fundamentals of petrophysics ch 1-5

CRAFT & HAWKINS DEPARTMENT OF PETROLEUM ENGINEERING

fundamentals of Petrophysics

TAMMY BOURGOYNE, PHD., P.E.

AND

JULIUS LANGLINAlS, PhD., P.E.

FIRST EDITION

PETROLEUM SERVICES INTERNATIONAL, PUBLISHING DIVISION

BATON ROUGE, LOUISIANA

2003

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© 2003 by Petroleum Services International. All rights reserved. No part of this document may be reproduced, in any form or by any means, without permission in writing from the publisher. Contact Petroleum Services International, Publishing Division at 225-766-6536.

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Table of Contents

1. INTRODUCTION .................................................. .4 9. ELECTRICAL RESISTIVITY OF ROCK ......... 66

2. REGARDING THE GREEK ALPHABET .•.••••..•. 5 FORMATION RESISTIVITY FACTOR ............................ 69 FORMATION RESISTIVITY INDEX ............................... 72

THE GREEK ALPHABET ............................................... 5 WATER SATURATION DETERMINATION ..................... 72

NOMENCLATURE ......................................................... 5 EFFECT OF CLAY IN ROCK ......................................... 74

3. DIMENSIONAL DECEPTIONS ........................... 7 10. ABSOLUTEPERMEABILITY ••••••••.••.•.••••••..•.•• 77

UNlTS AND DIMENSIONS : ........................................... 7 HISTORICAL BACKGROUND ....................................... 78

UNITS OF LENGTH AND MAss ..................................... 8 DARCY'S LAW EXTENDED ......................................... 79

UNIT CONVERSION ...................................................... 6 THE UNITS OF PERMEABILITY ................................... 83

UNITS OF FORCE .......................................................... 9 PERMEABILITY IN RESERVOIRS ................................. 85

WEIGHT .................................................................... 12 EFFECT OF NET OVERBURDEN .................................. 85

UNIT SYSTEMS .......................................................... 12 RELATIONSHIP TO RESISTIVITy ................................. 86

CONSISTENT UNITS & EQUATIONS ............................. 15 EFFECT OF CLAY DISTRIBUTION ............................... 87

4. PHYSICS OF FLUIDS REVISITED •...••••...••.••••. 20 11. APPLICATION OF DARCY'S LA W ................ 90

STATES OF MATTER .................................................. 20 LINEAR FLOW ............................................................ 90

DENSITY AND SPECIFIC GRAVlTY ............................. 20 RADIAL FLO\)' ........................................................... 91

IDEAL GAS LAW ............................. , .......................... 21 ApPARENT PERMEABILITY IN SERIES ........................ 92

REAL GAS LAW ......................................................... 22 ApPARENT PERMEABILITY IN P ARALLEL. .................. 94

IDEAL LIQUIDS .......................................................... 23 DIRECTIONAL NATURE OF PERMEABILITY ................ 93

SLIGHTLY COMPRESSIBLE FLUIDS ............................ 24 FORMATION VOLUME FACTOR .................................. 27

12. EFFECTIVE PERMEABILITY ...................... 100

HYDROSTATIC PRESSURE .......................................... 27 EFFECTIVE PERMEABILITY ........................................ 98 ARCHIMEDES' PRINCIPLE .......................................... 27 RELATIVE PERMEABILITY ....................................... 101

VISCOSITY ................................................................. 28 EFFECT OF SATURATION HISTORY .......................... 102 FLUID FLOW REGIMES .............................................. 28 POISELLE'sLAW ........................................................ 28

13. CAPILLARY FORCES ..................................... 107

5. PETROLEUM GEOLOGY REVISITED ........... 30 SURFACE TENSION (GAS-LIQUID) ........................... 105

INTERFACIAL TENSION (LIQUID-LIQUID) ................ 110 NATURE OF RESERVOIR ROCKS ................................. 30 WETTING AT LIQUID-SOLID INTERFACES ................ 112 PROCESS OF RESERVOIR FORMATION ........................ 31 RELATIVE WETT ABILITY ......................................... 113 RESERVOIR FLUID DISTRIBUTION ............................. 33 CAPILLARY RISE ..................................................... 115 RESERVOIRPRESSURE ............................................... 33

14. RESERVOIR CAPILLARY PRESSURE ....... 119 6. POROSITY ............................................................ 35

DISTRIBUTION OF FLUIDS IN RESERVOIR ................. 120

POROSITY DETERMINATION ...................................... 36 RELATIONSHIP TO WATER SATURATION ................. 125 SOURCES OF POROSITY ............................................ .43 EFFECT OF SATURATION HISTORY .......................... 124 FACTORS AFFECTING POROSITY .............................. .44 EFFECT OF HEIGHT ABOVE WATER LEVEL .............. 125

7. FLUID SATURATION •••••••••••••••••.••••••••••...•••••..••• 50 EFFECT OF PORE SIZE DISTRIBUTI(,)N ...................... l25

EFFECT OF DENSITY DIFFERENCE ........................... 126

IRREDUCIBLE WATER SATURATION .......................... 52 EFFECT OF INTERFACIAL TENSION .......................... 129

RESIDUAL OIL OR GAS SATURATION ........................ 52 EFFECT OF ROCK WETTABILITY .............................. 129

DISPLACEABLE PORE VOLUME ................................. 53 RECOVERY FACTOR .................................................. 53

15. PRACTICE & REVIEW PROBLEMS ........... 128

INITIAL OIL VOLUME ................................................. 54 BIBLIOGRAPHY .............................................. 145

8. COMPRESSIBILITY OF ROCK ........................ 55

NOMENCLATURE ........................................... 146

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Introduction An understanding of the properties of petroleum-bearing rocks and the interaction between these rocks'and the associated fluids provides the petroleum engineer with valuable insight when predicting and optimizing the production of oil and gas,

A s will be discussed in this course, petroleum-bearing rocks act as both the storage container and the initial transport medium or conduit for the production of oil and gas, Without a basic understanding of the properties of petroleum-bearing rocks and the fluids associated with them fluids

and their interactions, petroleum engineering calculations become meaningless and abstract, We first study rock properties or petrophysical properties (PETE 2031), then fluid properties (PETE 2032), and then these are then combined in reservoir engineering (PETE 4051 & 4052),

Furthermore, the science of petrophysics is directly applied in formation evaluation (PETE 4088) which strives to link rock properties which may be measured directly using logging and core analysis to other properties of interest to the petroleum engineer which may not be directly measured underground. In petrophysics, the properties associated with petroleum-bearing rocks and the phenomena associated which the interaction between fluids and the rock are studied.

This text is not intended to be an all encompassing survey course. The purpose is to firmly establish a foundation in petrophysics, which permits the development of the more advanced concepts of Petroleum Engineering.

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Regarding the Greek Alphabet As in mathematics and other sciences, Greek letters are a commonjy used short-hand notation to represent physical properties and concepts in engineering.

Scientific and technical fields such as engineering requite a large number of symbols to represent physical properties and concepts. The use of the Greek alphabet to provide a set of useful symbols in addition to the Roman alphabet used to communicate ideas in the English and other languages be

gan hundreds of years ago when Greek and Latin were commonly learned by scholars.

THE GREEK ALPHABET

Today Greek is rarely a topic of study in the conventional public education curriculum. The vast majority of us are introduced only to the Greek alphabet through their use as symbols to represent abstract ideas in mathematics. Those who go on to more advanced levels of study in technical fields such as engineering become familiar with their use as a short-hand notation to represent physical properties and concepts. Table 2.1 lists all of the letters in the Greek alphabet, upper-case and lower-case, with their names and pronunciations. The pronunciations listed are those commonly used today in English speaking countries.

