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Hydrocarbon porosity refers to the part of the porosity that contains

hydrocarbon. It is total porosity multiplied by the fraction of the pore

volume that contains hydrocarbon.

Two kinds of pores or voids:

Pore space, called interconnected or effective pore space

Dead pore space that consists of isolated or non-

interconnected pores or voids dispersed in the medium.

Only the effective or interconnected pore space contributes to the flow

of fluids through the porous medium.

Experimental methods commonly used to determine the porosity:

i) Imbibition method. The porous sample is saturated with a

preferentially wetting fluid by letting the fluid imbibe into the sample

under vacuum. The sample is weighed before and after imbibition. From

the two weights, the density of the fluid and the dimensions (bulk

volume) of the sample, the porosity can be calculated.

ii) Mercury injection method. As most materials are not wetted by

mercury, it will not penetrate into the pores unless a pressure is applied.After the sample is evacuated, mercury is forced to penetrate into the

sample under high pressure.

iii) Gas expansion method. The porous sample is enclosed in a vessel of

known volume V1 under known gas pressure P1. When the vessel isconnected to an evacuated vessel of known volume V2, the gas expandsinto this vessel and the gas pressure in the first vessel decreases to a

lower value P2. Applying the ideal gas law to the above process givesthe effective pore volume Vp of the sample:


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=

12

221

PP

PVVVV bp (1-2)

where Vb is the bulk volume of the sample which is determined in aseparate measurement.

iv)Density method. The bulk density of the sample b and the density of

the solid matrix of the sample s are determined. The total porosity of

the sample is calculated by

s

b

=1

(1-3)

Average porosity of a reservoir

The accuracy of the average porosity of a reservoir as found from core

analysis depends on the quality and quantity of the data available and on

the uniformity of the reservoir. The porosity is also calculated from

electric logs and neutron logs, often with the assistance of some core

measurements.

1.2 Permeability

1.2.1 Definition and Darcys law

- The property of a porous medium which allows a fluid to flowthrough it is called permeability.

- This parameter is determined entirely by the pore structure.Darcys law : The darcy permeability k is calculated by applyingDarcys law (Darcy, 1856) to a slow (creeping), one dimensional,

horizontal, steady flow of a Newtonian fluid:


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L

PAkq

=

, (1-4)

q volumetric flow rate (cm3

/sec)A cross-sectional area of the sample (cm

2)

L length of the sample in the flow direction (cm)P hydrostatic pressure drop (atm) viscosity of the fluid (cP)

Using these units in Darcys law results in the practical unit of

permeability the darcy (D). One darcy is equal to 0.987 (m)2

in SI units.

One darcy is a relatively high permeability, and for tight porousmaterials the unit millidarcy (mD) is used.

1.2.2 Measurement of permeability

Measurement of permeability is usually performed with one-

dimensional, cylindrically shaped samples. In the measurement various

flow rates of the fluid are recorded as a function of pressure drop.

Permeability is obtained by fitting a straight line to the data points.

Theoretically, this line should pass through the origin.

Figure 1. Effect of turbulent

flow on measured permeability


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Both liquids and gases are routinely used to measure permeabilities.

Liquids, however, may interact with the rock sample and thus change the

pore structure of the medium and therefore the permeability of themedium. For example, the flow of fresh water through porous samples

containing clay caused permeability reduction because of clay swelling

upon contact with fresh water.

1.2.3 Units of Darcys Law under different systems

The potential (datum pressure) of a fluid: zgp += Darcys law in general form:

ds

dkAq

=

or

+=

ds

dzg

ds

dpkAq

Darcy unit system:

+=

ds

dzg

xds

dpkAq

61001325.1

1

k darcy,A cm2, cp,p atm,s cm, gm/cm

3,g 980.7cm/sec,

q cm3/sec

SI unit system:

+=

ds

dzg

ds

dpkAq


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k m2,A m

2, kg/m-sec,p N/m

2(Pa),s m, kg/m

3,g 9.8

m/sec2, q m

3/sec

Field unit system:

+=

ds

dz

g

g

ds

dpkAq

c144001127.0

k md,A ft2, cp,p psia,s ft, lb/ft

3,g 32.17ft/sec

2, q

bbl/d

cgs unit system:

+=

ds

dzg

ds

dpkAq

k cm2,A cm

2, gm/cm-sec(poise),p dyne/cm

2,s cm,

gm/cm3,g 980.7cm/sec

2, q cm

3/sec

1.2.4 Gas Permeability

In using gas in measuring the permeability, the gas volumetric flow ratevaries with pressure. Therefore, the value ofq at the average pressure inthe core must be used. Assuming the used gases follow the ideal gas

behavior, Darcys law is in the following form:

bg

gbLp

ppkAq

2

)( 222

1 = (1-5)

qgb gas volumetric flow rate at base pressure (cm3/sec)

A cross-sectional area of the sample (cm2)

L length of the sample in the flow direction (cm)P1 inlet (upstream) pressure (atm)P2 outlet (downstream) pressure (atm)Pb base pressure (atm)g gas viscosity at mean pressure (cP)


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The Klinkenberg Effect

The permeability of a porous medium sample measured by flowing air is

always greater than the permeability obtained when a liquid is the

flowing fluid. This is because that gases exhibit slippage at the pore wallsurface. The gas slippage results in a higher flow rate for the gas at a

given pressure differential. For a given porous medium, the calculated

permeability decreased as the mean pressure (pm) increased. If a plot of

calculated permeability versus 1/pm is extrapolated to the point of 1/pm =

0 (or pm = infinite), this permeability would be approximately equal to

the liquid permeability (see Figure 2). The straight-line relationship in

Fig.2 can be expressed as:

+=

m

Lgp

ckk 1 (1-6)

kg = calculated gas permeability

pm = mean pressure (pm = (p1+p2)/2)

kL = Klinkenberg permeability

c = slope of the line

The magnitude of the Klinkenberg effect varies with the core

permeability and the type of the gas used in the experiment.

Figure 2. The Klinkenberg effect in gas permeability measurement


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Figure 3. Effect of gas pressure on measured gas permeability forvarious gases. (Calhourn, J., 1976)

1.3Pore Size and Pore Size Distribution

If )(D is the distribution of pore volume fraction as a function of pore

size,D, then the pore size is selected such that (Dullien, 1992)

=0 1)( dDD . (1-7)

Mercury porosimetry

-The volume of mercury penetrating the sample is measured as a

function of the pressure imposed on the mercury.

-Drainage capillary pressure curves obtained by mercury intrusion

porosimeter are customarily interpreted in terms of the bundle of

capillary tubes model.-The pore size is calculated from this pressure by Laplaces

equation of capillarity and, using the bundle of capillary tube

model of pore structure, the volume of mercury is assigned to this

pore size.


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In the case of interconnected pore space the body of each pore is

connected to the bodies of adjacent pores via necks or throats. The

sizes of both the pore bodies and the pore throats play an important

role in determining various macroscopic properties, such as permeability, capillary pressure curves of porous media, etc. Both the

body and the throat sizes can be measured using computer

reconstruction of pore structure from photomicrographs of serial

sections of the sample.

1.4Resistivity FactorA resistivity factor, often called formation resistivity factor, is

defined as

w

o

R

RF=

, (1-8)

Ro the electrical resistance of porous sample saturated with an ionicsolution,

Rw the bulk resistance of the same ionic solution occupying the samespace as the porous sample (same cross-section area and length).

In the case of nonconductive solids Ro

is always greater that Rw

and

therefore,Fis always greater than unity.

Correlation of the formation resistivity factor with porosity and a

cementation exponent m.

For clean and uniform sand 21

=F

Archie (1942) suggested:mF = 1.3 < m < 2.5

It is more common to express the formation factor as:maF =

where a, m are unique properties of sample.


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1.5 Isothermal Compressibility (coefficient)

The isothermal compressibility for a substance is:

dp

dv

vc

1=

or more generallyT

dp

dv

vc

=

1(1-9)

Where c = isothermal compressibilityv = volume

p = pressure

Rock-matrix compressibility cr:

The rock-matrix isothermal compressibility is the fractional change insolid rock volume with a unit change in pressure while the temperatureis held constant:

T

r

r

rdp

dv

vc

=

1(1-10)

Units are in reciprocal pressure units, psi-1

.

