 10 - 30% 5 - 40%

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.

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In summary, the amount of porosity is principally determined by shape and arrangement of sand grains and the amount of cementing material present, whereas permeability depends largely on the size of the pore openings and the degree and type of cementation between the sand grains. Although many formations show a correlation between porosity and permeability, the factors influencing these characteristics may differ widely in effect, producing rock having no correlation between porosity and permeability.

3.3. Subsurface Measurement Of Porosity

Porosity is measured directly from recovered rock samples as part of core analysis and also downhole by special tools which indirectly measure a property which can be related to the formation porosity. These downhole measurement techniques are very sophisticated in both their engineering and in their practice. For example, the porosity of a formation can be logged while the hole is being drilled, giving almost real time indications of the nature of the reservoir. Core analysis procedures will be reviewed later. In general the downhole porosity may be related to the acoustic and radioactive properties of the rock.

3.3.1 Density Log

The density log is derived from the response of the atoms in the minerals in the rock to bombardment with gamma radiation. The atoms accept energy of a specific frequency and emit energy of a different frequency; this energy is detected. The 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, rL, matrix density, rm , and the fluid density, rf , are related to the porosity, f

ρL = ρm (1- φ ) + ρf φ



φ =

ρ L − ρm 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

10

rQuartz rLimestone rWater

= 2.65 gcm-3 = 2.71 gcm-3 = 1.00 gcm-3

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3.3.2 Sonic Log

This log is similar in concept to the density log, however, it is acoustic energy which is radiated into the formation from sonic transducers in the logging tool. These produce compression waves which travel along the side of the borehole in the formation. The time taken for the wave to travel from the transmitter to the receiver (travel time) is related to the acoustic properties of the formation. As for the case of the density log, if the formation is known and its mineralogy is not changing, then variations in the travel time must result from the changes in the formation acoustic properties, the most significant of which is the density which is related to the porosity. As with the density tool, the density of the formation fluids in the pore spaces will affect the travel time and this must be accounted for. Calibration samples of different rock types have lead to an empirical relationship between the logged travel time, ∆TL , matrix travel time, ∆Tm , the fluid travel time, ∆Tf , and the porosity, f .

∆TL = ∆Tm (1- φ ) + ∆Tf φ



∆TL − ∆Tm ∆Tf − ∆Tm

φ =

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 water travel time are as follows

∆TQuartz = 55µs ft-1 ∆TLimestone = 47µs ft-1 ∆TWater = 190µs ft-1

3.3.3 Neutron Log

This is another radioactive logging technique which measures the response of the hydrogen atoms in the formation and can give an indication of the porosity. Neutrons of a specific energy are fired into the formation and they disrupt the steady state activity of hydrogen atoms. They then radiate energy which is detected by the tool: the energy returned is related to the number of hydrogen atoms which is related to the hydrocarbon and water in the pore spaces. By calibration, the porosity can be determined.

3.4 Average Porosity

Porosity is normally distributed and an arithmetic mean can be used for averaging. For unclassified data, n



φa =

∑φ i =1

n

i



(1)

where fa is the mean porosity, fi is the porosity of the ith core measurement and n is the number of measurements.

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4 PERMEABILITY 4.1 Darcy's Law

The permeability of a rock is the description of the ease with which fluid can pass through the pore structure. At one extreme, the permeability of many rocks is so low as to be considered zero even though they may be porous. Such rocks may constitute the cap rock above a porous and permeable reservoir and they include in their members clays, shales, chalk, anhydrite and some highly cemented sandstones. 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.

Q=

k∆P. A µL

Q A ∆P µ L k

= = = = = =

where:

(2)

flow rate in cm3/sec cross sectional area of sample in cm2 pressure different across sample, atm viscosity in centipoise length of sample in cm permeability in Darcy

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

12

Fundamental Properties of Reservoir Rocks S E V E N

p = 1 atmos Q = 1 cm 3 sec 1cm 2 µ = 1 cp L = 1 cm k = 1 darcy

Figure 8 Concept of permeable rocks

Darcy’s Law experiment consisted of a sandpack through which water flowed at a constant rate (Figure 9). Manometric heads of water

Length, L

Sand

Flowrate, Q

h1

h2

Flowrate, Q Area of the end of the sandpack

Figure 9 Schematic of Darcy’s experiment

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.

Q ∝ A, ∆h, or

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Q=k

1 L

A(h1 − h2 ) L

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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:

Q=

kA(h1 − h2 ) 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= 8µL

(3)

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 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.

4.2 Factors Affecting Permeability

Permeability along the flat surfaces will be higher, than the permeability in a direction perpendicular to the flat surfaces of the grains. In a reservoir, the permeability horizontally along the bed is usually higher than the permeability vertically across the bed because the process of sedimentation causes the grains to be laid down with their flattest sides in a horizontal position (minimising the area exposed to the prevailing currents during deposition). Figure 10 illustrates the concept. If sand grains of generally flat proportions are laid down with the flat sides nonuniformly 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.

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Fundamental Properties of Reservoir Rocks S E V E N

Horizontal permeability 400mD Vertical permeability 200mD

Horizontal permeability 900mD Vertical permeability 500mD

Porosity 16%

Porosity 32%

Figure 10 Directional Permeability

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. This directional perspective to any property is termed anisotropy. As shown above permeability is a directional property and gives rise to different permeabilities depending on the shape and depositional characteristics. Very dramatic anisotropy is generated if a rock is fractured. These anisotropic perspectives are illustrated in Figure 11. Porosity is a non directional property and therefore is isoptropic.

Sandstone

Fractured Core

Figure 11 Directional permeability.

4.3 Generalised Form Of Darcy’s Law

A three dimensional rock can be defined within the co-ordinate system illustrated in Figure 12.

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-Z

Vs

s 0

+x

+y +Z

Figure 12 Co-ordinate system for rock permeability

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,

Vs = -



k dp ρg dz ( − ) 6 µ ds 1.0133 x10 ds

(4)

h 2 )change in elevation head is equal to the sine of the angle to the horizontal A(h1 -the Q = k and L dz = sine q, where q is in degrees. ds kA(h1 - h 2 ) Q= The µL Darcy units are: Vs 4 k πr ∆P Q= m 8µL r g

= = = = =

velocity along path s - cms-1 permeability - Darcys viscosity - centipoise density of fluid - gcm-3 acceleration due to gravity - 980 cms-2

k dp rg dz gradient along s - atm cm-1 ) Vs = - ( - = pressure 6 µ ds 1.0133x10 ds

1.0133 x 106 converts from dynes cm-2 to atmospheres

dz = sin4.4 q , Dimensions where q is in degrees. Of Permeability ds

k dp ρg dz From dP Darcy’srgequation, K dz Vs = - µ ( ds − 1.0133 x10 6 ds ) the dimensions of each ) Vs = - ( µ ds 1.0133x106 ds term can be deduced in terms of length, L, mass, M and time, T

Vs =

L T

µ = 16

M LT

r =

M 3 L

Fundamental Properties of Reservoir Rocks S E V E N



Vs =

L T

µ=

M LT

P=

M LT 2

g=

L T2

ρ=

M L3

dP M = 2 2 ds L T

Therefore, the equation in terms of the dimensions (and keeping permeability as k) is

L kLT M ML ( 2 2 - 3 2) = T M LT LT L K = T LT

K = L2



(5)

It can be seen that the dimensions reflect the nature of the constant of proportionality and it should not be confused with, for example, the area open to flow, A, of the end of a core or a sand pack. In terms of metric units, since 1 atm = 14.73 psi = 1.013 bar and 1 cp = 10-3 Pas it follows that

1D = 9.87 x 10-13m2 ~ 1 x 10-12m2 -16 2 1mD = 9.87 x 10 m ~ 1 x 10-15m2

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.

4.5 Assumptions For Use Of Darcys Law

The simple Darcy Law, as used to determine permeability, only applies when the following conditions exist: (i) Steady state flow (ii) Laminar flow (iii) One phase present at 100% pore space saturation (iv) No reaction between fluid and rock (v) Rock is homogenous 1. Steady state flow, i.e. no transient flow regimes. This becomes unrealistic in terms of flow in a reservoir where the nature of the fluids and the dimensions of the reservoir may produce transient flow conditions for months or even years. For laboratory based tests, the cores are small enough that transient conditions usually last only a few minutes. 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 22/07/14

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Laminar Flow

Turbulent Flow

Q A

Q k . P = µ A L ∴K =

Q . L . µ P A

P L

Figure 13 Effect of Turbulent Flow on Measured Permeability

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.

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Fundamental Properties of Reservoir Rocks S E V E N

4.6 Applications of Darcys Law

To examine the applicability of this simple relationship, approximations to the type of flow encountered in a reservoir can be made: linear flow along a reservoir section and radial flow into a wellbore. More complex geometries cannot be analysed using this simple analytical equation and forms of approximating the geometry and flow are required. In the following expressions, the nomenclature is identical to that used above. (i) Horizontal, linear, incompressible liquid system (Figure 14)

A Q L P1

P2



Figure 14 Linear flow regime

From the basic Darcy equation

Vs = -

k dP ρg dz ( − ), 6 µ ds 1.0113 x10 ds

ρg dz = zero 6 1.0113 x10 ds

The flow rate and area open to flow are substituted for the flow velocity. The variables are separated and integrated over the length (for the flow rate) and the pressures P1 to P2 for the change in pressure. The pressure drop P2 minus P1 is negative and is corrected by the negative sign on the left hand side of the equation.

Q A kA dP Q = µ dx Vs = Vx =

L

P

kA 2 Q ∫ dx = dP µ P∫1 0 Q(L - 0) =

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kA ( P2 − P1 ) µ

kA( P1 − P2 ) µL



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(6)

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The final form is as formulated by Darcy and the permeability will have the units of Darcys if the other units are: flow rate, Q - cm3s-1 area open to flow, A - cm2 viscosity, µ - centipoise

pressure, P - atm length, L - cm

(ii) Horizontal, linear, compressible ideal gas system The flow regime is the same as for the linear liquid system and from the basic Darcy equation:

Vs = Vx = - k ( Q A kA dP Q = µ dx

dP ρg dz ρg dz − ), = zero 6 6 ds 1.0113 x10 ds 1.0113 x10 ds

Vs =



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. P1

P L A Core

P2

Valve

Pb

Qb

Flow measurement

Figure 15 Configuration for gas permeability measurements.

The flowrate measured, Qb at ambient pressure, Pb is related to the flowrate, Q in the core at the pressure in the core, P via the ideal gas law. If the assumption is made that the temperature is constant, then

QP = Q b Pb



Q =

Q b Pb P

and substituting into the equation, separating the variables and integrating produces

20

Fundamental Properties of Reservoir Rocks S E V E N

Q b Pb kA dP =µ dx P L

P

kA 2 Q b Pb ∫ dx = PdP µ P∫1 0 Q b Pb (L- 0 )= -



Qb =



k=

kA ( P22 − P12 ) 2 µ

kA( P12 − P22 ) 2 µLPb



2 µQb Pb L A( P12 − P2 2 )

(7)

(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.



Gas Q b =

kA( P12 − P22 ) kA( P1 − P2 ) Liquid Q = 2 µLPb µL

In certain circumstances, the mean flow rate, Q is measured at a mean pressure, P which, in the case of a laboratory experiment on a core, is the mean of the upstream and downstream pressure, i.e.

P=

P1 + P2 2

and P Q = Volume flow rate at P

P Q = P bQ b substituting this into the above gas equation 7.



Pb Q b = PQ =

kA( P12 − P22 ) 2 µL

and since 22/07/14

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1 1 kA (P1 − P2 )(P1 + P2 ) (P1 + P2 )Q = 2 2 µL Q=



kA( P1 − P2 ) µL



(9)

The ideal gas permeability can be calculated from the liquid equation using mean flowrate, Q measured at mean pressure. (iii) Horizontal, radial, incompressible liquid system (Figure 16) 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

re is the outer boundary radius rw is the inner boundary radius (well) Pe is the pressure at the external boundary Pw is the pressure at the inner boundary Starting from the basic Darcy expression again,

Vs = -

k dP ρg dz ρg dz ( − ), = zero 6 6 µ ds 1.0113 x10 ds 1.0113 x10 ds

Substituting for flow velocity, Vs = Vr =

Q A

In this case the direction of flow is in the opposite sense to the co-ordinate system, therefore

ds = -dr

For radial geometry, the area, A, is now radius dependent therefore

22

A = 2πrh

Fundamental Properties of Reservoir Rocks S E V E N

Substitution into the basic expression gives

Q k dP = µ − dr 2πrh

(10)

separating the variables and integrating r

P

Q e dr k e = ∫ dP 2πh r∫w r µ Pw



Q k (ln re − ln rw ) = ( Pe − Pw ) 2πh µ

which gives the final form

Q =

2πkh( Pe − Pw ) r µ ln e rw

(11a)

(iv) Horizontal, radial, compressible ideal gas system 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.

Vs = -



k dP ρg dz ρg dz ( − ), = zero 6 6 µ ds 1.0113 x10 ds 1.0113 x10 ds

Q k dP = 2πrh µ − dr

If the assumption is made that the temperature is constant, then

QP = Q b Pb



Q =

Qb Pb P

and substituting into the equation, 10

Q b Pb k dP = 2πrh µ dr P

separating the variables

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re



Reservoir Engineering

P

dr 2πkh e = PdP ∫ r µ Pw rw

Q b Pb ∫

and integrating produces

 r  2πkh  Pe 2 − Pw 2  Q bPb ln e  =  2 µ    rw 

Qb =

πkh Pe 2 − Pw 2 ) ( r  µPb ln  e   rw 



(11b)

4.7 Field Units

Measurements made in the field are often quoted in field units and to ensure compatibility with the Darcy equation, a conversion is required. The field units are usually as follows: Flow rate, Q - bbl/day or ft3/day Permeability, k - Darcy Thickness or height of reservoir, h - feet Pressure, P - psia Viscosity, m - centipoise Radius, r - feet Length, L - feet In order to convert the Darcy equation for liquid flow, Q =

KA( P1 − P2 ) µL

to oil field units, the following conversion factors are used:

Q

bbl 5.615 ft 3 1728in3 16.39cm 3 day hr ( )( )( )( )( ) 3 3 day bbl ft in 24hr 3600 s

929cm 2 atm ( K )( Aft )( )( ∆Ppsia)( ) 2 14.696 psia ft = 30.48cm ( µ )( Lft )( ) ft 2



to oil field units, the following conversion factors are used: and these produce the following version of Darcy’s equation in field units:

24

Q

bbl KA( P1 − P2 ) = 1.1271 µL day

(11c)

Fundamental Properties of Reservoir Rocks S E V E N

4.8 Klinkenberg Effect

Darcy’s Law would indicate that the permeability should be the same irrespective of the fluid transmitted, since viscosity is included in the equation. Measurements made on gas as against liquid for some conditions give higher permeabilities than the liquid. This phenomenon is attributed to Klinkenberg, who attributed the behaviour to the effect of the slippage of gas molecules along the solid grain surfaces. This occurs when the diameter of the capillary opening (pore throat diameter) approaches the mean free path of the gas (i.e. there is in effect only one gas molecule per capillary). Darcys Law assumes laminar flow and viscous theory specifies zero velocity at the boundary of the flow channel. This is not valid when the mean free path of the gas approaches the diameter of the capillary and the result is that low pressure permeability measurements are unrealistically high because there is insufficient gas molecules to form a zero velocity boundary layer at the edges of the pores and to form a mass of flowing gas within the pores. In this case, too many gas molecules flow through the pores and the permeability appears to be higher than it actually is: the effect reported by Klinkenberg. 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:

100

Gas Permeability: Millidarcies

80

60

40

Hydrogen Liquid permeability

Nitrogen

20

Carbon Dioxide

0

0

1

2

3

4

5

Reciprocal Mean Pressure: (Atm.)

Figure 17 Variation in gas permeability with reciprocal mean pressure. 22/07/14

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Pm is the mean pressure of the gas (the mean of the upstream and downstream pressures either end of the core P 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. The equation developed by Klinkenberg is of the form

kL =

kG b 1+ Pm

(12)

Where : kL = equivalent liquid permeability kG = permeability to gas Pm = mean flowing pressure 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.

4.9 Reactive Fluids

Darcys Law assumes that the fluid does not react with the formation. Many formation waters react with clays in the rock to produce a lower permeability to liquid than would be obtained with gas. Therefore the permeability to water in the formation may be much lower than would be determined to gas in the laboratory. Any water injected into the formation may severely reduce the permeability due to clay swelling. The change in permeability may be substantial, for example from several hundred millidarcys to less than one millidarcy.

4.10 Average Reservoir Permeability

Permeability is not normally distributed but has an exponential distribution, therefore a geometric mean is used to obtain an average reservoir permeability. The Geometric Mean of n numbers is the nth root of their product:

26

Fundamental Properties of Reservoir Rocks S E V E N

5 STRESS EFFECTS ON CORE MEASUREMENTS 5.1 Stress Regimes

In reservoir engineering the impact of reservoir stresses on reservoir flow and capacity parameters has been considered for a number of years but, increasingly, the interest in stress related measurement has grown. The effect of removing a core from the formation is to remove all the confining forces on the sample, allowing the rock matrix to expand in all directions, partially changing the shapes of the fluid-flow paths inside the core. 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.

Core plug for horizontal k measurement Core plug for vertical k measurement

Whole core

4 Inch

Formation

Figure 18 Trends in Reservoir Rock Characterisation

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.

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Major Principal Stress

Minor Principal Stress

(a)

Minor Principal Stress

Equal Stresses kh (b)

kv

(c)

Equal Stresses

Figure 19 Stress States in Reservoirs and Cores

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

28

Fundamental Properties of Reservoir Rocks S E V E N

the matrix will strain as the stress is altered. If the stress increases, the strain reduces the radius of the pore throats and reduces the volume of the pore space. This effect may be different for different rock types and even within the same rock type if the amount of cementing material is altered. The significant aspects of this phenomenon are when cores are removed from subsurface to the laboratory (since the overburden and pore pressure will change) and when the pore pressure in the reservoir changes due to local pressure conditions around the wells (drawdown) and within the reservoir as a whole as it is depressurised, for example. The impact of the net overburden stress which increases as the reservoir pressure ( pore pressure ) decreases is illustrated in Figure 20.

1.0

?Well Cemented

Permeability: Fraction of Original

.8

.6

?Friable

?Unconsolidated .4

.2

0

0

2000

4000

6000

8000

10000

Net Overburden Pressure: PSI

Figure 20 Permeability Reduction with Net Overburden Pressure

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. 22/07/14

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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 Zero stresses

No effect of the stress on the property

Hydrostatic stresses

The effect of the magnitude of the stresses are measured

Triaxial stresses The effect of stresses resolved in the three principal directions are measured



Real stress behaviour The effect of the magnitude and direction of the stresses are measured This topic is covered in more detail in the subsequent chapter.

5.2 Compressibility Of Porous Rock

As the rock matrix is subjected to a stress, it will deform and alter the pore space volume as the rock is compressed. For simplicity, the overburden will be considered to produce hydrostatic stress (called the compacting stress) on the reservoir, i.e. a grain-to-grain stress in the rock. Within the pores, fluid pressure acts on the surface of the grains and reduces the grain-to-grain (or compacting) stress. Therefore in a real reservoir there is a balance between the effect of the overburden stress and the pore pressure. This can be described by the relationship Pcompacting = Poverburden - Ppore pressure 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:

30

Fundamental Properties of Reservoir Rocks S E V E N

Surface Po

Cap Rock

Depth

Pf and Pc Reservoir

Pc Pc

Pc

Enlarged view of the pore space

Pf Pc

Pc

Grains

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 compacting 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.

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50 40

Porosity, f

Sandstone

30 20 10 0

Shale 0

3000 6000 Depth of burial (ft) or stress (psi)

Figure 22 Alteration in porosity with depth of burial (or stress)

The rate of change of pore volume with pressure change can be represented by an isothermal compressibility (assuming temperature is constant):

Cf = -



1 dv v dP

(15)

where Cf is the isothermal compressibility, v is the volume, dv is the change in volume and dP is the change in pressure (the negative sign accounts for the co-ordinate system: as the pressure increases, the volume decreases).

5.3. Types Of Compressibility

An issue with regard to the compressibility is: which part of the reservoir is being compressed and which part is significant in calculating the response of the reservoir. Three types of compressibility can be considered: (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;



Reservoirs composed of unconsolidated or very weakly consolidated material where the changes in porosity can be significant. The changes in the volume of the reservoir both in a vertical sense leading to subsidence and in a horizontal sense leading to shearing of the wellbore and the associated loss in integrity.

(iii) Pore volume compressibility - change in pore volume. This is of greatest interest since the pore volume affects the porosity which affects reservoir performance. For completeness, all aspects of the reservoir compressibility should be considered, however, in many problems only specific aspects of the compressibility may be required 32

Fundamental Properties of Reservoir Rocks S E V E N

such as in a well cemented sandstone reservoir where the bulk volume change is very small and the subsidence is negligible, but the pore compressibility is an important feature of the drive mechanism.

5.4 Measurement Of Pore Volume Compressibility

The measurement of pore compressibility is usually conducted in a coreholder which applies an equal compacting pressure around the core. An inner liner ensures the power fluid (usually hydraulic oil) does not contaminate the pores of the sample. The pore pressure is usually kept at ambient, i.e. the compacting pressure mimics the net effect of the overburden and the pore pressure in the reservoir. This makes the test simpler, however, there may be conditions where the compressibility of the grains themselves plays a significant role in the system and the test may require to be conducted at true overburden and pore pressure conditions. For the test at ambient pore pressure conditions, an outlet is connected to the core holder and this is lead to a pipette or a balance to measure the amount of pore fluid expelled. The pressure of the hydraulic oil is increased in stages and for each stage the amount of fluid expelled is measured after the rock has come to equilibrium. The data can then be analysed to indicate the change in porosity or pore compressibility. Figure 23 shows the concept. Pipette Sealed core

Pump

Pressure vessel

Figure 23 Measurement of the reduction in pore volume as the external stress (or compacting 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



Cp = -

1 dv p 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

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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. Pore compressibility 10-6 psi-1

10 9

Sandstone Limestone

8 7 6 5 4 3

0

10 Porosity %

20

Figure 24 Compressibility of Sandstones and Limestones

5.5 Effect of stress on permeability

As the effect of a stress on the rock matrix affects the pore volume, it also affects the pore throat radii and the permeability of the rock. In general, an increase in stress reduces the pore throat radii and the permeability declines. For most rocks subjected to an hydrostatic stress, this is the case as the stress is equal in all directions. Figure 25 shows typical permeability declines for increase in stress for sandstone.

34

Fundamental Properties of Reservoir Rocks S E V E N

Permeability stress sensitivity for various sandstones

1000

Permeability (mD)

100

10

1

0

20

40 60 Hydrostatic stress (MPa)

80

Figure 25 The reduction in permeability for a range of sandstone samples (the porosity is in the range 15% to 22%)

Unconsolidated material has larger absolute changes in permeability as the total strain is greater. 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).

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Fracture

Reservoir Engineering

sv Permeability sh maximum

Fracture closing under stress

Core

s

sh maximum

h minimum

sh maximum perpendicular to fracture

Fracture

sv Permeability

Core

sh minimum

sh maximum Fracture opening under stress

sh maximum

sh maximum parallel to fracture

Figure 26 Triaxial stresses applied to a fractured core

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. 36

Fundamental Properties of Reservoir Rocks S E V E N

1000

Oolitic Limestone

Sucrosic Dolomite

Permeability: Millidarcies

Reef Limestone

Well Cemented Hard Sand

100

Chalky Limestone

10

Intercrystalline Limestone and Dolomite Fine Grained Friable Sand 1.0

0

5

10

15

20

25

30

35

Porosity: Percent

Figure 27 Permeability and Porosity Trends for Various Rock Types (Core Laboratories Inc)

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.

7. SURFACE KINETICS

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).

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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.

7.1 Capillary pressure theory

The rise or depression of fluids in fine bore tubes is a result of the surface tension and wetting preference and is called capillarity. Capillary pressure exists whenever two immiscible phases are present, for example, in a fine bore tube and is defined as the pressure drop across the curved liquid interface. The equilibrium in force between the molecules of a single phase is disrupted at an interface between two dissimilar fluids. The difference in masses and the difference in the distances between the molecules of the different phases produces an initially unbalanced force across the interface. Figure 28 shows the interface between oil and water molecules. W W

O

W

O

O

W

Different mass. Different space between molecules.

O

W: water molecule O: oil molecule distance between molecules

Figure 28 Representation of an oil water boundary

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˚

38

Fundamental Properties of Reservoir Rocks S E V E N

Interfacial tension, s, defined as force / unit length σwo

Oil

θ

Contact angle, θ σso

σsw

Water

Solid

σ wo Interfacial tension between the water and oil σ

sw Interfacial tension between the solid and water

σ so

Interfacial tension between the solid and oil

Figure 29 Interfacial tension between oil, water and a solid

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. Octane

Napthenic acid 35°

30° Silica

Octane

Napthenic acid 106° 30° Calcite

Figure 30 The effect of a change in the surface on wetting properties

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, At = ssw - sso = swo cos qwo

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If a container of oil and water is considered as in Figure 31, the denser water lies below the oil. σcosθ

θ h

σ

OIL radius, r

Water

Oil water contact

Figure 31 Capillary rise in an oil/water system

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.

net force upwards = 2pr swocoss

(16)



net force downwards = (rwgh - rogh)pr2 = gh(rw - ro)pr2

(17)

the interface is at equilibrium, therefore 2pr swocosq=gh(rw - ro)pr2

(18)

The capillary pressure is the difference in pressure across an interface, therefore in terms of pressure (the pressure Pc, force acting on area pr2) gh( ρw − ρo )πr 2 2πrσ wo cos θ = = Pc πr 2 πr 2

40

gh( ρw − ρo ) =

2σ wo cos θ r

Fundamental Properties of Reservoir Rocks S E V E N

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.

Pc =

where Pc s q rc h rw ro

= = = = = = =

2σcos θ = gh( ρw − ρo ) rc

(19)

capillary pressure surface tension contact angle radius of the tube height of interface the density of water 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 free water level-the position of zero capillary pressure The oil -water contact The 100% water saturation at a distance above the free-water level due to the capillary action of the largest tube. The irreducible level representing the limit of mobile water saturation The different radii segregate the capillary pressure and therefore the height to which the water is drawn into the oil zone.

The zone of varying water saturation with height above the 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.

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q

oil

oil

oil q

oil q

h

FREE WATER LEVEL WATER

WATER

Irreducible Water

Pc

Oil

Oil water contact

Transition Zone

Figure 32 Capillary Rise in Distribution of Capillaries

Water OWC Water

0 FWL

0% 100%

Sw So

100% 0%

Free water level

Figure 33 Capillary Pressure Curve

It must be remembered that although concepts of capillary pressure were formulated in terms of fine bore tubes, application in practice deals with a complex network of interconnected pores in a matrix carrying surface chemical properties as illustrated in Figure 1 of the pore cast of the pore space. 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

42

Fundamental Properties of Reservoir Rocks S E V E N

inversely proportional to the tube radius and difference in fluid density. The smaller the pores ie. the lower the permeability, the higher the capillary pressure.

7.2 Fluid distribution in reservoir rocks

Water is retained by capillary forces as hydrocarbons accumulate in productive reservoirs. The water is referred to as connate or interstitial water and in water wet rocks it coats the rock surfaces and occupies the smallest pores, whereas hydrocarbons occupy the centre of the larger pores. The magnitude of the water saturation retained is related to the capillary pressure which is controlled by the rock fluid system. Rock Fluid Property

Wettability Rock / Fluid Property

2σ cosθ Sw _ ~ 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. Reservoir saturation reduces with increased height above the hydrocarbon-water contact. At the base of the reservoir there will usually be a zone of 100% water saturated rock. The upper limit of this is referred to as the water table or water oil contact (WOC). However, there is a non identifiable level, the free water level representing the position of zero capillary pressure. 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.

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Oil Sand Grain

Pc h

Transition Zone WOC

FWL

hD 0%

Water Saturation

100%

Water

Figure 34 Capillary Pressure Curve for Porous media

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.

44

Pc = (ρω - ρo) gh

Fundamental Properties of Reservoir Rocks S E V E N

High Pc needed to reach limiting water saturation.

Case 1

Irreducible (or non - communicating) water approach at low Pc

Case 2 I

h

Largest connecting pores about the same size. Therefore simular h D

h X

h

D

Irreducible water saturation

Water saturation

100%

Figure 35 Capillary Pressure Curves for Different Rocks

In addition to water transition zones, there can also be an oil/gas transition zone, but this is usually less well defined.

Height Above Water Level

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.

Interfacial Tension Effect

High Interfacial Tension A

Low Interfacial Tension

0 100 Water Saturation: Percent Pore Space

Figure 36 Interfacial Tension Effect

Saturation history influences the capillary pressure water saturation relationship and therefore the size of the transition zone. Drainage saturation results from the drainage of the wetting phase (water) from the rock as the hydrocarbons accumulate. It represents the saturation distribution which exists before fluid production. The level of saturation is dictated by the capillary pressure associated with the narrow pore and is able to maintain water saturation in the large pore below. Imbibition 22/07/14

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Height Above Water Level

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.

Drainage

Drainage

A

Imbibition

0 100 Water Saturation: Percent Pore Space

Imbibition

Figure 37 Saturation History Effect

Height Above Water Level

A large density difference between water and hydrocarbons (water-gas) suppresses the transition zone. Conversely, a small density difference (water-heavy oil) increases the transition zone. Figure 38 shows the differences in density for water/heavy oil and water/gas on capillary rise. Transition zones between oil and gas are not significant because of the large density difference between oil and gas.

Fluid Density Difference Effect

Small Density Difference (Water-Heavy Oil)

A

Large Density Difference (Water Gas)

0 100 Water Saturation: Percent Pore Space

Figure 38 Fluid Density Effect

7.3. Impact of Layered Reservoirs

A characteristic of reservoirs is the various rock types making up the reservoir section. Each rock type has its own capillary pressure characteristics. Wells penetrating such formations will show a water saturation distribution reflecting the specific capillary effects of each formation type. In some cases a 100% water saturation will be above a lower water saturation associated with a lower elevation material with a higher permeability, Figure 39.

46

Fundamental Properties of Reservoir Rocks S E V E N

For example well A would only indicate 100% water. Well B would penetrate the transition zone of the top layer then a region of 100% water saturation. The saturation profiles for well B and C are illustrated in Figure 39. The transition zone of the next layer 2, followed by an interfacial of 100% saturation associated with layers 2, 3 and 4 then into 100% for the next two layers. Well D penetrates through the top and next layer at the irreducible saturation level, into the transition zone for layer three, then into irreducible saturation for the 4th layer. B

C

D

SHALE

15

K=

25 0= 10 0= 0=

FWL 0

2

19 0

K=

K=

3

4

d

md d 5m

md 200

Free Water Level

100% 1

30

m 40

ONE DST

0=

K=

Water saturation profile well C only

SAN

Transition zone

RES

.

Water saturation profile Well B only

Height

A

SHALE

FWL

0%

Sw

100%

100% Water Level

Figure 39 Capillary Effects in Stratified Formations

8 EFFECTIVE PERMEABILITY 8.1 Definition

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.

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If measurements are made on two cores having different absolute permeabilities k1 and k2, there is no direct way of comparing the effective permeability kw and ko curves since for the two cores they start at different points k1 and k2. This difficulty is resolved by plotting the relative permeability krw and kro where Relative Permeability =

permeability to one phase when one or more phases are present permeability to one phase alone

kr =

ke k

Relative permeability is dimensionless and is reported as a fraction or percentage. On relative permeability plots the curves start from unity in each case, so direct comparisons can be made. 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

Relative Permeability

0.8 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

Figure 40 Relative permeability curves for water-oil system

48

1.0 Connate Water plus Residual Oil Saturation

Relative Permeability, Fraction of Absolute

Fundamental Properties of Reservoir Rocks S E V E N

0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

0

k rg k ro

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Liquid Saturation = SO + S W , %

Figure 41 Relative permeability curves for gas-oil system

The following points are to be noted: The introduction of a second phase decreases the relative permeability of the first phase: for example, kro 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.

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Relative Permeability, %

100 90

Water-Wet Drainage

80

Water-Wet Imbibition

70

Oil-Wet Drainage

(Decreasing S w )

(Increasing S w ) (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 characteristics are important in the displacement of hydrocarbons by water, and in the displacement of oil and water by gas. Such displacements occur during primary and secondary recovery operations, as well as during coring and core recovery. Relative permeability data when presented in graphical form are often referred to as drainage or imbibition curves. (Figure 42) Imbibition relative permeability is displacement where the wetting phase saturation is increasing. For example, in a water flood of a water wet rock, or coring with a water base mud.

50

Fundamental Properties of Reservoir Rocks S E V E N

Drainage relative permeability is where the non-wetting phase saturation is increasing. For example, gas expulsion of oil during primary depletion or gas expansion of fluids during core recovery, and the condition existing in the transition zone at discovery. 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.

8.2 Water displacement of oil

Prior to water displacement from an oil productive sand interstitial water exists as a thin film around each sand grain with oil filling the remaining pore space. The presence of water as previously stated has little effect on the flow of oil, and oil relative permeability approaches 100%. Water relative permeability is zero. 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.

8.2.1 Water-oil relative permeability

Accumulation of hydrocarbons is represented by drainage relative permeability curves as the water saturation decreases from 100% to irreducible. Water relative permeability reduces likewise from 100% to zero while oil relative permeability increases. Subsequent introduction of water during coring or water flooding results in a different set of relative permeability curves - these are the imbibition curves. The water curve is essentially the same in strongly water wet rock for both drainage and imbibition. The oil phase relative permeability is less during imbibition than during drainage. The oil remaining immobile after a waterflood is influenced significantly by the capillary pressure and interfacial tension effects of the system. It is of note that a high residual oil saturation is a result of the oil ganglia being retained in the large pores as a result of capillary forces. Figure 43 illustrates the pore doublet model illustrating how oil can be trapped in a large pore. The forces to displace this droplet have to overcome capillary forces and are too great to use pressure through pumping. The force required can be reduced by reducing the interfacial tension which is the basis for many enhanced oil recovery methods; for example, surfactant and miscible flooding.

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Water In

Oil

Advancing water

Water In

Water penetrating smaller pores due to capillary forces

Oil

Trapped oil

Water

Water In

Figure 43 Pore Doublet Model

An important perspective in a displacement process is the concept of mobility ratio. This relates the mobility of the displacing fluid relative to that of the displaced fluid. It is therefore a ratio of Darcy’s Law for each respective fluid at the residual saturation of the other fluid. In the context of water displacing oil.

M = mobility ratio =

where k'rw k'ro

krw ′ / µ w kro ′ / µ o

(20)

is the relative permeability at residual oil saturation is the relative permeability at the irreducible water saturation.

These relative permeabilities are sometimes referred to as end point relative permeabilities. When M is less than 1 this gives a stable displacement whereas when M is greater then 1 unstable displacement occurs.

52

Fundamental Properties of Reservoir Rocks S E V E N

This topic is covered extensively in the chapter on immiscible displacement

8.3 Gas displacement of oil and gas-oil relative permeability

Gas is a non-wetting phase and it initially follows the path of least resistance through the largest pores. Gas permeability is zero until a ‘critical’ or ‘equilibrium’ saturation is reached (Figure 41). 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.

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Relative Permeability: Percent

20

40

60

80

100

Gas

Figure 44 Gas Oil Relative Permeabilities ( Core Lab)

Characteristic Sand During Oil Displacement by Gas @ 5% Gas saturation

250 md. 183 md. 0.0md. 183/250 = 0.73 0.0/250 = 0.0

Gas Saturation: Percent Pore Space

0

Krg

Kro

Gas Saturation: 5% of Pore Space Specific Permeability (Ks): Effective Permeability to Oil (Ko): Effective Permeability to Gas (Kg): Relative Permeability to Oil (Kro) = Relative Permeability to Gas (Krg) =

0

20

40

60

80

Water

20

40

Krg

60

80

100

Characteristic Sand During Oil Displacement by Gas @ 20% Gas saturation

250 md. 52 md. 10md. 52/250 = 0.21 10/250 = 0.04

Gas Saturation: Percent Pore Space

0

Kro

Gas Saturation: 20% of Pore Space Specific Permeability (Ks): Effective Permeability to Oil (Ko): Effective Permeability to Gas (Kg): Relative Permeability to Oil (Kro) = Relative Permeability to Gas (Krg) =

0

20

40

60

80

100 Relative Permeability: Percent

54 20

40

60

Krg

80

100

Characteristic Sand During Oil Displacement by Gas @ 45% Gas saturation

250 md. 6.2 md. 70md. 6.2/250 = 0.025 70/250 = 0.28

Gas Saturation: Percent Pore Space

0

Kro

Gas Saturation: 45% of Pore Space Specific Permeability (Ks): Effective Permeability to Oil (Ko): Effective Permeability to Gas (Kg): Relative Permeability to Oil (Kro) = Relative Permeability to Gas (Krg) =

0

20

40

60

80

100 Relative Permeability: Percent

100

Oil

Petroleum Engineering Reservoir Engineering

Rock Properties Measurement E I G H T

32

LEGEND

30 Different reservoir sand sequence in a formation

28 26

Dimensionless Capillary Pressure Pc σ

K φ

24 22 20 18 16

Platen

14

Threaded end cap

Trapped tube

12 10

A

A

8

Core

Rubber sleeve

6

Aluminium cell body

4 2 0

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

(

0.9

Normalised Wetting Phase Saturation Sw* = Sw-Swc 1-Swc

(

Hydraulically pressured tubes

1.0

Maximum principal stress

σ3 3

3

3

3

2

2

1

σ2 1

3

2

3

2 3

2 2 1

σ2

1

2

2

3

3

3

3

3

Face of core plug

σ3

Section AA

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Rock Properties Measurement E I G H T

C O N T E N T S 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 6.2.2 Centrifuge method 6.2.3 Dynamic method 6.2.4 Mercury Injection 6.3 Use of Laboratory Capillary Pressure Data for Reservoir 6.4 Averaging capillary pressure data 7 EFFECTIVE PERMEABILITY

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LEARNING OBJECTIVES Having worked through this chapter the Student will be able to: • List the various types of recovered core. • Describe briefly the various methods of measuring porosity and permeability. • Briefly describe the various stress conditions that can be imposed on a rock sample. • Understand how to convert laboratory based capillary pressure measurement data to field related values of capillary pressure. • Be able to determine the saturation distribution in a well made up of different rock types given capillary pressure data. •

2

Understand the Leverett J function and be aware of the major tortuosity related assumption in its derivation.

Reservoir Engineering

Rock Properties Measurement E I G H T

1. INTRODUCTION 1.1 Core Analysis

In this chapter we will focus on the laboratory based methods used to determine some of the parameters outlined in the previous chapter. The topic is also covered in other modules of the overall Petroleum Engineering programme in the context of the specific module. Core recovery is covered in drilling and rock properties are also covered in the Formation Evaluation module. Cores obtained from the reservoir formation contain a considerable amount of information about the nature of the rocks themselves and various properties. They are also a source of material for investigating rock behaviour with respect to fluid displacement and its reaction to various fluid types. 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. A comprehensive document on the procedures for generating some of the rock properties through laboratory measurement is the API Recommended Practices for Core Analysis 2, APR RP40 which was revised in 1998. This API document goes into detail beyond that covered in this overview chapter. Routine Core Plug Analysis Slabbed Core • • • •

Photograph Sedimentology Lithology Samples

Government or Regulatory Board Sampling Curation

Porosity Permeability Grain Density As-Received Saturations

Special Core Analysis

Thin Sections • • • •

• • • •

• • • • • • • •

Detail Pore Structure Diagenesis Porosity Type Environmental Evidence

Preserved /Restored State Capillary Pressure Relative Permeabilty Electrical Properties Acoustic Properties Compressive Properties Clay Chemistry Effects Specific Tests

Small Samples • Grain Size Distribution • Mineral Analysis • X-Ray and SEM Analysis • Bio-Dating and Association

Calibration of Wireline Logs

Figure 1 Data Obtained From Cored Wells 1

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As covered in the previous chapter there are a number of properties in relation to measurements possible on the cores as shown in the Figure 1. In core analysis the measurements can be divided into two parts: routine measurements which cover fluid saturations, porosity and permeability; special core analysis which covers a wide range of measurements and special tests of special interest to the organisation commissioning the testing. In this chapter we will focus on routine core analysis and also cover briefly capillary pressure measurements.

1.2 Core definitions

Before examining some of the methods it is important to define the various core types used in examining rock properties and their reaction to the transmission of fluids. These definitions come from the API recommended practice RP 402. Fresh Core Any newly recovered core material preserved as quickly as possible at the wellsite to prevent evaporative losses and exposure to oxygen. The fluid type used for coring should be noted, e.g., fresh state ( oil-based drilling fluid), fresh state ( water -based drilling fluid). Preserved Core Similar to fresh core but some period of storage is implied. Preserved core is protected from alteration by a number of techniques, from simple mechanical stabilisation using bubble wrap or similar, freezing the core to lock in fluids which would otherwise evaporate ( in this case the freezing may alter some of the rock properties), enclosure in heat -sealable plastic laminates, and dips and coatings. Cleaned Core Core from which the fluids have been removed by solvents. The cleaning process (the specification and sequence of solvents, temperatures, etc) should be specified. Some solvents could damage the fabric of the rock and special cleaning procedures like critical point drying might be required for example with rocks containing friable clays (Figure 2).

Figure 2 Sandstone contains illite 4

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Rock Properties Measurement E I G H T

Restored - State Core This is core that has been cleaned and then reexposed to reservoir fluids with the intention of reestablishing the reservoir wettability condition. The conditions of exposure to the crude oil, especially initial water saturation, temperature and time. can all affect the ultimate wettablity. Pressure - Retained Core This is material that has been kept, so far as possible, at the pressure of the reservoir in order to avoid change in the fluid saturations during the recovery process.

2. SAMPLE PREPARATION 2.1 Whole Core Scanning

Prior to subdivision of the whole core for the various types of analysis a number of procedures can take place to record the characteristics of the whole core and to relate it to indirect down hole measurements. The purpose of this core examination and description is to recognise lithological, depositional, structural and diagenetic features of the whole core or slabbed core. Qualitative and quantitative core descriptions provide the basis for routine core analysis sampling, facies analysis, and further reservoir studies such as reservoir quality and supplementary core analysis. Besides visual examination and generating a photographic record, these techniques provide a means of relating to downhole measurements and to identify features of the core which might otherwise if undetected generate unrepresentative data in subsequent analysis. 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 a 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.

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Scintillometer Recorder Lead Shield Conveyor Belt Core

Figure 3 Natural gamma scan on whole core. (Corelab)

A number of X-Ray techniques can be used which include, fluoroscopy, x-radiography and computerised tomography (CT) scanning. In one method a continuous analysis is conducted where an attenuated x-ray beam directed through the core impinges on a fluorescent screen and the captured image is recorded by video camera. In x-radiography the attenuation of the beam is captured and recorded on sensitive film. In this procedure the core is stationary. The advances in CT scanning in medical applications have been used in CT scanning where the attenuated beam directed in multiple directions by a rotating beam enables a reconstruction of density variations within the core. The resolution of the image depends on the thickness of the beam and the size of pixel used to construct the image. A sketch of CT scanning and the principle on which it is based is shown in Figure 4 Sample for measurement

h

Io

I

Narrow incident beam

Attenuated beam

Particle or energy detector.

Shield I = Ioe -µh µ is a function of bulk density and atomic number

Figure 4(a) Computer aid tomography on whole core. Principle of attenuation

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Rock Properties Measurement E I G H T

Rotating energy source and detector

Intensity profiles Reconstruction algorithm in computer.

Figure 4(b) Reconstructed cross section

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 nucleus. 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.

2.2 Core Cleaning

Sample preparation is an important consideration in core analysis. Prior to samples or plugs being used for the determination of porosity or permeability they must be thoroughly cleaned to extract all of the oil and brine and then be properly dried, with the exception of saturation measurements for the determination of porosity. This is generally carried through flushing, flowing or contacting with various solvents to extract hydrocarbons, water and brine. 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).

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Measurement of collected water

Reservoir Engineering

Core plug

Figure 5 Dean Stark Apparatus

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 drying can be used to clear core containing delicate clays like illite (see Phase Behaviour chapter - section 8.1).

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Rock Properties Measurement E I G H T

3. POROSITY MEASUREMENTS 3.1 Methods

Figure 6 illustrates the methods used for routine determination of porosity.

Vacuum Gauge

Valve

Displacement Pump

Boyles Law Porosimeter Pore Volume Determination

Water

Oil

Gas

Porosity

Pressure Gauge Outlet Valve

Gas Inlet Valve Sight Glass

Sample in Place, Stopcock Open

Core Sample

Micrometer Scale

Mercury

Plunger

Washburn Porosimeter

Sample Chamber

Reference Volume

Valve

Kobe Porosimeter

Pressure Gauge Valve

Grain Volume Determination

Resaturation

Figure 6 Porosity measurement methods (Corelab)

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(a) Bulk Volume 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.

Adjustable fork

Thermometer

Reference mark Core plug Mercury vessel

Single pan balance _ + 0.01 gm

Weighted base

Figure 7 Archimedes mercury immersion apparatus (API)2

Pressure read-out

Volume read-out Sample chamber

Displacement plunger

Figure 8 Volumetric mercury displacement pump (API)2

(b) Summation of fluids 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.

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Rock Properties Measurement E I G H T

Thermocouple

Insulated Oven Heating Elements Sample Cup Screen Condensing Tube

Water Bath

Water Inlet

Temperature Controller

Receiving Tube

Figure 9 Oven retort (API)2

(c) Gas transfer (i) The Boyle's 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. P2

Sample chamber

Reference volume

P1

Gas in Pressure regulator

Figure 10 Boyle's law porosimeter 22/07/14

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(ii) The Washburn-Bunting method involves the vacuum extraction and collection of the gas contained in the pores of a prepared sample. The method measures pore volume. (d) Liquid resaturation 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. (e) Grain density 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.

3.2 Whole core versus conventional sidewall samples

As well as coring using a coring bit and core barrel, it is also possible to recover samples of the formation using wireline tools, these are termed sidewall coring. There are two types of sidewall coring devices. One is based on exploding a core plug shaped piece into the formation. Clearly samples recovered by this technique may be suitable for mineral description but are not so suited to porosity and permeability analysis as a result of the damage generated by the explosive force of the sampling device. Sidewall corers which cut into the formation do not suffer from such mechanical damage. 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 regime 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 transducers has made it possible to monitor these transient flow conditions. The other approach

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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.

4.2 Impact of Stress

Over recent years the impact of reservoir stresses on rock properties and therefore the interest in measuring rock properties under realistic stresses has grown in particular in relation to permeability. Stress effects also have an impact on other properties including porosity . In describing the various approaches to permeability measurement we will also look at various procedures for imposing stress on the samples. In Figure 19 of the previous chapter we identified the various stress directions in the context of permeability measurement. Figure 11 illustrates the core recovered from a vertical well and the natural stresses imposed. It is important to distinguish the different possible stress loadings that can be applied to core plugs and also the configuration of the stresses in the natural state. In the natural state the stresses can be considered to be resolved in three principal directions. The vertical direction being the major principal stress and the two horizontal directions the two minor principal stresses. Figure 11a

Core plug for horizontal k measurement

Major principal stress

Core plug for vertical k measurement

Whole core

Minor principal stresses

4 Inch

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.

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Major principal stress Major principal stress

Minor Minor principal principal stress stress

Minor principal stress Minor principal stress Major principal stress Major principal stress

Figure 11(b) Stress orientation for horizontal core plug. Major principal stress Major principal stress

Minor principal stress Minor principal stress

Figure 11(c) Stress orientation from vertical core plug

In requesting reservoir stresses to be applied to core plug measurements it is important to examine that the stresses applied actually represent those which the rock would be subjected to in the formation. The various modes of stressing a rock are shown in Figure 12 a-d 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).

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A

s1

D

L Sample

Isostatic stress

s1

s1

L

s1

D

Figure 12(a) Isostatic Stress

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 L3

L2

Figure 12(b) Triaxial Stress

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

Biaxial stress

σ2

σ2

L

σ1

D

Figure 12(c) Biaxial Stress 22/07/14

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Uniaxial Strain. Figure 12d. Uniaxial strain compression is a special case of biaxial stress loading; the stress applied to the circumference is just sufficient to maintain the diameter constant as the stress parallel to the cylinder axis is increased. Strain occurs only parallel to the axis of the cylinder. s1 L Sample

Uniaxial stress

s2

s2 L

s1

D

Figure 12(d) Uniaxial Stress

4.3 Steady State Permeability Methods

The most conventional permeability measurement approach has been to use the measurement of the pressure drop associated with a fixed flow rate. To determine specific permeability nitrogen or air is usually caused to flow through a prepared sample of measured dimensions. The pressure differential and flow rates are measured and the permeability calculated from the Darcy equation. A schematic set up is shown in the sketch below, Figure 13 End view showing radial stress Pressure transducer P1

Differential Pressure.

p

_

+

L Pressure regulator

P2

D

qr @ Pr, Tr

Pa

Flow meter

Sample holder

Figure 13 Schematic of steady state permeability measurement 2

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.

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Rock Properties Measurement E I G H T

Figure 14 shows a high pressure core holder designed to impose reservoir stresses. The slideable inlet tube enables the strain of the stressed core to be taken up. The stress loading for this arrangement is isostatic.

Cavity for Hydraulic Oil Slidable Inlet to Produce Confining Rubber Tube Sleeve Stresses

End Plug

Inlet Port for Confining Oil

Cylindrical Core Plug

Outlet Flow Tube Retaining Ring

End Plug

Figure 14 High pressure core holder for stress condition, isostatic 2

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.

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Inlet Flow Port

Rubber Sleeve

Port for Oil to Produce Radial Confining Stress, or Vacuum to Dilate Sleeve

Core Plug

Reach Rod, 4X

Outlet Ports

Port for Oil to Produce Radial Confining Stress, or Vacuum to Dilate Sleeve Large Piston of Axial Stress Intensifier Cavity for High Pressure Nitrogen for Axial Stress N2

Figure 15 High pressure core holder or biaxial loading 1

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 more realistic reservoir stress conditions.

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Rock Properties Measurement E I G H T

Platen Threaded end cap

Trapped tube

A

A Core

Rubber sleeve

Aluminium cell body

Hydraulically pressured tubes

Maximum principal stress

s3 3

3

3

3

2

2

1

s2 1

3

2

3

2

2

3

2 1

s2

1

2

2

3

3

3

3

3

Face of core plug

s3

Section AA

Figure 16 True trixial cell

Although liquids could be used in permeability measurements it is common to use a gas. Gas permeabilities need to be corrected for the Klinkenberg effect and reported as equivalent liquid permeabilities. 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.

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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 examination of the various levels of permeability measurement, (upscaling), have demonstrated the value of being able to measure the permeability over a small area which the probe permeameter affords. Figure 17 shows an arrangement of a typical probe permeameter. As well as back pack mounted version for use in outcrop studies they can also be laboratory mounted and can automatically scan the permeability variations in a slab of rock. Flow meter Pressure regulators

Pressure transducer

Rock being examined

ri

ro

Figure 17 Schematic of steady state probe permeameter

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.

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Rock Properties Measurement E I G H T

Rubber Gaskets

Calibrated Gas Burette

P1

Mercury Manometer

rw

Springs

re

P2 Pivot Ball Piston

L

Regulators Air Supply

Figure 18 Radial flow steady state permeameter 2

4.4 Unsteady State Permeability Measurements

The advent of high speed computers and data acquisition systems has enabled the application of unsteady state permeability measurements. The principles are similar to the behaviour of a well during a well test and the analysis of the pressures during the unsteady state draw down or build-up period. Figure 19 gives a schematic of a pressure-fall off system. An upstream gas reservoir of different volumes, to accommodate a wide range of permeabilities, is pressured and then released to atmosphere via flow through the core. The pressure just upstream of the core is accurately monitored. Full details of the calculation procedure presented by Jones are given in the API RP40 practise document. 2

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VT Fill

Vent

P1

VP

Pc

Hydrostatic confining pressure

Figure 19 Schematic of pressure - fall of gas permeameter 2

In the pulse decay method for permeability measurement a configuration of equipment is as shown in Figure 20. It consists of an upstream and downstream reservoir. The two gas reservoirs are filled to a pressure. When equilibrium is reached with all valves open, the joining valves are closed and the pressure in the upstream gas reservoir is increased by 2-3% of the pressure set in the vessels. The valve 1 is then opened and the pressure time behaviour of the transient flow behaviour is monitored. This procedure lends itself to very low permeability values, 0.1-millidarcies to 0.01 microdarcies. Calculation procedures are also given in the API practice document. Valve 2

Fill/vac.

+_

V1

Valve 1

p

P2 V2

VP Pc

Figure 20 Pulse decay apparatus axial flow of gas

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 surface prior to releasing the pressure the core in its container is frozen. It is then shipped and stored in 22

Rock Properties Measurement E I G H T

a frozen state. During controlled thawing of the core the fluids produced and retained enable downhole saturation to be obtained.

5.1 Gas saturation

Conventional and sidewall core samples have gas saturation measured by injecting mercury into the gas filled portions of the pores. The gas is compressed into a small volume or forced into solution in the liquids in the pores using a mercury pump. Measurement of the volume of mercury penetrated is a measure of the gas content of the sample.

5.2 Oil saturation by retort

Oil distilled at atmospheric pressure gives a measure of the oil content of the plug. The distillate is collected in a calibrated receiver. Temperatures up to 6500C are used (Figure 9).

5.3 Water saturation

Samples can have their water content determined by atmospheric distillation concurrently with the oil content determination. A distinction should be made between the pore water and the water of hydration or crystallisation. Water saturation can also be measured by a solvent refluxing method (Dean-Stark) (Figure 20). Toluene is the most commonly used solvent. The oil content of the sample is obtained by difference of the weight of the sample before and after extraction and drying less the weight of the water removed during solvent extraction.

Measurement of collected water

Core plug

Figure 21 Dean Stark Apparatus

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6. CAPILLARY PRESSURE 6.1 Introduction

The general laboratory procedure for capillary pressures is 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. The volume of the water expelled represents the lower capillary presure pores, so gives an indication of the number of pores still to be emptied at what must be a greater displacing 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, Pc =

2σ cos θ r

the curve can be calibrated to represent pore size versus percentage of pores less than the subject pore size.

6.2 Capillary Pressure Measurement Techniques

There are four main methods for capillary pressure measurement (i) Desaturation or displacement through a porous diaphragm. (ii) Centrifuge or centrifugal method. (iii) Dynamic capillary pressure method. (iv) Mercury injection method.

6.2.1 Porous Diaphragm (Figure 22)

In the porous diaphragm method there is a permeable membrane of uniform pore size distribution containing pores of such a size that the selected displacing fluid will not penetrate the diaphragm when the pressures applied to the displacing phase are below some selected maximum pressure of investigation. Pressure applied to the assembly is increased by small increments. The core is allowed to approach a state of static equilibrium at each pressure level. The saturation of the core is calculated at each point defining the capillary pressure curve. Any combination of fluids can be used: gas, oil and/or water.

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Rock Properties Measurement E I G H T

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. Nitrogen Pressure Saran Tube Crude Oil Neoprene Stopper

Nickel-Plated Spring Core Kleenex Paper

Diaphragm, e.g. ultra-fine fritted glass disk

Scale of Squared Paper

Seal of Red Oil

Brine

Figure 22 Porous diaphragm capillary-pressure system

6.2.2 Centrifuge method ( Figure 23)

The high accelerations in a centrifuge increase the field of force on a sample subjecting it to an increased gravitational force. The core plug is mounted in a modified centrifuge tube as shown and the desaturation of the sample is monitored with a strobe light. 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. Seal Cap

O-Ring

Core Holder Body

Core

Support Disk

Window

Tube Body

Figure 23 Centrifuge for determination of capillary pressure curves 5

6.2.3 Dynamic method ( Figure 24)

A dynamic method has been used where a simultaneous steady-state flow of two fluids is established in the core. The saturation is varied by regulating the quantity of each fluid entering the core and the pressure difference between the two fluids gives the capillary pressure.

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Gas outlet

po

Reservoir Engineering

Gas inlet pg

pc

Core

To atmosphere Oil burette

Oil inlet Porcelain plate

Figure 24 Dynamic capillary pressure equipment 5

6.2.4 Mercury Injection ( Figure 25)

The most common procedure for determination of capillary pressure is using mercury injection. The procedure was developed to accelerate the determination of the capillary pressure-saturation relationship. Mercury is the non-wetting fluid. The core sample is inserted into the mercury chamber of a mercury pump or a mercury porosimeter and evacuated. Mercury is then injected into the core under pressure. The volume of mercury injected at each pressure determines the non-wetting phase saturation. This procedure is continued until the core sample is filled with mercury or the injection pressure reaches some predetermined value. The procedure is used in a number of industries to determine the pore size characteristics of the porous media. 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.

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Rock Properties Measurement E I G H T

0-200 psi Pressure Guage 0-2,000 psi Pressure Guage

Regulating Valve Lucite Window

Lucite Window

To Atmosphere

Cylinder

U-Tube Manometer

Figure 25 Mercury injection porosimeter 5

6.3 Use of Laboratory Capillary Pressure Data for Reservoir Saturation Distribution

As we have noted above, laboratory capillary pressure tests can be made with a variety of fluids that differ from reservoir fluids. It is necessary therefore to convert laboratory based results to be applicable to the field where the fluids might be different. We will examine the procedure for converting air-mercury data to water-oil data for application in field determinations of saturation profiles. As shown previously, capillary pressure saturation data can be converted to height saturation data:

h=

Pc ( ρw − ρo )g

(1)

Air/mercury capillary pressure curves are comparable in shape to air/brine or oil/ brine capillary pressure curves. When converting capillary pressure curves to an equivalent height, the difference in interfacial tension and contact angle between the laboratory and reservoir systems must be accounted for. For example

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surface tension (s) of water = 70 dynes/cm surface tension (s) of mercury= 480 dynes/cm contact angle (q) water/solid = 0 degrees contact angle (q) mercury/solid = 140 degrees Pc =

2σcos θ r



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At corresponding saturations therefore



Pcair / mercury 480cos140 = ≅5 Pcair / water 70cos0

Pc air/mercury = 5 Pc air/water

(3)

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. Pc air / mercury = 10 Pc water / oil

(4)

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.



PcL (σ cos θ )R (σ cos θ ) 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) (sCosq)R = interfacial tension cosine of the contact angle (initial reservoir conditions) (sCosq)L = interfacial tension cosine of the contact angle (laboratory conditions) rw = density of water at initial reservoir conditions rh = density of hydrocarbon at initial reservoir conditions It should be noted that the interfacial tension of an oil/water system is approximately 10 times greater than that for an oil/gas system and that consequently capillary forces are more important for the former system.

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Rock Properties Measurement E I G H T

EXERCISE 1 Calculation of water saturation distribution in a layered reservoir. 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 1. Determine the Free Water level and locate it on Figure E2. 2. Construct the water saturation profile. 3. Estimate permeabilities 4. Which intervals would you recommend for completion based on the criteria Sw<50% and k>0.1mD? 5. What is the net pay (cumulative thickness having Sw>50%)?

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h (ft)

(mD) 35 Pc. (%) 15 (psi)

15 10

.2 10

Reservoir Engineering

5 5

.02 5

250

type A rock type B rock 200

150

100

50

0

0

50 Pore space unoccupied by mercury Figure E1 Capillary pressure curves from nearby well

30

100%

Rock Properties Measurement E I G H T

Rock type

Porosity 10

h (ft) 15% (1 cm for 10 ft)

5

0

Saturations Oil

100

Water

Unit No.

100% 0

k (mD)

A 8% B

12%

10%

A 15%

B

9%

6%

A

5% 8% 13%

B

10%

A 14%

B

10%

100 Sw in B type rock found at this level 22/07/14

15%

Figure E2 Opposite Institute of Petroleum Engineering, Heriot-Watt University

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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

Figure 26 Model of flow for Leverett J Function

The laminar flow of fluid through a pipe is given by Poiseuille’s equation:

q=

πr 4 ∆P 8µL cap

(6)

For n tubes

qn =

nπr 4 ∆P 8µL cap

(7)

The porosity of the bundle of tubes is

nπr 2 φ= A

(8)

and the permeability is

32

k=

qµL core A∆P

(9)

Rock Properties Measurement E I G H T

If φA is substituted for nπr and then

r2 =



8k L cap 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)

and substituting for r in the definition of capillary pressure gives:

Pc =

2σcos θ k  constant   φ

or

1 2



(12)

1

k 2 Pc   φ 1 = =J constant σ cos θ



(13)

Sometimes the J function is written without the cos q term. The capillary pressure measurements can therefore be normalised for differences in permeabilities, porosities and fluids and used to measure the capillary pressure, i.e. the J function is obtained independent of k, f, s and q. 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. CAPILLIARY PRESSURE vs WATER SATURATION (Sw)

Sample No. 1 2 3 4 5 6 7 8 9

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Permeability mD 0.55 76.10 3.59 0.85 2.82 1,100.00 649.00 457.00 58.10

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Porosity % 15.2 27.4 20.2 20.3 22.0 25.3 23.2 22.4 19.7

33

Petroleum Engineering

Reservoir Engineering

17 16 15 14 13 12 11 7 6 8

9

2

5

3

1

4

Pc (PSIG)

10 9 8 7 6 5 4 3 2 1 0

10

20

30

40

50

Sw %

60

70

80

90

100

Figure 27 Set of capillary pressure curves

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 E I G H T

1200

1100

1000

900

800

_ 1 k Pc( _ )2

700

600

500

400

300

200

100

10

10

20

30

40

50

Sw %

60

70

80

90

100

Figure 28 Leverett J Function

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

(Sw − Swc ) a better (1 − Swc )

correlation is obtained suggesting that the Swc reflects the tortuosity variations within the various rocks (Figure 29).

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Reservoir Engineering

32

LEGEND

30 Different reservoir sand sequence in a formation

28 26

Dimensionless Capillary Pressure Pc σ

K φ

24 22 20 18 16 14 12 10 8 6 4 2 0

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

(

0.9

Normalised Wetting Phase Saturation Sw* = Sw-Swc 1-Swc

1.0

(

Figure 29 Modified Leverett J Function Curves

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 E I G H T

In the unsteady state method, a displacement process is set up where one displaces another and the rates and pressure drops are monitored as a function of time rate process. The saturations are obtained by calculating the remaining for a volumes of the respective It is more 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. Oil recycle system Differential pressure transducer ∆P

Differential pressure transducer ∆P

Oil Brine

Composite core

Oil - water separator and production monitor

Brine recycle system

Pressure control system

Figure 30 Steady state relative permeability

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.

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SOLUTION TO EXERCISE EXERCISE 1 – Calculation of water saturation distribution in a layered reservoir. 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 1.

Determine the Free Water level and locate it on Figure E2.

2.

Construct the water saturation profile.

3.

Estimate permeabilities

4.

Which intervals would you recommend for completion based on the criteria Sw<50% and k<0.1mD.

What is the net pay (cumulative thickness having Sw<50%). SOLUTION 1. The first step is to convert the air-mercury capillary pressure data to oil-water. Pc air/mercury = 10Pc water/oil (equation 4, page 26) PcR = h (rw - ro) g (equation 5, page 27) Conversion values:

Pc



38

(

lb f in 2

(



Pc(air/mercury) = 10 Pc (water oil) -

lb f 144in 2 lbm = h ftx (1.03 − 0.8) x 62.4 3 xg 2 2 in ft ft

 lb 144in 2  lbm Pc oil / water  2f  = h( ft ) x (1.03 − 0.8) 62.4 3 xg 2  in ft  ft

Reservoir Engineering

Rock Properties Measurement E I G H T



1 lbf = 1 lbm xg Pc oil/water psi = 0.1 psi/ft oil/water Therefore Pc air/mercury = 1 psi/ft oil/water The capillary pressure curves can now be rescaled. Figure E3. Plotting hft = Pc air/mercury (psi) versus 0 - 100 % water saturation. 2. Free water level

This occurs in rock type B. f = 15%. From capillary pressure curve 100% water saturation at 15 psi i.e. 15 ft.

Free water level is 15 ft below this position, as indicated on Figure E4.

The free water level now provides the basis for the water saturation profile determination.

3. Water Saturation Profile

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% f to type B 10% f



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. 5. Completion intervals according to the criteria Sw<50% and k>0.1mD are shaded on the Figure E4. 6. Net pay adds up to around 125ft.

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250'

h (lt)

(mD) 35 Pc. (%) 15 (psi)

15 10

.2 10

Reservoir Engineering

5 5

.02 5

250

type A rock type B rock 200'

200

150'

150

100'

100

50'

50 40' 20'

0'

0

20' 15 psi 0

50 Water saturation

Figure E3 Capillary pressure curves from nearby well

40

100%

Rock Properties Measurement E I G H T

Rock type

h (ft)

15%

Porosity 10

5

230

200

A

150

B

A

100

B

50

Water

100% WL

0

k (mD)

0.02

16

0.08

15

21

14

15

13

0.2

12

35

11

12

10

0.12

9

0.03

8

0.02

7

10

6

26

5

15

3

10 mm

15'

100

100%

17

4

A

B

Saturations Oil

Unit No.

A

B

0

0.2 32

2

15

1

35

FWL Figure E4

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REFERENCES 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

Reservoir Engineering

Permeability – Its Variations N I N E

Gas Limited flaring Re-inject qo sales

Separation

Injection Pump qwi

qo + qwp

qwp Purify / Dump / inject

Sea Level Permeability Distributions

10

Seawater for injection

20

30

40 Log (Permeability mD)

20 40 60

Sea Bed

Normal log scale

80 Thickness (ft)

Reservoir

1000 2000 3000 4000 5000 Permeability (mD) 20 40 60

Linear scale

80 Thickness (ft)

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Petroleum Engineering

Permeability – Its Variations N I N E

C O N T E N T S 1 INTRODUCTION 2 AVERAGE PERMEABILITIES FOR SEVERAL LAYERS 2.1 Beds in Parallel 2.2 Layers in Series – Linear Flow 3 MODELLING HETEROGENEOUS SYSTEMS

Reservoir Engineering 22/07/14

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LEARNING OBJECTIVES Having worked through this chapter the Student will be able to: • Appreciate and understand that permeability is an anisotropic property. • Derive equations to enable the calculation of average permeability for: layers in parallel-linear and radial flow, layers in series – linear and radial flow. • Describe briefly the impact of layered reservoirs in the context of modeling water oil displacement in heterogeneous reservoirs.

2

Reservoir Engineering

Permeability – Its Variations N I N E

1 INTRODUCTION Although we have defined permeability as a rock property, seldom, if ever, is a homogeneous reservoir encountered in actual practice. In many cases the reservoir will be found to contain several distinct units or layers of varying rock properties. Even on a local scale the value of permeability is not necessarily the same in all directions. Permeability is an anisotropic property, (Figure 1) i.e. its value is dependent on direction. Porosity is an isotropic property however. Ky

Kx

Kz

Figure 1 Permeability Anisotropic Property.

The sedimentary nature of rocks is such that vertical permeability is less than horizontal 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 macroscale behaviour. Figure 2 represents the variation in permeability observed using a mini-permeameter in samples from the Leman gas field.

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2

1.0 15.5

Scale 1 (cm) 0

1.0

1.0

12.0

17.5

2.5

>0.5

36.5

"Sample plug" Poros 17.5% Perm. 19 mD

15.0

21.0

Location of minipermeability measurement Permeability in mD

b c d

1.0 <0.5

a

e

38.5 f >0.5 20.0

g h

Figure 2 Effect of scale of observation and measurement on permeability data from a Rottliegende aeolian sand cross bed set in the Leman gas field 1.

Although the conditions for use of Darcy's Law state that the rock should be homogeneous 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.

4

Permeability – Its Variations N I N E

Statistical Analysis of Rock Property Evaluation K

Log Permeability

K

300

180

25

160

200

Figure 3 Statistical Analysis of Rock Property Evaluation

Waren and Price2 demonstrated that the most probable behaviour of a heterogeneous systems tends towards that of the geometric mean.