NOMENCLATURE

Familiarity with the Greek Alphabet will enhance your ability to express and to interpret technical concepts in the study and practice of engineering. In general, the property or concept represented by each symbol is not universal. Each discipline has its own customary usage for the Greek alphabet. For example, in petro

leum engineering the Greek letter ~ (pronounced "FEE'') traditionally represents the physical property of a rock called porosity. However, the same letter in thermodynamics, the study of heat transfer, represents a property referred to as availability which has nothing at all to do with the porosity of a rock. Therefore, to avoid miscommunication good engineering practice requires that the particular use

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of each symbol is defined on a case -by -case basis whenever symbols are used to

convey ideas. This is commonly done in textbooks and technical papers as a listing of each symbol used along with a description of the property or idea it represents. This listing is commonly called the Nomenclature. Nomenclature is given as needed throughout this text. An abridged listing of the nomenclature used in this text is also given in Appendix C.

TABLE 2.1: LETTERS OF THE GREEK ALPHABET

Upper- Lower- Name Pronunciation Example of Usage case case (Upper/Lower)

A a ALPHA "AL-fuh" / Angle in geometry

B ~ BETA "BAY-tuh" / Angle in geometry

r y GAMMA ~'GA1vI-uh" Torque/Specific gravity

L'. 8 DELTA "DEL-tuh" Change / partial derivative

E E EPSILON "EP-sil-on"

Z 1; ZETA "ZAY-tuhP

H 11 ETA "AY-tuh" /Mobility Ratio

<9 e THETA "THA Y -tuh" / Contact Angle

I t IOTA "eye-OH-tuh"

K K KAPPA "KAP-uh"

A Ie LAiVlBDA "LAM-duh" /Mobility Ratio

M f.L MU "1vfYOO" /Viscosity

N V NU "NOO" /velocity

'" C, XI "KS-EYE"

0 0 OMICRON "OM-i-KRON"

IT n PI "PIE" /3.142

P P RHO "ROW" /Density

L IT SIGMA "SIG-muh" Summation /Surface Tension

T 1: TAU "TAW' /Tortuosity

y tJ UPSILON "OOP-si-LON"

<D cI> PHI "FEE" Flow Po\ential/Porosity

X X CHI "I <-EYE"

'¥ 'l' PSI "SIGH"

n OJ OMEGA "Oh-MAY-guh" Ohm (unit)/

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Values combined

with units convey a physical quantity_

©2003. All RI

Dimensional Deceptions Numerical values alone do not convey a pqysical quantity. Both a numerical value and a unit must be stated together to convey meaningful pqysical information.

A n inherent source of error and confus-ion for the engineer is the use of dimensional unit systems, particularly inconsistent ones. The engineer must use both metric and English units because of the differences between science and technology. Routine conversions between these units

are only mildly troublesome, but the intermixing of gravitational and absolute unit systems can easily complicate matters. For example, when pounds force and pounds mass appear in the same equation, confusion will frequently arise. To avoid these complications, the differences between these quantities - force and mass - should be clearly understood, and their units and dimensions watched closely in all formulas and calculations.

UNITS AND DIMENSIONS

Conveying the magnitude of a physical quantity requires 2 elements. First, a dimension which identifies the nature of the quantity is required. The dimension, such as lengrh or time, determines the choices available for the unit of measurement to be used. A unit provides a basis of comparison or a standard reference. For example, the dimension length might be measured in inches, feet, miles, meters, etc. For convenience and accuracy, the unit chosen usually depends qpon the scale of the dimension being measured. A relatively large distance might be reported in miles or feet while a relatively small distance might be measured in inches or centimeters.

Second, a numerical value which conveys how many of the standard reference units are in the physical quantity is also required to convey meaning. The unit of measurement chosen determines the magnitude of the numerical value. For example, 12 inches, 1 foot, and 0.3048 meters all represent the same distance, but the numerical value attached defining the number of unit lengths in the distance are not equal because the standard of reference is not equal. As a result, any

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A sufficient choice of primary quantities for most engineering work is mass, length, time, and temperature. All other quantities can be expressed in terms of these four.

©2003. ALL RIGHTS -RESERVED.

statement of a physical quantity having dimension is incomplete or meaningless without explicitly including the unit of measurement.

Consider the following scenario. In response to the question regarding the volume of oil produced, one student answers 42,000 and the other answers 5,615. The volume in question is 1,000 barrels. Neither student states what unit of volume he/she is using. Without stating the units, the answer is incomplete so its accuracy cannot be judged without further information. When prompted, the first student answered 42,000 gallons and the other student answered 5,615 cubic feet. As it turns out, both of these quantities are equivalent. That is, both represent the same volume and are also equivalent to 1000 barrels. Both students were correct and in agreement, but the numerical value alone could not convey the volume of oil produced. For this reason, units must always be stated explicitly in an engineering calculation.

The majority of physical quantities have dimensions which can be expressed in terms of the dimensions of a limited number of fundamental quantities. Velocity is defined as the time rate of change of position. Velocity can also be expressed as a length divided by a time interval. Thus, velocity will require a unit for length combined with a unit for time. Likewise, other physical quantities may be expressed in terms of a set of primary quantities which cannot be broken down any further. A sufficient choice of primary quantities for most engineering work is mass, length, time, and temperature. All other quantities can be expressed in terms of these four primary quantities. Any quantity which is not one of these four primary quantities is referred to as a secondary quantity.

UNITS OF LENGTH AND MASS

The fundamental standards of length and mass designed for the entire world are the distance between two lines inscribed on a particular platinum-iridium bar and the mass of a certain block of platinum. Known as the international meter bar and kilogram mass, these standards are maintained at the International Bureau of Weights and Measures in Sevres, France. At present, the commonly used unit of time is the second, was defined as 1/86,400 part of a mean solar day. Customary English units of length and mass for the U. S. are defined by reference to the standard yard and avoirdupois pound. But, these have been statidardized by the U. S. Congress in terms of the international meter and kilogram mass, as:

1 yard = 0.9144 meters (i.e., 1 inch = 2.54 em exactly)

1 pound mass = 0.4535924277 kilogram mass

UNIT CONVERSION

A measurement in a particular set of units can be converted to an equivalent quantity using a different set of units of the same dimensions. This algebraic manipula-

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Learn and apply this technique to

greatly reduce unit conversion errors.

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tion is accomplished by multiplying the measurement by an appropriate "well chosen value of one".

Example Problem 3.1 A car is traveling at a speed of 90 miles per hour. Express its speed in (a) feet per second, and (h) meters per minute.

1hr First, we make use of the fact 1

60 min

1 hr ---, and any expression can be multi-60 min

plied by 1 without changing its value. Other values of 1 are used as needed.

Solution:

mile 0 mile hr min 5280 ft (a) 90--=9 --x x x

hr hr 60 min 60 sec (90)(5280) ft = 132~

(60)(60) sec sec mile

mile 90 mile hr 5280 ft 0.3048 meter (h) 90--= --x x x-'..:.-'--'-- 2414 meter

mm hr hr 60 min mile ft

When this procedure is used, the probability of error is greatly reduced. Each conversion involves the multiplication of the original quantity by a series of well chosen, dimensionless factors, each having a numerical value of unity (one). To verify this last statement, notice that the numerator of each factor is equal to its denominator. That is, the fraction [ (1 hour) / (60 min)] has a value of unity, and therefore can be used as a multiplier since multiplying by 1 does not change the value of the original fraction.

A listing of some of the more commonly used conversion ratios or "well-chosen l's" are listed in Table 3.1. Using these ratios and the technique illustrated in Example 3.1

UNITS OF FORCE

It would be difficult, if not impossible, to maintain a fixed quantity of force as an unvarying physical standard. For by its nature, the very existence of a force can only be inferred from the changes it produces in the state of motion of a material body. By Newton's law of motion, the acceleration, a, produced in a body of constant mass, m, is attributed to the action of a force, F, whose magnitude is proportional to the mass and acceleration of the body, or:

F oc rna

9

Length

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Since this is a proportionality, we can make the equation an equality by introducing a constant of proportionality, g" and write:

1 F=-ma

g, (3.1)

This equation involves four quantities, force, mass, length, and time. Only three need be fundamental, thus defining the fourth. If mass, length, and time are recognized as the primary quantities, Equation (3.1) defines the dimensions of the derived quantity, force, as a mass times a length divided by the square of a time interval.