Rock-bulk compressibility cb:

T

b

b

bdp

dv

vc

=

1(1-11)

Pore volume compressibility (Formation isothermal compressibility)cf:At any value of external-internal pressure difference, the change in porevolume per unit of pore volume per unit change in pressure. Note that

pore volume increases with the pore pressure:


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T

p

p

fdp

dv

vc

=

1(1-12)

OrT

fdp

dc

=

1(1-13)

Example: The values ofcffor limestone and sandstone reservoir rocks:2x 10

-6to 25x 10

-6psi

-1.

When the internal fluid pressure within the pore spaces of a rock isreduced:

the volume of the solid rock material increases the pore volume decreases

Both of these volume changes act to reduce the porosity of the rockslightly, of the order of 0.5% for a 1000 psi change in the internal fluid

pressure (e.g., at 20% porosity to 19.9%).

2. Fundamentals of Capillarity

2.1 Surface (interfacial) tension

Interface is a boundary between two immiscible phases and is taken as a

mathematical line or mathematical surface.

Types of interfaces

liquidGas (LG)

LiquidLiquid (LL)

LiquidSolid ( LS)GasSolid (GS)

Surface Tension, , and Interfacial tension, , are basic properties of

interfaces. The stress causing this spontaneous decrease of surface area


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is called surface or interfacial tension. Surface tension may be thought of

either as a force per unit length in a surface (dyne/cm, or equivalently,

mN/m), or a free energy per unit surface area (erg/cm2, or equivalently,

mJ/m2

). The term Surface Tension is reserved for interfaces involving a

liquid in equilibrium with their own vapour, or for liquid-air type

of interfaces.

The term Interfacial Tension (IFT) is used for L-L or L-S types ofinterfaces.

The physical meaning of force per unit length and energy per unit area is

understood by the following examples.

Figure 4: Example of film of soap

Work = Force Length or displacementdxW = )( l

dAW =

where dxdA = l and denotes the change in interface area.


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Typical IFT values

H2O-air 72 mN/mHg-air 450 mN/m

Hydrocarbons-air ~2050 mN/n

H2O-Oil ~20 mN/m

(H2O + NaCl)-Oil ~30 mN/m

(H2O + Soap)-Oil ~0.1mN/m to 110-3

mN/m

2.2 Wettability and Contact Angle

The relative wettability of a porous medium by two fluids is

characterized by the contact angle . The contact angle is defined as

the angle between the tangent to the liquid/solid boundary constructed at

a point on the three-phase line of contact and the tangent to the

gas/liquid boundary constructed at the same point.

Figure 5 Contact angle of a liquid on a solid surface

Youngs equation:

cosgssg ll = , (1-14)where gl and sg are the surface tensions of liquid and solid,

respectively, andls is the interfacial tension between the liquid and the

solid.


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The contact angle may have any value between 0 and 180. It is

customary to simply classify fluids into two categories: wetting and non-

wetting fluid. For wetting fluids, 0 < 90, and for non-wetting fluids,90 < 180. In the pores of porous media, the curved interfaces

between two immiscible fluids may take different shapes and directions

of the curvature.

Figure 6 Menisci in (a) a water- wet and (b) an oil-wet capillary.

Terminology Used in Contact Angle Measurements A, Advancing contact angle: The contact angle when the interface

is forced by human action or otherwise to move in the direction of

the wetting phase displacing the nonwetting phase. A is usually

measured through the wetting phase and the solid surface. R, Receding contact angle: The contact angle when the wetting

phase is receding ( e.g. when the nonwetting phase displaces the

wetting phase). Ris also usually measured through the wetting

phase and the solid surface.

E, Equilibrium or intrinsic Contact Angle: the contact angle whenthe system was attained equilibrium position over time.

Equilibrium Contact Angle - Approach to EquilibriumRock- fluid interactions are time dependent. They significantly affect the

measurement of contact angles. Salinity of water, pH, type of ions,

temperature, and polar groups in the crude oil, all have an effect on

equilibrium contact angle A.


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2.3 Capillary Pressure

Immiscible fluids in porous media are separated from each other by

curved interfaces across which there exists a pressure difference or astep change in pressure. This pressure difference, called capillary

pressure PC, is balanced at equilibrium by a pressure difference P atany points of contact between the two fluids:

wnwC PPPP = (1-15)where subscripts w and nw represent wetting and non-wetting phase,respectively. As, at equilibrium, Pnw > Pw, the capillary pressure is, by

definition, always positive.

Equation 1-15 shows that the capillary pressure always tends to

compress the non-wetting phase relative to the wetting phase. It is

helpful to remember that the pressure in the phase on the concave side of

the surface is always greater than the pressure in the phase on the convex

side. Laplaces equation for capillary pressure in a capillary tube is:

m

CrPP

2

== , (1-16)

Where rm is the mean radius of the curvature of the meniscus.

Typical application of Eq. 1-16 is in a circular capillary of a very small

radius. If the contact angle of a liquid on the capillary walls is zero, the

meniscus can be approximately thought to be hemispherical. Thus, the

mean radius of the curvature is equal to the radius of the capillary.

rPC

2=, (1-17)

where ris the radius of the capillary.


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The general case is when the liquid/gas surface or liquid/liquid interface

meets the circular capillary wall through contact angle , as illustrated in

Fig. 2-3.

If the meniscus is a spherical cap and both r and R are positivequantities, the following geometric relation holds:

cos

rR =

, (1-18)

and Eq. 2-7 becomes

cos2r

PC = . (1-19)

Figure 6. The relationship between the radius of the

curvature of the meniscus

and the radius of the

capillary tube.

Equation 1-19 is the most widely used form of Laplaces equation in

porous media.


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2.4 Capillary Rise

The best illustration of the phenomenon of capillary rise may be the

capillary rise method for measuring the surface tension of a liquid, suchas water in a clean glass capillary. The contact angle of some fluids on a

clean glass surface is near zero. Eq. 1-19 can be used directly. At

equilibrium, the P, orPC, in Eq. 2-9 must equal to the hydraulicpressure drop in the liquid column in the capillary. This is P = gh,where is the difference in density between liquid and gas phase and g

is the gravitational constant. Eq. 1-19 becomes

rgh

2

= . (1-20)

Figure 7. Capillary rise in a glass tube


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2.5 Capillary Pressure Curves

Capillary pressure-saturation relationship

When a porous sample saturated with a single fluid, e.g. air, is

submerged in a second fluid, e.g., water, one of the followingphenomena may be observed:

1.The water may spontaneously imbibe into the sample and thusdisplacement of air from the sample takes place.

Water wetting phase

Air non wetting phase

Porous sample water wet

Displacement free imbibition

2.The water may not penetrate at all into the sample unless it isplaced under an externally applied pressure.

Water non wetting phase

Drainage capillary pressure curveThe quasi-static displacement of wetting phase, e.g., water, by a non-

wetting phase, e.g., oil or air, is defined by the set of increasing values of

the capillary pressure, Pc, and a corresponding set of decreasing values

of water saturation. The wetting phase is drained out of the sample.

Imbibition capillary pressure curveThe capillary pressure verses saturation relationship for the process of

wetting phase displacing non-wetting phase in quasi-static steps.

Capillary pressure hysteresisBecause of the difference between the conditions for configurational

stability of the fluid-fluid interfaces during the drainage process and the

imbibition process, the capillary pressure curves display hysteresis.The advancement of the non-wetting phase into the medium is

controlled by the neck pores.

The advancement of the wetting phase is controlled by the bulge

pores.


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Figure 8. Typical capillary pressure curves

The definitions of main terms of capillary pressure curves:

Irreducible (wetting phase) saturation Swi:The reduced volume of the wetting phase retained at the highest

capillary pressure where the wetting phase saturation appears to be

independent of further increases in the externally measured

capillary pressure.

Residual (non-wetting phase) saturation Snwr:The reduced volume of the nonwetting phase that is entrapped

when the externally measured capillary pressure is decreased from

a high value to zero.


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Primary drainage curve:The relationship characteristic of the displacement of the wetting

phase from 100% saturation to the irreducible saturation.

(Secondary) Imbibition curve:The relationship characteristic of the displacement of the

nonwetting phase from the irreducible saturation to the residual

saturation

Secondary drainage curve:The relationship characteristic of displacement of wetting phase

from residual saturation to the irreducible saturation

Hysteresis loopExperimental evidence has indicated that the irreducible saturation

obtained by initial drainage is the same as that obtained by

secondary drainage. When the residual saturations are the same,

the imbibition after secondary drainage will follow exactly the

imbibition curve obtained after primary drainage. The secondary

drainage curve and the secondary imbibition curve constitute a

closed and reproducible hysteresis loop.

