Reservior Rock Properties Essentials
January 17, 2023 | Author: Anonymous | Category: N/A
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RESERVOIR ROCK
PROPERTIES
Essentials Edited by Dr. Nayef Alyafei
Special thanks: Abdulsaboor Khan Afsha Shaikh Maryam Al-Suwaidi Mariam Obidan Mashaal Al-Salem Mohammed Al-Mohannadi Mohamed Idris Rand Alagha Sara Al-Marzouqi Tarek Saad Wajd Wa jdii Al ou oush sh
1
Contents
2
Conversion of units Porosity Rock compressibility Liquid permeability Gas permeability Flow in layered systems
4 5 6 7 8 9
Fluid saturation Electrical properties Wettability Capillary pressure Relative permeability Volumetric estimation of hydrocarbons
11 12 13 14 17 18
What is this booklet? This booklet summarizes the dierent topics covered in the Fundamentals of Reservoir Rock Properties book in a simple and easy-to-follow format with a one-page article covering each concept. The booklet highlights each property with a basic knowledge that intends to convey a general understanding of the covered topics. The purpose of this booklet is to give a quick overview of the reservoir rock properties. The Fundamentals of Reservoir Rock Properties book can be downloaded from the following link: https://www.qscience.com/content/books/9789927137273 Why studying Reservoir Rock Properties is important? Reservoir rock properties are part of petrophysics, which is the study of rock properties and rock-uid properties. Reservoir rock properties are essential aspects of understanding hydrocarbon reservoirs. Those properties should help in estimating the amount of hydrocarbons in reservoirs as well as understanding the production behavior of hydrocarbons. DOI: https://doi.org https://doi.org/10.5339/Reservoir_Ro /10.5339/Reservoir_Rock_Properties_ ck_Properties_Essentials Essentials
3
CONVERSION OF UNITS Length
Area
1 ft = 0.3048 m = 12 in
1 ft2 = 0.092903 m2 = 144 in2
1 m = 3.281 ft = 39.37 in = 100 cm
1 m2 = 10.7649 ft2 = 10000 cm2
Mass
Force
1
1 lbf = 4.44822 N = 32.2 lbm.ft/s2
= 0.45359 kg
2
1 kg = 2.2046 lbm = 1000 g
1 N = 0.2248 lbf = 1 kg.m/s
Interfacial Tension
Permeability
1 N/m = 1000 mN/m = 1000 dyne/cm
1 D = 1000 mD = 9.869233 x 10-13 m2
Volume 1 ft3 = 0.02831 m3 = 28.3168 L = 0.178 bbl = 0.178 RB 1 m3 = 35.29 ft3 = 1000 L
Pressure 1 atm = 101.3 kPa = 1.013 bar = 14.696 lbf/in2 (psia) 1 psia = 6.89 kPa = atm/14.696 1 Pa = 1 N/m2 = 1 kg/m.s2 = 10-5 bar = 1.450 x 10 -4 lbf/in2 = 10 dyne/cm2 psia = psig +14.7
Density 1 g/cc = 1000 kg/m3 = 62.427 lb/ft 3 = 8.345 lb/gal = 0.03361 lb/in3
Viscosity 1 cP = 0.01 poise = 0.01 g/cm.s = 0.001 kg/m.s = 0.001 n.s/m2 = 0.001 Pa.s = 0.01 dyne.s/cm2 = 6.72 x 10-4 lbm/ft.s = 2.09 x 10 -5 lbf.s/ft2
Metric Prexes Prex
Symbol
Multiplication Factor
giga
G
109
mega
M
106
kilo
k
103
centi
c
10-2
milli
m
10-3
micro
µ
10-6
nano
n
10-9
Oileld Prexes
4
Prex
Symbol
Multiplication Fa Factor
Thousand
M
103
Million
MM
106
Billion
MMM or B
109
Trillion
T
1012
POROSITY What is Porosity? Porosity (ф) is the fraction or percentage of empty volume that exists in a rock over the total volume. This is where water, oil and gas reside in a rock. Think of the reservoir like a sponge, the way a sponge absorbs water in the empty spaces is the same way that the uids occupy the pore spaces in a rock. The general formula for porosity is:
φ
=
V p V b
=
Factors Aecting Porosity 1. Primary porosity – developed during the burial of the rock: Sorting
V b − V m V b
where: ф = porosity [fraction or %] V p = pore volume [cm3 or ml] V b = bulk volume [cm3 or ml] V m = matrix volume [cm3 or ml]
2. Secondary porosity (induced) – developed after deposition due to external factors:
Types of Porosity There are two types of porosity:
a. Compaction due to stress (decreases porosity):
• •
Well sorted
Poorly sorted
Total porosity (фt ) Eective porosity (фe)
The total porosity takes all the pore space in a rock into consideration. Whereas, eective porosity, is the fraction of the volume of interconnected space in a rock over the bulk volume. As petroleum engineers, we mostly care about the eective porosity. Imagine a school with an area of 100 m2, 25 m2 of those 100 m 2 are occupied by ve hallways, and for simplicity let us say all the hallways have the same area. Now imagine that two of those hallways are closed. Students then can only walk through three of the hallways, or 15 m2. In the same sense, uids in the rock can only ow through pores that are not closed but are connected.