More complex quantities, such as pressure and energy, are defined in terms of a force. We do not reduce such quantities to the primary quantities as a matter of convenience to avoid needless complexity. Instead, force is used as a convenient secondary dimension. The size of the unit force is determined by an arbitrary choice of acceleration effect it must produce in a given mass. Once these standard effects have been specified, the proportionality ~onstant, &' is adjusted to preserve the numerical and dimensional equality of the equation (3.1) expressed above.

TABLE 3.1: C01-hI10NLY USED CONVERSION RATIOS OR tlWElL-CHOSEN 1 'S".

Mass Area Viscosity Volume

1 foot = 12 inches 1 kg - 1 000 grams 1 Acre = 43,560 feet' 1000 cp - 1 Pa-s 1 gallon - 231 in'

, I

1 yard - 3 feet 1 pound - 453.6 grams 1 barrel - 42 gallons I 1 mile = 5280 feet 1 Acre ft - 43,560 ft' , 1 inch - 2.54 cm_

100 cm = 1 meter

Example 3.2 A particular unit of force, called a dyne, is defined as that force required to accelerate a gram of mass at the rate of one centimeter per second squared. Calculate the value of the constant, gc, to be used with set of units.

Solution:

When the above definition is introduced into Equation 2.1, we have, by definition

1 1 2 F = -rna = -(1 gm)(1 cm/sec ) = 1 dyne

gc gc

To preserve dimensional as well as numerical equality, gc must have the value:

g = 1 gm(cm/sec2

)

c dyne

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I

Remember that the constant, ge, is not an accelera .. tion term, but a conversion factor between the unit of force and its assigned primary units.


Thus, we see that the constant go serves the twofold purpose of (1) defining the standard effects expected of a certain nnit force, and (2) identifying this unit by name. Since mass and force are not independenr quantities, the conversion factor go will appear in all problems that involve dynamics, for it allows the simultaneous use of both quantities in the same equation.

Example 3.3 Develop an expression for the kinetic energy Ek of a body of mass m, moving at a constant velocity v using the definition of Work = force x disrance.

Solution:

Assume the body to be originally at resr. Upon application of a steady unbalanced force (F) acting through a sufficient distance (s), the body will acquire the required velocity (v). In accordance with Equation (3.1), this force gives the body of mass m a constant accel-eration a:

1 F=-ma

g,

and the distance s traveled by the body while acquiring the velocity v is relared to its acceleration as follows:

Unreliable mental conversions are eliminated by the presence of the constant gc. The definition of each force unit is contained within the definitiDn of gc for each system of units.

The energy imparted to the body is then

Ek =work=FS=(maJ(~J= mv2

= gm (em/sec): g, 2a 2g, 2gmem/sec

dyne

mv2

--dynecm 2

This last equation, as given in elementary physics, probably did not contain go. But the dimensions of the result, (m L' / t') or (gm-cm2/s2 ), are not associated with energy until one makes the mental conversion 1 dyne:= 1 gm-em / 52, Unreliable mental conversions are eliminated by the presence of go.

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. WEIGHT

The force of attraction which the earth exerts on a material body is called the weight of the body. Being a force quantity, the weight, W, of a body of mass, m, may be calculated using Newton's law in the form

1 W=-mg

g, (3.2)

where g is the acceleration which gravity would impart to the body in a free fall. Unfortunately, the term "weight" is sometimes used improperly as a synonym for mass. Therefore, it cannot be over-emphasized that weight has the dimensions of force.

UNIT SYSTEMS

Each time an arbitrary choice of basic units is made for mass, length, and time, a new system of units is established together with' an opportunity for defining a new unit of force. Unit systems commonly used in scientific and engineering calculations are listed in Table 3.2. While each system is self-consistent, the various systems are not compatible with each other. The units of force are defined by the following statements:

• •

•

•

•

A force of 1 dyne will accelerate a one gram mass at the rate of 1 cm/ S2.

A force of 1 gram force (gf) will accelerate a one gram mass (gm) at the rate of 980.665 ft/ S2.

A force of 1 pound force (lbf) will accelerate a one pound mass (Ibm) at the rate of 32.174 ft/ S2.

A force of 1 pound force will accelerate a mass of one slug at the rate of 1 ftls'.

A force of 1 Newton (N) will accelerate a mass of one kilogram (kg) at the rate of 1 mls'.

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Remember that the constant gc is

not an acceleration term but is a conversion factor between the unit

offorce and its assigned primary units.

•

Quantity

IvIass

Length

Time

Force

Conversion

constant, gc

Velocity

Accelera-tion, gravity

Area

Volume

Density

Flow Rate

Specific Volume

Absolute Viscosity

Energy (work)

Power

Pressure

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TABLE 3.2: C01:1J\[ON PHYSICAL QUAl'\lTITIES AND THEIR DIlYIENSIONS AND UNITS.

Symbol CGS SI Engineer - Engineer- Technical

Metric English English

M gram, g kilogram, Gram (mass), Pound (mass), Slug kg gm Ibm

L centimeter, Meter, m centimeter, foot, ft Foot, ft em em

T second, s second, s second, s second, S Second, s

F; ML dyne Newton, gram (force), Pound (force), Pound

r' Nt gf lbf (force), lbf

ML gem 1 kg m gem 32.17lbmft slugft g, = FT' 1 1--980.7--2 Ibf s'

dynes 2 Nts2 • gf s lbf S2

v=L/T em/s m/s em/s ft/s ft/s

g= L/T' g=980.7 g = 9.81 g = 980.7 g =32.17 g = 32.17 em/52 mis' em/52 ft/s' ft/s'

U em' m' em' ft' ft'

L' em3 m3 em3 ft3 ft3

p=M/L' g/em3 kg/m3 gm/em3 lbm/ ft3 slug/ft3

Q = L3/T cm3/sec m3/sec cm3/sec ft3/sec ft3/sec

V =L3/lbm ----- ---- ----- ft3/lbf ft3/lbf

-g- ~ gm lbm slug M --fi= LT em s ms em s ft s ft s

(pa s) (poise)

E=FL Dyne-em Nt-m gkm Ft-Ibf Ft-Ibf (erg) Goule)

P=FL/T erg/s J/s gf-em /s Hp = 550 Hp=550 (Watt) ft-Ibf/s ft-Ibf/s

P=F/L' Dyne/em2 Nt/m' gf/cm 2 lbf/ft' Ibf/ft2

(pascal)

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Even though the numerical values of mass and weight are often equal, the dimen

sions and units of the 2 quantities are not.

To eliminate confusion, the pound mass is often ab .. breviated as Ibmthe

pound force as Ibf, and the gram mass asgm.


FOOD FOR THOUGHT Frequently, the terms )ve;{ghtand mass are often used interchangeably. Notice that in those unit systems wherein the acceleration due to gravity was included "in the definition (i.e., gram force, pound force) the mass of an object located on the surface of the earth is numerically equal to the weight. In other systems of units, the force unit is defined such that the value of the conversion constant gc is 1. For these reasons, the conversion constant, g, is typically dropped from equations relating mass and force. However, when this conversion constant is 1, weight and mass are not numerically equaL Even though the numerical values of mass and weight (when measured on the surface of the earth) are often equal, the dimensions and units of the 2 quantities are not.

Inclusion of the constant gc serves as a reniinder of the relationship between the quantities of mass and weight and their units. In addition, its inclusion is required when the object is not strictly located on the surface of the earth! Furthermore, a "weight" reported in kilograms is really a mass and is NOT numerically equal to its true weight in Newtons in the SI system of units. For example, a mass of 1 kg weighs 9.80 N on the surface of the earth.

