introduction to reservoir petrophysics
DESCRIPTION
Petroleum Reservoir PetrophysicsTRANSCRIPT
-
INTRODUCTION TO
RESERVOIR
PETROPHYSICS
MUHAMMAD SHOAIB (ROLL # 22)
HAFIZ MUHAMMAD BILAL (ROLL # 04)
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 2
INDEX
S.NO
TITLE
PAGE NO
1
POROSITY
04
2
COMPRESSIBILITY
13
3
SATURATION
26
4
PERMEABILITY
41
5
FORMATION RESISTIVITY
72
6
FORMATION VOLUME FACTOR
85
7
SOLUTION GAS RATIO
92
8
RESERVES IN PLACE
94
9
WETTABILITY
103
10
CAPILLARY PRESSURE
108
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 3
RESERVOIR PETROPHYSICS
It is basically the study of physical, mechanical, electrical and thermal
properties of reservoir rocks that describe the behavior and occurrence
of rock soil and fluids and their interaction.
RESERVOIR ROCK
Porous and permeable rock containing oil and gas trapped within void spaces of this rock
PHYSICAL PROPERTIES OF RESERVOIR ROCK
Porosity
Permeability
Fluid saturation
Compressibility
Wettability
Capillary pressure
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 4
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 5
POROSITY
It is the ratio of pore volume to the bulk volume
Porosity = = Vp / Vb = (Vb Vgr)/ Vb
Where;
Vb= bulk volume
Vgr= Sand grain volume
Vp= pore volume
According to this definition, the porosity of porous material could have any value, but porosity of most
sedimentary rocks is generally less than 50%.
FACTORS GOVERNING THE MAGNITUDE OF POROSITY
Uniformity of grain size
If small particles of silt and clay are mixed the large sand particles, the effective porosity will be
constantly reduced as shown in figure. These reservoirs are referred to as dirty or shaly.
Degree of cementation
The highly cemented sandstone has low porosity whereas the soft unconsolidated rocks have high
porosities. Cementation takes place within the void spaces which reduces porosity.
Amount of compaction
Generally porosity is lower in deeper, older rocks due to overburden pressure with increase in depth.
Method of Packing
With increase in over burden pressure, poorly soughed angular grains show a progressive change from
random packing to a closer packing due to the crushing and deformation of sand particles.
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 6
ENGINEERING CLASSIFICATION OF POROSITY
Absolute Porosity
It is the ratio of total void spaces in the sample to the bulk volume of that sample.
Effective Porosity
It is the ratio of interconnected void spaces in the sample to the bulk volume of that sample.
GEOLOGICAL CLASSIFICATION OF POROSITY
Primary Porosity
It is the porosity of rock sample that was developed at the time of deposition of sediments. It can be
classifies as;
Inter crystalline:
These are voids between cleavage planes of crystals, voids between crystal lattice. These voids are sub
capillary that is less than 0.002mm in diameter, this porosity is called microporosity.
Inter Granular:
These are voids between grains that are interstitial voids of all kinds in all types of rocks. These openings
range from sub-capillary to super capillary that is voids greater than 0.5mm.
Bedding Plane:
These voids are parallel to bedding planes, bedding plane voids are caused by the differences in
deposition of sediments, particle size and arrangement and the environment of deposition.
Secondary Porosity
It is the result of geological processes after the deposition of sediments.
It can be classified as;
Solution Porosity:
Channels due to the solution of rocks by circulating warm or hot solution, openings are caused by
weathering such as enlarged joints and later enlarged by solutions.
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 7
Dolomization:
The process by which limestone transforms into dolomite as:
2CaCO3 + Mg+2
CaMg (CO3)2 + Ca+2
When the magnesium ions pass through the pore spaces of limestone, it displaces calcium ions of
limestone and converts into dolomite. Since the ionic size of Mg is smaller as compared to ionic size of
Ca so calcium ion replaced by magnesium ion occupies less space and porosity increases.
It can mathematically be explained as;
Porosity = (VBULK VGRAIN) / VBULK
Since the bulk volume remains same but the grain volume decreases so porosity increases.