Leverett J-Function

Because the capillary pressure is a function of pore size, which is a

function of porosity and permeability, it varies with reservoir porosity

and permeability. It is not practical to measure a separate capillary

pressure curve for each value of porosity and permeability, all capillary

pressure curves for a particular reservoir can be combined into one curve

by use of the Leverett J-function given by:

kpJ c

cos= (1-21)


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If the capillary pressure curves are measured for several different

porosities and permeabilities, all the data can be plotted on one curve by

use of J-function.

3. Steady Multiphase Flow in Porous Media3.1 Introduction

When two immiscible fluids flow simultaneously through a porous

sample under steady conditions, there will be a pattern of occupancies of

the two fluids which greatly influence the effective permeabilities of the

two fluids in the sample.

The distribution of the two fluids depends

the saturations the wettability conditions of the pore surface the interfacial tension fluid viscosities the pore velocity

The case of the most interest corresponds to

one of the two fluids wets the pore surface preferentially the interfacial tension is large the velocities are low

Flow of immiscible fluids in porous media:

Steady-state All macroscopic properties of the system are timeinvariant at all points

Unsteady-state Properties change with time Co-current Both phases flow in the same direction Counter-current Different phases flow in opposite directions Steady state flow VS. displacement (imbibition, drainage)


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3.2 Saturations

Saturation is defined as the fraction, or percent, of the pore volume (PV)

occupied by a particular fluid (oil, gas, or water):

Oil, gas, and water saturation are:

PV

VS oo = ,

PV

VS

g

g = ,PV

VS ww =

Where Vo, Vg and Vw are oil, gas, and water volumes, respectively. The

saturation of each individual phase ranges between zero and 1.0. The

sum of the saturations is 1.0:

So + Sg+ Sw = 1.0.

3.3 Relative permeabilities

- A macroscopic (phenomenological ) description ofmultiphase flow in porous media

Equations of relative permeabilities

Darcys Law: L

PkAq

=

Conditions of steady cocurrent flow are usually established by injecting

both fluids at constant rates and allowing time for the discharge rates to

become equal to the injection rates. Under these conditions it has been

found in experiments that the saturation and the capillary pressure are

approximately independent of position.

An extended Darcys law for two-phase flow in porous media, under

steady-state conditions:


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L

PAkq i

i

ii

=

, i = o, w (1-22)

qi the volumetric flow ratePi the pressure dropi the viscosity of phase i.ki the effective permeability of the porous medium to phase i.

It is coustomary to express ki as relative permeability, i.e., as a fractionof the absolute permeability of the porous medium k

k

k

k

i

ri=

i = o, w (1-23)

and write Eq. 1-22 as

L

PAkkq i

i

rii

=

i = o, w (1-24)

As the saturation of a particular phase decreases, the effective

permeability of that phase also decreases. The sum of the effectivepermeabilities is always less than or equal to the absolute permeability,

i.e.,

kkkk wgo ++

Or 1++ rwrgro kkk


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Figure 9 Schematic of relative permeability curves of primary

drainage and secondary imbibition (Dullien, 1988).

Experimental methods of measuring relative permeabilities

The methods available for the measurement of relative permeabilities

can be divided into two categories: steady and unsteady flow tests.

Steady flow tests

In a typical steady-state method the two liquids are injectedsimultaneously as a fixed ratio and know, metered flow rates. The

criterion of steady state is determined by the condition that the inflows

equal the outflows and /or the constant pressure drop has been reached

across the sample. The attainment of steady state may take anywhere


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from 2 to 40 hours or even longer, depending on the sample permeability

and the method used.

Figure 10 Schematic diagram of steady state relative permeability

apparatus

Procedure: (for waterflooing)

Saturate the porous sample with the wetting phase (water) Displace the wetting phase with the non-wetting phase (oil) to

irreducible water saturation

Starting from irreducible water saturation, increase watersaturation step by step to determined oil and water permeabilities

at different water saturations until the residual oil saturation

reached.

Unsteady flow methods

In the unsteady-state (external drive) method, one phase is displacedfrom a core by pumping in the other phase and the relative

permeabilities are calculated by history matching the production data

(produced fluid ratio and pressure drop across the core sample) using

reservoir simulation.


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Advantages of the unsteady-state method:

Fast The relative permeabilities measured in this fashion will exactly

reproduce the experimental oil recovery data (from which they

were derived).

Procedure: (for waterflooding, SPE 99763 by Wang and Dong, 2006)

The absolute permeability is first measured by injecting singlewater phase through a 100% water-saturated sandpack.

Initial water saturation is established by injecting oil until waterwas no longer produced, which is followed by the determination ofeffective oil permeability at irreducible water saturation.

The relative permeability test is conducted at a constant waterinjection flow rate.

After waterflooding, the average oil saturation in the core sampleis measured using a Dean Stark glass distillation assembly, which

is applied to check the recorded production data.

Relative permeabilities for a fluid pair can be affected

pore structure Wettability capillary forces saturation history viscosities of fluid pair

Drainage and imbibition relative permeabilitiesWhenever capillary forces are controlling, there is also a hysteresis

effect of the relative permeability curves, as shown schematically in Fig.

9, i.e., imbibition relative permeability curves are different from

drainage relative permeability curves.


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Effect of Wettability

The difference of procedures in determining relative permeability curves driving the sample down to irreducible water saturation and

measuring the relative permeabilities at increasing water

saturations.

in a water-wet samples imbibition-type relative permeabilitycurves are determined

in an oil-wet sample drainage-type relative permeability curves aredetermined.

The most noticeable difference in relative permeability characteristics ofthe two types of rocks is the difference in water saturations at which

water and oil relative permeabilities are equal: greater than 50% for

water-wet and less than 50% for oil-wet.

Summary of relative permeability curves

Water-wet Oil-wetConnate water saturation Usually greater than 20 to

25%

Generally less that 15%,

frequently less that 10%

Saturation at which oil

and water relative

permeabilities are equal

Greater that 50% water

saturation

Less than 50% water

saturation

Relative permeability to

water at maximum water

saturation; i.e., floodout

Generally less that 30% Greater than 50% and

approaching 100%


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Introduction to Reservoir Engineering

Chapter 2 Reservoir Fluids Properties

2.1 Reservoir Types Defined in Reference of Phase Behavior

The types of reservoirs can be defined by the phase behaviour of the initial res-ervoir temperature and pressure with respect to the two-phase (gas and liquid)region on pressure-temperature (PT) phase diagrams.

Pressure temperature phase diagrams:

Typical pressure-temperature phase diagram of a pure substance

Vapor-pressure line Critical point Triple point

Melting-point line Sublimation-pressure line


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Typical pressure-temperature diagram of a multi-component mixture

Bubble point line Dew point line Two-phase region two phases, gas and liquid, coexist. Two phase

region - enclosed by the bubble-point and dewpoint lines. Constant liquid volume line -The curves within the two-phase region

show the percentage of the total hydrocarbon volume that is liquid forany temperature and pressure.

Constant liquid volume line Cricondentherm, or maximum of two-phase temperature (250F for thepresent example)

Cricondenbar, or maximum of two-phase pressurePhase diagram depends on the composition of the mixture. Initially each hydro-carbon accumulation will have its own phase diagram, which depends only on thecomposition of the accumulation.


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Examples of paths of production and phase change:

(Textbook Page 7)

Point A: Fluid at 300F and 3700 psia. One-phase state (gas) in reservoir duringproduction.

A-A1: It will remain in the single-phase or gaseous state as the pressure declinesalong path A-A1.

A-A2: The fluid produced through the wellbore and into surface separators mayenter the two-phase region due to the temperature decline (A-A2).This accounts

for the production of condensate liquid at the surface from a gas in the reservoir.

Point B: fluid at temperature of 180oF and an initial pressure of 3300 psia.

One-phase state (gas), where the reservoir temperature exceeds the critical tem-perature.

Point B1: The dew-point pressure is reached at 2700 psia.

B1-B2: Below dew-point pressure a liquid condenses out of the reservoir fluid asa fog or dew. This type of reservoir is commonly called a dew-point (retrograde)

reservoir. The term retrograde is used because generally vaporization, rather thancondensation, occurs during isothermal expansion.


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B2-B3: Vaporization of the retrograde liquid occurs from B2 to the abandonment

pressure B3. Low reservoir temperature and high abandonment pressure willcause more loss of the condensate.

Point C: Fluid at 2900 psia and 75F. One-phase state (liquid). The temperature

is below the critical temperature. This type is called a bubble-point (undersatu-rated) reservoir.

C-C1: Pressure declines, the bubble point will be reached (at 2550 psia) at pointC1.