Unpacked
packed
b. Fractures (increase porosity):
c. Dissolutions (Vugs) (increase porosity):
5
ROCK COMPRESSIBILITY Porosity is the fraction of a rock that consists of void space. A key aspect in studying porosity is studying rock compressibility.
Types of Compressibility Compressibility plays a major role on reservoir performance. It is divided into three types:
Rock Compaction Porosity is reduced by compaction, and the reduction is a function of the burial depth of the rock. Isothermal compressibility is dened as the change of volume per unit pressure change at constant temperature.
• • •
Matrix Compressibility, c m Pore or formation Compressibility, c p/f Bulk Compressibility, c b
F ov
Eects of Compaction • Changes in packing • Deformation of rock fragments
Simplifying Assumptions • Reservoirs are isothermal • As pressure increases, material volume decreases • As pressure decreases, material volume increases These assumptions are used to derive the compressibility equation shown below
c =
−
F m
F f
For static conditions; The summation of forces should equal zero
1 dV
F ov ov = F m + F f
V dP
Where F ov ov is the overburden force, F m is the matrix force resisting deformation, and F f is the uid force.
where:
= coecient of isothermal compressibility c = [1/psia] V = volume [ft3] P = pressure [psia]
Similarly for pressures, Pov = Pm + P f
As uids are produced from a reservoir, the uid pressure in the pores decreases; however, the overburden pressure remains constant causing the following eect: 1. Force on matrix increases 2. Bulk volume decreases 3. Pore volume decreases
This eect pushes some uids out of the reservoir, as the pore volume is decreasing with compaction. Also, this might cause subsidence, where formation thickness decreases to compensate for the produced uids.
6
LIQUID PERMEABILITY What is Liquid Permeability? Pores usually contain water in them referred to as brine. Brine is water that contains more dissolved salt than typical seawater. In other words, brine is a solution of salt in water.
Permeability is a measure of the uid conductivity of the rock, which means that permeability is the term used for uid transport.
Permeability can be calculated using Darcy’s law for 1-D, single phase liquid (in Darcy’s units):
q
=
kA dP µ dx
Where k = = permeability [D]
Understanding Permeability It is important to understand permeability, as it inuences the production of hydrocarbons. As previously discussed hydrocarbons are stored in the pores of the reservoir rocks. Now you might be wondering what the relationship between porosity and permeability is. Permeability is the ability of uids to be transported through porous rocks. If the pore spaces in rocks are not connected, then the rock is not permeable. The uid then, will be trapped
and unable to move. If the pore spaces in rocks are connected, then uid can be transported so it is permeable. The gure below provides an illustration of how low and high permeability look like.
dP dx
= pressure gradient [atm/cm]
A A = = cross sectional area [cm 2] µ µ = = viscosity of liquid [cP] q = volumetric ow [cm 3/s] When nding the permeability using Darcy’s law, there are certain conditions that must be assumed for Darcy’s law to be valid: •
The pore spaces should be fully saturated with a single phase; water, oil, or gas.
• •
The ow must be laminar. The ow must be at steady-state condition.
