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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.
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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.
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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
Introduction to Reservoir Engineering
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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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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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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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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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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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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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
Introduction to Reservoir Engineering
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. 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
Introduction to Reservoir Engineering
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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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. 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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@ 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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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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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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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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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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Introduction to Reservoir Engineering
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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Introduction to Reservoir Engineering
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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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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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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4.2 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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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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Introduction to Reservoir Engineering
Chapter 7 Single Phase Flow in Reservoirs II
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Introduction to Reservoir Engineering
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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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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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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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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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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Introduction to Reservoir Engineering
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
/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
/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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