Below C1: Bubbles, or a free-gas phase, will appear. The free gas evolved begins

to flow to the well bore, and in ever increasing quantities. Other names for thistype of liquid (oil) reservoir are depletion, dissolved gas, solution gas drive, ex-

pansion, and internal gas drive.

Point D: Fluid at 2000 psia and 150F. A two-phase reservoir, consisting of a

liquid or oil zone overlain by a gas zone or cap. The composition of the gas andoil are different from each other. The oil is at its bubble point and will be pro-duced as a bubble-point reservoir with the presence of the gas cap.

2.2 Gas Properties1. Ideal Gas Law

Equations of State: Describe the pressure-volume-temperature (PVT) behavior

of a fluid.

The ideal gas law:

PV = nRT (1.4)

p - absolute pressureV - volumen - molesT - absolute temperature

R' - the gas constant

Units: p - psia, V - ft3, n - lb-moles, T R

o(Rankine) , R' = 10.73

p - atm, V - cm3, n - g-moles, T K (Kelvin), R' = 82.05

R

o

= 460 +

o

FK = 273.15 +oC


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Standard conditions: 14.7 psia and 60F.Volume of gas at standard conditions: SCF (standard cubic feet), MSCF (1000standard cubic feet), MMSCF.

Example: Calculate the moles, pounds and molar volume at standard conditions

of ethane in a 500 cu ft tank at 100 psia and 100oF.

n=pV/RT = (100 x 500)/(10.73 x 560) = 8.32 lb-moles

m = Mn = 8.32 x 30.07 = 250.2 pounds

Vmolar=10.732x(60+460)/14.7 = 379.4 SCF/Mole

2. Specific Gravity

Density of gas, g, at a given temperature and pressure:

TR

Mwg '

=

Mw = molecular weight

Specific gravity - the ratio of the density of a gas at a given temperature andpressure to the density of air at the same temperature and pressure, usually 14.7

psia and 60F.

The specific gravity of a gas (at standard conditions):

====

2997.28

'

97.28

' ww

w

air

g

g

MM

TR

xp

TR

pM

(1.5)

3. Real Gas Law

pVa = znRT (1.7)

Va - the actual gas volume.


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Compressibility factorz(gas deviation factor) a measure of the deviation from

the ideal behavior. It is a dimensionless quantity and varies usually between 0.70and 1.20.

Definition of the gas deviation factor:

Figure 1.3 Gas deviation factors of two gases (Textbook 16)

Determination of the gas deviation factor:

The gas deviation factor is different for each gas or mixture of gases and for eachtemperature and pressure of that gas or mixture of gases.

1). The gas deviation factor is commonly determined by measuring the volume ofa sample at desired pressures and temperatures, and then measuring the

volume of the same quantity of gas at atmospheric pressure and at a tem-perature sufficiently high so that all the material remains in the vapor phase.


34/129

Example: A gas has a volume of 364.6 cu cm at 213F and 3250 psia. At 14.80

psia and 82F it has a volume of 70,860 cu cm.

The deviation factor at 3250 psia and 213F is:


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2). Estimation from specific gravity:

(Textbook P.17)

The correlations of the two curves are:

ppc = 756.8 131.0g-3.62g (1.8)

Tpc = 169.2 + 349.5g-74.02

g (1.9)

3). Pseudocritical pressure and temperature methodPseudocritical pressure and temperature are defined as:

=j

cjjpc TyT and =j

cjjpc pyp

yj - mole fraction of jth component

Tcj andpcj are critical temperature and pressure ofjth component.

Effects of CO2 and H2S on calculation of pseudoreduced properties (Wichert andAziz)

= 120 (A0.9

A1.6

) + 15(B0.5

B4) (1.15)

A - sum of the mole fractions of CO2 and H2S in the gas mixtureB - mole fraction of H2S in the gas mixture

The modified pseudoproperties are given by:

Tpc = Tpc (1.15a)Ppc = ppcTpc/[Tpc + B(1-B) ] (1.15b)


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4. Gas Formation Volume Factor Bg

Bg = [volume of gas in the reservoir]

[volume on the surface (standard conditions)]

Bg= psczT/Tscp (1.16)

Bg = 0.02829 zT/p cu ft/SCF

=0.00504 zT/p bbl/SCF (1.17)

(The constants in Eqs. (1.17) are for 14.7 psia and 60F)

Example: For Bell Field gas at 3250 psia, 213F, and z = 0.910:

Discussion:1) 1 SCF (at 14.7 psia and 60F) gas occupies 0.00533 ft

3or 0.000949 bbl of

space in the reservoir at 3250 psia and 213F.2) 1000 ft

3of reservoir pore volume in the Bell Field gas reservoir at 3250

psia contains

G = 1000/0.00533 = 188MSCF

5. Gas Isothermal Compressibility cg

Definition: dp

dv

vcg

1=

1). Isothermal compressibility of ideal gases:


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2). Isothermal compressibility of real gases:

Discussion:

I. For an ideal gasz = 1.0 and dz/dp = 0 , the compressibility is the reciprocalof the pressure. That is:

cg= 1/p (for ideal gas)

II. For real gases, cgis a function of both pressure andz-factor.a. At low pressures, thez-factor decreases as pressure increases. dv/dp

is negative and cgis larger than in the case of an ideal gas.

b. At high pressures, thez-factor increase with pressure. dv/dp is posi-tive and cgis less than in the case of an ideal gas.


38/129


39/129

3) Isothermal compressibility of gas mixtures:

p = ppcppr and dp = ppcdppr,

Equation (1.9) can be written in the following form:

prpcprpc

gdp

dz

zpppc

11=

(1.20)

Define: cr= cgppc

There is:

prpr

rdp

dz

zpc

11=

(1.21)

Correlations presented in Figs 1.7 and 1.8 can be used to find the isothermalcompressibility for gas mixtures.

2.3 Crude Oil Properties

1. Specific Gravity and API Gravity

Oil gravity, o, is defined as the ratio of the density of the oil to the density ofwater, both taken at the same temperature and pressure, mostly at standard con-ditions.

w

oo

=

API gravity is defined as:

5.1315.141=

o

oAPI

Where o is the specific gravity of the oil at standard conditions.


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2. Solution Gas-Oil Ratio,Rso

Rso = Solution gas volume measured at standard conditions (SCF)

Stock tank oil volume (STB)

The solubility of natural gas in crude oil depends on the pressure, tem-perature, and the composition of the gas and the crude oil.

The quantity of solution gas increases with pressure. The quantity of solution gas decreases with temperature. The quantity of solution gas increases as the compositions of the gas and

crude oil approach each other.

Saturated oil: A oil is said to be saturated with gas if on a slight reduction in

pressure some gas is released from solution.Undersaturated oil: if no gas is released from solution as pressure is reduced,

the crude oil is said to be undersaturated.

Fig. 1.11 shows the change ofRso of an oil with pressure (textbook).

Determination ofRso

Laboratory analysis with reservoir fluids Estimation using correlations

o Charts (McCain, 1990)o Equations


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3. Formation Volume Factor of Oil,Bo (FVF)

The definition of FVF:The volume in barrels that one stock tank barrel with solution gas occupies in

the formation at reservoir conditions.

Bo=

Volume of reservoir fluid (STO + solution gas) at reservoir conditions

Volume of oil entering stock tank at standard conditions

The reciprocal of FVF is calledShrinkage factor:

bo = 1/Bo

Example of FVF change with pressure and temperature is shown in Fig. 1.12.

Fig. 1.12 FVF of Big Sandy Field oil (Text book page 34)


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Two-phase formation volume factor (Total formation volume factor) Bt

Definition: The volume in barrels one stock tank barrel and its initial dissolvedgas occupies at any pressure and reservoir temperature.

Bt= Bo + Bg(Rsoi Rso) (1.28)

Discussion of Eq. (1.28) (see figure for this equation): For one barrel of stock tank oil plus its dissolved gas. The volume of oil at the lower pressure isBo. The quantity of gas evolved isRsoi Rso, measured at standard conditions. The volume of free gas at reservoir conditions isBg(Rsoi Rso). The total volume is equal to the total formation volume factor,Bt. Above or at the bubble-point,Rsoi= Rso, Bt= Bo. Below the bubble point,Bo decreases and (Rsoi Rso) increases as pressure

decreases, two-phase factor increases.