Low permeability reigon
High permeability reigon
7
GAS PERMEABILITY Introduction Often dry gas (He, N2, air) is used in the laboratory for permeability determination since it is readily available and is relatively less reactive with the rock. Gas Permeability Since gases are compressible, the average pressure Pm needs to be used, where: P m =
P 1 + P 2
2
Considering an ideal gas behavior of gas, we can assume:
Klinkenberg Eect Klinkenberg (1941) noticed that permeability measurements using liquids and gases gave dierent results. Permeability with air gave a higher value than permeability measurement with water. This phenomenon was attributed to gas slippage at the pore walls. Liquids have zero velocity at the sand grain surface; however, gases have some nite velocity at the sand grain surface. The slippage results in a higher owrate for a given pressure dierential for the gases. Klinkenberg related the liquid permeability k L to gas permeability k g by the following equation: kg = c
P 1 q 1
=
P 2 q 2
=
P q m
1 P m
+ kL
m
Thus, Darcy’s law equation becomes:
P q sc sc
sc sc =
P q m
kA(P 12 − P 22 ) q sc sc = 2µg LP sc sc where, qsc = gas owrate at standard conditions [cm3/s] Psc = pressure at atmospheric conditions [≈ 1atm] k = = absolute permeability [D] A = A = cross sectional area [cm 2] P1 = inlet pressure [atm] P2 = outlet pressure [atm] µ g = gas viscosity [cP] L = length of core [cm]
8
Slippage
m
k g
Absloute or Liquid Permeability
1 Pm
If we observe from the graph, the higher the value of Pm, the lower the value of 1/P 1/Pm. At extremely high pressures (innite pressure), gases are believed to behave like liquids. Thus if we extrapolate our 1/P 1/Pm to zero, it would mean the permeability would be approximately equal to the liquid permeability. The slope c of of the line is a function of absolute permeability, type of gas, and average radius of the rock capillaries.
FLOW IN LAYERED SYSTEMS The objective of understanding the ow in a layered system is to nd the average permeability of the system and its various beddings. The average permeability calculation will be centered around Darcy’s law but will vary based on the conguration of the layers as described in the following cases: Case 1: Linear Flow in Parallel Consider the three-layer system below where the layers are in parallel and each layer has a dierent permeability. In this case, the pressure drop in each layer is the same. The ow rate and thickness, however, are unique and sum up to the total ow rate and thickness of the system. Combining these relationships with Darcy’s law results in the average permeability formula for this system.
¯ k =
L
n
Li i=1 ki
Case 3: Radial Flow in Parallel The ow in parallel systems for liner and radial cases is similar. Consider the system below, the overall ow rate and total thickness is the summation of the ow rate and thickness value of each layer. The pressure drop, however, is common. Combining these conditions with Darcy’s law for radial ow results in a calculation of average permeability for this case.
r e r W
q1
P1
P2
k 1
h1
q1
k 2
h2
q2
k 3
h3
q 2
k 1 q
h
h1
k 2
q3
q
h2
q3
h k 3
h3
¯ k =
L
¯ k =
kh i
i
i
h
i=1
n
i
h
i=1
Case 2: Linear Flow in Series In this case, each bedding still has a dierent permeability but they are in series to each other. As a result, the ow rate that passes through each rock is the same but the pressure drop across each rock is dierent. Furthermore, the total length of the system is composed of the length of each rock. These conditions can be combined to calculate the average permeability of the system.
Case 4: Radial Flow in Series Finally, consider the radial system below. In this case, the ow rate across each layer is the same, however the pressure dierence in each layer varies. Combining these conditions allows for the average permeability for this system.
r e r 1
r W
h
P1
2 k
k 1
P2
k 1
q
kh n
W
P1 L1
k 2
k 3
P2
q
ln
P3
L2
L3
W L
h
¯ = k
re rw
r
n i=1
ln
(i+1) ri
ki
9
y g o l o h t i L
y t i s o r o P
g o L ) c i n o s ( c i t s u o c A
L f T T
d n u o S
l n o a i c t i g a c o l fi o i s e s G a l C
∆ ∆ − −
m m T T
∆ ∆ =
φ
g o L n o r t u e N
s n o r t u e N
g o L y t i s n e D
y a R a m m a G
y r a d n o c e S
r a l u n a r g a r t n I
y r a m i r P
m
V b − V b
V
r a l u n a r g r e t n I
y l t c e r i D d a e R
=
φ
p
b
V V =
φ
n o i t a r u t a S d i u l F
s a G
c i n a g r O
c i n a g r o n I