The units and numerical value of g, are obtained by substitution of the defining statements of the force units into the general form of Newton's equation as previously illustrated in Example 3.2. These values are given along with the primary and derived units for each system are listed in Table 3.2. Remember that the constant & is not an acceleration term but is a conversion factor between the unit of force and its assigned primary units. This is illustrated in Examples 3.4.1 and 3.4.2. Note in each example that each ratio has a value of 1, so the equality is preserved. In this way, the conversion factor between any quantity within or among any system of units can de determined using a few "well-chosen ones" either from memory or using Table 3.1 and Table 3.2.

Example 3.4.1 Calculate a factor for converting pounds of force to equivalent dynes of force.

Solution:

1 Ibf = 1 Ibf X 32.17 Ibm ft x 453.6gm x 2.54cm x 12in x dyne sec' Ibf sec' Ibm in ft I gm em

Ilbf= (32.17)(453.6)(2.54)(12) dyne = 444,773.7 dynes (I)

14

The equation is in balance dimensionally or is dimensionally coherentwhich must be true for any valid equation.


Example 3.4.2 Calculate a factor for converting pounds of force to equivalent Newtons (N) of force.

Solution:

llbf= 11bfx 32.17 Ibm ft 453.6 gm 1 kg 2.54 em 12 in m N s'

x x x x--x x---Ibfs' Ibm 1000 gm In ft 100 em I kgm

Ilbf= (32.17)(453.6)(2.54)(12) N = 4.447737 N (1000)(100)

CONSISTENT UNITS & EQUATIONS

Rather than an arbitrary choice of units for each secondary quantity, mechanical units not listed in Table 3.2 are derived from the units of mass, length, time, and force specified for each system. For example, area is defined as the square of length. As a result, the consistent unit adopted for area in the engineering English system is the square foot since the foot has already been selected as the unit of length. Likewise, the consistent unit choice for area in the CGS system is the square centimeter.

Similarly in the engineering English system, the consistent unit for volume is cubic feet, for pressure is pounds force per square foot (ps£), for volumetric flow rate is cubic feet per second, and for work or energy is foot pounds force. In the deriva- . tion of a physical equation, exclusive use of a consistent system of units, including the ubiquitous gO' has a distinct advantage. The resulting expression can be used with any other system of consistent units without further modifications.

Example 3.5 The pressure drop required to maintain a fluid in laminar flow through a horizontal pipe is given by the equation:

where p denotes pressure, fl denotes viscosity of the fluid, L denotes length of the pipe, V

denotes velocity of the fluid, and d denotes diameter of the pipe. Referring to the second column of Table 3.2, let us investigate the dimensional properties of this equation:

15

A good engineer always knows the units of each value given or calCUR

lated. Units may always be used as an error check or as a reminder of the definition or dimensions of any secondary quantity.


As shown above, the dimensions of the pressure term on the left shown to be the same as the combined dimensions of all terms on the right. Therefore, the equation is in balance dimensionally or is dimensionallY coherent which must be true for any valid equation.

This dimensional balance may also be shown by substituting consistent units from any sys· tern of units. Referring to Table 3.2, repeat the process substituting units from the engineering English system:

lbl Ibm ft --=--x-x ft' ft s 1

ft Ibf s' 1 lbl -x x--=--s Ibm ft ft' ft'

As a consequence, consistent units from ~ system may be used in this equation as given, without any other conversion factors or additional qualifying remarks.

Empirical equations based on experimental data rather than derived mathematically from fundamental engineering or scientific principles, as well as physical equations which have been modified to accept certain quantities expressed in more Hpractical" units, are usually inconsistent. These equations cannot stand alone, but must be ac!,ompanied by a detailed listing of particular units to be used with each different quantity.

Example 3.6 Modify the equation for pressure drop presented in Example 2.5 to accept the following "field units" or preferred set of inconsistent units:

Quantity: LIp !l L v d

Desired Unit: lbf/in' , psi centipoise (cp) ft ftls in

Solution:

Since the original equation as given will accept any consistent set of units, the desired "field" units will be converted to consistent engineering English units for substitution into the original equation.

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.

Quantity: "'p fl L v d

Desired Unit: Ibf/in2 , psi centipoise (cp) ft ft/s In

Consistent Unit: Ibf/fe Ibm/ (ft s) ft ft/s ft

Let us adopt the convention that the original variable (i.e., "'p) represents a value for the quantity expressed in the consistent unit while the same vatiable primed (i.e., "'p') represents the correspondingvalue for the quantity expressed in the desired unit. Note that the numerical values are not equal, but the physical quantity represented by the value and unit combination are equal. For example, !'J.p '" !'J.p' numerically, but the physical pres

sure drop represented is the same (1Ibf/ft2 = 1/144Ibf/in2).

'" Ibf = '" ,fbi X (12in )' .p ft' .p in' (ft)' ('" ,lbl)(i 44 in' ) = 144('" ') lbl

.p in' ft' .p ft'

Ibm, Pa s (N/m') kg ill Ibm • 1000 gm ill 2.54 em 12 in f.l --=J..l cp x X x--

2-x _ X X X x--

ft s 1000 ep Pa N s 453.6 gm kg 100 em in 1ft

Ibm, 1 1 1000 2.54 1 12 ( p' ) Ibm Ji fts =Ji cpx lOOO x 453.6 x-I-x-I-x IOO x-I = 1488.2 fig

Lft=L'ft

ft , ft V -=v -

s s

d ft = d' in x J..!!... = (£) ft 12 in 12

These variables representing values in the desired units have now been expressed in terms of the consistent unit. Because these expressions are in terms of a set of consistent units, they may now be substituted into the original equation.

( ') (32) P LV

(144!'J. ') = 1488.2

.p ( ')' 32.17 ~2

(32)(12)' jlL'v'

(32.17)(1488.2) (df

!'J. ,_ 0.09625 p'L'v' P - 144 (d')'

=> !'J.p' = 0.00066840 p'L'v' ( d')'

0.09625 p'L'v' (d'?

where all values must be in specified "desired" units. Our final equation is no longer consistent, for all quantities must now be expressed in the desired units specified, or else the equation will give a wrong answer.

17


FOOD FOR THOUGHT One might ask, "\'V'hy must we go through all the trouble of altering the equation to fit our desired units? ... Wouldn't it be straight forward to simply convert the values to the proper consistent umts before inserting them into the equation? ... \'7hy alter the equation and then be forced to remember the "practical" units for the altered form of the equation?"

In answer, a hint is given in the common names for these inconsistent sets of units which are often referred to as "practical" or "field" units. These "practical" or "field" units, as the names imply, are the units in which quantities are measured by the practicing engineer or technician on site in the oil field or laboratory. Suppose that as an engineer or technician or field hand, you wish to estimate the pressute drop through section of horizontal pipe in the field. The pressute itself in the section of pipe of interest cannot be readily measured, but you can easily measure or already know the viscosity of the fluid in centipoise (cp), the length of the pipe section in feet (ft), the velocity of the flowing fluid inside the pipe in feet per second (ft/ s), and the diameter of the pipe in inches (in). Thus, you have all the information you need to estimate the pressure drop . . Now further consider that this is a routine calculation which you perform often. You have !\Va options:

• Convert each measured value to its corresponding consistent Engineering English unit each time you need to calculate the pressure drop and then convert the calculated pressure drop in Ibf/ ft' to the more familiar Ibf/in2 or "psi" so that the value will have physical meaning to you since psi is the pressure unit that you work with on a daily basis, OR

Convert the equation itself for use with your set of units as shown in Example 2.6 and use the measured values directly in the equation without need of conversion. In addition, the computed answer js in the desired units of "psi".

Keeping in mind that this calculation is one that you make often, the second option is obviously preferable.