1/ n

kG = (k1 x k2 x k3 ...kn )

(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.

2 AVERAGE PERMEABILITIES FOR SEVERAL LAYERS Simple geometry systems can be dealt with as follows:

2.1 Beds in Parallel

The horizontal system reflects the sedimentary nature of the rock: the rock material may have been segregated as it was deposited giving different sizes, shapes etc. to different layers in the formation. The reason for determining an average permeability is in rationalising the permeability measured on small samples in the laboratory with the measurements made for example by well test analysis. Well test analysis cannot test small sections of the reservoir (which would be uneconomical) on the same scale as the laboratory tests. The results are therefore representative of flow through several layers rather than only one.

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Linear Flow

Consider the simple linear beds in parallel. Figure 4. P2

P1 Q1

A1

Q2

A2

Q3

A3

K1

h1 K2

h2 K3

h3

L

Figure 4 Linear Flow in Parallel

The average permeability can be developed using the Darcy flow equation:

QT = Q1 + Q2 + Q3

(2)

k A (P − P ) k A (P − P ) Q1 = 1 1 1 2 2 2 1 2 ...etc. (3) µL Q 2 = µL (4) QT = k Ai ( P1 − P2 ) / µL

∑

k1 A1 ( P1 − P2 ) k2 A2 ( P1 − P2 ) k3 A3 ( P1 − P2 ) + + µL µL µL



=



k=

∑k A ∑A i

i

i



(5)

(6)

If all the beds have the same width the A ∝ h so k is the arithmetic average:

kA =

∑k h ∑h i i i

(7)

This equation is commonly used to determine the average permeability of a reservoir from core analysis.

Radial Circular Flow

This is the case of several superimposed layers flowing simultaneously in the well. Each layer supplies a rate of Qi. The total rate of flow is QT = Σ Qi. In Figure 5:

6

Permeability – Its Variations N I N E

re

rw

pe

pw h1

Q1 k1

h2

Q2 k2

h3

Q3 k3

hT

Figure 5 Radial Flow in Parallel

Qi = Q = ∑ Qi =



k=

2πhi ki ( Pe − Pw ) (8) re µ ln rw

2 πhT k ( Pe − Pw ) 2 π( Pe − Pw ) ( k1h1 + k2 h2 + k3h3 + .......) = r r µ ln e µ ln e rw rw

∑h k hT

i i

(9)

(10)

This value can be compared with that obtained through well flow tests or pressure build-up tests.

2.2 Layers in Series - Linear Flow

In this case the reservoir may have been severely folded or faulted and the originally horizontal layers are now vertical. It is assumed that the flow is now through each of the layers towards the well. In this case, assuming a three layered system, the total flow rate is constant through all of the layers and the total pressure drop is now the sum of the pressure drops across each layer. Similarly, the total length is the sum of the lengths of the individual layers, and the area open to flow is constant. Figure 6 illustrates.

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P1

P2

Reservoir Engineering

P4

P3

Q A K1

K2

K3

L1

L2

L3

Figure 6 Linear Flow in Series

The average permeability of linear beds in series is obtained by adding the pressure drop across each bed.

(P1 - P4) = (P1 - P2) + (P2 - P3) + (P3 - P4)



P1 − P4 =

QµL1 QµL2 QµL3 QµL + + = k1 A1 k2 A2 k3 A3 kA

(11)

(12)

For beds of equal cross-sectional area then:

L

(13) k= Li ki

∑

This mean is the harmonic average permeability kH.

Radial Circular Flow

In addition to natural lateral variations in permeability, wellbore damage can reduce the permeability in the vicinity of the wellbore; also cleaning techniques, such as acidising can increase the permeability in the vicinity of the wellbore. For example in Figure 7:

8

Permeability – Its Variations N I N E

re

r1 rw pw k1

P1

k2

Pe

Figure 7 Radial Flow in Series

Q1 =

Q2 =

2πk 1 h( P1 − Pw ) r µ ln 1 rw

(14)

2πk2 h( Pe − P1 ) r µ ln e rw

(15)

Total flow:

2πkavg h (Pe − Pw ) r µ ln e rw



QT =



Total pressure drop:



(16)

i.e.



22/07/14

( Pe − Pw ) = ( Pe − P1 ) + ( P1 − Pw ) QT µ ln 2πk

re r r Q2 µ ln e Q1µ ln 1 rw rw r1 = + 2πk2 h 2πk1h h



At steady-state flow:



QT = Q1 = Q2

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(17)

(18)

(19) 9

Petroleum Engineering



ln re / rw ln r1 / rw ln re / r1 = + k k1 k2



k =

ln re / rw ln r1 / rw ln re / r1 + k1 k2

Reservoir Engineering



(20)



(21)

3 MODELLING HETEROGENEOUS SYSTEMS Permeability variations can often be traced from well to well throughout the reservoir, thereby 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 10

Tarbert

Upper ness

Etive

Rannoch

9 8 7

Permeability (Darcies)

6 5 4 3 2 1 9200

9300

9400

Measured Depth (Feet)

9500

Figure 8 Permeability Variation in North Sea Field

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: • • •

10

Reservoir sands are homogeneous Random variations in reservoir properties occur across the sand Ordered distributions of the properties observed in one well do not correlate with other wells.

Permeability – Its Variations N I N E

Representation of permeabilities as a function of depth used to be presented on a log permeability -versus- depth scale. Such a representation can lead to an observation that an average permeability might be appropriate when in reality when presented in a true linear form the differences in permeability are more distinct. Fluid behaviour in reservoirs obeys a linear law, as against log permeability! 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 Normal log scale

60 80 Thickness (ft)

1000 2000 3000 4000 5000 Permeability (mD) 20 40 60

Linear scale

80 Thickness (ft)

Figure 9 Permeability Distribution on a Log and Linear Scale.

The difference in, for example, the behaviour of a waterflood when modelled as a homogeneous system as against a layered system is illustrated as follows. This process is shown schematically by Figure 10. The significance in these simplifications is demonstrated in the context of modelling the behaviour of a waterflood, when modelling as a homogeneous system as against a layered system. The illustration presented in the subsequent example is that of the waterflooding of a reservoir with the permeability contrasts as illustrated in Figure 9. Figure 10 illustrates the process where injected water sweeps through the reservoir. In the example given the mobility ratio of the process is less than 1. That is the injected water is less mobile than the displaced oil. If the reservoir was homogeneous then the permeability can be considered as an average value and the displacement simulated on a one dimensional basis. The key issues for the reservoir engineer are: when does the water injected arrive at the producing wells and secondly what is the oil recovery. 22/07/14

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Gas Limited flaring Re-inject qo sales

Separation

Injection Pump qwi

qwp Purify / Dump / inject

qo + qwp

Sea Level

Seawater for injection

Sea Bed

Reservoir

Figure 10 Water Injection Process

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.

_ k average

Water

Figure 11 1-D Displacement Process

12

Oil

Permeability – Its Variations N I N E

WATER CUT 1.0 0.8 0.6

1D

0.4 0.2 0

2000

4000

6000

4000

6000

TIME (days)

OIL RATE 1.0 1D 0.5

0

2000

RECOVERY FACTOR 0.5 0.4 0.3

TIME (days)

49.7%

1D

0.2 0.1 0

2000

4000

6000

TIME (days)

Figure 12 Oil Rate, Water Cut and Recovery

Figure 13 a,b, c and d shows the process for a layered case where the permeability variations have been included in the simulation and the process is now two dimensional. The permeability distribution is as in Figure 9 with the highest permeability in the centre. Figure 13a shows the simulation after 600 days and the values on the contour lines are values of water saturation. The Figure clearly identifies the water moving through the high permeability zone. In Figure 13b after 1200 days water has broken through into the producing well and this well is 'cutting' water over a short interval. After 2400 days, Figure 13c, the high permeability layer is taking the majority of the water and a larger interval of the producing well is 'cutting' water.

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Reservoir Engineering

0 20 40 60

50 70

600 Days

50 1000

0

2000

3000

µw = 0.5 cp kv = kh

50

20 70

1200 Days

40 50 60 0 50% 20 70%

2400 Days

40 60

Impact of gravity 50% 1000

0

2000 50%

3000

µw = 0.5 cp kv = kh

Attic oil

20 70% 40

4800 Days 50%

60

Figure 13 (a,b,c and d) Displacement simulation in layered reservoir

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 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.

14

Permeability – Its Variations N I N E

WATER CUT 1.0 0.8 0.6 0.4

1D

2D

0.2 0

2000

4000

6000

µw = 0.5 cp

OIL RATE 1.0

2D =

2000

RECOVERY FACTOR 0.5

0.3

kv = kh kv = 0.1kh

2D

0

0.4

{

1D =

1D 0.5

TIME (days)

4000

6000

49.7%

1D

TIME (days)

48.8% 2D

39.0%

0.2 0.1 0

2000

4000

6000

TIME (days)

Figure 14 Predictions for Homogeneous 1D and Heterogeneous Flooding 2D

Clearly the main impact is that the field starts producing water after around 1200 days around half the time of the 1D prediction and the recovery at breakthrough is only around 25%. The example clearly demonstrates that for these heterogeneous systems with large permeability contrasts, during water flooding water flows preferentially down the high permeability section. Premature water breakthrough occurs resulting in deferment of oil production anticipated from an average value of permeability.

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Reservoir Engineering

With such a result the reaction is what can be done to improve the process? The immediate 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 and flows 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. 0 20 40

1000

2000

3000

With high permeability layers not complete

50% 70% 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. 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 geological 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.

16

Permeability – Its Variations N I N E

Coarser Sediment in Shallow Turbulent Water

Sea Level

Fine Sediment in Deeper Quiet Water

Increasing Increasing Wave Wave EnergyEnergy

Sea Level

Fine Sediment in Deeper Quiet Water

Coarser Sediment in Shallow Turbulent Water ain

e

Siz

n itio os eSpize D sin n oura Inc itioAdvance neG os of Bar ltiang ep s u a D e s Sirm Advance ou Inc ne of Bar lta u Sim r gG

in

s rea

GR Profile (A) GR Profile (A)

Permeability

GR Profile (B)

Injection Injection

Permeability Depth Depth

GR Profile (B)

Oil Oil

Water Water

Production Production

(a) Favorable (a) Favorable

Figure 16(a) Effect of Favourable Permeability in Waterflooding. Levee

Channel Migration

nk nk Ba Ba n of n of ErosioErosio

Channel Migration

Permeability

Depth Depth

Permeability

Strong Current Strong rse a Current Co e ars Co

Weak Current n Weak sitio o ep Current D n nd itio Sa os ep D nd Sa

e Levee

Fin

e

Growth Fin of Point Bar

tion ) e cre Ac e Lin Growthalof (Tim n r tio ) e t e r Point Bar La face Acc Line r Su l me a (Ti ter La face r Su

GR GR

Injection Injection Oil Oil

(a) Unfavorable (a) Unfavorable

Water Water

Production Production

Figure 16(b) Effect of Unfavourable Permeability in Waterflooding 3

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REFERENCES 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 2. Warren, J.E., Price, K. S., "Flow in Heterogeneous Porous Media" Society of Petroleum Engineering Journal 1961. 153 - 169. 3. Archer, J. S., Wall, C. C., Petroleum Engineering. Graham and Trotman 1988 London

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Fluid Flow in Porous Media T E N

10 vertical tubes, 100mm diameter, arranged linearly Tube number 1

2

3

4

5

6

7

8

9

10 Initial water profile

Outlet (constant flowrate) Profile after time t

Interconnecting, small diameter pipes

height of water in tubes

top of tubes

0 -50

time, t after start of flow

-100

t=0 t=1 t=2 t=3 t=4 t=5 t=6 t=7

-150 -200

bottom of tubes -250 1

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3

4

7 6 5 tube number

8

9

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Fluid Flow in Porous Media T E N

C O N T E N T S 1 INTRODUCTION 2 CHARACTERISATION AND MODELLING OF FLOW PATTERNS 2.1 Idealised Flow Patterns 2.2 General Case 2.2.1 Linear Horizontal Model of a Single Phase Fluid 2.2.1.1 Linearisation Of Partial Differential Flow Equation For Linear Flow 2.2.1.2 Conditions of Solution 2.2.2 The Radial Model 2.2.2.1 Range Of Application And Conditions Of Solution 2.3 Characterisation of the Flow Regimes by their Dependence on Time

4 THE CONSTANT TERMINAL PRESSURE SOLUTION 5 SUPERPOSITION 5.1 Effects of Multiple Wells 5.2 Principle of Superposition and Approximation of Variable – Rate Pressure Histories 5.3 Effects of Rate Changes 5.4 Simulating Boundary Effects (Image Wells) 6 SUMMARY SOLUTIONS TO EXERCISES

3 BASIC SOLUTIONS OF THE CONSTANT TERMINAL RATE CASE FOR RADIAL MODELS 3.1 The Steady State Solution 3.2 Non-Steady State Flow Regimes and Dimensionless Variables 3.3 Unsteady State Solution 3.3.1 General Considerations 3.3.2 Hurst and Van Everdingen Solution 3.3.3 The Line Source Solution 3.3.3.1 Range of Application and Limitations to Use 3.3.4 The Skin Factor 3.4 Semi-Steady-State Solution 3.4.1 Using The Initial Reservoir Pressure, Pi 3.4.2 Generalised Reservoir Geometry: Flowing Equation under Semi-Steady State Conditions 3.5 The Application of the CTR Solution in Well Testing

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LEARNING OBJECTIVES: Having worked through this chapter the student will be able to:

2

•

Understand the nature of fluid flow in a porous medium and the relation between time, position and saturation

•

Understand the assumptions used in the derivation of the diffusivity equation

•

Describe the characterisation of the reservoir flow regime on the basis of time

•

Apply the solutions of the diffusivity equation to steady state flow, semi-steady state flow and transient flow

•

Calculate the pressure in a reservoir at a specific radius and a specific time under transient flow conditions

•

Calculate the effect of multiple wells and multiple flow rates on reservoir pressure

•

From measured bottom hole pressures use the line source solution to determine the reservoir permeability and skin factor

•

Describe the application of semi-steady state solutions to determine reservoir boundaries and their influence on flow rates.

Reservoir Engineering

Fluid Flow in Porous Media T E N

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 radial reservoir with a well at the left side (Figure 1). The model represents a radial slice of reservoir from wellbore to external boundary.

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10 vertical tubes, 100mm diameter, arranged linearly Tube number 1

2

3

4

5

6

7

8

9

10 Initial water profile

Outlet (constant flowrate) Profile after time t

Interconnecting, small diameter pipes

height of water in tubes

top of tubes

0 -50

time, t after start of flow

-100

t=0 t=1 t=2 t=3 t=4 t=5 t=6 t=7

-150 -200

bottom of tubes -250 1

2

3

4

7 6 5 tube number

8

9

10

Figure 1 Model of a radial 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 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. 4

Fluid Flow in Porous Media T E N

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 semisteady 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.

2

CHARACTERISATION AND MODELLING OF FLOW PATTERNS

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 There are essentially two possibilities available to cope with complexities of actual flow properties. (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. 22/07/14

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(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. The analytical solutions will be examined in this chapter.

2.1 Idealised Flow Patterns

There are a number of idealised flow patterns representing fluid flow in a reservoir: linear, radial, hemispherical, spherical. The most important cases are the linear and radial models since both of them can be used to describe the water encroachment from an aquifer into a reservoir, and the radial model can be used to describe the flow of fluid around the wellbore. 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.

2.2 General Case

Consider the co-ordinate system shown in Figure 2. The X and Y coordinates form a horizontal plane with the Z coordinate perpendicular to this plane. The flow velocity, U, is a vector with components Ux, Uy, Uz. Z Y

U Ux

Uz

Uy X

Figure 2 The specification of the flow velocity in a Cartesian co-ordinate system

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)

6

Fluid Flow in Porous Media T E N

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. The flow equations are:

 k   ∂P  U x = −    (2.2a)  µ   ∂x 



δ (Ux ρ)  δρ  = −φ ;0 ≤ x ≤ L  δt  δx

(2.2b)

where kx = permeability (in the X direction), (m2) ρ = density, (kg/m3) Ux = 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):

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flowrate, qout dx

area, A x=L

x+dx porosity, φ

X axis

x flowrate, qin

x=0 isometric view

dx flowrate, q in

flowrate, qout

x=0

x

x+dx

x=L

X axis plan view

Figure 3 Flow into and out of a cuboid of porous rock.

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,

8

Fluid Flow in Porous Media T E N

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.

(qρ ) x + dx = (qρ ) x +

δ ( qρ ) * dx δx

Therefore the change in mass flow rate = −

δ ( qρ ) * 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 The rate of change of mass is equal to hence −

δ ( qρ ) 1 δρ =φ δx A δt

δρ Aφdx δt

since the flow velocity, U = q/A, this becomes

or

−

δρ δ (Ux ρ) =φ δt δx

δ(Ux ρ) δρ = −φ δt δx

(2.2b)

Substituting the parameters of equation 2.2a in 2.2b gives

δ kx δ P δρ = −φ (2.3) δx µ δx δt

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:

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c=−



Reservoir Engineering

1  δV  V  δP  T

where V is the volume (m3) and P is the pressure (Pa).

 

The density equals mass per unit volume ρ =

c=−

m , hence: V

ρ δ ( m / ρ ) 1 δρ = (Quotient Rule, constant mass system) m δP ρ δP

(2.4)

Since then

δρ δρ δP δP = = cρ (from above) δt δP δt δt





(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 and φcρ. The pressure dependence of the coefficients must be removed before µ

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.

2.2.1.1 Linearisation Of Partial Differential Flow Equation For Linear Flow

Assuming that the permeability and viscosity terms do not depend on location (i.e. distance along the cuboid), then

δ  δP   φµcρ  δP (2.6) = ρ δx  δx   k  δt

The left hand side can be expanded to:

10

 δ 2P  δρ δP + ρ 2   δx  δx δx

Fluid Flow in Porous Media T E N

Using equation 2.4 and since

δρ δρ δP = δx δP δx

the above becomes cρ(δP/δx)2 + ρ(δ2P/δx2) Usually (δP/δx)2 is neglected compared to δ2P/δx2 since the pressure gradient is small, and substituting gives

δ 2 P  φµc  δP = (2.7) δx 2  k  δt

This is termed the linear diffusivity equation The assumption is made that the compressibility is small and constant, therefore the coefficients

k φµc are constant and the equation is linearised. In equation (2.7) is φµc k

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:

c = coSo + cwSwc + cf (2.8)

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

2.2.1.2 Conditions of Solution

The solution of the equation requires initial conditions and the boundary conditions. (i) Initial Solution Condition. At time t = 0, the initial pressure, Pi, must be specified for every value of x. (ii) Boundary Conditions. At the end faces x = 0 and x = L, the flow rate or pressure must be specified for every value of time, t. 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. 22/07/14

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2.2.2 The Radial Model

Figure 4 illustrates the geometry of this model in which the flow occurs in horizontal planes perpendicular to the Z axis (i.e. in planes parallel to the XY plane) within a layer of constant height, h. The flow is radial and is either towards the Z axis or away from it. Z

Y

h

rw

re

X

Z radial element

h

wellbore

qρr+dr

qρr

dr r section in the XZ plane

Figure 4 Radial horizontal flow geometry geometry

At a distance r from x-axis, the flow velocity, U is now radius dependent:

U = q/2πrh (2.9)

From Darcy’s Law (taking account of the flow direction and the co-ordinate direction):

U=

k δP (2.10) µ δr

The mass balance gives:

12

δ ( qρ ) δρ = 2πrhφ (2.11) δr δt

Fluid Flow in Porous Media T E N

Eliminating U and q through equations 2.9 to 2.11 gives the non-linear equation:

1δ k δP δP = φcρ (2.12) ρr δt r δr µ δr

Making assumptions as for linear flow, linearises the equation to:

(2.13)

This is termed the radial diffusivity equation

2.2.2.1 Range Of Application And Conditions Of Solution

The two main systems to which the radial diffusivity equation can be applied are water influx and wellbore production although there are others. (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 corresponds to the wellbore radius, rw, the outer boundary to the boundary of the drainage area. In general the wellbore radius, rw is a mathematical concept, however, the following are widely treated as valid: Open hole, drilled close to gauge Well cased, cemented and perforated Slotted liner with gravel pack Out-of-gauge hole

: : : :

rw = 1/2 drill bit diameter rw = 1/2 drill bit diameter rw = 1/2 outer diameter (OD) of the liner rw = average radius from caliper log

The solution of the equation requires the initial conditions and the boundary conditions. (i) Initial Solution Condition. At time t=0, the initial pressure, Pi, must be specified for every point of the range of equation 2.13, i.e. in the reservoir or in the aquifer. (ii) Boundary Conditions The boundaries consist of the outer and inner boundaries. The number of solutions depend on the number of boundary conditions, but in the main there are a few sensible conditions representing the majority of reservoir performance. 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 reservoir pressure, Pi. 22/07/14

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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. (a) Constant Terminal Rate Case (C.T.R.) This can be applied to a wellbore in which the production rate of the well is held constant and the pressure varies through time. It can also be applied to water encroachment in which the influx rate of water from the aquifer into the reservoir across the initial oil-water contact is constant. (b) Constant Terminal Pressure Case (C.T.P.) Applied to a wellbore, the flowrate is varied to maintain a constant bottom hole pressure in the producing well. In the case of water influx, the pressure at the initial oil water contact of the reservoir remains constant and the flow rate varies.

2.3 Characterisation of the Flow Regimes by their Dependence on Time

To apply the diffusivity equation to real reservoirs requires careful consideration of the boundary conditions. It will be shown that for most practical purposes, the solutions to the diffusivity equation can be grouped according to the flow regime that they represent: steady-state, semi-steady-state (pseudo steady state) or unsteady state (transient). Steady-state refers to the situation in which the pressure and the flow 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.

3

BASIC SOLUTIONS OF THE CONSTANT TERMINAL RATE CASE FOR RADIAL MODELS

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. 14

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The radial constant terminal rate case is determined by the following system of equations:

1 δ δP φµc δP (r ) = ; a ≤ r ≤ b (3.1) r δr δr k δt



 2πrkh   δP  q= ;r = a (3.2)   µ   δr 

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

r = b, t ≥ 0:

δP = 0 (3.5b) δr

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 CS Matthews and DG Russell, SPE Monograph Volume 1. These are too complex for most practical applications and asymptotic solutions which are fair approximations of the general solution are used, i.e. simple solutions which approximate certain flow regimes can be used. The problem is to identify accurately which flow regime and therefore which asymptotic solution should be used. The steady state solution is the simplest and is the same as Darcy’s Law. The non-steady state solutions involve a time element and are conveniently expressed in dimensionless form.

3.1 The Steady State Solution

If a well is produced at a constant flow rate, q, and if the pressure at the external radius, re is maintained constant, flow will finally stabilise to steady state conditions.

δP

= 0 for all values of radius, i.e. flowrate, q = constant and the pressure gradient, δ t r and time, t

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therefore,

Reservoir Engineering

dr  2πkh  δP dP and the flow equation becomes q = =  dP r  µ  δr dr

integrating between the limits rw and r gives:

qµ   r  P − Pw =  ln  2πkh   rw  (3.6)



Integrating between the limits rw and re gives:

qµ   re  Pe − Pw =  ln (3.7)  2πkh   rw 



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.

initial pressure profile

wellbore

Pi P pressure profile due to production rate, q

h Pwf rw re

element of volume, dV, at radius, r and at pressure, P

Figure 5 Pressure distribution around a well



re

1 P = ∫ PdV (3.8a) V rw

where dV = 2πrhφdr 16

(3.8b)

Fluid Flow in Porous Media T E N

The volume of the well’s drainage zone, V = π(re2-rw2)hφ and considering rw <
2 P= 2 r e

re

∫ Prdr (3.9)

rw

from equation 3.6,

qµ   r  P = Pw +  ln  2πkh   rw  P=

2 re  qµ  ln r  rdr P + ∫ w  2πkh   rw  r 2e rw

P - Pw =

2  qµ  re  r  ∫ ln  rdr r 2e  2πkh  rw  rw  re

re 1  r 2  2 qµ   1 2 r  P - Pw = 2  r ln − ∫   dr  rw  rw rw r  2  r e  2πkh   2



P - Pw =

assuming

r 2w is negligible 4

P - Pw =



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2  qµ    r 2e re r 2w rw   r 2e r 2w   −   ln − ln  −  r 2e  2πkh    2 rw 2 rw   4 4  

2  qµ   r 2e re r 2e   ln −  r 2e  2πkh   2 rw 4 

qµ   re 1  P - Pw =  ln −  2πkh   rw 2 

(3.10)

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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 formation volume factor for oil, Bo net thickness of formation, h, viscosity of reservoir oil, µ wellbore radius, rw external radius, re initial reservoir pressure, Pi bottomhole flowing pressure, Pwf qreservoir = qstock tank x Bo 1bar = 105 Pa

19% 1.3rm 3/stm 3 (reservoir cubic metres per stock tank cubic metre) 40m 22x10-3 Pas 0.15m 350m 98.0bar 93.5bar

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, formation volume factor for oil, Bo net thickness of formation, h viscosity of reservoir oil, µ wellbore radius, rw external radius, re average reservoir pressure, P bottomhole flowing pressure, Pwf

23% 1.36rm3/stm3 23m 14x10-3 Pas 0.15m 210m 132.6bar 125.0bar

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 formation volume factor for oil, Bo net thickness of formation, h viscosity of reservoir oil, µ wellbore radius, rw external radius, re average reservoir pressure, P bottomhole flowing pressure, Pwf permeability, k

18

28% 1.41rm3/stm3 15m 21x10-3 Pas 0.15m 250m 83.0bar 75.0bar 125mD

Fluid Flow in Porous Media T E N

3.2 Non-Steady State Flow Regimes and Dimensionless Variables

As mentioned previously, dimensionless forms of the diffusivity equation have found wide application in the description of flow through porous media. They “normalise” the equation for use with many different reservoirs and allow general solutions to be found which can be applied to specific data to determine the specific solution for a particular reservoir. In such a way, general plots of, for example, the difference in pressure from the reservoir to the wellbore through time can be constructed which can then be used to determine the actual pressure difference for a specific reservoir. It should be noted that solutions for a radial flow reservoir can only be sensible if the dimensionless variables and diffusivity equation have been developed for a radial flow reservoir. If the dimensionless variables are defined as: dimensionless radius, rD

:

rD =

r rw kt

tD = dimensionless time, tD : φµcr 2

w

dimensionless pressure, PD (at a dimensionless radius and at a dimensionless time)

:

 2π kh  PD ( rD ,t D ) =   ( Pi − Pr,t )  qµ 

where r = radius in question rw = wellbore radius k = permeability t = time in question φ = porosity µ = viscosity c = compressibility h = thickness of the reservoir Pi = initial reservoir pressure Pr,t = pressure at the specified radius and time then the radial diffusivity equation becomes

1 δ  δPD  δPD (3.11)  rD = rD δrD  δrD  δt D

There are other definitions of dimensionless variables, such as a dimensionless external radius, which may be used in particular instances.

3.3 Unsteady State Solution

The constant terminal rate (CTR) solution can be obtained in several forms, using different assumptions and methods of mathematical analysis. The various solutions overlap, and all of them have particular uses and limitations.

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3.3.1 General Considerations

flowrate, q

zero flowrate

bottomhole flowing pressure, Pwf

time (a)

Pi

transient late transient semi - steady state

(b)

time

Figure 6 Wellbore pressure response to a change in flowrate

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), 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.

20

Fluid Flow in Porous Media T E N

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.

3.3.2 Hurst and Van Everdingen Solution

The constant terminal rate solution for all values of the flowing time was presented by Hurst and van Everdingen in 19492. They solved the radial diffusivity equation using the Laplace transform for both the constant terminal rate and constant terminal pressure cases. The full equation contains, as one of its components, an infinite summation of Bessel functions which are required to describe the complex wellbore pressure response during the late transient period. Simple solutions can be obtained for the transient and semi-steady state flow. 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. If the reservoir is fixed in size, i.e. reD is a particular value, then the dimensionless pressure drop, PD, is a function of the dimensionless time, tD and dimensionless radius, rD. The pressure in a particular reservoir case can then be calculated at any time and/ or radius. One of the most significant cases is at the wellbore since the pressure can be measured routinely during production operations and compared to the theoretical solutions. The determination of a reservoir pressure at a location remote from a well may be required for reasons of technical interest, but unless a well is drilled at that location, the actual value cannot be measured. At the wellbore radius, r=rw (or rD=1.0) PD = f(tD, reD) (3.13) i.e. − t 2 ∞ e α m D J1 (α m reD ) 2t D 3 PD ( t D ) = 2 + lnreD − + 2 ∑ 2 2 (3.14) 2 reD 4 m =1 α m J1 (α m reD ) − J1 (α m ) 2



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(

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)

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where αm are the roots of J1 (α m reD )Y1 (α m ) − Ji (α m )Y1 (α m reD ) = 0 J1 and Y1 are Bessel functions of the first and second kind. 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. Table 2

3

Presents

Valid for

i

PD as a function of tD <1000 (from table)

infinite acting reservoirs

ii

PD ≅ 2

iii

2 PD ≅ 0.5 (IntD + 0.80907) for 100< tD <0.25 reD (an extension of the table)

infinite acting reservoirs

iv

PD as a function of tD >0.25 reD2 (from table)

finite reservoirs

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.

ii

PD =

tD for tD <0.01 (an extension of the table) π

4 4 2 2( t D + 0.25) (3reD − 4reD lnreD − 2reD − 1) − 2 2 2 reD − 1 − 1) 4( reD

infinite acting reservoirs

finite reservoirs

2 < tD 25 < t D and 0.25reD

iii

PD ≅

3 2t D 2 + lnreD − for reD >> 1 2 4 reD

finite reservoirs

Table 1

The solutions summarised in table 1 are applicable to a well flowing at a constant rate or to a reservoir and aquifer with a constant flowrate across the oil water contact. Most problems involving flow at a well involve relationship 2(iii) and 3(iii). Table one should be used starting at the top. You then look at the conditions in the reservoir until you reach the relationship that fits them. You then use that relationship. It can be seen that in using these solutions, the pressure can be calculated anywhere in the reservoir as long as the flow rate is known. If the pressure in the reservoir at a location where the flow rate is unknown is required then an alternative solution is needed (the Line Source solution).

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Fluid Flow in Porous Media T E N

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 EXERCISE 5 An experiment on a cylindrical sand pack is conducted to examine the wellbore pressure decline. The sand pack is filled with pressurised fluid which is withdrawn from the wellbore at a constant flowrate of 0.1m3/d. There is no flow at the external boundary. Calculate the wellbore pressure at times 0.001 hour, 0.005 hour and 0.1 hour after the start of production. The Figure below indicates the sand pack. fluid production

flow to the wellbore

sand pack with fluid filled pore space

closed top, bottom and side

Data porosity, φ net thickness of formation, h viscosity of fluid, µ wellbore radius, rw external radius, re initial reservoir pressure, Pi permeability, k compressibility, c

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25% 0.2m 2x10-3 Pas 0.2m 2m 2bar 1200mD 0.15x10-7Pa-1

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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

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=  (3.15)  µ   δr  r = r w

and for a line source, the following boundary condition must hold:

lim δp qµ 2y = for time, t > 0. y → 0 δy 2πkh

Using the Boltzman Transformation 24

φµcr 2 and substituting into the diffusivity equation y= 4kt φµc δP   1 δ δP (r ) =  r δr δr k δt 

Reservoir Engineering

Fluid Flow in Porous Media T E N

gives

y



d 2 p dp + (1 + y) = 0 dy 2 dy

with the boundary conditions p → pi as y → ∝

lim δp qµ 2y = y → 0 δy 2πkh



dp dy then dp′ y + (1 + y)p′ = 0 dy

If p′ =

Separating the variables and integrating gives lnp' = -lny - y +C i.e. p′ = dp = C1 e − y (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



then C1 =

qµ and equation 3.16 becomes 4πkh

dp qµ e − y which is integrated to give = dy 4πkh y



y

qµ e − y p= dy + C 2 4πkh ∫∞ y



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or

qµ e − y p=− dy + C 2 4πkh ∫y y ∞

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which can be rewritten as

p=

qµ Ei(-y) + C 2 4πkh

Applying the boundary condition that p → pi as y → ∝ then C2 = pi and the line source solution is obtained:

p i − p(r,t) = −

qµ  φµcr 2  Ei() (3.17) 4πkh  4kt 

The term Ei(-y) is the exponential integral of y (the Ei function) which is expressed as

e−y Ei( − y) = − ∫ dy y y ∞

It can be calculated from the series



 yn  Ei( − y) = γ + lny −    n!n 

where γ = 0.5772157 (Euler’s Constant). On inspection of the similarities in the Ei function and the ln function, it can be seen that when y <0.01, Ei(-y) = g + lny and the power terms can be neglected. Therefore,

Ei( − y) = ln(1.781y) = ln(γy)

(γ = 1.781 = eγ = e 0.5772157 )

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 limitation

25φµcr 2 . that y<0.01 corresponds to time, t, from the start of production t > 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



P = Pi +

qµ  γφµcr 2   ln  4πkh  4kt 

or, since -ln(y) = ln(y-1) 26

P = Pi −

qµ  4kt   ln  (3.18) 4πkh  γφµcr 2 

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Fluid Flow in Porous Media T E N

and if the pressure in the wellbore is of interest,

Pwf = Pi −

qµ  4kt   ln  (3.19) 4πkh  γφµcrw2 

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.

3.3.3.1 Range of Application and Limitations to Use

The Ei function has limitations on its application: it cannot represent the initial flow into a wellbore since the assumption that the wellbore is a line is obviously not the case and some time has to elapse for the relative size of the wellbore to have a negligible effect on the flow and expansion of the fluid in the majority of the reservoir. The reservoir must also be infinite acting. Analysis of real reservoir performance has shown that the Ei function is valid for: (i)

flowing time, t >100 φµcrw2/k (3.20)

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). (ii)

t < φµcre2/4k (3.21)

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.

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EXERCISE 7 A well and reservoir are described by the following data: Data porosity, φ formation volume factor for oil, Bo net thickness of formation, h viscosity of reservoir oil, µ compressibility, c permeability, k wellbore radius, rw external radius, re initial reservoir pressure, Pi well flowrate (constant) skin factor

19% 1.4rm3/stm3 100m 1.4x10-3 Pas 2.2 x10-9Pa-1 100mD 0.15m 900m 400bar 159stm3/day = 0

159 stm3/second 24x3600

Determine the following: (1) the wellbore flowing pressure after 4 hours production (2) the pressure in the reservoir at a radius of 9m after 4 hours production (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

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 formation volume factor for oil, Bo net thickness of formation, h viscosity of reservoir oil, µ compressibility, c permeability, k wellbore radius, rw external radius, re initial reservoir pressure, Pi well flowrate (constant) skin factor

28

25% 1.32rm3/stm3 33m 22.0x10-3 Pas 0.6x10-9Pa-1 340mD 0.15m 650m 270bar 20stm3/day 0

Fluid Flow in Porous Media T E N

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 formation volume factor for oil, Bo net thickness of formation, h viscosity of reservoir oil, µ compressibility, c permeability, k wellbore radius, rw external radius, re initial reservoir pressure, Pi well flowrate (constant) skin factor Distance well 1 to well 2

18% 1.21rm3/stm3 20m 0.8x10-3 Pas 43x10-9Pa-1 85mD 0.15m 1950m 210bar 20stm3/day 0 50m

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 formation volume factor for oil, Bo net thickness of formation, h viscosity of reservoir oil, µ compressibility, c permeability, k wellbore radius, rw external radius, re initial reservoir pressure, Pi well flowrate (constant) skin factor

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16% 1.13rm3/stm3 10m 5x10-3 Pas 14x10-9Pa-1 10mD 0.15m 780m 86bar 2stm3/day 0

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EXERCISE 11 A well is put on production at 15stm3/day. The following well and reservoir data are relevant. Data porosity, φ formation volume factor for oil, Bo net thickness of formation, h viscosity of reservoir oil, µ compressibility, c permeability, k wellbore radius, rw external radius, re initial reservoir pressure, Pi well flowrate (constant) skin factor

21% 1.2rm3/stm3 23m 5x10-3 Pas 22 x10-9Pa-1 130mD 0.15m 800m 120bar 15stm3/day 0

Determine the following: (1) the wellbore flowing pressure after 2 hours production (2) the pressure in the reservoir at a radius of 10m after 2 hours production (3) the pressure in the reservoir at a radius of 20m after 2 hours production (4) the pressure in the reservoir at a radius of 50m after 2 hours production

3.3.4 The Skin Factor

The analysis of fluid flow encountered thus far has assumed that a constant permeability exists within the reservoir from the wellbore to the external boundary. In reality, the rock around the wellbore can have higher or lower permeability than the rest of the reservoir. This results from formation damage which may occur during drilling 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.

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bottomhole following pressure, Pwf

Fluid Flow in Porous Media T E N

pressure profile if no skin zone was present

Pwf (no skin) P (skin) Pwf (skin)

actual pressure profile through skin zone skin zone permeability, Ks rw

permeability, K

rs radius, r

P skin = Pwf(no skin) - Pwf(skin)

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.

∆Ps =

r  r  r  qµ qµ qµ k ln s  − ln s  = ( − 1)ln s  (3.22) 2πk s h  rw  2πkh  rw  2πkh k s  rw 

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:

Pi − Pwf = −

k  rs  qµ qµ  Ei( − y) + ∆Ps = − Ei( − y) − 2 − 1 ln  (3.23) 4πkh 4πkh   k s  rw 

If at the wellbore the logarithm approximation can be substituted for the Ei function, then:

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Pi − Pwf = −

k  r  qµ  γφµcrw2 ) − 2 − 1 ln s  (3.24) ln( 4πkh  4kt  k s  rw 

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A skin factor, s, can then be defined as:

k  r s =  − 1 ln s (3.25)  k s  rw

and the drawdown defined as:

 qµ  γφµcrw2 Pi − Pwf = − ) − 2s (3.26) ln(  4πkh  4kt 

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. 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, formation volume factor for oil, Bo net thickness of formation, h, viscosity of reservoir oil, µ compressibility, c permeability, k wellbore radius, rw external radius, re initial reservoir pressure, Pi bottomhole flowing pressure after 6 minutes well flowrate (constant)

32

28% 1.39rm3/stm3 8.5m 0.8x10-3 Pas 2.3 x10-9Pa-1 100mD 0.15m 6100m 227bar 192bar 286stm3/day

Fluid Flow in Porous Media T E N

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 formation volume factor for oil, Bo net thickness of formation, h viscosity of reservoir oil, µ compressibility, c permeability, k wellbore radius, rw external radius, re initial reservoir pressure, Pi bottomhole flowing pressure after 1.5 hours well flowrate (constant)

23% 1.36rm3/stm3 63m 1.6x10-3 Pas 17 x10-9Pa-1 243mD 0.15m 4000m 263.0bar 260.5bar 120stm3/day

3.4 Semi-Steady-State Solution

Once the initial pressure perturbation produced by bringing a well onto production has moved through the reservoir and met the boundaries, the infinite-acting nature of the fluid changes to become finite acting. As stated previously, this is termed pseudo steady state or semi steady state because the pressure drop with time is the same at all points around the flowing well, i.e.

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.

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Reservoir Engineering

well with constant flow rate, q

calculated average pressure

pressure profile in reservoir

Pi Pe

Pwf rw

flowing pressure, P

height of formation

initial pressure

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.

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.

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Fluid Flow in Porous Media T E N

q=

dV dt

dV qdt dt = =q dP dP dP since c = −

q = − cV



1  dV  V  dP  T

dP dt

dP q =− dt cV

(3.28)

which, for the drainage of a radial cell, can be expressed as

dP q =− (3.29) dt cπre2 hφ

Substitution of equation 3.29 in the radial diffusivity equation gives



1 δ δP φµc δP (r ) = r δr δr k δt

1 δ δP φµc q (r ) = − r δr δr k cπre2 hφ

which is



1 δ δP qµ (r ) = − 2 πre hk r δr δr

Integration gives

dP qµr 2 r =− + C (3.30) dr 2πre2 kh dP = 0 therefore at the outer boundary the pressure gradient is zero, i.e. r dr qµ and substitution into equation 3.30 gives C= 2πkh

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Reservoir Engineering

dP qµ  1 r  =  −  (3.31) dr 2πkh  r re2 

When integrated, this gives

or



[P]

Pr Pwf

r

qµ  r2  = lnr − 2πkh  2re2  r

w

Pr − Pwf =

qµ   r2   rw2  − − − lnr lnr w 2πkh   2re2   2re2 

Pr − Pwf =

qµ  r r2  − 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),



Pe − Pwf =

qµ  re 1   ln − + s (3.33) 2πkh  rw 2 

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.

P=

2 r2e

re

∫ Prdr (3.9)

rw

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  P − Pwf = 2 − r ln   dr re 2πkh r∫w  rw 2re2 

and integrating gives

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Fluid Flow in Porous Media T E N

(i)

re

re

r

e  r  r2 r 1 r2 r ln dr = ln −   2 r  ∫ rw  ∫ r 2 dr w  rw  rw  rw

re re re 2  r  r2 2 qµ r  r  r2  r =  ln  e −  P − P2wf = 2 ∫ r ln − 2  dr 2 qµ  4r r−w r 2  dr re 2πkh rw  rw 2re  P − Pwf = 22 rw  r∫w r ln re 2πkh r  rw 2re  re2 re wre2 ln − ≈ re re re  r   r 2 2 rw 4 1 r2 r  re re r ln − r ln dr dr =     e 2 2  r   r ∫ r  ∫  1 rrw  2 2 r r   w rw rw ∫r r   ln rw   dr =   2 ln rw   r r−w r∫ r 2 dr (ii) r r w w e w  r 2  r 2  e r  r r r e =  ln  −   re e  r22  r 2  e r   r4  r3  2 rw  rw  4  rw e ln −   =   ≈ dr =  8r 2  ∫r 2re2  2 r  4 w  rw rw  e  rw  8 w re2 re re2 ≈ ln − re2 re re2 2 ≈ ln −inclusion of the skin factorrwgives4 and substitution into equation 3.32 2 with rw 4 r e 3  r 4  r e r 2 r re re dr =  2  ≈ e r 3 qµ   rr4e  3 re2 ∫ 2re2  8re  r w 8 rw  2  − ≈+ s (3.34) P − Pwf∫ = 2 dr = ln 2r 8r 8   rwe r w4  e2πkh  rw

qµ 

r

3



e The pressure differences (Pr - Pwf ), (Pe- Pwf), ( P -P whereas − P) do= not change ln with − time, + s re 3  wfwf 2πkh  qµ 4 r   w Pr, Pe, Pw and P − doPchange.  ln − + s wf =

2πkh 

rw

4



 q  P = Po +   (t o − t )  q   cV  P = Po +   (t o − t )   cV A well has been on production in a reservoir which  is qt in  a semi-steady state flow  = Pi −  flowing regime. For the following data, calculate the Pbottomhole pressure, Pwf  qt   cV  P = Pi −    cV  Data EXERCISE 14

formation volume factor for oil, Bo 1.62rm3/stm3 qµ  re 3 2kt  net thickness of formation, h 72m   P P = − ln +  wf  r  viscosity of reservoir oil, µ 1.2x10 qµ  re 3 i-3 Pas2kt 2πkh 4 φµcr 2   w   e P P = − ln + i wf permeability, k 123mD 2πkh  4 φµcre 2   r0.15m  w wellbore radius, rw external radius, re 560m qµ  re 3  average reservoir pressure, 263.0bar P − P =  ln −  wf qµ  re 216stm 3  3/day 2πkh  rw 4  well flowrate (constant) P − P = ln −   wf skin factor 2πkh  r 0 4  w

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 re 3 −   ln Pi  rInitial 3.4.1 Using The 3 Reservoir Pressure, 4 r e  w −  initial pressure conditions is ln from If the pressure drop required then equation 3.27  rw 4 2    may be written as: 3  12 re 3  1  2  re  1   re    2ln −    =   ln    3  −  = ln  3  1   re  2  3  1  r  r=w 1  2ln  re 2  −  lnrw e 2   2  2    rw   2lnq e −  =  ln −       2 2 2 r r r 2 2 P = P + − t t    (  ) (3.35) o    w    w    cV w o   r  2   e  2   r   1   rw    e   = ln  3 r    w  1 2   2  Institute of Petroleum Engineering, Heriot-Watt University 37 = ln  3   e    2  2       e 

Petroleum Engineering



qt P = Pi −   (3.36)  cV 

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 3.36 from equation 3.34 gives

Pi - Pwf =

2 kt  qµ  re 3 (3.37) ln − + 2πkh  rw 4 φµcre 2 

3.4.2 Generalised Reservoir Geometry: Flowing Equation under SemiSteady State Conditions

The key aspect of the radial flow equation under semi-steady state conditions is that the boundary of the reservoir has an effect on the flow regime. The pressure decline is influenced by the fact that there is a finite limit to the amount of fluid present in the reservoir. The equations developed have been for radial geometries. However, the semi-steady state flow regime in non-radial reservoirs can be examined by the radial equation if the shape of the reservoir can be attributed to a factor which encapsulates the relative position of a producing well in a volume of reservoir fluid. This nonsymmetrical geometry can be described by the Dietz shape factor (given the symbol CA ) as follows. Using the average reservoir pressure and assuming no skin factor, the pressure drop is described by equation 3.34 as

P − Pwf =

qµ  re 3   ln −  (3.34) 2πkh  rw 4  

Expressing the terms  ln



38

re 3  −  as rw 4 

Reservoir Engineering

Fluid Flow in Porous Media T E N

2 2  23   3  1   re  re 3  1   re  1 2ln − ln ln = − = − ln          e  rw 2  2   rw  2  2   rw  2   

 r  2   e   r  1 = ln  w3  2    e2      



    2 1  (πre )  = ln 2   2 23     πrw e      

The area drained (for a radial geometry) is πre2 therefore the logarithm term becomes



     2  (4A)  ( 4πre )  =   3   1.781 x 31.6 x rw2 )  (  2 2    4πrw e      

_ where A is the area drained g = 1.781 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

P − Pwf =

 qµ  1 4A + s (3.38)  ln 2 2πkh  2 γC A rw 

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

Pi − Pwf =

qµ 1 4A 2πkt + ( ln ) (3.39) 2 2πkh 2 γC A rw φµcA

Pwf = Pi −

qµ 1 4A 2πkt + ( ln ) (3.40) 2 2πkh 2 γC A rw φµcA

or

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In a convenient dimensionless form, this can be expressed as

or

 kt  rw2 2πkh 1 4A + 2π  (P - Pwf ) = ln  qµ 2 γC A rw2  φµcrw2  A PD(t D =



4A rw2 1 ln 2 t + π (3.41) D 2 γ C A rw2 A

(

The term involving the wellbore radius can be accommodated by using the following modified dimensionless time

t DA = t D



rw2 A

in which case

PD(t D =

(



1 4A ln + 2πt DA 2 γ C 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

t < t DA



φµcA 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

t > t DA



φµcA 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 40

t > t DA

φµcA k

Reservoir Engineering

Fluid Flow in Porous Media T E N

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.

3.5 The Application of the CTR Solution in Well Testing

The study of fluid flow so far has related the pressure drop expected as a result of a flow rate from a well in a reservoir. If the appropriate parameters, such as porosity, permeability and fluid viscosity are known, then for a particular flow regime, such as unsteady state, the pressure drop at a certain distance from the well at a certain time after production starts can be calculated. 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.

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Reservoir Engineering

EXERCISE 16 A well is tested by producing it at a constant flow rate of 238stm3/day (stock tank) for a period of 100 hours. The reservoir data and flowing bottomhole pressures recorded during the test are as follows: Data porosity, φ formation volume factor for oil, Bo net thickness of formation, h viscosity of reservoir oil, µ compressibility, c wellbore radius, rw initial reservoir pressure, Pi well flowrate (constant) Time (hours)

18% 1.2rm3/stm3 6.1m 1x10-3 Pas 2.18 x10-9Pa-1 0.1m 241.3bar 238stm3/day Bottomhole flowing pressure (bar)

0.0 1.0 2.0 3.0 4.0 5.0 7.5 10.0 15.0 30.0 40.0 50.0 60.0 70.0 80.0 90.0 100.0

241.3 201.1 199.8 199.1 198.5 197.8 196.5 195.3 192.8 185.2 180.2 176.7 173.2 169.7 166.2 162.7 159.2

1

Calculate the effective permeability and skin factor of the well.

2

Make an estimate of the area being drained by the well and the Dietz shape factor.

(Refer to solution to exercise 16 on page 93)

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Fluid Flow in Porous Media T E N

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, 1. calculate the permeability and skin factor of the well 2. estimate the shape of the drainage area Data porosity, φ formation volume factor for oil, Bo net thickness of formation, h viscosity of reservoir oil, µ compressibility, c wellbore radius, rw initial reservoir pressure, Pi well flowrate (constant)

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22% 1.3rm3/stm3 21m 1.9x10-3 Pas 4.3 x10-9Pa-1 0.15m 378.7bar 200stm3/day

Time (hours)

Bottomhole flowing pressure (bar)

0.0 1.1 2.1 3.2 4.3 5.4 8.0 10.7 16.1 21.4 32.1 42.8 53.5 64.2 74.9 85.6 96.3 107.0

378.7 326.41 324.7 323.8 323.1 322.1 320.5 318.8 315.5 312.2 305.6 300.8 296.0 291.2 286.3 281.5 276.7 271.9

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4

THE CONSTANT TERMINAL PRESSURE SOLUTION

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. Mathematically, all linear differential equations fulfil the following conditions: (i) If P is a solution, then C x P is also a solution, where C is a constant. (ii) If both P1 and P2 are solutions, then P1 + P2 is also a solution.

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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.

5.1 Effects of Multiple Wells

In a reservoir where more than one well is producing, the effect of each well’s pressure perturbation on the reservoir is evaluated independently (i.e. as though the other wells and their flow rate/ pressure history did not exist), then the pressure drop calculated at a particular well at a particular time is the simple addition of all of the individual effects superimposed one effect upon the other. Consider 3 wells, X, Y and Z, which start to produce at the same time from an infinite acting reservoir (Figure 9). Well X

Well Y

Well Z

Flowrate, qx

Flowrate, qy

Flowrate, qz

rxy

rzy

P caused by well X independent of well Y or well Z

P caused by well Z independent of well Y or well Z

Initial Pressure, Pi

No Barrier Detected

P caused by well Y independent of well X or well Z

No Barrier Detected

Actual well pressure, Pwf

Pressure in well Y after flowing time, t

Figure 9 The superposition of pressure changes from several wells

Superposition shows that: (Pi-Pwf)Total at Well Y

= (Pi -P)Due to well X + (Pi-P)Due to 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:

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(Pi − Pwf )total at well Y =



Reservoir Engineering

2   − q Y µ   γφµcrwY ln  − 2SY    4πkh  4kt  

+

2  − q X µ   φµcrXY Ei    4πkh  4kt  

+

2  − q Z µ   φµcrZY Ei    4πkh  4kt  

(5.1)

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, formation volume factor for oil, Bo net thickness of formation, h, viscosity of reservoir oil, µ compressibility, c permeability, k wellbore radius, rw (both wells) initial reservoir pressure, Pi Well 1 flowrate (constant) Well 2 flowrate (constant) skin factor around both wells

46

21% 1.4rm3/stm3 36m 0.7x10-3 Pas 8.7 x10-9Pa-1 80mD 0.15m 180.0bar 500stm3/day 150stm3/day 0

Fluid Flow in Porous Media T E N

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.

flowrate, q

q2 q1

(q2 - q1)

bottomhole flowing pressure, Pwf

t1

time, t

continuation of the effect of q1 in the reservoir initial reservoir pressure

∆P associated with (q1 - 0) ∆P associated with (q2 - q1)

t1

time, t

Figure 10 Effect of flowrate changes on the bottomhole flowing pressure

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

∆P( t ) = Pi - Pwf =∆PD ( t )

q 1µ (5.2) 2πkh

where ∆PD(t) is the dimensionless pressure drop at the well for the applicable boundary condition.

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Reservoir Engineering

For times greater than t1, the pressure drop is described by

∆P( t ) =



qµ (q − q 1) 1 µ ∆PD ( t - t1 ) (5.3) ∆PD ( t ) + 2 2πkh 2πkh

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:

0 ≤ t ≤ t1: ∆P(t) = ∆PD (t)



t > t1 : ∆P(t) =

q 1µ (5.4) 2πkh

qiµ q − q1 µ ∆PD (t - t1 ) (5.5) ∆PD (t) + 2 2πkh 2πkh

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

∆PD =

Pi − Pwf 1 −φµcr 2 w = − Ei ( ) (5.6) qµ / 2πkh 2 4 kt

and the equation for time, t less than or equal to t1 would be as expected

∆P(t) = -

q1µ −φµcr 2 w Ei ( ) (5.7) 4πkh 4 kt

For times greater than t1 the additional pressure drop is added to give

q1µ −φµcr 2 w (q2 − q1 ) µ −φµcr 2 w ∆P(t) = Ei ( ) Ei ( ) 4πkh 4 kt 4πkh 4 k (t − t1 )

(5.8)

This approach can be extended to many flowrate changes as illustrated in Figure 11.

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Fluid Flow in Porous Media T E N

flowrate, q

different flow rates

q3

q4

q2 q1

Bottomhole flowing pressure, Pwf

time, t

pressure responses cused by rate changes

time, t

Figure 11 Multi rate pressure response in a wellbore

This leads to a general equation

∆P(t) =

or

q1µ (q − q1 ) µ (q − q 2 ) µ ∆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)

∆P(t) =

n  q1µ  q i − q i −1 ∆ + ∆PD (t − t i −1 ) (5.10) P (t) ∑ D  2πkh  q1 i=2 

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

Pi − Pws =

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n  q µ q1µ  q i − q i −1 + P (t) ∆ ∆PD (t n − t i −1 + ∆ t) − n ∆ PD ( ∆t) (5.11) ∑ D  2πkh  q1 i =2  2πkh

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where Pws is the shut in bottomhole pressure tn-1 is the total producing time before shut in ∆t is the closed in time from the instant of shut in.

5.3 Effects of Rate Changes

The application of superposition to a well with several rate changes is illustrated as follows. A well is known to have the flowrate history as presented in Figure 12. It is seen that the well is brought onto production at a flowrate, q1 and this is maintained constant until time, t1 at which the flowrate is decreased to q2. This second flowrate continues until time t2 when the flowrate is increased to q3. In terms of the reservoir, it is assumed that the reservoir is in unsteady state flow regime and the line source can be used to describe the pressure drop caused by the flowrate changes. In this case, the first flow rate change is when the well is brought on production, so the change from zero to q1 causes the first pressure perturbation to move into the reservoir. 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.

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real well flowrate history q3 flowrate, q

q1

q2

t1

time, t

t2

equivalent flowrate effects in the reservoir q1

flowrate, q

time, t

t1

q2 - q1

time, t

q3 - q2 t2

time, t

Figure 12 The equivalence of flowrate changes in a reservoir

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).



∆P1 = ( Pi − Pwf )1

 − q1µ   γφµcrw2  =  − 2s ln 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:

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∆P2 = ( Pi − Pwf )2 =

Reservoir Engineering

 −(q 2 - q1 ) µ   γφµcrw2  ln  − 2s 4πkh   4k(t - t1 )  

And finally the third pressure drop is:



∆P3 = ( Pi − Pwf )3 =

 −(q 3 - q 2 ) µ   γφµcrw2  ln  − 2s 4πkh   4k(t - t 2 )  

The total pressure drop at the wellbore caused by all of the flowrate changes is (Pi - Pwf )= ∆P1 + ∆P2 + ∆P3 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, formation volume factor for oil, Bo net thickness of formation, h, viscosity of reservoir oil, µ compressibility, c permeability, k wellbore radius, rw (both wells) external radius, re initial reservoir pressure, Pi Well 1 flowrate (constant) Well 2 flowrate (constant) skin factor around both wells

19% 1.2rm3/stm3 36m 1x10-3 Pas 10 x10-9Pa-1 110mD 0.15m 7000m 250.0bar 110stm3/day 180stm3/day 0

The wells are 350m apart. 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 52

Fluid Flow in Porous Media T E N

permeability, k wellbore radius, rw (both wells) external radius, re initial reservoir pressure, Pi flowrate (constant) skin factor around well

50mD 0.15m 6000m 180.0bar 60stm3/day 0

The observation well is 100m from the flowing well. 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, formation volume factor for oil, Bo net thickness of formation, h, viscosity of reservoir oil, µ compressibility, c permeability, k wellbore radius, rw (both wells) external radius, re initial reservoir pressure, Pi 1st flowrate (constant) 1st flowrate period 2nd flowrate (constant) 2nd flow period skin factor around well

21% 1.31rm3/stm3 20m 0.6x10-3 Pas 8 x10-9Pa-1 75mD 0.15m 5000m 200.0bar 25stm3/day 6days 75stm3/day 4days 0

5.4 Simulating Boundary Effects (Image Wells)

One of the intriguing possibilities of the application of the principle of superposition to reservoir flow is in simulating reservoir boundaries. It is clear that when a well in a reservoir starts production, there will be a period where the flow regime is unsteady while the reservoir fluid reacts to the pressure perturbation as if the volume of the reservoir was infinite (i.e. an infinite acting reservoir). Once the boundaries are detected, there is a definite limit to the volume of fluid available and the pressure response changes to match that of, for example, semi steady state or steady state flow. This assumes that the pressure perturbation reaches the areal boundary at the same time, i.e. if the well was in the centre of a circular reservoir, the pressure perturbation would reach the external radius at all points around the circumference at the same time (assuming homogeneous conditions). If the well was not at the centre then some parts of the boundary would be detected before all of the boundary was detected. This means that some of the reservoir fluid is still in unsteady flow whilst other parts are changing to a different flow regime. This would appear to render the use of the line source solution invalid, however, the effect of the nearest boundary in an otherwise infinite acting reservoir has the same effect as the interaction of the pressure perturbations of two wells next to each other in an infinite acting reservoir. 22/07/14

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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. Infinite Acting Reservoir

Well with pressure perturbations moving out into the reservoir

Pressure perturbations hit a fault at the edge of the reservoir

Real Reservoir

Infinite Acting Reservoir

Well with pressure perturbations moving out into the reservoir

Imaginary well mimics the effect of the fault

Imposition of an Imaginary Well

Figure 13 The pressure effect of the barrier in the real reservoir can be represented by an imaginary well Can be modelled as drainage boundary between wells L

reservoir boundary

L

L

actual well

image well

Figure 14 Representation of the boundary by a real well and an image well

This shows a plane-fault boundary in an otherwise infinite acting reservoir, as in the top of Figure 13. To determine the pressure response in the well, the line source solution can be used until the pressure perturbation hits the fault. Thereafter there are no solutions for this complex geometry. However, the reservoir can be modelled

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Fluid Flow in Porous Media T E N

with an infinite acting solution if a combination of wells in an infinite-acting system that limit the drainage or flow around the boundary is found. The bottom of Figure 13 indicates 1 image well with the same production rate as the actual well is positioned such that the distance between it and the actual well is twice the distance to the fault of the actual well. No flow occurs across the plane midway between the two wells in the infinite-acting system, and the flow configuration in the drainage area of each well is the same as the flow configuration for the actual well. Pressure communication crosses the drainage boundary, but there is no fluid movement across it and the problem of the flow regime has been resolved: the real well can be thought of as reacting to the flowrate in it and to the pressure drop produced by the imaginary well on the opposite side of the fault. The pressure drop is therefore:



Pi − Pwf = −

  −φµc(2L)2  qµ  γφµcrw2 qµ − ln( ) 2s Ei −      4πkh   4πkh  4kt 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. EXERCISE 22 A well in a reservoir is produced at 120 stm3 /day for 50 days. It is 300m from a fault. Using the data given, calculate the bottomhole flowing pressure in the well and determine the effect of the fault on the bottomhole flowing pressure. Data porosity, φ, formation volume factor for oil, Bo net thickness of formation, h, viscosity of reservoir oil, µ compressibility, c permeability, k wellbore radius, rw external radius, re initial reservoir pressure, Pi flowrate (constant) flowrate period, t distance to fault, L skin factor around well

19% 1.4rm3/stm3 20m 1x10-3 Pas 9 x10-9Pa-1 120mD 0.15m 4000m 300.0bar 120stm3/day 50days 300m 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.

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image well 1

Reservoir Engineering

L2

image well 3

L2

L1

L1 R3 L1

L1 L2

Actual Well

L2 image well 2

Boundary

Figure 15 Representation of a well at the intersection of two boundaries

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. 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 The number and position of image wells can become complex. actual well

image wells

i7

i6

i3

i2

i1

i4

i5

parallel equidistant boundaries

Figure 16 Representation of an actual well between two barriers

In the apparently simple geometry of an actual well surrounded by two equidistant barriers, such as illustrated in Figure 16, the flow can be balanced as before by defining image well, i1 on the right. On the left side, the barrier is balanced by image wells i2 and i3 (because seen from i2, there is a barrier with 2 wells on the other side - a real well and an image well). Now there is an imbalance in production across the right barrier, so image wells i4 and i5 are added. This unbalances the left barrier and image wells i6 and i7 are added. This should continue to infinity, however, since the line 56

Fluid Flow in Porous Media T E N

source solution is known to have little influence above a certain distance from the actual well, the number of image wells used can be fixed with no error in the approximation. 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.

Boundary

Figure 17 Representation of a well surrounded by 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

well

Data porosity, φ, formation volume factor for oil, Bo net thickness of formation, h, viscosity of reservoir oil, µ 22/07/14

120m

22% 1.5rm3/stm3 36m 1x10-3 Pas

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compressibility, c permeability, k wellbore radius, rw external radius, re initial reservoir pressure, Pi flowrate (constant) flowrate period, t distance to fault, L1 distance to fault, L2 skin factor around well

Reservoir Engineering

9 x10-9Pa-1 89mD 0.15m 6000m 240.0bar 100stm3/day 32days 70m 120m 0

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, formation volume factor for oil, Bo net thickness of formation, h, viscosity of reservoir oil, µ compressibility, c permeability, k wellbore radius, rw external radius, re initial reservoir pressure, Pi flowrate (constant) flowrate period, t distance to fault, L skin factor around well

25% 1.13rm3/stm3 23m 1.1x10-3 Pas 10.1 x10-9Pa-1 125mD 0.15m 6000m 210.0bar 80stm3/day 32days 80m 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. 58

Fluid Flow in Porous Media T E N

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. Figure 18 summarises the route taken through the analytical solutions for radial flow regimes examined in this chapter. The number of solutions is mathematically infinite; only a few are suitable for real reservoir problems. 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 Flow regimes based on reservoir geometry radial traditional assumption for most reservoirs

linear

spherical

hemispherical

specific reservoir aquifer influxes

short time application formation testing devices thin layers

short time application formation testing devices thin layers

constant terminal rate

constant terminal pressure Hurst and Van Everdingen solutions

pressure at a specified radius pressure at a specified time for a known flowrate

flowrate at a specified radius and time

outer boundary sealing

asymptotic solutions (radial geometry) based on flow regime

outer boundary non-sealing

unsteady state

semi-steady state

line source solution pressure based on wellbore flowrates

steady state initial reservoir pressure solution average reservoir pressure solution

Ei function ln approximation to Ei function sealed outer boundary average reservoir pressure solution initial reservoir pressure solution

Figure 18 Summary of basic solutions to diffusivity equation 22/07/14

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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

Pe − Pwf =



qµBo  re  ln  2πkh  rw 

k=

r  qµBo ln e  2π (Pe − Pwf )h  rw 

k=

15x22x10 −3 x1.3 350.00  ln (3.7) 5  24x3600x2πx(98.0 − 93.5)x10 x40 0.15 

= 341x10 −15 m 2 = 341mD

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, P 132.6bar bottomhole flowing pressure, Pwf 125.0bar 60

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SOLUTION EXERCISE 2 the steady state inflow equation (accounting for fluid flowrate at reservoir conditions in m3/s and pressure in Pa) is

P − Pwf =

qµBo  re 1   ln −  2πkh  rw 2 

k=

 re 1  qµBo  ln −  2π P − Pwf h  rw 2 

k=

13 x14 x10 −3 x1.36  ln 210.00 − 1  5 24 x 3600 x 2π (132.6 − 125.0) x10 x 23  0.15 2

(

)

k = 176 x10 −15 m 2

k = 176 mD

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, P 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

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P − Pwf = q=

q=

qµBo  re 1   ln −  2πkh  rw 2 

( P − P )2πkh wf

 r 1 µBo  ln e −   rw 2 

(83.0 − 75.0) x10 5 x 2π 125 x10 −15 x15 250.00 1  21x10 −3 x1.41x  ln −  0.15 2

q = 46 x10 −6 stm 3 / s



q = 4.0 stm 3 / day

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

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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.



reD =

re 150.00 = = 1000 rw 0.15

tD =

kt 140x10 -15 xt = = 0.148t φµcrw2 0.21x10x10 -3 x0.2x10 −7 x0.152

time (hour)

time (second)

0.01 0.10 1.00 10.00 100.00

36 360 3600 36000 360000

tD (0.148t) 5.3 53.3 532.8 5328.0 53280.0

PD 1.3846 2.4146 3.5473 4.6949 5.8462

expression table 2 table 2 table 2 0.5(lntD +0.80907) 0.5(lntD +0.80907)

the bottomhole flowing pressure, Pwf is found from re-arrangement of the dimensionless

 2πkh   ( pi − pwf ) . Accounting for the oil formation volume factor,  qµ 

pressure PD = 

Bo, the bottomhole flowing pressure, pwf, is:

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Pwf = Pi −

Reservoir Engineering

qµBo PD 2πkh

Pwf at 0.01hour = 83.0x10 5 − i.e. Pwf at 0.01 hour =82.1bar

23x10x10 −3 x1.13 x1.3846 = 82.1x10 5 Pa −15 24x3600x2π 140x10 x53

similarly for the rest of the times time (hour)

PD

Pwf (bar)

0.00 0.01 0.10 1.00 10.00 100.00

0 1.3846 2.4146 3.5473 4.6949 5.8462

83.0 82.1 81.4 80.7 80.0 79.2

EXERCISE 5 An experiment on a cylindrical sand pack is conducted to examine the wellbore pressure decline. The sand pack is filled with pressurised fluid which is withdrawn from the wellbore at a constant flowrate of 0.1m3/d. There is no flow at the external boundary. Calculate the wellbore pressure at times 0.001 hour, 0.005 hour and 0.1 hour after the start of production. The Figure below indicates the sand pack. fluid production

flow to the wellbore

closed top, bottom and side

DATA

sand pack with fluid filled pore space

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

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Fluid Flow in Porous Media T E N

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.



reD =

re 2.0 = = 10 rw 0.2

tD =

kt 1200 x10 −15 xt = = 4t φµcrw2 0.25 x 2 x10 −3 x 0.15 x10 −7 x 0.2 2 time

time

tD

(hour)

(second)

(4t)

PD

expression

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 0.25reD2 i.e. not infinite acting

0.100

360.0

1440.0

30.35

PD =

2t D 3 + lnreD − 2 reD 4

the bottomhole flowing pressure, Pwf is

Pwf = Pi −

Pwf

qµ PD 2πkh

at 0.001 hour

i.e. Pwf

= 2 x10 5 −

at 0.001 hour

0.1x 2 x10 −3 x1. 808 = 1 97 x10 5 Pa 24 x 3600 x 2π 1200 x10 −15 x 0.2

= 1.97bar

similarly for the rest of the times time (hour) 0 0.001 0.005 0.100

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PD 0 1.808 3.048 30.35

Institute of Petroleum Engineering, Heriot-Watt University

Pwf (bar) 2.00 1.97 1.95 1.53

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Reservoir Engineering

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 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.



reD =

re 900 = = 6000 rw 0.15

tD =

kt 150 x10 −15 x 5 x 60 = = 934 φµcrw2 0.17 x14 x10 −3 x 0.9 x10 −9 x 0.152 time

time

(minutes)

(second)

5

300

tD

PD

expression

934

3.826

table 2

the bottomhole flowing pressure, Pwf is

Pwf = Pi −

Pwf

at 5

i.e. Pwf

66

qµ PD 2πkh

5 minutes = 200 x10 −

at 5

minutes

2.9 x14 x10 −3 x1.27 x 3.826 = 199.39 x10 5 Pa 24 x 3600 x 2π 150 x10 −15 x 40

= 199.4bar

Fluid Flow in Porous Media T E N

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 24x3600 skin factor 0 Determine the following: (1) the wellbore flowing pressure after 4 hours production (2) the pressure in the reservoir at a radius of 9m after 4 hours production (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 (ii) The reservoir is infinite acting. Thereafter, the choice of Ei function or ln approximation to the Ei function has to be made. A. Check Ei applicability

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line source not accurate until

t>

100φµcrw2 k

100x0.19x1.4x10 -3 x2.2x10 −9 x0.152 t> (3.20) 100x10 -15



t >13.2s

time is 4 hours, therefore line source is applicable. B. Check reservoir is infinite acting the reservoir is infinite acting if the time, t < i.e.

t<



0.19x1.4x10 −3 x2.2x10 −9 x900 4x100x10 -15



t < 1185030s



t < 329 hours

φµcre2 (3.21) 4k 2

therefore line source solution is applicable. (1) The bottomhole flowing pressure after 4 hours production, Pwf at 4 hours (i) Check ln approximation to Ei function the ln approximation is valid if the time, t >

t>



25φµcrw2 k

25x0.19x1.4x10 −3 x2.2x10 −9 x0.15 100x10 -15

2

t > 3.3s

therefore ln approximation is valid. (ii) Pwf = Pi +

68

qµBo  γφµcrw2  ln  (taking account of the conversion from stock 4πkh  4kt 

tank to reservoir conditions via the formation volume factor for oil, Bo, flow rates in reservoir m3/s and pressures in Pascal).