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Example of calculatingBtat 1200 psia using Fig 1.13:

Fig 1.13 Demonstration ofBo andBtCalculation (Textbook page 36)

Knowns:

Based on 1STB Initial solution gas = 567 SCF/STB At 1200 psia and 160F the liberated gas has a deviation factor of 0.890 liquid phase shrinks to 1.210 bbl at 1200 psia At 1200 psiaRso = 337 SCF/STB

Solution:The gas volume factor with reference to standard conditions

Bg = 0.02829 zT/p = 0.02829x 0.890 x 620/1200 Eq. (1.17)= 0.01300 cu ft/SCF

= 0.002316 bbl/SCFRsoi Rso = 567 337 = 230Vg = Bg(Rsoi Rso) = 230 x 0.002316 = 0.5326 bbl

Bo = 1.210 bbl/STBBt = 1.210 + 0.002316 (567 - 337)

= 1.210 + 0.533 = 1.743 bbl/STB


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4. Isothermal Compressibility of Oil, co

Definition: dp

dv

vco

1=

dv/dp is negative, the negative sign converts the compressibility to a posi-tive number.

co changes with pressure.Average compressibilities can be estimated using the following equation:

21

211

pp

VV

Vco

= (1.31)

The reference V in Eq (1.31) can be V1, V2, or the average of the two.

5. Viscosity of Oil

Fig. 1.14 Viscosity of reservoir oils (Textbook Page 40)

The viscosity of oil under reservoir conditions is commonly measured inthe laboratory.

Viscosities of reservoir oilso Above bubble point, viscosity decreases with decreasing pressure.o Below bubble point, viscosity increases with pressure owing to the

release of solution gas. Correlations have been developed for both above and below the bub-

ble-point pressure.


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1


Chapter 3 Single-Phase Gas Reservoirs

3.1 Introduction

Reservoirs single-phase gas reservoirs

Reservoir fluid natural gas

Dry gas reservoir no condensate formed in the reservoir, but

condensate may form in the wells and on the surface

3.2. Gas in place by volumetric method

The gas pore volume Vg is related to the bulk, or total, reservoir volume

Vb by the average porosity and the average connate water Sw.

)1( wbg SVV =

The standard cubic feet of gas in a reservoir:

g

g

B

VG =

IfVb of the reservoir is in acre-feet, and the standard cubic feet of gas in

place, G, is given by:

g

wb

B

SVG

)1(560,43 =

(3.1)

Example (Bell Field):

Area = 1500 acres

Thickness = 40 ft

Vb = 60,000 ac-ft

Porosity = 22%


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2

Sw = 23%

Bg (@3250 psia) = 0.00533 cu ft/SCF

G = 43,560 x 60,000 x 0.22 x (1-0.23)/0.00533 = 83.1 MMM SCF

Bulk productive volume of reservoirs:

The volumetric method uses subsurface and, isopachous maps based on

the data from electric logs, cores, and drill-stem and production tests.

Subsurface contour map - lines of equal elevations on the top of a bed

Net isopachous map - lines connecting points of equal net formation thicknessIsopach lines - lines of equal thickness

Equations used to determine the approximate volume of the productive

zone:

)(3

11 ++ ++= nnnnb AAAAh

V (3.2)

Vb the bulk volume between n and n+1 isopach lines (acre-feet)

An the area enclosed by the lower isopach line (acre)An+1 the area enclosed by the upper lsopach line (acre)

h the interval between the lsopach lines (feet)

The volume between successive isopach lines, and the total volume is the

sum of these separate volumes.

Or use the trapezoid equation: )(2

1++= nnb AAhV

For a series of successive trapezoids:


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3

navgnnb AtAAAAAh

V ++++++= )2...22(2

1210 (3.3)

A0 the area enclosed by the zero isopach line

A1, A2, An the areas enclosed by successive isopach linestavg the average thickness above the top isopach line

h the isopach interval.

Connate (Intersticial) water in gas reservoirs: takes the amount of pore space available to gas affects gas recovery not uniformly distributed in the reservoir varies with the permeability, lithology, the height above water table

Calculation of average reservoir pressure after initial production

Well average pressure =n

pn

i0 (3.4)

Areal average pressure =

n

i

n

ii

A

Ap

0

0

(3.5)

Volumetric average pressure =

n

ii

n

iii

hA

hAp

0

0 (3.6)

Notes:n is the number of wells in (3.4) and of reservoir units in (3.5) and (3.6).

Volumetric average is used in the volumetric and material balance calculations.


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4

3. Recovery from Volumetric Gas Reservoirs

Gas-in-place in one unit (1 ac-ft) of bulk reservoir rock:

Vb = 1 ac-ft, Eq. 3.1 becomes:

g

w

B

SG

)1(560,43 =

(SCF/ac-ft) (3.7)

Gas at abandonment pressure:

ga

waB

SG

)1(560,43 =

(SCF/ac-ft) (3.8)

Bga the gas volume factor at the abandonment pressure.

For a reservoir under volumetric control, no change in the interstitial

(connate) water.

Unit recovery (Gas produced at abandonment pressure):

Unit recovery =

gagi

wBB

S11

)1(560,43 SCF/ac-ft (3.9)

Recovery factor =ga

gia

B

B

G

GG=

1

)((3.10)


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5

4. Recovery under Water Drive

Under water drive, after an initial decline, water enters the reservoir at a

rate to equal the production, and the pressure stabilizes. In this case thestabilized pressure is the abandonment pressure.

At the abandonment pressure, a unit (1 ac-ft) contains:

Volume of water: )1(560,43 grS ft3

Volume of residual gas in reservoir: grS560,43 ft3

Gas volume at surface:ga

gr

B

S560,43SCF

Sgr residual gas saturation

Unit recovery =

ga

gr

gi

wi

B

S

B

S1560,43 SCF/ac-ft (3.11)

Recovery factor =ga

gi

wi

gra

B

B

S

S

G

GG

=

11

)(

(3.12)

If the water drive is very active so that there is essentially no decline in

pressure, Eqs (3.11) and (3.12) become:

Unit Recovery =

gi

grwi

B

SSx

1560,43

Recovery Factor =

gi

grwi

B

SS1


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6

Discussion:

For the same residual gas saturation (or the same gas volume),higher abandonment pressure will retain more mass of gas, or therecovery will be greater for the lower stabilization pressure.

Generally gas recoveries by water drive are lower than byvolumetric depletion (the same conclusion does not apply to oil

recovery).

3.5 Material Balance

Derivation of Equation 3.15:

Material balance is used to find the gas in place for the reservoir as a

whole from production and PVT data.

For most gas reservoirs, the formation and water compressibilities are

negligible.

Assume: the pore space volume is constant

Vp = Vg + Vw (all are initial)

= GBgi + Wi

Vp = [gas volume] + [Initial water + water influx water produced]

Vp = (G Gp)Bg + Wi + We WpBw

G (Bg Bgi ) + We = GpBg + WpBw (3.15)


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7

If We = 0 and Wp = 0 Eq. (3.15) reduces to:

G (Bg Bgi ) = GpBg (3.16)

From Eq. (1.16),pT

zTpB

sc

scg =

We havepT

zTpB

sc

scg = and

isc

iiscgi

pT

TzpB = . SubstitutingBg andBgi gives:

pT

zTpG

pT

TzpG

pT

zTpG

sc

scp

isc

iisc

sc

sc = (3.17)

If the reservoir temperature is constant, T = Ti

p

zG

p

zG

p

zG p

i

i =

or

i

i

pi

i

z

p

GGz

p

z

p

+= (3.18)

Eq (3.18):

pi,zi, and G are constants p/z vs. Gp a straight line Slope =

Gz

p

i

i , y intercept =i

i

z

p

At anyp, Gp known, G can be found.


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8

Discussion of Fig. 3.6:

1) If p/z is set equal to zero, Gp = G2) The plot could be extrapolated to any abandonment p/z to find the

initial reserve.

3) A plot ofGp versusp is not linear (why?)4) In water-drive reservoirs, Gp vs.p/z is not linear. The pressure drops

less rapidly than in volumetric reservoirs.