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 8
LABORATORY MEASUREMENTS OF POROSITY
Two of the three parameters are required for calculating porosity.
1. Bulk volume
2. Matrix volume
3. Pore volume
BULK VOLUME
From dimensions
For cylindrical core; VB = r2h
Displacement method
VOLUMETRIC (measure volume)
Weigh core sample a
To prevent test liquid to go into sample, coat it with paraffin
Weigh paraffin coated sample b
Weight of paraffin = b-a
Volume of paraffin = weight /density of paraffin (0.9 gm/cc)
Place paraffin coated sample in test liquid and observe change in volume.
Change in volume of testing liquid will be the sum of bulk volume and paraffin volume.
Bulk volume = change in volume of testing liquid volume of paraffin
GRAVIMETRIC(measure mass)
Weigh saturated core sample (actual weight)
Immerse in liquid in Russel tube and its weight will reduce (apparent weight)
Weight of liquid displace = actual weight apparent weight (Archimedes principle)
Bulk volume = weight of liquid displaced / density of displacing liquid
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 9
MATRIX VOLUME
In case of clean formation like clean sand, densities of formation are known so
VM = dry weight/density of formation
Densities of formations (in gm/cc)
Sandstone 2.65, limestone 2.71, dolomite 2.87
These are matrix densities of core sample having 0% porosity.
Displacement method
Reduce core sample to particle size. VM
Place it in liquid to obtain change in volume
Change in volume will be matrix volume
Porosity obtained from this method will be total porosity
BOYLES LAW (Helium porosimeter)
P1 closed
He evacuated
Chamber 1 Chamber 2
Chamber 2 is evacuated
Chamber 1 is filled with He gas at pressure P1
Valve between the chambers is closed
Volume of chamber 1 is VX
Volume of chamber 2 is VY
State of gas at initial condition is; volume of gas = V1 = VX and pressure of gas = P1
Place core in chamber 2 and open valve
P2 open
Chamber 1 Chamber 2
Helium gas goes into chamber 2 and pressure decreases to P2 (it will be the pressure in both
chambers)
State of gas at final condition is; volume of gas = V2 and pressure of gas = P2
By using Boyles law P1 V1 = P2 V2, V2 can be find. V1 is known as it is the volume of first chamber
Both the pressures can be known from gages.
Note that for Boyles law pressure should be in absolute so convert it by adding 14.7
(atmospheric pressure).
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 10
Volume occupied by the gas at final condition is volume of chamber 1 + volume of chamber 2
matrix volume. Gas goes into the pores
V2 = VX + VY VM
VX + VY = VT (Total volume of both chambers)
VM = VT V2 (Matrix volume)
Note: In practical, chamber 1 was console, chamber 2 was matrix cup and valve was HV02.
PORE VOLUME
Gravimetric method
VP = (Saturation weight dry weight) / density of saturation fluid
BOYLES LAW
Hassler sleeve core holder is evacuated
Chamber 1 is filled with gas at pressure P1
Valve between them is closed
Volume of chamber 1 is VX
State of gas at initial condition is; volume of gas = V1 = VX and pressure of gas = P1
Place core sample in Hassler sleeve core holder which has same dimensions as that of core and
open valve.
Gas goes into pores of core sample and pressure decreases to P2
State of gas at final condition is; volume of gas = V2 and pressure of gas = P2
By using Boyles law P1 V1 = P2 V2, V2 can be find. V1 is known as it is the volume of first chamber
Both the pressures can be known from gages.
Note that for Boyles law pressure should be in absolute so convert it by adding 14.7
(atmospheric pressure).
Volume occupied by the gas at final condition is volume of chamber 1 + pore volume
V2 = V1 + VP
VP = V2 V1 (Pore volume)
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 11
CALCULATION OF MATRIX DENSITY
In order to determine density of solid portion of rock (its grain density), the rock has been crushed using
an impact crusher (not a grinder). An appropriate size pycnometer, whose volume is known, is dried and
weighed, and then volume and mass of a portion of sand grains is determined using pycnometer.