Fluid Flow in Porous Media T E N

qµBo 159x1.4x10 −3 x1.4 = = 28703 4πkh 24x3600x4πx100x10 −15 x100

φµcr 2 0.19x1.4x10 −3 x2.2x10 −9 r 2 = = 101597 x10 −9 r 2 -15 4kt 4x100x10 x4x3600



Pwf = 400x105 + 28703xln(1.781x 101597x10-9x0.152) = 400x105 - 356249 = 39643751Pa =396.4bar (2) The pressure after 4 hours production at a radius of 9m from the wellbore (i) Check ln approximation to Ei function

25φµcr 2 the ln approximation is valid if the time, t > k t>

25x0.19x1.4x10 −3 x2.2x10 −9 x92 100x10 -15



t > 11850s



t > 3.3hours

therefore ln approximation is valid. 2 qµBo  γφµcr  (ii) P = Pi + (taking account of the conversion from stock tank ln 4πkh  4kt 



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).

qµBo 159x1.4x10 −3 x1.4 = = 28703 4πkh 24x3600x4πx100x10 −15 x100



φµcr 2 0.19x1.4x10 −3 x2.2x10 −9 r 2 = = 101597 x10 −9 r 2 -15 4kt 4x100x10 x4x3600

P

= 400x105 + 28703xln(1.781x 101597x10-9x92) = 400x105 - 121209 = 39878791Pa = 398.8bar

(3) The pressure after 4 hours production at a radius of 50m from the wellbore (i) Check ln approximation to Ei function 22/07/14

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Reservoir Engineering

25φµcr 2 k −3 −9 2 25x0.19x1.4x10 x2.2x10 x50 t> 100x10 -15

the ln approximation is valid if the time, t >

t > 365750s



t > 101.6 hours

therefore ln approximation is not valid and the Ei function is used. (ii) P = Pi +

2 qµBo  φµcr  (taking account of the conversion from stock tank Ei − 4πkh  4kt 

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).

qµBo 159x1.4x10 −3 x1.4 = = 28703 4πkh 24x3600x4πx100x10 −15 x100

φµcr 2 0.19x1.4x10 −3x2.2x10 − 9x 502 = = 0.254 4kt 4x100x10 -15 x4x3600

P

= 400x105 + 28703xEi(-0.254)

Ei(-0.254) = -1.032 (by linear interpolation of the values in Table 4) 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 (i) Check ln approximation to Ei function the ln approximation is valid if the time, t >

t>

25x0.19x1.4x10 −3 x2.2x10 −9 x50 2 100x10 -15

25φµcr 2 k

t > 365750s t > 101.6 hours therefore ln approximation is not valid and the Ei function is used.

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Fluid Flow in Porous Media T E N

(ii) P = Pi +

2 qµBo  φµcr  (taking account of the conversion from stock Ei − 4πkh  4kt 

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). qµBo 159x1.4x10 −3 x1.4 = = 28703 4πkh 24x3600x4πx100x10 −15 x100



φµcr 2 0.19x1.4x10− 3 x2.2x10−9x 50 2 = = 0.020 4kt 4x100x10 -15 x50x3600

P

= 400x105 + 28703xEi(-0.020)

Ei(-0.020) = -3.355 P

= 400x105 +28703x-3.355 = 400x105 -96300 = 39903700Pa = 399.0bar

Summary time radius (hours) (m)

pressure (bar)

0 4 4 4 50

400.0 396.4 398.8 399.7 399.0

all 0.15 9.00 50.00 50.00

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 22/07/14

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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. Thereafter, the choice of Ei function or ln approximation to the Ei function has to be made. A Check Ei applicability line source not accurate until

t>

100φµcrw2 k

t>

100 x 0.25 x 22 x10 −3 x 0.6 x10 −9 x 0.152 340 x10 −15

t > 21.8s



time is 8 hours, therefore line source is applicable. B Check reservoir is infinite acting the reservoir is infinite acting if the time, t < i.e.

t<



φµcre2 4k

0.25 x 22 x10 −3 x 0.6 x10 −9 x 650 2 4 x 340 x10 −15

t < 1025184s t < 285 hours therefore line source solution is applicable. (i) check ln approximation to Ei function the ln approximation is valid if the time, t >

72

25φµcr 2 k

Fluid Flow in Porous Media T E N

t>

25 x 0.25 x 22 x10 −3 x 0.6 x10 −9 x 0.152 340 x10 −15

t > 5.5s



therefore ln approximation is valid. (ii) Pwf = Pi +

qµBo  γφµcrw2  ln  (taking account of the conversion from stock 4πkh  4 kt 

tank to reservoir conditions via the formation volume factor for oil).

qµBo 20 x 22 x10 −3 x1.32 = = 47677 4 π kh 24 x 3600 x 4 π x 340 x10 −15 x 33



φµcrw2 0.25 x 22 x10 −3 x 0.6 x10 −9 0.15 2 = = 1896 x10 −9 −15 4 kt 4 x 340 x10 x 8 x 3600

Pwf

= 270x105 + 47677xln(1.781x 1896x10-9) = 270x105 - 600663 = 26399337Pa = 264.0bar

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

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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. Thereafter, the choice of Ei function or ln approximation to the Ei function has to be made. A Check Ei applicability line source not accurate until

100φµcrw2 t> k 100 x 0.18 x 0.8 x10 −3 x 43 x10 −9 x 0.152 t> 85 x10 −15

t > 164 s

it is expected that the time will be in excess of 164 seconds therefore the line source solution is acceptable B Check reservoir is infinite acting the reservoir is infinite acting if the time, t < i.e.

t<

φµcre2 4k

0.18 x 0.8 x10 −3 x 43 x10 −9 x1950 2 4 x 85 x10 −15



t < 69250235s



t < 802 days

t < 19236 hours

therefore line source solution is applicable. C check ln approximation to Ei function the ln approximation is valid if the time, t >

74

25φµcr 2 k

Fluid Flow in Porous Media T E N

t>

25 x 0.18 x 0.8 x10 −3 x 43x10 −9 x 50 2 85 x10 −15

t > 1265 hours = 53 days

To start with, it is assumed that the ln approximation is valid.

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Now,

Pi − Pat 50 m

from well 1

Pi − Pat 50 m from qµBo − 4πkh



well 1

 P − Pat 50 m from e i qµBo  −   4πkh t=

t=

=−

qµBo  γφµcr502 m  ln  4πkh  4 kt 

 γφµcr502 m  = ln   4 kt 

 γφµcr502 m well  =  4 kt  

γφµcr502 m

 P − Pat 50 m from 4 ke i qµBo  −   4πkh

 well    

1.781x 0.18 x 0.8 x10 −3 x 43 x10 −9 x 50 2   5 0 5 x 10 .   4 x 85 x10 −15 xe  20 x 0.8 x10 −3 x1.21 −   24 x 3600 x 4π 85 x10 −15 x 20 



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. 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 76

Reservoir Engineering

Fluid Flow in Porous Media T E N

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. Thereafter, the choice of Ei function or ln approximation to the Ei function has to be made. A Check Ei applicability line source not accurate until



t>

100φµcrw2 k

t>

100x0.16x5x10 -3 x14x10 −9 x0.152 10x10 -15

t > 2520s t > 0.7 hours

time is 2 years, therefore line source is applicable. B Check reservoir is infinite acting

φµcre2 the reservoir is infinite acting if the time, t < 4k i.e.

t<

0.16x5x10 −3 x14x10 −9 x780 2 4x10x10 -15

t < 170352000s t < 5.4 years

therefore line source solution is applicable. 22/07/14

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(i) check ln approximation to Ei function

25φµcr 2 the ln approximation is valid if the time, t > k

25x0.16x5x10 −3 x14x10 −9 x0.152 t> 10x10 -15



t > 630s

therefore ln approximation is valid. (ii) Pwf = Pi +

qµBo  γφµcrw2  ln  (taking account of the conversion from stock 4πkh  4kt 

tank to reservoir conditions via the formation volume factor for oil).

qµBo 2x5x10 −3 x1.13 = = 104077 4πkh 24x3600x4πx10x10 −15 x10



0.16x5x10 −3 x14x10 −9 0.152 φµcrw2 = = 99.89 x10 −9 4kt 4x10x10 -15 x2x365x24x3600

Pwf

= 86x105 + 104077xln(1.781x 99.89x10-9) = 86x105 - 1617567 = 6982433Pa = 69.8bar

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

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Determine the following: (1) The wellbore flowing pressure after 2 hours production (2) The pressure in the reservoir at a radius of 10m after 2 hours production (3) Thepressure in the reservoir at a radius of 20m after 2 hours production (4) Thepressure in the reservoir at a radius of 50m after 2 hours production 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. Thereafter, the choice of Ei function or ln approximation to the Ei function has to be made. A Check Ei applicability line source not accurate until

t>

100φµcrw2 k



100x0.21x5x10 -3 x22x10 −9 x0.152 t> 130x10 -15



t >

400s

time is 2 hours, therefore line source is applicable. B Check reservoir is infinite acting the reservoir is infinite acting if the time, t < i.e.

t< t < t <

φµcre2 4k

0.21x5x10 −3 x22x10 −9 x800 2 4x130x10 -15 28430769s 7897 hours

therefore line source solution is applicable. (1) The bottomhole flowing pressure after 2 hours production, Pwf at 2 hours

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(i) Check ln approximation to Ei function

25φµcrw2 the ln approximation is valid if the time, t > k t>



25x0.21x5x10 −3 x22x10 −9 x0.152 130x10 -15

t > 100s

therefore ln approximation is valid. (ii) Pwf = Pi +

qµBo  γφµcrw2  ln  (taking account of the conversion from stock tank 4πkh  4kt 

to reservoir conditions via the formation volume factor for oil, Bo).

qµBo 15x5x10 −3 x1.2 = = 27724 4πkh 24x3600x4πx130x10 −15 x23



φµcr 2 0.21x5x10 −3 x22x10 −9 r 2 = = 0.0062r 2 -15 4kt 4x130x10 x2x3600

Pwf

= 120x105 + 27724xln(1.781x 0.0062x0.152) = 120x105 - 230117 = 11769883Pa = 117.70bar

(2) The presure after 2 hours production at a radius of 10m from the wellbore (i) Check ln approximation to Ei function

25φµcr 2 the ln approximation is valid if the time, t > k

25x0.21x5x10 −3 x22x10 −9 x10 2 t> 130x10 -15



t > 444231s t > 123hours

therefore ln approximation is not valid and the Ei function is used. 2 qµBo  φµcr  (ii) P = Pi + (taking account of the conversion from stock Ei − 4πkh  4kt 



80

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).

Reservoir Engineering

Fluid Flow in Porous Media T E N

qµBo 15x5x10 −3 x1.2 = = 27724 4πkh 24x3600x4πx130x10 −15 x23



φµcr 2 0.21x5x10 −3 x22x10 −9 r 2 = = 0.0062r 2 = 0.0062 x10 2 = 0.62 -15 4kt 4x130x10 x2x3600

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 (i) Check ln approximation to Ei function

the ln approximation is valid if the time, t >

t>



25x0.21x5x10 −3 x22x10 −9 x20 2 130x10 -15

25φµcr 2 k

t > 1776923s t > 493hours

therefore ln approximation is not valid and the Ei function is used. 2 qµBo  φµcr  Ei − (taking account of the conversion from stock tank (ii) P = Pi + 4πkh  4kt 



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).

qµBo 15x5x10 −3 x1.2 = = 27724 4πkh 24x3600x4πx130x10 −15 x23



22/07/14

φµcr 2 0.21x5x10 −3 x22x10 −9 r 2 = = 0.0062r 2 = 0.0062 x 20 2 = 2.48 4kt 4x130x10 -15 x2x3600

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Petroleum Engineering

P

Reservoir Engineering

= 120x105 + 27724xEi(-2.48)

Ei(-2.48) = -0.026 (by linear interpolation between adjacent values in the tables) P

= 120x105 +27724 x -0.026 = 120x105 -721 = 11999279Pa = 119.99bar (4) The pressure after 2 hours production at a radius of 50m from the wellbore (i) Check ln approximation to Ei function

the ln approximation is valid if the time, t >

t>



25x0.21x5x10 −3 x22x10 −9 x50 2 130x10 -15

25φµcr 2 k

t > 11105769s t > 3085hours

therefore ln approximation is not valid and the Ei function is used. (ii) P = Pi +

2 qµBo  φµcr  Ei − (taking account of the conversion from stock tank 4πkh  4kt 

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).

qµBo 15x5x10 −3 x1.2 = = 27724 4πkh 24x3600x4πx130x10 −15 x23



φµcr 2 0.21x5x10 −3 x22x10 −9 r 2 = = 0.0062r 2 = 0.0062 x 50 2 = 15.5 -15 4kt 4x130x10 x2x3600

P

= 120x105 + 27724xEi(-15.5)

Ei(-15.5) is less than 1.56x10-6 therefore is assumed zero 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.

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Fluid Flow in Porous Media T E N

Pressure v Distance 120.0

Pressure (bar)

119.5 119.0 118.5 118.0 117.5 117.0

0

10

20

30

40

50

60

Distance from centre of well (m)

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 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. Thereafter, the choice of Ei function or ln approximation to the Ei function has to be made. A Check Ei applicability

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line source not accurate until

t>

100φµcrw2 k



100x0.28x0.8x10 -3 x2.3x10 −9 x0.152 t> 100x10 -15



t > 11.6s

time is 6 minutes, therefore line source is applicable. B Check reservoir is infinite acting the reservoir is infinite acting if the time, t > i.e.

φµcre2 4k



0.28x0.8x10 −3 x2.3x10 −9 x6100 2 t< 4x100x10 -15



t < 47926480s t < 555 days

therefore line source solution is applicable. (i) check ln approximation to Ei function



25φµcrw2 k −3 −9 25x0.28x0.8x10 x2.3x10 x0.152 t> 100x10 -15



t > 2.9s

the ln approximation is valid if the time, t >

therefore ln approximation is valid. (ii) Pi - Pwf = −

84

 qµBo   γφµcrw2   − 2s (taking account of the conversion from ln 4πkh   4kt  

stock tank to reservoir conditions via the formation volume factor for oil).

Fluid Flow in Porous Media T E N

−

qµBo 286x0.8x10 −3 x1.39 =− = −344610 4πkh 24x3600x4πx100x10 −15 x8.5

−3 −9 2 γφµcrw2 1.781x0.28x0.8x10 x2.3x10 x0.15 = = 143371x10 −9 -15 4kt 4x100x10 x6x60



Pi - Pwf = (227-192)x105Pa = 35 x105Pa

 γφµcrw2  Pi - Pwf + ln  qµBo  4kt  4πkh (227 -192)x10 5 2s = + ln(143371x10 -9 ) 344610 2s = 10.2 − 8.9 2s =



s = 0.65

qµB 4πkh ∆Ps = 2x0.65x344610 = 447993Pa = 4.5bar ∆Ps = 2s

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 1.5 hours 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 22/07/14

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(ii) The reservoir is infinite acting. Thereafter, the choice of Ei function or ln approximation to the Ei function has to be made. A Check Ei applicability line source not accurate until



100φµcrw2 k 100x0.23x1.6x10 -3 x17x10 −9 x0.152 t> 243x10 -15



t > 58s

t>

time is 6 minutes, therefore line source is applicable. B Check reservoir is infinite acting the reservoir is infinite acting if the time, t < i.e.

t<



φµcre2 4k

0.23x1.6x10 −3 x17x10 −9 x4000 2 4x243x10 -15

t < 102979424s t < 1192 days

therefore line source solution is applicable. (i) check ln approximation to Ei function



25φµcr 2 k −3 −9 2 25x0.23x1.6x10 x17x10 x0.15 t> 243x10 -15



t > 14.5s

the ln approximation is valid if the time, t >

therefore ln approximation is valid. (ii) Pi - Pwf = − 86

 qµBo   γφµcrw2   − 2s (taking account of the conversion from ln 4πkh   4kt  

stock tank to reservoir conditions via the formation volume factor for oil).

Fluid Flow in Porous Media T E N

qµBo 120x1.6x10 −3 x1.36 − =− = −15710 4πkh 24x3600x4πx243x10 −15 x63



−3 −9 2 γφµcrw2 1.781x0.23x1.6x10 x17x10 x0.15 = = 47762 x10 −9 4kt 4x243x10 -15 x1.5x3600

Pi - Pwf = (263.0-260.5)x105Pa = 2.5 x105Pa

2s =

 γφµcrw2  Pi - Pwf + ln  qµBo  4kt  4πkh

(263.0 - 260.5)x10 5 + ln( 47762x10 -9 ) 15710 2s = 15.9 − 10.0

2s =

s = 2.95

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

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SOLUTION EXERCISE 14 Substitute the values into the semi-steady state flow equation



P − Pwf =

qµBo   re  3  ln  − + s 2πkh   rw  4 

Pwf = P -

qµBo   re  3  ln  − + s 2πkh   rw  4 

Pwf = 263x10 5 −

216x1.2x10 −3 x1.62    560.00  3 ln − + 0 −15    24x3600x2xπ 123x10 x72  0.15 4 

Pwf = 25647120Pa Pwf = 256.5bar 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.

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SOLUTION EXERCISE 15 1 Circle For infinite acting reservoirs, time,

t < t DA



φµcA k

t < 0.1x

0.2x1x10 −3 x1.45x10 −9 x1618370 100x10 -15

t < 469327s t < 130hours

2 Square For infinite acting reservoirs, time,

t < t DA



φµcA k

t < 0.09x

0.2x1x10 −3 x1.45x10 −9 x1618370 100x10 -15



t < 422395s t < 117hours

4

Quadrant of a square

For infinite acting reservoirs, time,

t < t DA



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φµcA k

t < 0.025x

0.2x1x10 −3 x1.45x10 −9 x1618370 100x10 -15

t < 117332s t < 33hours

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EXERCISE 16 A well is tested by producing it at a constant flow rate of 238stm3/day (stock tank) for a period of 100 hours. The reservoir data and flowing bottomhole pressures recorded during the test are as follows: 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 Time (hours)

0.0 1.0 2.0 3.0 4.0 5.0 7.5 10.0 15.0 30.0 40.0 50.0 60.0 70.0 80.0 90.0 100.0

Bottomhole flowing pressure (bar) 241.3 201.1 199.8 199.1 198.5 197.8 196.5 195.3 192.8 185.2 180.2 176.7 173.2 169.7 166.2 162.7 159.2

1

Calculate the effective permeability and skin factor of the well.

2

Make an estimate of the area being drained by the well and the Dietz shape factor.

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 reservoir limits. SI units will be used at reservoir conditions, therefore flowrates are 90

Fluid Flow in Porous Media T E N

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:

Pwf = Pi −

or

 qµ  4kt  + 2s  ln 2 4πkh  γφµcrw  

Pwf = m lnt + c

(3.26)

Examining the data, the following are constant: initial pressure, Pi, permeability, k, γ , porosity, φ, viscosity, µ, compressibility, c, wellbore radius, rw, and skin factor, s. Both permeability and skin factor are unknown (but they are known to be constant). Therefore in equation 3.26, there is a linear relationship between the bottom hole flowing pressure, Pwf and the logarithm of time, lnt, the slope of the relationship, m, equal to

m=

qµ 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:



2s =

 4kt  Pi − Pwf − 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 Ex16 and Figure Ex16 illustrates this.

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Time (hours)

0.0 1.0 2.0 3.0 4.0 5.0 7.5 10.0 15.0 30.0 40.0 50.0 60.0 70.0 80.0 90.0 100.0

Reservoir Engineering

Bottomhole flowing pressure (bar) 241.3 201.1 199.8 199.1 198.5 197.8 196.5 195.3 192.8 185.2 180.2 176.7 173.2 169.7 166.2 162.7 159.2

In time

0.0 0.7 1.1 1.4 1.6 2.0 2.3 2.7 3.4 3.7 3.9 4.1 4.2 4.4 4.5 4.6

Table Ex16

Bottom hole flowing pressure, Pwf (bar)

Pressure - time data (log to base e) 210 slope = 1.98 bar/unit

200 190

P

180 170 160 150

0

1

2 3 4 In flowing time, t (hours)

5

Figure Ex16a

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:

92

Fluid Flow in Porous Media T E N

k=



qµBo 238 x 1.2 x 1x10 −3 = = 218x10 −15 m 2 = 218mD 4πmh 24x3600x4πx1.98x10 5 x 6.1

(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 =

 4kt  241.3 − 201.2   Pi − P1 hour 4x218x10 -15 x3600 − ln = − ln  2  1.781x0.18x1x10 −3 x2.18x10 −9 x0.12  m 1.98  γφµcrw 

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 relationship 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 Ex16b

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Pressure- time data

210

Bottom hole flowing pressure, Pwf (bar)

Reservoir Engineering

200

190

slope = 0.35 bar/hour

180

170

160

150

0

20

40

60

80

100

120

Flowing time, t (hours)

Table Ex16b

From the plot, the gradient is determined to be -0.35bar/hour or -9.72Pa/s. This is related to the volumetric compressibility of the reservoir, i.e.

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,

A=−



A=−

qBo dP chφ dt 238 x 1.2 24 x 3600 x 2.18x10 -9 x 6.1 x 0.18 x - 9.72

A = 142076m2

The semi steady state inflow equation is



Pwf = Pi −

qµ 1 4A 2πkt ( ln + + s) 2 2πkh 2 γC A rw φµcA

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: 94

Fluid Flow in Porous Media T E N

Pi − P0 =

i.e.

 qµ  4A  ln 2 − lnC A + 2s 4πkh  γrw 

 4 x 142076  (241.3 − 194.2) x10 5 = 1.98 x10 5  ln − + lnC 2 x 3.62 A 2   1.781x0.1 

17.28 + 7.24 - 23.79 = 0.73 = lnCA CA = 2.08 From Table 5, this is close to the configuration in the Figure below.

1

2

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, 1. calculate the permeability and skin factor of the well 2. estimate the shape of the drainage area

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

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Time (hours)

Bottomhole flowing pressure (bar)

0.0 1.1 2.1 3.2 4.3 5.4 8.0 10.7 16.1 21.4 32.1 42.8 53.5 64.2 74.9 85.6 96.3 107.0

378.7 326.41 324.7 323.8 323.1 322.1 320.5 318.8 315.5 312.2 305.6 300.8 296.0 291.2 286.3 281.5 276.7 271.9

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

Pwf = Pi −



 qµBo   4kt  2s + ln   4πkh   γφµcrw2  

y = c1 + m y = mx + c

X

+ c2

therefore, m is the gradient of the line Pwf versus lnt. Calculate the values as in the table below and plot Pwf versus lnt to obtain the straight line section when the well is in transient flow.

96

Fluid Flow in Porous Media T E N

Time

Bottomhole flowing pressure (bar)

In time

(hours) 0.0 1.1 2.1 3.2 4.3 5.4 8.0 10.7 16.1 21.4 32.1 42.8 53.5 64.2 74.9 85.6 96.3 107.0

378.7 326.4 324.7 323.8 323.1 322.1 320.5 318.8 315.5 312.2 305.6 300.8 296.0 291.2 286.3 281.5 276.7 271.9

0.1 0.8 1.2 1.5 1.7 2.1 2.4 2.8 3.1 3.5 3.8 4.0 4.2 4.3 4.4 4.6 4.7

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. Bottom hole flowing pressure versus In time

pressure (bar) bottomhole flowing

330.0 320.0 310.0 y = -2.4161x + 326.6

300.0 290.0

slope

intercept

280.0 270.0

0.0

1.0

2.0

3.0

4.0

5.0

In time

(i) Permeability From this the slope is 2.42 bar/log cycle therefore

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m=

Reservoir Engineering

qµBo 4πkh

qµBo 200x1.9x10 −3 x1.3 = = 89.5x10 −15 m 2 k= 5 4πmh 24x3600x4π 2.42x10 x21 k = 90mD



ii) Skin factor Extrapolation of the line to a time of 1 hour gives the pressure, P1 hour as 326.6bar.



2s =

Pi - Pwf(1hour)  4kt  − ln  m  γφµcrw2 

2s =

  4x90x10 -15 x1x3600 378.7 − 326.6 − ln −3 −9 2 2.42  1.781x0.22x1.9x10 x4.3x10 x0.15 



2s = 21.5 - ln(17993.4)



2s = 21.5 - 9.8 = 11.7 s = 5.9 (2) Area drained

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. Bottom hole flowing pressure versus time

bottomhole flowing pressure (bar)

390.0 370.0

slope

350.0

y = -0.45x + 320.05

intercept

330.0 310.0 290.0 270.0

0.0

50.0

100.0

150.0

time (hours)

The linear section of the data appears to be present after about 50 hours, therefore this section is used to determine the slope and the extrapolated initial pressure. 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. 98

Fluid Flow in Porous Media T E N

dP qBo =− dt cAhφ A=−



qBo dP chφ dt

dP -0.45x1x10 5 = −0.45bar / hour = = −12.5Pa / s dt 1x3600 200x1.3 = 12118m 2 A=− -9 24x3600x4.3x10 x21x 0.22x -12.5

The semi-steady state pressure decline is



Pwf = Pi −

 qµ  1 4A 2πkt + + s  ln 2 2πkh  2 γC A rw φµcA 

and extrapolation of the line to small values of time gives a pressure, Po of 320.05bar. Insertion of these values at time = 0 gives



Pi − P0 =

 qµ  4A  ln 2 − lnC A + 2s 4πkh  γrw 

i.e.

  4 x 12118 (378.7 − 320.05) x10 5 = 2.42 x10 5  ln − lnC A + 2 x 5.9 2  1.781x0.15 

58.7x105 = 2.42x105 (14.01- lnCA +11.8) lnCA =14.01+11.8-24.27 = 1.54 CA = 4.7 which from Table 5 is close to

1

2

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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 (ii) The reservoir is infinite acting. A Check Ei applicability line source not accurate until

100φµcrw2 (3.10) k 100x0.21x0.7x10 -3 x8.7x10 −9 x0.152 (3.21) t> 80x10 -15 t>

t >36s

time is 50 days, therefore line source is applicable. B Check reservoir is infinite acting the reservoir is infinite acting if the time, t < 100

φµcre2 4k

Fluid Flow in Porous Media T E N

i.e.

t<

0.21x0.7x10 −3 x8.7x10 −9 x4000 2 4x80x10 -15

t < 63945000 t < 740 days

therefore line source solution is applicable. 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 (A) At 10 days, contribution to pressure drop from production from Well 2 check ln approximation to Ei function



25φµcr 2 k −3 −9 25x0.21x0.7x10 x8.7x10 x0.152 t> 80x10 -15



t > 9s

the ln approximation is valid if the time, t >

therefore ln approximation is valid.

qµBo  γφµcrw2  Pwf = Pi + ln  4πkh  4kt  Pi - Pwf = −

−



qµBo  γφµcrw2  ln  4πkh  4kt 

qµBo 150x0.7x10 −3 x1.4 = −47011 =− 4πkh 24x3600x4πx80x10 −15 x36

−3 −9 2 γφµcrw2 1.781x0.21x0.7x10 x8.7x10 x0.15 = = 185 x10 −9 -15 4kt 4x80x10 x10x24x3600

Pi - Pwf = -47011x ln(185x10-9) Pi - Pwf = -47011x -15.5 Pi - Pwf =728671Pa

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(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 check ln approximation to Ei function t > k



25x0.21x0.7x10 −3 x8.7x10 −9 x400 2 t> 80x10 -15



t > 63945000s t > 740 days

therefore ln approximation is not valid and the Ei function is used.

Pi - Pwf at Well2 caused by Well 1 = −

2  qµBo  φµcr1-2 Ei −  4πkh  4kt 

qµBo 500x0.7x10 −3 x1.4 = 156704 − =− 4πkh 24x3600x4πx80x10 −15 x36



2 0.21x0.7x10 −3 x8.7x10 −9 x 400 2 φµcr1-2 = 0.148 = 4kt 4x80x10 -15 x50x24x3600



Ei(-0.148) = -1.476

Pi - Pwf at Well 2 caused by Well 1 = -156704x-1.476 Pi - Pwf at Well 2 caused by Well 1 = 231295Pa Pwf Well2 = 180.0 - 7.3 - 2.3 Pwf Well2 = 170.4bar 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-1 102

Reservoir Engineering

Fluid Flow in Porous Media T E N

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 The wells are 350m apart. 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 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 line source not accurate until



100φµcrw2 t> k 100x0.19x1x10 -3 x10x10 −9 x0.152  t> 110x10 -15



t >

39s

time is 13 days, therefore line source is applicable. B Check reservoir is infinite acting

φµcre2 the reservoir is infinite acting if the time, t < 4k i.e. 22/07/14

t<

0.19x1x10 −3 x10x10 −9 x7000 2 4x110x10 -15

t<

211590909s

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t < 2449 days

therefore line source solution is applicable. C PRESSURE DROP AT WELL 2 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 A) At 13 days, contribution to pressure drop from production from Well 2 check ln approximation to Ei function the ln approximation is valid if the time, t >



t>

25φµcr 2 k

25x0.19x1x10 −3 x10x10 −9 x0.152 110x10 -15

t > 10s

therefore ln approximation is valid.

Pwf = Pi +

qµBo  γφµcrw2  ln  4πkh  4kt 

Pi - Pwf = − −



qµBo  γφµcrw2  ln  4πkh  4kt 

qµBo 180x1x10 −3 x1.2 = −50238 =− 4πkh 24x3600x4πx110x10 −15 x36

−3 −9 2 γφµcrw2 1.781x0.19x1x10 x10x10 x0.15 = = 154 x10 −9 -15 4kt 4x110x10 x13x24x3600

Pi - Pwf = -50238x ln(154x10-9) Pi - Pwf = -50238x -15.7 Pi - Pwf =788737Pa = 7.9bar (B) At 13 days, contribution to pressure drop at Well 2 from production from Well 1 check ln approximation to Ei function the ln approximation is valid if the time, t >

104

25φµcr 2 k

Fluid Flow in Porous Media T E N



t>

25x0.19x1x10 −3 x10x10 −9 x350 2 110x10 -15

t > 52897727s t > 612 days

therefore ln approximation is not valid and the Ei function is used.

Pi - Pwf at Well2 caused by Well 1 = −

2  qµBo  φµcr1-2 Ei −  4πkh  4kt 

qµBo 110x1x10 −3 x1.2 − =− 4πkh 24x3600x4πx110x10 −15 x36



2 0.19x1x10 −3 x10x10 −9 x350 2 φµcr1-2 = 4kt 4x110x10 -15 x40x24x3600



Ei(-0.153) = -1.447

Pi - Pwf at Well 2 caused by Well 1 = -30701x-1.447 Pi - Pwf at Well 2 caused by Well 1 = 44424Pa = 0.4bar Pwf Well2 = 250.0 - 7.9 - 0.4bar Pwf Well2 = 241.7bar D PRESSURE DROP AT WELL 1 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 (A) At 40 days, contribution to pressure drop from production from Well 1 check ln approximation to Ei function



25φµcr 2 k −3 −9 2 25x0.19x1x10 x10x10 x0.15 t> 110x10 -15



t > 10s

the ln approximation is valid if the time, t >

therefore ln approximation is valid.

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Pwf = Pi +

qµBo  γφµcrw2  ln  4πkh  4kt 

Pi - Pwf = −

−



qµBo  γφµcrw2  ln  4πkh  4kt 

qµBo 110x1x10 −3 x1.2 = −30701 =− 4πkh 24x3600x4πx110x10 −15 x36

−3 −9 2 γφµcrw2 1.781x0.19x1x10 x10x10 x0.15 = = 50.1x10 −9 -15 4kt 4x110x10 x40x24x3600

Pi - Pwf = -30701x ln(50.1x10-9) Pi - Pwf = -30701x -16.8 Pi - Pwf =515777Pa = 5.2bar (B) At 40 days, contribution to pressure drop at Well 1 from production from Well 2 check ln approximation to Ei function

25φµcr 2 the ln approximation is valid if the time, t > k

t>

25x0.19x1x10 −3 x10x10 −9 x350 2 110x10 -15

t > 52897727s t > 612 days

therefore ln approximation is not valid and the Ei function is used.

Pi - Pwf at Well1 caused by Well 2

−



106

2  qµBo  φµcr1-2 =− Ei −  4πkh  4kt 

qµBo 180x1x10 −3 x1.2 = −50238 =− 4πkh 24x3600x4πx110x10 −15 x36

2 0.19x1x10 −3 x10x10 −9 x350 2 φµcr1-2 = = 0.471 4kt 4x110x10 -15 x13x24x3600

Reservoir Engineering

Fluid Flow in Porous Media T E N



Ei(-0.471) = -0.597

Pi - Pwf at Well 1 caused by Well 2 = -50238x-0.597 Pi - Pwf at Well 1 caused by Well 2 = 29992Pa = 0.3bar Pwf Well1 = 250 - 5.2 - 0.3bar Pwf Well1 = 244.5bar 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 The observation well is 100m from the flowing well. 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. Therefore, the pressure effect of 6 days production lasts 7 days, the pressure effect of 1 day shut in lasts 1 day. 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 line source not accurate until

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100φµcrw2 t> k <



t

100x0.19x0.4x10 -3 x3x10 −9 x 0.152 50x10 -15

t>10.3s time is 6 days, therefore line source is applicable. B Check reservoir is infinite acting the reservoir is infinite acting if the time, t < i.e.



t<

φµcre2 4k

0.19 x 0.4 x10 −3 x10 x10 −9 x 6000 2 4 x110 x10 −15

t < 41040000s t <475 days

therefore line source solution is applicable. The pressure drop at the observation well is described by



Pi − Pobs well = −

 φµcr 2    φµcr 2  µBo   + (q 2 − q1 )Ei − q1Ei −   4kt  4πkh   4k(t − t1 )  

Checking for the validity of the ln approximation,



25φµcr 2 k −3 −9 2 25x0.19x0.4x10 x3x10 x100 t> 50x10 -15



t > 1140000s t > 13 days

the ln approximation is valid if the time, t >

therefore ln approximation is not valid.

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−

µBo 0.4x10 −3 x1.3 =− = - 35982857 4πkh 4πx50x10 −15 x23

2 0.19x0.4x10 −3 x3x10 −9 x100 2 φµcr1-2 = = 0.019 4kt 4x50x10 -15 x7x24x3600



2 0.19x0.4x10 −3 x3x10 −9 x100 2 φµcr1-2 = = 0.132 4k(t - t1 ) 4x50x10 -15 x(7 - 6)x24x3600



Ei(-0.019) = -3.405 Ei(-0.132) = -1.576

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 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

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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 line source not accurate until



100φµcrw2 t> k t>

100x0.21x0.6x10 -3 x8x10 −9 x0.152 75x10 -15

t > 30.3s

time is 10 days, therefore line source is applicable. B Check reservoir is infinite acting the reservoir is infinite acting if the time, t < i.e.

t<

φµcre2 4k

0.21x0.6x10 −3 x8x10 −9 x5000 2 4x75x10 -15



t < 84000000s



t < 972 days

therefore line source solution is applicable. Checking for the validity of the ln approximation,

25φµcr 2 the ln approximation is valid if the time, t > k

t>

25 x 0.21x 0.6 x10 −3 x 8 x10 −9 x 0.152 75 x10 −15

t > 7.6s

therefore ln approximation is valid.

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Fluid Flow in Porous Media T E N

 γφµcrw2    γφµcrw2  µBo  Pi − Pwf = −  + (q 2 − q1 )ln q1ln   4kt  4πkh   4k(t − t1 )   −

µBo 0.6x10 −3 x1.31 =− = −41698595 4πkh 4πx75x10 −15 x20

−3 −9 2 γφµcrw2 1.781x0.21x0.6x10 x8x10 x 0.15 = = 155.8 x10 −9 -15 4kt 4x75x10 x10x24x3600

1.781x0.21x0.6x10 −3 x8x10 −9 x 0.152 γφµcrw2 = = 389.6 x10 −9 -15 4k(t - t1 ) 4x75x10 x(10 - 6)x24x3600 (75 - 25) 25 xln(155.8x10 -9 ) + ln(389.6x10 -9 ) Pi − Pwf = −41698595 24x3600  24x3600  Pi - Pwf = -41698595x(-0.00454 -0.00854) Pi - Pwf = 545418Pa Pwf = 200.0 - 5.5 = 194.5bar EXERCISE 22 A well in a reservoir is produced at 120 stm3 /day for 50 days. It is 300m from a fault. Using the data given, calculate the bottomhole flowing pressure in the well and determine the effect of the fault on the bottomhole flowing pressure.

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

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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 (ii) The reservoir is infinite acting. A Check Ei applicability line source not accurate until



100φµcrw2 t> k 100x0.19x1x10 -3 x9x10 −9 x0.152 t> 120x10 -15

t >32s

time is 50 days, therefore line source is applicable. B Check reservoir is infinite acting the reservoir is infinite acting if the time, t <

t<

0.19x1x10 −3 x9x10 −9 x4000 2 4x120x10 -15

φµcre2 4k

t < 57000000s t <660 days

therefore line source solution is applicable. Checking for the validity of the ln approximation, for the real well the ln approximation is valid if the time, t >

112

25φµcr 2 k

Fluid Flow in Porous Media T E N



25x0.19x1x10 −3 x9x10 −9 x0.152 t> 120x10 -15

t > 8s

therefore ln approximation is valid. Checking for the validity of the ln approximation, for the image well

25φµc(2L)2 the ln approximation is valid if the time, t > k −3 −9 2 25x0.19x1x10 x9x10 x600 t> 120x10 -15

t > 128250000s t> 1484 days

therefore ln approximation is not valid. 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. 2 qµBo  γφµcrw2  qµBo  φµc(2L)  Pi − Pwf = − ln Ei − − 4πkh  4kt  4πkh  4kt 



qµBo 120x1x10 −3 x1.4 = −64473 =− 4πkh 24x3600x4πx120x10 −15 x20 −3 −9 2 γφµcrw2 1.781x0.19x1x10 x9x10 x0.15 = = 33.1x10 -9 -15 4kt 4x120x10 x50x24x3600 2 0.19x1x10 −3 x9x10 −9 x600 00 2 φµc(2L) = = 0.297 4kt 4x120x10 -15 x50x24x3600



Ei(-0.297) = -0.914



Pi − Pwf = −64473x ln(33.1x10 -9 ) − 64473x − 0.914

−

Pi - Pwf = 1110466 + 58928 =1169394Pa = 11.7bar Pwf = 300.0 - 11.7 = 288.3bar The fault 300m away pulled the bottomhole flowing pressure down by an extra 58928Pa or 0.6bar.

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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

well

120m

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

L1

70m

fault L2

well

114

r3

120m

L2 image well 2

Fluid Flow in Porous Media T E N

The three image wells balance the effect of the flow (and therefore the pressure disturbance) from the real well. The pressure disturbances are superposed onto the real well, i.e. 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 wells. 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 line source not accurate until

100φµcrw2 t> k



100x0.22x1x10 -3 x9x10 −9 x0.152 t> 89x10 -15

t > 50s

time is 32 days, therefore line source is applicable. B Check reservoir is infinite acting the reservoir is infinite acting if the time, t <

t<

0.22x1x10 −3 x9x10 −9 x6000 2 4x89x10 -15

φµcre2 4k

t < 200224719s t < 2317 days

therefore line source solution is applicable. Checking for the validity of the ln approximation, for the real well

25φµcr 2 the ln approximation is valid if the time, t > k −3 −9 2 25x0.22x1x10 x9x10 x0.15 t> 89x10 -15

22/07/14

t > 13s

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therefore ln approximation is valid. Checking for the validity of the ln approximation, for the image well 1 the ln approximation is valid if the time, t >

t>

25φµc(2L1)2 k

25x0.22x1x10 −3 x9x10 −9 x140 2 89x10 -15

t > 10901124s t > 126 days

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

Pi -Pwf = ∆Pwell + ∆Pimage well 1 + ∆Pimage well 2 + ∆Pimage well 3

Pi − Pwf = −

qµBo  γφµcrw2  ln  4πkh  4kt 

2 qµBo  φµc(2L1)  Ei − −  4πkh  4kt 



−

2 qµBo  φµc(2L2)  Ei −  4πkh  4kt 

−

qµBo  φµcr32  Ei −  4πkh  4kt 

evaluating the groups

−



116

qµBo 100x1x10 −3 x1.5 =− = −43120 4πkh 24x3600x4πx89x10 −15 x36

−3 −9 2 γφµcrw2 1.781x0.22x1x10 x9x10 x0.15 = = 80.6 x10 −9 4kt 4x89x10 -15 x32x24x3600

Fluid Flow in Porous Media T E N

ln(80.6x10-9) = -16.3



φµc(2L1)2 0.22x1x10 −3 x9x10 −9 x140 2 = = 0.039 4kt 4x89x10 -15 x32x24x3600



Ei(-0.039) = -2.706



φµc(2L2)2 0.22x1x10 −3 x9x10 −9 x240 2 = = 0.116 4kt 4x89x10 -15 x32x24x3600



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

Pi - Pwf = -43120 x -16.3 -43120 x -2.706 -43120 x -1.689 -43120 x -1.436 Pi - Pwf = 702856 + 116683 + 72830 + 61920 Pi - Pwf = 954289Pa = 9.5bar Pwf = 240.0 - 9.5 = 230.5bar The effect of the boundary is to pull the bottomhole flowing pressure down by an extra 2.5bar. 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

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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 line source not accurate until

100φµcrw2 k 100x0.25x1.1x10 -3 x10.1x10 −9 x0.152 t> 125x10 -15 t>

t > 50s

time is 10 days, therefore line source is applicable. B Check reservoir is infinite acting



φµcre2 4k 0.25x1.1x10 −3 x10.1x10 −9 x6000 2 t< 4x125x10 -15



t < 199980000s t < 2315 days

the reservoir is infinite acting if the time, t <

therefore line source solution is applicable. Checking for the validity of the ln approximation, for the real well

25φµcr 2 k −3 −9 25x0.25x1.1x10 x10.1x10 x0.152 t> 125x10 -15

the ln approximation is valid if the time, t >

t > 13s

therefore ln approximation is valid.

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Reservoir Engineering

Fluid Flow in Porous Media T E N

Checking for the validity of the ln approximation, for the image well

25φµc(2L)2 the ln approximation is valid if the time, t > k

25x0.25x1.1x10 −3 x10.1x10 −9 x160 2 t> 125x10 -15

t > 14220800s t> 165 days

therefore ln approximation is not valid.

 qµBo  φµc(2L)2  qµBo   γφµcrw2  Pi − Pwf = − Ei −  − 2s − ln 4πkh   4kt  4kt   4πkh  −

qµBo 80x1.1x10 −3 x1.13 = −31857 =− 4πkh 24x3600x4πx125x10 −15 x23

−3 −9 2 γφµcrw2 1.781x0.25x1.1x10 x10.1x10 x0.15 = = 257.6 x10 −9 -15 4kt 4x125x10 x10x24x3600

φµc(2L)2 0.25x1.1x10 −3 x10.1x10 −9 x160 2 = = 0.165 4kt 4x125x10 -15 x10x24x3600

[

]

Pi − Pwf = −31857 ln(257.6x10 −9 ) − 2x5.0 − 31857xEi( −0.165)

Ei(-0.165) = -1.383

Pi - Pwf = (-31857x[-15.2 - 10]) - (31857x -1.383) Pi - Pwf =802796 +44058 Pi - Pwf = 846854Pa = 8.5bar Pwf = 210.0 - 8.5 = 201.5bar

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tD 0 0.0005 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01 0.015 0.02 0.025 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

Reservoir Engineering

pD

tD

0 0.15 0.0250 0.2 0.0352 0.3 0.0495 0.4 0.0603 0.5 0.0694 0.6 0.0774 0.7 0.0845 0.8 0.0911 0.9 0.0971 1.0 0.1028 1.2 0.1081 1.4 0.1312 2.0 0.1503 3.0 0.1669 4.0 0.1818 5.0 0.2077 6.0 0.2301 7.0 0.2500 8.0 0.2680 9.0 0.2845 10.0 0.2999 15.0 0.3144 20.0 30.0 40.0 50.0

pD

tD

0.3750 60.0 0.4241 70.0 0.5024 80.0 0.5645 90.0 0.6167 100.0 0.6622 150.0 0.7024 200.0 0.7387 250.0 0.7716 300.0 0.8019 350.0 0.8672 400.0 0.9160 450.0 1.0195 500.0 1.1665 550.0 1.2750 600.0 1.3625 650.0 1.4362 700.0 1.4997 750.0 1.5557 800.0 1.6057 850.0 1.6509 900.0 1.8294 950.0 1.9601 1000.0 2.1470 2.2824 2.3884

pD 2.4758 2.5501 2.6147 2.6718 2.7233 2.9212 3.0636 3.1726 3.263 3.3394 3.4057 3.4641 3.5164 3.5643 3.6076 3.6476 3.6842 3.7184 3.7505 3.7805 3.8088 3.8355 3.8584

Table 2 pD vs. tD - Infinite radial system, constant rate at inner boundary.

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Fluid Flow in Porous Media T E N

reD = 1.5 tD pD 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20 0.22 0.24 0.26 0.28 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80

0.251 0.288 0.322 0.355 0.387 0.420 0.452 0.484 0.516 0.548 0.580 0.612 0.644 0.724 0.804 0.884 0.964 1.044 1.124 1.204 1.284 1.364 1.444

reD = 2.0 tD pD 0.22 0.24 0.26 0.28 0.30 0.32 0.34 0.36 0.38 0.40 0.42 0.44 0.46 0.48 0.50 0.60 0.70 0.80 0.90 1.0 2.0 3.0 5.0

0.443 0.459 0.476 0.492 0.507 0.522 0.536 0.551 0.565 0.579 0.593 0.607 0.621 0.634 0.648 0.715 0.782 0.849 0.915 0.982 1.649 2.316 3.649

reD = 2.5 tD pD 0.40 0.42 0.44 0.46 0.48 0.50 0.52 0.54 0.56 0.58 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 2.0 3.0 4.0 5.0

0.565 0.576 0.587 0.598 0.608 0.618 0.628 0.638 0.647 0.657 0.666 0.688 0.710 0.731 0.752 0.772 0.792 0.812 0.832 1.215 1.506 1.977 2.398

reD = 3.0 tD pD 0.52 0.54 0.56 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.0 1.2 1.4 1.6 2.0 3.0 4.0 5.0

0.627 0.636 0.645 0.662 0.683 0.703 0.721 0.740 0.758 0.776 0.791 0.806 0.865 0.920 0.973 1.076 1.328 1.578 1.828

reD = 3.5 tD pD 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.25 2.50 2.75 3.0 4.0 5.0 6.0

0.802 0.830 0.857 0.882 0.906 0.929 0.951 0.973 0.994 1.014 1.034 1.083 1.130 1.176 1.221 1.401 1.579 1.757

reD = 4.0 tD pD 1.5 1.6 1.7 1.8 1.9 2.0 2.2 2.4 2.6 2.8 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 8.0 9.0 10.0

0.927 0.948 0.968 0.988 1.007 1.025 1.059 1.092 1.123 1.154 1.184 1.255 1.324 1.392 1.460 1.527 1.594 1.660 1.727 1.861 1.994 2.127

Table 3 pD vs. tD - Finite radial system with closed exterior boundary, constant rate at inner boundary

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reD = 4.5 tD pD 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 3.2 3.4 3.6 3.8 4.0 4.5 5.0 5.5 6.0 7.0 8.0 9.0 10.0 11.0 12.0 13.0 14.0 15.0

1.023 1.040 1.056 1.702 1.087 1.102 1.116 1.130 1.144 1.158 1.171 1.197 1.222 1.246 1.269 1.292 1.349 1.403 1.457 1.510 1.615 1.719 1.823 1.927 2.031 2.135 2.239 2.343 2.447

reD = 5.0 tD pD 3.0 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 4.0 4.2 4.4 4.6 4.8 5.0 5.5 6.0 6.5 7.0 7.5 8.0 9.0 10.0 11.0 12.0 13.0 14.0 15.0

1.167 1.180 1.192 1.204 1.215 1.227 1.238 1.249 1.259 1.270 1.281 1.301 1.321 1.340 1.360 1.378 1.424 1.469 1.513 1.556 1.598 1.641 1.725 1.808 1.892 1.975 2.059 2.142 2.225

Reservoir Engineering

reD = 6.0 tD pD 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10.0 11.0 12.0 13.0 14.0 15.0 16.0 17.0 18.0 19.0 20.0 25.0 30.0

1.275 1.322 1.364 1.404 1.441 1.477 1.511 1.544 1.576 1.607 1.638 1.668 1.698 1.757 1.815 1.873 1.931 1.988 2.045 2.103 2.160 2.217 2.274 2.560 2.846

reD = 7.0 tD pD 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10.0 11.0 12.0 13.0 14.0 15.0 16.0 17.0 18.0 19.0 20.0 22.0 24.0 26.0 28.0 30.0

1.436 1.470 1.501 1.531 1.559 1.586 1.613 1.638 1.663 1.711 1.757 1.810 1.845 1.888 1.931 1.974 2.016 2.058 2.100 2.184 2.267 2.351 2.434 2.517

reD = 8.0 tD pD 8.0 8.5 9.0 9.5 10.0 10.5 11.0 11.5 12.0 12.5 13.0 13.5 14.0 14.5 15.0 17.0 19.0 21.0 23.0 25.0 30.0 35.0 40.0 45.0

Table 3 (continued)

122

1.556 1.582 1.607 1.631 1.653 1.675 1.697 1.717 1.737 1.757 1.776 1.795 1.813 1.831 1.849 1.919 1.986 2.051 2.116 2.180 2.340 2.449 2.658 2.817

reD = 9.0 tD pD 10.0 10.5 11.0 11.5 12.0 12.5 13.0 13.5 14.0 14.5 15.0 15.5 16.0 17.0 18.0 19.0 20.0 22.0 24.0 26.0 28.0 30.0 34.0 38.0 40.0 45.0 50.0 60.0 70.0

1.651 1.673 1.693 1.713 1.732 1.750 1.768 1.786 1.803 1.819 1.835 1.851 1.867 1.897 1.926 1.955 1.983 2.037 2.096 2.142 2.193 2.244 2.345 2.446 2.496 2.621 2.746 2.996 3.246

reD = 10.0 tD pD 12.0 12.5 13.0 13.5 14.0 14.5 15.0 15.5 16.0 17.0 18.0 19.0 20.0 22.0 24.0 26.0 28.0 30.0 32.0 34.0 36.0 38.0 40.0 50.0 60.0 70.0 80.0 90.0 100.0

1.732 1.750 1.768 1.784 1.801 1.817 1.832 1.847 1.862 1.890 1.917 1.943 1.968 2.017 2.063 2.108 2.151 2.194 2.236 2.278 2.319 2.360 2.401 2.604 2.806 3.008 3.210 3.412 3.614

Fluid Flow in Porous Media T E N

-Ei(-y),0.000<0.209,interval=0.001 y 0 1 2 3 0.00 + ∞ 6.332 5.639 5.235 0.01 4.038 3.944 3.858 3.779 0.02 3.355 3.307 3.261 3.218 0.03 2.959 2.927 2.897 2.867 0.04 2.681 2.658 2.634 2.612 0.05 2.468 2.449 2.431 2.413 0.06 2.295 2.279 2.264 2.249 0.07 2.151 2.138 2.125 2.112 0.08 2.027 2.015 2.004 1.993 0.09 1.919 1.909 1.899 1.889 0.10 1.823 1.814 1.805 1.796 0.11 1.737 1.729 1.721 1.713 0.12 1.660 1.652 1.645 1.638 0.13 1.589 1.582 1.576 1.569 0.14 1.524 1.518 1.512 1.506 0.15 1.464 1.459 1.453 1.447 0.16 1.409 1.404 1.399 1.393 0.17 1.358 1.353 1.348 1.343 0.18 1.310 1.305 1.301 1.296 0.19 1.265 1.261 1.256 1.252 0.20 1.223 1.219 1.215 1.210

4 4.948 3.705 3.176 2.838 2.590 2.395 2.235 2.099 1.982 1.879 1.788 1.705 1.631 1.562 1.500 1.442 1.388 1.338 1.291 1.248 1.206

5 4.726 3.637 3.137 2.810 2.568 2.377 2.220 2.087 1.971 1.869 1.779 1.697 1.623 1.556 1.494 1.436 1.383 1.333 1.287 1.243 1.202

6 4.545 3.574 3.098 2.783 2.547 2.360 2.206 2.074 1.960 1.860 1.770 1.689 1.616 1.549 1.488 1.431 1.378 1.329 1.282 1.239 1.198

7 4.392 3.514 3.062 2.756 2.527 2.344 2.192 2.062 1.950 1.850 1.762 1.682 1.609 1.543 1.482 1.425 1.373 1.324 1.278 1.235 1.195

8 4.259 3.458 3.026 2.731 2.507 2.327 2.178 2.050 1.939 1.841 1.754 1.674 1.603 1.537 1.476 1.420 1.368 1.319 1.274 1.231 1.191

9 4.142 3.405 2.992 2.706 2.487 2.311 2.164 2.039 1.929 1.832 1.745 1.667 1.596 1.530 1.470 1.415 1.363 1.314 1.269 1.227 1.187

-Ei(-y),0.000<2.09,interval=0.01 0.0 + ∞ 4.038 3.335 0.1 1.823 1.737 1.660 0.2 1.223 1.183 1.145 0.3 0.906 0.882 0.858 0.4 0.702 0.686 0.670 0.5 0.560 0.548 0.536 0.6 0.454 0.445 0.437 0.7 0.374 0.367 0.360 0.8 0.311 0.305 0.300 0.9 0.260 0.256 0.251 1.0 0.219 0.216 0.212 1.1 0.186 0.183 0.180 1.2 0.158 0.156 0.153 1.3 0.135 0.133 0.131 1.4 0.116 0.114 0.113 1.5 0.100 0.099 0.097 1.6 0.086 0.085 0.084 1.7 0.075 0.074 0.073 1.8 0.065 0.064 0.063 1.9 0.056 0.055 0.055 2.0 0.049 0.048 0.048

2.681 1.524 1.076 0.815 0.640 0.514 0.420 0.347 0.289 0.243 0.205 0.174 0.149 0.127 0.109 0.094 0.081 0.071 0.061 0.053 0.046

2.468 1.464 1.044 0.794 0.625 0.503 0.412 0.340 0.284 0.239 0.202 0.172 0.146 0.125 0.108 0.093 0.080 0.070 0.060 0.052 0.046

2.295 1.409 1.014 0.774 0.611 0.493 0.404 0.334 0.279 0.235 0.198 0.169 0.144 0.124 0.106 0.092 0.079 0.069 0.060 0.052 0.045

2.151 1.358 0.985 0.755 0.598 0.483 0.396 0.328 0.274 0.231 0.195 0.166 0.142 0.122 0.105 0.090 0.078 0.068 0.059 0.051 0.044

2.027 1.309 0.957 0.737 0.585 0.473 0.388 0.322 0.269 0.227 0.192 0.164 0.140 0.120 0.103 0.089 0.077 0.067 0.058 0.050 0.044

1.919 1.265 0.931 0.719 0.572 0.464 0.381 0.316 0.265 0.223 0.189 0.161 0.138 0.118 0.102 0.088 0.076 0.066 0.057 0.050 0.043

2.0
2 3.72x10-2 1.01x10-2 2.97x10-3 9.08x10-4 2.86x10-4 9.22x10-5 3.02x10-5 9.99x10-6 3.34x10-6

2.959 1.589 1.110 0.836 0.655 0.525 0.428 0.353 0.295 0.247 0.209 0.177 0.151 0.129 0.111 0.096 0.083 0.072 0.062 0.054 0.047

3 3.25x10-2 8.94x10-3 2.64x10-3 8.09x10-4 2.55x10-4 8.24x10-5 2.70x10-5 8.95x10-6 3.00x10-6

4 2.84x10-2 7.89x10-3 2.34x10-3 7.19x10-4 2.28x10-4 7.36x10-5 2.42x10-5 8.02x10-6 2.68x10-6

5 2.49x10-2 6.87x10-3 2.07x10-3 6.41x10-4 2.03x10-4 6.58x10-5 2.16x10-5 7.18x10-6 2.41x10-6

6 2.19x10-2 6.16x10-3 1.84x10-3 5.71x10-4 1.82x10-4 5.89x10-5 1.94x10-5 6.44x10-6 2.16x10-6

7 1.92x10-2 5.45x10-3 1.64x10-3 5.09x10-4 1.62x10-4 5.26x10-5 1.73x10-5 5.77x10-6 1.94x10-6

8 1.69x10-2 4.82x10-3 1.45x10-3 4.53x10-4 1.45x10-4 4.71x10-5 1.55x10-5 5.17x10-6 1.74x10-6

9 1.48x10-2 4.27x10-2 1.29x10-3 4.04x10-4 1.29x10-4 4.21x10-5 1.39x10-5 4.64x10-6 1.56x10-6

Table 4 Values of the exponential integral, -Ei(y) 22/07/14

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In Bounded Reservoirs

Less Than 1% Error Exact for tDA> for tDA>

Use Infinite System Solution With Less Than 1% Error for tDA<

31.62

0.1

0.06

0.10

31.6

0.1

0.06

0.10

27.6

0.2

0.07

0.09

27.1

0.2

0.07

0.09

21.9

0.4

0.12

0.08

0.098

0.9

0.60

0.015

30.8828

0.1

0.05

0.09

12.9851

0.7

0.25

0.03

4.5132

0.6

0.30

0.025

3.3351

0.7

0.25

0.01

1

21.8369

0.3

0.15

0.025

1

10.8374

0.4

0.15

0.025

1

4.5141

1.5

0.50

0.06

1

2.0769

1.7

0.50

0.02

1

3.1573

0.4

0.15

0.005

60º

1/3

CA

Reservoir Engineering

1 4

3

2

2

2

2

2

Table 5 Shape factors for various single-well drainage areas.10

124

Fluid Flow in Porous Media T E N

In Bounded Reservoirs

CA

Less Than 1% Error Exact for tDA> for tDA>

Use Infinite System Solution With Less Than 1% Error for tDA<

0.5813

2.0

0.60

0.02

0.1109

3.0

0.60

0.005

1

5.3790

0.8

0.30

0.01

1

2.6896

0.8

0.30

0.01

1

0.2318

4.0

2.00

0.03

1

0.1155

4.0

2.00

0.01

2.3606

1.0

0.40

0.025

1 2 1 2

4

4

4

4

1 5

Table 5 (continued)

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Drive Mechanisms E L E V E N

Gas cap present initially Oil at interface is at Pb

Gas cap

Oil

Reservoir Pressure

Oil Oil may be above P b G.O.R Prod With production - Gas cap expansion Solution gas liberation Oil Prod

G.O.R

Reservoir Pressure

Time-Year

Reservoir Engineering 22/07/14

Petroleum Engineering

Drive Mechanisms E L E V E N

C O N T E N T S 1 DEFINITION 2 NATURAL DRIVE MECHANISM TYPE 2.1 Depletion Drive Reservoirs 2.2 Water Drive 2.3 Compaction Drive 2.4 Gravity Drainage 2.5 Depletion Type Reservoirs 2.5.1 Solution Gas Drive 2.5.2 Gas Cap Drive 2.6 Water Drive Reservoirs 2.7 Combination Drives 3 RESERVOIR PERFORMANCE OF DIFFERENT DRIVE SYSTEMS 3.1 Solution Gas Drive 3.1.1 Solution Gas Drive, Oil Production 3.1.2 Solution Gas Drive, Gas/Oil Ratio 3.1.3 Pressure 3.1.4 Water Production, Well Behaviour, Expected Oil Recovery and Well Location 3.2 Gas Cap Drive 3.2.1 Oil Production 3.2.2 Pressure 3.2.3 Gas/Oil Ratio 3.2.4 Water Production, Well Behaviour, Expected Oil Recovery and Well Locations 3.3 Water Drive 3.3.1 Rate Sensitivity 3.3.2 Water Production, Oil Recovery 3.3.3 History Matching Aquifer Characteristics 3.3.4 Well Locations 4 SUMMARY 4.1 Pressure and Recovery 4.2 Gas/Oil Ratio

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Petroleum Engineering

LEARNING OBJECTIVES Having worked through this chapter the Student will be able to: • Define reservoir drive mechanism. • Describe briefly with the aid of sketches a depletion drive reservoir. • Describe briefly with the aid of sketches a water drive reservoir. • Describe briefly with the aid of a sketch a gravity drainage reservoir. • 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 gas cap drive . • Describe briefly with the aid of sketches the reservoir performance characteristics of a solution gas drive reservoir. • Describe briefly with the aid of sketches the reservoir performance characteristics of a gas cap drive reservoir. • 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.

2

Reservoir Engineering

Drive Mechanisms E L E V E N

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 preoccupations 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.

2 NATURAL DRIVE MECHANISM TYPES 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. 22/07/14

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Gas Oil Water

Figure 1 Depletion reservoir: No aquifer. Isolated sand lenses

Gas Oil Water

Figure 2 Depletion reservoir: Aquifer limited by faults

2.2 Water Drive

Gas Oil Water

Figure 3 Water drive: Active aquifer 4

Drive Mechanisms E L E V E N

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

Figure 4 Reservoir having artesian water drive.

2.3 Compaction Drive

Figure 5 illustrates another drive mechanism, compaction drive. Although not a common drive energy, the characteristics of its occurrence can be dramatic. Compaction drive occurs when the hydrocarbon formation is compacted as a result of the increase in the net overburden stress as the reservoir pore pressure is reduced during production. The nature of the rock or its degree of consolidation can give rise to the mechanism. For example a shallow sand deposit which has not reached its minimum porosity level due to consolidation can consolidate further as the net overburden stresses increase as fluids are withdrawn. The impact of the further consolidation can give rise to subsidence at the surface. This phenomenon of compaction with increasing net overburden stress is not restricted to unconsolidated sands, since chalk also demonstrates this phenomenon. One of the spectacular occurrences of compaction drive is that associated with the Ekofisk Field, in the Norwegian sector of the North Sea. This is a very undersaturated chalk reservoir. The field was developed on the basis of using depletion drive down to near the bubble point and then to inject sea water to maintain pressure above the bubble point. During this period of considerable pressure decline, the net overburden stress was increasing, causing the formation to compact 22/07/14

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Reservoir Engineering

to an extent that subsidence occurred at the seabed. In an offshore environment such uniform subsidence can go undetected, as was the case for Ekofisk. The magnitude of the subsidence has been such that major jacking up of the structures has been required. Old land surface New land surface

Oil

Figure 5 Compaction drive

2.4 Gravity Drainage

Gravitational segregation or gravity drainage can be considered as a drive mechanism. Figure 6 illustrates a situation where the natural density segregation of the phases can be responsible for moving the fluids to the well bore. Gravity drainage is where the relative density forces associated with the fluids cause the fluids, the oil, to drain down towards the production well. The tendency for the gas to migrate up and the oil to drain down clearly will be influenced by the rate of flow of the fluids as indicated by their relative permeabilities. Gravity drainage is generally associated with the later stages of drive for reservoirs where other drive mechanisms have been the more 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.

6

Drive Mechanisms E L E V E N

Closed in

Z

1000

Initial GOC Present GOC ΟWC

Gas Oil Water

Inactive aquifer Gravity drive typically is active during the final stage of a depletion reservoir.

Figure 6 Gravity drive

2.5. Depletion Type Reservoirs

In depletion drive reservoirs the energy comes from the expansion of the fluids in the reservoir and its associated pore space. There are two types of depletion drive reservoirs, solution gas drive reservoirs and gas cap drive reservoirs. In solution gas drive reservoirs there are two stages of drive mechanism where different energies are responsible for fluid production. 2.5.1. Solution Gas Drive In solution gas drive reservoirs the initial condition is where the reservoir is undersaturated, i.e. above the bubble point. Production of fluids down to the bubble point is as a result of the effective compressibility of the system. When considering pressure volume phase behaviour, in the chapter on phase behaviour, we observed a small increase in volume of the oil for large reductions in pressure, for oil in the undersaturated state. Associated connate water also has a compressibility as has the pore space within which the fluids are contained. This combined compressibility provides the drive mechanism for depletion drive above the bubble point. Perhaps this part of the depletion drive should be called compressibility drive. The low compressibility causes rapid pressure decline in this period and resulting low recovery. Of the three compressibilities, although it is the oil compressibility which is the larger, the impact of the other compressibility components, the water and the pores, should not be neglected. As pressure is reduced, oil expands due to compressibility and eventually gas comes out of solution from the oil as the bubble point pressure of the fluid is reached. The expanding gas provides the force to drive the oil hence the term solution gas drive. It is sometimes called dissolved gas drive (Figure 7). Gas has a high compressibility compared to liquid and therefore the pressure decline is reduced. Solution gas drive only occurs once the bubble point pressure has been reached.

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Initially no gas cap and Oil above Pb

Figure 7 Solution gas drive reservoir

2.5.2. Gas Cap Drive Another kind of depletion type is where there is already free gas in the reservoir, accumulated at the top of the reservoir in the form of a gas cap (Figure 8), as compared to the undersaturated initial condition for the previous solution gas drive reservoir. This gas cap drive reservoir, as it is termed, receives its energy from the high compressibility of the gas cap. Since there is a gas cap then the bottom hole pressure will not be too far away from the bubble point pressure and therefore solution gas drive could also be occurring. The gas cap provides the major source of energy but there is also the expansion of oil and its dissolved gas and the gas coming out of solution. The oil expansion term is very low and is within the errors in calculating the two main energy sources. The two significant sources of driving energy are ;

8



(1)

Gas cap expansion



(2)

Expansion of gas coming out of solution

Reservoir Engineering

Drive Mechanisms E L E V E N

Gas cap present initially Oil at interface is at Pb

Gas cap

Oil

Oil may be above Pb With production - Gas cap expansion Solution gas liberation

Figure 8 Gas cap drive reservoir

2.6 Water Drive Reservoirs

Water drive reservoirs are also of two types. There is an edge water drive reservoir. The reservoir is thin enough so that the water is in contact with the hydrocarbons at the edge of the reservoir (Figure 9). The other type of water drive reservoir is the bottom-water-drive reservoir; where the reservoir is so thick or the accumulation so thin that the hydrocarbons are completely underlain by water (Figure 10).