If Eq (3.16) is expressed in terms of the initial gas pore volume, Vi

Vi = GBg,

pT

zTpB

sc

sc

g

=

isc

iscgi

pT

TzpB =

We have Eq (3.20)

zT

pV

Tz

Vp

T

Gpi

i

ii

sc

psc = (3.20)

Or

Tz

p

Tz

p

T

Gp

V

f

f

i

i

sc

psc

i

=(3.20-1)


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9

3.6. Gas Equivalent of Produced Condensate and Water

To modify Gp to include the condensate liquid production atsurface

To find gas and GE volume produced from low pressure separatorsusing correlations

GE of 1stock tank barrel (STB) condensate liquid:

SC

SCSC

p

TRnVSTBGE

')1( == (3.21)

wo

o

wo MM

massn

350== (1bbl of water = 350.5 lb)

For a three-stage separation system:

Gp = Gp(surf) + GE(Np) = Gps + Gss + GSt + GE(Np) (3.23)

For a two-stage separation system:

Gp = Gp(surf) + GE(Np) = Gps + Gst + GE(Np) (3.24)

Two correlations (diagrams) for the amount of gas and in low pressure

separators:

Gp = Gps + Veq (Np) (3.25)


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1


Chapter 4 Material Balance for Gas-Condensate

Reservoirs above Dew Point

1. Introduction

Gas-condensate reservoirs generally produce:

Liquids with gravities above 45oAPI (light color or colorless)

Gas-oil ratios in the range of 5,000 to 100,000 SCF/bbl

2. Calculation of Initial Gas and Condensate

To calculate initial gas and condensate from generally available field

data collected at pressures above dew point pressure

Derivation of Equation (4.1):

On the basis of 1STB and (R1 + R3) SCF gas, the mass of the well fluid

is:

Mass of the well fluid = Mass of the gas + Mass of the liquid

ooW RRxRR

m

350)(0764.0350380

29)(3311

3311++=+

+=

Total moles

woo

o

ot MRR

M

RRn /350)(00264.0

350

380

)(31

31

++=+

+=


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2

The molecular weight of the total well fluid

woo

o

t

WW

MRR

RR

n

mM

/350)(00264.0

350)(0764.0

31

2211

++

++==

The specific gravity of the well fluid is Mw/29, then

woo

oWW

MRR

RRM

/000,133)(

4600)(

29 31

3311

++

++== (4.1)

The molecular weight of the stock tank oil is given by the following

correlation:

o

o

APIo

woM

=

=

008.1

43.42

811.8

5954

,

(3.22)


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3


57/129

4


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5

3. Recombination of surface fluids composition known

The surface liquid and gases must be recombined in order to determine

the properties of the gas in the reservoir. Once the composition of the

reservoir gas has been calculated, its PVT properties can be predicted.

Known:

Compositions of the stock tank liquid, separator gas, and stocktank gas

The producing gas oil ratiosSteps:

Convert gas-oil ratios in SCF/STB to lb-mole gas/lb-mole stock

tank liquid (the density and molecular weight of the liquid must becalculated).

The gas-oil ratios of lb-moles are used to combine thecompositions of separator gas, stock tank gas and stock tank oil in

the proper ratios.

Example 4.2 A gas-condensate reservoir produces through a

separator at 250 psia and 70oF to a stock tank. The separator produces

86,000 SCF/STB and the stock tank produces 550 SCF/STB. The stocktank liquid gravity is 56

oAPI. Compositions of the separator gas, stock

tank gas and stock tank liquid is given below. Calculate the composition

of the reservoir gas.

Component SP gas composition ST gas composition ST liquid composition

C1 0.852 0.323 0.001

C2 0.085 0.184 0.006

C3 0.043 0.262 0.036

i-C4 0.005 0.046 0.068

n-C4 0.008 0.082 0.042i-C5 0.003 0.032 0.048

n-C5 0.002 0.036 0.035

C6 0.001 0.029 0.011

C7+ 0.001 0.006 0.753

1.000 1.000 1.000

Properties of C7+:

Specific gravity = 0.812

MW = 124 lb/lb mole


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Solutions:

MSTO = 108.5 lb/lb-mole

STO = 0.755

STO = 264.1 lb/STB

Mole ratios:

93.0 lb mole SP gas/lb-mole STO

0.595 lb mole SP gas/lb-mole STO

Composition of combined reservoir gas:

Calculation Results based on 1 lb-mole STO

ComponentSP gascomposition

lb moleSP gas

ST gascomposition

lb mole STgas

ST liquidcomposition

Total lbmoles

Core

C1 0.852 79.236 0.323 0.192 0.001 79.429

C2 0.085 7.905 0.184 0.109 0.006 8.020

C3 0.043 3.999 0.262 0.156 0.036 4.191

i-C4 0.005 0.465 0.046 0.027 0.068 0.560

n-C4 0.008 0.744 0.082 0.049 0.042 0.835

i-C5 0.003 0.279 0.032 0.019 0.048 0.346

n-C5 0.002 0.186 0.036 0.021 0.035 0.242

C6 0.001 0.093 0.029 0.017 0.011 0.121

C7+ 0.001 0.093 0.006 0.004 0.753 0.850

1.000 93.000 1.000 0.595 1.000 94.595


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1


Chapter 5 Undersaturated Oil Reservoirs

1. Introduction

Undersaturated oil reservoirs:

Initial reservoir pressure is higher than the bubble point pressureof the oil (no gas cap).

Free gas develops afterp < pb. There may be water influx.

2. Oil in Place and Oil Recoveries

Under initial conditions a unit (1 ac-ft) oil reservoir contains:

Interstitial Water wS758,7

Reservoir oil )1(758,7 wS

Stock tank oiloi

w

B

S )1(758,7

(1 ac-ft = 7758 barrels)

For oil reservoirs without water influx, at abandonment conditions 1 ac-ft of bulk rock contains:

Interstitial water wS758,7

Reservoir gas gS758,7

Reservoir oil )1(758,7 gw SS

Stock tank oilo

gw

B

SS )1(758,7

Sg = the gas saturation

Bo = oil volume factor at abandonment


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2

Recovery per ac-ft (in STB):

Recovery =

o

gw

oi

w

B

SS

B

S )1()1(

758,7 (5.1)

Recovery factor =o

oi

w

gw

B

B

S

SS

)1(

)1(1

(5.2)

For reservoirs under hydraulic control (no appreciable pressure decline),

the oil remaining at abandonment in barrel per acre-foot, is:

Reservoir oil orS758,7

Stock tank oiloi

or

B

S758,7

Sor= residual oil saturation after water displacement

Recovery =oi

orw

B

SS )1(758,7 (5.3)

Recovery factor = )1(

)1(

w

orw

S

SS

(5.4)

Correlation of oil recovery by water-drive recovery (residual oil

saturation) in sandstone reservoirs:

RF = 0.114+0.272log k+0.256 Sw -0.136 logo-1.538-0.00035 h (5.5)


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3

3. Material Balance

@ p > pb

1)Neglecting change in porosity of rocks with the change of internalfluid pressure

2)Zero or negligible water influx3)Initially undersaturated (initially only connate water and oil, with

their solution gas)

4)Solubility of gas in reservoir water negligible5)Water production small and negligible6)FromPi toPb, the reservoir oil volume constant, and oil produced

by liquid expansion.

Incorporating the above assumptions and considering volume balance,

Eq (5.7) can be derived as follows:

Reservoir oil volume = Constant

[Initial reservoir oil volume] = [Reservoir oil volume @ p > pb]

NBoi = (N Np)Bo

@p > pb, Bt= Bo and Boi = Bti (see Fig. 5.2), rearranging the aboveequation gives:

N(Bt Bti) = NpBt (5.7) (textbook)

N(Bo Boi) = NpBo (5.7)

Fractional recovery:

t

tit

o

oiop

B

BB

B

BB

N

NRF

=

== (5.8)


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4

@ p < pb

Below bubble point pressure, a free gas phase develops. The

hydrocarbon pore volume remains constant, that is

Voi = Vo + Vg (5.9)

Terms in Eq (5.9):

Voi = NBoiVo = (N Np)BoVg= [Initial total gas] - [solution gas in res.] [produced gas]

= [NRsoi (N - Np) Rso NpRp]Bg

Eq (5.9) can be expanded to:

NBoi = (N Np)Bo + N(Rsoi Rso)Bg + Np(Rso Rp)Bg


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6

4. Calculations Including Formation and Water Compressibilities

The effect of compressibilities on calculations forNis examined.