18


Common Oil Field Units and Measurements

Linear and Distance Measurements

Depths and lengths in the U.S. are measured in feet, but world wide, the meter is typically used. By definition, 1 inch = 2.54 em, or

. (2.54cm) 1£t= 12m in = 30.48 em

or 1 ft = 0.3048 meters. Inversely, 1 meter = 3.28 ft

For example, a 10,000 ft well is

(lmeter) 10,000 = 3048 meters 3.28 ft

Tubing diameter is measured in inches or em, e.g., 2 ;3/8 inch aD tubing has a 1.995 inch

. (2.54cm) ID, and 2.375m in = 6.0325 em (60.325 mm) and 1.995 in ID = 5.067 em

(50.67 mm)

Area

5280 2 (ft/mD2 An acre is defined as 1/640 of a square mile or 2 = 43,560 ft2/acre (/1.

640 acres/mi section of land is defined as a mile by a mile in area (mile2), and again by definition, the square mile of area contains 640 acres).

Thus 1 acre = 43,560 ft2 (1 m/3.28 ft)2 = 4048.9 m2

Since a hectare is 10,000 m2 (100 m x 100 m), then 1 acre = 0.405 hectares,

or,1 hectare = 1/0.405 = 2.47 acres

Flow Rates

Liquid Flow Rate

Typically, flow rate is measured in barrels, which is defined to be 42 gallons. A gallon is defined to be 231 in3, or a barrel is

1 bbl(42 gal)(231 in3

J( ft3

J=5.615ft 3

bbl gal 123 in3 ,

19

©2 __ 003. All RIGHTS RESERVED.

(1 bblJ( lday J( lhr ) . 1 bbl/day = -- --- . =0.000694 bbls/mm = 0.029 gal/min day 24 hrs 60 mm

( 5.615ft3 J(0.3048m)3

For flow rates in m3/day, start with 1 bbl, or = 0.159 m3/bbl bbl ft

Inversely, 0.159 m3/bbl implies 6.29 bbls/m3. Thus, a well flowing 500 m3/day of oil is a rate of 3,145 bbls/ day. A well flowing 400 bbls/day is flowing 63.6 m3/day.

Gas Flow Rate

Gas flow rate is given in standard cubic ft per day, or scf/ d. We also use msef/ day for 1000 scf/ day, and mmscf/ d for 1,000,000 scf/ day. To convert flow rate to a different pressure, we use

P, v, PV --=--1; Z, TZ

where

P - pressure, psia

v - Volume, ft3

Z - gas deviation factor (Z = 1 at standard conditions)

s - refers to standard conditions, eg, 14.7 psia and 60 deg F (520 deg Rankine)

For example, a well flowing 10 mmscf/d has a flow rate at 1000 psia and 100 deg F of

14.7(10,000,000) (1000)V ----''----''-'--- or V = 140,260 ft3. Therefore, the actual flow

(460+60)(1) (460+100)(0.886)

rate will be 140.26 mef/day (at 100 deg F and 1000 psia).

Density:

In addition to standard measures of density, such as Ibm/ ft3, petroleum engineers also use densities such as Ibm/gal. For example, fresh water has a density of 62.4Ibm/ft3• Converting this to Ibm/gal (ppg)

20


(62.4lbmJ(_ft_J(231in

3J =8.34 lbm/ al ft 3 12in3 gal g

Specific gravity is a very useful concept when comparing densities. In the case of liquids and solids, their densities are compared to the density of water, or

density of material Specific gravity = ---"-----

density of water

For example, a specific gravity = 2 implies the material is twice as heavy as water (twice the density). A 16 ppg mud has a specific gravity of 16/8.34 = 1.92. A cubic foot of this mud would have a mass of 1.92 (62.4) = 119.7 Ibm, which would weight 119.7(g) / g, = 119.7Ibf.

Oil density is typically given in degrees API, which is a scale varying from 0 upwards to values such as 60 or so (larger the number, the less dense the oil). The relationship between specific gravity and API is

141.5 Specific Gravity = -----

131.5+ API

Pressure

Engineers typically measure pressure in Ibf/in2, but the engineering system of units requires Ibf/ ft2. A factor of 144 is needed, that is,

1 lbf ( 144in2 J= 144 Ibf

in2 l ft 2 ~ ft 2

The SI u~t of pressure is the Pascal, or Nt/m2• to convert from SI to psi, we use

1.1~f ( in )2 [4.4477 Nt](100em)2 = 6,894Nt/m2 In 2.54 em Ib f m

This is a rather large number, so typically we USe the unit of kilopascal, or 1000 pascals. Thus, 1 psi = 1 Ibf/in2 = 6.894 kilopascals. For example, a pressure of SilO psia is 500 (6.894) = 3447 kpa .

21

©2003. ALL RI

Physics of Fluids Revisited A general understanding of some basic definitions and concepts in fluid mechanics is helpful bifore discussing the nature of petroleum accumulations and the rocks associated with them.

M uch of the material presented in this chapter is a review of material covered in previous courses in physical science or physics. Specifically, several fundamental definitions and concepts related to fluid mechanics will be revisited before proceeding with a discussion of the nature of

petroleum accumulations and the rocks associated with them. In addition, a few definitions specific to petroleum engineering have been included.

STATES OF MATTER

Matter exists is 3 ordinary states which are solid, liquid, and gas. The 3 states are distinguished by the degree to which each maintains its shape and volume. Except under crushing forces, a solid maintains its shape under an applied force. Aliquid is also able to maintain its volume almost constant under pressure, but its shape is dictated by the container or other forces acting on it. A gas maintains neither a fixed shape or volume. A gas expands or compresses as needed to exactly fit its container. A substance which is able to flow is referred to as a fluid. Thus, a fluid can either be a liquid or a gas.

DENSITY AND SPECIFIC GRAVITY

Density is defined as mass per unit volume and is often denoted by the lower case of the Greek letter ''RHO'', p. Density may be expressed mathematically as follows:

m p=

V (4.1)

Specific gravity of a liquid or solid is defined as the ratio of the density of the solid or liquid to the density of water at 40°C. Specific gravity is a .dimensionless num-

22

©2003. ALL RIGHTS RESERVED

ber often denoted in petroleum engineering by the lower case of the Greek letter "GANIMA" and may be expressed mathematically as follows:

p . Y=

Pw@40'C (4.2)

Similarly, the specific gravity of a gas is defined as the ratio of the density of the gas to the density of air at a given pressure and temperature. Gas specific gravity is most often defined at standard conditions. It is also a dimensionless number and is also often denoted in petroleum engineering by the lower case of the Greek letter "GAtvIMA" with the subscript "g" to denote gas and may be expressed mathematically as follows:

IDEAL GAS LAw

Pg@p,T Y '=

g @ T Pair p,

(4.3)

The Ideal Gas Law is a combination of Charles' and Boyle's laws which expresses the changes in g;g; volume as a result of changes in pressure and or temperature. The Ideal Gas Law may be used for real gases at pressures close to atmospheric. The Ideal Gas Law may be expressed mathematically as follows where p is the absolute pressure, V is the volume, n is the number of moles of gas, R is the uruversal gas constant and T is the absolute temperature.

pV = nRT (4.4)

"When pressure is measured using a gauge the pressure reading is relative to atmospheric pressure. A gauge indicating a pressure of Zero on the surface of the earth would be indicating an absolute pressure equal to atmospheric pressure. Atmospheric pressure depends upon elevation and also changes slightly from day to day. It is customary to use a standard or reference value for atmospheric pressure when determining a gas volume at standard conditions. A standard atmosphere of 14.7 psi is generally used when applying the gas law. This value will be approximately equal to atmospheric pressure at sea level. However, when working with gases in a laboratory, measuring and using the actual barometric pressure in the laboratory is necessary to accurately determine the absolute pressure from the pressure gauge readings.

The pressure indicated by the gauge is often referred to as the gauge pressure. Absolute pressure is equal to the gauge pressure plus a pressure equal to 1 atmosphere. For example when pressure is in units of pounds force per square inch (psi), absolute pressure is equal to gauge pressure plus 14.7 psi.