Procedure
Fill the pycnometer with a liquid (hydrocarbon or water), and obtain its mass (Mpyc +M1)
Empty and dry the pycnometer
Place a sample of crushed rock in the pycnometer (about one-half the volume of pycnometer)
and determine the mass (Mpyc +Mgrains)
Fill the pycnometer (containing sand grains) with the liquid used in step 1 and determine the
mass (Mpyc +Mgrains + M1)
The sand grain density is calculated from data as follows:
Sample calculation
Known volume of pycnometer (Vpyc = 10.0cm3)
Mass of pycnometer that is measured (Mpyc = 16.57gm)
Obtain the mass after filling the pycnometer with liquid (Mpyc +M1 = 26.58gm)
Mass of liquid which is filled in known volume of pycnometer (10.0cm3) is obtaines by;
M1 = (Mpyc +M1) Mpyc = 26.58 -16.57 = 10.01gm
Density of liquid is obtained by; 1 = M1/V1 where volume of liquid is volume of pycnometer
1 = M1/Vpyc = 10.01/10.0 = 1.01gm/cm3
Fill the empty pycnometer with crushed sand grains and measure
Mpyc + Mgrains = 20.59gm
The space that is left in pycnometer is filled by adding same liquid that was used before and
measure mass i.e.
Mpyc + Mgrains + M2 = 29.175gm
Where M2 is mass of liquid that is added above crushed grains
Mass of grains can be calculated by;
Mgrains = (Mpyc +Mgrains) Mpyc = 20.59 -16.57 = 4.02gm
Mass of liquid that is added above crushed grains can be calculated by;
M2 = (Mpyc +Mgrains + M2) (Mpyc + Mgrains) = 29.175 20.59 = 8.585gm
Volume of liquid that is added above crushed grains can be calculated by;
V2 = M2/1 (1 is used as liquid is same)
V2 = 8.585/1.01 = 8.50cm3
Volume of grains can be calculated by;
Vgrains = Vpyc V2
Because volume of pycnometer is occupied by grains and liquid
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 12
Vgrains = 10.0 8.50 = 1.50cm3
Density of grains (matrix density) can be calculated by;
(grains) = Mgrains / Vgrains = 4.02/1.50 = 2.68gm/cm3
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 13
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 14
TYPES OF FLUID
COMPRESSIBLE FLUID
A fluid in which volume changes with respect to pressure is called compressible fluid.
dV/dP 0
A fluid in which density changes with respect to pressure is called compressible fluid.
d/dP 0
With increase in pressure, volume decreases but with increase in pressure density increases
Density is mass per unit volume. With less pressure P1 , molecules are far so mass in unit volume is less
while mass in unit volume in case of high pressure P2 is greater so, density is less in case of pressure P1
and density is high in case of pressure P2.
INCOMPRESSIBLE FLUID
A fluid in which volume does not change with respect to pressure is called incompressible fluid.
dV/dP = 0
A fluid in which density does not change with respect to pressure is called incompressible fluid.
d/dP = 0
COMPRESSIBILITY
Compressibility is basically the measure of ability of matter to be compressed under the action of
pressure.
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 15
If we increase pressure, the volume get decrease and if we decrease pressure, the expansion will take
place. Compressibility is helpful in studying the drive mechanism of reservoir.
ISOTHERMAL COMPRESSIBLITY
It can be defined as:
The fractional change in volume per unit change in pressure at constant temperature
Our reservoirs are at constant temperature but varying pressure state.
Mathematically;
C = - (dV/dP)T / V
Reason for taking fractional change in volume
It is because change in volume depends upon initial volume of fluid which can be understand by
example that if we take 1scf from reservoir and place it in chamber of volume 1ft3 and 2ft3. If we apply
unit pressure on both chambers then volume decrease in 2 ft3 chamber will be more as compared to 1
ft3 chamber but if we calculate V/V (fractional change in volume) then it will be same in both cases and
formula become generalize i.e. if we calculate compressibility of reservoir (infinite extent) or if we
calculate compressibility of core sample, both will be same.