n i g i r O
10
) C H ( n o b r a c o r d y h g n i r : u c c M U o E y l L l O a r R u T t E a P N
y t i y s t i o s r o r o o P P
= φ
S A G
l i O
s l a m i n a & s t n a l p
m m
ρ ρ
D I U Q I L
l i O & s a G
y r a d n o c e S
− −
E R U S S E R P
d e m r s i o s f r y l e a p h : n a t p G a t a e N I & d d f G n l a o G o c i i O t h n o L i s p i i t E u a c r N q g n I c u L a o f E e R e g a I h o T f a s W
s n o i t c a e r l a c i m e h c
b ρ f ρ
F A =
P
S E T I R E P O R P S D I U L F
y t i s o c s i V
y t i s n e D
a P r o ² m / N
μ
m V =
ρ
S E I T R E P O R P R I O V R E S E R
P c
g s r i n o t t c c a e F ff
e d i m l o u v o f v o l k o u i t b a r R e v o
% 0 4 5 : e g n a R
g n n o i i r t e a e c fi n i i g s s n a E l C
g l n i i a t n r e e t a m e M C
n e e d r r u u s b r s e e r v P O
& e r s u g t u c V a r
s t y r n e o t m a r e o r b u a s L a e M
y r a m i r P
g n i k c a P
t n e d i m e u l c a F l p s i D
n o i s d n o a p h x t E e s M a G
s e i k t c r o e R p o r P
2 2
P
−
2 1
P
A L µ k 2 = a
q
? S C I S Y H P O R T E P Y H W
o t s e r c e n e n e i i c g s n l e a t m n u e e l m a o r t d e n u p F
t n u o m a g n i t C a H m f i t o s e n i s p l e H
g n i d n a t s r w e o d fl n d i u u n i fl s p l e H
y t i l i b a e m r e P
ff E
s
T
F
p
V P d d
p 1 V
A
: w a s L e s s ’ a y g c r r a o f D
e v i t c e
c c / g
s e d i s e r C H : e R I r e O h V w R e E c S a E l R P
l a t o T
k c h o c r i h s w u o e r s o a p e h e g h t u f o r o h t e r w u o s fl a e s d M i u fl
D , D m , ² m
: k
−
g n i t r o S
) N O E I R T , A f p O M c P R O F (
y t i l i b i s s e r p m o C k c o R
T
V P d d
1 V −
e g k a n p i h p r i l a i g S g h : s s g r o e r b c n a e e s c k a a g f n r i l f u K o s
d e d e n s i n o i t c e r r o C
2
P − 1
P
(
A L k µ =
q
e l b i s r e v e r r I
K L b U c B
c
X I m R c T A M
: i l e h l l i h a r k 1 a P n = i r a = e k n i ¯ L
: l e l l a r a P l e i d a R
i
h
i h
k
1
=
n
i
= ¯ k
1 i
d n e m i r e t w e s o y l a y F L s
e c a f r d u e s t k a c r u o r t a S h t i % w 0 t 0 c 1 w a e e o r b l F t o . o t d e t w n s i a d u t o l s e e fl S - F o e e r y a d n n d i d k o a m i c h e u o i t t a l S L • • F • R • w
)
w : a s d L i s u ’ q y i l c r r a o D f
f / p
c
=
: s k e L 1 i r L = e n i s r a e = n i ¯ L k i i
=
s e
n r i u t w c a o r l F F
2
2
h 1
= k
+ i r k i r : s e w e i r r n r l e 1 s n l l = n i e i d a R =
¯ k
s l e n i n n w a o l h F C
2
r
8
= k
FLUID SATURATION What is Saturation? Pores usually contain more than one uid in them. The fraction of pore space occupied by a specic uid is called saturation. Saturation is the ratio of uid volume to the pore volume. V S w
=
w
V p
V o
S o
=
S g
=
V p V g V p
Important Denitions ): it it is the minimum • Connate water (S w wc c ): water saturation which would remain adhered to the pores and not become mobile. During deposition, water saturation is 100% which declines by the migration of oil/ gas to (20-30%) and that is the connate water saturation. • Irreducible water saturation (S wir ): it it wir ): is the minimum water saturation in a rock forming a layer of water stuck to the rock surface and that does not move. Therefore, S w wc c = S wir wir ): it it is the oil left behind • Residual oil (S or or ): (could not be produced) after immscible displacement. This oil is trapped in the pore spaces and cannot move.
where: S w w = water saturation [fraction or %] S o = oil saturation [fraction or %] S g = gas saturation [fraction or %] w = water volume [cm3 or ml] V w V o = oil volume [cm3 or ml] V g = gas volume [cm3 or ml] 3 V p p = pore volume [cm or ml]
Types of Measurements There are two methods to determine or measure uid saturation • •
We could then notice that saturations are fractions, and collectively in pores the summation of all saturations should equal 1:
Look at the gure below for an idea on how uids look like in a pore
Direct: using Direct: using core samples through: - Retort distillation - Solvent extraction (Dean-Sta (Dean-Stark rk method) Indirect: capillary pressure and well log Indirect: capillary analysis (resistivity logs)
Factors Aecting Saturation There are many factors that aect saturation. For example, core samples from reservoir get aected by ltrate from drilling mud. The mud can usually be water or oil based and depending on the wettability of the rock could aect saturation dierently.