@ p > pb

Derivation of Equation (5.21)

1. Recall the derivation of Eq. (5.7):

Reservoir oil volume = Constant

[Initial reservoir oil volume] = [Reservoir oil volume @ p > pb]

NBoi = (N Np)Bo

2. Consider the water influx and water production (without considering

compressibility of water and formation):

Reservoir volume = Constant @ p > pb

NBoi + W = (N Np)Bo + (W + We WpBw)Oil Water Oil Water

3. Consider the compressibilities of water and formation

Volume of pore space decreases pcS

NB

pcV fwi

oi

ff

=

1

Expansion of water pcSS

NBpcSVpWc wwi

wi

oiwwifw

==

1


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7

NBoi + W pcV ff = (N Np)Bo + (W + We WpBw) + pWcw

Then

pwope

wi

fwiw

oioio WBBNWpS

cScNBBBN +=+

++

1)( (5.21)

Solve for N:

pS

cScBBB

WBWBNN

wi

fwiw

oioio

pweop

++

+=

1

(5.22)

Oil compressibility,co, is often used with the following definition:

pB

BB

ppv

vvc

oi

oio

ioi

oioo

=

=

)(

Therefore,

pcBBB ooioio += (5.23)

Expansion of oil pcSS

NBpcSVpcNB oo

wi

oioofooi

==

1

(Above the bubble point,Bo = Bt, andBoi = Bti)


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8

Substituting (5.23) into (5.21) and rearranging gives:

pweop

wi

fwiwoo

oi WBWBNp

S

cScScNB +=

++

1 (5.25)

Effective fluid compressibility,ce, is defined as:wi

fwiwoo

eS

cScScc

++=

1

Eq. (5.25) can be written as

pweopeoi WBWBNpcNB += (5.27)

For volumetric reservoir, We =0, and Wp is negligible, and from (5.27):

oi

o

e

p

B

B

pc

NN

=

(5.28)

Ifcf= cw = 0, ce = co, and oiooio BBpBc = , then (5.28) reduces to(5.8):

t

tit

o

oiop

B

BB

B

BB

N

N =

= (5.8)

@ p < pb,

pS

cScBBB

WBWBRRBNN

wi

fwiw

titit

pwegsoiptp

++

++=

1

)(

(5.29)


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Change in Oil Volume:

= = --

Oil volume change = NBoi (NNp)Bo (2.1)

Change in Free Gas Volume:

Define:

oi

gi

NB

GB

volumeoilreservoirInitial

volumegasfreereservoirInitialm ==

Initial free gas volume = GBgi = mNBoi

Free gas in the reservoir at t =

(in SCF)

( ) sopppsoioi

oi RNNRNNRB

mNBttimeatgasFree

+= (in SCF)

( ) gsopppsoioioi BRNNRNNR

B

mNBttimeatvolumegasfreeservoir

+=Re

( )gsopppsoi

oi

oi

oiBRNNRNNR

B

mNBmNBvolumegasfreeinChange

+= (2.2)

Change in

oil volume

Initial reservoir

oil volume

Oil volume

at p

Initial gas

free +dissolvedGas

produced

Gas remaining

in solution


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Change in Water Volume:

Change in water volume =

+ (all in SCF)

pWcBWWvolumewaterinChange wwpe = (in bbls) (2.3)

Change in the Void Space Volume:

pcVspaceporeinChange ff = (2.4)

Initial void space can be expressed as:wi

oioif

S

mNBNBV

+=

1

Initial water reservoir volume: wiwi

oioiwif S

S

mNBNBSVW

+==

1

Combination of the changes in water and rock volumes:

+

= pcS

mNBNBpcS

S

mNBNBBWW f

wi

oioiwwi

wi

oioiwpe

++

++

11

= pS

cScNBmBWW

wi

fwiw

oiwpe

+++

1)1( (2.5)

Water

influx

Water

produced

Water

expansion

Change in

water

volume

Change in

pore

volume


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Equating the changes in the oil and free gas volumes to the negative of the changes in the

water and rock volumes:

+ = + +

sogpsoggppgsoi

gi

goi

oiopooi RBNRNBBRNBNRB

BmNBmNBBNNBNB ++

++

=p

S

cScNBmBWW

wi

fwiw

oiwpe

+++

1)1(

gsosoiot BRRBB )( +=

And tioi BB =

[ ]

+++

gi

g

tigsoiptpttiB

BmNBBRRBNBBN 1)()(

= pS

cScNBmBWW

wi

fwiw

tiwpe

+++

1)1( (2.6)

Rearranging Equation 2.6 gives:

ewi

fwiw

tigiggi

ti

tit WpS

cSc

NBmBBB

mNB

BBN +

+

+++ 1)1()()(

= wpgsoiptp BWBRRBN ++ )( (2.7)

Change inoil volume

Change infree ga s

volume

Change inwater

volume

Change inpore space

volume


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5

Discussion of Eq. (2.7):

On the left-hand side:

The first two terms account for the expansion of any oiland/or gas zones.

The third term accounts for the change in pore volume, whichincludes pore space decrease and expansion of the connate

water.

The fourth term is the amount of water influx that hasoccurred into the reservoir.

On the right-hand side:

The first term represents the production of oil and gas. The second term represents the water production.


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1


Chapter 6 Saturated Oil Reservoirs

1. Introduction

Reservoirs with an initial gas cap:

The oil is initially saturated. The energy stored in solution gas is supplemented by that in the gas

cap.

Recoveries from gas cap reservoirs are generally higher than fromthose without caps, other things remaining equal.

Gas cap retards pressure decline and t herefore the l iberation ofsolution gas within the oil zone. Expansion of gas cap displaces oil downward toward the wells and,

therefore, vertical component of fluid flow is important for

gravitational segregation.

Recoveries from volumetric gas cap reservoirs could range fromthe recoveries for unders aturated reservoirs up to 70 to 80% of the

initial stock tank oil in place.

Good gravitational segregation characteristics include: pronouncedformation structure, low oil visc osity, high permeability, and low

oil velocity.

Water drive/hydraulic controlled recovery process:

The vol ume of t he r eservoir i s constantl y reduced by t he wat erinflux.

The re servoir p ressure is re lated to th e ra tio o f wate r in flux tovoidage.

Under favorable conditions, the o il recoveries are high and rangefrom 60 to 80% of the oil in place.


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4. Laboratory PVT Studies (Reservoir Fluid Studies, PVT Studies, Phase

Behavior Studies)

Reservoir fl uid s tudies pr ovide values of physi cal properti es of oi l

required in material balance calculation and reservoir simulation.

Physical properties of oil and gas

Bubble point pressure Formation volume factor of oil Solution gas-oil ratio Total formation volume factor Coefficient of isothermal compressibility of oil Oil viscosity Z factor of gas Formation volume factor of gas Gas viscosity Solubility of gas in oil as a function of pressure

The five major tests

Composition measurement Flash liberation Differential liberation Separator tests Oil viscosity measurement

Flash liberation process: (Constant composition expansion) All the gas

evolved during a reducti on i n pres sure re mains in con tact a nd in

equilibrium with the liquid phase from which it is liberated.

Differential process: T he gas evol ved duri ng a pressure reduction is

removed from contact with the liquid phase as rapidly as it is liberated.

Separator test: Simulation of a separation process.


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Chapter 7 Single Phase Flow in Reservoirs

1. Introduction Flow of fluids in reservoirs and the models that are used to relate

reservoir pressure to flow rate.

Introduction to well testing. Discussion limited to single-phase flow.

2. Darcys law of single phase flow

Darcy's law (1856): sPkAq =

k= permeability, darcy (D)

= fluid viscosity, cp

A = cross-sectional area of the porous medium sample, cm2

p = pressure, atm

s = Length of the porous medium sample, cm

Definition of the permeability of a porous medium:

If a fluid of1 cp flow through a porous medium of a cross-sectional area

of 1cm2

and a length of1 cm at a flow rate of1 cm3/sec under a

pressure gradient of1 atm/cm, the permeability of the porous medium is

1 darcy.

1 D = 1000 millidarcies (md)


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2

3. The classification of reservoir flow system

Three types of reservoir fluids:

1) incompressible Simplifies many equations and sufficiently accurate for many

purposes

No true incompressible fluids2) slightly compressible

Describes nearly all liquids by the following equation (c is constant):

)( ppc

RReVV

= (7.2)

R= reference conditions

Equ (7.2) can be approximate as:

)](1[ ppcVV RR += (7.3)

3) compressible

All gases are in this category.

Volume change with pressure asp

TznRV

'

= (1.7)

cg change with pressure as

dp

dz

zp

cg11

= (1.19)


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3

4. Steady-state flow

4.1 Liner flow of incompressible fluids

L

ppkAq

)(001127.0 21

=

(bbl/d)

or

L

pp

B

kAq

)(001127.0 21

=

(STB/day) (7.7)

Example:

pressure differential = 100 psi

permeability = 250 md

viscosity = 2.5 cpformation volume factor = 1.127 bbl/STB

length = 450 ft

cross-sectional area = 45 sq ft

q = 1.0 STB/day


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4

4.2 Linear flow of slightly compressible fluids

Recall Eq (7.3) )](1[ ppcVV RR +=

There is:)](1[ ppcqq

RR+=

Darcys Law is written as:

dx

dpkppc

A

q

A

qR

R

001127.0)](1[ =+==

Separating variables and integrating give Eq (7.9):

4.3 Linear flow of compressible fluids

Units: For gas flow rate, we want to use SCF/day; In Darcys law under

field unit system, flow rate is in bbl/day.