GAS COMPRESSIBLITY
The relation of gas compressibility and reservoir pressure (or simply the pressure) is given
below:
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 16
The graph shows that at any particular value of pressure, how much the gas compressible is. As it is seen
from the graph that at a state of low pressure, the gas is highly compressible means with unit change in
pressure V/V (fractional change in volume) will be large as molecules are so far apart. At moderate
pressure, the compressibility is moderate and at high pressure, the compressibility is negligible (gas
behaves as incompressible) i.e. with unit change in pressure V/V (fractional change in volume) will be
small as molecules are very close to each other.
DERIVATIONS
For ideal gas
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 17
For real gas
GRAPH OF Z (Gas deviation factor)
The graph is drawn at particular temperature. Initially when the pressure increases Z decreases as
Z=Vactual/ Videal, Vactual decreases with increase in pressure. But the time come when molecules
come very close to each other and repulsion starts and Vactual starts increasing thus Z increases.
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 18
NUMERICAL
Find the compressibility of gas at 1000psi, 2500psi and at 4500psi assuming
Gas to be ideal
Gas to be real
SOLUTION
Ideal gas
At 1000psi: Cg = 1/1000 sip
At 2500psi: Cg = 1/2500 sip
At 4500psi: Cg = 1/4500 sip
Real gas
Cg = 1/P (dZ/dP)T /Z
At 1000psi: Cg = 1/1000 (-127x10-6)/0.83 = 1.15x10-3 sip
At 2500psi: Cg = 1/2500 0 = 1/2500 sip
At 4500psi: Cg = 1/4500 (110x10-6)/0.9 = 1x10-4 sip
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 19
COMPRESSIBLITY OF OIL
The graph of compressibility of oil vs. pressure (reservoir pressure) is given below which can be studied
in two phases:
Above Bubble point pressure
Below Bubble point pressure
The graph is plotted in the same manner as that was plotted in case of gas compressibility such that we
take samples at different pressure and apply unit change in pressure to the sample to obtain V/V.
Above Bubble Point Pressure
When the reservoir pressure is above bubble point pressure such that in under saturated condition, the
gas remains dissolved in it and as the pressure is depleted, the free space is left which is soon occupied
by oil due to gas expansion in it hence change in volume or we can say increase in oil volume will be
there as pressure is decreasing due to expansion of dissolved gas in oil. The change in volume will not be
significant if the pressure change is occurring far above bubble point pressure but as soon as the
pressure reaches near to the bubble point pressure, the increase in oil volume will be seen significantly
showing a significant rise in compressibility value as we can see from graph.
At bubble point pressure as we can see the dramatically change of graph; this is due to variable changes
in the values of compressibility of oil at this pressure due to sudden evolution of gas.
Below Bubble Point Pressure
Below bubble point pressure, a gas cap starts forming above the oil. In this phase, as pressure is reduced
more and more gas will evolve from the oil and the volume of gas cap will increase and the volume of
liquid oil will decrease but the volume should increase with decrease in pressure and compressibility
should increase and this is what the graph is showing so the case is that however the volume of oil is
decreasing due to decrease in pressure but the volume of mass that was originally liquid occupying the
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 20
reservoir volume (oil + dissolved gas) is increasing so the total volume change in reservoir below bubble
point will be the sum of volume change of oil and volume change of free gas (we will be taking the
volume of both because at one instant both were the same entity, gas and oil were the same thing), so
as pressure will decrease, increase in the volume will be seen rapidly with every pressure decrement
due to escaping of gas due to which the total volume of the mass will rise.
The following diagram shows how the volume increases by decreasing the pressure below bubble point
pressure.
In the above figure, V1 is the volume of mass that was originally (oil + dissolved gas) but as the pressure
is decreased below bubble point the volume increases to V2 of the same mass but now the same mass
contains both oil (with dissolved gas) and free gas and it is seen clearly that due to pressure drop the
mercury level will drop and volume increases by the mass expanding i.e. V2>V1. Volume of oil is
decreased but the volume of mass i.e. oil (with dissolved gas) + free gas increases.