Another factor that aects uid saturation would be the dierences in pressure and temperature that are caused by bringing the core sample from the reservoir to the surface. Those dierences in pressure and temperature cause uids such as gas trapped in oil to be released. This is especially important to know when conducting direct measurements to nd the uid saturation using core samples. The reason uid saturation is very important is that it helps in estimating the amount of hydrocarbons stored in the reservoir.
Rock
water
Oil or gas
11
ELECTRICAL ELECTRICAL PROPERTIES The formation water (brine) contains dissolved salts which make up the electrolytes and help it conduct current within. These salts dissolve and break up into cations (Na +, Ca2+), and anions (Cl- and SO42-). The higher the salt concentration, the better it is at conducting electricity. Resistivity Resistivity is a measure of the material’s ability to resist the ow of electricity. The unit of resistivity is ohm.m [Ω.m]. The range for a poorly consolidated sand is 0.2 ohm.m to several ohm.m. For well consolidated sands the resistivity ranges from 1 ohm.m to 1000 ohm.m. Notations for resistivity Rw = formation water resistivity [Ω.m] Ro = resistivity of rock with 100% water saturation [Ω.m] Rt = true formation resistivity [Ω.m] Factors that aect resistivity • Water Resistivity • Formation Porosity • Formation Lithology • Type and Amount of Clay Formation Factor Formation factor relates Rw to Ro based on an empirical formula/relation developed by Archie (1942). He found out that this correlation is related to three parameters: a, ф, and m as shown in the following equation:
Saturation Equation Archie also found out the saturation equation (resistivity index) which relates Rt to Ro. He found out that this correlation is related to water saturation (S (S w ) and n is the saturation exponent. I r
Rt =
Ro
=
−n S w
Archie’s Equation Archie’s equation is the combination of the formation factor and the saturation equation since Ro is common in both equations. It is used to calculate the water saturation of un-invaded zones near the bore hole using wireline resistivity logs. Archie’s equation:
S w
n
aRw
=
m
φ Rt
Resistivity of Earth Materials
Rock y t i v i t s i s e R
Gas Oil Fresh Water
F
=
R R
o
m
=
aφ−
w
where: a = constant ≈ 1.0 for most formations ф = porosity [fraction] m = cementation factor ≈ 2 for most formations
12
Salt Water
C o n d u c t i v i t y
WETTABILITY What is Wettability? Wettability is a uid’s tendency to adhere on a solid surface, while co-existing with another immiscible uid.
Wetting phase fluid Oc Occu cupy py sm smal alle lest st por pores es
Non-wetting phase fluid Oc Occu cupy py larg larges estt po pore ress
Has low mobility
Usually of higher mobility
Immiscibility
Immiscible uids are uids that do not mix with one another. This is due to an imbalance in the molecular forces between two uids in contact with one another. These molecular forces present at the interface of two uids are known as the interfacial tension. Interface between Immiscible Fluids
How to determine the wettability of a rock? If a water droplet is surrounded by oil then; Water-Wet os > ws
Oil-Wet os < ws
AT > 0 (positive) 0 < < 90
AT < 0 (negative)
90 < < 180
What aects wettability? • Composition of pore-lining minerals • Composition of the uids • Saturation history
What are the types of wettability? • Oil- or water-wet • Neutral- or intermediate-w intermediate-wet et • Fractional- or mixed-wet
Adhesion Tension A key factor in determining the wettability of a rock is adhesion, also known as the dierence between two solid-uid interfacial tensions, is expressed with the following equation:
Wettability Aects • Capillary Pressure • Residual oil saturation • Relative permeability
AT = σsw
− σso = σwo cos θ
If the Adhesion tension is positive the denser uid will wet the rock becoming the wetting phase uid. For instance, water is the wetting phase in the gure below.
σ wo
θ
-x
σ so
σ sw
σ
+xso
13
CAPILLARY PRESSURE: PART 1 What is Capillary Pressure? Most of hydrocarbon reservoirs contain a minimum of two immiscible uids. For water and oil, the interface at which the separation happens is called the oil-water contact. Capillary pressure is dened as the pressure dierence between the two immiscible uids at the interface.