Edge water

Figure 9 Edge water drive reservoir

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Reservoir Engineering

Bottom water Water coning

Figure 10 Bottom water drive reservoir

2.7 Combination Drives

‘Pure’ types of reservoirs are those reservoirs where only one drive system operates, for example, depletion drive only - no water drive or water drive only - no gas drive. It is rare for reservoirs to fit conveniently into this simple characterisation. In many of them a combination of drive mechanisms can be activate during the production of fluids. Such reservoirs are called combination drives (Figure 11). In the case in Figure 11, which is not unusual, we have a gas cap with the oil accumulation underlain by water providing potential water drive. So both free gas and water are in contact with the oil. In such a reservoir some of the energy will come from the expansion of the gas and some from the energy within the massive supporting aquifer and its associated compressibility.

Gas Cap

Water

Oil zone Original condition

Water

Gas Cap

Water

Oil zone 50 % Depleted

Water

Figure 11 Combination water and gas - cap drive

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 E L E V E N

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.

3 RESERVOIR PERFORMANCE OF DIFFERENT DRIVE SYSTEMS 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. The performance of different types of reservoirs in relation to the daily production, gas/oil ratio and water production can give some indication of the type of drive mechanism operative in the reservoir.

3.1 Solution Gas Drive

In the first part of solution gas drive, in what we termed compressibility drive, within the reservoir no production of gas occurs and the fluid moves as a result of decompression of the three components oil, water and pore space. The pressure reduction is rapid in relation to volumes produced. The gas to oil ratio produced at the surface is constant since the reservoir at this stage is above its bubble point pressure. Once the bubble point is reached gas comes out of solution. Initially the gas bubbles are small and isolated. The size and number of the bubbles increase until they reach a critical saturation when they form a continuous phase and become mobile. At this stage the gas has relative permeability. The impact of the first bubbles of gas on the oil is very significant. The relative permeability to the oil is reduced by the presence of the non wetting gas. (See gas-oil relative permeabilties in chapter 7. Figure 44) As the increase in saturation of gas increases at the expense of oil saturation, the relative permeabilties move in the same directions giving rise to reduced well productivity to oil and increased productivity to gas, Figure 12. That is the oil relative permeability decreases and the gas relative permeability increases. The gas although providing the displacing medium is effectively leaking out of the system. Not only does the gas progress to the wellbore, depending on vertical permeability characteristics it will move vertically and may form a secondary gas cap. If this occurs it can contribute to the drive energy. Well location and rate of production can be used to encourage gas to migrate to form such a gas cap as against being lost through production from the wellbore. Vertical gas migration

Rs< Rsi

Gas relative permeability

Rs< Rsi

<

Rs Rsi

Oil relative permeability

Figure 12 Schematic of solution gas drive.

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We will now review the various production profiles, specific to the drive mechanisms but before doing so we will review the various phases of production.

Production

Plateau phase

0

Decline phase Production build up Abandonment Time

Figure 13 Phases in production.

Production Phases (Figure 13) The first phase, production build up, which may exist or not depending on the drilling strategy is the increased production as wells are brought on stream. Clearly, as in some cases, wells might be predrilled through a template and then all brought on stream together when connected to production facilities, such a build up of production will, therefore, not occur. 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 E L E V E N

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 initial 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. Reservoir Pressure

Oil G.O.R Prod

Oil Prod

G.O.R

Reservoir Pressure

Time-Year

Figure 14 Production for solution gas drive

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!

3.1.2 Solution Gas Drive, Gas/Oil Ratio

The distinctive characteristic of the solution gas drive mechanism is related to the producing gas to oil ratio. When the reservoir is first produced the GOR being produced may be low corresponding to the RSi value of the reservoir liquid. If the reservoir is highly undersaturated there will be a period when a constant producing GOR occurs 1-2 in Figure 15. 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.

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As the pressure further reduces the released gas becomes mobile and moves at a velocity greater than its associated oil due to the relative permeability effects. Oil enters the well bore, with its below bubble point solution GOR value, but also gas enters the well bore from oil which has not yet arrived. The net effect is that at the surface the producing GOR increases rapidly as free gas within the reservoir, which has come out of solution, moves ahead of the oil 3-4 in Figure 15.

Producing GOR.

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 point pressure

1

Rsi

2

4

5

3 Pb

Pressure

Figure 15 Producing GOR for solution gas drive reservoir

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.

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Drive Mechanisms E L E V E N

Secondary gas cap

Oil water contact

Figure 16 Well location for solution gas drive reservoir.

3.2 Gas Cap Drive

Whereas for a solution gas drive reservoir where we have a reservoir initially in an undersaturated state, for a gas cap drive reservoir, Figure 17, the initial condition is a reservoir with a gas cap. Since the gas oil contact will be at the bubble point pressure the pressures within the oil accumulation will not be higher than this only so far as relates to the density gradient of the fluid. It is the gas cap, with its considerable compressibility, which provides the drive energy for such fields, hence the name. To get flow in the wells it is likely that gas will come out of solution in the near well bore vicinity and therefore some degree of solution gas drive will also take place. A good analogy for this type of reservoir is the plastic chemical dispenser fitted with a pump to maintain gas pressure above the dispensed liquid.

Gas Cap

Water

Oil zone Original condition

Water

Gas Cap

Water

Oil zone 50 % Depleted

Water

Figure 17 Gas-cap drive

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3.2.1 Oil Production

The producing characteristics for a gas cap drive reservoir are illustrated in Figure 18. Although the production may be high as in the solution gas drive, the oil production still has a significant decline but not as rapid as for solution gas drive. This decline in oil production is due to the reducing pressure in the reservoir but also from the impact of solution gas drive on the relative permeability around the well bore. If the well is allowed to produce at too fast a rate, the very favourable mobility characteristics of the gas, arising from its low viscosity compared to the oil, are such that preferential flow can cause gas breakthrough into the wells and the well is then lost to oil production. Indeed it is this condition which will determine well abandonment.

3.2.2 Pressure

500

Pressure

G.O.R

Oil Prod (1000)

With an associated gas cap a loss of volume of fluids from the reservoir is associated with a relatively low drop in pressure because of the high compressibility of the gas. In solution gas drive much of the driving gas is produced, but with a gas cap the fluid remains till later in the life of the reservoir. The pressure drop for a gas cap system therefore declines slowly over the years. The decline will depend on the relative size of the gas cap to the oil accumulation. A small gas cap would be 10% of the oil volume whereas a large gas cap would be 50% of the volume.

10 Gas Break through

Pressure

Oil Prod Rate

5

250

5000

2500

G.O.R

BSW % 20 10

0

0

1

2

3

4

Time-Year

5

6

7

0

Figure 18 Reservoir performance gas - cap drive.

3.2.3 Gas/Oil Ratio

During the early stages of replacement of oil by gas a 100% replacement takes place. Later on gas by-passes oil and causes a reduced displacement efficiency. In the early stages the GOR remains relatively steady increasing slowly as the impact of solution gas drive generates gas from oil still to reach the well bore. The increasing mobility of the gas is such that there is an increasing GOR both from dissolved gas and by-pass gas and eventually the well goes to gas as the gas cap breaks through.

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Drive Mechanisms E L E V E N

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!

3.3.1. Rate Sensitivity.

The characteristic features of natural water drive reservoirs are strongly influenced by the rate sensitivity of these reservoirs. If oil production from the formation is greater than the replacement flow of the aquifer then the reservoir pressure will drop and another drive mechanism will contribute to flow, for example solution gas drive. 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.

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Outcrop of sand

Oil well

Water flow

Pi

Reservoir pressure Oil production rate

Production GOR

Rsi

Water production Time

Figure 19 Producing characteristics for artesian water drive.

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 E L E V E N

Pi

Rsi

Oil production rate

Reservoir pressure

Production GOR

Water production Time

Figure 20 Producing characteristics for water drive (confined aquifer).

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 reservoir 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.

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2000 10000

Oil production

50

250

GOR

25 Bsw 69

70

71

72

Water

73

74

75

0

ratio Producing gas / oil

B/d Oil production rate

0

PROD

Water production

psi Reservoir pressure 0

GOR 500

Reservoir pressure

Ps 1000 5000

BSW

0

Figure 21 Reservoir performance - Water drive.

3.3.2 Water Production, Oil Recovery Because there is a large aquifer associated with the oil reservoir unlike depletion drive systems, water production starts early and increases to appreciable amounts. This water production is produced at the expense of oil and continues to increase until the oil/water ratio is uneconomical. Total fluid production remains reasonably steady. The expected oil recovery from a water drive reservoir is likely to be from 35 to 60% of the original oil-in-place. Clearly these recovery factors depend on a range of related aspects , including reservoir characteristics for example the heterogeneity as demonstrated by large permeability variations in the formation.

3.3.3. History Matching Aquifer Characteristics.

Predicting the behaviour of water drive reservoirs in particular the rate of water encroachment is not straightforward. The topic is covered in a later chapter, but a significant perspective as mentioned previously is that data is required of the aquifer to carry out the calculations. In particular the size and geometry of the aquifer and its permeability and compressibility characteristics. Since such information is generally not available during the exploration and development phase, the characteristics of the aquifer are only determined once production has been operational and the support from the aquifer can be calculated from production and pressure data. (History Matching). Getting such information may require producing a significant proportion of the formation say 5% of the STOIIP. RFT surveys have provided a very effective way of determining the aquifer strength as well as the communicating layers of the formation. Pressure depth surveys taken in an open hole development well after production has started will give indications of pressure support in the formation 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 E L E V E N

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.

3.3.4. Well Locations

Well locations for water drive reservoirs are such that they should be located high in the structure to delay water breakthrough.

4 SUMMARY The following summaries and tables give the main features associated with the various drive mechanisms.

4.1 Pressure and Recovery 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 oilin-place.

4.2 Gas/Oil Ratio

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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.

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SOLUTION GAS DRIVE RESERVOIRS 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

GAS CAP DRIVE RESERVOIRS Characteristics

1. Reservoir Pressure 2. Gas/oil ratio 3. Production Rate 4. Water Production 5. Well Behaviour Cap 6. Expected Oil Recovery



Trend

Falls slowly and continuously Rises continuously First high, then declines gradually Absent or negligible Long flowing life depending on size of gas cap 20 to 40% of original oil-in-place

WATER DRIVE RESERVOIRS 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 E L E V E N

Reservoir pressure trends for reservoirs under various drives.

Reservoir pressure - percent of original

100 80

Water drive

60 Gas cap drive

40 20 0

Dissolved gas drive 0

20 40 60 80 Oil produced - percent of original oil in place

100

Figure 22

5

Reservoir gas - oil ratio trends for reservoirs under various drives.

GOR MCF /BBL

4 Dissolved gas drive

Gas cap drive

3 2 1 0

Water drive 0

20 40 60 80 Oil produced - percent of original oil in place

100

Figure 23

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Vapour Liquid Equlibria T W E L V E

CUSTOMERS

ONSHORE FACILITIES

OFFSHORE FACILITIES

OFFSHORE FACILITIES

V WELLS

WELLS

RESERVOIR

F

yj

T&P

RESERVOIR

zj

L

xj

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Vapour Liquid Equlibria T W E L V E

C O N T E N T S 1 INTRODUCTION – THE IMPORTANCE OF VAPOUR-LIQUID EQUILIBRIUM 2 IDEAL SOLUTIONS 2.1 Raoult's Law 2.2 Dalton's Law 2.3 Ideal Equilibrium Ratio 3 NON IDEAL SYSTEMS 4 EQUATIONS FOR CALCULATING EQUILIBRIUM RELATIONS 4.1 Vapour – Liquid Calculations 4.2 Dew – Point Calculation 4.3 Bubble Point Calculation 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

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LEARNING OBJECTIVES Having worked through this chapter the Student will be able to: • Define equilibrium ratio. • Derive equations for vapour-liquid equilibrium calculations for real systems and explain the application of the equations. • 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 on the gas-oil ratio and oil formation volume factor.

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Vapour Liquid Equlibria T W E L V E

1 INTRODUCTION - THE IMPORTANCE OF VAPOUR-LIQUID EQUILIBRIUM The multiphase perspectives of hydrocarbon mixtures in and produced from reservoir accumulations are an important aspect in reservoir management, well productivity, facility design and pipeline transport. 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. The behaviour in reservoirs of multicomponent mixtures and their production to surface has provided one of the most rigorous challenges to design engineers because of the complex and unique nature of the fluids and in many cases their behaviour near the critical point. Figures 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

OFFSHORE FACILITIES

OFFSHORE FACILITIES

WELLS

WELLS

RESERVOIR

RESERVOIR

Figure 1 The total system.

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CHEMICAL FEEDSTOCK

Reservoir Engineering

FUEL

OIL SALES & TAXES

TERMINAL

Composition = (TA*CA)+(TB*CB)+(TC*CC) TA + TB + TC

Throughput = TA + TB + TC

FIELD A

FIELD C

Composition CA Throughput TA

Composition CC Throughput TC

FIELD B Composition CB Throughput TB

Figure 2 Complexity of allocation of produced oil to supply fields.

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 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. 4

Vapour Liquid Equlibria T W E L V E

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 :

Bo =



Vo 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 :

Rs =



Vg Vst

The volume factors can be determined directly in the laboratory or from equilibrium calculations. In addition to separator calculations, vapour liquid equilibrium data can be used for: •

Reservoir calculations

•

Two phase pipeline flow calculations

•

Process calculations

Although phase behaviour considerations are required throughout the production process from reservoir to refinery the context of this particular chapter is in relation to reservoir predictions. When reservoir fluids undergo phase alteration as a result of changes in pressure, temperature or composition it is considered that these changes are

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slow and therefore the resulting separate phases are in equilibrium, i.e the properties of the phases are not changing with respect to time. For multicomponent phase behaviour predictions, thermodynamic principles have been applied to provide the predictive tools. Many works have been written which provide the foundation of the topic. Danesh1 in his text provides a good review of the topic both with respect to the foundation principles and the equations used. Vapour-liquid equilibrium calculations have been somewhat restricted to analysis of behaviour with the separate areas, i.e. reservoir and wells, surface separation, pipelines, onshore treatment and refinery operations. Increasingly in the modern multidisciplinary approach to technical management there is an interest in the integrated perspective of vapour liquid equilibrium. For example, what is the impact on the quality of product exiting from a multifield oil transport line of a pressure change in field X. Such integrated perspectives provide a considerable technical and commercial challenge to the various technical disciplines which have been separately involved.

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.

2.1 Raoult’s Law

Raoult’s equation states that the partial pressure of a component in the gas is equal to the product of the mole fraction, xj in the liquid, multiplied by the vapour pressure of the pure component pvj. pj = xjpvj (1) where pj is the partial pressure of component j in the liquid with a composition xj and pvj is the vapour pressure of the pure component j.

2.2 Dalton’s Law

Dalton’s law states that the partial pressure of a component pj is equal to the mole fraction of that component in the gas, yj times the total pressure of the system p, i.e. pj = yjp (2) where yj is the composition of the vapour and p is the pressure of the system

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2.3 Ideal Equilibrium Ratio

By combining the above two laws, yjp = xjpvj (3)

y j pvj (4) = xj p



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. zjn = xjnL + yjng (5) where nL and ng are the moles of liquid and gas such that nL + ng = n From equation 4.

z j n = x j nL + x j

∴ xj =



pvj ng p

zjn p nL + vj ng p

(6)

Σxj by definition = 1.0

zjn = 1.0 where c is the number of components (7) p j =1 n + vj .n L g p

c

c

∴∑ xj = ∑ j =1

Similarly:

c

zjn = 1.0 (8) p j =1 n + . n g L pvj c

∑y = ∑ j =1

j

If a basis of one mole of mixture is used i.e. n

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∑x = ∑ j

g

+ n L = 1.0

zj = 1.0 (9)  pvj  − 1 1 + ng   p 

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∑y = ∑ j

zj

 p  − 1 1 + nL   pvj 

= 1.0

Reservoir Engineering

(10)

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.

Kj =

yj (11) xj

Clearly Kj is defined at a particular pressure and temperature. 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,

φ=

fi Pzi

(12)

All systems behave as ideal gases at very low pressures, therefore φ > 1 when P is a few tens of psia When fugacities are not 1 , then this gives an indication of non-ideality. Fugacity has been imagined (Danesh)1 as a measure of the escaping tendency of molecules from one phase to an adjacent phase. In multicomponent systems, if the fugacity of a component in adjacent phases is the same, the two phases will be in equilibrium with no net transfer of molecules from one phase to another. At equilibrium therefore fg = fL. (13) The fugacity coefficient, φ of a pure component can be calculated from the following general equation (Danesh).

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Vapour Liquid Equlibria T W E L V E



p

 z − 1 1  RT ln φ = ∫  − P dv (14)  dp = ( z − 1) − ln z + ∫    p  RT ∞ v o v

The ratio of the fugacity at the state of interest to that at a reference state is called the activity εi = fi/fio The activity is a measure therefore of the fugacity contribution or activeness of the component in a mixture. fi = εifio . The ratio of activity to concentration is called the activity coefficient Θi, where Θi = εi/xi

Therefore fi=Θixifio

(15)

3 NON IDEAL SYSTEMS Ideal solution assumptions cannot be applied to the systems relevant to multicomponent hydrocarbon fluids in reservoir flow, transport and processing conditions. The ideal assumptions only apply to low pressures and moderate temperatures, chemically and physically similar components and behaviour below the critical point. Different methods have been developed for treating vapour-liquid equilibrium for non ideal systems. The previous K value is based on both ideal and non 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.

4 EQUATIONS FOR CALCULATING EQUILIBRIUM RELATIONS 4.1 Vapour-Liquid Calculations

The calculations for determining the amount of liquid and vapour present in a mixture when the pressure and temperature are known are obviously important, for example, in optimising the performance of a separator process. 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. Figure 3 represents such a separation element.

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V

yj F

T&P

zj

L

xj

Figure 3 Vapour-liquid separation in an element.

F L V zj xj yj

= = = = = =

total moles of system both liquid and gas total moles of material within liquid phase total moles of material within vapour phase mole fraction of jth component in the mixture mole fraction of jth component in the liquid mole fraction of jth component in the vapour

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 zjF = xjL + yjV

For F = 1.0 mole

zj = xjL + yjV (17) The equilibrium ratio:

yj (18) xj



Kj =



By definition:



10

m

m

m

j =1

j =1

j =1

∑ x j = ∑ y j = ∑ z j = 1 (19)

Vapour Liquid Equlibria T W E L V E

where m is the number of components. Replacing yj by Kjxj in (17) zj = xjL + xjKjV zj = xj (L + KjV) dividing both sides by L + KjV

zj L + K jV

xj =

and:

m

m

(20)

zj

∑ x = ∑ L + K V = 1.0 (21) j =1

j

j =1

j

similarly:

zj = 1.0 (21a) j =1 V + L K j

m

m

∑ yj = ∑ j =1

by multiplying (21) by V we get:

m

∑L j =1

V

zj + Kj

= V (22)

and (21a) in the same way:

m

∑ j =1

zj = V (22a) L +1 K jV

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 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.) (2) Assume a vapour liquid split i.e. V&L such that V + L = 1.0; 22/07/14

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(3) Calculate either V, L, ∑xj or ∑yj from equations 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. It can be understood therefore that in a compositional reservoir simulator a considerable amount of computational time is taken up because of these iterative calculations at each grid block. In a black oil simulator no such iteration takes place; the specific pressure and temperature provide the direct phase values either from a PVT report or an empirical black oil correlation. 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.

4.2 Dew-Point Calculation

The dew-point is when the first drop of liquid begins to condense. At this point the composition of the liquid drop is higher in heavier hydrocarbons whereas the composition of the vapour is essentially the composition of the system: Figure 4. Vapour yj = Zj Liquid Xj

Vapour Liquid

Figure 4 Conditions at the Dew Point.

At the dew point therefore: zj = yj or:

(22)

zj = xjKj

The mixture at the dew-point is therefore in equilibrium with a quantity of liquid having a composition defined by the above equation. Clearly:

m

zj = 1.0 (23) j =1 K j m

∑ xj = ∑ j =1

Similarly for the bubble point.

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Vapour Liquid Equlibria T W E L V E

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

Figure 5 Conditions at the Bubble Point.

At the bubble point therefore: zj = xj or:

zj =

(24)

yj Kj

The mixture at the bubble point is therefore in equilibrium with a quantity of vapour having a composition defined by the above equation. Also:

m

m

j =1

j =1

∑ y j = ∑ z j K j = 1.0 (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 In a separator a stream of fluid is brought to equilibrium at separator temperature and pressure. Vapour and liquid are separated within the unit and continue as separate streams. Several separators can be operated in series each receiving the liquid phase from the separator operating at the next higher pressure.

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Each condition of pressure and temperature at which vapour and liquid are separated is called a stage-separation. Hence a process using one separator and a stock tank is a two stage process a three stage process has two separators and one stock tank. (Figure 6). Separator calculations are performed to determine the composition of products, the oil formation-volume factor and the volume of gas released per barrel of oil and to determine the optimum separator conditions for the particular conditions of fluid. Using equilibrium calculations already derived we can calculate the separation achieved at each stage, the composition of the phases separated, the gas/oil ratio, and the oil formation volume factor. Vapor

Vent

Feed To pipeline / tanker Liquid Separator at P sep and T sep

Stocktank at P st and T st

Two-stage separation

1st stage vapor

2nd stage vapor

Vent

Feed To pipeline / tanker Liquid 1st stage separator at (P sep ) 1 and (T sep ) 1

Liquid 2nd stage separator at (P sep ) 2 and (T sep ) 2

Stocktank at P st and T st

Three-stage separation

Figure 6 Schematic drawing of separation process.

5.1 Gas/Oil Ratio

Gas is removed from each stage so that the solution GOR can be calculated for each stage or combination of stages.

14

Total gas / oil ratio =

sum of gas volumes (SCF) = RT volume of stock tank oil (bbl)

Vapour Liquid Equlibria T W E L V E

(a) Calculation for Stock Tank Oil, STO. If n1 moles enter first stage, moles of liquid entering 2nd = n2 = n1L1. where L1 = separation in stage one based on basis of one mole feed. Number of moles entering third stage n3 = n2L2 = L2L1n1 If third stage is the stock tank then: nST = L3n3 = L3L2L1n1 nST is the moles of liquid in stock tank for n1 moles into first separator:

m

∴ nST = n1 ∏ Li (26) i =1

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 (27) i =1

m



n ST = ∏ Li i =1

= mole fraction of STO in the feed.

(b) Calculation of Total Gas ngi = number of moles off stage i ng1 = V1n1 ng2 = V2n2 = V2L1n1 ng3 = V3n3 = V3L2L1n or in general for total gas:

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m

m

i −1

i =1

i =1

j =1

ngT = ∑ ngi = n1 ∑ Vi ∏ L j



∴ If nj = 1



n gT = ∑ Vi ∏ L j (28)



ngT = mole fraction of total gas in the feed

m

i −1

i =1

j =1

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Total gas volume per mole of feed =ngT Vmcu ft where Vm is the molar volume

5.2 Oil Formation Volume Factor

Volume of stock tank oil per mole of feed

(VST )m =

n ST MST (29) ρST

MST = molecular weight of stock tank oil

(V ST )m =

nST = moles of STO pet mole ST M ST ρST



ρST = density of STO at standard conditions lb/bbl



RT =

Total gas to oil ratio RT =

n gT Vm n gT Vm ρST = (VST )m n ST MST

(30)

where RT is the total gas - oil ratio. 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.

ρ res. = density of feed (lb/bbl)



Volume of reservoir oil per mole = Vres = Mres/ρres



Bo =

Oil formation volume factor Bo =

(Vres ) = VST

Mres ρST (31) ρres MST nST

where and

nST =

Mres = molecular weight of reservoir fluid =

lb.res. fluid lb.mole. fluid

lb.mol. stock tank fluid lb.mol.res. fluid

5.3 Optimum Pressure of Separator System

The operating conditions of pressure and temperature of a separator influence the amount of gas and stock tank oil produced. Changes in these valves will change the GOR and the Bo. In quoting these values therefore it is important to keep note of the associated separation conditions of pressure and temperature. A number of units in series also influence these parameters. It is the role of the process designers to optimise the operating conditions of such limits and the number of units required. 16

Vapour Liquid Equlibria T W E L V E

1.36

33.4

580

1.34

560

1.32

540

1.30

33.2

33.0

32.8

32.6

32.4

API g ravity

600

Separator + stock tank gas-oil ratio, scf/bbl

33.6

520

1.28 io -oil rat

Total gas

500

480

e olum Formation v 0

20

100 40 60 80 First-stage separator pressure, psi

or fact

120

Formation volume factor

stock tank gravity, ˚API at 60˚F

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.

1.26

1.24 140

Figure 7 Effect of separator pressure in a two-stage separation process (Amyx, Bass & Whiting).

Figure 7 illustrates the influence of a change of pressure for a two-stage separation process on GOR, Bo and the density of stock tank oil. 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.

5.4 Example of Separator Problem (McCain)4

The following example is taken from McCain’s text on Petroleum Fluids and the values for K used in the calculations come from the GPSA sources5. Calculate the gas-to-oil ratio, stock-tank oil gravity and formation-volume factor which will result from a two-stage separation of the hydrocarbon mixture below. Use separator conditions of 76˚F and 100 psig. Assume that the mixture is a liquid at its bubble point at reservoir conditions of 2,695 psig and 220˚F.

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Component

Mole Fraction

Methane Ethane Propane i-butane n-butane i-pentane n-pentane Hexanes Heavier

0.3378 0.0694 0.0982 0.0133 0.0299 0.0125 0.0193 0.0299 0.3897 1.0000

Reservoir Engineering

Gravity °API at 60°F 60°/60°

Molecular Weight

0.8859

263

28.2

Step 1: Calculate the composition and quantities of separator gas and liquid using equation 21.



zj

∑ x = ∑ L + K V = 1.0 j

Component of feed to separator

j

Component, mole fraction zj

C1 C2 C3 i-C4 n-C4 i-C5 n-C5 C6 C7+ *

0.3378 0.0694 0.0982 0.0133 0.0299 0.0125 0.0193 0.0299 0.3897 1.000 * Used K for C9

Equilibrium ratio at 114.7 psia and 76°F Kj 20.8 4.07 1.16 0.495 0.343 0.142 0.108 0.0334 0.00150*

x=

V=0.42 0.03626 0.03031 0.09202 0.01688 0.04129 0.01954 0.03086 0.05033 0.67117 0.98866

zj L+KjV V=0.43

0.03551 0.02991 0.09188 0.01699 0.04167 0.01985 0.03131 0.05117 0.68291 1.0015

V=0.4291 0.03557 0.02995 0.09189 0.01698 0.04164 0.01978 0.03127 0.05109 0.68184 1.00001

The summation equals 1.0 when V = 0.4291 and L = 0.5709 and the compositions of the separator gas and liquid are: Component C1 C2 C3 i-C4 n-C4 i-C5 n-C5 C6 C7+

xj =

zj L+KjV

0.0356 0.0299 0.0919 0.0170 0.0416 0.0198 0.0313 0.0511 0.6818 1.0000

yj=kjxj 0.7399 0.1219 0.1066 0.0084 0.0143 0.0028 0.0034 0.0017 0.0010 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.

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Vapour Liquid Equlibria T W E L V E

Component of feed to separator

Component, mole fraction zj

C1 C2 C3 i-C4 n-C4 i-C5 n-C5 C6 C7+*

0.0356 0.0299 0.0919 0.0170 0.0416 0.0198 0.0313 0.0511 0.6818 1.0000

Equilibrium ratio at 14.7 psia and 76°F, Kj 161 30.7 8.15 3.27 2.30 0.90 0.675 0.20 0.0089*

V=0.13 0.00163 0.00615 0.04763 0.01313 0.03559 0.02006 0.03286 0.05703 0.78264 0.99654

zj L+KjV V=0.14

V=0.1351

0.00152 0.00580 0.04593 0.01290 0.03519 0.02008 0.03279 0.05755 0.79164 1.00340

0.00157 0.00597 0.04674 0.01301 0.03538 0.02007 0.03274 0.05729 0.78720 1.00000

xj=

* 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 0.0016 0.0060 0.0467 0.0130 0.0354 0.0201 0.0327 0.0573 0.7872 1.0000

yj = Kjxj 0.2534 0.1831 0.3810 0.0425 0.0814 0.0181 0.0221 0.0115 0.0070 1.0001

Step 3: Calculate the density and molecular weight of the stock tank oil. Component of stock tank oil

Component, mole fraction xj

Molecular weight

C1 C2 C3 i-C4 n-C4 i-C5 n-C5 C6 C7+ Total

0.0016 0.0060 0.0467 0.0130 0.0354 0.0201 0.0327 0.0573 0.7872 1.0000

16.0 30.1 44.1 58.1 58.1 72.2 72.2 86.2 263

Weight

Mj

xjMj 0.026 0.181 2.059 0.755 2.057 1.451 2.361 4.939 207.034 220.863 ∑C3+ = 220.656 ∑C2+ = 220.837 MSTO = 220.863 lb lb mole

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Liquid density at 60°F and 14.7 psia ρoj 31.66 35.12 36.45 38.96 39.35 41.30 55.25

Liquid volume at 60°F and 14.7 psia xjMj / ρoj 0.0650 0.0215 0.0564 0.0372 0.0600 0.1196 3.7472 4.1069 cu.ft. C3+ lb.mole STO

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220.656 lb = 53.73 cu ft Density of propane plus = 4.1069 0.181 = 0.001 Weight fraction ethane in ethane plus = 220.837 0.026 = 0.0001 Weight fraction methane in STO = 220.863

ρSTO = 53.73

From chapter 6, figure 13.

53.73 = 0.861 62.4

γ STO = o

lb cu.ft.

API =

141.5 − 131.5 = 32.8 0.861

Step 4: Calculate gas to oil ratio R SP =



V1  lb moles sep gas   SCF sep gas   5.615 lb mole STO  × ρSTO 379    L1L ST  lb mole STO   lb mole sep gas   M STO STB 

RSP = 2130

Similarly:

RST =

V1ρSTO L1 LST MSTO

2130V2 ρSTO L2 MSTO

RT = RST + RSP



RSP =

SCF (2130)(0.4291)(53.73) = 450 STB (0.5709)(0.8649)(220.9)

RST =

SCF (2130)(0.1351)(53.73) = 81 STB (0.8649)(220.9)



RT = 531

SCF STB

Step 5: Calculate the density and molecular weight of the reservoir liquid at reservoir conditions.

20

Vapour Liquid Equlibria T W E L V E

Component Component, of reservoir mole liquid fraction xj

Molecular weight Mj

C1 C2 C3 i-C4 n-C4 i-C5 n-C5 C6 C7+

16.0 30.1 44.1 58.1 58.1 72.2 72.2 86.2 263

0.3378 0.0694 0.0982 0.0133 0.0299 0.0125 0.0193 0.0299 0.3897 1.0000

5.405 2.089 4.331 0.773 1.737 0.902 1.393 2.577 102.491 Mor =121.699 lb / lb mole res oil ∑C2+ = 116.294 ∑C3+ = 114.205

Liquid density at 60°F and 14.7 psia ροj 31.66 35.12 36.45 38.96 39.35 41.30 55.25

Liquid volume at 60°F and 14.7 psia xjMj / ροj 0.1368 0.0220 0.0477 0.0232 0.0354 0.0624 1.8550 2.1825 cu.ft.C3+ lb.mole res oil

114.205 lb = 52.33 2.1825 cu ft



Density of propane plus =



Weight fraction ethane in ethane plus =



Weight fraction methane in reservoir oil =

ρpo = 49.1 lb.cu ft.

2.089 = 0.018 116.294 5.405 = 0.044 121.699

From figure 13, Chapter 6.



From figure 14 chapter 6 compressibility correction 49.1 + 0.8 = 49.9 at 60˚F and 2710 psia.



From figure 15 chapter 6 thermal expansion correction 49.19 - 3.86 = 46.04 at 220˚F and 2710 psia.



ρor = 46.04 lb/cu ft.



Step 6: Calculate formation volume factor using equation:





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Weight xjMj

Bo =

M res ρSTO ρ res M STO L1L ST

Bo =

(121.7)(53.73) (46.04)(220.9)(0.5709)(0.8649)

Bo = 1.302

res bbl STB

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Integration of the Black-Oil and Compositional Approach The example above illustrates the combination of the compositional based prediction of phase volumes and associated properties and that based around the black-oil model, centred around parameters of oil formation volume factor and gas-oil ratio. By such a combination, the weaknesses of the simple two component black-oil model which is at the heart of describing oil field parameters, can be overcome by using compositional derived values rather than using perhaps inappropriate empirical correlations and charts. 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.

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Vapour Liquid Equlibria T W E L V E

REFERENCES (1) Danesh, A, "PVT and Phase Behaviour of Petroleum Reservoir Fluids." 1998 Elsevier. pp 105-206 (2)Prausnitz,J.M., Lichtenthaler,R.N., and d Azevedo,E.G. “ Molecular Thermodynamic of Fluid -Phase Equilibria’ 2nd Edition, Prentice Hall Inc, NY., (1986) (3) Smth,J.M. and Van Ness,H.C. “ Introduction to Chemical Engineering Thermodynamics”, Third Edition, McGraw-Hill ( 1975) (4) McCain,W.D. “The Properties of Petroleum Fluids” Petroleum Publishing Co. Tulsa 1973 (5) GPSA: Engineering Dat Book, Gas Processors Association. Tulsa Oklahoma, Gn Education 1972

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Equilibrium Ratio Prediction and Calculation T H I R T E E N

10 1,000

2

30

4

50

Pressure, PSIA

6 7 8 9 100

2

300 4 500 6

7 8 9 1,000

9 8 7 6

2

3,000 4

6 7 8 9

Plotted from 1947 tabulation of G. G. Brown, University of Michigan. Extrapolated and drawn by The Fluor Corp. Ltd. in 1957.

5 4

9 8 7 6 5 4

3

3

2

2

100

100

9 8 7 6

9 8 7 6

5

5

4

4

3

3

Temperature °F 50 40 0 0

2

2

30

25

10

10

0

0

9 8 7 6

9 8 7 6

0

20

K= y/x

10,000 1,000

10 0 80

5 4

5

60 40 20

3 2

3 2

0 -2 0 -4 0

1.0

9 8 7 6

1.0 9 8 7 6

-6

0

5

5

4

4

3

3

2

2

0.1

9 8 7 6

5 4 3

Kj - 1.0

2

2

30

4

50

6 7 8 9 100

2

300 4 500 6

Pressure, PSIA

7 8 9 1,000

Component 1

9 8 7 6

4 3

CONSTANT TEMPERATURE

0.1

In Kj

5

10

K= y/x

4

2

2

3,000 4

6 7 8 9

ETHANE

10,000

CONV. PRESS. 5000 PSIA

Component 2

In P

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In Pc

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Equilibrium Ratio Prediction and Calculation T H I R T E E N

C O N T E N T S 1 INTRODUCTION 1.1 Importance of the Critical Point 1.2 Equilibrium Ratio 2 EMPIRICAL PREDICTION METHODS FOR EQUILIBRIUM BEHAVIOUR 2.1 Black-Oil System 2.2 Correlation of Experimental Data Through K Values 2.2.1 Method of Calculating Convergence Pressure 3 METHODS BASED ON THERMODYNAMIC PRINCIPLES 3.1 Methods Based on Empirical Equations of State of Fluid Phase Theory 3.2 Prediction of Vapour-Liquid Equilibrium 4 THE COMPOSITION DATA REQUIREMENTS FOR THE APPLICATION THERMODYNAMIC EQUILIBRIUM PREDICTION METHODS TO MULTI-COMPONENT HYDROCARBON MIXTURES

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LEARNING OBJECTIVES Having worked through this chapter the Student will be able to:

2

•

Comment briefly on the approach to vapour equilibrium ( VLE ) calculations in terms of: the black oil approach, empirical K values and EOS.

•

Describe briefly the application of convergence pressure based K values in VLE calculations.

•

Comment on the evolution of EOS in VLE application. (The student would not be expected to reproduce the equations.)

Reservoir Engineering

Equilibrium Ratio Prediction and Calculation T H I R T E E N

VAPOUR-LIQUID EQUILIBRIUM PREDICTIONS OF PHYSICAL PROPERTIES 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 In petroleum engineering applications the situation is complicated in that often very little detailed information is available on the hydrocarbon mixture. That is, either individual components are not identified, or, because it is a multi-component mixture certain component concentrations are difficult to measure accurately even though they can greatly influence the properties of the mixture. 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 addition for petroleum engineering applications the calculation method used should be: 1 applicable to multi-component mixtures; 2

be accurate in predicting thermodynamic equilibrium and volumetric properties over a wide range of conditions of temperature and pressure and specifically be accurate around and beyond the critical point;

3

preferably require only pure component data or binary data which is either readily available from the literature or derivable from available data.

1.1 Importance of the Critical Point

The prediction of the true critical properties of a multi-component system is an important aspect of the general problem of predicting the overall phase behaviour of the system. The critical state is the unique condition about which the liquid and vapour phases are defined, and hence it has theoretical as well as practical significance. In hydrocarbon processing and producing operations, a knowledge of the critical condition is of particular significance because many of these operations take place under conditions which are at or near the bubble point or upper dew point regions and 22/07/14

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are frequently accompanied by isobaric or isothermal retrograde phenomena. Fluid property predictions and design calculations in this region are often the most difficult to make, and a knowledge of a precise location of the critical point for the system under study is of the utmost assistance. From a theoretical point of view, the derivatives of many of the thermodynamic and transport properties take on a special significance as the critical state is approached. In an empirical way the critical state has formed an integral part of many useful generalised correlations such as those based on the theorem of corresponding states or the convergence pressure concept in vapour-liquid equilibrium calculations. 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.

1.2 Equilibrium Ratio

In the previous chapter we presented the concept of equilibrium ratio, the measure of how a component is distributed between different phases however we didn’t explain how these distribution values are obtained. The literature abounds with methods and equations associated with phase equilibrium prediction. The methods can be broadly classified as: 1 methods which involve empirical curve fitting to experimental data; 2 methods which are based on thermodynamic principles.

2

EMPIRICAL PREDICTION METHODS FOR EQUILIBRIUM BEHAVIOUR

There are a range of approaches to expressing the vapour liquid distribution behaviour of reservoir fluid hydrocarbon systems, as outlined below.

2.1 Black-Oil System

The black-oil system as described in the chapter on liquid properties treats the fluids as two components, stock tank oil and solution gas. This concept and associated calculations were covered in that chapter.

2.2 Correlation of Experimental Data Through K Values

A complicating factor in the application of equilibrium ratios in the context of reservoir fluids is that the distribution of a component between phases is not only influenced by the temperature and pressure but is also influenced by the composition. Calculating 4

Reservoir Engineering

Equilibrium Ratio Prediction and Calculation T H I R T E E N

therefore how a unique fluid like that from a new exploration well performs in these 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:

where: K j =

yj 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. Figures 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. The effect of composition on the K values is achieved by applying the concept of convergence pressure. The concept of convergence pressure arose from the observation that for light hydrocarbon mixtures the isothermal component K-values converge to unity at a specific pressure known as the convergence pressure Pc, see Figures 3 for a binary mixture and 4 for a light oil. This convergence pressure is a measure of the composition of the mixture and can be calculated from a knowledge of the effective boiling point of the lightest and heaviest components in the mixture. 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. 22/07/14

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5

Petroleum Engineering

Reservoir Engineering

100

M

et ha

10

Et

K=1

He

Pr op

an es

ne

an e

ta ne s

xa ne s

e H

pt an es

Pe n

Bu t

ha ne

0.1

0.01 10

an d

He av ie

r

100

1,000 Pressure, psia

10,000

Figure 1 Equilibrium Ratios at 200˚F for Low-Shrinkage Oil(Amyx Bass & Whiting).1

40 20 10 8 6

Et ha

Equilibrium Ratio, K

4 2

1 0.8 0.6 0.4 0.2 0.1 0.08 0.06

Pr o Bu

ne

pa ne

tan es

Pe nta ne s He xa nes

0.04 He pta ne sa 0.02 nd 0.01 100

M et ha ne

Heavier

1,000 Pressure, psia

10,000

Figure 2 Equilibrium Ratios at 200˚F for a Condensate Fluid (Amyx Bass & Whiting).1 6

Equilibrium Ratio Prediction and Calculation T H I R T E E N

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.

M

10.0

et ha ne

Equilibrium Constant, K

h

Et an Pr e op Bu an ta e ne Pe s nt H an ex es an es

Variations of Equilibrium Constant With Pressure at 200º F

1.0

H 0.1

0.01

10

ep ta ne s

an

d

He av ie

r

100

1,000

10,000

Pressure, psia

Figure 4 Variation of K Values with pressure for a crude oil containing Light Hydrocarbons.

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Petroleum Engineering

The apparent convergence pressure is related to the composition of the fluid. This empirical basis has found favour in the industry for many years as a source of K value data despite the ability to calculate the information based on equations of state. However with the increasing availability of computational power and greater confidence in equations of state the use of convergence pressure based K values is diminishing. A comprehensive set of equilibrium ratios is published in the GPSA Engineering Data Book2. The K values for single components are presented in the NGAA book for a range of convergence pressures. K values for a convergence pressure of 5,000 psia are at the back of this chapter for 2 pure components. Figure 9 and 10.

2.2.1 Method of Calculating Convergence Pressure

As was indicated in figures 2 and 3 different mixtures present different K values as represented by different convergence pressures. The challenge is therefore to select the convergence pressure value appropriate to the fluid for which we are seeking to determine its phase separation characteristics. 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. It has been established that the convergence pressure of systems as encountered generally in natural gas processes is a function of the temperature and the composition of the liquid phase. This presumes that the liquid composition had already been known from a flash calculation using a first approximate guess for convergence pressure. Therefore, the method of calculating convergence pressure is an iterative procedure. This calculation is suggested: 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 1.0 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. 8

Reservoir Engineering

-300

100

200

300

400

500

1,000 900 800 700 600

Triple Phase Locus

en

Nitrog

2,000

-200

-100

Triple Phase Locus

Met h an e

3,000

yle ne

Eth

4,000

e 0

Eth an

5,000

CO2

C1-nC9

an e

10,000 9,000 8,000 7,000 6,000

op yle ne

Pr

-C 2 H2 100

Pr op

-C 2 H2 I-B

uta ne

C -Kens ol 1

H2S

7

SO2

7

C3-nC

C -nC 2 C63-nC6

NH3

C -n 1 C

10

C -nC 1

C nC 4-n 10

200

300 Temperature, ºF

10

C2 -nC

NBu ta ne IP en ta ne N Pe nt an e NH ex an e

20,000

400

NH e pt an e N -O ct an e

Institute of Petroleum Engineering, Heriot-Watt University Benzene

500

600

nC9

nC7-nC19

Methyl- Toluene Cyclohexane

ne

-C 3 H2 ca

22/07/14 nC 6 H 2NDe

Convergence Pressures for Hydrocarbons (Critical Locus)

800

e

nC15 nC14 n ca de

nC13

xa He n-

nC12

700

nC11

nC7-nC23

nC5-nC24 nC5-nC22 nC5-nC18 nC5-nC16

H2O

nC17

900

Kensol nC18

Equilibrium Ratio Prediction and Calculation T H I R T E E N

Figure 5 Convergence pressure for Hydrocarbons.2

9

Convergence Pressure, PSIA

Petroleum Engineering

Reservoir Engineering

20,000

C thru' Mid-Continent 1 Crude

10,000 9,000 8,000 7,000 6,000

Absorber O

5,000 4,000

C thru' Gaso lin 1

3,000

e

,C ne

ru de C en t nt in

Le an

O

il

M ol .

wt .1

Cr

200

61

ud

e

300

M id -C o

Li gh t

Di st illa te

Absorbers, Crude Flashing Towers

400

G as ol in e

500

tha

1,000 900 800 700 600

1

2,000

Me

Convergence Pressure, PSIA

verhead

C 1 thru' Distillate Crude

100 -100

0

100

200

300

400

500

600

700

800

900

Temperature, ºF

Figure 6 Pseudocritical properties of Heptane Plus. (NGPSA 5th. edition 1957).

Step 5: Read the convergence pressure (ordinate) at the temperature (abscissa) corresponding to that of the desired flash conditions, from Figure 5. Step 6: Using the convergence pressure determined at Step 5, and making reference to the system under question, 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 calculations 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 T H I R T E E N

1700 Pseudocritical temperature, °R

1600 1500

Specific gravity of heptanes plus = 1.0

1400

.95

1300

.85

1200

.75

1100

.90 .80 .70

1000 900 100

150 200 250 Molecular weight of heptanes plus

300

500

Pseudocritical pressure, °R

450 400

1.0 = Specific gravity of heptanes plus

350

.95 .90

300

.85 .80

250

.75 .70

200 150 100

100

150 200 250 Molecular weight of heptanes plus

300

Figure 7 Pseudocritical properties of Heptane Plus6. 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 Example: McCain5. The gas-liquid equilibrium of a high-shrinkage crude oil has been calculated. The composition of the liquid phase formed at 75˚F and 100 psia is given below. A convergence pressure of 2000 psia was used to determine the equilibrium ratios for the calculations. What value of convergence pressure should have been used for this mixture?

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Petroleum Engineering

Component

Reservoir Engineering

Composition of liquid mole fraction

Methane Ethane Propane i-butane n-butane i-pentane n-pentane Hexane Heavier*

0.0356 0.0299 0.0919 0.0170 0.0416 0.0198 0.0313 0.0511 0.6818 1.0000

* Molecular weight and gravity of C7+ assumed to be 263 and 0.886

Solution First, the composition of the liquid must be expressed as weight fraction. Component

Composition of liquid mole

C1 C2 C3 i-C4 n-C4 i-C5 n-C5 *C6 *C7+

0.0356 0.0299 0.0919 0.0170 0.0416 0.0198 0.0313 0.0511 0.6818 1.0000

Molecular weight Mj

16.0 30.1 44.1 58.1 58.1 72.2 72.2 86.2 263

Weight xMj

0.5696 0.9000 4.0528 0.9877 2.4170 1.4296 2.2599 4.4048 179.3134 196.3348

Composition of liquid, weight fraction xjMj/ΣxjMj 0.0029 0.0046 0.0206 0.0050 0.0123 0.0073 0.0115 0.0224 0.9133 0.9999

* Molecular weight and gravity of C7+ assumed to be 263 and 0.886 respectively

Second, adjust weight fraction to exclude methane and calculate weighted-average critical properties. Component Composition excluding

Critical °R temperature wjtcj °R Tcj

psia pressure, wjpcj psia pc psia pcj

C2 C3 i-C4 n-C4 i-C5 n-C5 C6 *C7+

549.8 665.7 734.7 765.3 828.8 845.4 899.3 1360

707.8 616.3 529.1 550.7 490.4 488.6 445.4 240

methane, wj

0.0046 0.0207 0.0050 0.0123 0.0073 0.0115 0.0225 0.9160 0.9999

* Critical properties of C7+ from figure 7.

12

2.53 13.78 3.67 9.41 6.05 9.72 20.23 1245.76 wt avg Tc= 1311°R, 851°F

Critical

3.26 12.76 2.65 6.77 3.58 5.623 10.02 219.84 wt avg pc= 265 psia

Equilibrium Ratio Prediction and Calculation T H I R T E E N

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 calculation 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.

3

EQUATION OF STATE BASED EQUILIBRIUM CALCULATIONS

3.1 Methods Based on Empirical Equations of State of Fluid Phase Theory

The thermodynamic properties of a pure fluid may be represented by an equation of state of the generalised form:

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:

 ∂2 P   ∂P    =  2 = 0  ∂V  T  ∂V  T



Equations of state can be used for the following purposes: 1 representation of PVT data to assist data smoothing and improve interpolation; 2

prediction of vapour-liquid equilibria of mixtures especially at high pressures;

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.

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Reservoir Engineering

3.2 Prediction of Vapour-Liquid Equilibrium

An equation of state capable of predicting behaviour in the liquid phase and vapour phase is sufficient in itself for vapour-liquid equilibrium predictions. Unfortunately as we have indicated although modifications to the Van der Waals type equation have predicted vapour properties the equations in general have not been so accurate in predicting the liquid behaviour. Many vapour-liquid prediction methods have therefore gone to an equation of state, for example the Redlich-Kwong equation, for the prediction of the vapour phase fugacity and have employed a liquid theory to evaluate the liquid phase properties. We will now go through the steps in the application of equation of states in vapour liquid equilibrium calculations. It is worth noting that the steps described are those which would be taking place within each grid block of a compositional reservoir simulator. The implication therefore is a very large computational load for large numbers of grid blocks. The application of equation of states is based on the fact that at equilibrium, the fugacity of the gas and liquid phases are identical i.e: fiv = fiL

(2)

where fiv and fiL are the fugacities of component i in the vapour and liquid phases respectively. The ratio of the fugacity to pressure is called the fugacity coefficient where

φi = fi/(pzi)

(3)

where zi is the composition of the component in the system. The fugacities can be expressed for a vapour and liquid therefore by: fiL = φi xip and: fiv = φi yip p - is the system pressure; yi & xi - are the mole fractions of i in the vapour and liquid phase. Therefore:



14

fiL xp φ y Ki = iL = i = i fiv φig ni (n ) st yi p

Equilibrium Ratio Prediction and Calculation T H I R T E E N

The fugacity coefficients for the liquid and gas can be calculated using the following equation

V  1  RT  ∂p    ln φi = − dV − ln z   RT ∫∞  V  ∂ni  T , V , n  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. The Peng Robinson equation is

p=

RT aT − 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. and b =

∑yb i

(6)

i i

aT = ∑ ∑ yi y j (aTi aTj )0.5 (1 − kij ) i

(7)

j

where kij are the binary interaction coefficients, and kii = kjj = 0 and kij = kji. The value of bi and aTi for the individual components are calculated as follows and



22/07/14

bi = 0.07780

RTci Pci

aTi = aciαi and

ac i = 0.45724

(8) (9)

R2 T 2 ci Pci

Institute of Petroleum Engineering, Heriot-Watt University

(10)

15

Petroleum Engineering

Reservoir Engineering

αi is a temperature dependant factor where

α i 0.5 = 1 + m(1.0 − Tri0.5 )



(11)

where

m = 0.37476 + 1.54226ω i − 0.26992ω i2



(12)

The Peng Robinson equation can be written as a cubic equation in terms of z, the compressibility factor, z3 - (1-B)z2 + (A-2B -3B2)z - (AB - B2 -B3) =0

(13)

where

A=



aP R2 T 2

B=

bP 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.

ln φi = − ln( z − B) + ( z − 1) Bi′ −

 z + (21.5 + 1) B  A A − B ln ′ ′ ( ) i i  z − (21.5 + 1) B  21.5  

Where B'i = bi/b

(15) (16)

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. fLi = xipφLi and fvi = yipφvi When fLi=fvi then equilibrium is achieved and calculations are complete. 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. Step 1: Estimate the K values of the system. The Wilson11 equation is good for this purpose. 16

Equilibrium Ratio Prediction and Calculation T H I R T E E N



p  T   Ki =  ci  exp 5.37(1 + ω i )1 − ci     p T  

(18)

wi = acentric factor for component i

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 iterate 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 5: Repeat steps 3 & 4 using the vapour phase compositions. Step 6: Calculate fgi and fLi fgi=yipφi and fLi= xipφi. Check if fgi= fLi. If the discrepancy in absolute 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.

Rather than set up the tolerance check on fugacity equivalence the tolerance can be

εi = fLi - fgi



based on K values. A value of ε of 0.001 can be used for the sum of the errors.



Ki =



εi

φ Li φ gi

(K =

E i

(19)

− KiC )

KiE KiC

2



(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.

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Petroleum Engineering

Reservoir Engineering

An example follows to illustrate the calculation process. Start Input zi, P, T, Component Properties Estimate Ki, Using Eq.(18) Calculate xi, yi, Using Figs. 20-22 (Ch 12) Set-Up EOS For Liquid

Set-Up EOS For Vapour

Calculate ZL

Calculate ZV

Calculate fiL

Calculate fiV

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

Figure 8 Flow chart of flash calculations using equation of state.2

18

Equilibrium Ratio Prediction and Calculation T H I R T E E N

EXERCISE 1. Calculate the liquid and vapour phase composition when the mixture with the composition 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 Methane n-Pentane n-Decane n-Hexadecane

Composition mole fraction 0.55100 0.11400 0.14600 0.18900

Table 2: Binary interaction parameters of Peng- Robinson Eq. No. 1 2 3 4

4

Component Methane n-Pentane n-Decane n-Hexadecane

1 0.0000 0.0236 0.0489 0.0600

2

3

4

0.0000 0.0000 0.0070

0.0000 0.0000

0.0000

THE COMPOSITION DATA REQUIREMENTS FOR THE APPLICATION OF THERMODYNAMIC EQUILIBRIUM P R E D I C T I O N M E T H O D S TO M U LT I - C O M P O N E N T HYDROCARBON MIXTURES

In designing a versatile phase equilibrium prediction computer program, the available experimental data for the mixture whose performance is to be simulated, must be considered. Most laboratory analyses of crude oil are predominantly concerned with the composition. This is normally reported as a percentage weight. This presents few problems for lighter hydrocarbons such as methane which can have no isomers nor for other light hydrocarbons which do have isomers, for example butane, since the number of hydrocarbons having the same number of carbon atoms is relatively few. However, for heavier hydrocarbons the number of hydrocarbons which possess the same number of carbon atoms can be very large due to the presence of not only aliphatic or straight chain compounds but also of cyclic compounds such as aromatics or naphthenes, or combinations of these structural types. Obviously the structure of molecules greatly influences the forces between them and hence their deviations from ideality. This is most significant when binary interaction parameters are employed in the mixing. The analysis of the heavier components in crudes is further complicated by the fact that most of the individual components are present in very small quantities which makes them difficult to identify and quantify. 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 22/07/14

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19

Petroleum Engineering

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. Most of the work carried out on fluid phase equilibria has concerned itself solely with binary or tertiary systems and occasionally mixtures containing pure components whose behaviour is quite well understood. Much of the published work on the adaptation of experimental data for prediction techniques has been related to petroleum refinery operations.

20

Reservoir Engineering

Equilibrium Ratio Prediction and Calculation T H I R T E E N

10 1,000

2

30

4

50

Pressure, PSIA

6 7 8

9 100

2

300 4 500 6

7 8 9 1,000

9 8 7 6

2

3,000 4

6 7 8 9

Plotted from 1947 tabulation of G. G. Brown, University of Michigan. Extrapolated and drawn by The Fluor Corp. Ltd. in 1957.

5 4

9 8 7 6 5 4

3

3

2

2

100

100

9 8 7 6

9 8 7 6

5

5

4

4

3

3

Temperature °F 50 40 0 0

2

2

30

25

10

10

0

0

9 8 7 6

9 8 7 6

0

20

K= y/x

10,000 1,000

10 0 80

5 4

5 4

60

3 2

40 20

3

-4

1.0

2

0 -2 0

1.0

0

9 8 7 6

9 8 7 6

-6

0

5

5

4

4

3

3

2

2

0.1

0.1

9 8 7 6

9 8 7 6

5

5

4

4

3

3

2

2

10

K= y/x

2

30

4

50

6 7 8 9 100

2

300 4 500 6

Pressure, PSIA

7 8 9 1,000

2

3,000 4

6 7 8 9

ETHANE

10,000

CONV. PRESS. 5000 PSIA

Figure 9 K value chart for Ethane.

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21

Petroleum Engineering

Reservoir Engineering

Pressure, PSIA 10 1,000

2

30

4

50

6 7 8 9 100

300 4 500 6

2

7 8 9 1,000

9 8 7 6

3,000 4

6 7 8 9

Plotted from 1947 tabulation of G. G. Brown, University of Michigan. Extrapolated and drawn by The Fluor Corp. Ltd. in 1957.

5 4

10,000 1,000 9 8 7 6 5 4

3

3

2

2

Temperature °F

100

100

50

9 8 7 6

0

45

9 8 7 6

0

40 0 38 0 36 0 34 0 32 0 30 0 28 0 26 0 24 0 22 0 200

5 4 3

2

10

K= y/x

2

9 8 7 6 5 4 3

5 4 3

2

10 9 8 7 6 5 4 3

18

0

2

K= y/x

2

16

0

14

0

1.0

9 8 7 6

1.0

12

9 8 7 6

100

5

80

3

0

5 4 3

2

4

2

60 40

0.1

9 8 7 6

0.1 9 8 7 6

20

5

5

4

4

0

3

3

2

2

-20 10

2

30

4

50

6 7 8 9 100

2

300 4 500 6

Pressure, PSIA

7 8 9 1,000

Figure 10 K value chart for Octane.

22

2

3,000 4

6 7 8 9

OCTANE

10,000

CONV. PRESS. 5000 PSIA

Equilibrium Ratio Prediction and Calculation T H I R T E E N

SOLUTION TO EXERCISE EXERCISE Calculate the liquid and vapour phase composition when the mixture with the composition 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 Methane n-Pentane n-Decane n-Hexadecane

Composition mole fraction 0.55100 0.11400 0.14600 0.18900

Table 2: Binary interaction parameters of Peng- Robinson Eq. No. 1 2 3 4

Component Methane n-Pentane n-Decane n-Hexadecane

1 0.0000 0.0236 0.0489 0.0600

2

3

4

0.0000 0.0000 0.0070

0.0000 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 aci = 0.45724 Pci RT bi = 0.07780 ci Pci

(10) (8)

If ω <0.5 : m= 0.37464 + 1.54226 ω - 0.26992 ω2

(12)

If ω >0.5 : m= 0.3796 + 1.485 ω - 0.1644 ω2 - 0.01667ω3 w - acentric factor from thermodynamic property tables for pure components

(

(

)

2

α i = 1.0 + m 1.0 − T r )

ai = ci αi

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(11) (9)

23

Petroleum Engineering

Component Tci K

Pci atm

C1 n-C5 n-C10 n-C16

45.39 33.26 20.82 13.82

190.6 469.7 617.7 723

ωi acentric factor 0.0115 0.2515 0.4923 0.7174

Reservoir Engineering

aci

αi

ai

bi

2.4632 20.4230 56.4253 116.4577

0.7112 1.1686 1.5327 1.9176

1.7518 23.8663 86.4825 223.3236

.0268 .0902 0189 .3340

Step 1: Select initial values for equilibrium ratios (K-values) and calculate the trial composition of liquid and vapour phases at equilibrium. The procedure is iterative as follows. Iteration One Initial K-values can be calculated from Wilson Equation:

Pci T Exp (5.37)(1.0 + ω i )1.0 − ci )) P T

Ki =

Component Methane n-Pentane n-Decane n-Hexadecane

Feed (Zi) 0.55100 0.11400 0.14600 0.18900

(15)

K-Values (Ki)-Wilson Eq. 4.7644 4.290 E-02 7.9917 E-04 1.7768 E-05

Step 2: Determine the trial value of phase fraction (vapour fraction here) by solving the vapour equilibrium equation:

N

∑L i=l

V

Zi + KL

=V

(5)

Vapour Fraction (Vf) = 0.4378 Use the determined vapour fraction (Vf) to calculate the liquid and the vapour phase composition from material balance: Component Methane n-Pentane n-Decane n-Hexadecane

Feed (Zi) 0.55100 0.11400 0.14600 0.18900

Liquid (xi) 0.2081 0.1962 0.2595 0.3362

Vapour (yi) 0.9914 0.0084 0.0002 0.000006

Step 3, 4 and 5: Calculate composition dependent coefficients for compressibility factor (Z - factor) calculations for both liquid and vapour: Liquid: N

24

N

( ) (1. − k )

a = ∑ ∑ xi x j ai a j i=l J =l

0.5

ij



(7)

Equilibrium Ratio Prediction and Calculation T H I R T E E N

N

b = ∑ xi bi



i=l

N

N



N

( ) (1. − k )

a = ∑ ∑ yi y j ai a j

Vapour:



(8)

i=l J =l

b = ∑ yi bi i=l

A= Phase Liquid Vapour

0.5

(7)

ij

(8)



aP R2 T 2

B=

bP RT

a 74.88118 1.83396

b 0.18470 0.02737

(14) A 10.86872 0.26619

B 0.82089 0.12165

Calculate Z - factors of the liquid and vapour phases (Peng - Robinson EOS): Z3 - (1 - B)Z2 + (A - 2B - 3B2)Z - (AB - B2 - B3) = 0.

(13)

Vapour Phase, take the highest root: Zv = 0.92034 Liquid Phase, take the smallest root: Zl = 0.96602 Step 6: Calculate the fugacity of each component in the liquid and vapour phase:

ln φi = − ln( Z − B) + ( Z − l )

 bi  bi A  1  0.5 N 0.5 − 2 ai ∑ y j a j l − kij  −  x  b 2 2B  a  j =l  b

(

 Z + ( 2 + 1) B 

ln  Z − ( 2 − 1) B   

(15)

Fugacity of components in the vapour phase: Fugacity of components in the Liquid phase: Component Methane n-Pentane n-Decane n-Hexadecane

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)

Liquid φil fil .23811E+01 .67429E+02 .66312E-01 .17708E+01 .19023E-02 .67189E-01 .43075E-04 .19707E-02

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fi v = yi Pφi v fi l = xi Pφi l

Vapour φiv fiv .90058E+00 .12150E+03 .36516E+00 .41838E+00 .14686E+00 .41455E-02 .51625E-01 .41965E-04

25

Petroleum Engineering

Reservoir Engineering

End of Iteration One Next Iteration: Ki =

φil φiv

(16)

Iterate till the fugacity of every component in the liquid phase is equal to that of vapour phase (iterate from Step 2.): Final Iteration Component Methane n-Pentane n-Decane n-Hexadecane

Feed (Zi) 0.55100 0.11400 0.14600 0.18900

K-Values (Ki) 2.44097 0.19991 1.6774 E-02 1.25099 E-03

Then: Vapour Fraction (Vf) = 0.2710 Component Methane n-Pentane n-Decane n-Hexadecane

Feed (Zi) 0.5510 0.1140 0.1460 0.1890

Liquid (xi) 0.3962 0.145665 0.199039 0.259147

Vapour (yi) 0.967236 0.029100 0.003339 0.000324

Calculate composition dependent coefficient for z - factor calculations for both liquid and vapour: Liquid:

N

N

( ) (1. − k ) 0.5

a = ∑ ∑ xi x j ai a j i=l J =l

ij

N

(7)

b = ∑ xi bi



i=l

N

(8) N

( ) (1. − k )

a = ∑ ∑ yi y j ai a j

Vapour:

i=l J =l

0.5

N



b = ∑ yi bi



A=

ij

(4)



(8)

i=l

Phase Liquid Vapour

aP R2 T 2

B=

a 48.03656 2.11326

bP RT b 0.14799 0.02929

(14) A 6.97233 0.30673

B 0.6577 0.1301

Calculate z - factors of Liquid and Vapour and Vapour Phase (Peng - Robinson EOS):

26

Equilibrium Ratio Prediction and Calculation T H I R T E E N

Z3 - (1 - B)Z2 + (A - 2B - 3B2)Z - (AB - B2 - B3) = 0.

(3)

Vapour Phase, take the highest root: Zv = 0.90039 Liquid Phase, take the smallest root: Zl = 0.81539 Calculate the fugacity of each component in the liquid and vapour phase:

ln φi = − ln( Z − B) + ( Z − l )

 bi  bi A  1  0.5 N 0.5 − a y a l − k 2  ∑ i j j ij  − x b 2 2 B  a  j =l  b

(

)

 Z + ( 2 + 1) B  ln    Z − ( 2 − 1) B  Fugacity of components in the vapour phase:

fi v = yi Pφi v

Fugacity of components in the liquid phase: f l = x Pφ l i i i Component Methane n-Pentane n-Decane n-Hexadecane

Molecular volume



P2

Liquid φil fil .22032E+01 .11881E+03 .66155E-01 .13105E+01 .19873E-02 .53832E-01 .45319E-04 .15983E-02

Vm 2

PM 2 RT

Phase Liquid Vapour

2 099039 .81539

Vapour φiv fiv .90269E+00 .11882E+03 .33093E+00 .13106E+01 .11847E+00 .53831E-01 .36226E-01 .1598E-02

2 RT P

Vm gr mole/cc 0.05451 0.04936

Ma g/mole/cc 103.86 18.165

Ps/cc 0.56613 0.08967

Compositions are for final iteration Component Methane n-Pentane n-Decane n-Hexadecane

22/07/14

Feed (Zi) 0.5510 0.1140 0.1460 0.1890

Liquid (xi) 0.3962 0.145665 0.199039 0.259147

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Vapour (yi) 0.967236 0.029100 0.003339 0.000324

27

Petroleum Engineering

REFERENCES 1

Amyx, J.W, Bass, D.M, Whiting, RL, Petroleum Reservoir Engineering, McGraw Hill. New York 1960

2

Gas Processors Suppliers Association. “Engineering Data Book 9th Edition”. Tulsa

3 Danesh, A, PVT and Phase Behaviour of Petroleum Reservoir Fluids. 1998 Elsevier. pp 66-77 4

Wilson,G: “A Modified Redlich-Kwong EOS, Application to General Physical Data Calculations”, Paper Nu 15C, presented at the AIChE 65th National Meeting (May 1968)

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.

7

Ahmed, T. Hydrocarbon Phase Behaviour, Gulf Pub., 1989.

8

Peng, D.Y., Robinson, D.B., I.E.C. Fundamentals 1976, 15, 1, 59-64.

9

Jhaveri B.S. and Youngren G.K. Three-Parameter Modification of the PengRobinson Equalising State to Improve Volumetric Predictions SPE 13118 (1984)

10 Soave, G., Chem. Eng. Sci., 1972, 27, 1197-1203. 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

Reservoir Engineering

PVT Analysis F O U R T E E N

Gas

Oil

Gas Oil

P = Pb

P1 < Pb

P = P1

P < P1

Valve 1 Gas closed Oil Mercury / displacing fluid

At each stage - volume of oil and gas, density and P composition of gas measured. Remaining oil is called RESIDUAL OIL

V Tres

P Pb Vb

V

PVT Cell Pump

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Petroleum Engineering

PVT Analysis F O U R T E E N

C O N T E N T S 1 SCOPE

9 RETROGRADE CONDENSATION

2 SAMPLING 2.1 Subsurface Sampling 2.2 Surface Sampling

10 UNDERSTANDING PVT REPORTS

3 SAMPLING WET GAS AND GAS CONDENSATE SYSTEMS 3.1 Introduction 3.2 Phase Behaviour 3.3 Sampling Gas And Gas Condensate Reservoirs 3.4 Separator Sampling Points 3.5 Sample Details

12 OIL SAMPLE PVT STUDY 12.1 Separator Tests of Reservoir Fluids 12.2 Fluid Properties at Pressures Above The Bubble Point Pressure 12.3 Total Formation Volume of Original Oil Below The Bubble Point Pressure 12.4 Differential Liberation Tests 12.5 Calculation of Gas - Oil Ratios Below The Bubble Point. 12.6 Calculation of Formation Volume Below The Bubble Point 12.7 Viscosity Data

4 APPARATUS 5 RECOMBINATION OF THE SURFACE OIL AND GAS SAMPLES 6 PVT TESTS 6.1 Flash Vaporisation 6.2 Differential Vaporisation 6.3 Separator Tests 6.4 Viscosity 6.5 Hydrocarbon Analysis

11 PURPOSE OF THE PVT REPORT

13 GAS CONDENSATE PVT REPORT 14 HIGH PRESSURE / HIGH TEMPERATURE, HP/HT, FLUIDS 15 MERCURY

7 WAX AND ASPHALTENES 7.1 Wax Crystallization Temperature 8 SUMMARY OF RESULTS PROVIDED BY AN OIL SAMPLE PVT TEST 8.1 Interfacial Tension IFT 8.2 IFT Measurements Methods

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Petroleum Engineering

LEARNING OBJECTIVES Having worked through this chapter the Student will be able to:

2

•

Describe briefly the scope of PVT analysis.

•

Describe the sampling options for oil systems.

•

Describe the impact of well flow interruption on the sampling of ‘wet’ and gas condensate systems.

•

List the main items of equipment in PVT analysis.

•

List and describe the five main PVT tests for oils systems and their application.

•

Be able to use an oil PVT analysis report to determine:

•

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.