Recall Eq (1.17): pT

TzpB

SC

SCg

615.5= (bbl/SCF)

q (SCF/day)x Bg (bbl/SCF) = qBg (bbl/day)

pT

TzqpqB

SC

SC

g 615.5=

(q in SCF/day)

Darcys law:


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5

Variation of viscosity and z-factor:

Fig. 7.4 Isothermal variation ofz with pressure

Variation ofz vs. pressure of a real gas:

Forp < 2000 psia, nearly constant

Forp > 2000 psia, z/p is nearly constant

For T, k, z orz/p are constant, there is Eq (7.10):

(7.10)


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Chapter 7 Single Phase Flow in Reservoirs II


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1


Chapter 8

Water Influx

1. Introduction Many reservoirs are bounded on a portion or all their perimeters by

water bearing rocks aquifers.

As reservoir fluids are produced, a pressure differential developsbetween the surrounding aquifer and the reservoir. The aquifer

reacts by encroaching across the original hydrocarbon-water

contact.

Aquifers retard pressure decline in reservoirs by providing a sourceof water influx We.

We is a function of time (production). We is dependent on the size of aquifer and the pressure drop from

the aquifer to the reservoir.

2. Steady-state method

Schilthuis Steady-state method is the simplest model for water influx.

Water influx is proportional to the pressure drawdown (pi p):

)( ppkdt

dWi

e = (8.1)

Integrating Eq (8.1) gives

=t

ie dtppkW0

)( (8.2)

Where k = water flux constant, bbl/day-psip = pressure at the original oil-water contact

pi = initial pressure at the external boundary of the aquifer.


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2

Calculation of k and We from production data:

In a reasonably long period, if the production rate and reservoir pressure

remain substantially constant, there is:

Reservoir voidage rate = Water influx rate

(Volumetric withdraw rate)

dt

dWB

dt

dNBRR

dt

dNB

dt

dW pw

p

gso

p

oe ++= )( (8.3)

where

dt

dNp

= daily oil production rate, STB/day

dt

dNRR

p

so )( = daily free gas production rate, SCF/day

dt

dWp = daily water production rate, STB/day

Equation (8.3) can be rearranged to:

(by adding and subtractingdt

dNBR

p

gsoi )

dt

dWB

dt

dNBRR

dt

dNB

dt

dW pw

p

gsoi

p

te ++= )( (8.4)

k can be found from Equation (8.1)

)(

1

ppdt

dWk

i

e

= (8.5)


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3

If the pressure stabilizes and the withdraw rates are not reasonably

constant, water influx in the pressure stabilized period tcan becalculated from the total productions of oil, gas and water within t:

pwgpsoippte WBBNRGNBW ++= )( (8.6)

Then k can be found from the following equation:

)( ppt

Wk

i

e

= (8.7)

For an under-saturated oil reservoir and at pressures higher than the

bubble point pressure, Equation can be simplified to:

dt

dWB

dt

dNB

dt

dW pw

p

te += (8.8)


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3. VEH unsteady-state method

Van Everdingen and Hurst solutions to the single-phase unsteady-state flow equation is used to calculate water influx.

The hydrocarbon reservoir is the inner boundary condition and isanalogous to the well in Chapter 7 and the aquifer is the flowmedium analogous to the reservoir in Chapter 7.

Properties of aquifer are assumed homogeneous and constant. Reservoir and aquifer are assumed cylindrical in shape.

Water flux is calculated by the following equations:

j

n

j

eDje pWBW = =1

(8.9)

In Where eDW is given as a function of dimensionless time Dt and

dimensionless radius Dr (see Tables 8.2 and 8.2 and Figures 8.7-8.10):

)( , DDeD trfW = (8.10)and

360119.1 2

hrcB Rt= (8.11)

re

rR

re

rR


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6

The dimensionless time and dimensionless radius are defined as

20002637.0

Rt

Drc

tkt

=

(8.12)

R

eD

r

rr =

(8.13)

In Equations (8.9) and (8.11), is the angle subtended by a pie-shaped

cylindrical reservoir. For example:

For a circle, = 360o

For a half circle, = 180o.


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WeD as a function of tD and rD.


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10

Values forpj are determined from measure pressures. The pressure

changes are calculated as follows to approximate the pressure-time

curve:

)(2

111 ppp i = (8.14)

)(2

122 ppp i = (8.15)

)(2

133 ppp i = (8.16)

and )(

2

12 jjj ppp = (8.17)

Pressure steps used to approximate the pressure-time curve.


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4. Calculate the water influx during a time step

=

ei

ni

W

tjp

Rnn

i

eie epp

p

WW 1)( 1 (2.22)

1np = average aquifer pressure at the end of (n-1)th time step

5. Calculate the total cumulative water influx at the current time

=

=n

j

eje WW1

(2.23)

6. Calculate the average aquifer pressure at the end of the current

timestep

=

ei

ein

W

Wpp 1 (2.24)

7. Repeat Steps 3 to 6 for next time step.


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1


Chapter 9

Decline Analysis

1. Introduction The most widely used method of forecasting future production

form oil and gas fields since Arps developed the technique in 1945.

The technique has few fundamental theoretical foundations. The simplicity and success of its forecasts responsible for its

general acceptance and use.

Decline analysis involves the use of curves of flow rate versus timeof the well.

2. Arps Decline Model

Arps decline model relates the production rate with time:

n

i

it

tnd

qq

/1)1( +

= (9.1)

Where qi = initial oil production rate

qt= production rate at time t

di = initial decline rate

n = hyperbolic decline exponent

The decline rate is defined as the rate of change of the natural logarithm

of the production rate with respect to time:

dt

qd

dt

dq

qd

)(ln1== (9.2)

The minus sign has been added because dq and dthave opposite signs

and it is convenient to have dalways positive.


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Arps recognized the following three types of rate decline behaviour:

1. Exponential decline, n = 0td

it

i

eqq

= (9.3)The cumulative production at time tis given by

i

tip

d

qqN

=

(9.4)

2.Hyperbolic decline, 0 < n < 1

ni

it

tnd

qq

/1)1( +

=(9.5)

and

( )ntnin

ip qq

n

qN

= 11

)1( (9.6)

3.Harmonic decline, n = 1

)1( td

q

qi

i

t+= (9.7)

and

=

t

i

i

ip

q

q

d

qN ln

(9.8)

Example 9.1 -- Production forecastingUsing the following production data of a well to determine the

production rate and cumulative production at t = 3 years using (1)

exponential decline, (2) hyperbolic decline (n = 0.6), and (3) harmonic

decline:

qi = 500 STB/d , di = 1.8% /month


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3. Parameter Estimation for Decline Analysis

1. Exponential decline

The most widely used of the three methods, primarily because it is easyto determined the parameters and gives a more conservative estimation

of the future reservoirs for given parameters.

Ifn = 0, the decline is exponential.

qi and di need to be determined.

Taking the natural log of both sides of the equation (9.3) gives:

tdqq iit = lnln (9.9)

A plot of lnqtvs. tgives a straight line:

slope m = -di,

y-intercept b = lnqt

Figure 1. Exponential decline curve: lnqtvs. t (from examples 1)

5.2

5.4

5.6

5.8

6

6.2

6.4

0 5 10 15 20 25 30 35 40

t (month)

lnq

(STB/d)


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3.Hyperbolic decline

All real wells have hyperbolic decline. Approaches to calculation of parameters of hyperbolic decline

include trial-and-error methods, type-curve overlays, graphical

methods, mathematical or statistical analysis.

Hypobolic decline-curve analysis using linear regression (Towler and

Bansal, J. Petroleum Science and Engineering, 8, 257, 1993)

Method 1: Using Equation (9.5)

Equation (9.5) can be rearranged to:

logqt=logqi 1/n x log (1+ndit) (9.12)

1)Assume a value ofndi;2)Plot logqtvs. log (1+ndit);3)Try ndi to obtain a straight line or a maximum regression

coefficient (R2).

Slope m =1/n

y-intercept b = logqi

di = (ndi)/n

Figure 3. Linear regression of hyperbolic decline data, ndi = 0.0108

(month-1

)

y = -1.6667x + 2.699

R2

= 1

2.4

2.6

2.8

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16

Log (1+ndit)

logqt

.


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(reservoir engineering course)

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