DERIVATION
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 21
NUMERICAL
The volume of sample of oil at reservoir condition at 5000psig was 59.55cc. The volume was 60.37cc at
4000psig. Calculate Co?
SOLUTION
BUBBLE POINT
Bubble point is a point at which first vapor escapes from oil. Bubble point pressure changes with
temperature.
PHASE DIAGRAM
Phase diagram is a conditions of temperature and pressure at which different phases exist.
Phase diagram for pure substance
Critical point is an upper limit of vapor pressure line and at this point, liquid and gas states are identical.
Triple point is a point at which three phases exist.
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 22
Oil and gas reservoirs are differential on the basis of critical point. If the reservoir temperature is higher
than critical temperature then it is gas reservoir and if the reservoir temperature is less than critical
temperature then it is oil reservoir.
Bubble point pressure changes with temperature. At higher temperature T2, bubble point pressure PB2 is
greater as compared to bubble point pressure PB1 at low temperature T1.
This two phase diagram can be made after collecting samples as critical point and different parameters
are different for different reservoirs. The application of this phase diagram is that by samples, we know
phase at that time but as we know pressure of reservoir declines so by phase diagram we can predict
the phase of reservoir after some time.
Our work is not on pure substance and with reservoir, phase diagram also changes.
Retrograde reservoir is a reservoir that converts from gas to oil after some time.
FORMATION COMPRESSIBILITY (PORE COMPRESSIBILITY)
Our reservoirs are present at some depth from the surface under a pressure of overlying rocks called
overburden pressure. The overburden pressure is balanced by two pressures which are:
Fluid pressure or pore pressure (also called reservoir pressure)
Rock matrix pressure
These two pressures overcome the overburden pressure. Initially the reservoir is in the geological
equilibrium such that no deformations has occurred and the overburden pressure is balanced by the
above two mentioned pressures. As we take production, the fluid expels out from the pores and the
pore pressure such that the pressure that fluid due to its mass was applying on the rock strata or the
reservoir pressure decreases, but since the overburden pressure is constantly applying and exist there
so the pressure has to be balanced for which the rock matrixes will bear extra pressure due to which the
deformation of rock grains will take place which could either be from any of the following:
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 23
1) For platy grains (like in clay) and non- platy grains (like in quartz, feldspar), due to compaction rotation
and closer packing takes place such that grains come closer to one another to reduce porosity.
2) In ductile grains, the deformation of grains take place and some grains may elongate under pressure to
block pore spaces.
3) If grains are brittle, then large grains may break into smaller ones and reduce porosity.
4) There may be a condition of pressure solution happen which basically means that due to overburden
pressure of rocks, there might be dissolution of rock grains such that they may dissolve into each other
or gets mixed.
The deformation of grains will reduce the porosity and as the pore volume decreases such that grains
are compacted, the bulk volume also reduces with approximately same amount provided that negligible
increase in the volumes of grains that are expand as a result of compression.
So we can say that matrix compressibility is nearly zero and pore, formation and bulk compressibility are
same such that (Cp=Cb=Cf (in general)).
Due to overburden pressure, volume in axial direction decreases and volume in lateral direction
increases in small amount but since the reservoir is completely covered by the surroundings so the
change in area is negligible and the only significant change that will occur will be in thickness which will
cause subsidence.
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 24
DERIVATION
NUMERICAL
Calculate the porosity at 4500psi when the initial pressure and porosity are 5000psi and 18%
respectively? Pore compressibility is 10 x 10-6sip.
SOLUTION
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 25
REASON
Q. Why compressibility of water is negligible?
Compressibility of oil is due to presence of gas in it. Gas in oil expands and contract and volume of oil
with change in pressure increases or decreases. But solubility of water is negligible as it is polar
compound and contains partial positive and partial negative ion,
H+ OH-
While gas is non- polar compound so polar compound (water) and non-polar compound (gas) cant be
soluble. On the contrary, gas and oil both are non-polar compounds so they are soluble and the
solubility of gas in oil is the reason of compressibility of oil.