Absolute p = Gauge p + 1 atmosphere (4.5)

23


Often a gauge pressure reported in petroleum engineering practice in units of psi will be reported in "units" of "psig" where the "g" serves as a reminder that the pressure is a gauge pressure reading not an absolute pressure. Likewise, absolute pressures are often reported in "units" of "psia". Note that each pressure is actually in ~he same pressure units of "psi" and that the "g" or "a" serve only as a shorthand reminder to convert the pressure from a gauge reading relative to atmospheric pressure to an absolute pressure when needed.

The Society of Petroleum Engineers (SPE) now discourages this practice and suggests that the word "pressure" be preceded by the qualifier "gauge" or "absolute" in the text or heading. For example, a pressure should be reported as an absolute pressure of 1000 psi rather than a pressure of 1000 psia. Likewise, pressure reading should be reported as a gattge pressure of 1 000 psi rather than a pressure of 1000 psig.

The unit of absolute temperature most frequently used in petroleum engineering is degrees Rankine ~R) which is related to degrees Fahrenheit by the following relationship:

(4.6)

The numerica1 value of the universa1 gas constant (R) depends upon the units chosen for the other variables. For practica1 petroleum engineering units where pressure is in psi, volume is in cubic feet, and temperature is in oR, the va1ue an units of R are as follows:

R = 10.73 psi ft3 oR

Recall that moles (n) are defined as follows:

m

(4.7)

n=- (4.8) M

where m is the mass of the gas and M is the molecu1ar weight of the gas. Molecular weight is a physica1 property which is constant for a given gas ..

REAL GAS LAW

Natura1 gases do not follow the Idea1 Gas Law closely enough to allow it's use at the higher pressures commonly found in the subsurface of the earth. To more closely approximate the behavior of natura1 gases at high pressure, a deviation factor is added to the Idea1 Gas Law. This modified form of the Gas Law referred to as the Real Gas Law. The deviation factor is commonly designated by the letter Z and is commonly referred to in petroleum engineering as the :cfactor.

24


pV=znRT (4.9)

. This deviation factor is not a constant but is a function of pressure, temperature, and the gas composition. The z -factor is the ratio of the actual volume of a gas to the volume of the gas as if it were behaving ideally at a given temperature and pressure. At low pressures, natural gases approach ideal behavior and as a result the z-factor approaches 1.0. The z-factor is defined as follows:

Actual volume ocuppied by n moles of gas @ p, T z = ----------"-"---"-----"'---"'-'-''----

Ideal volume ocuppied by n moles of gas @ p, T (4.1 0)

Since the value of z-factor is a function of pressure and temperature for a given gas composition, these values must be determined experimentally. Tables, charts, and correlations (equations) based on these experimental measurements have been compiled and published in various handbooks and other literature. An example of one of these correlations for estimating the z-factor for a natural gas is listed as a set of equations in Table 4.1. •

IDEAL LIQUIDS

Under isothermal (constant temperature) conditions, an ideal liquid is one which has a constant compressibility factor, c. The compressibility factor expresses the change in either volume or density with changes in pressure. In equation form, the compressibility factor of an ideal liquid is defined as follows:

1 dV 1 dp C = ----;or:c =--

Vdp pdp (4.11)

where V is the original liquid volume, p is the liquid density and p is the pressure. A useful form of this equation may be obtained by separating the variables and integrating between a pressure where the density is known and a pressure where the density is desired. Putting this result in exponential form we get the following:

p = poec(p-po) (4.12)

By inspection of Equation (4.11) or (4.12), the compressibility factor for a fluid whose volume or density is constant as a function of pressure is equal to zero. A value of zero for the variable c in Equation (4.12) would yield a density equal to the original density regardless of pressure. A liquid with a compressibility factor of zero is called incompressible. As the term implies, an incompressible liquid cannot be compressed. Neither volume nor density will change no matter how much or litde pressure is applied.

25


'L\BLE 4.1: EXAMPLE Z-FACTOR CORRELATION

PARAi\fETER NAME SYNIBOL i\.t"'lDjOR EQUATION

PSUEDO-CRITlCAL Ppc PRESSURE;

PSUEDO-REDUCED p PRESSURE: Ppr=-

Ppc

TfuvfPERATURE, oR: T PSUEDO-CRITlCAT.

Tpc TEMPERATURE, oR:

PSUEDO-REDUCED T TENfPERATURE; Tp'=-TpC

Z-FACTOR (Z) Z = A+(l- A)e(-B) +C(p~) where:

A =1.39~I;" -0.92 -0.36I;" -0.101

B=B1+B2+B3 . B1= pp,(0.62-0.23I;,,)

B2 = 2 [0.066 _ 0.037] PP

' (I;,r -0.86)

6

B3=0.32( . PP' ) J20.723 (Tl'-ll]

C = 0.l32- 0.32LoglO (I;,,)

D=eG

G = 0.715 -1. 128(I;" ) + 0.42(T),)

SLIGHTLY COMPRESSIBLE FLUIDS

A fluid with a very small compressibility factor (c) is referred to as slightly compressible. Liquids such as oil and water are slighdy compressible. Expanding the exponent term in Equation (4.12) by using Maclauren's series, we get: .

(4.13)

The series in Equation (4.13) may be truncated after the second term in the parenthesis to get a very useful equation for the density of slighdy compressible liq-

uidsas follows: p= Po(l+c(p-po)) (4.14)

26


FORMATION VOLUME FACTOR

The petroleum producing industry uses the formation volume factor, B, to relate fluid volumes in the reservoir to fluid volumes at surface conditions. This factor includes both compressibility and solubility changes with the change in pressure and temperature. The factor is defined simply as the volume occupied by the fluid at the reservoir pressure and temperature divided by the volume it would occupy at the surface pressure and temperature (most often taken to be standard conditions). Subscripts w, 0, and g are used to signiry water, oil and gas. The factor should be dimensionless, but this is not always the case --- especially for gas --- so beware of units!

The formation volume factor for oil is .defined as follows:

liquid oil volume in the reservoir B =--~------------------

o liquid oil volume in the stock tank (4.15)

The formation volume factor for gas is defined as follows:

B = volume ocuppied by gas in the reservoir

g volume occupied by gas at standard conditions (4.16)

HYDROSTATIC PRESSURE

Suppose for a moment that you are swimming in the ocean. As you swim deeper below the surface of the water, the pressure on Y0ll! body increases. Just as in the deep end of a swimming pool, Y0ll! ears begin to hurt if you go deeper than about 10 feet. The pain is caused by the increase in pressure as you submerge in a body of water. As the water depth increases so does the pressure. This pressure due to a change in elevation within a motionless (static) body of water is referred to as hydrostatic pressure. Hydrostatic pressure is the reason that a maximum limit exists for safe scuba diving. If you go too far down in the water, the pressure becomes greater than your body can withstand safely.

Hydrostatic pressure is due to the weight of the overlying water. Recalling that pressure is defmed as a force applied per unit area and that the weight of the overlying water can be related to the density of the water, the following expression for hydrostatic pressure can be found:

F W mg (pV)g p(Ah)g pgh p 0" - = - = - = -"-.-'-"-

A A Ag, Ag, Ag, g, (4.17)

27

Pressure change due to a change in elevation within a motionless (static) body of water is referred to as hydrostatic pressure.

~ I water

r······ )~· .... ~~···rIz:;:±;l i.1 hI

: ................................ "."" ....... i

water


Thus, hydrostatic pressure is directly related to the density (p) of the water (or other fluid) and the change in elevation (h) where the surface of the water has an elevation of zero and downward is positive.

The change in hydrostatic pressure per foot of change in elevation is called the hydrostatic gradient. By inspection, the hydrostatic gradient in a particular liquid is constant as long as the liquid's density is constant. The hydrostatic gradient in fresh water is about 0,433 pounds force per square inch per foot. The hydrostatic gradient in salt water (brine) is slightly higher since salt water is heavier than fresh water. The hydrostatic gradient in some sea water is about 0,465 psi/ft.

Example 4.1 A manometer is an instrument which measures pressure using the concept of hydrostatic pressure. Determine the pressure difference existing across the wall of the soap bubble formed on the end of a pipe which is attached to a manometer as shown if the difference in elevation between the 2 water levels shown in the manometer is 1.2 mm.