•

Calculate oil formation volume factors above the bubble point.

•

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.

•

Explain the two basic PVT tests for gas condensates, constant mass and constant volume depletion.

Reservoir Engineering

PVT Analysis F O U R T E E N

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. PVT-analysis of a reservoir fluid comprises the determination of: a the correlation between pressure and volume at reservoir temperature; b various physical constants that enter into reservoir engineering calculations, such as viscosity, density, compressibility etc; c

the effect of separator pressure and temperature on oil formation volume factor, gas/oil ratio etc;

d

the chemical composition of the most volatile components.

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.

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2 SAMPLING The value to be attached to the laboratory determinations depends on whether the sample investigated is representative of the reservoir contents or at least of the drainage area of the well sampled. Therefore, the composition of the fluid entering the well and of the samples taken should be identical with that of the fluid at all other points in the drainage area. It is therefore desirable to take samples early in the life of the reservoir in order to minimise errors caused by differences in relative movement of oil and gas after solution gas has been liberated. The taking of samples can be accomplished either by sub-surface sampling or by surface sampling. There have been considerable advances in recent years in this area aimed at taking samples under down hole pressure conditions. A brief survey of these methods is given below, for detailed descriptions of the techniques, which is outside the scope of this section, service companies providing sample collection service should be consulted.

2.1 Subsurface Sampling

In this case a subsurface sampler is lowered into the well and kept opposite the producing layer for a sufficiently long time, Figure 1. Many types of bottom hole samplers have been devised and described in the literature. 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. In recent years there have been considerable advances on downhole fluid sampling. 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. Subsurface sampling is generally not recommended for gas-condensate reservoirs nor for oil reservoirs containing substantial quantities of water. 4

Reservoir Engineering

PVT Analysis F O U R T E E N

Lubricator

To wire line truck

Ground level

Tubing

Packer

Producing Formation

Bottom hole sampler run in tubing

Subsurface sampling

Figure 1 Subsurface Sampling.

2.2 Surface Sampling

A sample of oil and gas is taken from the separator connected with the well (Figures 2 - 5 give sketches of vertical and horizontal separators and the arrangement for collecting different fluid samples). The surface oil and gas samples are recombined in the laboratory on the basis of the producing GOR. Particular care, therefore, must be exercised in the field to obtain reliable samples and accurate measurement of the GOR and separator conditions. In the case of two or three stage separation the samples are taken from the high pressure separator.

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Petroleum Engineering

Reservoir Engineering

Gas outlet

Vertical Separator

Gauge Inlet

Sight glass

Liquid outlet

Momentum absorber

Guage Gas outlet

Inlet

Sight glass

Horizontal Separator

Figure 2 Vertical and Horizontal Separators.

Sample Vessel

Figure 3 Separator Gas Sampling.

6

Liquid outlet

PVT Analysis F O U R T E E N

Sample Vessel

Figure 4 Separator Liquid Sampling by Gas Displacement.

Figure 5 Separator Liquid Sampling by Water Displacement.

3

SAMPLING WET GAS AND GAS CONDENSATE SYSTEMS

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.

3.2 Phase Behaviour

The behaviour characteristics of fluids are uniquely described by a phase diagram, Figure 6. The hydrocarbon mixture with its own unique composition will have its own phase diagram and phase envelope. Within the phase envelope the system is in two phases, whereas outside the phase envelope a single phase exists.

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Petroleum Engineering

Reservoir Engineering

Liquid

c e rv

Dewpoint curve

uid

ur

uid

Liq Liq ui d

%

20

5%

Li

q

10 0%

uid

Vap o

Liq

40 %

50 %

10 0% Liq 80% uid Liq uid 70 % Liq uid

Pressure

Bu bb le

po in t

cu

Vapour

Temperature

Figure 6 Phase diagram for a reservoir fluid.

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 its bubble point and the gas at its dew-point. Therefore, for example, in the separator, gas will be separated and produce gas at its dew-point and the oil separated will be at its bubble point. Gas at dewpoint

Separator

Oil at bubble point

Figure 7 Conditions in a 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.

8

PVT Analysis F O U R T E E N

3.3 Sampling Gas And Gas Condensate Reservoirs

The potential locations for sampling these reservoirs are shown in Figure 8; samples could possibly be taken in the reservoir, at the bottom of the well, at the wellhead or in the separator. The respective advantages and disadvantages are given in Table 1. The reservoir would be the most ideal sampling point but clearly this is impossible.

4. Separator

3. Wellhead

2. Bottom hole 1. Reservoir

Figure 8 Locations for sampling. 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

Table 1 Merits and disadvantages of sampling locations.

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 rise to a two phase situation and an unrepresentative collection of these two phases. 22/07/14

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9

Petroleum Engineering

Reservoir Engineering

Bottom hole samples are also more costly to collect. The wellhead from a cost point of view could be the most suitable location point, however, again the question of the representative nature of the sample is a concern. As a result of the lower pressure and temperature it is likely that the single phase fluid at the bottom of the well has gone into the two phase region at the wellhead and therefore the relative proportions of liquid to gas would be unknown and their sampling would be difficult to produce a representative sample. The most common sample location is the separator. Considerable care, however, has to be taken to ensure that the samples taken from the separator are those representative of the reservoir from which the fluids derive. 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 Flow must be sufficient to lift liquid to surface

Condensate

Gas

Retrograde liquid due to fall in pressure and temperature

Gas

Gas

Condensate

Figure 9 Flowing well.

10

PVT Analysis F O U R T E E N

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

Retrograde liquid rains down in the well

Reservoir gas flowing into vicinity of well sets up new equilibrium with condensed retrograde liquid.

Gas flow Pressure build up Condensate

Figure 10 Shut well in after flow

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, 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 fluid in the near wellbore region which also could have been influenced by the pressure and compositional changes experienced during the shut-in period. 22/07/14

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Petroleum Engineering

Reservoir Engineering

Gas

Condensate Lean gas moves to surface Higher GOR

Excess liquid in well moves to surface Lower GOR

Equilibrium gas flows to surface richer in heavier components. Lower GOR

Figure 11 Well flowed after shut in period

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.

3.4 Separator Sampling Points

A very practical aspect is often ignored in the design of separators and in particular in relation to the sampling points associated with them. These sampling points are often located primarily in relation to accessibility rather than the representative nature of the fluids which can be extracted from them. For example, in the gas line, Figure 12a the sampling valve might be located on the lower portion of the valve. It is likely that 12

PVT Analysis F O U R T E E N

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)

Figure 12 Location of sampling points

3.5 Sample Details

In taking samples there are some obvious aspects which are often ignored. Clearly details in relation to the sample should be taken in particular: the date and the time of a sample, the identification of the cylinder into which the sample is to be collected, the location of where the sample was taken, the temperature and pressure at which the sample was taken, details of any other aspects which will be important for those subsequently handling the sample, for example, the presence of any H2S etc. The details of the sample including, for example, the gas to oil ratio during the separation will be transmitted with a sample to the laboratory which will carry out the analysis. For example, if the sample identification cylinder has not been noted then if sample details become separated from the sample cylinder then the sample would be wasted. Prior to any liquid samples being transported to the laboratory it is important to reduce the pressure within the container to a value below the bubble point to ensure that a two phase mixture is transported. Very high pressures can occur as a result of a temperature rise on a single phase liquid sample. Such pressures could go over safe working pressures of the vessel!

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Reservoir Engineering

4 APPARATUS The apparatus required for PVT analysis consists of: 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. Analysis Separator

Viscometer Trap Bleed

Oil sample bottle

To liquid analysis

Vacuum gage

Gas meter Displacement fluid

Vacuum pump

To gas analysis

Pressure Guage PVT cell in thermal bath

Bleed

Precision pressure gage

Dead weight tester

Displacement pump

Figure 13 Equipment for the Study of Subsurface Samples Analysis Separator

Viscometer Trap

Recombination cell Gas sample bottle

Displacement fluid

Vacuum gage

Bleed To liquid analysis

Liquid sample bottle

Gas meter

To gas analysis

Vacuum pump Pressure Guage PVT cell in thermal bath Bleed Precision pressure gage Displacement pump

Figure 14 Equipment for the Study of Subsurface Samples

14

Dead weight tester

PVT Analysis F O U R T E E N

Analysis

Gas Sample Bottle

Recombination Cell

Separator Trap To liquid analysis

Vacuum gage Displacement Fluid

Gas meter Vacuum pump Liquid Sample Bottle

Gas condensate cell in thermal bath

To gas analysis

Pressure Guage

Bleed Precision pressure gage

Dead weight tester

Displacement Pump

Figure 15 Equipment for the Study of Gas Condensate Samples

5

RECOMBINATION OF THE SURFACE OIL AND GAS SAMPLES

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 stock 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 the stock tank in the field i.e. at the same pressure and temperature. The collected stock tank oil is weighed and the density determined, after which the volume of oil is known. The volume and the density of the liberated gas (stock tank gas) are also determined. From the above measurements, the stock tank gas/stock tank oil ratio is also known. If the volume of stock tank oil is lower than the corresponding quantity of separator oil we speak of shrinkage, in the opposite case of expansion; shrinkage occurs when a large quantity of gas is produced and expansion occurs when a small quantity of gas is dissolved in the separator oil. When the shrinkage or the expansion is known, recombination can take place on a stock tank-oil basis.

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6

Reservoir Engineering

PVT TESTS

At this stage we should remind ourselves of the main applications of the PVT data. The three main application areas are; • To provide data for reservoir calculations, • To provide physical property data for well flow calculations • For surface facility design. Although all are cited as users of the data, the reservoir calculation requirement has provided the main driving force for the tests to be carried out. In surface facility design for example the more simplistic black oil approach around which the PVT analysis is structured is considered too limiting, and the main data for this application is the compositional analysis of the fluids. 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 five main PVT tests for oil systems plus associated compositional analysis: i The flash vaporisation or relative volume tests. ii The differential test. iii The separator tests. iv Viscosity measurements v Compositional measurements. A simple layout of a PVT facility is given in Figure 16. PVT Facility P

Valve 1

T

V1

Ps1

L1

Ts1 Valve 2

Displacement fluid or piston

V2

Ps2

L2

Ts2

PVT Cell in temperature controlled bath

Figure 16 Simple Schematic of PVT Facility for Oil Tests.

16

PVT Analysis F O U R T E E N

6.1 Flash Vaporisation (Relative Volume Test)

By flash vaporisation is meant the determination of the correlation between pressure and volume of a reservoir liquid at constant temperature (reservoir temperature) from high pressure to the lowest possible pressure. The gas liberated below the point of saturation remaining in equilibrium with the oil throughout the experiment. That is, the system remains constant (Figure 17). The vaporisation process occurring in the reservoir cannot be duplicated in the laboratory, since in the reservoir below the bubble point the system does not remain constant as the increased mobility of the gas causes it to move away from its associated oil. 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 Hg

Thermal expansion V2 − V1 β= V2 (T2 − T1 )

V1 = volume of oil at T1

V2 = volume of oil at T2

Figure 17 Flash Vapourisation. Determination of the relationship between P & V of a reservoir liquid at constant (reservoir) temperature. Liberated gas remains in equilibrium with the oil.

By plotting the volume of the system versus pressure a break is obtained in the slope. This occurs at the Bubble Point pressure. To carry out a relative volume test run, the PV cell is set up as in Figure 18.

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Valve 1 closed

P

V Tres

P Pb Vb

V

PVT Cell Pump

Figure 18 PVT Set Up for Flash Vapourisation /Relative Volume Test

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. During heating, it is necessary to maintain pressure by increasing the content volume of the PV cell. When the pressure remains constant, the temperature in the cell is equal to that of the bath. The thermal expansion factor (β) can then be calculated from the volume withdrawn. It is equal to:



β=

V2 − V1 V2 (T2 − T1 )

where: V2 = volume of the oil at reservoir temperature T2 V1 = volume of the oil at room temperature T1 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). The compressibility of the oil phase above the bubble point can now be calculated from the graph.



18

c=

V2 − V1 V2 ( P2 − P1 )

PVT Analysis F O U R T E E N

where: V2 = volume at pressure P2 V1 = volume at pressure P1 The compressibility is expressed in reciprocal atmospheres, psi, etc. After the reservoir liquid has expanded to its maximum volume (dependent on the capacity of the PV cell), the gas cap is removed at constant pressure (lowest possible flash expansion pressure). Volume, density, gas expansion and gas compressibility factor of the liberated gas are then determined successively. 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.

6.2 Differential Vaporisation

When the reservoir pressure falls below the bubble point the process of gas liquid separation in the reservoir is one of a constant changing system. A PVT process has been designed in an attempt to provide a means of in part simulating the changing systems as separation occurs within the reservoir below 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

Gas Oil

P = Pb

P1 < Pb

At each stage - volume of oil and gas, density and composition of gas measured. Remaining oil is called RESIDUAL OIL

P = P1

P < P1

Gas Oil Mercury / displacing fluid

Figure 19 Differential Vaporisation. In differential vaporisation. At each successive pressure drop liberated gas is removed.

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 22/07/14

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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. Differential liberation is considered to be representative of the gas-liquid separation process in the reservoir below the bubble point pressure. 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.

6.3 Separator Tests

The object of these tests are to examine the influence of separator pressure and temperature on formation volume factor, gas/oil ratio, gas density and tank-oil density. These tests are not driven by those who will be responsible for the optimised separation process if the field ultimately is developed. They are carried out to give an indication of the oil shrinkage and GOR which occurs when the fluids are produced to surface conditions. It should be emphasised at this stage that there is not a unique value for the formation volume factor and solution gas-oil ratio. It depends on the stages and conditions of separation through which the fluids pass. With the equipment available, a single test or a multiple separation test can be carried out.

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PVT Analysis F O U R T E E N

The system is set up as shown in the schematic below, Figure 20. Valve 1 Pb

V1

Ps1

L1

Ts1

Valve 2 Vres

Tres

V2

Ps2

L2

Ts2

60 F & 14.7 psia

PVT Cell Pump

GOR = (V1+V2) / L2 Bo = Vres / L2

Figure 20 Schematic of a Two Stage Separator Test

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. In the sketch Figure 20 oil formation volume factor Bob is equal to Vres/L2 reservoir volumes/stock tank volumes. 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.

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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.

6.5 Hydrocarbon Analysis

A hydrocarbon analysis is made of the methane to an upper paraffin fraction of the recombined surface (or subsurface) sample. Historically this upper limit was C6 and the remainder lumped as a C7+ fraction. Higher C numbers are now used as analytical methods enable even higher levels of characterisation. The plus component having been separated in a distillation column needs to be characterised by its specific gravity and its apparent molecular weight. The latter is achieved using a depression of freezing point method. The different components are determined by means of gas/liquid chromatography. If the description is based solely on paraffins then non paraffinic components are added to the next higher paraffin. The value of a higher characterisation is particularly helpful to process engineering considerations, say up to C30 because at lower temperatures long chain hydrocarbons will form a solid phase (such as wax) and adhere to surfaces. At reservoir conditions they are in the liquid phase and therefore do not effect the reservoir flow process. 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.

7

WAX AND ASPHALTENES

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.

7.1 Wax Crystallization Temperature

Different techniques can be used for this. In the following procedure a sample of separator oil is transferred to a vessel and then pre filtered by passing through a 0.5 22

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PVT Analysis F O U R T E E N

micron filter. The crystallisation temperature (WCT) is measured by carrying out a series of flow experiments where the oil is flowed through a fine filter (15 microns) across which the pressure differential is measured. Prior to reaching the filter the oil is flowed through a temperature equilibrium coil at the same temperature as the filter at a temperature of 140°F. The temperature of the bath is gradually lowered and the pressure difference measured. A sudden increased in pressure indicates the onset of wax crystals building up on the filter giving an indication of the WCT. From this rough indication constant temperature flows are taken at temperatures just above and below the indicated WCT to give a more precise value. A plot of differential pressure vs. flow indicates a more precise value of the WCT. Figures 21 & 22 below from a PVT report provided by Core Laboratories (UK) Ltd gives the plots used to determine the WCT for a separator oil sample. The appearance temperature is considered to be affected by super cooling, whereas the disappearing temperature is considered to be the equilibrium value. There is a concern that the appearance temperature may not be so accurate as would be the case for the disappearance temperature. 60

Differential Pressure, psig

50

40

30

Approximate Cloud Point

20

10

0

75

85

95

105

115

125

135

145

Temperature, °F

Figure 21 Graph of differential pressure v temperature during constant cooling. (Core laboratories) 22/07/14

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45 40

Differential Pressure, psig

35 30 90°F 88°F 86°F

25 20 15 10 5 0

0

5

10 15 20 Cumulative Volume Flow, cc

25

30

Figure 22 Graph of Cumulative Volume Flow Across Filter v Differential Pressure. (Core laboratories)

8

SUMMARY OF RESULTS PROVIDED BY AN OIL SAMPLE PVT TEST

The PVT analysis as described above furnishes the following data: 1 Saturation pressure. ie. bubble point. 2

Compressibility coefficient



c =−

3

Coefficient of thermal expansion



β=−

4

Relative total volume of oil and gas: vt

5

Cumulative relative gas volume: vg

6

Relative oil volume: vo

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.

24

1 dV (specify unit of pressure change) V dP 1 dV (specify unit of temperature change) V dT

PVT Analysis F O U R T E E N

8

Gas compressibility factor: z defined by Pv = zn RT.

9

Specific gravity of gases (air = 1)

10 Liquid density: ρ 11 Viscosity of the liquid phase at different pressures: µ 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 liberated 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. 14 Shrinkage of separator oil to tank oil. 15 Tank-gas/tank-oil ratio. 16 Hydrocarbon analysis of the reservoir and produced fluids. 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.

Vt

Vt Cb

V

V

V

Bt

Cb Vo P (a)

I

Vt

Pb P

Vi Pi

I

P

Vo Cb (b)

l Vt Vi Cb Cb Cb Pb P Pi

Bo

l P

(c)

Bb Pb

Bt P

Bt Pi

Figure 23 Nomenclature PVT SYMBOLS

a b c

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All volumes relative to oil volume at bubble point Same as (a) using black oil notations All volumes relative to oil volume at standard surface conditions

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Reservoir Engineering

8.1 Interfacial Tension, IFT

Over recent years, particularly with respect to gas condensates, the impact and importance of interfacial tension has developed. 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. The determination of the interfacial tension of vapour-liquid systems is very important therefore for these systems where such phase separation occurs in the reservoir. For these forces will impact on the effective recovery, in particular the balance between gravity forces. Recent research has shown that liquid drop out in retrograde condensation is not necessarily immobile as previously considered. Recovery by gravity drainage, when IFT's are low can therefore be significant. IFT's are very low for those fluids near critical conditions as the value of zero is approached.

8.2 IFT Measurement Methods

The most common approach of measuring IFT has been using the Pendant Drop Method. In this method the liquid phase is suspended from a capillary tube in a vessel containing its equilibrium vapour. Figure 24. Tube Tip

ds

Gas

de

Oil

Figure 24 Pendant Drop Method For IFT Measurement.

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:

σ=

gde2 L (ρ − ρ V ) ) l

Where s = interfacial tension (dyne/cm) g = acceleration due to gravity de = equitoral diameter drop ds = diameter of drop measurement at height de above the bottom of the droplet rL, rV liquid and vapour phase densities l = shape factor, which is = ds/de 26

PVT Analysis F O U R T E E N

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

Figure 25 Rising film method for IFT measurement

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:

σ=

gde2 L (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.

9

RETROGRADE CONDENSATION

In the phase behaviour chapter we described the characteristics of gas condensates, i.e., those reservoir fluids which show the phenomenon of retrograde condensation, i.e. liquid is formed from a vapour when the pressure is reduced isothermally. In this case the system saturation pressure is the dew-point pressure (the pressure at which condensation starts) instead of the bubble point pressure for oil systems.

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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. This is carried out for two reasons: a The dew-point pressure of most condensates cannot be detected by a sharp change in the pressure-volume relation as the bubble point pressure can; b

The liquid phase constitutes only a small part of the total volume in the PVcell. 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. There are five main aspects of a gas condensate study. • Compositions of gas, liquid and reservoir fluids. • Constant mass (composition) expansion. • Constant volume depletion. • Specialised tests. e.g. Interfacial Tension Measurements. • Compositions of gas, liquid and reservoir fluids. There are different approaches to compositional measurements in gas condensates. It is only recently that it has been possible to measure the compositions of liquid and gas phases at reservoir conditions and only now these direct sampling methods have limited pressure applications. Compositions of condensate phases are often obtained by blow down techniques and then recombining the compositions of the produced gas and liquid in the proportion of these two phases. For liquid compositions, the fluid is flashed at 120°F to obtain gas and liquid components. The resulting gas and liquids are then analysed by gas chromatography techniques and the compositions recombined in proportion to the ratio of gas to liquid produced. For gas compositions gas chromatography is used. The composition of the recombined reservoir fluid is obtained by recombining the liquid and gas compositions in proportion to the condensate to gas ratio of the fluids.

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Figure 26 gives a schematic of the two main gas condensate tests. (a) Constant Mass Study

Gas Oil Hg

Vg

Measurement of volume of gas and oil (condensate) as a function of preessure.

Vo

Dew point when first condensate appears. No fluids removed from the cell.

P = Pd

(b) Constant volume depletion Gas removed at each step

Gas

Vi

Gas Oil

Hg P = Pd

Hg P1 < Pd

V>Vi

Gas Oil Hg P = P1

Vi

Pressure is reduced below dew point by expanding system. Volume is put back to original volume by injection of mercury at constant pressure and removing some gas. 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 an 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. 22/07/14

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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

Figure 27 Constant Volume Depletion Liquid Drop Out Curve.

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.

10 UNDERSTANDING PVT REPORTS 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.

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11

PURPOSE OF THE PVT REPORT

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 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 high as C29. 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.

12 OIL SAMPLE PVT STUDY 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 22/07/14

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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 principal observations. The source of these principal 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 is 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 separator test, at first stage of 300 psig and 2nd stage 0 psig and both at 75oF?

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. 32

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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 Rsb, 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.

12.2 Fluid Properties at Pressures Above The Bubble Point Pressure

The source of information for calculations for conditions above the bubble point is a combination of two tests, the flash vaporisation test (relative volume test ) and the separation tests. The results of the flash test are presented in page 4 of 15, titled Pressure -Volume Relations at 220˚F. 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. 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.

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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:

Bo =

vol reservoir oil vol bubble point oil vol reservoir oil = × 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:



ρoi =

1 1 = voi vb ⋅ vrel

ρoi =

1 = 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:



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 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:

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co = −

1  ∆v    vavg  ∆p  T

For example, to get Co at 4500 psig using relative volume values of 500 psi on each side of 4500 psig:

co =



co =

1 (0.9639 − 0.9771) 0.9639 + 0.9771 (5000 − 4000) 2 1 0.0219 = 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, 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. 22/07/14

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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.

Formation Volume Factor - B

Bt

Bti

(Rsi-Rs)Bg 5.61

Bo

Boi p

pb

Pressure p

pi

Figure 28 Shape of Total Formation Volume Factor Bt and Oil Formation Volume Factor Bo

Above pb it makes no difference whether we consider the formation volume to be a total formation volume or an oil formation volume.

12.4 Differential Liberation Tests.

Previously we have considered what happens when reservoir fluid comes to the surface and is separated into surface gas and oil products. In determining what happens in the separator test we consider it as flash equilibrium conditions because we believe that the action going on in the separator is essentially one in which the whole system entering the trap immediately separates into two components – separator gas and liquid. This constitutes the elements of a flash separation. However as we discussed in section 6 in the reservoir the separation below the bubble point is different and the differential test has been devised to enable calculation of the appropriate volume factors and GOR’s. The standard PVT report includes data referred to as “the differential data”. These are gas solubility and phase volume data taken in a manner to model what some people believe happens to the oil phase in the reservoir during pressure decline. Basically, the argument that differential liberation tests model the subsurface behaviour comes primarily from two things: 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. 2

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because of the relative permeability characteristics of reservoir rock-fluid systems, the gas phase moves toward the well at a faster rate than the liquid phase. Of consequence, the overall composition of the entire reservoir system is changing. These two ideas promote the idea that a test procedure modelled

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on a differential process should be used to study subsurface behaviour. Because of experimental limitations and time-cost considerations the laboratory cannot perform a true differential procedure. Instead, they perform a series, usually about ten, of stepwise flashes at the reservoir temperature, commencing at the bubble point. Of course, the greater the number of steps, the closer the true differential process is modelled. 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.

b c Volume

Initial Bubble Point

a 2620 psig

b' 200° F

c'

2350 psig

60° F

Pressure Resudual Oil

Figure 29 Volume Changes During Differential Liberation

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 22/07/14

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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 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 volumes at pressure p are expressed per unit volume of residual oil. Again, these relative volumes must not be confused with formation factor volumes because formation factor volumes are specified per barrel of stock tank oil. Note on page 5 that relative volumes start at 1.000 at 0 psig/60˚F and that the value of 1.075 at 0 psig/220˚F is the thermal expansion of 35.1˚API residual oil from 60˚F to 220˚F. 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.

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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.

12.5 Calculation of Gas – Oil Ratios Below The Bubble Point.

If we examine the separator and differential tests there seems to be a confusing result since the initial and final pressures are the same. 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 situations, 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. 854 ft3/B residual oil

200°F

795 ft3/B stock tank oil

Rs ft3 /bbl

Differential

Two stage surface flash

2620 psig Pressure

Figure 30 Comparison of data from flash and differential tests.

EXERCISE PVT 3 Calculate the solution GOR at 1850 psig using the 300/0psig separator data.

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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 - 612. 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. If we let (∆Rs)diff be liberated gas-oil ratio by differential vaporisation, we can convert this to a (∆Rs)flash as follows:



(∆Rs)diff = ft3/B Residual Oil

(1)

ft 3 B Residual Oil ft 3 = ⋅ B Residual Oil B Bubble Point Oil B Bubble Point Oil

(2)

ft 3 B Bubble Pointoil ft 3 = ⋅ B Bubble Point Oil B Stock Tank Oil B Stock Tank Oil

(3)

ft 3 = ( Rs ) flash B Stock Tank Oil

(4)



In equation (2)

B Residual Oil 1 = B Bubble Point Oil 1.600



In equation (3)

B Residual Oil = 1.495 B Stock Tank Oil

Therefore: and



1.495

(∆Rs ) flash = (∆Rs )diff ⋅ 1.600

1.495

= 795 − ( ∆Rs ) flash = 795 − 242 ⋅ ( Rs )1850 flash 1.600 = 795 − 226.1= 569 ft 3 / BSTO ( Rs )1850 flash

Solution gas to oil ratio at 1850 psi = 569 scf/STB. For those who prefer equations, this can be generalised as:



40

Bob b / VR

( Rs ) flash = ( Rsb ) flash − (∆ Rs) diff ⋅ V

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12.6 Calculation of Formation Volume Below The Bubble Point

Examination of the formation volume factors between the bubble point pressure and surface pressures also show a distinct difference between the flash tests and the differential data. This is illustrated by the Figure (Figure 31). The bubble point state has a relative oil volume of 1.600 B/B residual oil. We also have the formation volume factor at the bubble point state, Bob, with a value of 1.495 B/B stock tank oil (300/0 separator combination).

Volume Factor

1.600

1.495

Differential @ 200º F

1.0

Two Stage Surface Flash

2620 psig

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. 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: V/VR = relative oil volume at pressure p, B/B residual oil. Then:

B Saturated Oil B Residual Oil ⋅ B Residual Oil B Bubble Point Oil =

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Therefore:



Bo = V VR .

Bob 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



B1850 1 B residual B Bubble point x x 1.495 = 1.479 Bresidual 1.600 B at bubble point B stock tank oil

Bo1850 = 1.3819 B/B stock tank oil

12.7 Viscosity Data

Page 6 of 15 presents the viscosity data for the fluid measured for the oil and calculated for the gas. It should be noted that the pressure for the data below the bubble point are the same as those for the differential tests, since the viscosity is also measured below the bubble point having generated the fluid pressure through a differential process.

13 GAS CONDENSATE PVT REPORT 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 compositional analysis of these condensate fluids is far more comprehensive than the oil report results. This is because the heavy fraction has a significant impact on the properties of the condensate and therefore it is important to have the data for modelling the behaviour of these fluids. 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 42

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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 then 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 psig 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. 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.

14 HIGH PRESSURE/HIGH TEMPERATURE, HP/HT, FLUIDS 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. 22/07/14

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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 its 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 principles are the same, where for example the mercury is replaced by a rigid piston driven by a safe fluid e.g. Water.

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PVT Analysis F O U R T E E N

SOLUTIONS TO EXERCISES 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 Rsb, 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.

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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:

Bo =

vol reservoir oil vol bubble point oil vol reservoir oil = × 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:



r oi =

1 1 = voi vob ⋅ vrel

r oi =

1 = 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:



Co = −

1  ∆v    v avg  ∆p  T

For example, to get Co at 4500 psig using relative volume values of 500 psi on each side of 4500 psig: 70

Reservoir Engineering

PVT Analysis F O U R T E E N

1 (0.9639 − 0.9771) Co = − 0.9639 + 0.9771 (5000 − 4000 ) 2



Co =

1 0.0219 = 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. If we let (∆Rs)diff be liberated gas-oil ratio by differential vaporisation, we can convert this to a (∆Rs)flash as follows:

(∆Rs)diff = ft3/B Residual Oil

ft 3 B Residual Oil ft 3 ⋅ = B Residual Oil B Bubble Point Oil B Bubble Point Oil

(1) (2)

ft 3 B Bubble Point Oil ft 3 ⋅ = (3) B Bubble Point Oil B Stock Tank Oil B Stock Tank Oil

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ft 3 = ( ∆Rs ) flash B Stock Tank Oil

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(4)

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In equation (2)

B Residual Oil 1 = B Bubble Point Oil 1.600



In equation (3)



Therefore:

B Bubble Point Oil = 1.495 B Stock Tank Oil



(∆Rs ) flash = (∆Rs )diff ⋅ 1.495 1.600

and

(R )

flash

(R )

flash

s



s

1850

1850

= 795 − ( ∆R s )flash = 795 − 242 ⋅

1.495 1.600

= 795 − 226.1 = 569ft 3 / BTSO

Solution gas to oil ratio at 1850 psi = 569 scf/STB. For those who prefer equations, this can be generalised as:



( Rs ) flash = ( Rsb ) flash − (Rs ) diff ⋅

Bob 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

Bo1850 =1.479



72

B1850 1 Bresidual B Bubble point x x1.495 Bresidual 1.600 Bat bubble point B stock tank oil

B1850 =1.3819 B/B stock tank oil o

Material Balance Equation F I F T E E N

FLUID VOLUMES Prim. Gas cap

Initial volumes at pressure Pi

Oil + originally dissolved gas

Fluid expansions down to pressure P

Connate water Water influx We

PORE VOLUMES

New pore volume at pressure p

Reduction in total pore volume down to pressure P Total pore volume at pressure Pi

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Material Balance Equation F I F T E E N

C O N T E N T S 1 INTRODUCTION 2 LIST OF SYMBOLS 3 MATERIAL BALANCE FOR GAS RESERVOIRS 3.1 Dry gas, no water drive 3.2 Dry gas reservoir with water drive 3.3 Graphical Material Balance 3.4 Wet Gas Reservoirs 3.5 Gas Cap Expansion

10 SOURCES OF DATA TO BE USED IN THE MATERIAL BALANCE 11 LIMITATIONS OF THE MATERIAL BALANCE 12 CONCLUSION

4 MATERIAL BALANCE FOR OIL EXPANSION 4.1 Above The Bubble Point 4.2 Gas Liberation Below the Bubble Point 4.3 Material Balance with Gas Cap and Water Drive 4.4 Effect of Pore – Volume Changes 4.4.1 Compressibility Effects 4.4.2 Overburden Pressure 4.4.3 Connate Water 5 THE GENERAL MATERIAL BALANCE EQUATION 6 MODIFICATIONS TO THE GENERAL EQUATION 7 DERIVATION OF THE MATERIAL BALANCE EQUATION BY EQUATING SUBSURFACE VOLUME OF PRODUCED FLUIDS TO EXPANSION OF ORIGINAL FLUIDS PLUS PORE VOLUME REDUCTION 8 ASSUMPTIONS IN MATERIAL BALANCE EQUATION 9 SIGNIFICANCE AND USAGE OF THE MATERIAL BALANCE EQUATION

reservoir evaluation and management Reservoir Engineering 22/07/14

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LEARNING OBJECTIVES Having worked through this chapter the Student will be able to:

2

•

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.

•

Given the equation be able to identify the component parts of the MB equation, eg. gas cap expansion etc.

•

Comment briefly on the assumptions, significance , use, data and limitations of the MB equation.

Material Balance Equation F I F T E E N

1 INTRODUCTION

In the chapter on Drive Mechanisms we reviewed qualitatively the various drive energies responsible for hydrocarbon production from reservoirs. In this and subsequent chapters we will introduce some reservoir engineering tools used in calculating 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 material 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 factors can only be expressed volumetrically. The principle of conservation underpins the equation: Mass of fluids originally in place = fluids produced + remaining reserves. 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. The basic ‘material balance’ equation is presented as a volumetric reservoir balance as follows: The reservoir volume of original fluids in place = reservoir volume of fluids produced + volume of remaining reserves. 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 pore volume of the reservoir will become smaller •

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The connate water will expand

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•

Oil, if still undersaturated, will expand

•

Iil, if at or already below bubble point, will shrink while gas will come out of solution

•

Free gas, if present, will expand

•

Water may start flowing into the reservoir, for instance, across the original oil/ water contact (OWC).

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: a At the initial pressure pi, 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 approaches, the first examining the equation as applied to specific reservoir types and then a simple volumetric expansion approach. The nomenclature to be used for the various terms is given below: NOTE: In the following derivations, volumes at standard conditions will be converted into subsurface volumes and vice versa. Remember that to convert a volume from standard conditions to reservoir conditions, one must multiply by a formation volume factor (B) and to convert from reservoir into standard conditions one must divide by a formation volume factor.

2

LIST OF SYMBOLS

Symbols Bg Bo Bt Bw cf cw G G p

4

Units

Gas formation volume factor bbl/SCF Oil formation volume factor bbl/STB Total formation volume factor bbl/STB Water formation volume factor bbl/STB Pore compressibility vol/vol/psi Water compressibility vol/vol/psi Initial gas-cap volume SCF Cumulative gas produced = Gps + Gpc SCF

Units SI M3/SCM M3/STM3 M3/STM3 M3/STM3 vol/vol/MPa vol/vol/MPa SCM SCM

Material Balance Equation F I F T E E N

Gps Gpc m N N p p pi Rp Rs S w We Wp

Cumulative solution gas produced Cumulative gas cap produced Ratio initial reservoir free gas volume to initial reservoir oil volume Stock tank oil initially in place Cumulative tank oil produced Average reservoir pressure Initial reservoir pressure Cumulative gas/oil ratio Solution gas/oil ratio Average connate water saturation Cumulative water influx Cumulative water production

SCF SCF

SCM SCM

bbl/bbl STB STB psi psi SCF/STB SCF/STB fraction bbl or STB bbl or STB

M3/M3 STM3 STM3 MPa MPa SCM/STM3 SCM/STM3 fraction M3 or STM3 M3 or STM3

Other subscripts i at initial conditions b at bubble point Bw may be unknown. In this case assume it equals 1.0. Bw may be omitted from material balance equations and the reader must assume its value equals 1.0.

3

MATERIAL BALANCE FOR GAS RESERVOIRS

The simplest material balance equation is that applied to gas reservoirs. The compressibility 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.

3.1 For a dry gas reservoir - no water drive: Figure 1

G.Bgi = (G-Gp) Bg

(1)

Bgi - based on zi, pi, Ti Bg - based on z, p, T Gp G Bgi

Pi

=

(G - Gp)Bg

P

Figure 1 Material Balance For a Dry Gas Reservoirs no Water Drive. 22/07/14

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N.B:

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 also be produced. Figure 2 below illustrates the contact with a supporting aquifer. Because the mobility 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.

3.2 For a dry gas reservoir with water drive

With water drive water will enter pore volume originally occupied by gas and some water may be produced. Figure 2 Gp GBgi

(G - Gp) Bg =

Water

Wp

We - Wp Water

Figure 2 Material Balance For a Dry Gas With Water Drive.

GBgi = (G-Gp)Bg + We - Wp

(2)

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.

3.3 Graphical Material Balance

One can use a graphical form of the material balance equation to analyse a gas reservoir and predict its behaviour especially if no water drive is present.

6

Material Balance Equation F I F T E E N

G. Bgi = (G − Gp ) Bg From equation 25 in Behaviour of Gases chapter where Bg =  0.00504 zi T  G  = ( G − Gp ) pi   G

0.00504 zi T p

 0.00504 zT    p  

zi z = ( G − Gp ) pi p

p  Gz  Gp = G −    i   z   pi 



(3)

hence plot of Gp vs p/z should give a straight line G X

Gp Cumulative gas production

O

X X

p/z

Pi/Zi

Figure 3 Gp vs. p/z.

If gas was ideal a plot of Gp vs p would be a straight line. It is often practice to do this and get a relatively straight line, but caution has to be taken, since deviation from a straight line could indicate additional energy support. - when p/z = 0 Gp = G the original gas in place - when Gp = 0 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. 22/07/14

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This simple analysis method for gas reservoirs has gained wide acceptance in the industry as a history matching tool, to determine for example an estimate of initial gas reserves based on production data. This Figure, (Figure 3 ), can then be compared to estimates from exploration methods. It can also give indications of gas to be produced at abandonment pressures. The following example exercise from Slider’s2 reservoir engineering text illustrates the application of the method

EXAMPLE A dry gas reservoir has produced as follows: Data 07-Jan-65 07-Jan-66 09-Jan-67 10-Jan-68 11-Jan-69

Cumulative production MM SCF 0 1,800 3,900 5,850 9,450

Static Res. Pressure psia 3,461 3,370 3,209 3,029

Data Reservoir Temperature T = 100°F Gas Gravity SG = 0.68 1

Determine the original pressure and original gas in place.

2 What will be the average reservoir pressure at the completion of a contract calling for delivery of 20 MM SCFD for 5 years (in addition to the 9,450 MM SCF produced to 11-Jan-69?)

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 The pseudo-critical parameters are found to be: Pseudo critical pressure (psia) Ppc = 667.5 psia Pseudo critical temperature (oR) Tpc = 385.0 °R Graphical material balance Cumulative production MM SCF 0 1,800 3,900 5,850 9,450

8

Reservoir Pressure psia

Pseudo reduced Pressure Temp. Pr Tr

P/Z Z (From Fig. 2)

3,461 3,370 3,209 3,029

5.19 5.05 4.81 4.54

0.796 0.790 0.778 0.765

1.45 1.45 1.45 1.45

4,348 4,266 4,125 3,959

Material Balance Equation F I F T E E N

The P/Z vs. G, plot is shown in the following Figure: Graphical Material Balance

4500

y=-0.0522x + 4448.3 R2 = 0.9904

4400

P/Z (psia)

4300 4200 4100 4000 3900 3800

0

2,000

4,000

6,000

8,000

10,000

Cumulative gas production (MM scf)

12,000

14,000

Figure 4 (a) Graphical Material Balance

4500 4000

y=-0.0522x + 4448.3 R2 = 0.9904

P/Z (psia)

3500 3000 2500 2000 1500 1000 500 0

0

10

20 30 40 50 60 Cumulative gas production (MMM scf)

80

90

Figure 4 (b)

From the straight line of Figure 4a Slope = -0.0522 Intercept = 4,448.3 Equation

=

P/Z= -0.0522 Gp + 4448.34

(1a)

Initial Pressure From Figure 4, at Gp = 0: Pi/Zi

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=

4,448.3 psia

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( Pi / Zi ) = Pri

Now, dividing by Pc ,

i.e. Pri/Zi = 6.6642

Pc

Zi

From Figure 5: Zi= 0.81 Therefore: Pi= 3,603 Original Gas in Place "G" G can be calculated directly from Eq. (1a), when P/Z = 0 G= 85,217 MM SCF Pressure at the Completion of the Contract Gas production rate (contract) = 20 MMSCFD Duration (contract) t = 5 Years Cumulative volume (contract) = 36,500 MM SCF At the end of the contract: Total cumulative production Gp = 45,950 MM SCF P/Z = 2,050 psia from equation 1a Again, by dividing by Pc Pr/Z = 3.07 and from Figure 5

Z

= 0.775

Finally,

P

= 1,589 psia

10

1.05 1.10 1. 1.2015 1.25 1.30 1.35 1.40 1.45 1 . 1 50 1 .6 0 1 .7 1.9.800 2 0 .0 2.2.20 0 40

Material Balance Equation F I F T E E N

1.75 1.65 1.55 1.45 1.35

Used to obtain p and z when analyses result in p/z answers.

(pr/z) = (p/z)/pc

z can be read from graph.

p = (pr/z) z pc

2.60 2.80 3.00

1.25 Gas deviation factor, z

1.15

Reduced temperature, Tr

1.05

3.00 2.80 2.60 40 2. 2.20 2.00 1.90 1.80 1.70 1.60 1.50 1.45 1.40 1.35 1.30 1.25

0.95 0.85 0.75 0.65 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)

Figure 5 Gas Deviation Factor z vs. Pr/z (Slider2).

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 then 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. (b)

(a) 3500

3500

P/Z

P/Z

O

Gp

G

G'>G

2700

Gp

Figure 6 p/z Plots For A Water Drive Gas Reservoir3.

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We will consider this topic later. If there is water drive then the equation; GBgi=(G-Gp)Bg +(We-Wp)Bw

(4)

applies.

3.4 Wet Gas Reservoirs

Another aspect that needs to be considered with gas reservoirs is the treatment of wet gas reservoirs. In these reservoirs production also includes liquids as well as gas, although in the reservoir the liquids were in a gaseous state, Figure 7. In the application of the material balance equation to these reservoirs it is important to convert oil production to gas equivalent Figures to add to the gas production Figures. Pi, Ti

Liquid

Bubble point line

Single phase

P 2 phase

Sep.

Mixture

Dew point line

Gas T

Figure 7 Phase Diagram For a Wet Gas System.

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). The volume of 1 STB of condensate of molecular weight Mo and specific gravity γo follows from equation.



V=

znRT P

z = 1.0 at p = 14.7 psia and T = 520˚R (Density of water = 62.4 lb/ft3)

∴ V / STB = 10.73

V = 133,000

12

psia SCF 520°R lb mole lb cu ft x x x62.4γ o x5.615 lb mole°R 14.7 psia M o lb cu ft STB

γ o SCF M o STB

Material Balance Equation F I F T E E N

3.5 Gas Cap Expansion

If a gas reservoir is attached to an oil reservoir (Figure 8 ), a similar equation to (2): 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

Figure 8 Gas Cap Expansion.

Change in gas cap volume is

(G - Gpc)Bg - GBgi

(5)

4. MATERIAL BALANCE FOR OIL RESERVOIRS 4.1 Above the Bubble Point

Above the bubble point the production of the reservoir is due to the expansion of the liquid (including water) in the reservoir as pressure declines, and the reduction in pore volume due to the decrease in pressure. Assuming that the production is due entirely to liquid oil expansion, a material balance for an oil reservoir is similar to that for a gas reservoir. NBoi = (N - Np)Bo

(6)

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 development. Pore volume changes will be considered later in the context of pore space and connate water.

4.2 Gas Liberation Below the Bubble Point

When the oil in the reservoir reaches the bubble point pressure, gas will be liberated and will continue to be liberated as the pressure declines. This is the mechanism of solution gas drive described previously. 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 released free gas will be produced to the surface, some will remain in the reservoir. 22/07/14

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The free gas in the reservoir can be written: Free gas in reservoir = original gas in solution - remaining gas in solution – produced gas(Gps).

= NRsi - (N - Np)Rs - Gps SCF



= (NRsi - (N - Np)Rs - Gps)Bg

bbl

(7)

The volume of free gas and the remaining oil can now be added to the original oil volume. NBoi = (N - Np)Bo + (NRsi-(N - Np)Rs - Gps)Bg

(8)

The equation can be written in terms of the original stock-tank volume in the reservoir.

N=

N p Bo + Bg (G ps − N p R s ) Bo − Boi + (R si − R s )Bg

(9) Np

NBoi

=

Gps

Free Gas (NRsi-(N-Np)Rs-Gps)Bg Oil (N-Np)Bo

Figure 9 Material Balance For Solution Gas Drive.

4.3 Material Balance with Gas Cap and Water Drive

The equation just developed assumes no change in reservoir volume. If fluid encroaches into original oil bearing volume either from an expanding gas cap or an encroaching water drive there will be loss to the reservoir volume. Change in volume due to gas cap expansion. = (G - Gpc)Bg - GBgi

(10)

If we don,t knowBw, we assume it is 1.0. In some equations, the Bw is left off the terms. But we should remember the We, Wp and Wi values are stock tank volumes that need a formation volume factor to become reservoir volumes. Change in volume due to water encroachment = (We-Wp) Bw

(11)

∴ Total change in volume = original oil volume - (oil volume + free solution gas)

14

Material Balance Equation F I F T E E N

Np NBoi

=

Gpc

Gps

Wp

Gas cap expansion Oil volume and free solution gas Net water encroachment

Material balance with gas cap and water drive

Figure 10 Material balance with gas cap and water drive.

That is Bw (We-Wp) + (G - Gpc)Bg - GBgi

(

= NBoi - (N - Np)Bo + (NRsi-(N - Np)Rs - Gps)Bg

N=

)

(12)



(

)

N p Bo + Bg (G ps − N p R s )- (G-G pc ) Bg − GBgi ) − (We − W p ) Bw Bo − Boi + (R si − R s )Bg



(13)

The gas production is separated into gas cap and solution gas, Gpc and Gps. However, the two can be combined so that Gp=Gpc+Gps then:

N=

(

)

N p Bo + Bg (G p − N p R s )-G Bg − Bgi − (We − W p )Bw Bo − Boi + (R si − R s )Bg



(14)

4.4 Effect of Pore-Volume Changes 4.4.1 Compressibility Effects

The compressibility of water is about 10-6 psia-1 compared to oil at around 10-5 psia-1 and although it is a low value it can contribute significantly to the hydrocarbon pore volume change when pressure declines. Another contributing effect to reduction of pore volume available to hydrocarbons is the compressibility of the pore volume itself. The reduction in pore volume with decline in pressure is due to two factors: • Reduction in the bulk volume of reservoir • Increase in volume of the reservoir grains

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Figure 11 below illustrates the impact of overburden stress. Overburden pressure

Pore pressure

- sand grains - pore space

Figure 11 Cross Section of Sandstone Influence of Overburden and Pore Pressure2.

4.4.2 Overburden Pressure

A reservoir is subjected to an overburden pressure caused by the weight of the formation above the reservoir. It is equivalent to about 1 psi/ft of depth. These pressures simply apply a compressive force to the reservoir rock. The pressure in the pore space, the pore pressure, does not normally approach that of the overburden pressure and is normally about 0.5 psi/ft. If the reservoir sands are highly unconsolidated then this pressure could be higher as the overburden pressure is transmitted to the fluids in the pore space. When the pore pressure is reduced then the effective opposing pressure is increased and the bulk volume (Vb) 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 (Vr). hence reduce Vb) ) and increase Vr )

- decrease in porosity ø

Compressibility of rock cf =

∆Vp = Cf ∆P Vp

1 ∆Vp . Vp ∆ p

Where Vp is the volume of the pores.

16

(15)

Material Balance Equation F I F T E E N

4.4.3 Connate Water

Expansion of connate water can contribute to the reduction in pore volume for the hydrocarbons. The compressibility of the water can be expressed as

cw = -

1 ∆Vpw Vpw ∆p

(16)

Where Vpw is the volume of water in the pores Vpw = Vp x Swc

∴ ∆Vpw = Cw∆PSwcVp

(17)

Total pore volume change:

∆Vp = ∆Vpw + ∆Vpr



∴ ∆Vp = (cf + cwSwc) ∆p Vp

(18)

This term can be added to the material balance equation in the same way that water encroachment can be considered. Pore volume: Vp =

NBoi 1 − Swc

Compressibility effect of water and pores

∆Vpw + pores =

NBoi (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



Vp = (1 + m)

NBoi 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.

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5

THE GENERAL MATERIAL BALANCE EQUATION

Combining all the effects of gas cap expansion, water encroachment, pore volume changes and equating these to the volume changes associated with the oil gives the general material balance equation below.

net waterinflux



porre volume reduction (C f + Cw Swc ) pNBoi Bw (We − W p ) + (G − G pc ) Bg − GBgi + (1 − Swc ) gas cap expansion

original oil

(

= NBoi − ( N − N p ) Bo + ( NRsi − ( N − N p ) Rs − Gp ) Bg N=



oil and free solution gas

)

N p Bo + Bg (G p − N p Rs ) − G ( Bg − Bgi ) − (We − W p ) Bw Boi Bo − Boi + ( Rsi − Rs ) Bg + (C f + Cw Swc ) p 1 − Swc

The above equation is the general material balance equation. 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. Present Oil Volume (N-Np)Bo

=

Original Oil Volume NBoi

Free Solution Gas [NRsi-(N-Np)Rs-Gps]Bg

Gas Cap Expansion (G-Gpc)Bg-GBgi

Net Water Influx (We-Wp) Bw

Rock(pore) and connate water expansion NBoi (1+m)∆p

cwSwc+cf 1-swc

Injected volumes WinjBw+GinjBg

Figure 12 Elements of the MB Equation. 18

Material Balance Equation F I F T E E N

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. •

total formation volume factor Bt



where Bt= Bo + (Rsi-Rs)Bg

•

m=

•

GBgi NBoi

Gp = NpRp

Substituting these factors the general material balance equation because:

6

N=

N p ( Bt + ( Rp − Rs ) Bg ) − (We − W p ) Bw (21) ( Bg − Bgi ) Bt − Bti + (c f + cw Swc ) p Bti / (1 − Swc ) + mBti Bgi

MODIFICATIONS TO THE GENERAL EQUATION

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 composition 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, associated 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 22/07/14

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Wi. Clearly when any of these three drive supports, natural or otherwise, are not active 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. We have developed the equation by considering the impact of the various elements involved in fluid production. An alternative derivation is based on the perspective that the equation is an expression of the total compressibility of the reservoir system.

7

DERIVATION OF THE MATERIAL BALANCE EQUATION BY EQUATING SUBSURFACE VOLUME OF PRODUCED FLUIDS TO EXPANSION OF ORIGINAL FLUIDS PLUS PORE VOLUME REDUCTION

The simple definition of compressibility, c, is

c=−



1 dV 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. C g Co Cw Cf

= 500x10-6 = 10x10-6 = 3x10-6 = 1x10-6

to to to to

1500x10-6 psi-1 20x10-6 psi-1 5x10-6 psi-1 25x10-6 psi-1

It is better to consider rock compressibilities in the context of the reservoir behaviour in terms of pore volume compressibility, since it is the pore volume which is available for containing fluids. If the compressibility is in terms of the change in pore volume per unit bulk volume, dividing it by the porosity changes the numeric value when expressed as pore volume per pore volume. For example, at a porosity of 20%, a compressibility of 1x10-6 pore volume per bulk volume is 5x10-6 pore volume per pore volume per psi. Even these pore compressibilities are small but their significance should not be neglected particularly above the bubble point. This compressibility perspective as an alternative way to derive the material balance equation is based on the following philosophy. Visualise again a reservoir at initial pressure pi containing oil plus dissolved gas, a primary gas cap and connate water (Figure 13)

20

Material Balance Equation F I F T E E N

FLUID VOLUMES Prim. Gas cap

Initial volumes at pressure Pi

Oil + originally dissolved gas

Fluid expansions down to pressure P

Connate water Water influx We

PORE VOLUMES

New pore volume at pressure p

Reduction in total pore volume down to pressure P Total pore volume at pressure Pi

Figure 13 Material Balance By Equating Sub Surface Expansion To Fluid Production.

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 13. 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 13. 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: Reservoir volume at pressure p of the produced fluids = expansion of primary gas cap + expansion of oil plus originally dissolved gas + expansion of connate water + water influx + reduction of total pore volume.

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Or put in another way by Dake15 Underground withdrawal = expansion of the system + cumulative water influx. The individual components in this equation can be quantified as follows: Reservoir Volume at p of Produced Fluids Production consists of: Np STB of oil, Gp or Np RpSCF of gas and Wp STB of water. The subsurface volume at pressure p of oil is Np Bo res. bbl. including an amount of dissolved gas equivalent to Np Rs SCF. We have produced Np Rp SCF, therefore the equivalent of Np Rp – Np Rs SCF must exist as free gas in the reservoir at pressure p. Its subsurface volume is Np (Rp-Rs)Bg res.bbl. Hence the reservoir volume at pressure of produced hydrocarbons ,HCPV is: produced HCPV (p)

= Np Bo + Np (Rp-Rs)Bg (res.bbl.) = Np (Bo + (Rp-Rs) Bg) (res.bbl.) (22)

Reservoir volume of produced water is about equal to WpBw barrels, hence: Res. volume at p of produced fluids

= Np(Bo+(Rp-Rs)Bg) + Wp Bw (res. bbl.)

(23)

Expansion oil + originally dissolved gas The original volume of the oil is:NBoi, when the pressure is reduced the oil shrinks and gas is liberated. Volume change of oil is N(Bo-Boi). During this process gas comes out of solution whose reservoir volume is:

N(Rsi-Rs)Bg.

Therefore the total change in volume is:

N[(Bo - Boi) + (Rsi- Rs)Bg]

Expansion of primary gas cap The original hydrocarbon pore volume of the gas cap is: mNBoi The surface volume is;



22

mN

Boi scf Bgi

(24)

Material Balance Equation F I F T E E N

At a lower pressure,p, the reservoir volume is:



mNBoi

Bg Bgi

The expansion of the gas cap is therefore:

B  mNBoi  g − 1  Bgi 

(26)

Expansion of connate water Connate water has a low compressibility but is significant in undersaturated conditions The compressibility of the water is:



cw =

1 dVw Vw dP

The expansion of the connate water with pressure decline is therefore: 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 − S wc )

The pore volume of connate water is:



(1 + m) NBoi Swc (1 − Swc )

The expansion of the connate water is therefore:

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(1 + m )NBoiSwc c w p (1 − Swc )

(27)

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Pore Volume Changes The impact of pressure reduction on the pore volume is to reduce volume available for hydrocarbons and therefore can be treated as an expansion term alongside the expansion terms associated with the oil, gas and connate water. The change in volume of the pores associated with the total pore volume is therefore:

(1 + m )NBoiCf 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 Bw (res.bbl). If we don't know Bw, we assume it equals 1.0. 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.

[

]

reservoir volume of produced fluids N p Bo + ( Rp − Rs ) Bg + Wp Bw

B  + mNBoi  g − 1  Bgi 

expansion of gas cap

B  Bgi

[



g pore+volume mNBoi reduction − 1 +



]

N (Bo - B o i + ( Rsi- Rs ) Bg

(

Expansion of oil and dissolved gas

(1 + m) NBoi (cw Sw + c f )∆p + We (1 − Swc )

(22) (24)

(26)

(27 + 28)

(This includes encroached water, We) This simplifies to:

[

]

N p ( Bo + ( Rp − Rs ) Bg =  ( B − Boi ) + ( Rsi − Rs ) Bg   Bg   c S + cf   NBoi  o − 1 + (1 + m) w wc  + m  ∆p Boi  1 − Swc     Bgi   +(We − Wp Bw )

29)

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 F I F T E E N

8

ASSUMPTIONS IN 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 production and injection wells. In order to apply the equation at a particular time an average pressure has to be selected 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, conditions, 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 including 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 application of other equations. Representative PVT data The PVT measurements should be made in an attempt to reflect the behaviour in the reservoir. Although this may not be totally possible conditions as near to the real situation are used, for example in the differential test to reflect below bubble point conditions. Good production data It is important in the application of the MB equation to have reliable production data not only oil and gas but also water.

9

SIGNIFICANCE AND USAGE OF THE MATERIAL BALANCE EQUATION

The material balance is roughly a relation between four quantities: • Oil and gas in place (N, m or G) • Production (Np, Rp, Wp) • Water influx (We) • Average reservoir pressure (pressure dependent PVT parameters and in pore/ water compressibility term).

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This means that if three of these quantities are known, the fourth can be calculated. Some examples illustrate this: •

If production and pressure 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: (a) Deciding whether or when artificial lift facilities will be needed. (b) Estimating the reserve of the reservoir down to a certain abandonment pressure, also as function of the cumulative gas oil ratio Rp.

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. Should be known Potential unknown Np N Rp We Wp p cw Bo, Bg, Rs Swc m Bw cf 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 petrophysical related information. The material balance generated result provides a more effective value since it would not include volumes in undrained or low permeability areas of the reservoir. 26

Material Balance Equation F I F T E E N

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 the influx (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 a 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. 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.

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10 SOURCES OF DATA TO BE USED IN THE MATERIAL BALANCE 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. PVT Data From PVT reports of individual wells. Averaging and correcting PVT data prior to use in and M.B. may also be required Production Data From well and reservoir records (data banks) or the subject of calculation. Oil and Gas in Place From volumetric estimates or subject of the calculation. Connate Water Saturation Sw : from petrophysics Water Compressibility cw : at oilfield temperatures and pressures: Should be determined. 4 to 5 x 10-5 atm-1 = about 3 x 10-6 psi-1 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.

11

LIMITATIONS OF THE MATERIAL BALANCE

A material balance is a zero dimensional mathematical model, in which fluid properties and pressures are averaged over the entire reservoir. Variations in initial fluid properties, for instance, a change in bubble point either laterally or as function of depth, as a result of compositional variations, cannot be handled adequately. The degree to which the results of an M.B. calculation are invalidated depends on the magnitude of such variations. 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.

28

Material Balance Equation F I F T E E N

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 situation will be achieved in 1, 10 or 100 years. By combining the material balance results with other methods for example well productivity equations time information can be added to the production / pressure predictions from MB methods.

12 CONCLUSION Summarising: the material balance is an important and indispensable reservoir engineering tool. As with other reservoir engineering tools it has its limitations of which the user should be aware. Viewed against a somewhat wider background the following quotation from Muskat (Reservoir Engineering News Letter September 1947) is still applicable:

“The materials balance method is by no means a universal tool for estimating reserves. In some cases it is excellent. In others it may be grossly misleading. It is always instructive to try it, if only to find out that it does not work, and why. It should be a part of the stock in trade of all reservoir engineers. It will boomerang if applied blindly as a mystic hocus-pocus to evade the admission of ignorance. The algebraic symbolism may impress the old timer and help convince a Corporation Commission, but it will not fool the reservoir. Reservoirs pay little heed to either wishful thinking or libellous misinterpretation. Reservoirs always do what they ought to do. They continually unfold a past which inevitably defies all man-made laws. To predict this past while it is still the future is the business of the reservoir engineer. But whether the engineer is clever or stupid, honest or dishonest, right or wrong, the reservoir is always right.”

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 Gas Material Balance with no water drive GBgi = ( G-Gp)Bg (5x108x0.001)=(5x108 -Gp)x0.0011 Gp = 4.55E+07 scf

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equation 1

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REFERENCES 1

Schilthuis, R.J., Active Oil and Reservoir Energy, Trans AIME, 118:33-52, 1936.

2

Slider, H.C., Petroleum Reservoir Engineering Methods, Petroleum Publishing Co. Tulsa, 1976.

3

Drake, L.P., The Practise of Reservoir Engineering. Elsevier Ams. 1994

4 Archer, J.S., and Wall,C.G. Petroleum Engineering, Principles and Practise, Graham&Trotman .Ldn 1986 5

30

Drake, L.P., Principles of Reservoir Engineering. Elsevier 1978

Material Balance Equation Application S I X T E E N

5000 4000 3000 2000

Volumetric Gp vs p

1000 0

Initial gas in place

Pressure or P/Z

Water drive Volumetric Gp vs p/z

Wrong extrapolation 1

2 3 4 5 g F/E Cumulative Production, MMMSCF

Strong water drive 6

7

Moderate water drive Volumetric depletion

G = initial gas in place Gp

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Material Balance Equation Application S I X T E E N

C O N T E N T S 1 INTRODUCTION 2 LINEAR FORM OF MB EQUATION 2.1 Short Hand Version of MB Equation 2.2 No Water Drive and No Original Gas Cap 2.3 Gas Drive Reservoirs, No Water Drive and Known Gas Cap 2.4 Gas Drive Reservoirs with No Water Drive, N and G Are Unknown 3 DEPLETION DRIVE OR OTHER 4 GAS FIELD APPLICATIONS OF THE MATERIAL BALANCE EQUATION 5 MATERIAL BALANCE EQUATION APPLIED TO OIL RESERVOIRS 5.1 Depletion Drive Reservoirs 5.1.1 Above The Bubble Point 5.1.2 Solution Gas Drive 5.1.2.1 Instantaneous Gas – Oil Ratio 5.1.2.2 Oil Saturation Equation 5.2 History Matching 5.3 Performance Prediction Methods For Solution Gas Drive Reservoir Reservoirs 5.3.1 Solution Gas Drive Characteristics 5.3.2 Performance Prediction – Schilthuis Method 5.3.3 Tarner's Methods 5.3.4 Tracy's Form of Tarner Method 5.3.5 Muskat Method 5.4 Data For Solution Gas Drive Predictions 5.5 Gas Drive Reservoirs 5.6 Average Reservoir Pressure 6 RESERVOIR PERFORMANCE AS A FUNCTION OF TIME

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LEARNING OBJECTIVES Having worked through this chapter the Student will be able to:

2

•

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.

Reservoir Engineering

Material Balance Equation Application S I X T E E N

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.

2

LINEAR FORM OF MB EQUATION

2.1 Short Hand Version of MB Equation

The material balance equation in itself is not a difficult concept to understand, the difficulties lie in the application of the equation to real reservoir problems. The problem which generally faces the engineer is the inadequate understanding of the reservoir preventing knowing the extent of the driving mechanism or mechanisms. 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 selection of the variable depending on the drive mechanism.

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Reservoir Engineering

Their technique is useful in that if a linear relationship does not exist for a particular interpretation of the reservoir, then this deviation from linearity suggests that the reservoir itself is not performing as anticipated and other mechanisms are involved. 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. The material balance equation can be written in the following form:

[

]

[

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 (1 − Swc )  Bgi 

] (1)

In some instances water formation volume factors are not included, i.e. Wp, We and in some they are WiBw Havlena and Odeh simplified the equation into a short hand form:

F = NEo + NmEg + NEfw +We (2)

The left hand side of equation 2 represents the production terms in reservoir volumes and are denoted by F, i.e.

F = Np[Bo + Bg (Rp - Rs)] + Wp...bbl (3)

The right hand side includes: (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:

E fw = (1 + m)

Boi (c f + Sw cw )∆p.....bbl / STB 1 − Sw

(5)

(iii) expansion of the free gas Eg where:

4

B  Eg = Boi  g − 1 ....bbl / STB  Bgi 

(6)

Material Balance Equation Application S I X T E E N

With the above terms the material balance equation can be written:

F = NEo + NmEg + NEfw + We (7)

This equation as presented above neglects water or gas injection terms. Using this equation as a basis, Havelena and Odeh manipulated the equation making different assumptions to produce a linear function.

2.2 No Water Drive and No Original Gas Cap In this condition We and Nm are zero and the equation becomes:

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.

F

N

O

Eo

Figure 1 F vs. Eo No Water Drive and No Gas Cap.

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:

F = N(Eo + mEg) (9)

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.

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m

o

to

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l

al

sm

m too

O

N large

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.



E F = N +G g Eo Eo

(10)

where:



G = Nm

Boi Bgi

A plot of F/Eo vs Eg/Eo should be linear with an intercept of N and a slope of mN. Figure 3.

F/Eo

mN

N

F = N + G Eg Eo Eo

Eg /Eo

Figure 3 F/Eo vs. Eg/Eo

6

Material Balance Equation Application S I X T E E N

2.5 Water Drive Systems

Water influx is discussed in greater detail in a later chapter and we will examine this linearisation of the MB equation in that context then.

3

DEPLETION DRIVE OR OTHER

Determining the drive energies responsible for production is important in understanding and predicting the future behaviour of a reservoir. The material balance equation can be used in this short hand form to get an indication of whether a field is depleting volumetrically or there is other energy supporting the system for example, water drive or gas cap expansion. If we consider a reservoir which does not have an aquifer but there is the possibility of a supporting aquifer. In this case the short hand version of the MB equation is;

F = N(Eo + Efw) + We.....bbl (11)

If both sides are divided by Eo +Efw Then the following equation results:

F We =N+ ....STB (12) Eo + E fw Eo + E fw

F Eo + Efw

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 x+

A

N O

Np, p or time

Figure 4 Determining Drive Mechanism.

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 that the left and right hand side of the equation are constant, ie. We = 0 and the pore 22/07/14

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compressibility is constant. This is a solely depletion drive where the energy comes from the compressibility of the oil and its originally dissolved gas. The intercept on the y axis is also the constant term, N, the oil in place, the STOIIP. The other plots indicate that pressure support is also coming from elsewhere, water drive, or abnormal compaction or a combination of both. Perhaps in C there is a water drive from an infinite aquifer, where the aquifer boundary has still to feel the pressure. Curve B might be for a finite aquifer, where later in production, there is less support from the aquifer. Another feature of this presentation is that back extrapolation of the B and C curves also gives the in place oil volume, N.

4

GAS FIELD APPLICATIONS OF THE MATERIAL BALANCE EQUATION

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 3000 2000

Volumetric Gp vs p

1000 0

Initial gas in place

Pressure or P/Z

Water drive Volumetric Gp vs p/z

Wrong extrapolation 1

2 3 4 5 Cumulative Production, MMMSCF

6

7

Figure 5 Comparison of the Gp vs p/z and p for Volumetric and Water Drive Gas Reservoirs. (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.

8

Material Balance Equation Application S I X T E E N

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 S

w wc

+ cf

1 − Swc

) ∆p + W

e



(13)

Using the short hand version of Havlena and Odeh results in

F = Gp Bg + Wp Bw res cu ft

(

Eg = Bg − Bgi

E fw = Bgi

(c S

)

w wc

rcf / scf + cf

1 − Swc

) ∆p

rcf / scf (14)

The short hand gas material balance equation becomes

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. The equation results in

F = GEg + We (15)

Dividing both sides by Eg gives

F W = G + e (16) Eg Eg

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.

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Strong water drive F/Eg

Moderate water drive Volumetric depletion G = initial gas in place Gp

Figure 6 Material Balance Application to Determine G and Drive Mechanism.

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. The mobility ratio for water displacing is



M=

krw ′ krg ′ / µw µg

Where: krw′ and krg′ are the end point relative permeabilities for water and gas. M can be as low as 0.01 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. Gas initially in place =G = Vφ(1-Swc)/Bgi (17) The pore volume, PV, of the reservoir, Vφ can be expressed in terms of the gas in place.



10

PV = GBgi

1 1 − Swc

(18)

Material Balance Equation Application S I X T E E N

the movable gas volume, MGV, by water flooding is;

MGV = PV (1 − Sgr − Swc ) = GBgi

(1 − Sgr − Swc ) (19) 1 − Swc

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

1 − Swc

)

(20)

The volume of trapped gas within this swept proportion is therefore:



α ( PV )Sgr / Bg = α

GBgi Sgr S p/z = αG gr 1 − Swc pi / zi Bg 1 − Swc

(21)

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. The p/z approach is long established in gas reservoir engineering as a method to determine gas in place and recovery at field abandonment. Gas produced from Reservoir = Gas initially in reservoir - Gas remaining in reservoir (scf)



c + cwc Swc   Gp = G −  GBgi − GBg f ∆p − We Bw  / Bg 1 − Swc  

(22)

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:

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 Gp  1 −   p pi G = z zi  We Bw / Bgi  1 −    (24) G

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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 −  (25) z zi  G

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 15.) If there is any other pressure support the curve will deviate from linear as demonstrated in Figure 7 below by Dake1. pi/zi

Material balance at abandonment

Strong waterdrive

Depletion p/z

Moderate waterdrive

Abandonment pressure Surface compression

0

Gp

Gp=6

Figure 7 P/z vs Gp Plots For Depletion and Water Drive (Dake1).