Solubility depends upon pressure (increase with pressure), temperature (may increase or decrease) and
nature of solvent; hydrophilic (water loving) and hydrophilic (water hating).
APPLICATIONS OF COMPRESSIBILITY
1) Porosity can be calculated
2) Volume can be calculated
3) Understanding reservoir drive mechanism
If formation compressibility is more, initial production will be more. Recovery will be due to drive
mechanism and also due to compressibility.
4) For abnormally pressured reservoirs
Abnormal pressure reservoirs are those reservoirs whose pressures are higher than normal because of
various geological factors. Importance of this is for reservoir engineer is that production is higher in early
stages of life of field because as we produce formation compresses so we have production because of
two factors. One is Darcy forces and other is due to compaction but the production due to compaction
will deplete after certain time period. If we estimate reserves in early life, we would have wrong (large)
reserves.
5) Compressibility also helps to calculate that if reservoir pressure is depleted, the stresses may be
increased so much that it will break rock and sand starts producing.
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 26
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 27
SATURATION
Saturation is defined as fraction or percentage of the pore volume occupied by a particular fluid OR
It is the relative volume of fluids in a porous medium
We assume that reservoir is in state of phase equilibrium i.e. equilibrium between the phases has been
achieved; gas at top, oil at middle and water at bottom.
Phases are three i.e. gas, oil and water. Components are different like, oil with dissolved gas. We take
free water as insoluble as water is polar so its solubility is negligible.
Due to secondary migration, oil and gas displace water as we assume that our reservoir is formed in
marine environment. Initially, water is present in source rock which is displaced by oil and gas due to
difference in specific gravities in reservoir rock. When gas and oil displace water, some water remains as
water which is present around grains (stick to grains) cant be displaced as grains are rough and water is
polar compound so the saturation of water which left around grains is known as irreducible water
saturation or connate water saturation.
It means oil and gas will contain initial water saturation.
CONNATE WATER
The word connate means from birth or beginning. Connate water is the irreducible water present in oil
and gas zone. It depends on lithology like; connate water saturation in high permeability zone is low.
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 28
INTERSTITIAL WATER
Water present in the rock, whether it is present in pores, grains or around grains, is called interstitial
water. In interstitial water, pore water, connate water and bound water all three are included. Pore
water is free water present at water zone, connate water is water present around grains and bound
water is water which dissolves in grains in crystal lattice is called bound water or water of crystallization.
Grains possess many crystals and the crystals contain water. Like in shale, water dissolves in its grains
and shale expands. Saturation of interstitial water in oil and gas zone is 10 to 40% while in free water
zone, saturation is 100%.
TRANSITION ZONE
The area between oil and water at inter-phase is known as transition zone. This zone contains both oil
and water.
In transition zone, with increase in depth water saturation increases
At the top of transition zone, connate water saturation is present but as depth increases (in graph it
increases downward), saturation of water increases and it reaches to maximum value i.e. 100% at the
bottom of transition zone and remains 100% in free water zone. If difference in density is more than
transition zone will be less like in case of mercury and water transition zone will be less as mercury and
water are nearly insoluble.
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 29
CRITICAL OIL SATURATION
It is the minimum saturation below which oil will not flow in reservoir. Assume dry core sample,
If oil is entered into dry core sample at 10% saturation then it will first adhere to rock grains due to
wettability and if saturation is increased to 15% then oil will store in between grains but if further
saturation is increased like up to 20% then the coming oil will provide energy to already present oil so oil
flows so critical oil saturation is 20% which is the minimum saturation for the oil to flow in reservoir.
CRITICAL GAS SATURATION
The gas is non-wettable but the case is same as that in oil i.e. if saturation is increased pressure
increases such that pressure reaches the value at which flow starts so saturation at which flow of gas
starts is known as critical gas saturation. It will obviously be very small as compared to critical oil
saturation.