Solution:

The hydrostatic pressure in a column of fluid is equal along lines of constant elevation. Moving downward in a column of fluid causes pressure to increase while moving upward causes pressure to decrease. At atmospheric pressure, the change in hydrostatic pressure due to a change in elevation in the air may be neglected. Since the system is open to the atmosphere at both ends, the pressure difference across the bubble can be determined as follows:

Pout = PlItm

P.=P +pgh,_pgh2=p +pg(h-h) In aim aim 12

gc gc K

= + pg t:Jz = + (lgml ee)(980.7eml S2) (1.2mm X lem ) Pm Pat", Patm gm em 10

g, 980.7 mm gfs2

= + 1(~) gf = 0 12 gf Pin Patm 2 Patm +. 2 10 em em

I'1p = Pin - Pau, = Patm +0.12 gf2 - Pol'" = 0.12 gf2 em em

Notice that a manometer inherently measures a pressure difference directly as a height or elevation difference. Pressure expressed as an elevation difference is often referred to as a head. In this example, the pressure difference is a water head of 1.2 mm.

28


ARCHIMEDES' PRINCIPLE

Archimedes' Principle states that the apparent weight loss of a solid immersed in liquid is equal to the weight of the liquid displaced by the solid. In other words, the buoyant force on a body immersed in liquid is equal to the weight of the liquid displaced by the body. Thus, by immersing a solid of unknown volume or density into a liquid of known density and measuring the apparent weight loss of the immersed solid, the volume or density of the solid may be calculated. Determining the volume or density of an object by fluid displacement is especially useful when the volume of an object may not be easily or accurately determined by measurement of its physical dimensions.

As the story goes, Archimedes was inspired one day while taking a bath and pondering how he might discover the authenticity of the king's new "gold" crown. Archimedes knew that the specific gravity of pure gold is 19.3. Archimedes also knew that he could easily determine the specific gravity of the crown if he knew its volume. However, directly calculating the volmne of an irregularly-shaped crown is not easily accomplished. Archimedes realized that the buoyant force acting on a body immersed in liquid is equal to the weight of the liquid displaced by the body. Using this principle, Archimedes was able to determine the specific gravity of the crown.

Example 4.2 Suppose that the king's crown weighed 14.7 kg when measured in air and that it appeared to weigh 13.4 kg when submerged in \-vater.

Solution:

Note that the "weights" are reported in kilograms which means that these quantities are actually mass and "apparent mass". The mass does not really change, but only appears to change due to the lifting effect of buoyancy. Recalling that the buoyant force (Ph) acting on a body immersed in liquid is equal to the weight of the liquid displaced ('\fIf) by the body and using the relationships among weight, mass, and density, an expression relating weight to mass and to density can be written as follows:

mog Po~g W gc w;, gc Po

Ll~pparent (mo -m;)g Wf PfVfg Pf gc gc

29


w rno 1 Po ;:=Po =Yo m, -m, Pj Pw .6. ~pparent

14.7 kg

(14.7 kg - 13.4kg)

Yo = 11.3

Thus, the specific gravity (y) of an object submerged in water is equal to the ratio of the objects true weight to its apparent loss in weight. As shown, the same is also true with regards to the object's mass. The specific gravity of the crown in this case is not equal to 19.3, the specific gravity of gold, but rather is equal to the specific gravity of lead.

VISCOSITY

Viscosity is a property of fluids which is a measure of the internal friction existing between the different layers of a fluid in motion. In liquids, this internal friction is due to the cohesive forces between the molecules of the liquid. In gases, the internal friction results from the collisions occurring among the gas molecules. In general, the internal friction for a liquid in motion is much greater than the internal friction for a gas in motion. Thus, the viscosity of a liquid is generally greater than the viscosity of a gas. Viscosity is often denoted by the lower case Greek let

ter "MU" (fl) in petroleum engineering.

FLUID FLOW REGIMES

Fluid flows in 2 distinct regimes. Fluid that is flowing smoothly along orderly paths called streamlines is said to be in laminar flow. Streamlines never cross one another and are well-defined paths or flow boundaries which may be predicted mathematically. Fluid that is flowing in unpredictable and complex eddies similar to tiny whirlpools is said to be in turbulent flow. The fluid velocity is one of the factors which determines the flow regime. For a given fluid in a given flow system of a given geometry, a particular velocity exists above which the fluid flow regime will change from laminar to turbulent.

POI SELLE'S LAw

In the early 1800's a French physicist named]. Poiseuille was interested in describing blood circulation. In the course of his work, Poiseuille developed a relationship to predict the flow rate of a viscous fluid in a capillary tube. The relationship is now known as Poiseuille's Law and can be expressed as follows for a tube of

radius, r, transmitting a fluid with viscosity, fl under a pressure gradient (dp/ dl):

30

_,,-r 4 dp

q = 81' dl


(4.18)

Note that the volumetric flow rate, q, is directly proportional to the pressure gradient and inversely proportional to the viscosity of the fluid. The viscosity unit poise is named after Poiseuille. Later in this course, a law developed for the flow of fluids in petroleum reservoirs will be discussed which will be very similar to Equation (4.18) .

31

©2003. ALL RI

Petroleum Geology Revisited A general understanding of the process in which hydrocarbonbearing rocks were formed provides insZght into the study of rock properties and the natural distribution of fluids within the rock.

A common misconception is that oil and gas (hydrocatbons) accumulate and exist as "latge undergtound pools" which are surrounded by "solid" rock. This misconception is fueled by the notion that all rock is a "solid" (non-porous) material. In reality, rock which is capable of bearing hydro

carbons is porous like a sponge.

The "solid" portion of the rock is referred to as the rock matrix and the often microscopic void spaces in the rock are referred to as pores. The hydrocarbons reside within the pores of the rock itself. The particular volume of rock wherein the oil and natural gas resides is referred to as the rese1710ir: Understanding of the process in which hydrocarbon-bearing rocks were formed provides insight into the study of petrophysics (rock properties) and the natural distribution of fluids within the reservoir.

NATURE OF RESERVOIR ROCKS

Hydrocatbon-bearing rocks are generally sedimentary rocks. As the name implies, the rocks ate composed primarily of sediment or pieces of broken rock eroded and carried by water. The sediment is washed away from its source and into lakes and rivers which transport and deposit the sediment downstream, eventually ending in the ocean. Sedimentaty rocks may also be composed of chemicals which result from a chemical process or prectpztate in the water, such as calcium carbonate. Note that in either case water is involved in the formation of sedimentary rocks.

32

The primary type of hydrocarbon-bearing rock is the sedimentary rock.

A particular layer of rock deposited at a particular time is often referred to as a formation.

©2003. All RIGHTS RESERVED

Successive !.lyers of sediment are deposited one on top the other in the depositional environment. The weight of the overlying layers acts to compress and compact the underlying sediments while the precipitation of minerals from the water ill-which the sediment was deposited acts to cement the sediment grains together. The result is a layer of sedimentary rock. The rock matrix is composed of some combination of rock grains and cement and the pores are originally filled with water. A particular layer of rock deposited at a particular time is often referred to as a formation. Because a formation is a unit of rock created at the same time and in the same environment, the physical properties of a formation are generally consistent over a wide area.

Types of sedimentary rock which commonly bear hydrocarbons include sandstone, limestone, and dolomite. As the name implies, the primary ingredient in sandstone is sand or quartz. Likewise the primary ingredient in limestone is lime (calcium carbonate). Dolomite is a secondary form of limestone.

PROCESS OF RESERVOIR FORMATION

Although the exact process by which hydrocarbons are created is not fully known, the geologic conditions favorable to forming petroleum reservoirs are known. The theory explaining petroleum reservoir formation may be summarized as follows. When organic material becomes buried to a sufficient depth, the elevated temperature and pressure cause it to be converted to petroleum. Tectonic forces in the earth's crust and the huge pressures generated by the weight of overlying rock layer caused the petroleum to ooze from its source and migrate upward following a meandering and tortuous path through pores of the overlying rock layers toward the surface until a barrier to flow is encountered.