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 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. 12

Material Balance Equation Application S I X T E E N

low summer rate

p/z high winter rate

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, 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). At abandonment gas production Gp = G -Trapped Gas - By-passed gas.



S   Gp = G − αGBgi gr + (1 − α ) GBgi  / Bgab (26) 1 − Swc  

α is the volumetric sweep efficiency at the abandonment pressure, Bgab, gas formation volume factor at abandonment. When this equation is expressed in terms of p/z it becomes



 Gp  1 −   p pi G = zab zi  Sgr 1−α  α + (27) α  1 − Swc

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 .

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5

Reservoir Engineering

MATERIAL BALANCE EQUATION APPLIED TO OIL RESERVOIRS

5.1 Depletion Drive Reservoirs

We will now examine the application of the material balance to oil reservoirs and present some methods using the material balance equation to predict reservoir performance. 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.

5.1.1 Above the Bubble Point

Above the bubble point, in the undersaturated state, the system is all liquid and without any other drive mechanism therefore the fluid production is purely a result of the total compressibility of the system. Although it appears complex, the material balance equation is a sophisticated version of the compressibility definition:

∆V = C x V x ∆p

Production = Expansion of reservoir fluids. If we take the full MB equation developed in the previous chapter then the equation simplifies above the bubble point to:

 ( B − Boi ) (cw Swc + cf )  N p Bo = NBoi  o + ∆p (28) 1 − Swc  Boi 

•

no gas cap

•

aquifer small in volume We =Wp =0

•

Rs=Rsi=Rp - all gas at surface dissolved in oil in reservoir.

Oil compressibility is

co =

( Bo − Boi )

Boi ∆p (29)

Replacing this for the oil term gives:

14

(

 cw Swc + c f N p Bo = NBoi co + 1 − Ssw 

) ∆p 

(30)

Material Balance Equation Application S I X T E E N

Above the bubble point only oil and connate water exists , therefore: So + Swc = 1 The MB equation for application above the bubble point becomes:

c S + c S + cf N p Bo = NBoi  o o w wc 1 − Swc 

or:

  ∆p 

(31)



N p Bo = NBoi ce ∆p (32)



ce =

1 (co So + cw Swc + c f ) (33) 1 − Swc

ce is the effective saturation weighted compressibility of the reservoir system. The recovery at bubble point pressure is therefore:

N p Boi = (ce ∆p) (34) N Bob

Clearly this fractional recovery expressed as Np/N can be determined from a simple function of the change in oil formation volume factors, and water and pore compressibilities for the conditions above the bubble point where Np = cumulative oil produced N = original oil in place 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: NpBo + WpBw = NBoice∆p (35) 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 is a basis for determining the time availability for water injection installations to arrest the pressure decline.

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EXERCISE: Solution Gas Drive – Undersaturated Oil Reservoir Determine the fractional oil recovery, during depletion down to bubble point pressure, for the reservoir whose PVT parameters are listed in PVT Chapter 14. Assume a separator pressure of 300 psi. Other reservoir fluid properties are listed in table below.

Initial reservoir pressure Reservoir temperature Water compressibility Rock compressibility Water saturation Separator pressure

pi = T = cw = cf = Swc = Psep =

3,000 220 3.0E-06 8.6E-06 0.22 300

psi ºF psi-1 psi-1 psi

5.1.2 Solution Gas Drive

When the reservoir pressure drops below the bubble point, then solution gas drive is effective, then the analysis is more complex as gas comes out of solution. Solution gas drive is the most common drive mechanism in oil reservoirs but unfortunately it is also very inefficient. It is often associated with other drive mechanisms for example water drive and gas cap drive. In this next section however we will ignore supplementary drive from these two sources. In order to use the material balance equation to predict further production as a function of pressure decline, we need to develop other independent equations, the instantaneous producing gas-oil ratio equation and the saturation equation.

5.1.2.1 Instantaneous Gas – Oil Ratio

The instantaneous gas- oil ratio, R, is the ratio of gas production to oil production at a particular point in the production time of the reservoir, that is at a particular reservoir pressure. The gas-oil ratio equation is based on the Darcy flow equation. The instantaneous producing gas-oil ratio is:

R=

Gas producing rate, SCF / Day Oil producing rate, STB / Day (36)

The gas production can come from gas in solution in the reservoir and from production of free gas in the reservoir which has come out of solution. Free Gas =

16

qg Bg

Solution gas = QoRs

Material Balance Equation Application S I X T E E N

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 Total gas production rate:

Qg =

qg + Qo Rs (37) Bg

where: Qg = total gas producing rate. Oil producing rate is:

Qo =

qo Bo (38)

where: qo = reservoir oil flow rate, res.bbls/day Combining equations 36,37 and 38 gives:

since:





qg + Qo Rs Bg R= (39) qo / Bo

Qo =

qo Bo

(40)

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:



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qg =

2πkeg h∆p µ g ln re / rw

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and:



qo =

2πkeo h∆p µo ln re / rw

Therefore in (41):



2πkeg h∆p B µ lnr / r R = g g e w + Rs 2πkeo h∆p Bo µo lnre / r w

R=

Bo keg µ o + Rs Bg keo µ g

(42)

This is the usual form of the instantaneous gas-oil ratio equation.

Instantaneous GOR

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.

Rsi

3

4 1

Pb

2

Np

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

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Material Balance Equation Application S I X T E E N

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 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.

1 Rp = Np



Np

∫ RdN o

(42)

p



ΣR ∆N Rp = i pi (43) Np

whereRi is the average instantaneous GOR over the period that ∆Npi of oil was produced. Another equation which is required in conjunction with the instantaneous gas-oil ratio equation is the Oil Saturation Equation.

5.1.2.2 Oil Saturation Equation

This equation, the oil saturation equation, provides an average oil saturation for a reservoir at any time. It is defined as:



So =

oil volume remaining total pore volume

So =

( N − N p ) Bo NBob / (1 − Swi )

(44)

where So = oil saturation at any time Swi = connate water saturation N is the oil in place at the bubble point. Np is the cumulative oil production below the bubble point. This equation can be rearranged to



So = (1 −

Np Bo ) (1− Swi ) N Bob

(45)

the oil saturation equation.

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Equation 45 is important because the oil saturation as a function of pressure must be known to determine the relative permeability associated with the instantaneous GOR equation. 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.



So ′ =

( N − N p ) Bo (NBob − m NBob ) / (1 − Swi )

So ′ =

( N − N p ) Bo (1 − Swi ) (NBob 1 − m )

N p  Bo  (1 − Swi ) 1 −   N  Bob So ′ = 1− m

(46)

where: So' = oil saturation in 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.

5.2 History Matching

A common tool in reservoir engineering is history matching, which operates on the principle if your prediction model cannot be used to predict the past then its application for the future will be in serious question. When past performance cannot be calculated then either the input data is wrong or the model is incorrect or both are not correct. It is likely that there are errors associated with the input data. The instantaneous GOR equation can be used to provide a history matched set of relative permeability data. 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.

20

Keg Bµ = ( R − Rs ) g g (47) Keo Bo µo

Material Balance Equation Application S I X T E E N

One of the most difficult aspects in predicting reservoir performance of solution gas drive reservoirs is having relative permeabilities which represent the reservoir. Laboratory values are specific to small samples of rock which clearly cannot be representative of the whole field. There is even doubt on the laboratory procedures used and the preparation procedures to obtain ‘reality’ of rocks and fluids. The above equation provides where data is available a procedure to make use of past production data to generate field relative permeabilities for use in performance predictions. 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).

5.3 Performance Prediction Methods For Solution Gas Drive Reservoirs

There are two phases to a solution gas drive reservoir, the period above the bubble point where reservoir fluids remain single phase, with the gas remaining in solution. The other phase is that below the bubble point, where gas comes out of solution and the productivity of the formation is influenced by the relative properties of the oil and gas phases. In predicting reservoir performance, these phases of production are treated separately. The production and pressures from the initial pressure down to the bubble point, and then the production and pressure profile from the bubble point pressure down to the reduced pressure as the reservoir is produced.

5.3.1 Solution Gas Drive Characteristics.

Before looking at the various predictions it is worthwhile reviewing the characteristics of solution gas drive reservoirs. When the subject was considered in the drive mechanism chapter it was indicated that these reservoirs do not yield good characteristics from an oil recovery perspective. They are characterised by: 1 Rapid pressure decline 2 Water-free production 3 Rapidly increasing gas-oil ratio 4 Low ultimate recovery 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 point 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.

5.3.2 Performance Prediction – Schilthuis Method

Schilthuis rearranged the material balance equation to the following form:

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Np =

[

]

N Bo − Bob + ( Rsi − Rs ) Bg + G( Bg − Bgi ) − Gp Bg Bo − Bg Rs

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(48)

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Petroleum Engineering

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. The gas production, Gp is obtained from the producing GOR and cumulative oil production. This GOR however varies as the reservoir is depleted, so an incremental procedure is used based on

j =n

Gp = ∑ Rj ∆N pj (49) j =1

Rj is the average producing GOR over the increment of production. The GOR from the instantaneous gas-oil ratio equation is also a function of relative permeability.

R=

Bo keg µo + Rs (42) Bg keo µ g

and the relative permeability is related to saturation which is related to cumulative oil production through the saturation equation.



N  B  So = 1 − p  o (1 − Swi )  N  Bob (45)

Schilthuis’s procedure is as follows: 1 Set a pressure below Pb- Bubble point pressure 2 Assume Gp and calculate Np from the MB equation above. 3 Calculate So from oil saturation equation above, using Np determined in 2. 4 Determine relative permeability ratio from laboratory or field generated rel. perm vs So data. 5 Calculate Gp using instantaneous GOR eqn. and Gp equation above. Ri and Ri+1 are the instantaneous GOR at the start and end of the production interval. 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. Tarner approach is not too different from the Schilthuis method.

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Material Balance Equation Application S I X T E E N

5.3.3 Tarner’s method5

In Tarners5 method the material balance equation is arranged as follows in a form to calculate Gp, where subscripts refer to the saturation condition.

N p Rp =

N ( Bo + ( Rsi − Rs ) Bg − Bob ) − N p ( Bo − Rs Bg ) Bg

NpRp = cumulative gas produced SCF.

(50)

The procedure is of a trial and error approach using the two independent equations, instantaneous GOR and MB: 1 Starting at the bubble point pressure. 2

Select a future pressure and assume a value of Np at that pressure. It may be convenient to express Np as a function of N. Solve the MB equation for NpRp which is the cumulative gas production.

3

Using the assumed Np solve the oil saturation equation-equation 45 for So. These enable keg/keo to be determined.

4

Calculate the instantaneous gas-oil ratio R from equation 42.

5

Calculate the gas produced during the pressure drop period, i.e.



( Ri + Ri +1 ) N p1 2



where Ri = instantaneous gas-oil ratio at start of period.

(51)

Ri+1 = instantaneous GOR at end of period Np1 = cumulative oil produced at end of the period. The assumption of the equation is that a plot of R vs Np is linear. Since it is unlikely that such a relationship exists small pressure drops should be used. A Figure of 50 psi. is suggested. 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

Step Two 1 A second pressure is selected and a new value of Np assumed Np2. 2

Solve MB for Np2. This is the cumulative gas production at the end of the second pressure.

G2 = N p 2 Rp 2 − N p1 Rp1 = 22/07/14

N ( Bo + ( Rsi − Rs ) Bg − Bob ) − N p 2 ( Bo − Rs Bg ) − N p1 Rp1 (52) Bg

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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.

5

Instantaneous GOR calculated keg/keo above ratio determined.

6

Gas produced during second step:

(R i +1 + R i + 2 ) (N p2 − N p1 ) = G 2 2 (53)



where G2 is the total gas produced during step two. 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.

5.3.4 Tracy's Form of Tarner Method Tracy6 took the Schilhuis MB equation

N=

N p (Bo − R s Bg ) + G p Bg − (We − Wp ) Bo − Boi + (R si − R s )Bg + mBoi (Bg − Bgi ) / Bgi (54)

and suggested a shorthand version, where:

φn =

Bo − R s Bg Bo − Boi + (R si − R s )Bg + mBoi (Bg − Bgi ) / Bgi

(55)

φg =

Bg Bo − Boi + (R si − R s )Bg + mBoi (Bg − Bgi ) / Bgi

(56)

φw =

1 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:

φn =



φg =



φw =

24

Bo − Rs Bg Bo − Boi + ( Rsi − Rs ) Bg (58) Bg Bo − Boi + ( Rsi − Rs ) Bg (59) 1 (60) Bo − Boi + ( Rsi − Rs ) Bg

Material Balance Equation Application S I X T E E N

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:

N = Npφn + Gpφg - (We-Wp) φw (61)

If we assume no water encroachment or production:

N = Npφn + Gpφg

(62)

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. For the kth pressure:

N = Npkφnk + Gpkφgk (63)

If N is set at 1.0 then the equation takes a fractional recovery form:

1.0 = Npkφnk + Gpkφgk (64)

also:

1.0 = (Npj+∆Npk)φnk + (Gpj+∆Gpk) φgk

(65)

and:

1.0 = (Npj+∆Npk)φnk + (Gpj+R′avg x ∆Npk) φgk (66)

where R′ avg is the estimated average gas-oil ratio between pressure Pj and Pk. i.e.

Ravg ′ =

R j + R′k (67) 2

R′k can be estimated by extrapolating the plot of R versus pressure calculated at a higher pressure. Equation 66 can also be written in the form:

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1.0 = Npjφnk+Gpjφgk+ ∆Npk(φnk+φgk R′avg) (68)

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solving for ∆Npk gives:

∆N pk =

1.0 − N pjφnk − Gpjφ gk φnk + φ gk R′ avg

(69) 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.



Rk =

Bo keg µo + Rs Bg keo µ g

keg/keo is estimated from the liquid saturation since keg/keo=f(So) So being obtained from the oil saturation equation:



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. 100 80 60

0.10 .08 .06 .04

40

.02

20

.010 .008 .006

10 8 6

.004

φg

φn

2

.002 .001 0

4

5 10 15 20 25 Pressure, hundreds of psia

1

Figure 10 φg and φn Functions (Tracy6).

Using the above equations the step by step procedure of Tracy is as follows.

26

Material Balance Equation Application S I X T E E N

Tracy’s Procedure Set pressure step below Pb 1

Estimate R'k • above bubble point - Rs • from extrapolation of previous trend • from instantaneous GOR equation.

2

Estimate R´avg (Equation 67)

3

Determine functions φn and φg (Equation 55 and 56).

4

Using equation 69 determine ∆Npk & Σ∆Np.

5 Using Σ∆Np determine So from saturation equation (equation 45) and thereby keg/keo from So vs keg/keo data. 6

Calculate Rk from instantaneous GOR equation 42.

7

Compare Rk with R´k. of step 1. If (Rk - R´k) is greater than tolerance, tolerance limit set equal to R´k = Rk and repeat steps 1 - 6.

8 Estimate ∆Gp and Σ∆Gp. ∆Gp = ∆Np x Ravg. 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'. 10 Set next pressure and repeat step 1-9.

5.3.5 Muskat Method

Morris Muskat, according to Dake1 did more in the 40’s and 50’s to evolve reservoir engineering techniques in terms of well established physical principles and their mathematical foundation. Muskat7 considered the oil remaining in the reservoir, Nr, from which Np of oil had been produced. The remaining oil is:

Nr = N − N p =

Vp So Bo

(70)

Vp is the pore volume in rb (res bbl). The change in this volume is;

dNr 1 dSo S dBo = Vp − Vp 2o dp Bo dp B o dp

(71)

The total volume of dissolved and free gas is:

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Gr = Vp

So Rs Vp + (1 − So − Swc ) (72) Bo Bg

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The change in this volume with pressure is:



 S dRs Rs dSo Rs So dBo 1 − So − Swc dBg  dGr = Vp  o + − 2 −  dp B o dp B2 g dp  (73)  Bo dp Bo dp

The current producing GOR can be obtained by dividing eqn 73 by equation 71.

R=



∆G p ∆G r ∆G r / ∆p dG r / dp = = = ∆N p ∆N r ∆N r / ∆p dN r / dp

So dR s R s dSo R sSo dBo 1 dSo 1 − So − Swc dBg + − 2 − − Bo dp Bo dp B o dp Bg dp B2g dp ( 75) R= 1 So So dBo − 2 Bo dp Bo dp

The producing GOR can also be obtained through the instantaneous GOR equation

R=

Bo keg µo + Rs (42) Bg keo µ g

This equation is equated to the eqn. 75 above to yield:



So Bg dRs So krg µo dBo (1 − So − Swc ) dBg + − Bo dp Bo kro µ g dp Bg dp dSo = (76) krg µo dp 1+ kro µ g

This equation can be expressed in an incremental form to yield:

∆So = ∆p

In which :

So X ( p) + So Υ ( p)krg / kro − (1 − So − Swc ) Z ( p) (77) k µ 1 + rg o kro µ g



Χ( p) =

Bg dRs (78) Bo dp



Υ( p) =

1 dBo µo (79) Bo dp µ g



Ζ( p) =

1 dBg (80) Bg dp

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Material Balance Equation Application S I X T E E N

These are all pressure related functions and can be obtained from PVT data.

So =



oil remaining ( N − N p ) Bo = (1 − Swc ) (81) one PV NBoi

Which gives

Np So Boi =1− 1 − Swc Bo N



(82)

Dake1 proposes structuring a table with the following steps: Steps: (i)

(ii) (iii) (iv) (v) (vi) (vii) (viii)

Pressure X,Y,Z(p) So krg/kro (psia)

∆So So Np/N GOR

/psi PV PV PV scf/stb

i Pressure in steps below bubble point ii Table of X,Y &Z(p) values, calculated at the average pressures between the steps in (i). iii So prior to the pressure drop ∆p. iv Relative permeability ratio at last value of So. v ∆So determined using eqn. 77. vi The lower value of So at reduced pressure. vii The fractional recovery from bubble point. Equation 82 viii The GOR obtained from GOR eqn using krg/kro value obtained from new So 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.

5.4 Data for Solution Gas Drive Predictions

Prior to carrying out the MB procedure it is important to test the data, if there is past production data available. The methods can then be used to examine if they are capable of predicting past performance. Clearly if the ‘cold data’ does not enable past performance predictions to be matched, then there is an opportunity to adjust some of the data to obtain a history match. The data can then be used with better confidence to predict future performance.

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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. a

Reservoir fluid information from PVT analysis: i Oil formation-volume factor Bo ii Gas solubility, Rs iii Compressibility factor, z z can be calculated from gas composition, or gas gravity. iv Reservoir oil and gas viscosity, µo and µg It is convenient to plot the data as a function of pressure.

b

Past production: i Oil production ii Gas production iii Water production iv Net water influx

Again these are plotted versus pressure. c

Core analysis data: i Laboratory relative permeability keg/keo vs So or kew/keo vs So

d

Geological, petrophyisical and simple core data: i Original oil in place, N ii Size of gas zone, m iii Initial water saturation, Swi iv Porosity.

5.5 Gas Drive Reservoirs

As discussed in chapter 10, Drive Mechanisms, Gas Drive is also a depletion type reservoir. From the nature of the pressure gradients within the oil column it is also likely that solution gas drive is also active when depleting a gas cap drive reservoir. Cole8 has pointed out that gas drive is essentially a frontal drive displacing mechanism. In this respect the high mobility of the displacing gas to that of the displaced oil is such that in depleting the oil reservoir it is important to minimise the rate, to reduce by-passing of oil by the advancing gas oil contact. The density differences of gas and oil clearly help to offset the advancing mobility ratio effect. 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.

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Material Balance Equation Application S I X T E E N

In Tarner's method the equation for Gas Production, Gp, becomes:

 B − Bgi  N ( Bo + ( Rsi − Rs ) Bg − Bob ) + mBob  g  − N p ( Bo − Rs Bg )  Bgi  (83) N p Rp = Bg

5.6 Average Reservoir Pressure

The material balance approach is sometimes considered as the ‘tank model’. In the application of the MB equation we are assuming that the pressure is uniformly distributed across the reservoir. If there is uniform pressure decline in all the wells in the reservoir then this pressure decline gives confidence for application of the MB tool. Dake1 pointed out that if this equilibrium is not achieved, the MB approach can still be used. He suggested that an average pressure can be determined to represent a reservoir where there are large differential pressures across the reservoir. He presents an averaging procedure for reservoirs where pressure equilibrium has not been achieved.

Pressure

In the Figure 11 from Dake1 are presented the pressures for equilibrium conditions and the well positions and boundaries for a non equilibrium condition.

Time

Figure 11(a) Individual Well Pressures -Equilibruim In The Reservoir.

pj,qj,Vj

Figure 11(b) Well Positions and Drainage Boundaries.

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Pressure

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Time

Figure 11c Well pressure for Non Equilibrium Wells.1

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. The volume weighted average pressure is therefore:

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 (85) dt dt

For a constant compressibility this becomes

Vj ∝ q j / p′ j (86)

Therefore equation 85 can be expressed as:



∑ p q / p′ p= ∑ q / p′ j

j

j

32

j

j

j

j

(87)

Reservoir Engineering

Material Balance Equation Application S I X T E E N

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:

∑ p ∆UW / ∆p p= ∑ UW / ∆p j

j



j

j

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.

6

RESERVOIR PREDICTIONS AS A FUNCTION OF TIME

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. On the basis of productivity index the reservoir performance as a function of time can be predicted. The productivity index of a well is:

J=

where:

Qo STB / psi (89) pe − pw

Qo = Oil Producing rate, STB/day Pe = Reservoir pressure, psia Pw = Well bore producing pressure, psia Replacing Qo in equation 89 with the Darcy radial flow equation results in the following equation:

J=

7.07keo h( pe − pw ) (90) µo ln(re / rw ) Bo ( pe − pw )

For most reservoir producing conditions, the height and drainage area remains constant and therefore:

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7.07h = Constant = C (91) lnre / rw

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Combining equations 90 and 91, and eliminating the identical (pe - pw) terms gives:

J=

or:

C=

Cko (92) µo Bo Jµo Bo 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: Solution Gas Drive - Undersaturated Oil Reservoir Determine the fractional oil recovery, during depletion down to bubble point pressure, for the reservoir whose PVT parameters are listed in PVT Chapter 14. Assume a separator pressure of 300 psi. Other reservoir fluid properties are listed in table below. Data

Initial reservoir pressure Reservoir temperature Water compressibility Rock compressibility Water saturation Separator pressure

pi = T = cw = cf = Swc = Psep =

3,000 220 3.0E-06 8.6E-06 0.22 300

psi ºF psi-1 psi-1 psi

SOLUTION (A) From PVT data Bubble point pressure Pb Oil formation factor at Pi: Boi Oil formation factor at Pb: Bob

= = =

2,620 psi 1.4844 bbl/STB 1.4950 bbl/STB

Separator Pressure FVF psig bblb/STB 300 1.495

34

Material Balance Equation Application S I X T E E N

Oil Compressibility:

co =

Bob − Boi Boi ∆P

co = 1.88E - 05 psi-1

Pressure psig 3,000 2,620

Relative Volume* bbl/bblb 0.9929 1.0000

FVF Bo bbl/STB 1.4844 1.4950

Effective, saturation-weighted compressibility of the system:



Ce =

(co So + cw Swc + cf ) (1 − Swc )

ce = 3.06 x 10-5 psi-1 Finally, the fractional recovery at bubble point pressure is:

Np N

= ce ∆P

Boi Bob

Np N = 0.0115 or 1.15% of the original oil in place

If rock and water compressibilities are ignored in undersaturated conditions then a significant error results. Using the previous exercise calculate this error. SOLUTION (B) When water and rock compressibility is ignored:

Np Bo − Boi = N Bo Np 1.4950 − 1.4844 = 1.4950 N Np = 0.00714 N

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Np = 0.00714 or 0.71% N

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REFERENCES

1

Dake, L.P. “The Practise of Reservoir Engineering” Elsevier, Ams 1994

2. Schilthuis, R.J. Active Oil and Reservoir Energy. Trans., AIME,118:33-52, 1936. 3

Havelena, D and Odeh, A.S. The Material Balance Equation as an Equation of a Straight Line. J.Pet.Tech. Aug. 896-900. Trans AIME. 1963

4 Craft and BC Hawkins M.F. Applied Petroleum Reservoir Engineering. Prenticetiall, New Jersey, 1959 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

7

Muskat, M. “The Production Histories of Oil Producing Gas-Drive Reservoirs” J.Applied Physics,16,147, 1945.

8

Cole, F.W “Reservoir Engineering Manual” Gulf Publishing, Houston 1961.

9

Havlena, D. and Odeh, A.S. “The Material Balance as the Equation of a Straight Line- Oart II Field Cases”, JPT July1964

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AVERAGE BOUNDARY PRESSURES

Water Influx S E V E N T E E N

Pi P1

P1

P2

P1 = 1/2 (Pi - P1)

P2 = 1/2 (Pi - P2)

P2

P3 = 1/2 (P1 - P3)

P3 P3

P4 = 1/2 (P2 - P4)

P4

P5

P4 0

1

2

3

4

5

TIME PERIODS

θ

Aquifer OWC

Reservoir

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Water Influx S E V E N T E E N

C O N T E N T S 1 DRIVING FORCE FOR WATER DRIVE 2 MODELS FOR WATER ENCROACHMENT 2.1 The Diffusivity Equation 2.2 States of Flow 2.3 Schilthuis Steady State 2.4 Hurst Modified Steady State 2.5 Van Everdingen and Hurst unsteady-state 3 RESERVOIR PERFORMANCE PREDICTION 3.1 Unsteady-State Model – Van Everdingen & Hurst 3.2 Application to a Declining Pressure 3.3 What are the correct values ∆p? 4 HISTORY MATCHING WATER INFLUX 4.1 Water Drive, No Gas Original Cap 4.2 Water Drive, Gas Cap of Known Size 5 RESERVOIR PERFORMANCE PREDICTION USING THE UNSTEADY STATE EQUATION AND THE MATERIAL BALANCE EQUATION 6 FETKOVITCH METHOD FOR WATER INFLUX DETERMINATION 7 CARTER-TRACY WATER INFLUX PROCEDURE

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LEARNING OBJECTIVES Having worked through this chapter the Student will be able to:

2

•

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.

•

Sketch the progressive pressure profile for a constant boundary pressure.

•

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.

Reservoir Engineering

Water Influx S E V E N T E E N

RESERVOIR PERFORMANCE PREDICTION - 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. A large proportion of the world’s hydrocarbon reservoirs have an associated aquifer which depending on production rates can provide a major part of the energy for producing the oil. The consideration is that the original reservoir system was occupied with water and hydrocarbons have migrated in, displacing some of the water. The hydrocarbon and aquifer are therefore part of the same reservoir system responding to the various pressure changes resulting from the production of fluids. As pointed out during the drive mechanism chapter if the aquifer bounds the edges of the oil zone it is called edge water drive and if the water bounds the base of the oil reservoir it is called bottom water drive. The characteristics of water drive are usually the most efficient displacing agent naturally available in the reservoir. The most significant characteristics of a water drive system are: (1) Pressure decline is very gradual. (2) Excess water production occurs in structurally low wells. (3) The gas-oil ratio normally remains steady during the life of the reservoir. (4) A good recovery of oil can be anticipated

1

DRIVING FORCE FOR WATER DRIVE

The driving force for water drive comes from the response to pressure being lowered 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 representing 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. Water encroaches (moves ) into a reservoir in response to pressure reduction resulting from well production. This pressure reduction causes: (a) Expansion of the water due to pressure drop within the aquifer (b) Expansion of hydrocarbons in the aquifer, if any (c) Expansion of rock, which decreases porosity (d) Artesian flow, if any, where the outcrop is located structurally higher than the hydrocarbon accumulation, and the water is replenished at the surface Clearly the amount of expansion or fluid encroachment is proportional to: (a) The size of the aquifer (b) The porosity and permeability of the rock (c) The presence of any artesian water support

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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 It is clear therefore in predicting the performance of an aquifer a whole range of aquifer reservoir characteristics are required. 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 generate reservoir characteristics. 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 uncertainties 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 In relation to artesian aquifer support it is considered that their occurrence is probably rare, although there are some reservoirs which purportedly have this type of drive. Due to faulting or pinchouts, most hydrocarbon reservoirs do not communicate to a surface outcrop. In addition, for artesian flow to contribute substantially to water influx into a reservoir there must be sufficient ground water moving into the outcrop to replace the fluid withdrawals from the reservoir, and this water must move through the entire distance from outcrop to reservoir at this same rate. Thus it is believed that water influx is usually the result of expansion as a result of pressure drop. 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 reduction in the pore volume of the reservoir as withdrawals continue. 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 increasing 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. 4

Reservoir Engineering

Water Influx S E V E N T E E N

Effective water compressibilities which have been used in reservoir engineering calculations with good results vary from 1.0 x 10-6 bbl/bbl/psi to 1.0 x 10-4 bbl/bbl/psi. The Figure below illustrates an aquifer supported oil reservoir.

We

AQUIFIER Wi

OIL

Oil/Water Contact

Figure 1 Aquifer supported oil reservoir.

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: Wi pi p

= initial volume of water in aquifer (function of geometry) = initial aquifer/reservoir pressure = current reservoir pressure. Generally assumed to be pressure at original oil water contact. c = total aquifer compressibility. cf +cw where cf = pore compressibility cw = water compressibility

The main problem facing the reservoir engineer is determining the characteristics of the aquifer; its geometry, size and flow characteristics. The following example illustrates the water influx impact of a relatively low compressibility oil reservoir aquifer system.

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Petroleum Engineering

Reservoir Engineering

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 Reservoir radius ro Water compressibility cw Rock compressibility cf Reservoir thickness h Pressure drop ∆P Porosity φ Water saturation swc

2

= 35,000 ft. = 3,200 ft. = 3.0E-06 psi-1 = 5.0E - 06 psi-1 = 45ft. = 1,100 psi = 0.22 = 0.25

MODELS FOR WATER ENCROACHMENT

As we have indicated water influx arises as a result of pressure changes decompressing the aquifer. If the pressure in an aquifer can be calculated then the resulting volumetric changes can be determined from the pressure/volume compressibility relationship The Figure 2 below illustrates the pressure profiles in a reservoir aquifer system. It is important to appreciate that these profiles are generated as a result of decompression of the oil and aquifer as a result of well bore production. (i) Aquifer boundary not affected by decreasing pressure (ii) Aquifer boundary influenced by pressure decline.

(i) (ii)

Pressure

OIl RESERVOIR Wellbore

AQUIFER

Reservoir outer boundary

Aquifer Outer Boundary

Figure 2 Pressure distribution in an oil reservoir aquifer system.

Before examining the different models we will review the development of equations which enable the pressure, time and distance solution to be obtained.

6

Water Influx S E V E N T E E N

2.1 The Diffusivity equation

The diffusivity equation results from a combination of the continuity equation and Darcy’s law.

Well

p q

+

p

d p+

dq

q

h r

dr

Figure 3 Radial flow segment

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: (i) Expansion of the fluid q due to pressure drop dp over element dr (ii) Expansion of the fluid in the element due to pressure changing with time dp/dt The expansion of (i) is too small and can be neglected. Volume of element

V = 2πrh φ dr

(2)

Change in volume dV due to pressure drop dp is:



dV = - cV dp = - c 2πrh φ dr dp

dq =

(3)

dv ∂p = -2πrh φcdr dt ∂t (4)

∂q ∂p = − 2πrh φc (5) ∂r ∂t

Darcy's law for radial flow is

q=−

2πrkh ∂p (6) µ ∂r

Differentiating

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∂q −2πkh = ∂r µ

Reservoir Engineering

 ∂ 2 p ∂p  r ∂r 2 + ∂r  (7)  

Equating equation 5 and 7 gives



−2πrh φc

dp −2πkh = dt µ

 d 2 p dp  r dr 2 + dr   

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. This group of terms is referred to as the diffusivity constant,η where

1 φµc = η k

(9)

The equation becomes:

∂ 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. In the radial diffusivity equation when applied to an aquifer hydrocarbon reservoir system the inner boundary is the extent of the hydrocarbon reservoir and the outer boundary is the limit of the aquifer. In the analysis of the flow and pressures in the hydrocarbon reservoir the inner and outer boundaries are the radius of the well bore and the radius of the reservoir.

2.2 States of flow

The diffusivity equation developed above shows that the pressure is a function of time. As long as this exists, the pressure change with time δp/δt is not constant, and the flow is termed unsteady state. All states of flow at the start are unsteady state. During this period we need to analyse the pressure at elements across the radial symmetry and from that determine the resulting expansion. After a time period,δp/δt becomes constant; when this occurs the system is termed pseudo steady state and fluid expansion can be obtained from a tank model concept.

8

Water Influx S E V E N T E E N

All aquifers are finite in size, however there is a period of time when a pressure disturbance 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 modelling, 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.

2.3 Schilthuis steady-state model

The simplest model is that due to Schilthuis. The assumptions associated with this model are that the aquifer is very large such that the pressure remains constant, and it has a high permeability such that there is no pressure gradient across the aquifer. Craft and Hawkins2 present a useful hydraulic analogue of this steady state model as shown in Figure 4 below.

Pi

p Sand-filled pipe

Aquifer

Production

Figure 4 Hydraulic analogue of the Schilthuis steady state model.2

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.

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Initially both aquifer and reservoir tanks are at the original reservoir pressure and as the reservoir is produced at constant rate the pressure in the reservoir drops. At any instant, when the reservoir pressure has dropped to a value p, the rate of water influx by Darcy’s law, will be proportional to the permeability of the sand in the pipe, the cross-sectional area of the pipe, and the pressure drop (pi - p); and inversely proportional to the water viscosity and the length of the pipe, provided the pressure of the aquifer tank remains constant. The maximum rate of influx occurs when p=0. If this rate is greater than the reservoir production rate then at some intermediate pressure the rate of influx will be equal to the rate of production and the pressure will stabilize, at a steady-state value. This is an analog of steady-state water influx into a reservoir as expressed analytically by the Schilthuis equation in which the constant, C, depends upon the permeability and dimensions of the aquifer rock and the average viscosity of the water in the aquifer. Schilthuis equation for this is:

t

We = C ∫0 ( pi − p )dt

(11)

C is the aquifer constant and contains the unchanging components of Darcy’s Law, (units vol/time/pressure). Expressed in terms of rate of water influx:

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.

2.4. Hurst Modified Steady State

The analogue of this is if the tank is neither very large nor replenished, then as production proceeds, the level in the aquifer tank falls and the potential of the aquifer declines. If this decline is exponential then it may be represented by Hurst’s following equation in which the declining value of C is represented by C1/log at, and a is a time conversion constant. t

We = c1 ∫ o



10

( pi − p)dt log e at

dWe c1 ( pi − p ) = dt log e at

(13)

Reservoir Engineering

Water Influx S E V E N T E E N

2.5 Van Everdingen and Hurst unsteady-state

The following equation which we will develop later is the Van Everdingen and Hurst unsteady-state equation, which is a model which has become generally accepted 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

Figure 5 Hydraulic Analogue of unsteady State Water Influx2 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 dimensionless 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 representing 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.

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T = 0 = Time 1

T

=

Pi

Reservoir Pressure

Reservoir Engineering

T

=

10 T

P1

00 =1

T=

00 10

P10 P100 P1000 RW

RE

Radius

Figure 6 Pressure Distributions in an aquifer at several time periods, for a constant rate of water influx2 at Rw

1

T = 0 = Time T=

Reservoir Pressure

P1

T

=

10

T=

0 10

T=

P RW

10

00

T=∞

Radius

RE

Figure 7 Pressure distributions in an aquifer at several time periods for a constant boundary pressure2 Rw

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 compressibility in a pressure decline from pi to zero.

12

Water Influx S E V E N T E E N

Reservoir

1

2

3

4

1

2

3

4

Aquifer

Reservoir

Figure 8 Cylindrical element in an aquifer surrounding a reservoir.2

3

PERFORMANCE PREDICTION

Although a consideration of the nature of various aquifers lends support to the analytical 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. Examination of the mechanics of water expansion into a hydrocarbon reservoir shows that it must be an unsteady state process. However, for combination drive reservoirs, where the water influx rate is small compared to the other driving forces, the use of the Schilthuis steady state equation can usually be used with reliable results. The water influx constant, C, can be determined from past production data, and then this same value of C can be used to aid in predicting reservoir performance. 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).

3.1 Unsteady-State Model – Van Everdingen & Hurst

From the analog above it is clear the exact solution for water influx is the unsteady state solution, where to access the water production we need to determine the pressure time distance profile across the aquifer.

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Petroleum Engineering

Reservoir Engineering

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:

δ 2 p 1 δp φµc δp + = k δt δr 2 r δr



(8)

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:

η=



k µφc

An exact analytical solution of the diffusivity equation for specified boundary and initial conditions define this pressure time profile and therefore will allow the calculation of the rate of water influx into a reservoir, provided the proper data are available. Van Everdingen & Hurst did this in 1949.4 Their analysis was for two cases: (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. The dimensionless form of the diffusivity equation is

1 δ  δpD  δpD (14)  rD = rD δrD  δrD  δtD

where:

14

tD =

kt r 2πkh∆p , rD = e , pD = (15) 2 µφcro ro qµ

Water Influx S E V E N T E E N

Since only basic equations have been utilised, the units for the above quantities are: tD = time, dimensionless t = time, seconds k = permeability, darcy µ = viscosity, centipoise φ = porosity, fraction c = effective aquifer compressibility, vol/vol/atmosphere ro = reservoir radius, centimetres Converting equation 15 to more commonly used units of t = days; k = millidarcies; µ= centipoises; φ = fraction; c = vol/vol/psi and r = feet; results in:

kt µφcro2 (16)

t D = 6.323 x10 −3

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 (17) 2πkh∆p

q D (t D ) =

qD(tD) = dimensionless influx rate at rD=1. 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. Examining equation (17) from t = 0 to t t

t



D dt µ qdt = q D (t D ) dt D ∫ ∫ 2πkh∆p o dt D o

(18)



We µ φµ cro2 = 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.

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Petroleum Engineering



2πφcro2 h∆pQ( t ) We = = 1.119φcro2 h∆pQ( t ) 5.615

Reservoir Engineering

(21)

where; We = cumulative water influx, barrels ∆p = pressure drop, psi Q(t) = dimensionless water influx 5.615 = conversion factor, cubic feet to barrels. Equation (21) may also be expressed as:

We = B x ∆pQ(t)

(22)

where: B = 1.119φcr2oh (23) This term B can be considered to be an aquifer characteristic, where the terms do not change during the decline. Van Everdingen and Hurst's 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, 9a-e.

16

Water Influx S E V E N T E E N

Dimensionless water influx and dimensionless pressures for infinite radial aquifers (courtesy of SPE)3 tD 1.0 x 10-2 5.0 x 10-2 1.0 x 10-1 1.5 x 10-1 2.0 x 10-1

Qt 0.112 0.278 0.404 0.520 0.606

pD 0.112 0.229 0.315 0.376 0.424

tD 1.5 x 103 2.0 x 103 2.5 x 103 3.0 x 103 4.0 x 103

Qt 4.136 x 102 5.315 x 102 6.466 x 102 7.590 x 102 9.757 x 102

tD 1.5 x 107 2.0 x 107 2.5 x 107 3.0 x 107 4.0 x 107

Qt 1.828 x 106 2.398 x 106 2.961 x 106 3.517 x 106 4.610 x 106

tD 1.5 x 1011 2.0 x 1011 2.5 x 1011 3.0 x 1011 4.0 x 1011

Qt 1.17 x 1010 1.55 x 1010 1.92 x 1010 2.29 x 1010 3.02 x 1010

2.5 x 10-1 3.0 x 10-1 4.0 x 10-1 5.0 x 10-1 6.0 x 10-1

0.689 0.758 0.898 1.020 1.140

0.469 0.503 0.564 0.616 0.659

5.0 x 103 6.0 x 103 7.0 x 103 8.0 x 103 9.0 x 103

11.88 x 102 13.95 x 103 15.99 x 103 18.00 x 103 19.99 x 103

5.0 x 107 6.0 x 107 7.0 x 107 8.0 x 107 9.0 x 107

5.689 x 106 6.758 x 106 7.816 x 106 8.866 x 106 9.911 x 106

5.0 x 1011 6.0 x 1011 7.0 x 1011 8.0 x 1011 9.0 x 1011

3.75 x 1010 4.47 x 1010 5.19 x 1010 5.89 x 1010 6.58 x 1010

7.0 x 10-1 8.0 x 10-1 9.0 x 10-1 1.0 1.5

1.251 1.359 1.469 1.570 2.032

0.702 0.735 0.772 0.802 0.927

1.0 x 104 1.5 x 104 2.0 x 104 2.5 x 104 3.0 x 104

21.96 x 102 3.146 x 103 4.679 x 103 4.991 x 103 5.891 x 103

1.0 x 108 1.5 x 108 2.0 x 108 2.5 x 108 3.0 x 108

10.95 x 106 1.0 x 1012 1.604 x 107 1.5 x 1012 2.108 x 107 2.0 x 1012 2.607 x 107 3.100 x 107

7.28 x 1010 1.08 x 1011 1.42 x 1011

2.0 2.5 3.0 4.0 5.0 6.0 7.0 8.0 9.0 1.0 x 101

2.442 2.838 3.209 3.897 4.541 5.148 5.749 6.314 6.861 7.417

1.020 1.101 1.169 1.275 1.362 1.436 1.500 1.556 1.604 1.651

4.0 x 104 5.0 x 104 6.0 x 104 7.0 x 104 8.0 x 104 9.0 x 104 1.0 x 105 1.5 x 105 2.0 x 105 2.5 x 105

7.634 x 103 9.342 x 103 11.03 x 104 12.69 x 104 14.33 x 104 15.95 x 104 17.56 x 104 2.538 x 104 3.308 x 104 4.066 x 104

4.0 x 108 5.0 x 108 6.0 x 108 7.0 x 108 8.0 x 108 9.0 x 108 1.0 x 109 1.5 x 109 2.0 x 109 2.5 x 109

4.071 x 107 5.032 x 107 5.984 x 107 6.928 x 107 7.865 x 107 8.797 x 107 9.725 x 107 1.429 x 108 1.880 x 108 2.328 x 108

1.5 x 101 2.0 x 101 2.5 x 101 3.0 x 101 4.0 x 101

9.965 1.229 x 101 1.455 x 101 1.681 x 101 2.088 x 101

1.829 1.960 2.067 2.147 2.282

3.0 x 105 4.0 x 105 5.0 x 105 6.0 x 105 7.0 x 105

4.817 x 104 6.267 x 104 7.699 x 104 9.113 x 104 10.51 x 105

3.0 x 109 4.0 x 109 5.0 x 104 6.0 x 109 7.0 x 109

2.771 x 108 3.645 x 108 4.510 x 108 5.368 x 108 6.220 x 108

5.0 x 101 6.0 x 101 7.0 x 101 8.0 x 101 9.0 x 101

2.482 x 101 2.860 x 101 3.228 x 101 3.599 x 101 3.942 x 101

2.388 2.476 2.550 2.615 2.672

8.0 x 105 9.0 x 105 1.0 x 106 1.5 x 106 2.0 x 106

11.89 x 105 13.26 x 105 14.62 x 105 2.126 x 105 2.781 x 105

8.0 x 109 9.0 x 109 1.0 x 1010 1.5 x 1010 2.0 x 1010

7.066 x 108 7.909 x 108 8.747 x 108 1.288 x 109 1.697 x 109

1.0 x 102 1.5 x 102 2.0 x 102 2.5 x 102 3.0 x 102

4.301 x 101 5.980 x 101 7.586 x 101 9.120 x 101 10.58 x 101

2.723 2.921 3.064 3.173 3.263

2.5 x 106 3.0 x 106 4.0 x 106 5.0 x 106 6.0 x 106

3.427 x 105 4.064 x 105 5.313 x 105 6.544 x 105 7.761 x 105

2.5 x 1010 3.0 x 1010 4.0 x 1010 5.0 x 1010 6.0 x 1010

2.103 x 109 2.505 x 109 3.299 x 109 4.087 x 109 4.868 x 109

4.0 x 102 5.0 x 102 6.0 x 102 7.0 x 102 8.0 x 102 9.0 x 102 1.0 x 103

13.48 x 101 16.24 x 101 18.97 x 101 21.60 x 101 24.23 x 101 26.77 x 101 29.31 x 101

3.406 3.516 3.608 3.684 3.750 3.809 3.860

7.0 x 106 8.0 x 106 9.0 x 106 1.0 x 107

8.965 x 105 10.16 x 106 11.34 x 106 12.52 x 106

7.0 x 1010 8.0 x 1010 9.0 x 1010 1.0 x 1011

5.643 x 109 6.414 x 109 7.183 x 109 7.948 x 109

Table 1 Dimensionless Qt and PD vs. tD for infinite reservoir 4.

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Reservoir Engineering

Table 4-7 Dimensionless water influx for finite outcroping radial aquifer (courtesy of SPE)3

tD

rD = 1.5

Qt

tD

rD = 2.0

Qt

tD

rD = 2.5

Qt

tD

rD = 3.0

Qt

5.0 x 10-2 0.276

5.0 x 10-2 0.278

1.0 x 10-1 0.408

3.0 x 10-1 0.755

1.00

7.0 x 10-2 0.330

1.0 x 10-1 0.404

2.0 x 10-1 0.599

5.0 x 10-1 1.023

1.40

9.0 x 10-2 0.375

1.50 x 10-1 0.507

3.0 x 10-1 0.758

7.0 x 10-1 1.256

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

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

5.5 x 10-1 1.083

1.50

1.997

6.0 x 10-2 0.304

8.0 x 10-2 0.354

1.1 x 10-1 0.414 1.3 x 10-1 0.446

7.5 x 10-2 0.345

1.25 x 10-1 0.458

2.00 x 10-1 0.597

1.5 x 10-1 0.509 2.5 x 10-1 0.681

4.0 x 10-1 0.897

4.0 x 10-1 0.895

6.0 x 10-1 1.143

9.0 x 10-1 1.465

1.60 1.80

Qt

1.571

2.00

1.940

2.40

1.761

2.111

2.60

rD = 4.0

Qt

2.442

3.196

2.893

4.0

3.859

3.00

3.170

5.0

4.454

2.40

2.715

3.50

3.493

6.0

4.986

2.80

2.976

4.00

3.792

7.0

3.098

4.25

3.932

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

1.8 x 10-1 0.507 1.9 x 10-1 0.517

3.75 x 10-1 0.848

7.5 x 10-1 1.229

2.50

5.00

7.5

5.684

8.5

6.089

4.068

8.0

4.323

9.0

6.276

2.886

4.25

3.742

6.00

4.779

10

6.621

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

2.5 x 10-1 0.559

6.00 x 10-1 1.070

1.2

1.679

4.25

3.317

2.8 x 10-1 0.574

7.00 x 10-1 1.143

1.4

1.811

4.75

3.439

2.4 x 10-1 0.554

2.6 x 10-1 0.565

3.0 x 10-1 0.582 3.2 x 10-1 0.588 3..4 x 10-1 0.594

5.50 x 10-1 1.028

6.50 x 10-1 1.108 7.50 x 10-1 1.174

8.00 x 10-1 1.203 9.00 x 10-1 1.253

1.1

1.3 1.5 1.6

2.106

1.8

4.0 x 10-1 0.606

1.2

1.358

2.2

4.5 x 10-1 0.613

5.0 x 10-1 0.617 6.0 x 10-1 0.621 7.0 x 10-1 0.623

8.0 x 10-1 0.624

1.3 1.4 1.6 1.7 1.8

2.0

2.5

3.0

4.0

5.0

1.382 1.402 1.432

2.4 2.6 2.8

1.444

3.0

1.468

3.8

1.453

1.487

1.495

1.499

1.500

1.924

2.0

1.295 1.330

1.870

1.975

1.00 1.1

1.747

1.7

3.6 x 10-1 0.599 3.8 x 10-1 0.603

1.605

3.4

4.2

4.6

5.50 6.00

3.491

9.50

5.917

18

8.50

3.767

9.50

4.749 4.846 4.932

11 12

6.035 6.246 6.425

9.200

5.321

17

7.004

38

11.00

3.951

14.00

16.00

22.00

3.928 3.967

3.985

3.993

13 14 15

16

17

3.997

18

3.999

25

20.00

3.999

24.00

4.000

5.385 5.435 5.476

5.506

5.531

5.551

16 18

20 22

24

26

30

6.922

7.076

7.189 7.272

7.481

100

7.464

30

5.621

42

7.490

40

5.625

50

7.497

5.624

46

50

80

34

35

46

7.434

5.579

38

42

60

20

5.611

34

7.332

7.377

7.494

8.809

28

12

5.241

8.365

8.968

3.894

11

7.880

24

9.00

3.843

6.712

7.680

8.611

22

30

14

7.457

20

6.825

5.078

Table 2 Dimentionless Qt Vs. tD for finite reservoir.

18

10

15

2.599

2.624

4.639

7.50

6.930

8.060

5.138

18.00

2.622

7.00

6.453

16

10.00

2.551

9.0

15

5.790

3.809

7.00

12.00

2.619

5.653

9.00

26

2.444

2.613

8.50

4.516

6.580

10.00

2.525

4.378

6.50

14

13

2.340

2.491

6.00

5.504

5.009

8.00

2.380

8.00

9.00

2.241 2.294

4.222

3.656

7.50

11 13

8.00

2.178

4.982

5.343

5.50

4.047

6.50

7.50

3.581

3.717

7.0

10.0

5.00

3.381

6.50

2.570

8.0

4.50

3.247

2.022

5.0

6.0

4.00

3.850

9.5

5.892

3.00

4.50

4.560

5.464

8.5 x 10-1 1.395

2.990

5.50

5.231

4.25 x 10-1 0.905

3.25

3.628

6.5

2.0 x 10-1 0.525

9.0 x 10-1 1.440

4.00

3.507

4.727

8.0 x 10-1 1.348

4.50 x 10-1 0.932

2.772

3.75

4.50

3.645

5.5

4.165

4.00 x 10-1 0.877

2.1 x 10-1 0.533

2.75

2.646

3.242

3.75

3.334

3.00 3.25

2.849

3.25

2.184 2.353

2.60

2.574

4.5

3.537

2.427

1.75

2.00

3.5

3.034

6.0 x 10-1 1.140 6.5 x 10-1 1.195

Qt

2.835

3.0

2.748

3.00 x 10-1 0.751 3.25 x 10-1 0.785

rD = 4.5

2.598

1.5 x 10-1 0.474 1.6 x 10-1 0.486

2.5

tD

2.80

2.20

2.273

2.20

tD

5.0 x 10-1 1.024

2.75 x 10-1 0.715

1.791

1.20

rD = 3.5

2.50 x 10-1 0.678

1.4 x 10-1 0.461

1.25

tD

70

90

9.097 9.283

9.404 9.481 9.532 9.565

9.586

9.612

9.621

9.623

9.624

9.625

Water Influx S E V E N T E E N

4.0

reD = 3.5 reD = 4.0

3.5

reD =

= r eD

∞

3.0

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.

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20

Reservoir Engineering

reD = 7.0 reD = 8.0

18

reD =

∞

r eD =

16

6.0

14

Qt

reD = 5.0

12 10

reD = 4.5

8

reD = 4.0

6

reD = 3.5 reD = 3.0

4 2 0

0

10

20

30

40

50

60

70

tD Figure 9b Graphical form of Qt Vs. tD for infinite and finite reservoirs. Dake.1

20

Water Influx S E V E N T E E N

110

re D

=

∞

100

90

80

Qt

70

60

50

reD = 10.0

40

reD = 9.0 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. tD for infinite and finite reservoirs. Dake.1

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22 Qt

10-1 10-2

2

4

1.0 8 6

2

4

10 8 6

2

4

6 8 10-1 2

4

6

tD

8

1.0

2

4

6 8 10

FINITE LINEAR AQUIFER

reD = 1.5

reD = 2.0

reD = 6.0 reD = 5.0 reD = 4.5 reD = 4.0 reD = 3.5 reD = 3.0 reD = 2.5

Petroleum Engineering Reservoir Engineering

Figure 9d Graphical form of Qt Vs. tD for infinite and finite reservoirs.

Water Influx S E V E N T E E N

10 6 8 104 4

reD = 2.5

reD = 3.0

102 reD = 4.0 reD = 3.5

reD = 5.0 reD = 4.5

reD = 6.0

∞

2

=

10 1

2

4

10 8 6

2 Qt

4

102 8 6

2

4

6 8 102

2

r eD

∞ =

4

tD

6 8 103

r eD

reD = 15.0

) t 'd on (c

reD = 7.0

reD = 10.0 reD = 9.0 reD = 8.0

103

USE THIS SCALE FOR LINE reD = ∞ (CONT'D) ONLY

Figure 9e 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:

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B = 1.119φcr2ohf (24)

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where: f = fraction of the reservoir periphery into which water is encroaching, Figure 10

θ

Aquifer OWC

Reservoir

Figure 10 Segment of radial water influx.

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 constant. Dake' has indicated that this maximum value of Q(t) depends on the size of the aquifer and is equal to: Q(t) max = 0.5 (reD2-1) for radial systems And Q(t) max = 1 for linear systems

(25) (26)

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 (27)  ro 

Examination of this equation indicates that it is the total water influx resulting 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 using the equation below: Dake1.

24

We = 2 hw

φkct × ∆p (ccs) πµ

(28)

Water Influx S E V E N T E E N

in field units this is:

φkct × ∆p (bbls) µ



We = 3.26 × 10 −3 hw



t in the above equation is in hours.

(29)

Table 3 lists the summary of expressions for Hurst and Van Everdingen for both radial and linear system1. Radial Geometry

ro

re f= θ/360°

θ Oil reservoir Aquifer

Linear Geometry

Aquifer

Oil reservoir

h

L

W

Figure 11 Parameters for radial and linear geometry.

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DARCY UNITS

RADIAL SYSTEM r0 = radius of reservoir

kt tD = φµcr 20

B = 2 π φ hcr 02 f

Reservoir Engineering

FIELD UNITS

t D = const × constant t - hrs. t - days t - yrs.

kt φµcr 20

= 0.000264 = 0.00634 = 2.309

B = 1.119 fφhcr 20 (bbl/psi) LINEAR SYSTEM

tD =

w = width of aquifer L = length of aquifer boundary to oil reservior boundary

kt φµcL2

B = wLhφc

t in secs and B in cc/atm

t D = const ×

kt φµ c L2

const. as for radial system

B = 0.1781wLhφc B-bbls/psi

Table 3 Summary of equations and constants for Van Everdingen and Hurst water influx model

EXERCISE 2 Water Influx – Infinite Aquifer Extent 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

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.

26

Water Influx S E V E N T E E N

3.2 Application to a Declining Pressure

The application of the Van Everdingen & Hurst model to water influx modelling is by application of their constant terminal pressure solution; that is the boundary at the reservoir aquifer contact is constant. For this constant pressure solution the rate and the cumulative water influx is calculated. In the example above this was the condition and the effects of a fixed pressure drop were determined. In reality however the pressure at the reservoir/aquifer boundary is declining continuously. How can we apply this fixed terminal pressure drop solution to a situation where there is a declining pressure? Figure 12 below gives such a declining pressure at the aquifer / reservoir boundary 3800

3780

Pi

P1

3750

Boundary Pressure - Psia

3750

P2 3700

3700

P3

3650 3620 P4

3600

3550

3500

0

3

6

9

12

15

18

21

24

Time - Months

Figure 12 Declining pressure profile at the oil / aquifer boundary.

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.

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The method is illustrated in the Figure 13 where the progressive impact of a series of fixed pressure drops is illustrated. owc Pi

owc Pi

Pi

T=0

We0 = 0 owc Pi

P1

owc P1

Pi

P1

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) owc Pi

P1

P2

P3

owc P3

We = We1+We2

P3

P2

P1

Pi

We(1) = B∆P1Q(T3) We(2) = B∆P2Q(T3-T1)

T=T

3

We(3) = B∆P3Q(T3-T2) We = We1+We2+We3

Figure 13 Progression of fixed pressuredrops through the aquifer.

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 13 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 throughout 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 pressure 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.

28

Water Influx S E V E N T E E N

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 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. In calculating cumulative water influx into a reservoir at successive intervals, it is necessary to calculate the total water influx from the beginning. This is required because of the different times during which the various pressure drops have been effective. The aquifer term, B is usually a constant for a given reservoir. Thus where the water 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: We1 = B x ∆p1 Q(t)1 We2 = B x ∆p2 Q(t)2 Wen = B x ∆p3 Q(t)n Combining the above equations:

We = B(Σ∆p x Q(t))

(30)

This equation is the form usually used to calculate water influx.

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3.3 What are the correct values ∆p?

AVERAGE BOUNDARY PRESSURES

By considering the gradual pressure drop to be a series of series of fixed pressure drops we need a method which will go towards representing this. Clearly shorter time periods are an advantage together with pressure drops which overlay the curve as shown in the Figure 14 below.

Pi P1

∆P1

P2

∆P1 = 1/2 (Pi - P1)

∆P2 = 1/2 (Pi - P2)

∆P2

∆P3 = 1/2 (P1 - P3)

∆P3 P3

∆P4 = 1/2 (P2 - P4)

∆P4

∆P5

P4 0

1

2

3

4

5

TIME PERIODS

Figure 14 Calculation of time period pressure drops.

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.

30

Water Influx S E V E N T E E N

EXERCISE 3 Water Influx -Infinite Aquifer - Declining Pressure Profile. Data Table 1 Reservoir Radius ro Effective water permeability kw Effective water compressibility ce Water viscosity µw Reservoir thickness h Porosity φ Initial reservoir pressure Pi Water encroaching factor f

= = = = = = = =

2,500 ft 105 mD 1.0E06 psi-1 0.8 cp 30 ft 0.22 3,800 psi 1

The pressure time relationship is as follows: Period

Time Months

Pressure psi

0 6 12 18 24

3,800 3,780 3,750 3,700 3,620

1 2 3 4

i.e. as Figure 12.

4

HISTORY MATCHING WATER INFLUX

In carrying out water influx calculations it has become clear that a number of parameters have a big impact on the pressure support from an aquifer. Not least the relative size and geometry of the aquifer when compared to the hydrocarbon reservoir. In reservoir predictions many of these parameters are not available to the reservoir engineer. Clearly it is not until production has commenced and the reservoir reacts to fluid production can one determine the pressure support coming from various drive energies. It is during this phase that the aquifer characteristics can be determined. In the previous chapter on the material balance equation applications we discussed the Havelena and Odeh5 approach to history matching using a linearisation approach. We will now continue this in the context of reservoirs with a water drive, using their basic equation:

F = NEo + NmEg + NEfw + We

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.

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Water Influx = Initial water volume x Pressure Drop x Aquifer Compressibility We = (cw +cf ) Wi x ∆p (31) As we have seen this equation is generally not sufficient to describe water influx behaviour, 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 perspective, where they use the dimensionless water influx term to express We, ie: We = B∑∆pQt (30) They then apply their short hand MB equation to examine different reservoir senarios as follows:

4.1 Water Drive, No Gas Original Cap Havlena and Odeh’s equation can be re-arranged as:



F ∑ ∆pQt =N+B Eo Eo (32)

A plot of F/Eo vs.

∑ ∆pQ Eo

t

should give a straight line, as shown in Figure 15.

∑∆pQt correct

all

B

W e

to o

sm

incorrect geometry

F/Eo

We

large

too

F Eo

= N+B

∑∆pQt Eo

N ∑∆pQt Eo

Figure 15 Plot of F/Eo vs .

32

∑ ∆pQ Eo

t

Water Influx S E V E N T E E N

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 too 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 reservoir 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 16 below the curves show the results for a range of dimesionless radii for a field in the North Sea6. The results show a straight line fit with an infinite aquifer. 16

o

x x

NPBo + Wρ - Wi , MMMSTB Eo

14

Re/Rw = 10

Re/Rw = 20

xx x xx x x xx x

12 10

6

4

x x x

2

x

x

x

x x

x

xx

x

x

x x x

x x x xx

x x

x x x

x

x

o x o x o x o x

Re/Rw = 00

x x xx xx

x x x

8

ox o oxx x oxxx

x

x

x

6 MONTHS 0 0

1.0

2.0

3.0

4.0

5.0

∑∆PQ(Qt) , MPSI Eo

Figure 16 Havalena and Odeh approach applied to Piper Field.6

Very small Aquifers For small aquifers, the assumption might be made that steady state flow exists as the pressure drop is quickly transmitted through the aquifer. In this case; We = B′ ∆p′

22/07/14



where: ∆p′ = pi - p



and: B′ = Wicw

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(33)

(34) 33

Petroleum Engineering

Reservoir Engineering

Wi is the water volume in the aquifer. The equation now becomes

F ∆p′ = N + B′ Eo Eo

(35)

A plot of F/Eo vs ∆p'/Eo should give a straight line of slope B′ and intercept N. Havlena and Odeh point out that the points on this graph will plot backwards as in Figure 17 below. This is because Eo increases faster than ∆p, therefore ∆p/Eo decreases as the pressure decreases. In some situations a steady state water influx sets in after a certain period. In this case the points plotted for the unsteady state period will plot in forward sequence but when the steady state exists then the plotted points reverse the plot backwards.

B'

F/Eo

F Eo

= N + B'

∆p' Eo

N ∆p' Eo

Figure 17 Plot of F/Eo Vs. ∆p'/Eo for small aquifers.

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:

34

F ∑ ∆pQt =N+B Eo + mEg Eo + mEg

(36)

Water Influx S E V E N T E E N

where We = B∑∆pQt Again this gives a straight line function ,Figure 18, 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.

B

F Eo + mEg F Eo + mEg

=N+B

∑∆pQt Eo + mEg

N ∑∆pQt Eo + mEg

Figure 18 Havalena and Odeh plot for water drive and known gas cap.

Small Aquifers, Gas Cap of Known Size For small aquifers, as for the system without a gas cap, a plot of the lefthand side and the B term of the following equation should result in a straight line, the points plotting in reverse sequence. Figure 19.

22/07/14

F ∆p ′ = N + B′ Eo + mEg Eo + mEg

(37)

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Reservoir Engineering

F Eo + mEg F Eo + mEg

= N + B′

Eo + mEg

N

Eo + mEg

Figure 19 Havlena and Odeh plot for small aquifers and known gas cap.

5

RESERVOIR PERFORMANCE PREDICTION USING THE UNSTEADY STATE EQUATION AND THE MATERIAL BALANCE EQUATION

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. The procedure is as follows: (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 calculation 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: F = NEo + NmEg + NEfw + We We(MB) = F - N( Eo +mEg + Efw) 36

F are production terms.

Water Influx S E V E N T E E N

Using the unsteady state equation and knowing the time and pressure drop,we can calculate

∑∆pQ(t)

The value of the aquifer characteristic, B, is then obtained by division:

B=



We ( mb ) ∑ ∆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.

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6

Reservoir Engineering

FETKOVITCH METHOD FOR WATER INFLUX DETERMINATION

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 Everdingen 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. The inflow equation used is:

qw =

dWe = J ( pa − p) dt



(39)

where: qw = water influx rate J = aquifer productivity index p = reservoir pressure (at oil or gas water contact) pa = average pressure in the aquifier We = cumulative water influx volume For a finite aquifer the total water influx arising from a total pressure drop is:

We = cWi ( pi − pa )



(40)

where: pi = is the initial pressure in the aquifer Wi = initial volume of water in the aquifer and reservoir c = total aquifer compressibility = cw + cf therefore: or: where:

 We  pa = pi 1 −  (41)  cWi pi   W  pa = pi 1 − e  (42)  Wei 

Wei = cWipi is the maximum possible expansion of the aquifer arising from a pressure drop of Pi.

38

Water Influx S E V E N T E E N

Differentiating equation 42 with respect to time gives:

dWe W dpa = - ei (43) dt p i dt

substituting in equation 39 gives:

dpa Jp = − i dt (44) pa − p Wei

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:

ln( pa − p) = −

Jpi t +C (45) Wei

C= constant of integration which from initial conditions= 1n(pi-p) therefore:

( pa − p) = ( pi − p)e

− Jpi t / Wei

(46)

substituting in equation 39 gives:



− Jpi t / Wei dWe = J ( pi − p)e (47) dt

Integrating this equation yields:



We =

− Jpi t / Wei Wei pi − p) 1 − e ( pi

(

)



(48)

As t tends to infinity then:

We =

Wei ( pi − p) = cWi ( pi − p) pi



(49)

which is the maximum amount of water influx that could occur once the pressure drop has moved through the aquifer.

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The equations developed so far assume a constant inner boundary pressure. Therefore to apply to a declining pressure a superposition principle should be applied as for van Everdingen and Hurst. Fetkovitch however showed that a difference form of the equation can be used. The water influx for a time step ∆t1.

∆We1 =

Wei ( pi − p1 ) 1 − e − J pi ∆t1 / Wei pi

(

)



(50)

where p1 is the average reservoir boundary pressure during the first time step. Similarly for ∆t2

W − J ∆t / W ∆We 2 = pei ( pa1 − p 2 ) 1 − e pi 2 ei (51) i

(

)

where pa1 is the average aquifer pressure at the end of the first time interval.

 ∆We1  pa1 = pi 1 −  (52) Wei  

The general solution for the nth time step is:

∆Wen =

Wei − J ∆t / W pan−1 − pn 1 − e pi n ei pi

(

)(

)

(53)

where:



pa( n−1)

n −1   ∆Wej  ∑  = pi 1 − j=1  (54) Wei    

The average reservoir boundary pressure Pn is calculated as in the Van Everdingen & Hurst method, i.e.:

pn =

pn −1 + pn 2

(55)

Equations 53 and 54 used in a stepwise technique by Fetkovitch have demonstrated that the water influx calculated match closely results using the van Everdingen and Hurst.

40

Water Influx S E V E N T E E N

The aquifer productivity index J values are listed below1. FLOWING CONDITIONS

RADIAL AQUIFERS J (cc/sec/atm)

Semi-steady State (drawndown expressed asPa - P)

2 πfkh  re 3  µ1n −   ro 4 

Steady State (drawndown expressed as Pi - P)

2 πfkh r µ1n e ro

Field Unit Factor

7.08 × 10 −3

LINEAR AQUIFERS J (cc/sec/atm)

3

khw µL

...................(56)

khw µL

...................(57)

1.127 × 10−3

Table 4 Values of the aquifer productivity index J. (Dake1)

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.

qw =

dWe = J ( pi − p)e − Jpi t / Wei (47) dt

Since steady state assumes Wei as infinite: i.e.

qw =

dWe = J ( pi − p) dt t

We = J ∫ ( pi − p)dt o





(58)

(59)

The expression of the above equation 57 is the same as the Schilthuis steady state equation, equation 11.

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Infinite Aquifers 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.