To further dene the capillary pressure, wettability needs to be recalled. In a heterogeneous system where one uid wets the surface (wetting phase) in preference to the other uid (non-wetting phase), the capillary pressure can be calculated as:
P c
=
P nw
− P
w
where: Pc = capillary pressure [Pa] Pnw = pressure of the non-wetting phase [Pa] Pw = pressure of the wetting phase [Pa] Understanding Capillary Pressure from Capillary Tubes To simplify the study of capillary pressure, porous media in a reservoir can be modeled as a collection of capillary tubes as shown below.
The height of the water in the capillary tube can be calculated using the following formula: ∆h =
2σwa cos θ ∆ρwa gr
where: σ wa wa = surface tension between air and water [N/m] g = = gravitational acceleration [9.8 m/s 2] Δρwa wa = density dierence between air and water [kg/m3] θ = contact angle between the surface and water [˚] Combining the two previous equations, capillary pressure can also be computed using: P c =
2σwa cos θ
r
Capillary Curves pressure is very As can be Pressure inferred, capillary dependent on the wetting phase saturation. To graphically represent the capillary pressure, the capillary pressure curves can be drawn as a function of this saturation. To do that, two uid ow processes need to be dened over which the capillary pressure varies. These are drainage and Imbibition. Drainage: The process through which the saturation of the non-wetting phase increases. +
For a water-gas system in a collection of capillary tubes, like in a reservoir, water (wetting phase) rises lling smaller pores and leaves the larger pores to be lled by air (non-wetting phase). The height of the water (shown below) in a capillary tube is a function of the adhesion tension between the air and water, the radius of the tube (shown below), and the density dierence between air and water (the two immiscible uids present). r θ
Δh Air
Drainage Pc
0 -
Imbibition
0
S wir
S w
Pct
S m 1
Imbibition: The process through which the saturation of the wetting phase increases. Parameters used in the gure above are the following: • S wir wir = irreducible wetting phase saturation [fraction or %] • S m = 1 – residual non-wetting phase saturation [fraction or %] • Pct = threshold capillary pressure, the
h = 0 Water
14
minimum pressure to force wetting uid into therequired largest pore [Pa]nonThe shapes of these curves are aected by reservoir quality. This includes the pore size distribution, grain sorting, and permeability.
CAPILLARY PRESSURE: PART 2 Measurement Capillary pressure laboratory measurements are a part of special core analysis (SCAL) and can be measured through three main methods: • Porous Plate Technique: Technique: Uses a low permeability water wet ceramic disc toboth retain inwater) a water wet core sample (containing oil oil and while allowing water to pass through. In this setup, the inlet pressure will be the oil pressure and the outlet pressure will be the water pressure. The dierence between the inlet and outlet pressure is the capillary pressure. • Mercury Injection Capillary Pressure (MICP): (MICP): The core sample is placed in a pressure chamber and mercury is injected through it. Mercury acts as the non-wetting phase while air acts as the wetting phase. The volume of mercury injected is calculated and can be converted to a non-wetting phase saturation. The wetting phase saturation will be one minus the non-wetting phase
Hydrostatic Pressure and Repeat Formation Tester (RFT) As can be inferred, capillary pressure is very dependent on the wetting phase saturation. To graphically represent the capillary pressure, the capillary pressure curves can be drawn as a
function of this need saturation. To do that, uidthe ow processes to be dened overtwo which capillary pressure varies. These are drainage and Imbibition. Drainage: The process through which the saturation of the non-wetting phase increases.
Pressure
( P g , Z g )
Gas Free oil level (FOL)
( Po , Z o )
h t p e D
Oil Free water level (FWL)
( Pw , Z w )
Water
saturation. • Centrifuge: Centrifuge: This This method can measure the drainage and waterood capillary pressure curves. For drainage, a core sample lled with the wetting phase is placed in a cell surrounded by the non-wetting phase. The cell is centrifuged at various speeds and the uid displaced is recorded at each speed. At each increment, the rotation speed is converted to pressure using specic equations.
Porous Plate Technique • The most accurate method • Very time consuming and can take a few hours to a few weeks • Non-destructive
MICP • The fasted method • Requires an extra conversion step and does not use reservoir fluids • Damaging to the core
Z F OL =
P o − P g + ρg gZ g − ρo gZ o (ρg − ρo )g
Z F W L =
P w − P o + ρo gZ o − ρw gZ w (ρo − ρw )g
Centrifuge • An intermediate method in terms of time and accuracy • Can use reservoir fluids
15
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