FOOD FOR THOUGHT You might be asking yourself, "Why does the petroleum migrate upward? Why not flow in other directions?" In response, consider that the pores of sedimentary rocks are originally filled with water according to the theory of sedimentary rock formation (discussed in the previous section). Also recall that oil and natural gas are generally less dense (lighter) than water. As a result, gravity will tend to cause gas, oil, and water to segregate (separate) with the lightest fluid on top. This mechanism is often referred to as gravity segregation. In processes driven by gravity segregation, the fluids will redistribute until the lighter fluid (petroleum) floats on top of the heavier fluid (water).

A barrier to flow will halt upward migration of the oozing droplets of petroleum. As a result, the petroleum droplets will begin to coalesce (combine) and accumulate beneath the barrier. Without some type of barrier, the petroleum would continue to percolate upward through the subsurface rock layers until it reached the surface of the earth. Thus, one of the necessary geologic conditions for the formation of a petroleum reservoir is the presence of a barrier which prevents further

33

One of the necessary geologic conditions for the accumulation of hydrocarbons in the

subsurface is a trap.


migration. The combination of the barrier and the rock layers containing the ac~ cumulated petroleum is referred to as a trap. According to this definition, the res~ ervoir rock itself is part of the trap.

Traps are classified as either .rtTl4ctttra~ Jtratigraphic or combination. As the name im~ plies, structural traps are usually associated with some type of deformation of the original configuration of the rock layers, such as folding or faulting. Examples of structural traps include anticlinal traps, fault traps, and dome plug traps. \'{Ihereas stratigraphic traps are usually associated with a condition or process associated with the deposition of the sediments. Examples of stratigraphic traps include un~ conformities and lenticular traps.

An unconformity can be thought of ",S a "time gap in the geologic record" (Van Dyke, 1987). When a previously formed layer of rock is brought to the surface, it will begin to erode. If after a period of erosion, a new layer or sediment is depos~ ited over the eroded surface, an unconformity will result. Lenticular traps derive their name from the shape of the reservoir roGk which resembles a lens. Often these reservoirs result from an uneven deposition of sand and clay as will often occur in the formation of river sandbars. Once buried, the sandbar may form a lenticular trap.

Structural traps are often associated with a structural anomaly or irregularity in the subsurface. In the earliest days of petroleum exploration before the mechanics of reservoir formation were understood, petroleum prospectors looked for distinc~ rive geologic structures outcropping at the surface. Experience had shown these prospectors that oil and gas reservoirs could often be found by drWing around to~ pographic features such as salt domes. Salt domes often are indicated on the sur~ face by a "dome~shaped" hill. A hill in the middle of a low~lying coastal marsh area is a very distinctive indication of a structural anomaly (irregularity) in the sub~ surface rock layers. In the past, the structure of the subsurface had to be inferred from an analysis of the surface. With the advent and advance of seismic technol~ ogy, much more is known about the subsurface structures associated with the formation of petroleum reservoirs and traps.

RESERVOIR FLUID DISTRIBUTION

Reservoir rock typically contains 2 or 3 types of fluid in varying proportions within the pore space. These 3 fluids are water, oil, and/or natural gas. The pres~ ence of oil and/or gas is required for the rock to be classified as a reservoir. The presence of the water along with the oil and/or gas and the location and configu~ ration of the fluids within the pores is a result of the following:

(1) the process .in which sedimentary rocks are formed, (2) the properties of the reservoir rock, (3) the properties of the fluids within the rock, and (4) the interaction between the roc)< & the fluids contained within it.

34

Normal pressure in the Gulf Coast re

gion of the United

States is 0.456

pounds force per square inch per foot of depth.


Typically, water coats all grains of the reservoir rock and exists as a flim between the inner surfaces of the rock's pores and the oil and/or gas. \'V'hen both oil and gas are present, gas occupies the largest pore spaces while the oil occupies the intermediate pore spaces. The smallest pore spaces remain completely filled with water. This configuration, while not always true, is mote common than the alternative. The reasons and mechanisms behind both the typical and alternative distribution of fluids within the reservoir rock are too complex to be discussed here. Explaining these mechanisms will be the purpose behind many of the future topics of this course in petrophysics.

RESERVOIR PRESSURE

Petroleum reservoirs are generally under pressure. In other words, the pressure within the pore space of the reservoir rock is greater than the pressure at the surface of the earth. The level of pressure within a particular reservoir is a function of the geologic conditions in which the rock was created. The pressure exhibited by a particular reservoir is strongly dependent upon the depth of the reservoir below the surface of the earth. As depth increases so does pressure.

The vast majority of reservoirs are normally pressured while some exhibit a pressure greater than would be expected at a particular depth and are abnormally pressured. As the name implies, normal pressure is that level of pressure which can be expected at a particular depth. What is considered normal.varies regionally. In the Gulf Coast region of the United States, the normal pressure is 0.465 pounds force per square inch per foot of depth. Consider again the process which creates sedimentary rocks. Sediments are transported and deposited in water then buried and compacted by successive layers of sediment deposited over time. Since the sediment is deposited in water, water occupies all of the spaces between the individual particles of sediment. As the sediments are buried and compacted, some of the water is pushed our. As long as the water has a escape path to the top of the sediments (the sea floor), the sediments will not be under an additional or abnormal pressure. That is, the pressure within the rock that these sediments eventually form will exhibit a normal pressure. If the water's path of escape to the surface is removed or sealed, the water will not be able to escape as the sediments are compacted. This trapped and compressed volume of water will lead to an abnormally high pressure or abnormal pressure within the rock.

FOOD FOR THOUGHT What does this value tell us about the subsurface pressure in the area of the Gulf of Mexico? This means that for every foot down below the surface of the earth in the area of the Gulf of Mexico, the pressure will increase by 0.465 pounds force per square inch. At a

. depth of 10,000 feet below the surface of the earth on the campus ofLSU, one would expect to find a pressure of (0.465 psi/ft)(10,000 ftl = 4,650 psi.

35

The hydrostatic gradient in sea water is aboUt. 0.442 psi/ft.


The hydrostatic pressure gradient in the sea water is 0.442 psi/ft. Notice that the change in pressure per foot of depth observed in the subsurface rock layers in the normally pressured areas of the Gulf Coast is approximately equal to that observed in a body of sea water. Why is that?

Imagine that you have an open tank filled with clean sea water. Leaving the tank open, gravel is placed into the water-filled tank until the gravel reaches the top of the tank and no more gravel will fit into the tank. What will happen as the gravel is placed into the tank? Since the tank is open, a volume of water equal to the volume of the gravel in the tank will slosh over the sides of the tank. Once the gravel has settled to the "bottom" of the tank, what will be the hydrostatic pressure gradient in the tank?

Consider the following facts:

(1) the hydrostatic pressure gradient is related to the density of the fluid,

(2) all of the water in the tank is still in contact or communication with itself (i.e. none of it is sealed off or trapped), and

(3) the gravel particles have settled so that they are resting one on top of the other and are not suspended in the water.

The water occupies all of the space in the tank among the pieces of gravel except for where the pieces of gravel touch each other. Because the gravel is not suspended in the water, the density of the sea water within the tank remains the same as before. The weight of the bottom-most layer of gravel is suppotted by the bottom of the tank. The weight in the next layer of gravel upward is supported by the bottom-most layer of gravel. Likewise, each successive layer of gravel is supported by the layer directly below it. Thus, the presence of the gravel in the tank under these conditions does not affect the hydrostatic gradient in the tank. It is the same as before, 0.442 psi! ft. This illustrates conceptually why the normal pres~ure gradien~ in subsurface layers of rock is essentially equal to the hydrostatic pressure gradient of the sea water in which they were originally deposited.

36

fundamentals of petrophysics ch 1-5

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