7

CARTER & TRACY WATER INFLUX PROCEDURE

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 − W ′  e ( t Dj −1 ) p( t Dj ) ( t Dj )  We t D = We t + ( Dj −1 )  p − t p′ t ( )  Dj Dj − 1  ( t Dj ) 



( )

B We tD

(t

Dj

)

- tDj-1

(60)

= aquifer constant as previously = 1.119fφhcr2o (bbl/psi) = cumulative water influx (bbls)

= dimensionless time = 2.309

kt where t is in years φµcro2

P(tD) = dimensionless constant terminal rate solution of diffusivity equation P'(tD) = time derivative dPtD/dtD. 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; p(tD)=a0 + a1tD + a2lntD + a3(lntD)2

42

Water Influx S E V E N T E E N

The regression coefficients are given in the table 5 below Carter-Tracy influence functions regression coefficients for the constant terminal rate case (17) reD

Regression coefficients a0

1.5 2.0 3.0 4.0 5.0 6.0 8.0 10.0 ∞

0.10371 0.30210 0.51243 0.63656 0.65106 0.63367 0.40132 0.14386 0.82092

a1 1.66657 0.68178 0.29317 0.16101 0.10414 0.06940 0.04104 0.02649 -3.68 x 10-4

a2

a3

-0.04579 -0.01599 0.01534 0.15812 0.30953 0.41750 0.69592 0.89646 0.28908

-0.01023 -0.01356 -0.06732 -0.09104 -0.11258 -0.11137 -0.14350 -0.15502 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 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 Reservoir radius ro Water compressibility cw Rock compressibility cf Reservoir thickness h Pressure drop ∆P Porosity φ Water saturation swc

= = = = = = = =

35,000 ft. 3,200 ft. 3.0E-06 psi-1 5.0E - 06 psi-1 45ft. 1,100 psi 0.22 0.25

SOLUTION (A) Total aquifer compressibility ce 8.0E-06 psi-1

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Aquifer Volume

Wi = π* (rw2 - ro2)*h*φ



Wi = 3.78E+10 ft.3

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We =ce * Wi * ∆P

We = 3.325E+08 ft.3

SOLUTION (B) Initial hydrocarbon volume Vhc = π* ro2 h* φ(1-Swc)

Vhc = 2.389E+08 ft.3

The water influx is capable of replacing the hydrocarbons pore volume. EXERCISE 2 Water Influx – Infinite Aquifer Extent 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 Effective water permeability Effective water compressibility Water viscosity Reservoir thickness Porosity Initial reservoir pressure Current reservoir pressure Water encroaching factor

ro kw ce µw h φ Pi P f

= = = = = = = = =

2,400 ft 120 mD 1.0e-06 psi-1 0.8 cp 35 ft 0.22 4,500 psi 4,490 psi 1

SOLUTION Water influx constant

B = 1.119 φ ce ro2 h f



B = 49.63 bbl/psi

Dimensionless time

td = 6.323x10-3

After 1 year, i.e.

t = 365 days tD = 273 days

44

(24)

k t (16) φ µ ce ro2

Water Influx S E V E N T E E N

and the Dimensionless influx rate Q(t); => is extracted from the table for infinite aquifers Q(t)

= 98.0

Water influx We: Results Summary

We = B ∆P Q(t) We = 48,626 bbl

Time (t) Years Days

tD

Q(t) Current (Interpolation) Pressure (psi)

Pressure Drop (psi)

Water Influx We (bbl)

After : 1 Year => 2 Years => 3 Years =>

1 2 5

365 730 1825

273 546 1,366

98.0 175.3 383.9

4,490 4,490 4,490

10 10 10

48,626 86,988 190,524

EXERCISE 3

WATER INFLUX -INFINITE AQUIFER - DECLINING PRESSURE PROFILE. Data Table 1 Reservoir Radius Effective water permeability Effective water compressibility Water viscosity Reservoir thickness Porosity Initial reservoir pressure Water encroaching factor

ro kw ce µw h φ Pi f

= = = = = = = =

2,500 ft 105 mD 1.0E06 psi-1 0.8 cp 30 ft 0.22 3,800 psi 1

The pressure time relationship is as follows: Period

i.e. as Figure 12.

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1 2 3 4

Time Months

Pressure psi

0 6 12 18 24

3,800 3,780 3,750 3,700 3,620

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SOLUTION Water influx constant

B = 1.119 φ ce ro2 h f



B = 46.16 bbl/psi

Dimensionless time

td = 6.323x10-3 φ µ c r 2 (16) e o

(24)

k t

Dimensionless influx rate Q(t); => is extracted from the table for infinite aquifers

Σ

We = B*( ∆P*Q(t))

Water influx

(30)

1 Water influx at the end of 6 months B = 46.15875 bbl/psi t = 182.5 days tD = 110.15 => Q(t) = Period

1

Time (t) (Duration) Months Days 6

182.5

tD

110.1

46.63 (By interpolation) Pressure Effective ∆p Q(t) (psi) Pressure Drop (From Table (psi) Infinite Aquifers) 3,780

10 *

∆P*Q(t) Water Influx (psi) We (bbl)

46.63

466

Σ(∆P*Q(t)) =

466

21,522

* Where:



∆P1 =

1 ( Pi − P1 ) 2

2 Water influx at the end of 12 months 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 effective for one year, but the second pressure 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)) will need to be used for each pressure drop.

46

Water Influx S E V E N T E E N

Calculating Water Influx for a year: Period

Time (t) (Duration) Months Days

1 2

12 6

365.0 182.5

tD

220.3 110.1

Pressure Effective ∆p Q(t) (psi) Pressure Drop (From Table (psi) Infinite Aquifers) 3,780 3,750

10 25*

82.06 46.63 Σ(∆P*Q(t)) =

∆P*Q(t) Water Influx (psi) We (bbl) 821 1,166 1,986

91,681

* Where:



∆P2 =

1 1 ( Pi − P1 ) + ( P1 − P2 ) 2 2

∆P2 =

1 ( Pi − P2 ) 2

3 Water influx at the end of 18 months 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) Months Days

1 2 3

18 12 6

547.5 365.0 182.5

tD

330.4 220.3 110.1

Pressure Effective ∆p Q(t) (psi) Pressure Drop (From Table (psi) Infinite Aquifers) 3,780 3,750 3,700

10 25 40*

114.87 82.06 46.63 Σ(∆P*Q(t)) =

∆P*Q(t) Water Influx (psi) We (bbl) 1,149 2,051 1,865 5,065

233,800

* Where:



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∆P3 =

1 1 ( P1 − P2 ) + ( P2 − P3 ) 2 2

∆P3 =

1 ( P1 − P3 ) 2

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4 Water influx at the end of 24 months 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 second pressure drop, ∆P2, has been effectinve for eighteen months, the third pressure 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) Months Days

1 2 3 4

24 18 12 6

730.0 547.5 365.0 182.5

tD

440.6 330.4 220.3 110.1

Pressure Effective ∆p Q(t) (psi) Pressure Drop (From Table (psi) Infinite Aquifers) 3,780 3,750 3,700 3,620

* Where:



48

∆P4 =

1 1 ( P2 − P3 ) + ( P3 − P4 ) 2 2

∆P4 =

1 ( P2 − P4 ) 2

10 25 40 65*

146.23 114.87 82.06 46.63 Σ(∆P*Q(t)) =

∆P*Q(t) Water Influx (psi) We (bbl) 1,462 2,872 3,282 3,031 10,647

491,452

Water Influx S E V E N T E E N

REFERENCES 1

Dake, L.P. “Fundamentals of Reservoir Engineering” Elsevier ISBN 0-44441667-6. 1978

2

Craft, B.C. & Hawkins, M.F. “Applied Petroleum Reservoir Engineering” Prentice-Hall,NY. 1959

3

Pirson, S.J. “Elements of Oil Reservoir Engineering” 2nd Edition McGraw-Hill Book Co. Inc (NY) 1958 pp608.

4

Van Everdingen, A.F. & Hurst,W, “The Application of Leplace Transformation to Flow Problems in Reservoirs”. Trans AIME (1949) Vol i86, 305.

5

Havlena, D. and Odeh,A.H. “The Material Balance as an Equation of a Straight Line” J. Pet.Tech Aug. 896-900. Tarns AIME 1963

6

Stewart, L. “Piper Field- Reservoir Engineering”. EUR 152. Paper presented at the SPE European Offfshore Petroleum Conference. London .Oct 1980

7

Fetkovitch. M.J. “A Simplified Approach to Water Influx Calculations-Finite Aquifer Systems”, J.Pet.Tech., July,814-828, 1971

8

Carter, R.D. and Tracy,G.W. “An improved Method for Calculating Water Influx”, Trans AIME. Vol 219, p 415-417., 1960.

9

Dake, L.P. “The Practise of Reservoir Engineering” Elsevier ISBN 0-44482094-9, 1994

10 Fanchi, J.R. “Analytical Representation of the van Everdingen-Hurst Aquifer Influence Functions for Reservoir Simulation”, SPE-Reservoir Engineering, June 1985.

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Immiscible Displacement E I G H T E E N

lug Core P

Water

Saturations: Mobility: Water 1- Sor Krw'/µw Oil Sor O

Reservoir Engineering 22/07/14

Oil

Saturations: Mobility: O Water Swi Oil 1- Swi Kro'/µo

Petroleum Engineering

Immiscible Displacement E I G H T E E N

C O N T E N T S 1 INTRODUCTION 2 THE REASON FOR WATER INJECTION 2.1 Zone Isolation 2.2 Permeability 2.3 Oil Viscosity 2.4 Undersaturated Reservoirs 2.5 Overpressured Reservoirs 2.6 Reservoir Depth 2.7 Facility Design 2.8 Thermal Fracturing 2.9 Water Handling 3 BASIC WATERDRIVE THEORY 3.1 Introduction 3.2 Water-Oil displacement at Microscopic and Macroscopic levels 3.3 Relative Permeability 3.4 Fractional Flow

6 COPING WITH HETEREOGENEITY 6.1 Introduction 6.2 Vertical Heterogeneity 6.3 Areal Hetereogeneity 6.4 Vertical Sweep Displacement Calculations for Layered Reservoirs 6.5 Ordering of the Layers 6.6 Impact of Capillary Pressure in Homogeneous Systems 6.7 Impact of Permeability Distribution on Waterflooding 7 APPLICATION TO FIELD PERFORMANCE 8 IMMISCIBLE DISPLACEMENT IN GAS DRIVE SYSTEMS 8.1 Mobility Ratio for Gas Oil Systems. 8.2 Gravity Segregation 8.4 Other parameters

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. 5 TWO DIMENSIONAL BEHAVIOURSEGREGATED FLOW 5.1 Introduction

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LEARNING OBJECTIVES Having worked through this chapter the Student will be able to: • • • • • • • •

• • • • • • • • • • •

2

Describe briefly the various reasons for water injection. Present a simple equation for the fractional flow of water in terms of water and oil flow rate. (equation 9) Comment briefly on the impact of angle of dip, capillary pressure, and velocity on the fractional flow. Plot a set of relative permeabilties and identify end-point relative permeabilities. (Figure 11) Be able to define mobility ratio and present an equation for it and calculate its value given relative permeability and viscosity data. Generate a fractional flow curve given relative permeability and viscosity data for injected and displaced fluids. Understand the significance of the Buckley-Leverett Frontal Advance equation. ( not expected to derive the equation ) Show the shape of the fractional flow curve and its associated derivative curve and the progressive saturation displacement profile for the following three types of displacement; • Water displacing viscous oil • Water displacing very light oil. • Water displacing medium denisty oil • Determine the breakthrough fractional flow, saturation and time for a diffuse flow displacement process given the necessary equations Calculate the oil recovery at breakthrough for a water displacing oil process. Calculate the oil recovery after breakthrough, using an equation or the Welge graphical construction. Comment on the impact of oil viscosity on the fractional flow curve and the displacement process. Show how the relative permeability curves for segregated flow are two straight lines between end point relative permeabilities and saturation values. Generate a set of pseudo relative permeability values for segregated flow and use them to generate a fractional flow curve for such flow conditions. Calculate breakthrough and subsequent fractional flow, saturation and recovery for segregated flow conditions. Present an equation expressing the recovery in terms of vertical and areal recovery efficiency. ( eqn. 77 ) Generate pseudo relative permeabilties and subsequent fractional flows for layered reservoir systems for both layers where cross flow exists and where no cross flow occurs. Sketch the water flood profiles for different permeability distributions. Describe how a reservoir fractional flow curve can be generated and used to justify increased water injection to improve oil recovery. Be aware of the application of immiscible displacement theory in gas oil and dry gas wet gas displacement processes.

Reservoir Engineering

Immiscible Displacement E I G H T E E N

1 INTRODUCTION In previous chapters we have examined the various fundamental properties associated 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 principal 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 operations. The modelling of water drive in reservoirs in relation to understanding the displacement behaviour and associated recovery is generally carried out using computer based numerical reservoir simulation at dimension scales considerably greater compared to the scale at which the physics of immiscible displacement takes place. In this chapter we will review and consider some of the important properties important to predicting the displacement and then examine analytical techniques which can be used to provide predictions of behaviour in immiscible displacement processes. The methods presented are not intended to displace the use of reservoir simulation but as an encouragement to use the methods to understand the contribution of the various parameters involved rather than blindly use the numerical simulation results where the large sizes of the ‘grid blocks’ inevitably disperse behaviour which occurs at pore size, centimetre or a few metres scale. 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.

2

THE REASON FOR WATER INJECTION

Water injection is the main intervention method used in reservoir development, primarily because of the associated recovery achieved and also the availability of the injection fluid. Historically it is termed a secondary recovery process in recognition of the application of using it after a reservoir has been depleted by its natural energy, 22/07/14

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and the pressure has dropped below the bubble point. The perspective here is using the injected water to displace some of the remaining oil and thereby recover more oil. Water injection can have two benefits, and for this reason it being termed as a secondary recovery process causes some confusion. Allowing a reservoir to fall below the bubble point we have seen leads to solution gas drive and resulting low recoveries. Keeping the reservoir above its saturation pressure by the injection of another fluid maintains the energy of the process providing good well productivity and more important keeps the reservoir fluid in single phase. This voidage replacement, pressure maintenance process using water injection has been common practise in major offshore oil sectors where there is plentiful supply of clean injection material. Dake1, in his text, outlines the engineering benefits of water drive in the context of the North Sea, a major area where water injection has been practised for over twenty five years. 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 environmental 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:

2.1 Zone Isolation

Although a field might be supported by an active aquifer providing natural pressure support, in some cases faulting within the reservoir structure, can result in zones being isolated from pressure support, Figure 1. If no intervention was used the zone would produce by its natural energy with a rapid loss of pressure and resulting poor recovery resulting from solution gas drive. Also as we saw in the section on water influx, in order to predict aquifer pressure support a considerable amount of aquifer characteristics are required. The cost of collecting this information prior to production are high and therefore it is not until oil production starts can the strength of any aquifer be determined. In many offshore fields therefore water injection is planned as part of the development since the associated facilities required and the implication on platform design are such that a delayed decision to implement water injection once production has started is very costly.

Zone Isolation

OIL

Sealing Fault

OIL OIL

AQUIFER AQUIFER

Figure 1 Zone isolation. 4

Immiscible Displacement E I G H T E E N

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.

2.3 Oil Viscosity

As discussed in the chapter on rock properties the displacing characteristics at pore scale are influenced by the relative mobility of the two fluids, the fluid being displaced and the fluid displacing. The ratio of the mobility of the displacing fluid to the displaced fluid is a ratio of Darcy’s law as applied to the system. The different parameters being the permeability of the one fluid in the presence of the irreducible saturation of the other, the end point relative permeability, and the viscosity of the fluids. The mobility ratio for water displacing oil is expressed as, M where:

M=

krw ′ kro′ / 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 mobility 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 viscosities are higher, then the higher velocity of the water causes an increasing instability and water fingers through the oil and breaks through 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 Impact Resistability Ratio onwill Horizontal heterogeneities and the on influence of gravity have aDisplacement big impact on the reservoir flooding behaviour.

Injection

Production Water

Oil

M<1 Stable Displacement

Injection

Production

M>1 Unstable Displacement

Figure 2 Impact of mobility ratio on horizontal displacement. 22/07/14

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2.4

Reservoir Engineering

Undersaturated Reservoirs

As pointed out in the material balance chapter, when a reservoir is above its bubble point, i.e. it is undersaturated, then if there is no pressure support from an aquifer the pressure declines rapidly. This pressure decline can be detected using pressure surveys in an open hole well as the dynamic behaviour of the reservoir is reflected in the various layers making up the formation. This is demonstrated in the context of the Montrose field3, where following production the pressure depth profile was determined for successive development wells through the oil zone and basal aquifer. The pressure profile did not follow the original water pressure gradient established during the evolution of the field but reflected the permeability variation and communication between the various sand layers, Figure 3. As more development wells are drilled pressure surveys continue to confirm the layering of the formation (Figure 4). This powerful application of pressure surveys to determine the communication characteristics of a reservoir enables waterfloods to be planned and simulated much more effectively. Sw% 100 0

θ% 0 50

Reservoir pressure - psig 2500 Top paleocene Perforations

Layer 1 Layer 2

True vertical subsea depth - metres

2500

3000

3500

4000 8100

Original pressure gradient

8200

Layer 3

8300

2550 8400

Layer 4

8500

2600

8600 Layer 5

2650

8700

8800 14

24 18 22 16 20 Reservoir pressure - MPa

26

Figure 3 Pressure depth profile for Montrose Field well.3

6

True vertical subsea depth - feet

Gr% 0 100

Immiscible Displacement E I G H T E E N

2.5 Overpressured Reservoirs

In those areas where reservoirs are overpressured (Chapter 2), the overpressure provides extra energy support. This additional pressure enables high production rates in the early time period and also information from the reservoir of the dynamics of the various units making up the system particularly if pressure depth surveys are carried out during this period. The overpressure provides an opportunity to ‘feel’ the reservoir during the early production period without the reservoir dropping below the bubble point and reducing oil recovery. Reservoir pressure - psig 3000 3400 3200

True vertical subsea depth - metres

A15 A11 A17 A18

2500

A6

A8

8000

Original pressure gradient

8100 8200 8300

2550

8400 8500

2600

8600

2650

symbol

2700

18

?Well number 22/17-A6 A8 A11 A15 A17 A18

20

Date 05/04/77 27/01/78 20/12/77 15/08/78 02/11/78 28/03/79

26 22 24 Reservoir pressure - MPa

8700 8800

True vertical subsea depth - feet

2450

8900 28

9000

Figure 4 Pressure depth profiles for Montrose Field wells.3

2.6 Reservoir Depth

The cost of offshore production facilities are such that it is important to maximise the functionality of each platform. If waterflooding is carried out then the water flooding wells are generally at the extremities of the formation. The well slots on the platform therefore have to be capable of reaching these limits, Figure 5. The deeper accumulations provide the drillers with an easier task to reach the outer limits of the reservoir using deviated and vertical wells. The application of horizontal wells in recent years also enables shallower accumulations to be reached.

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Sea Level Sea Bed

Crestal Producer

OIL

WATER Injector

Figure 5 Application of deviated wells from one structure to reach limits in the reservoir.

2.7 Facility Design

In planning water injection, at least two important considerations are required. The injection perspective: where should injection take place in relation to the various zones of the formation and the ability to inject in relation to formation characteristics? Secondly and equally important, the time and associated cost of handling the water when it eventually arrives at the production wells.

2.8 Thermal Fracturing

It is beyond the scope of this text to go into detail, but in recent years, as experience in large waterflood operations has been obtained, new insight is developing on how reservoirs have reacted. Of great significance is the phenomenon of thermal fracturing. In waterfloods where large injection flow rates are required large pumps are utilised which can handle the necessary capacity. In many offshore zones where the injection water is cold, the reduction in temperature around the injection well reduces the natural fracture gradient and the pumps capable of overcoming the resistance of flow through the formation generate a pressure greater than the fracture gradient causing the formation to successively fracture. This generates a high surface area for flow and therefore injectivities have been maintained compared to those expected from predictions using simple radial flow around a well. This thermal fracturing phenomenon enabling good injectivities to be maintained is causing some companies to consider using forced fracturing associated with water injection where temperature gradients in warmer regions or with warm injection fluids will not reduce the natural fracture gradient. Such could be the case when reinjected produced water is being used. 8

Immiscible Displacement E I G H T E E N

Although good injectivities can be achieved due to fracturing a greater understanding of the stress sensitivity of the formation is required. The fracture will follow the natural direction according to the natural stresses and strength characteristics of the formation. A concern is that such fractures will cause the injected water to by-pass the desired flood front and cause premature water breakthrough. ( Figure 6). Thermal Fracture

Producer Produces Premature Water Breakthrough

Injector

Desired Water Injection Flood Front

Figure 6 Impact of fracture on water injection flood profile.

2.9 Water Handling

The handling of water is a major technical challenge in the oil industry particularly in offshore operations, where many operators as fields mature find themselves handling more water than oil. This technical challenge is also increasing as water disposal options in relation to reducing oil emissions become more limited. Those involved in providing associated production and treating facilities require important information from the reservoir engineer. A schematic layout of a typical offshore water injection scheme is shown in Figure 7. Some key information required is: when will water breakthrough to the producing wells, and how much water will be increasingly be produced? The water handling facilities required are not insignificant and therefore good forecasts are important. A more demanding challenge to the reservoir engineer and outside the scope of this text is how can we manage the reservoir to reduce water production.

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WATER INJECTION DESIGN Injection Water Treatment

Gas/Oil/Water Separation

Injection Pump

Gas-sales or re-inject

q o Sales

Treat/ qw Dump/ Inject q o + qw

qi

Sea Level

Seawater

Sea Bed

Reservoir

Figure 7 Schematic of ffshore facilities for Water Injection.

Dake2 points out an equation which links the reservoir engineer to the production engineer which is:

qwi = qoBo + qwpBw (rb/d)

(2)

where qwi = injection rate (assume Bw = 1) qo + qwp = produced fluids requiring separation qwp = produced water for disposal or reinjection He points out that this simple equation is fundamental to the process. In water injection, the injection rate, qwi is maintained constant, since this is the drive in water drive, and is therefore under engineering control. The right hand side, the fluids requiring facilities for treatment are under the control of the reservoir. Dake points out that this equation is not just a statement of material balance but it can be regarded as a ‘platform equation’ since it contains the key elements associated with topside capacities. 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.

10

Immiscible Displacement E I G H T E E N

Equations were presented by Dake2 which he used to illustrate the impact of these reservoir considerations on production capacities. The injected water in the reservoir provides two functions maintaining pressure and displacing oil. Until breakthrough, only oil is produced, after water breakthrough an increasing watercut occurs. This watercut or fractional flow is defined as:



fws =

qwp where ‘s’ denotes surface conditions. qo + qwp

(3)

Expressing the equation in terms of water production and substituting in (2) gives: and

qwp = qo

fws 1 − fws

 B f  qwi = qo  Bo + w ws  1 − fws  

(4)

(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, applied at a reservoir scale.

3

BASIC WATERDRIVE THEORY

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 review 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.

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As in many reservoir engineering processes we are combining properties, measurements and application over a huge range of physical scales. Such an example of this is in immiscible displacement calculations in oilfield oil recovery predictions. It is important to keep this relative scale perspective in mind so as not to make an unrealistic "jump" in application of data beyond its significance. In water-oil displacement considerations we are dealing with a process which takes place at a range of scales. At pore level or the microscopic scale, where the isolation and movement of the respective phases is dependant on fundamental properties such as interfacial tension, wettability, viscosity, pore size and shape to name the obvious. At a significant larger scale, the macroscopic scale, we measure behaviour and generate properties at the laboratory level where fluid movement and displacement are examined at core plug scale, such as permeability, relative permeability and capillary pressure. The field scale,or behavioural scale, where the impact of characteristics at another quatum leap level of scale will impose behaviour on those measured at microscopic and macroscopic scale. For example the heterogeneous characteristics of the various layers of the formation giving rise to different mobilties within the layers and the large thicknesses of the layers resulting in vertical segregation perspectives. An illustration of these different perspectives is shown in Figure 8, where the oil water displacement process is illustrated at 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 compared to the engineering of an oil refining plant. In the later, the process takes place in vessels and pipes of centimetres 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 Geoscience and Reservoir Simulation modules.

3.2 Water-Oil displacement at Microscopic and Macroscopic levels

Figure 8(a) illustrates the remaining oil at microscopic level following displacement by water. This remaining or residual oil is held by the competition between the interfacial tension forces and the viscous flow forces associated with fluid flow. This topic was covered briefly in the rock chapter, when the pore doublet model was presented, explaining how the continuous phase of oil is broken leaving oil ganglia held by capillary forces. The residual oil saturation, Sor, in the water swept rock can be in a range of 10-40% of the pore space (Figure 8(a)). At the field displacement level the nature of the reservoir formation and well locations causes some of the rock to be unswept by the water. This leads to two residual oil saturations, in the swept portions oil at residual oil saturation and in the unswept portions oil at original oil saturation.

12

Reservoir Engineering

Immiscible Displacement E I G H T E E N

Swept Zone

(a)

Water Oil Rock Grains

(b)

Water Injection Wells

Oil at Residual Oil Saturation By-Passed Oil

Oil at Original Oil Saturation

Oil Producers

Figure 8 (a) microscopic displacement (b) Residual oil remaining after a water flood.

This microscopic behaviour illustrates its effects in macroscopic properties of relative permeability, and capillary pressure curves.

3.3 Relative Permeability

Permeability is a macroscopic property of the rock describing its resistance to flow in terms of fluid velocity, fluid viscosity, pore size and shape, and pressure gradient. This flow resistance term comes from Darcy’s law:

u=

Q _ k ∆p = A µ ∆l



(6)

in relation to Figure 9.

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P Superficial Fluid Velocity

A

Q

distance

Figure 9 Darcys' law for permeability

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. For example the relative permeability of water, krw is expressed as follows;

krw =

kew 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



14

qo = −

kkro A  ∂po  + ρo gSinθ     ∂x µo

qw = −

kkrw A  ∂pw  + ρw gSinθ    µ w  ∂x

(8)

Immiscible Displacement E I G H T E E N

qt (Cross section view) qi

h

qo qw y x

z

ø X

w

O

X

(Plan view)

Production

Injection

O

X

O

Figure 10 Configuration of water injection in a reservior

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 permability.

k'rw

kr

0

Swc

Irreducible Water Saturation

Sw

Water Saturation

1-Sor

1.0

Residual Oil Saturation

Figure 11 Relative permeability curves for an oil-water system 22/07/14

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Identified on the curves are the two conditions at the limiting saturation of the respective 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 oil 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 with 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 engineering 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 saturations 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, 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 ;

M=

krw ′ kro′ / 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.

3.4 Fractional Flow

Considering flow in a core plug or a reservoir, the ratio of the flow of water at any point is termed the fractional flow, fw ,where:



fw =

qw qw + qo

(9)

The oil rate qo can also be expressed as;

16

qo = qt - qw

(10)

Immiscible Displacement E I G H T E E N

If the Darcy equations for water and oil are subtracted and rearranged (using field units P in atmos.) the equations become;



where;

 µ µ  qµ ∆ρg sin θ   ∂P qw  w + o  = t o + A c −   ∂x 1.0133 × 10 6   kkrw kkro  kkro

(11)

∂Pc ∂po ∂pw = − ∂x ∂x ∂x which is the capillary pressure variation in the direction of flow and, Dr = rw - ro is the density difference of water and oil. If values for flow rates using Darcy’s Law are now substituted in fraction flow equation (equation 9) it becomes;



fw =

1+

kkro A  ∂Pc ∆ρg sin θ  −   qt µo  ∂x 1.0133 × 10 6  k µ 1 + ro w krw µo

(12)

Dake2 has also simplified this equation as;

fw =



1− G µ k 1 + w ro µo krw where G is a positive gravity number;

G = 4.886 × 10 −4

kkro A ∆ γ sin θ in field units. qt µ o

(13)

(14)

where ∆g is the specific gravity difference relative to water. The above term not only considers gravity effects but also includes a velocity term v, which is qt/A The impact of the various components of this equation is worthy of consideration. The angle of dip. If water is being injected up dip then 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.

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The impact of capillary pressure, is illustrated from the slope of the capillary pressure and saturation with distance curves, Figure 12 since;

1-Sor Pc - dPc

Sw + dSw

- dSw

Swf + dX

Swc Swc

Sw

1-Sor

X

Figure 12 Capillary pressure curve and saturation distribution as a function of distance



∂Pc dPc ∂Sw = ⋅ ∂x dSw ∂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;



18

fw =

1 µw k ro 1+ µo k rw

(15)

Immiscible Displacement E I G H T E E N

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. 1.0

ƒw

Swc

Sw

1-Sor

Figure 13 Fractional flow curve

EXERCISE 1 Plot the water-oil relative permeability from the following data set. Indicate the end point relative permeabilities. Sw 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80

Krw 0.0000 0.0021 0.0095 0.0210 0.0347 0.0536 0.0788 0.1050 0.1386 0.1785 0.2184 0.2636 0.3150

Kro 0.8800 0.6710 0.5170 0.4070 0.3135 0.2420 0.1793 0.1320 0.0891 0.0550 0.0297 0.0110 0.0000

Table E1

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Capillary Pressure Height Above Free Water Level

An interesting presentation was given by Mayer-Gurr4 as illustrated in Figure 14 where the capillary pressure, relative permeability and fractional flow curves are presented. The impact of various well locations are considered. Oil & Connate Water

C

0

50

100

% Relative Permeability

A

Transition Zone B

100

Free Water Level

100% Water

k'rw k'ro

50

kro 0

krw 50

100

50

100

% Water Cut

100

50

0

% Water Saturation

Figure 14 The relationship between capillary pressure, relative permeability and fractional flow in a reservoir4

The capillary pressure curve 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 20

Immiscible Displacement E I G H T E E N

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. EXERCISE 2 Water is to be injected into a horizontal core, with the relative permeability characteristics of table E1, to displace oil. Determine the mobility ratios, and the fractional flow curves for the following three cases. Case

Oil Viscosity "µo" (cp)

Water Viscosity "µw" (cp)

1 2

35.0 4.5

0.5 0.5

3

4

0.4

1.0

DISPLACEMENT THEORIES

4.1 Introduction

To model the displacement process a number of theories have been successfully applied. 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. This will occur if the following reservoir flows exists; qt=qo+qw=qi where qt = the total flow rate in reservoir volumes/time. qo = the oil flow rate in reservoir volumes/time. qw = the water flow rate in reservoir volumes/time. qi = the water injection flow rate in reservoir volumes/time. 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 22/07/14

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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.

4.2 Buckley- Leverett Theory

The theory that has established itself in reservoir engineering for displacement calculations is that by Buckley and Leverett in 19425. Their theory is for linear, immiscible, one dimension displacement, in which the total flow rate is constant in every cross section, (incompressible). The theory determines the velocity of a plane of constant water saturation moving through a linear system, such as a core in a water flood test, Figure 15. The theory is well founded on the conservation of mass principle. dx Water

qw rw x + dx

q w rw x

A

Water + Oil

ø Porosity

L

Figure 15 Mass flow through a linear core.

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.



q wρ w

x

qw ρw

x + dx

= Aφdx

or

qw ρw x - ( qw ρw x+

∂ qw ρw d x ) = A φ dx ∂ ( ρw Sw ) ( ∂x ∂t

(



∂ ( ρ w Sw ) ∂t

(16)





(16b)

This becomes

22

∂ ∂ qw ρw ) = − Aφ ( ρw Sw ) ( ∂x ∂t

(17)

Immiscible Displacement E I G H T E E N

Since we are assuming incompressible flow, ρw is a constant. Therefore;

∂qw ∂S = − Aφ w ∂x t ∂t x

(18)

The differential of water saturation is



dSw =

∂Sw ∂S dx + w dt ∂x t ∂t x

(19)



We are examining the advancement of a particular saturation value. Since Sw is constant dSw=0. Then Also



∂Sw ∂S dx =− w ∂t x ∂x t dt sw

(20)

 ∂q ∂S  ∂qw =  w ⋅ w ∂x t  ∂Sw ∂x  t

(21)

Inserting equations 20 and 21 in equation 18 gives;

∂qw dx = Aφ dt Sw ∂Sw t

(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:



v Sw =

dx q ∂fw = t dt S w Aφ ∂Sw

Sw



(23)

where vSw is the velocity of the plane of saturation, Sw. This is the Buckley-Leverett equation, and is also the equation of characteristics. It indicates the velocity of a plane of saturation moving through the linear system. It enables the calculation of Sw as a function of time and distance and indicates its dependance on the derivative of the fractional flow curve.

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Chierici6 has presented a very thorough analysis of the displacement process for three fractional flow curves. In understanding the use of the equation it is important to appreciate the initial boundary conditions, for our injection process. These are; Sw = Swi for 0 < x ≤ L,t = 0 Sw = 1 - Sor for x = 0, t ≥ 0

(24)

That is the system is at its initial connate water saturation If the initial conditions at t=0 are applied to the general equation;



vSw =

dx q ∂fw = 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.



[ x(S )] w

t

= x0 ( Sw ) +

qt  dfw    t φA  dSw  sw

(25)

This equation describes a series of straight lines, the characteristics, with an initial ordinate value of x0(Sw) and a slope of

qt  dfw    φA  dSw  Sw

Chierici considers three cases Case 1 For viscous oils In this case the viscosity of the displaced phase, the oil, is considerably greater than the injected water phase. The fractional flow curve has a concave downward shape, Figure 16A and its gradient fw′ increases from Sw=1-Sor to a maximum value at Sw=Swi+∆Swi. Figure 16B

24

Immiscible Displacement E I G H T E E N

x

1 A

C

Swi

fw

w Sw

0 B

0 Swi S1 S2

Sw

0

i

1 - S or

xo(S2)

1

Swi+ Sw

S2

xo(S1) Sor

dfw dSw

S1

xo(Swi)

dfw dSw

qt Αθ

t

0

Figure 16 Displacement of viscous oil by water6. A = Concave downwards fractional flow curve B = Velocity of water saturation C = Characteristics of water saturations 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 producing end. As can be seen the breakthrough saturation is just greater than Swi and explains why for a very viscous oil breakthough occurs with low water saturations and then gradually increases until the saturation reaches an unacceptable level. Water Injector 1

Sor

Sw

0

Producer

t=0

t1

t2

tBT

Swi 0

X

L

Figure 17 Progressive saturation profile for a concave downwards fractional flow curve6.

Case 2, Very Light Oils In this case when the oil is very light with a low relative viscosity and large gravitational effects for example with a highly dipping structure, and with very low velocity, a concave upward fraction flow curve is generated, Figure 18A, resulting in a fw′ curve decreasing from its value at Sw=1-Sor to a minimum value at Sw=Swi, Figure 18B. 22/07/14

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x

1

C

A fw

S wi

xo(Swi)

S1

0

S2

xo(S1)

B dfw dSw

0

Swi S1 S2

Sw

qt Αθ

dfw dSw

Swi+ Sw

1

-S or

xo(S2) Sor

w

0

1

0

t1

t2

t3 t

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 saturation. C = characteristics of water saturations Sw.

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 1

Sor

Sw

0

Producer

t=0

t2

t1

t3

tBT

Siw 0

X

L

Figure 19 Progressive saturation profile for a concave upwards fractional flow curve.6

Case 3 Typical medium density oils. Figure 20A presents the fractional flow curve for a medium density and viscosity oil, where the displacement velocities are not unlike field values. The S shaped curve generates the two curvatures we have considered in case 1 & 2. With the corresponding derivative values, fw'. The slope in the fw curve increases from its starting value, Sw=1Sor and then decreases. Figure 20B 26

Immiscible Displacement E I G H T E E N

x

1 A fw

xo(Swi) xo(S1)

0 B

dfw dSw

C S wi

S1

S wf

S2

xo(Sw,f) xo(S2)

Sor 0

Swi S1 S2

Sw

xo(1-Sor) 0 0

1

w

1 - S or

t1

t2

t3

t

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

The development of the saturation would be such that there would be a steady increase in the velocity of the increasing saturation, but this would reach a maximum at a saturation Swf, where Swi<Swf<(1-Sor). Behind this the velocities would decrease with decreasing Sw, Figure 20C. The impact on the process is such that a shock front is developed, at the value Swf, the saturations greater than this moving at a lower velocity; behind this shock front there is a steady increase in the saturations moving at decreasing velocity. This process is illustrated in Figure 21, which shows that water free oil is produced until breakthrough at a saturation of Swf,and a breakthrough fractional flow of fwbt. The saturation then climbs until it reaches the irreducible oil saturation level when only water is produced. Water Injector 1

Sor

Sw

0

Producer

t=0

t1

t2

Siw 0

t3

tBT

Sw,f X

L

Figure 21 Progressive Saturation for an S-shaped Saturation Curve6.

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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 Front at time t

Front at time t + dt

q w,L

q w,R

S w,L

dxf

S w,R

x

Figure 22 Conditions ahead (R) and behind (L) Water Front6 If the velocity of the front vf=dxf/dt, then

qw,Ldt-qw,Rdt = Af(Sw,L - Sw,R)dxfs

(26)

Since

qw= qtfw

dx f

q

(f (

w, L

− fw , R )

vf = = t dt Aφ Sw , L − Sw , R Since qt=Aut Then

vf =

)



dx f ut ( fw, L − fw, R ) = dt φ ( Sw, L − Sw, R )

(27)

(28)

This is the Rankine-Hugoniot condition for the frontal velocity of shock fronts for physical systems. If we specify our limiting conditions then Sw,L=1-Sor fw,L=1 Sw,R=Swi fw,R=0 Therefore:

28

Immiscible Displacement E I G H T E E N

(f (S

w, L



w, L

− fw, R )

− Sw , R )

=

1

1 − Sor − Swi

= tan α

(29)

Where tan α is the angle on the Sw vs. fw curve joining (Swi,0) and (1-Sor,1) Figure 23. 1

fw

0

α 0

Swi

sw

1-Sor 1

Figure 23 Calculation of Front Velocity using Rankine-Hugonist conditions6

From the Buckley-Leverett equation, equation 23

vSw1− Sor =

dx u ∂f = t w dt 1− Sor φ ∂Sw 1− S

or

Therefore from eqns. 23,28 & 29 we have

vS1− S = or

dx u ∂f u ( fw , L − fw , R ) ut u 1 = 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)

If we apply in a similar fashion the Rankine-Hugoniot condition to case 3 for the shock front we have for our limiting conditions : Sw,L=Swf fw,L= fw(Swf) Sw,R=Swi fw,R=0 In equation 30, this gives us: 22/07/14

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vSwf =

dx u df = t w dt wf φ dSw

= wf

Reservoir Engineering

fwf ut ( fw , L − fw , R ) ut u = = t tan β (31) φ ( Sw, L − Sw , R ) φ Swf − Swi φ

Examination of these equations generates a convenient graphical procedure to determine the conditions at the shock front.

tan β =

dfw dSw

= wf

fwf Swf − Swi

(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. 1

fw(Swf)

tan β =

dfw dSw Sw,f

fw

0

β Swi

Swf

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 linear system we generate a useful equation which we will use later in water injection performance calculations.

tbt =

L Lφ Swf − Swi = vf ut fw S wf

where L is the distance from injector to producer.

30

(33)

Immiscible Displacement E I G H T E E N

4.3 Welge Analysis

In 1952 Welge(7) presented a method to obtain the average saturation behind the shock front, which is useful in determining the oil recovery. Figure 25 gives the saturation profile as the shock front breaks through at the producing end, a distance L from the injection end. Water is being injected at a rate qw Time for breakthrough tbt



tbt =

L ALφ Swf − Swi = Vf qw fw

1-Sor

Sw

Sw

Swf

Swi

X1

X

X2

Figure 25 Water Saturation as a Function of Distance before Breakthrough.

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))

N pbt = qw tbt = ALφ Therefore:

Sw − Swi =

Swf − Swi = ALφ ( Sw − Swi ) fwf

Swf − Swi fwf

(34) (35)

from equation 32 this can be written as:

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Sw − Swi =

1

 dfw     dSw  Swf

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(36)

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Combining equations 35 and 36 gives

1 − fwf = 1 −

Swf − Swi Sw − Swf  df  = = Swbt − Swf  w  Sw − Swi Sw − Swi  dSw  Swf

[

]

(37)

rearranging this equation becomes:

Swbt = Swf +

1 − fwf  dfw     dSw  Swf



(38)

This is Welge's equation for average saturation and combining with equation 36 gives:



dfw dSw

= S wf

(1 − f ) = w S wf

Sw − Swf

1 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 .. The construction is illustrated in the Figure 26 Sw

1.0 fwf

Swf, f fw

Swi

Swf

wf

1-Sor

Figure 26 Determination of Average Saturation Behind Front

32

Immiscible Displacement E I G H T E E N

EXERCISE 3 Fractional Flow – Diffuse Flow Conditions Oil is being displaced by water in a horizontal, direct line drive under the diffuse flow condition. The rock relative permeability functions for water and oil are listed in table E.1. Pressure is being maintained at its initial value for which Bo = 1.36 bbl/STB and Bw = 1.01 bbl/STB. Data Summary Oil formation volume factor Water formation volume factor Initial water saturation

Bo = 1.36 bbl/STB Bw = 1.01 bbl/STB Swc = 0.20

Compare the values of the producing water cut ( at surface conditions) and the cumulative oil recovery at breakthrough for the following combinations. Case

Oil Viscosity "µo" (cp)

Water Viscosity "µw" (cp)

1 2 3

35.0 4.5 0.4

0.5 0.5 1.0

4.4 Calculations for Oil Recovery

The objective of Welge’s work was to enable oil recoveries to be determined. We will now develop the equations and methods to calculate oil recovery over the displacement period both before water breakthrough and subsequent to it. We should not forget at this stage we are discussing the displacement process and associated recovery in a relatively small core plug, of oil by water. In analysing recoveries it is convenient to express the volumes of fluids injected and recovered in terms of pore volumes, (PV). For our core plug with a length L, cross section area, A, and porosity, φ, the pore volume is.

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

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qid is the water injection rate expressed in pore volumes, ie.



qid =

qi AφL

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(42)

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tbt is the time taken for the water to breakthrough, which is;

Wid bt



tbt =



Sw bt average water saturation at water breakthrough

qid

(43)



At breakthrough, when x=L in the Buckley Leverett equation (equation 23), the following convenient result exists;



xSwbt = L =

Therefore ;



Wi dfw Aφ dSw

1 Wi Wid = = dfw LAφ dSw

S wbt

(44)



(45)



S wbt

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 increases 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

1 - Sor

Sw Swe Swbt= Swf

Swbt

0

X

L

Figure 27 Water Saturation profiles after breakthrough

34

Immiscible Displacement E I G H T E E N

From Welge’s equation 38 where the saturation value is now Swe, the average saturation remaining is now



(

Sw = Swe + 1 − fw

S we

) / dSdf



w

(47)

w S we

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

(48)

id

The oil recovered associated with this average saturation is given by;

(

N pd = S w −Swi = (Swe − Swi ) + 1 − fw S

we

)W

id



(49)

Equation 47 also gives a useful construction to determine the average saturation. Since;



dfw dSw

= S we

(1 − f ) (S

w

w S we

(50)

− Swe )

As for breakthrough the average saturation is the intersection at fw =1 of the tangent drawn to the fractional flow curve at the exit fwe saturation value. Figure 28 Sw

fw= 1

Sw,bt

Sw

1-Sor

(1-fwe) Swe,fwe (Sw - Swe)

fw

dfw dSw Swbt

Swe

=

1-fwe Sw - Swe

Figure 28 Welge construction for Determination of Average Water SaturationS after breakthrough

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The procedure for the oil recovery calculations is summarised below. 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

(

)

(ii) N pd = Wid = qid tbt = Sw − Swi = 1 bt bt bt (iii) The time for breakthrough is

tbt =

3

Wid bt qid

dfw dSw



(41, 45)

S wbt

(43)



After breakthrough, select a saturation value greater than the breakthrough value

and (i) fw=fwe, Sw=Swe and extrapolation of line to fw=1 gives the average water saturation value. Swe (ii) Oil recovery at this point is;





(

N pd = Sw − Swi = ( Swe − Swi ) + 1 − fw Wid =

Wi 1 = LAφ dfwe dSwe

(iii) Time te =

Wide qide

S we

)W

id







(49) (45)

(43)

Step 3 is then repeated for increasing values of Swe up to Swe = 1-Sor. 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 breakthrough saturation.

36

Immiscible Displacement E I G H T E E N

Breakthrough Point

Np

Time

Figure 29 Oil Recovery up to and after breakthrough.

4.5 The Impact of Viscosity.

The various analyses demonstrate the impact of the fractional flow curve on the recovery before and after displacement. Earlier we demonstrated the various shapes of the curve as impacted by the relative viscosities of the fluids. (Figure 30). 1.0

µo = 100 µw A

fw

µo =1 µw

B

C

Swi

Sw

µo = .01 µw

1-Sor

Figure 30 Fractional Flow Curves for different Oil/Water Viscosity Ratios.

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 22/07/14

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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.

1-Sor Sw

C A M>>1

M<1

M around 1 B

Swi

Figure 31 Water Saturation Distributions in Systems for Different Oil/Water Viscosity Ratios.

EXERCISE 4 Oil Recovery Prediction for a Waterflood – Diffuse Flow Conditions Water is being injected at a constant rate of 1,200 bbl/d/well in a direct line drive in a reservoir whose rock and fluid properties, as well as the flood pattern geometry, are listed in the data summary table below. The relative permeabilities for oil and water are the same as for the previous exercise. Data Summary Injection rate Qi Water viscosity µ w Oil viscosity µ o Initial water saturation Swc Residual oil saturation Sor Porosity φ Dip angle θ Reservoir thickness h Distance between injection wells w Distance – injectors and producers L

38

= = = = = = = = = =

1,200bbl/d/well 0.5cp 4.5cp 0.20 0.20 0.22 0 degrees 50ft 800ft 2,000ft

Immiscible Displacement E I G H T E E N

Assuming that diffuse flow conditions prevail and that the injection project starts simultaneously with oil production from the reservoir, determine: 1 the time when breakthrough occurs 2

the cumulative oil production as a function of both the cumulative water injected and the time.

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

TWO DIMENSIONAL BEHAVIOUR- SEGREGATED FLOW

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 perspectives 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.

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lug Core P

Water

Saturations: Mobility: Water 1- Sor Krw'/µw Oil Sor O

Oil

Saturations: Mobility: O Water Swi Oil 1- Swi Kro'/µo

Figure 32 Displacement under segregated flow.

In segregated flow, it is assumed therefore that ahead of the displacement interface, only oil is flowing with water present at its irreducible saturation Swc. Whereas behind the displacement front only water is flowing in the presence of irreducible oil. It can be considered there is a piston like displacement pushing oil ahead. In reality there would not be a sharp interface but a transition zone, associated with the imbibition capillary curve. In the present analysis we are assuming that this transition zone is small in thickness relative to the thickness of the displacement layer. In segregated flow using criteria presented by Coats9, vertical equilibrium, is considered to exist. In vertical equilibrium the following conditions would exist; • A large density contrast, between the displacing fluid and oil • A high vertical permeability • Low oil viscosity • Strong capillary forces • Low fluid velocity A useful indicator of the dominating forces is that in relation to the ratio of viscous forces to capillary forces:

Nvc =

uµo σCosθ

(51)

and viscous to gravity forces

Nvg =

uµo k0 g( ρw − ρo )

Where Nvc and Nvg are the capillary and gravity numbers. 40

(52)

Immiscible Displacement E I G H T E E N

For vertical equilibrium, both of these are low, and indicate that as the displacement 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. The various flow profiles are illustrated in Figure 33.

dx

a)

t2 β

-dy t 1

t0 θ t2 b)

t1 t0

Figure 33 Stable (a) and Unstable Flow (b) in Segregated Displacement.

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



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uo = ut = −

kkro′  δpo  + ρo g sin θ  µo  δx

Institute of Petroleum Engineering, Heriot-Watt University

(53)

41

Petroleum Engineering

and for the water

uw = ut = −

Reservoir Engineering

kkrw ′  δpw  + ρw g sin θ (54)  µ w  δx

where uw and uo and ut are the oil water and total velocities at the interface. Combining these yields;



 µ µ  δ ut  o − w  = − ( po − pw ) + ∆ρg sin θ δx ′   kkro′ kkrw

(55)

The condition for stable displacement is that the angle between the fluids interface remains constant. The capillary equation is dPc = d(po - pw) = Drg cos qdy

(56)

Equation 55 above becomes



 µ µ  dy ut  o − w  = ∆ρg cos θ + sin θ    dx ′   kkro′ kkrw

(57)

Expressed in terms of total flow rate the equation becomes



′   kkrw ′ A∆ρg sin θ  1 dy   kkrw  µw +1  kkro′ − 1 =  tan θ dx  µ w qt   µo  

(58)

We have previously expressed the mobility ratio and gravity number as



M=

krw kk ′ A∆ρgsin θ ′ kro′ / , and G = rw , equation 1 and 14 µ w µo qt µ w

in equation 58 above therefore

dy 1 M − 1 = G + 1  dx tan θ 

(59)

dy M − 1 − G tan θ = − tan β =    dx G

(60)

The angle of the interface of the front can therefore be expressed as;

42

Immiscible Displacement E I G H T E E N

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 qcrit =



kkrw ′ A∆ρg sin θ µw ( M − 1 )



(61)

In field units this becomes; (bbl, psi)

qcrit =



4 .9 × 10 −4 kkrw ′ A ∆ γ sin θ µ w (M − 1)

where ∆γ is the specific gravity difference.(62)

The various aspects of stability of the front can be summarised in the context of equation 60.

β

Oil

Water θ β

Oil

Water θ β

Oil

(a) M<1

(b) M=1

(c) M>1

Water θ Figure 34 Water-Oil front for 3 stabilised flow conditions.

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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 34a.

2

M = 1 The front is always stable regardless of G. From Equation 60 above tanβ/tanθ = -1 therefore β=θ. This is illustrated in Figure 34b.

3

M > 1. The front will only be stable if G > M-1. In this case β<θ. This is illustrated in 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 the conditions when M is greater than 1 gravity will stabilise flow up to a limiting flow rate given by the critical flow rate qcrit.

5.2 Displacement Calculations for Segregated Flow

Figure 35 below illustrates the conditions in our segregated flow system. Clearly when we place our core plug scale system on the diagram the associated relative permeability values come into focus. Indeed the core plug cannot be identified since its size is so small!

Oil

h

Sw=Swi So=1-Swi

1- hw hw Water

X

Sw=1-Sor So=Sor

Figure 35 Segregation Flow of Oil and Water.

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. The average saturation over this formation thickness is;

44

Sw = hw (1 − Sor ) + (1 − hw )Swc

(63)

Immiscible Displacement E I G H T E E N

Solving for hw gives

hw =

Sw − Swc 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:





krw

= hw krw

Sw

( Sw =1− Sor )

+ (1 − hw )krw

( Sw = S wc )

krw = 0 at Swc and krw = k rw at Sw = 1 − Sor . krw

(65)





(66)

= hw krw ′

Sw 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 )



kro = kor′ at Swc and kro = 0 at Sw = 1 − Sor .

Therefore kro

Sw

(67) (68)

= (1 − hw )kro′

k′ro is the relative permeability to oil in the presence of irreducible water, the end point relative permeability to oil. If hw is now replaced by the expression



hw =

Sw − Swc 1 − Sor − Swc



(69)

we have



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 S − Swc  krw S =  w ′  krw w  1 − Sor − Swc 

Institute of Petroleum Engineering, Heriot-Watt University

(70)

45

Petroleum Engineering

Reservoir Engineering

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′ro k′rw

kro Sw

Swi

Sw

krw Sw

1-Sor

Figure 36 Linear thickness, averaged relative permeabilities for segregated flow.

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. The procedure in this case is to; (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. (iii) Apply the one dimensional calculations to determine breakthrough and recovery. Dake1 amplifies further the topic with respect to recovery calculations for unstable flow in a horizontal situation and generates the following relations.

46

Immiscible Displacement E I G H T E E N

Where volumes are expressed as moveable pore volumes MOV where MOV = PV (pore volume) x (1 - Sor - Swc).

N pD =

1 2 wiD M −WiD − 1 M −1

(

)



(72)

At water breakthrough, the pore volume recovered is:

N pDbt =

1 M

(73)

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 max = M

(74)

For the case of unstable displacement in a dipping reservoir where G<M-1, then



N pDbt =

1 M −G

(75)

and the maximum recovery of N pDbt=1 is when

Wid max =

M G +1

(76)



EXERCISE 5 Displacement Under Segregated Flow Conditions 1) Re-work example 4 for the same water injection rate (1,200 bbl/d/well) and using precisely the same reservoir and fluid property data, but assuming that displacement takes place under segregated flow conditions. Compare the results with those obtained in exercise 4. Additional Data Absolute permeability Specific gravity of water in the reservoir Specific gravity of oil in the reservoir

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K = γw = γo =

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500 1.02 0.83

mD

47

Petroleum Engineering

6

Reservoir Engineering

COPING WITH HETEREOGENEITY

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 influence on the recovery of fluids in an injection/ displacement process. The total recovery Np of the fluids in place N, is a combination of the fraction recovered in the respective dimensions. ie.



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 saturation at the irreducible saturation level and that unswept is at the original saturation level. Figure 37. Oil at Original Water Saturation, Swc Producer

Producer

Injector

Producer

Producer

Aerial Sweep Efficiency

Vertical Sweep Efficiency

Residual Oil

Figure 37 Aerial and vertical sweep. 48

Immiscible Displacement E I G H T E E N

6.2 Vertical Heterogeneity

The process of generation of a reservoir rock system clearly generates variations of rock properties as a result of various depositional processes. The major impact is on the variation of permeability, which can vary considerably over a short vertical distance. This permeability variation has a considerable impact on water drive performance and therefore it is vital in the exploration phase or early in the life of a development that this hetereogeneity perspective is well characterised. A layer in the formation of very high permeability compared to the rest of the formation can have a very serious impact on the oil recovery process. In the early development of the North Sea where sea water injection was used for pressure maintenance, early breakthrough of water came as a surprise as relatively thin but very high permeability layers had not been identified. Another important perspective in the vertical sweep behaviour is the connectivity between the layers. 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 production of a field these pressure surveys are able to distinguish the various layers in a formation and also their communication characteristics. These pressure surveys 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

Depth

Original water pressure gradient.

b)

Pressure gradient post production. Pressure

Depth

Original water pressure gradient.

Pressure gradient post production.

Figure 38 Pressure profiles pre and post production (ref Dake) (a) layers in communication (b) non communicating layers. 22/07/14

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Clearly in actual reservoirs there could be some layers in communication and others not.

6.3 Areal Hetereogeneity

The areal hetererogeneity similarly reflects the impact of a number of reservoir parameters related to flow on the sweep efficiency. Other imposed impacts also have an influence for example, well spacing and configuration. Dake2 considered that there is more uncertainty in the areal sweep perspective as against the vertical sweep behaviour. He suggests that vertical sweep calculations be done which in combination with production data of recovery, can be used to history match the areal sweep behaviour.

6.4 Vertical Sweep Displacement Calculations for Layered Reservoirs

The methods we have used for diffuse flow and segregated flow can also be applied to layered systems, both with respect to those where there is communication between the layers, cross flow, and those where there is no communication, no cross flow. 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.

6

Water Sori

Oil Swci

5 4 3 2 1

Figure 39 Layered reservoir in natural order.

(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 equations apply. When the nth layer of a total N layers has flooded with water:

50

Immiscible Displacement E I G H T E E N

(

n

Swn =



)

N

∑ hjφ j 1 − Sorj + j =1

∑ hφ S

j = n +1

N

∑h φ j

1

j

j wc j

(78)

j

Subscript j refers to the ordering of the flooding out of the layers. n

krwn



S wn

=

∑ h k k′ j =1

j

j rw j

N

∑h k j

1

and



(79)



(80)

j

N



kron

S wn

=

∑ h k k′

j= n +1 N

j

j

ro j

∑ hj k j

j= 1

(iv) These pseudo-relative (thickness averaged) permeabilities are then used to generate the fractional flow curve. (v) The problem can now be solved using the 1-D approach using the Welge procedure.

6.5 Ordering of the Layers

The ordering of the layers is where the reservoir engineer has some control. 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.

6 5 4 3 2 1 Flooding Out Sequence

Figure 40 Flooding out sequence for strong vertical equilibrium.

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The other extreme is where there is a total lack of cross flow and then the layers will flood out according to the velocity of the displacement in each layer. The velocity of each layer i is given by



vi α

ki krw ′i

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. Layers According to Velocity 4 1 6 2 3 5

Figure 41 Stiles reordering (layers reference) Figure 39.

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.

6.6 Impact of Capillary Pressure in Homogeneous Systems

So far we have neglected the impact of capillary pressure both with respect to diffuse flow where the saturation distribution is uniform over the thickness or the other extreme where in segregated flow we have an oil water contact. If capillary pressure is significant relative to the reservoir thickness, then the thickness of the associated transition zone cannot be ignored. If the reservoir is thin relative to the transition zone thickness then diffuse conditions can be considered to apply. If however the transition zone is negligible then the displacement can be considered segregated.

52

Immiscible Displacement E I G H T E E N

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 relative permeabilities. Dake1 gives a thorough account of this. It is summarised briefly below. 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) k'ro

Segregated Flow Intermediate (Finite Capillary Transition Zone)

krw

kro

k'rw

0

0

.6

Sw

1.0

B

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).

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A

1.0 .8

Reservoir Engineering

.8

Oil

.6 krw

Oil

.6

.4

kro

Water

krw

.2 0

B

1.0

.4

Water

kro

.2 0

.2

.4

.6

.8

Sw

0

1.0

0

.2

.4

.6 Sw

.8

1.0

Figure 43 Relative permeabilty curves for three layered system (a) high permeability at top (b) high permeabilty at base (Dake1).

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 breaks through. Whereas with the high permeability on the top a better breakthrough profile is obtained. 1.0 .8

High Permeability at the Top

.6 fw

High Permeability at the Base

.4 .2 0

0

.2

.4

.6 Sw

.8

1.0

Figure 44 Fractional flow curve for layered systems high permeability at the top, high permeability at the base (Dake1).

6.7 Impact of Permeability Distribution on Waterflooding

The example above illustrated the impact of the relative value and position of different permeability layers. It is interesting to note that the absolute value of permeability is not so significant as the relative contrast of the permeability values. Clearly these permeability variations are generated as a result of the depositional environment. These depositional environments have been illustrated by both Dake1 and Archer11 and the sketches from Archer11 illustrating how such extreme permeability profiles can result are shown in Figure 45&46. This topic was also covered in the chapter on 54

Immiscible Displacement E I G H T E E N

permeabilty variations. The various saturation profiles resulting from four permeablity distributions are shown in Figure 47. Levee

Eros io

n of Ba

Channel Migration

e

Fin

Weak Current

nk

Strong Current

n

rse

Sa

on

siti

po

e dD

Growth of Point Bar

al ter La face r Su

a

Co

tion ) e cre Ac e L i n m (Ti

GR

Permeability

Injection

Depth

Oil

(a) Unfavorable

Production

Water

Figure 45 Effect of unfavourable permeabilty distrubution in waterflooding (Archer11). Coarser Sediment in Shallow Turbulent Water Increasing Wave Energy

Sea Level

Fine Sediment in Deeper Quiet Water

ain

r gG

in

s rea

Inc

Si

po

e sD

u

eo

an

lt mu

e

Siz

ion

sit

GR Profile (A)

Permeability

Advance of Bar

GR Profile (B)

Injection Oil

Depth

Water Production (a) Favorable

Figure 46 Effect of favourable permeabilty distrubution in waterflooding (Archer11).

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k

Super-Homogenous

D

k

Worst Case

D

k

D

Tunnel k

D

Random

Figure 47 Permeabilty distribution and resulting waterflooding profile.

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.

56

Immiscible Displacement E I G H T E E N

B

A 1.0

1.0

M=16.1

M=16.1

M=1

0.8

0.8

M=1 bt

0.6 fw

0.6 fw

0.4

0.4

0.2

0.2

0

0

0.2

0.4

Sw

0.6

0.8

1.0

0

0

0.2

0.4

Sw

0.6

0.8

1.0

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 mobility 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 water breakthrough then a rapid increase in water cut over a much shorter period.

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1.00

ing en rs

ning

M = 16.1 M = 16.1

Do wn

0.40

Coar se

0.60

fw

Up

Co a

Do wn

0.80

Piston-Like Recovery

se n

ing

0.20

ar Co 0.10

0.20

M=1 Mov = 0.594 HCPV 0.30

0.40

0.50

0.60

NPD (HCPV)

Figure 49 Reservoir fractional flow of water as a function of the vertical sweep efficiency11

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.

7

APPLICATION TO FIELD PERFORMANCE

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. Using the recovery equation developed earlier



(

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. Figure 50 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.

58

Immiscible Displacement E I G H T E E N

0.9 0.8

Watercut

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

0

50

100

150

200

Cumlative Production (MMstb)

250

300

350

Figure 50 Watercut development for a fractured limestone reservoir (Dake2).

1 Core Data at Sor

0.8 0.6

fwe

0.4 0.2 0.525 0 0.1

0.2

0.3

0.4

0.5

0.65 0.6

0.7

0.8

Swe (PV)

Figure 51 Reservoir fractional flowcurve.

Extrapolation of the fractional flow curve for the last few data points to the fw=1 position 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.

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Allowing the water cut to go up to the value given of 0.98 would yield an average saturation of 0.65 and a 31% increase in recovery compared to the last data point position. Such an improvement however has to be considered against the further factor of 2.5 for the time taken to achieve such recovery. However another option is to increase the injection rate if facility upgrades are possible and therefore reduce the time to achieve the recovery increase. 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 generated 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.

8

IMMISCIBLE DISPLACEMENT IN GAS DRIVE SYSTEMS

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.

8.1 Mobility Ratio for Gas Oil Systems.

The gas to oil ratio for gas displacing oil is very unfavourable because of the large viscosity variation between the phases. For example



M=

krg′ kro′ 0.5 1 = / / = 20 µ g µo 0.025 1

(82)

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. The fractional flow equation for horizontal gas drive is:



60

fgh =

1 µ k 1 + g ro µo krg



(83)

Immiscible Displacement E I G H T E E N

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.

1.0

k'ro

k'rg

kr

0 Sg

1-Sor-Swc 1.0

fg

0

1-Sor-Swc Sg

1.0

Figure 52 Laboratory-measured, one- dimensional gas-oil relative permeabilities and fractured flow (Dake2).

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.

8.2 Gravity Segregation

This is the dominant perspective in gas drive and is illustrated in Figure 53. Figure 53a shows the displacement front with vertical gas drive where a stable interface is shown. This would be the characteristic of gas cap drive behaviour. In this gas injection perspective the vertical velocities are low. However in Figure 53 b where flow is along the bedding of the reservoir then the unfavourable mobility ratio together with the large gravity impact can cause the gas to override the oil. Its control therefore is difficult

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Gas Injection a) Gas Oil

Oil Production Gas Injection b) Oil

Gas

Figure 53 Vertical and parallel gas injection.

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 permeabilities 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



Sgn =

j

j =l

j

or j

N

∑h φ j =l

j

− Swc j

)



(85)

j

n



krgn =

∑ h k k′ j =l

j

j rg j

N

∑h k j =l

j



(86)

j

n



kron =

∑ h k k′

j j ro j j = n +l N

∑ hj k j j =l

62



(87)

Immiscible Displacement E I G H T E E N

These show the thickness averaged gas saturation, and corresponding pseudo-relative permeability after the nth layer out of a total of N layers has flooded with gas. From these equations the fractional flow curve is generated and it is important to note in this case the gravity impact is involved since

fg =

1− G µ k 1 + g ro µo krg

(88)



where G is a positive gravity number which in field units is



G = 2.743 × 10 −3

kkro A∆ρ sin θ vµo

(89)

∆ρ is the density difference between oil and gas at reservoir conditions. v is the average Darcy velocity,q/A of the gas injection front and whereas the actual velocity v' is equal to



v′ =

q 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. The following equations are used in the analysis



N pd =

(

)

Sg′ + 1 − fg GiD 1 − Swc



(91)

where S'g and fg are the averaged saturations and fractional flows determined on the fractional flow curve after breakthrough. Npd is the oil volume expressed in hydrocarbon pore volumes. GiD is the gas injected in pore volumes and is the reciprocal of the slope of the fractional flow curve.

8.3 Impact of heterogeneity

The depositional aspects of the formation can influence the fractional flow curve and therefore the flood front profile. Whereas with water displacing oil the coarsening downwards of the rocks gives high early breakthrough for water, with gas injection this improves the flood profile as the higher permeability in lower layers counteracts the density difference perspective.

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8.4 Other parameters

The stability of the flood front is influenced by a number of formation parameters including, k,q,µo,∆ρ and also the injection rate. Clearly the only parameter the engineer has control over is the rate. Figure 54 from Dake2 shows the impact of varying the velocity. Plot A is for the case where gravity is ignored. Case B shows the impact of reducing the velocity to 1 ft/day and curve C to 0.25ft/day. With curve A a very unfavourable displacement results whereas with B, the curve has an inflection point and therefore a breakthrough point. Curve C gives an even better curve with a higher breakthrough condition.

A fg

bt

A: Horizontal (fg = fgh) B: v = 1 ft/d C: v = 0.25 ft/d

B Sg = 0

C

Sg

1-Sor-Swc

Figure 54 Influence of the velocity of gas flood on the stability of frontal advance (Dake2).

In analysing gas produced in a gas injection process consideration has to be taken of the solution gas produced which is not part of the reservoir gas produced in these displacement calculations. Dake2 develops the equation for the GOR in terms of the fractional flow value and the respective formation volume factors.

GOR =

5.615

Bo Bg

1  − 1 f   g 

+ Rs

scf/stb

(92)

where Rs is the solution GOR at the flooding pressure. fg is the pseudo fractional flow for gas in consideration of the gravity term for inclined systems

64

Immiscible Displacement E I G H T E E N

where

fg =

1− G µk 1 + g ro µo krg

(93)

Dake2 provides a field example of the application of these equations for immiscible displacement in a heterogeneous reservoir.

8.5 Gas Cycling

In chapter 4 on phase behaviour we discussed the characteristics of retrograde gas condensate reservoirs, which if allowed to fall below the saturation pressure, the dew point, liquid condensate separates in the formation and is considered to be stranded in the formation. Maintaining pressure above the dew point, prevents this retrograde condensation behaviour. Gas cycling can be a considered option for reservoirs with a high condensate to gas ratio, where the separated gas is reinjected into the formation, and which with imported gas, to replace the gas equivalent of exported condensates, maintains the pressure. In partial gas cycling the pressure declines as a result of not supplementing the injection for the sold condensates and also where gas is exported during high demand periods. Figure 55 illustrates the process. Condensates to Export

Sea Level Gas Imported from another Field

Sea Bed

Wet Gas Production

Dry Gas Injection Dry Gas

Wet Gas

Figure 55 Gas cycling.

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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 breaks through into the wells and then to convert the reservoir into a gas producing system when it would be hoped gas export facilities were in place. Clearly dry gas displacing wet gas is a miscible process with interaction of components at the dry gas/wet gas interface. A unfavourable displacement would give early dry gas breakthrough and cesation of the project since the activity of circulating dry gas would not be very profitable! The common approach to modelling these processes is to use full compositional reservoir simulation. Coats12 has indicated that whereas for partial gas cycling compositional simulation is required more simple analysis can be used for depletion and full gas cycling. Dake2 in his second text goes through an example using the Buckley Leverett displacement approach to calculate the recovery characteristics for a full gas cycling process. For a miscible dry gas displacing wet gas scenario, the core relative permeabilties are simple linear functions since no residual wet gas saturation remains after contact with dry gas. Figure 56 1.0

Dry Gas

Wet Gas

Miscible Front

k'rgw

k'rgd

1.0

kr

Sgd

1-Swc

Figure 56 Dry gas recycling flooding experiment and resulting one-dimensional relative permeabilities (Dake2).

The other important parameter in the development of the fractional flow is the mobility ratio, which using example fluid data gives:

M=

krgd k′ ′ 1 1 / 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 66

Immiscible Displacement E I G H T E E N

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. Production Injection

Permeability

Thickness

a)

Dry

Wet

Production Injection

Permeability

b)

Wet

Thickness

Dry

Figure 57 Influence of permeability distribution of the efficiency of gas recycling under the VE condition (Dake2).

SOLUTIONS TO EXERCISES EXERCISE 1 Plot the water-oil relative permeability from the following data set. Indicate the end point relative permeabilities. Sw 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80

Krw 0.0000 0.0021 0.0095 0.0210 0.0347 0.0536 0.0788 0.1050 0.1386 0.1785 0.2184 0.2636 0.3150

Kro 0.8800 0.6710 0.5170 0.4070 0.3135 0.2420 0.1793 0.1320 0.0891 0.0550 0.0297 0.0110 0.0000

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SOLUTION

Relative Permeability

Plot 1.00 0.90 0.80 0.70 0.60 0.50 0.40 0.30 0.20 0.10 0.00 0.00

Krw Kro

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

Water Saturation "Sw"

Figure E1

End point relative permeabilities: k'ro = 0.88 @ Sw = 0.2 k'rw = 0.3150 @ Sor = 0.2 EXERCISE 2 Water is to be injected into a horizontal core, with the relative permeability characteristics of table 1, to displace oil. Determine the mobility ratios M, and the fractional flow curves for the following three cases. Case

Oil Viscosity "µo" (cp)

Water Viscosity "µw" (cp)

1 2 3

35.0 4.5 0.4

0.5 0.5 1.0

Table E2

SOLUTION

Mobility ratio =

k ʹrw /µ w kʹro /µo

fw for horizontal flow = fw =

68

(1)

1 µ w kro 1+ krw µo

(15)

Immiscible Displacement E I G H T E E N

Case

µw/µo

1 2 3

M

0.01 0.11 2.50

25.06 3.22 0.14

Table E3

The Fractional flow in the reservoir for the three cases can be calculated as follows:

Sw

0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80

Krw

0.000 0.002 0.010 0.021 0.035 0.054 0.079 0.105 0.139 0.179 0.218 0.264 0.315

Kro

0.880 0.671 0.517 0.407 0.314 0.242 0.179 0.132 0.089 0.055 0.030 0.011 0.000

Fractional Flow (fw) µw/µo (1) µw/µo (2) µw/µo (3) 0.01 0.11 2.50

Kro/Krw

Infinite 319.5238 54.4211 19.3810 9.0346 4.5149 2.2754 1.2571 0.6429 0.3081 0.1360 0.0417 0.0000

0.0000 0.1797 0.5626 0.7832 0.8857 0.9394 0.9685 0.9824 0.9909 0.9956 0.9981 0.9994 1.0000

0.0000 0.0274 0.1419 0.3171 0.4990 0.6659 0.7982 0.8774 0.9333 0.9669 0.9851 0.9954 1.0000

0.0000 0.0013 0.0073 0.0202 0.0424 0.0814 0.1495 0.2414 0.3836 0.5649 0.7463 0.9055 1.0000

Table E4

1.00

fw (rbbl/rbbl)

0.90

M = 25

0.80 0.70 0.60

M = 3.22

0.50 0.40 0.20 0.10 0.00 0.00

Case 1 Case 2 Case 3

M = 00.4

0.30

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

Sw Figure E2

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EXERCISE 3

Fractional Flow – Diffuse Flow Conditions

Oil is being displaced by water in a horizontal, direct line drive under the diffuse flow condition. The rock relative permeability functions for water and oil are listed in Table E.1. Pressure is being maintained at its initial value for which Bo = 1.36 bbl/ STB and Bw = 1.01 bbl/STB. Data Summary Oil formation volume factor Water formation volume factor Initial water saturation

Bo = 1.36 bbl/STB Bw = 1.01 bbl/STB Swc = 0.20

Compare the values of the producing watercut (at surface conditions) and the cumulative oil recovery at breakthrough for the following combinations. Case

Oil Viscosity "µo" (cp)

Water Viscosity "µw" (cp)

1 2 3

35.0 4.5 0.4

0.5 0.5 1.0

SOLUTION For horizontal flow the fractional flow in the reservoir is:

fw =

1 µ w kro 1+ krw µo

While the producing watercut at surface, fws, is:

fws =

qw

qw

Bw qo Bw + Bo

Combining the above two equations leads to an expression for the surface watercut as:

fws =

70

1  B  1 1 + w  − 1 Bo  fw 

Immiscible Displacement E I G H T E E N

Viscosity ratio µw/µo : Case 1 2 3

Case 1 2 3

µw/µo 0.01 0.11 2.50

Mobility Ratio 25.06 3.22 0.14

Fraction flow for case 1,2,3 calculated using fractional flow equation above and tabulated in table E5. Oil recovery, Npd - pore volumes Npd = Swavg - Swi Equation 41 in text. Sw

Krw

Kro

Kro/Krw

Fractional Flow (fw) µw/µo (1) µw/µo (2) 0.01 0.11

µw/µo (3) 2.50

0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80

0.000 0.002 0.010 0.021 0.035 0.054 0.079 0.105 0.139 0.179 0.218 0.264 0.315

0.880 0.671 0.517 0.407 0.314 0.242 0.179 0.132 0.089 0.055 0.030 0.011 0.000

Infinite 319.5238 54.4211 19.3810 9.0346 4.5149 2.2754 1.2571 0.6429 0.3081 0.1360 0.0417 0.0000

0.0000 0.1797 0.5626 0.7832 0.8857 0.9394 0.9685 0.9824 0.9909 0.9956 0.9981 0.9994 1.0000

0.0000 0.0013 0.0073 0.0202 0.0424 0.0814 0.1495 0.2414 0.3836 0.5649 0.7463 0.9055 1.0000

0.0000 0.0274 0.1419 0.3171 0.4990 0.6659 0.7982 0.8774 0.9333 0.9669 0.9851 0.9954 1.0000

Table E5

1.00 0.90

fw (rbbl/rbbl)

0.80 0.70 0.60 0.50 0.40 0.30

Case 1 Case 2 Case 3

0.20 0.10 0.00 0.00

0.10

0.20

0.30

0.40

0.50 Sw

0.60

0.70

0.80

0.90

1.00

Figure E3 Fractional flow plots for different oil - water viscosity ratios.

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Fractional flow plots for three cases are shown in Figure above, and the results obtained by applying Welge's graphical technique, at breakthrough are listed in the following table: Case 1 2 3

At breakthrough Swbt fwbt Fig.1.2 (reservoir) 0.28 0.52 0.45 0.66 0.80 1.00

fwsbt (surface) 0.59 0.72 1.00

Extrapolated Swavg Fig.1.2 0.37 0.57 0.80

Oil recovery Npdbt (PV) 0.17 0.37 0.60

Table E6 Welge's graphical solution.

EXERCISE 4 Oil Recovery Prediction for a Waterflood - Diffuse Flow Conditions Water is being injected at a constant rate of 1,200 bbl/d/well in a direct line drive in a reservoir whose rock and fluid properties, as well as the flood pattern geometry, are listed in the data summary table below. The relative permeabilities for oil and water are the same as for the previous exercise. Data Summary Injection rate Qi = Water viscosity µw = Oil viscosity µo = Initial water saturation Swc = Residual oil saturation Sor = Porosity φ = Dip angle θ = Reservoir thickness h = Distance between injection wells w= Distance - injectors and producers L=

1,200 0.5 4.5 0.20 0.20 0.22 0° 50 800 2,000

bbl/d/well cp cp

ft ft ft

Assuming that diffuse flow conditions prevail and that the injection project starts simultaneously with oil production from the reservoir, determine: 1 the time when breakthrough occurs 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:

fw = 72

1 µ k 1 + w ro krw µo

Immiscible Displacement E I G H T E E N

For this case, the water / oil viscosity ratio is:

µw/µo = 0.11

Results of the fractional flow calculations for this case are summarised in table E.7 Sw

Krw

Kro

Kro/Krw

fractional flow fw

0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80

0.000 0.002 0.010 0.021 0.035 0.054 0.079 0.105 0.139 0.179 0.218 0.264 0.315

0.880 0.671 0.517 0.407 0.314 0.242 0.179 0.132 0.089 0.055 0.030 0.011 0.000

Infinite 319.5238 54.4211 19.3810 9.0346 4.5149 2.2754 1.2571 0.6429 0.3081 0.1360 0.0417 0.0000

0.000 0.027 0.142 0.317 0.499 0.666 0.798 0.877 0.933 0.967 0.985 0.995 1.000

Table E7

1.00

Fractional flow for µw/µo

Fractional Flow (fw)

0.90

= 0.11

Water breakthrough

0.80 0.70

Extrapolated 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

Sw

0.60

0.70

0.80

0.90

1.00

Figure E4

From the above Figure, breakthrough occurs when: Water saturation at breakthrough Swbt = 0.45 Fractional flow at breakthrough fwbt = 0.665 Extrapolated Savg at fw = 1 Savg = 0.572

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Oil recovery

Npdbt = Widbt = Savg − Swc



Widbt = Npdbt = 0.37 (PV) 1

Calculation of the breakthrough time (in years) For a constant water injection rate the time, in years,

t=



Widbt * PV Qi * 5.615 * 365

Where: PV is the pore volume in ft3 :

PV = h*W*L*φ



PV = 1.76E+07ft3 t= 2.66 years or 972 days 2 Cumulative oil recovery The oil recovery after breakthrough, expressed in pore volumes, can be calculated using the equation:



N pd = ( Sw − Swc ) = ( Swe − Swc ) + (1 − fwe )Wid

Where:

Wid =

74

1

dfw dSw

Swe

Reservoir Engineering

Immiscible Displacement E I G H T E E N

Allowing Swe, the water saturation at the producing end of the block, to rise in increments of 5% (for Swe >=Swbt) the corresponding values of Wid are calculated in Table 2.2. Swe

fwe

Swe

0.45 fwe 0.5

0.45

0.665 0.55

0.5

0.799 0.6

0.55

0.875 0.65

0.6

0.932 0.7

0.65

0.965 0.75

0.7

0.981 0.8

0.75

0.994

0.665 <= At breakthrough Swe 0.05fwe 0.134fwe/ Sw2.68 e Swe* 0.799 0.05 0.076 1.52 <= At breakthrough 0.875 0.05 0.05 0.134 0.057 2.68 1.140.475 0.932 0.05 0.05 0.076 0.033 1.52 0.660.525 0.965 0.05 0.05 0.057 0.016 1.14 0.320.575 0.981 0.05 0.05 0.033 0.013 0.66 0.260.625 0.994 0.05 0.05 0.016 0.006 0.32 0.120.675 1.000 0.05 0.013 0.26 0.725

0.8

1.000

0.05

Swe

fwe

fwe/ Swe

E8fwe* 0.006Swe* Table 0.12 Swe* - Swc

Swe*

At breakthrough => 0.450 0.475 0.525 0.575

Swe* fwe* 0.475- Swc 0.275

Wid (PV)

Wid 0.475 (PV) 0.525

0.373 0.658

0.373 0.877 0.575 0.658 1.515 0.625 0.877 3.125 0.675 1.515 3,846 0.725 3.125 8.333 0.775 3,846

0.7751-fwe*8.333Wid

(from fig.2.1)

At breakthrough => 0.450

Swe*

0.665

0.7351-fwe*

(PV)

Wid 0.265

Time Npd

(PV)

(years)

0.372

Time 0.3720

Npd 0.373

t

t 0.3739

2.66 2.67

0.525

(from fig.2.1) 0.325 0.841(PV)

(PV) 0.159

0.658

(years) 4.71 0.4296

0.575

0.375

0.0905

0.877

0.4583

0.373 0.025

0.3739 0.5531 2.67 3.125

0.625

0.665 0.425

0.275 0.675

0.735 0.475

0.375

0.905

0.325 0.725

0.841 0.525

0.905 0.950

0.372 0.050

0.9880.159

0.658 0.012

0.9750.265 0.0905

0.877

0.3720 0.5008 2.66 1.515 0.4296 0.5712 4.71 3.846 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

6.28 10.84 22.36 27.52

Table E9

The values of fwe* in table E9, have been obtained from Figure E4, 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.

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Oil recovery for Qi = 1200 bbl/d/well

0.60

Npd (PV)

0.50 0.40 0.30 0.20 0.10 0.00

0

1

1

2

2 3 Wid (PV)

3

4

4

5

Figure E5

EXERCISE 5 Displacement Under Segregated Flow Conditions 1) Re-work example 4 for the same water injection rate (1,200 bbl/d/well) and using precisely the same reservoir and fluid property data, but assuming that displacement takes place under segregated flow conditions. Compare the results with those obtained in 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

Relative Permeability

From the relative permeability relations given in Figure E6, the end point relative permeability are:

1.00 krw 0.90 k'rw (end points) 0.80 kro 0.70 k'ro (end points) 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 Water Saturation (Sw)

Sw 0.20 0.80

76

K'rw

0.000 E6 Figure 0.315

K'ro 0.880 0.000

Relative Permeabilit

0.80 kro 0.70 k'ro (end points) 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

Immiscible Displacement E I G H T E E N

Water Saturation (Sw)

Sw 0.20 0.80

K'rw 0.000 0.315

K'ro 0.880 0.000

Table E10

End point relative permeability to water: k'rw = End point relative permeability to oil: k'ro = Since: Water viscosity µw = Oil viscosity µo =

0.315 0.880 0.5 cp 4.5 cp

The end poin Mobility Ratio "M" can be evaluated as:

M = 3.2216 Savg

Savg

low

d flow

1.200

ef



krw ′ / µw kro′ / µo

egate

Dif fus



M=

Segr

1.000

Segregated Flow

0.800

fw

0.600 Diffuse Flow

0.400

0.200

0.000

0.00

0.2

0.4

Sw

0.6

0.8

1.0

Figure E7

At the time of water breakthrough: Npd = Wid where: WiD is the cumulative water injection expressed in Pore Volume (PV): Savg 22/07/14

- segregation floow at breakthrough = 0.386

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N pd bt = Savg − Swi



Since: 1 MOV = PV (1 - Sor - Swc) Npdbt = 0.186

(PV)

WiD

(PV)

(PV)

At Breakthrough = >

0.186 0.420 N 0.720pD (PV) (PV) 0.960 1.200 At Breakthrough = > 0.186 0.186 1.440 0.420 0.352 1.680 0.720 0.468 1.933 Maximum Recovery => 0.960 0.524 1.200 0.561 1.440 0.584 Table E11 Oil Recovery for 1.680 0.596 1.933 0.600 Maximum 0.60Recovery = >

WiD

0.50 0.60

Npd(PV)

0.40

Npd(PV)

0.186 Time 0.352 t 0.468 (Years) 0.524

0.561 1.333 0.584 3.006 0.596 5.153 0.600 6.870 8.588 10.305 Qi = 1200 12.023 13.833

1.333 3.006 5.153 6.870 8.588 10.305 12.023 13.833

bbl/d/well

Oil Recovery for Qi = 1200 bbl/d/well

0.40

0.50

Diffuse Flow

Segregated Flow

0.30

Diffuse Flow

0.20

0.30

Segregated Flow

0.10

0.20

0.00

0.10 0.00

Time t (Years)

NpD

0

0

1

1

2

2

Wid (PV)

Wid (PV)

3

3

4

4

5

5

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.

78

Immiscible Displacement E I G H T E E N

REFERENCES 1

Dake,L.P., Fundamentals of Reservoir Engineering, Elsevier,Amsterdam, 1978

2

Dake,L.P., The Practise of Reservoir Engineering, Elsevier,Amsterdam, 1994

3

Bishlawi,M and Moore,R.l., Montrose Field Reservoir Management, SPE Europec Conference, London,(EUR166), Oct.1980

4

Mayer and Gurr

5

Buckley,S.E. and Leverett,M.C. Mechanism of Fluid Displacement in Sands. Trans,AIME,146,107-116, 1942

6

Chierici,G.L. Principles of Petroleum Reservoir Engineering, Vol 2, Springer Verlag, Berlin , 1995

7

Welge,H.J., A Simplified Method of Computing Oil Recovery by Gas or Water Drive, Trans AIME, 195,,91-98,1952

8

Dietz,D.N.,A Theoretical Approach to the Problem of Encroaching and ByPassing Edge Water, Proc. Ned.Akad Wet Ser B,56:83-92,1953

9

Coats,K.H.,Demsey,J.R. Henderson,J.H. The Use of Vertical Equilibrium in Two Dimensional Simulation of Three Dimensional Reservoir Performance,Soc. Pet.Eng J. Trans AIME, March.63-71,19971

10 Stiles,W.E., Use of Permeability Distribution in Water Flood Calculations, Trans. AIME, 186, 9-, 1949 11 Archer,J.S.,Wall,C.G., Petroleum Engineering Principles and Practise, Graham & Trotman, London,1986 12 Coats,K.L. Simulation of Gas Condensate Reservoir Performance,SPE Symposium Reservoir Simulation,New Orleans SPE 10512, 1982.

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Reservoir Engineering

Examination and Model Solutions

O N t N O no TI o A E: td n M TR bu o A N GIS E : on inati cti : RS se xam RE is R: U th e e hed EA AT te l th inis N ple unti f G m is SI l Co sea

RE

CO

Y

U

.:

8 Pages

Model Solutions to Examination

•

Date: Subject: INSTRUCTIONS TO CANDIDATES

• No.

graph made of P vs V and intersection of two slopes indicates bubble point pressure used to determine Pb and Bo above bubble point and BT over total

pressure

Mk.

1. Complete the sections above but do not seal until the examination is finished.

P

2. Insert in box on right the numbers of the questions attempted. 3. Start each question on a new page.

Gas

4. Rough working should be confined to left hand pages.

P

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

Oil Hg

Pb

PLEASE READ EXAMINATION REGULATIONS ON BACK COVER

Vb

V

b) Separator Test

• • • •

oil from PVT cell at reservoir temperature and at bubble point pressure released to surface conditions gas collected from each stage collected and final volume of oil at standard conditions measured volume of oil removed from PVT cell at bubble point and reservoir temperature measured used to determine GOR and Bob as a function of separator

conditions

11

Reservoir Engineering 22/07/14

Petroleum Engineering

Course Code: G137/G13W/X G127/G12W INSTITUTE OF PETROLEUM ENGINEERING HERIOT-WATT UNIVERSITY

DEGREE OF MSc / DIPLOMA IN PETROLEUM ENGINEERING DEGREE OF MSc / DIPLOMA IN RESERVOIR EVALUATION & MANAGEMENT RESERVOIR ENGINEERING – Module G11RE Monday 8 December 200X, 09:30 – 12:30 (3 hours) This is a closed book examination 15 minutes reading time is provided from 09:15 – 09:30 1.

This Paper consists of 2 Sections - A and B.

2.

Section A Section B -

Attempt all Questions Attempt 3 Questions from 6

3.

Section A Section B -

25% of marks 75% of marks

The allocation of marks within each question is shown in brackets in the right hand margin. This examination represents 100% of the class assessment. If more questions are attempted than stated above, they will be marked in the order they appear in the scripts until the requisite number has been marked. All remaining answers will be disregarded. 4.

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.

5.

Write your answers in the books provided as follows:A = Green B = Gold

6.

22/07/14 10/07/12

Unit Conversion Tables are included.

SECTION A A1

Explain briefly what you understand by:

(a)

the compositional model description and

(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)

A3

Derive the instantaneous gas-oil ratio equation. (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)

A5

(i) Explain briefly what is meant by (a)

a normal pressured reservoir

(b)

an overpressured reservoir (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: Compositions of Produced Fluids (Mole fractions) Gas C1 Methane C3 Propane C5 N-Pentane C8 Octane

0.89 0.07 0.04 —

Condensate — 0.21 0.61 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 o R 1 lb mole R

= = = =

5.615 cu ft 460+oF 379.4 SCF 10.73 psi cu ft/lb mole oR

See Attachment B1 (Table and Figure) (14 marks) 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. (a)

Relative Volume (Flash Vaporisation Test)

(b)

Separator Test

(c)

Differential Test (6 marks)

(ii) Explain briefly the constant volume depletion test for gas condensate (4 marks) 22/07/14

(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) Pressure psig

Volume cc

Pressure psig

Volume cc

5000 4500 4000 3500 3000 2500 2000 1900 1800 1700 1601

162.54 163.21 163.90 164.64 165.43 166.32 167.21 167.40 167.60 167.80 168.00

1591 1573 1555 1515 1435 1341 1234 1113 989 854 728

168.39 169.08 169.85 171.56 174.97 180.11 186.95 197.28 211.04 231.71 259.31

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 PVT Cell psig

Bubble Point 1400 1200 1000 800 0

Cumulative Gas Produced cc (standard conditions) 0 2044 4438 6732 9076 26,928 @ 60oC

Volume of Oil in Cell cc

184.80 182.35 179.37 176.52 173.67 140

(a)

Determine the bubble point pressure of the reservoir fluid at reservoir temperature.

(b)

The oil formation volume factor at 3650 psig (c) The solution gas-oil ratio at 3650 psig and 2700 psig (d) The solution gas-oil ratio at 1200 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. (b) Construct the water saturation profile in the space provided. (c) Calculate the oil-in-place per unit cross-section over the thickness of the reservoir. Data:

Specific gravity of water Specific of oil Density of water Air-mercury capillary Pressure

= = =

1.03 0.80 62.4 lb/ft3

=

10 x water-oil capillary pressure 1.22

Oil formation volume factor =

(12 marks) 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 Aquifer Radius 90,000 ft.

22/07/14

Oil Reservoir Radius 15,000 ft.

Over the first two years of production the pressure decline is expected to be as follows: Time (months) Pressure (psia)

0 6700

6 6688

12 6642

18 6584

24 6508

After the first 6 months 80,000 bbls of water were calculated to have influxed from the acquifer. The properties common to the reservoir and acquifer are as follows: K µw porosity water compressibility pore/rock compressibility

= = = = =

180 mD 0.6 cp 0.19 3 x 10-6 psi-1 4 x 10-6 psi-1

(a)

Calculate the thickness of the acquifer sands

(b)

Calculate the cumulative water influx at the end of 12 months, 18 months and 24 months.

The Hurst & van Everdinger equation for a full radial system is: We =

1.119∅cRo2h∆pWD

where We ∆p WD c Ro h ∅

= = = = = = =

cumulative water influx (bbls) pressure drop (psi) dimensionless water influx compressibility of the acquifer (psi-1) radius of oil reservoir (ft) thickness of acquifer (ft) porosity

Charts are supplied of dimensionless water influx WD versus dimensionless time tD (see 2 attachments) where: tD

=

t = k = µw =

2.309

kt µ w φcR 20

time (years) permeability (millidarcies) 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, _ formation volume factor for oil, Bo net thickness of formation, h viscosity of reservoir oil, _ compressibility, c permeability, k wellbore radius, rw external radius, re initial reservoir pressure, Pi well flowrate (constant) skin factor

25% 1.30rm3/stm3 50m 2.2x10-3 Pas 0.8x10-9Pa-1 120mD 0.15m 650m 270bar 200stm3/day 0 (15)

B6 A well has been on production in an oil reservoir. For the following data, calculate the bottomhole flowing pressure, Pwf for (i) steady state conditions (ii) semi steady state conditions

(6) (6)

and briefly describe the main differences in the flow regimes

(8)

Data formation volume factor for oil, Bo net thickness of formation, h viscosity of reservoir oil, _ permeability, k wellbore radius, rw external radius, re average reservoir pressure, P well flowrate (constant) skin factor

22/07/14

1.42rm3/stm3 60m 1.3x10-3 Pas 100mD 0.15m 530m 270.0bar 220 stm3/day 0

Compressibility Factors for Natural Gases as a Function of Pseudoreduced Pressure and Temperature. Pseudo Reduced Pressure, Pr 1.1

1

0

2

3

4

5

6

7

8

1.1

Pseudo Reduced Temperature 3.0 2.8 2.6 2.4 2.2 2.0 1.9 1.8

1.0

0.9

5

1. 4

1.

05

1.7

1.

1.6

1.7

1

0.8

1.3 1.1

1.0 1.05 1.2 0.95

1.

1.5

0.6

2

1.35 1.3

1.2

1.6 1.8

0.5 1.15

1.7 1.9

1.3

2.4 3.0

2.6

1.2

1.05

0.25 2.8

1.1

3.0

1.1 Compressibility of Natural Gases (Jan. 1, 1941)

2.6 2.4 2.2 2.0

1.0

1.8 1.7 1.6

0.9 7

1.4

2.2

2.0

1.1

0.3

1.5

4 1. 1.5

1.25

0.4

1.6

3

1.4

1.

0.7

1.

Compressibility Factor, z

1.45

1.9

1.2

1.0

1.1 1.4 1.3

8

1.05

0.9 9

10

11

12

13

Pseudo Reduced Pressure, Pr

14

15

Air/Mercury Capillary Pressure Curves

h(ft)

k(md) 55

25

5

0.05

0.01

φ(%) 15

10

10

5

5

150

C

Capillary Pressure (psi)

A

D

E

B

100

50

0

10

20

30

40

50

60

70

80

90 100

Pore Space Not Occupied by Mercury

22/07/14

h (ft) Scale 2cm = 10ft Rock Type

Top of Reservoir

A

D

E

C

A

B

A

100% Sw in Rock Type A Found at This Level

0

20

40 60 Oil Saturation

80

FWL 100

Water Saturation 100

80

60

40

20

0

22/07/14

N -PENTANE N -HEXANE

28.97 28.02

32.00 44.01

34.08

18.02

AIR NITROGEN

OXYGEN CARBON DIOXIDE

HYDROGEN SULPHIDE

WATER

128.3 142.3

72.15

72.15 86.17

ISO-PENTANE

N -NONANE N-DECANE

58.12 58.12

ISO-BUTANE N -BUTANE

100.2 114.2

30.07 44.09

ETHANE PROPANE

N -HEPTANE N -OCTANE

16.04

MOLECULAR WEIGHT

METHANE

COMPOUND

212.0

–76.5

–297.4 –109.3

–317.7 –320.4

303.4 345.2

209.2 258.2

96.9 155.7

82.1

10.9 31.1

–127.5 –43.7

–258.7

ºF

BOILING POINT AT 14.7 PSIA

3206

1306

732 1072

547 492

335 312

397 370

485 434

482

530 551

709 518

673

PRESSURE PSIA

1165

673

278 548

239 227

1073 1115

973 1025

847 915

830

733 766

550 666

344

TEMP ºR

CRITICAL CONSTANTS

0.9991

– –

–

0.7211 0.7333

0.8875 0.7082

0.8305 0.8837

0.6241

0.5625 0.5838

– 0.5072

–

Gm. PER cc

62.37

– –

–

48.02 44.78

42.92 44.09

39.36 41.43

38.96

35.12 36.43

– 31.66

–

POUNDS PER CU FT

LIQUID DENSITY 60º F, 14.7 PSIA

337.95 374.91

264.03 300.99

190.11 227.07

190.11

153.15 153.15

79.23 116.19

42.27

POUNDS PER WCF

17.78 18.30

21.89 19.55

27.83 24.34

27.38

30.59 31.75

– 36.35

–

CU FT GAS PER GALLON LIQUID

GAS DENSITY 60º F, 14.7 PSIA (PERFECT GAS)

PHYSICAL PROPERTIES OF THE PARAFFIN HYDROCARBONS & MISCELLANEOUS COMPOUNDS

15.88 14.30

19.95 17.77

26.36 22.83

26.17

29.70 30.77

– 35.78

–

LIQUID (ACTUAL)

CU FT GAS PER GALLON

4.0 reD = 3.5 reD = 4.0 3.5

reD =

= r eD

∞

3 .0

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

tD

6.0

7.0

20 reD = 7.0 reD = 8.0

18

reD =

∞

6 r eD =

.0

16

14

Qt

reD = 5.0

12

10

reD = 4.5

8

reD = 4.0

6

reD = 3.5 reD = 3.0

4

2

0 0

10

20

30

40

50

tD

22/07/14

60

70

110

re D

=

∞

100

90

80

70

Qt 60

50

reD = 10.0

40

reD = 9.0 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

22/07/14

Qt

10-1 10-2

2

4

1.0 8 6

2

4

6

10 8

2

4

6 8 10-1

2

4

6

tD

8

1.0

2

4

6 8 10

FINITE LINEAR AQUIFER

reD = 1.5

reD = 2.0

∞

reD = reD = 6.0 reD = 5.0 reD = 4.5 reD = 4.0 reD = 3.5 reD = 3.0 reD = 2.5

WD

4

6 8 102

2

4

6 8 103

2

r eD

=∞

reD = 5.0 reD = 4.5

reD = 6.0

reD = 7.0

reD = 9.0 reD = 8.0

reD = 10.0

1

2

4

tD

4

6

102

103

10 8 104

reD = 2.5

reD = 3.0

reD = 3.5

∞

6

2

=

'd) nt o (c

reD = 4.0

10

r eD

reD = 15.0

10 8

2

4

6

102 8

∞ (CONT'D) ONLY USE THIS SCALE FOR LINE reD =

22/07/14

Compressibility Factors for Natural Gases as a Function of Pseudoreduced Pressure and Temperature. Pseudo Reduced Pressure, Pr 1.1

1

0

2

3

4

5

6

7

8

1.1

Pseudo Reduced Temperature 3.0 2.8 2.6 2.4 2.2 2.0 1.9 1.8

1.0

0.9

5

1. 4

1.

05

1.7

1.

1.6

1.7

1

0.8

1.3 1.1

1.0 1.05 1.2 0.95

1.

1.5

0.6

2

1.35 1.3

1.2

1.6 1.8

0.5 1.15

1.7 1.9

1.3

2.4 3.0

2.6

1.2

1.05

0.25 2.8

1.1

3.0

1.1 Compressibility of Natural Gases (Jan. 1, 1941)

2.6 2.4 2.2 2.0

1.0

1.8 1.7 1.6

0.9 7

1.4

2.2

2.0

1.1

0.3

1.5

4 1. 1.5

1.25

0.4

1.6

3

1.4

1.

0.7

1.

Compressibility Factor, z

1.45

1.9

1.2

1.0

1.1 1.4 1.3

8

1.05

0.9 9

10

11

12

13

Pseudo Reduced Pressure, Pr

14

15

Air/Mercury Capillary Pressure Curves

h(ft)

k(md) 55

25

5

0.05

0.01

φ(%) 15

10

10

5

5

150

C

Capillary Pressure (psi)

A

D

E

B

100

50

0

10

20

30

40

50

60

70

80

90 100

Pore Space Not Occupied by Mercury

22/07/14

h (ft) Scale 2cm = 10ft Rock Type

Top of Reservoir

A

D

E

C

A

B

A

100% Sw in Rock Type A Found at This Level

0

20

40 60 Oil Saturation

80

FWL 100

Water Saturation 100

80

60

40

20

0

22/07/14

N -PENTANE N -HEXANE

28.97 28.02

32.00 44.01

34.08

18.02

AIR NITROGEN

OXYGEN CARBON DIOXIDE

HYDROGEN SULPHIDE

WATER

128.3 142.3

72.15

72.15 86.17

ISO-PENTANE

N -NONANE N-DECANE

58.12 58.12

ISO-BUTANE N -BUTANE

100.2 114.2

30.07 44.09

ETHANE PROPANE

N -HEPTANE N -OCTANE

16.04

MOLECULAR WEIGHT

METHANE

COMPOUND

212.0

–76.5

–297.4 –109.3

–317.7 –320.4

303.4 345.2

209.2 258.2

96.9 155.7

82.1

10.9 31.1

–127.5 –43.7

–258.7

ºF

BOILING POINT AT 14.7 PSIA

3206

1306

732 1072

547 492

335 312

397 370

485 434

482

530 551

709 518

673

PRESSURE PSIA

1165

673

278 548

239 227

1073 1115

973 1025

847 915

830

733 766

550 666

344

TEMP ºR

CRITICAL CONSTANTS

0.9991

– –

–

0.7211 0.7333

0.8875 0.7082

0.8305 0.8837

0.6241

0.5625 0.5838

– 0.5072

–

Gm. PER cc

62.37

– –

–

48.02 44.78

42.92 44.09

39.36 41.43

38.96

35.12 36.43

– 31.66

–

POUNDS PER CU FT

LIQUID DENSITY 60º F, 14.7 PSIA

337.95 374.91

264.03 300.99

190.11 227.07

190.11

153.15 153.15

79.23 116.19

42.27

POUNDS PER WCF

17.78 18.30

21.89 19.55

27.83 24.34

27.38

30.59 31.75

– 36.35

–

CU FT GAS PER GALLON LIQUID

GAS DENSITY 60º F, 14.7 PSIA (PERFECT GAS)

PHYSICAL PROPERTIES OF THE PARAFFIN HYDROCARBONS & MISCELLANEOUS COMPOUNDS

15.88 14.30

19.95 17.77

26.36 22.83

26.17

29.70 30.77

– 35.78

–

LIQUID (ACTUAL)

CU FT GAS PER GALLON

4.0 reD = 3.5 reD = 4.0 3.5

reD =

= r eD

∞

3 .0

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

tD

6.0

7.0

20 reD = 7.0 reD = 8.0

18

reD =

∞

6 r eD =

.0

16

14

Qt

reD = 5.0

12

10

reD = 4.5

8

reD = 4.0

6

reD = 3.5 reD = 3.0

4

2

0 0

10

20

30

40

50

tD

22/07/14

60

70

110

re D

=

∞

100

90

80

70

Qt 60

50

reD = 10.0

40

reD = 9.0 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

22/07/14

Qt

10-1 10-2

2

4

1.0 8 6

2

4

6

10 8

2

4

6 8 10-1

2

4

6

tD

8

1.0

2

4

6 8 10

FINITE LINEAR AQUIFER

reD = 1.5

reD = 2.0

∞

reD = reD = 6.0 reD = 5.0 reD = 4.5 reD = 4.0 reD = 3.5 reD = 3.0 reD = 2.5

WD

4

6 8 102

2

4

6 8 103

2

r eD

=∞

reD = 5.0 reD = 4.5

reD = 6.0

reD = 7.0

reD = 9.0 reD = 8.0

reD = 10.0

1

2

4

tD

4

6

102

103

10 8 104

reD = 2.5

reD = 3.0

reD = 3.5

∞

6

2

=

'd) nt o (c

reD = 4.0

10

r eD

reD = 15.0

10 8

2

4

6

102 8

∞ (CONT'D) ONLY USE THIS SCALE FOR LINE reD =

yyyy ;;;; ;;;; yyyy ;;;; yyyy ;;;; yyyy

Model Solutions to Examination

O N t N O no TI o E: RA td M T bu ion A n N GIS SE: at io in ct RE UR E: se am : R s ex d hi CO AR TU e t l the ishe E A et N pl unti fin G m is l Co sea

Y

SI

.:

8 Pages

Date: Subject:

Reservoir Engineering INSTRUCTIONS TO CANDIDATES No.

Mk.

1. Complete the sections above but do not seal until the examination is finished. 2. Insert in box on right the numbers of the questions attempted. 3. Start each question on a new page. 4. Rough working should be confined to left hand pages. 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

PLEASE READ EXAMINATION REGULATIONS ON BACK COVER

22/07/14

1

2

Model Solutions to Examination

SECTION A 1. Compositional Model

Based on the fluid description in terms of the paraffin series up to an upper C value eg C6. Then defining the rest as a C+ component, eg C7+. The properties of the C+ component are unique and therefore need to be measured for apparent molecular weight and specific gravity. Can also include non paraffin components in some cases PNA analysis

Black-oil model Based on the concept of a two component system of solution gas and stock tank oil. Results in black oil parameters of Bo and Rs. Values of these change as a result of changes in operating conditions of separators.

2. The variations in permeability can give rise to preferential flooding of the higher permeability layers and result in poor flooding characteristics, and reduced recovery, even when the mobility ratio for water displacing oil may be very favourable.

3. Instantaneous GOR = Producing gas/producing oil = Qg/Qo Qg & Qo surface production rates of gas and oil.

qg and qo down hole rates for gas & oil

22/07/14

3

qg =

k rg Adp µ g dx - D’arcy’s law

qo =

k ro Adp µ o dx

Surface flow of gas Qs

Q g = Q o R s + q g Bg Surface flow of oil Qo Q o = q o Bo

Qg k Adp = R s + rg µ g dxBg Qo

k ro Adp µ o dxBo

Instantaneous GOR = R s + 4.

krg µ oB o kroµ gB g

At dew point gas composition the same as system composition ie zj = yj Equilibrium ratio kj = yj/xj Therefore zj = kj x xj Composition of liquid xj = zj/kj Since ∑xj = 1.0 Then ∑xj = ∑ zj / kj = 1.0

eqn 1

At bubble point liquid composition the same as system composition ie zj = xj Equilibrium ratio kj = yj/xj Therefore zj = yj/kj Composition of gas yj = zj kj

4

Model Solutions to Examination

Since ∑yj = 1.0 Then ∑yj = ∑ zj * kj = 1.0

eqn 2

At dew point use equation 1. Estimate Pd to determine k values from whatever source. Check if eqn. 1 adds up to 1. If not select new values of Pd until convergence is reached.

At bubble point use equation 2. Estimate Pb to determine k values from whatever source. Check if eqn. 2 adds up to 1. If not select new values of Pb until convergence is reached.

5.

(i) a) Normal pressured reservoir

Normal pressured reservoir is where the pressure versus depth gradient for water zone intersects at zero depth with a pressure of zero ie 1 atmosphere absolute.

b) Overpressure reservoirs

Overpressured reservoir is where the pressure vs depth gradient for water zone intersects at a positive pressure at zero depth.

22/07/14

5

0

Pressure Normal Pressured

Depth

Gas Gradient

Oil Gradient

GAS

OIL

Water Gradient

WATER Over Pressured

6.

The steady state solution (essentially darcy’s Law) requires stable pressures at the outer boundary and the wellbore. In petroleum engineering, the requirement for steady state conditions may take a long time to develop, or may never develop (i.e. the flow may never change from transient conditions during the field’s economic life). Therefore to determine sensible values for permeability, short term tests were required and these required transient solutions which could be related to real measurements in the field taken over economically viable times, i.e. 1 day, 2 days etc.

7.

The radial diffusivity equation has a specific solution assuming that 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 equivalent undamaged well, an additional pressure drop must be imposed around the wellbore. It is assumed that since the effect of 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 in steady state, and a steady state inflow equation can be applied to the small region at most a few feet in extent in which the skin effect is produced. This can be used to calculate a pressure drop, and by comparison with the pressure drop required for the equivalent undamaged region, a skin factor can be defined. Thus the skin factor is a normalised pressure drop for a given flow rate.

Section B Question 1 Critical Point

(i)

Cricondenbar

P

Region of Retrograde Condensation

Dew Point Line Bubble Point Line Lines of Constant L/V

Cricondentherm

T

Gas Cycling: Gas cycling is the reinjection of separated gas back into the reservoir - in some cases with gases from other near reservoirs. 22/07/14

7

It is used to keep the reservoir above the dew point line - preventing condensation in the reservoir. (ii) Comp

MW lb/lb ml

Gas

C1

16.04

0.89

-

C3

44.09 72.15

0.07 0.04

0.21 0.61

31.66 39.36

9.2589 44.0115

0.29 1.12

-

0.18

44.09 Σ

20.5560 73.826

0.47 1.88

C5 C8

GOR ratio

114.2

Liquid lb/cu ft

Density

-

Weight lb/lb ml

Volume cu ft

-

-

= 1 x 106 ÷ 80 STB

= 12500 SCF/STB

= 12500

SCF 1 STB lb mole gas 73.83 lb 1 cu ft × × × × STB 5.615 cu ft 379.4 SCF lb mole oil 39.34 lb oil

GOR

= 11.0127

Comp

C1

MW

lb mole gas lb mole oil

Gas

Liquid

lbMole zj Res. Fluid Mole Frac

Pc

Tc

zjPc

C3

16.04 44.09

0.89 0.07

0.21

9.8013 0.9809

0.8159 0.0817

673 618

344 666

549.11 50.46

280.67 54.38

C5 C8

72.15 114.2

0.04 -

0.61 0.18 Σ

1.0505 0.1800 12.0127

0.0874 0.0150 1.00

485 370

847 1025

42.41 5.54 647.53

74.07 15.36 424.48

Ppc

at 240˚F and 8500 psia

240 = 700˚R

8

zjTc

Tr = 700/424.48 = 1.6491

Tpc

Model Solutions to Examination

Pr=8500/647.53 = 13.1268

z = 1.3500

from Z vs. Tpr & Ppr charts

at 6750 psi

Pr = 6750/647 = 10.4243

z = 1.15

Vm = ZRT/P cu ft/lb mole

at 8500 psi Vm = 1.1918 cu ft/lb mole

6750 psi Vm = 1.2785 cu ft/lb mole

Volume of reservoir = 5 x 1011 cu ft

Water Saturation = 0.17 Gas Volume = 4.15 x 1011 cu ft

Reservoir moles @ 8500 psia = 3.48 x 1011 moles

Reservoir moles @ 6750 psia = 3.25 x 1011 moles

Moles of reservoir fluid produced = 2.36 x 1010 moles

Mole split = 11.0127 moles gas:1 mole liquid

Moles condensate produced 22/07/14

9

= 2.36 ×

1010 × 1 moles liq 12.0127 total moles

= 2.36 ×

1010 × 1 moles liq 73.83lb cu ft lbb × × × 12.0127 wt lb mole 39.34lb 5.615 wt

= 6.57 x 108 STB condensate

Gas produced

= 2.36 ×

11.0127 SCF × 379.4 12.0127 total moles lb ml

= 8.21 x 1012 SCF gas (or 6.57 x 108 STB Condensate x GOR)

QUESTION B2

PART 1

a) Rel. Volume Test

• • •

carried out in a PVT all at reservoir temperature no fluids removed from the cell measurement made of volume of fluids as a function of pressure (pressure determined by expanding the system)

•

graph made of P vs V and intersection of two slopes indicates bubble point pressure

10

Model Solutions to Examination

•

used to determine Pb and Bo above bubble point and BT over total pressure

P Gas P

Oil Hg

Pb

Vb

V

b) Separator Test

•

oil from PVT cell at reservoir temperature and at bubble point pressure released to surface conditions

•

gas collected from each stage collected and final volume of oil at standard conditions measured

•

volume of oil removed from PVT cell at bubble point and reservoir temperature measured

•

used to determine GOR and Bob as a function of separator conditions

22/07/14

11

c) Differential Test

• • •

oil in cell at reservoir temperature pressure dropped in stages to atmospheric pressure from Pb gas produced at each pressure step removed, measured and volume of oil remaining measured

•

used to determine Bo and Rs below bubble point pressure

PART 2

Constant Volume Depletion Test

• • •

gas in PVT cell at dew point pressure reduced in stages at each stage gas displaced to bring all back to original volume and volume of condensate measured

•

relative liquid volume curve produced to determine extent of retrograde condensation

Relative Volume Vol Condensate/Vol of Fluid in Cell

Max. liqd. Dropout

P

12

Pd

Model Solutions to Examination

PART 3

Draw graph and determine intersection of oil and oil and gas slope = occurs at p = 1595 psi and a volume of 168.08cc

(a)

Bubble point = 1595 psig

(b)

Bo at 3650 psi Vol @ 3650 psi = 164.42 cc

Bob from separator test = 34/27.4 = 1.241 bbl/stb

Bo @ 3650 = Bob x rel. vol. @ 3650/vol @ Bp = 1.02409 x 164.42/168.08

Bo @ 3650 = 1.2138 bbl/stb

(c)

GOR @ 3650 & 2700

GOR

= 4976 cc of gas per 27.4 cc

=

4976 cc or SCF 5.615 cu ft × 27.4 cc or STO Cu. Ft. bbl

GOR

= 1020 SCF/STB

22/07/14

13

Same value since both above Bp.

(d)

GOR @ 1200

Below Bp need differential result

Gas produced from Bp to 1200

= 4438 cc for 140 cc residual oil

∆R s

diff

=

= 177.99

4438 cc or SCF 5.615 cu ft × 140 cc or Cu. Ft. residual oil bbl

SCF bbl residual oil

to convert to flash units

= 177.99

SCF 140 cc residual oil 34 cc or bbl pt oil × . × bbl residual oil 184.80 cc or bbl Bpt oil 27.4 STO oil

= 167.33 SCF/STB

GOR @ 1200 = GOR @ Bpt – 167.33 = 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

22/07/14

15

(e) BT @ 3650 = Bo @ 3650 = 1.2138

BT @ 1200

= Bob x rel. oil @ 1200 = 1.24 x rel. volume @ 1200

vol at 1234 psi = 186.95 cc 1113 psi = 197.28 cc  197.28 − 186.95  × 34 at 1200 = 186.95 +  1234 − 1113  = 186.95 + 2.9026

vol @ 1200 = 189.85 cc

BT @ 1200

= 1.24

vol @ Bpt 189.85 × vol STO 168.08

= 1.3713

QUESTION 10

(i)

Free Water Level (FWL)

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 h

OWC FWL 0 0

100% Water Saturation

(ii)

Based on Pore Doublet Model

Oil

Trapped Oil

Water

As water rises capillary forces move ahead of natural level, more in 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 in the swept position of the reservoir.

(iii)

Recalibrate Pcvs. saturation curve air/Hg for oil/water

22/07/14

17

Pc = h∆pg

pc

lb f lbm = h(ft )∆p 3 × g 2 ft 1n

lbf = lbm x g

= Pc

lb m × g 144in 2 lb × = h ft × (1.03 − 0.8) × g × 62.4 m3 2 2 in ft ft

h=

Pc × 144 (1.03 − 0.8) × 62.4 =

2.30 0.23

h = 10.03 Pc oil water

Pc oil water = Pc air/Hg ÷ 10

h ft = Pc air/Hg psi

Rescale capillary pressure to this:

18

Model Solutions to Examination

Air/Mercury Capillary Pressure Curves

h(ft)

k(md) 55

25

5

0.05

0.01

φ(%) 15

10

10

5

5

150

C

Capillary Pressure (psi)

A

D

E

B

100 100 95.5 3 82'=18 5

85.5

69.5

49.5

50

36.75 23' 17.5

0

10

20

30

40

50

60

70

80

90 100

FWL

Pore Space Not Occupied by Mercury

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19

Rock Type

h (ft) Scale 2cm = 10ft Top of Reservoir 103.2'

91

A

(90)

45

95.5'

90

D

(44) 85.5'

10

43

82' (5)

E 69.5'

44

C (35) 83

49.5' A

(79) 36.75'

44

76

(22)

B

23' 23' 46

17.5' A (23) 100% Sw in Rock Type A Found at This Level

6' 0

20

40 60 Oil Saturation

80

FWL 100

Water Saturation 100 20

80

60

40

20

0

Model Solutions to Examination

(a)

Free water level = Rock A = 6ft below OWC

(b)

Construction of water profile on the shell

(c)

Oil in place – A x h x φ x SWAVG

Oil in Place: ∑h x So x φ

Zone A1 = (17.5 – 6)ft x 0.23 x 0.15

0.39675

B1 = (36.75 – 23)ft x 0.22 x 0.10

0.3025

A2 = (49.54 – 36.75)ft x 0.79 x 0.15

1.5156

C1 = (69.5 – 49.5)ft x 0.35 x 0.10

0.7

E = (85.5 – 82)ft x 0.44 x 0.5

0.00875

D = (95.5 – 85.5)ft x 0.05 x 0.05

0.22

A3 = (103.25 – 95.5) x 0.9 x 0.05

0.34875

3.4924 Oil in place = 3.4924 cu ft oil/sq ft Oil in place = 0.6220 res bbl/sq ft Oil in place = 0.5098 stb/sq ft 22/07/14

21

QUESTION 11

(i)

Rate Sensitive Water Reservoirs

Three aquifer characteristics

(a) Artesian well type whose oil production is less than aquifer influx then pressure at oil/water contact remains constant. If not get solution gas drive combination.

(b)Sealed aquifer system whose oil production is less than aquifer influx then pressure at oil/water contact will steadily decline. Again if oil production greater than aquifer influx combination drive of water drive and solution gas drive will result.

(c) Sealed system where oil production is greater then aquifer system will result in combination drive.

Type a

P Pressure at OWC

Type b Type c

Time

(ii)

Hurst and Van Everdingen unsteady state solution is for a fixed boundary pressure, i.e. We
= B∆pQt, where ∆P is constant and Qt is a function of dimensionless time tD.

22

Model Solutions to Examination

A declining pressure can be simulated by superimposing fixed ∆P’s to 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 to the water influx for the other ∆P’s. ∆P1=(Pi-P1)/2

Pi

P

∆P2=(Pi-P1)/2+(P1-P2)/2 =(Pi-P2)/2 ∆P3=(P1-P2)/2+(P2-P3)/2 =(P1-P3)/2

1

2

3

4

T

(iii)

Need to calculate h

We = 1.119φ cRo2h∆PQt

h is unknown but We is known

t D = 2.309 ×

kt µ w φcR 2o

calculate Qt for tD’s as a function of t

22/07/14

23

Time Yrs.

0

0.5

1.0

1.5

2.0

Pressure ∆P

6700 -

6688 6

6642 29

6584 52

6508 67

tD

0

1.1574

2.3148

3.4722

4.6296

Qt (Chart)

0

1.88

2.74

3.65

4.33

We=BΣ∆pQt

-

80,000

495,016

1,672,657

3.54x10

B = 1.119φ cRo2f where f = 140˚/360˚ = 139.52 x h

We = 80,000 = 139.52 x h x ∆PQt

h = 50 ft.

h is now used in years = 1, 1.5 and 2 B = 139.52 x h = 6976 h = thickness = 50 ft

Water influx at

1 year = 495,016 bbl 1.5 years = 1.672 x 106 bbls 2 years = 3.54 x 106 bbls

24

6

Model Solutions to Examination

QUESTION 12

(i)

The well is in the centre of a radial reservoir and the full thickness is perforated therefore there are no geometrical effects to influence the flow. The well is assumed to have a constant flow rate from the moment it is put on production therefore there is only 1 rate dependent pressure perturbation. The flow regimes may be described in terms of:

(a) an initial transient flow regime where the fluid acts as if it were infinite in extent. The pressure changes within the reservoir manifest themselves as a wave which moves from the wellbore out into the reservoir. The transient nature of the flow regime means that the pressure close to the wellbore falls at a greater rate than the pressure farther into the formation. Initially, the fluid at the outer boundary senses no pressure gradient and doesn’t expand (and flow).

(b) a flow regime that mimics steady state flow, termed semi-steady state. In this flow regime, the fluid pressure gradient in the reservoir from the wellbore to the outer boundary does not change through time, i.e. the differences in pressure between points in the reservoir remain the same (as in steady state flow) but the absolute value of pressure declines through time. In this case, the pressure wave has developed across the reservoir and it slowly sinks through time as the fluid is produced.

22/07/14

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 the line source solution to be accurate (ii) the reservoir is infinite acting.

Thereafter, the choice of Ei function or ln approximation to the Ei function has to be made.

A Check Ei applicability

line source not accurate until 100φµcrw2 t> k

t>

100x0.25x2.2x10 -3 x0.8x10 −9 x0.152 120x10 -15

t > 8.25s

time is 8 hours, therefore line source is applicable.

B Check reservoir is infinite acting

φµcre2 the reservoir is infinite acting if the time, t < 4k

26

Model Solutions to Examination

0.25x2.2x10 −3 x0.8x10 −9 x650 2 4x120x10 -15 t < 387292s

i.e. t <

t < 108 hours

therefore line source solution is applicable.

(i) check ln approximation to Ei function

the ln approximation is valid if the time, t > 25x0.25x2.2x10 −3 x0.8x10 −9 x0.152 t> 120x10 -15 t > 2.1s

25φµcr 2 k

therefore ln approximation is valid.

qµBo  γφµcrw2  ln  (taking account of the conversion from (ii) Pwf = Pi + 4πkh  4kt  stock tank to reservoir conditions via the formation volume factor for oil). qµBo 200x2.2x10 −3 x1.30 = = 87805 4πkh 24x3600x4πx120x10 −15 x50 −3 −9 2 φµcrw2 0.25x2.2x10 x0.8x10 0.15 = = 716x10-9 4kt 4x120x10 -15 x8x3600

Pwf = 270x105 + 87805xln(1.781x 716x10-9) = 270x105 - 1191726 = 25808274Pa = 258.1bar

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27

QUESTION 13 (i) Steady state flow regime:

Substitute the values into the steady state flow equation

P − Pwf =

qµBo  re 1   ln −  2πkh  rw 2 

Pwf = P −

qµBo  re 1   ln −  2πkh  rw 2 

Pwf = 270 −

220x1.3x10 −3 x1.42  ln 530 − 1  −15 24x3600x2π 100x10 x60  0.15 2 

Pwf = 270.0 - 9.6 Pwf = 260.4bar

(ii) Semi-Steady state flow regime:

Substitute the values into the semi-steady state flow equation

P − Pwf =

qµBo  re 3   ln −  2πkh  rw 4 

Pwf = P −

qµBo  re 3   ln −  2πkh  rw 4 

220x1.3x10 −3 x1.42  ln 530 − 3  −15 24x3600x2π 100x10 x60  0.15 4  Pwf = 270.0 - 9.3 Pwf = 270 −

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 pressure perturbation produced by opening the well has travelled through the reservoir to the oil water contact. There has been an influx from the aquifer of equal magnitude as that removed at the well, and the pressure profile that has developed in the reservoir is now constant and doesn’t vary with time.

In contrast, pressure perturbation in the reservoir in a semi-steady state flow regime has also reached the outer boundary, but there is no influx. The pressure at the boundary falls, as does the absolute pressure at every point in the reservoir. However, the pressure gradient at all points within the reservoir remains constant, mimicing the constant pressure gradient in the steady state flow regime. In this flow regime, the absolute pressure in the reseroir fluid declines through time, but the pressure gradient remains the same.

22/07/14

29

Reservoir Engineering Past Papers

Please note some questions in these past papers are no longer relevant, those questions have been highlighted in red.

22/07/14

Course:- 28117 Class:- 289013 HERIOT-WATT UNIVERSITY DEPARTMENT OF PETROLEUM ENGINEERING Examination for the Degree of MEng in Petroleum Engineering Reservoir Engineering 1 Tuesday 6th January 1998 09.30 - 13.30 NOTES FOR CANDIDATES 1.

This is a Closed Book Examination.

2.

15 minutes reading time is provided from 09.15 - 09.30.

3.

Examination Papers will be marked anonymously. See separate instructions for completion of Script Book front covers and attachment of loose pages. Do not write your name on any loose pages which are submitted as part of your answer.

4.

This Paper consists of 3 Sections:- A, B and C.

5.

Section A:Sections B & C:-

Attempt all Questions Attempt 4 numbered Questions with at least 1 Question from each Section

6.

Section A:Section B / C:-

20% of marks 80% of marks

Marks for Questions and parts are indicated in [brackets] 7.

This Examination represents 100% of the Class assessment.

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.

9.

Answers must be written in separate, coloured books as follows:Section A:Section B:Section C:-

22/07/14

Blue Green Yellow

SECTION A A1

Define:(i) (ii) (iii) (iv)

A2

(ii)

Compositional model description for the characterisation of a reservoir fluid. Black oil description for the characterisation of a reservoir fluid.

Bubble point and dew point lines Critical point Cricondentherm Lines of constant proportion of liquid-gas Region of retrograde condensation. [3]

Derive an equation for the average permeability, resulting from radial circular flow into a well from layers of different permeabilities and thicknesses. From where would such an average be obtained ?

A5

[3] Draw the pressure - temperature phase diagram for a gas condensate reservoir indicating the following:(i) (ii) (iii) (iv) (v)

A4

[2] Explain briefly what you understand by:(i)

A3

Gas formation-volume factor Oil formation-volume factor Total formation-volume factor Solution gas-oil ratio

[3] List the rock properties that should be determined in a rock-mechanicaloriented, special core analysis programme. [3]

A6

Describe the nature of a partially communicating fault and how pressure data may be used to identify the transmissibility index of such features. [3]

A7

Explain the difference between a steady-state and a semi-steady-state productivity index.

22/07/14

[3]

SECTION B B8 (a)

Methane is a significant component in reservoir fluids. Using a sketch for a binary of methane and n-decane (C10), illustrate the impact of methane on the critical point loci of C1-C10 binary mixtures. What is the significance of this diagram ?

(b)

[6] 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 The composition of the producing gas is given below:Component (MW)

Composition (Mole fraction) Gas Condensate

Methane (16.04) Ethane (30.07) Propane (44.09) N-butane (58.12) N-pentane (72.15) Hexane (86.17) Heptane plus (114.2) (equivalent to octane)

0.84 0.07 0.05 0.02 0.02 -

0.16 0.34 0.31 0.10 0.09

(i)

What is the composition of the reservoir gas?

(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? R = 10.72cu ft.psi/lb mole.°R 1lb mole = 379.4 SCF 1bbl = 5.615cu ft

[14]

B9 (a)

You have been given a PVT report for an oil sample. List the procedures and calculations you would take to determine:(i) (ii)

(b)

The bubble point pressure at reservoir temperature The solution gas-oil ratios and the oil formation volume factors above the bubble point and below the bubble point. [10]

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:(i) (ii) (iii)

Bo and R s at the bubble point. Bo, Bt, Bg, Rs and z at 1,700 psia and 145°F Bt at 2100 psia and 145°F [10]

B10 Using field examples, describe the various rock-mechanical phenomena which can be activated in producing reservoirs, and explain the impact of these phenomena on reservoir development and management strategies. [20]

22/07/14

B11 (a) For low permeability rocks, the measured permeability of the rock, using gas as the fluid is more than it is using a liquid. Comment briefly on this and how the permeability of a rock can be obtained using gas as the fluid. (b)

[4] The system below represents the common arrangement for measuring the permeability of a core plug using a gas.

Derive an equation to calculate the gas permeabilities of a rock using the above system. [6] (c)

What is the free water level? Explain briefly why, because of capillary pressure, it is possible to have water saturations of large values, up to 100%, above the oil-water contact and above layers with lower water saturations. [4]

(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:q=

πr 4 ΔP 8μL

to relate Capillary pressure (Pc), to Permeability (k), and Porosity (ø). 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 )Δp ⎢ f ⎥ − ( WinjBw + G injBg ) ⎣ 1 − Swc ⎦

with a line above the symbols of the equation e.g. ( We − Wp Bw ) = net water inf lux

[3] (b)

Simplify the material balance equation above so that it can be used for an undersaturated reservoir without waterdrive and water production. Modify the simplified equation so that it can be used to express the recovery at the bubble point in terms of the effective compressibility of the reservoir system.. [4]

(c)

To predict the performance of a solution gas drive reservoir we require both the instantaneous gas-oil ratio equation and an equation to express the average oil saturation. 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.

(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. Describe briefly, in a step by step procedure, the application of one of these methods.

22/07/14

[6]

[7]

SECTION C C13 A tight formation of horizontal permeability, ky, equal to 0.5md and thickness, h, of 150ft is to be developed for oil production. The choice lies between horizontal wells or vertical wells with hydraulic fracturing. The rock mechanics suggest that an infinite-conductivity fracture of halflength, xf, equal to 300ft will be feasible. Determine the length of a horizontal well, L, which will have a comparable semi-steady-state productivity index (PI) to the fractured, vertical well. Additional Well Data:Formation vertical permeability, kz = 0.01md Oil viscosity, μ0 = 1.5cp Oil formation volume factor, B0 = 1.25 Wellbore radius, rw = 0.3ft Equivalent cylindrical external radius of reservoir drainage compartment, re = 10,000ft. [20] C14 A particular formation has been shown to be amenable to acid treatment which can increase the rock permeability from the intrinsic value of 5md to an improved value of 17md due to its effect on interstitial clay. 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. Additional reservoir data:Connate water saturation, Swc = 0.25 Oil formation volume factor, B0 = 1.2 Oil viscosity, μ0 = 0.8cp Drainage area external radius, re = 5000ft Wellbore radius, rw = 0.3ft [20]

C15 Discuss how the wireline formation tester can give information on dynamic aquifer effects and indicate where other phenomena can give similar pressure-depth plots to those associated with a dynamic aquifer. Why is it important to know if dynamic aquifer effects are present?

End of Paper

22/07/14

[20]

22/07/14

22/07/14

Course:- 28117 Class:- 289013

HERIOT-WATT UNIVERSITY DEPARTMENT OF PETROLEUM ENGINEERING Examination for the Degree of MEng in Petroleum Engineering

Reservoir Engineering 1 Friday 8 January 1999 09.30 - 13.30 NOTES FOR CANDIDATES 1.

This is a Closed Book Examination.

2.

15 minutes reading time is provided from 09.15 - 09.30.

3.

Examination Papers will be marked anonymously. See separate instructions for completion of Script Book front covers and attachment of loose pages. Do not write your name on any loose pages which are submitted as part of your answer.

4.

This Paper consists of 3 Sections:- A, B and C.

5.

Section A:Sections B & C:-

Attempt all Questions Attempt 4 numbered Questions with at least 1 Question from each Section

6.

Section A:Section B / C:-

20% of marks 80% of marks

Marks for Questions and parts are indicated in [brackets] 7.

This Examination represents 100% of the Class assessment.

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.

9.

Answers must be written in separate, coloured books as follows:Section A:Section B:Section C:-

22/07/14

Blue Green Yellow

SECTION A A1

Define: (i) (ii) (iii) (iv)

The gas formation volume factor The oil formation-volume factor The total formation volume factor The solution gas-oil ratio [2]

A2

Derive an equation in terms of equilibrium ratios and composition to predict liquid and vapour ratios and compositions resulting from the flash separation of a reservoir fluid. Explain briefly the application of the equation when the reservoir fluid composition, temperature, and the pressure of the separation are known. [3]

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]

A4

With the aid of a diagram, comment on the fluid pressure gradients in an oil reservoir with a gas cap with a supporting aquifer for a normally pressured reservoir. Illustrate the gradient for an overpressured reservoir. [3]

A5

Briefly describe how the permeability sensitivity of rocks to stress can be measured in the laboratory, and draw a graphical, normalised, representation of the results you would expect to see for sandstones with high, medium and low permeabilties measured at ambient conditions. [3]

A6

Discuss the common mechanisms of formation damage occurring during drilling. [3]

A7

Discuss the concept of a critical rate in coning situations. [3]

SECTION B B8 (a)

Draw a Pressure-Temperature phase diagram to illustrate the phase behaviour of a gas condensate fluid. [2]

(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. The pore volume of the reservoir is considered to be 9.5 x 108 cu ft and the average water saturation is 19%. (i)

Calculate the composition of the reservoir fluid

(ii)

Calculate the inplace reserves in SCF of gas and STB condensate of the reservoir at the original pressure.

(iii)

Calculate the production of gas and condensate when the reservoir pressure has declined to 4260psia. 1bbl = 5.615 cu ft 0˚F = 460˚ R 1lb mole = 379.4 SCF R = 10.73 psi cu ft/lb mole° R Composition of Produced Fluids (Mole fraction) Gas C1 C3 C5 C8

Attachments

22/07/14

Methane Propane n-Pentane Octane

0.890 0.075 0.035 --

Condensate -0.215 0.620 0.165

[15]

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 Pressure(psia) 4640

6 4630

12 4612

18 4584

24 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: permeability k water viscosity μw porosity
effective water compressibility =

= 120 milli darcies = 0.8 cp = 0.2 -6 1 x 10 psi-1

(i)

Calculate the average thickness of the aquifer sands.

(ii)

The cumulative water influx after 12, 18 and 24 months.

The Hurst & van Everdingen equation for a full radial flow

system is: [10]

We = 1.119φcR 2o hΔpWD

where We
p WD c Ro h Ø

= = = = = = =

cumulative water influx (bbls) pressure drop(psi) dimensionless water influx compressibility of the aquifer (psi-1) radius of the oil reservoir (ft) average thickness of the aquifer sands. (ft) porosity

Charts are supplied of dimensionless water influx WD versus dimensionless time tD (see 2 attachments) where t D = 2.309

t k μw

kt μ w φcR o2

= time ( years) = permeability (millidarcies) = viscosity (cp)

Aquifer Radius 40,000ft Oil Reservoir Radius 8,000ft Angle - 135º

22/07/14

B10 (a)

Describe briefly the drive mechanisms associated with producing an undersaturated oil reservoir, without a supporting aquifer, down to a pressure well below the bubble point. [6]

(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. In a further test a differential liberation procedure was carried out on a sample of the oil at 220 ˚F and the volumes of oil remaining and standard volumes of gas liberated at each stage were recorded. The results are presented in Table 2. From the results of these tests assuming separator conditions of 500 psig and 160 ˚F and 0 psig and 60 ˚F determine: (i)

The bubble point pressure of the reservoir fluid at 220 ˚F

(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.

(v)

The total formation volume factor at 3300 psig. [10]

22/07/14

5500 193.00

3643 198.46

2453 234.39

Pressure (psig)

Volume of Oil in Cell (cc)

Pressure (psig)

Volume of Oil in Cell (cc)

Pressure (psig)

Volume of Oil in Cell (cc)

252.81

2186

199.21

3594

194.97

5000

274.35

1968

200.85

3493

196.08

4500

302.11

1769

201.66

3446

196.42

4350

339.97

1580

204.67

3293

196.81

4173

415.31

1344

208.00

3146

197.21

4000

Table 1 Constant Mass Study of Reservoir fluid at 220 ºF

213.29

2954

197.68

3800

221.82

2713

198.00

3675

Table 2 Differential test at 220 ºF Pressure (psig) in Cell

Cumulative Gas Produced

Volume of Oil

cc @ 60 º F & 0 psig

(cc at pressure)

0

177.25

2924

698

171.13

2265

1440

164.75

1605

2051

159.50

1095

2537

154.88

420

3322

147.13

0

3996

135.50

Bubble point pressure

@60ºF

125.00

B11 (a)

(b)

In reserve estimates what do you understand by Proven, Probable and Possible reserves. [6] 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.

(ii)

Construct the water saturation profile on the saturation height diagram. [7]

(iii)

Estimate the oil -in place per unit cross section area over the reservoir thickness.

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. Air/mercury capillary pressure = 10 x capillary pressure water/ oil. 1 bbl = 5.615 cu. ft 1 lbf = 1 lbm x g. psi = lbf / sq inch

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[2]

[5]

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:

(a) the relevant rock mechanical phenomena to be considered, and their impact on reservoir performance (b) the data set required to conduct the screening process (c) the software tools necessary

[9] [9] [2]

Section C 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. CA =

4A ⎡ ⎤ ⎢⎛ 4 π ⎡ re 3 ⎤⎞ ⎥ θro ln − + ln γ exp ⎢⎜ 2 ⎢ ⎥⎟ πθo ⎞ ⎥⎥ ⎛ ⎢⎝ θ ⎣ ro 4 ⎦⎠ 2 π sin ⎝ θ ⎠ ⎥⎦ ⎢⎣

where

A=

θ θ πre2 = re2 2π 2

or Derive the Hawkins equation for the damage skin factor in an openhole 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]

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Closed Outer Boundary

16 Well "Super Well" in a Large Circular Reservoir

C14

A two layer reservoir is produced commingled as shown in the diagram and it is desired to balance the production by the use of variable chokes; here balanced production implies that the rate from a layer is proportional to its thickness i.e. q1 /h1 = q2 /h2 . This is the basis of “smart” wells used to optimise production behaviour. Each choke diameter may be continuously varied from fully open to closed. Given a vertical lift curve for the well i.e. a plot of bottom-hole flowing prerssure, pwf
, versus well total flow-rate, q
, where q = q1 + q2
, show how the choke settings may be calculated to give maximum balanced production. Assume that the choke performance is given by an equation of the form: q i = Cch At

2p  1




2 

where p = pressure drop over choke q i = flow-rate through device Cch = discharge coefficient (constant) 2 At = throat area = D t /4 Dt  = D p Dt = throat diameter Dp = pipe diameter

C15

Describe the phenomenon of “supercharging” and the procedures which may be used to obtain a useful WFT survey in a low permeability reservoir [20]

End of Paper

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[20]

q = q1 + q2

q

q 2

q1

S2

S1

k2 h2

k1 h1

Layer 2

Layer 1

Balanced Production Using Variable Chokes

1

Q14

p2

p

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BEHAVIOUR OF OIL FIELD HYDROCARBON SYSTEMS Pseudo Reduced Pressure, Pr 1.1

1

0

2

3

4

5

6

7

8

1.1

Pseudo Reduced Temperature 3.0 2.8 2.6 2.4 2.2 2.0 1.9 1.8

1.0

0.9

5

1. 4

1.

05

1.7

1.

1.6

1.7

1

0.8

1.3 1.1

1.0 1.05 1.2 0.95

1.

1.5

0.6

2

1.35 1.3

1.2

1.7 1.6 1.9 1.8 .2 2.0 2

0.5 1.15

3.0

2.6

1.2

1.05

0.25 2.8

1.1

3.0

1.1 Compressibility of Natural Gases (Jan. 1, 1941)

2.6 2.4 2.2 2.0

1.0

1.8 1.7 1.6

0.9 7

1.4

1.3

2.4

1.1

0.3

1.5

4 1. 1.5

1.25

0.4

1.6

3

1.4

1.

0.7

1.

Compressibility Factor, z

1.45

1.9

1.2

1.0

1.1 1.4 1.3

8

1.05

0.9 9

10

11

12

13

Pseudo Reduced Pressure, Pr

14

15

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