Consider a core saturated with oil and dissolved gas. When the pressure reaches bubble point pressure
then first bubble escapes from oil but bubble has not enough pressure to flow but as the pressure
further declines, its saturation increases and bubbles after escaping from oil starts accumulating and
results in larger bubble which has high pressure so the pressure of gas increases with increase in
saturation although the pressure of reservoir is declining and finally it reaches to saturation when the
pressure of gas increases from capillary pressure (minimum pressure for flow) which is known as critical
gas saturation.
DRIVE MECHANISMS
Solution gas drive
Rock and fluid expansion
Gas cap drive
Water drive
Combination drive
Gravity drainage
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 30
RESIDUAL OIL SATURATION
The remaining oil left in the reservoir after displacing process (after drive mechanism) is called residual
oil and the saturation is known as residual oil saturation.
There are three types of forces in reservoir;
Gravity force
Darcys force
Capillary force
Gravity forces are due to different phases as densities of fluids are different. Darcys forces are based on
pressure difference. In case of more pressure difference, flow will be more.
Initially we discover reservoir and perforate against oil zone. At that time, gravity forces and Darcys
forces are equal.
But as we start production, pressure difference between the reservoir and well bore increases so,
Darcys forces increases and gravity forces decreases. Due to difference in pressure, oil flows. As we now
that viscosity of oil is greater than water and gas so with same pressure difference, water and gas will
flow more as compared to oil so, gas and water starts coming towards perforation which is known as gas
coning and water coning respectively. In this way, gas and water will produce and in the end oil remains
in the reservoir which is known as residual oil.
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 31
MOVEABLE OIL SATURATION
The oil saturation that could be produced by primary drive mechanism is called moveable oil saturation.
Movable oil saturation is given by;
Somovable = 1 Swconnate Socritical
Where Swconnate is connate water saturation and Socritical is critical oil saturation which is the minimum
saturation of oil for flow.
Consider oil reservoir in which 30% is the saturation of connate water then saturation of oil will be 70%.
Out of 70%, 15% is the critical saturation then movable oil saturation is 55%.
Somovable = 1 Swconnate Socritical
Somovable = 1 0.3 0.15 = 0.55 = 55%
Unproducable means which cant be produced like we cant produce tight gas as permeability is low.
Techniques are required which are not economical in current situations.
RESOURCES
DISCOVERED
PRODUCABLE
RESERVES CUMMULATIVE PRODUCTION
UNPRODUCABLE
UNDISCOVERED
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 32
SATURATION
Saturation of oil, gas and water is given by;
So = Vo / Vp
Sg = Vg / Vp
Sw = Vw / Vp
Vp is the pore volume of reservoir and it will be same in all three cases. Vw is the volume of free water
and connate water both. Vo is volume of oil only which we obtained by subtracting connate water
volume which we get by different techniques. So obtained will include critical oil saturation.
HOW INITIAL SATURATION IS CALCULATED?
In development phase, when we want to know initial saturation say saturation of oil then we should find
Vo and Vp.
CALCULATION OF Vp
To calculate Vp, we require Vb and as Vp = Vb
Initial calculation of bulk volume
Initial bulk volume of reservoir is obtained by knowing thickness and area for which we have different
methods;
Isopachous method
Contour map
Planimeter
Initial calculation of porosity
Log methods
Neutron porosity
Density porosity
Sonic porosity
Core analysis
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 33
Now, we can calculate pore volume as
Vp = Vb
CALCULATION OF Vo, Vg and Vw
For that we should know gas oil contact and oil water contact for which we use geophysical properties.
OIL WATER CONTACT
Conductivity of water is more due to salinity i.e. due to presence of Na+ and Cl- ions. To get oil water
contact, we do resistivity log and as water is conductive than oil so resistivity of water will be less.
This resistivity log technique is known as Quick Look Technique.
FOR GAS OIL CONTACT
To obtain oil gas contact, we do neutron logging and density logging.
Neutron logging and density logging
As hydrogen content in gas is less as compared to oil so we do neutron logging. We do density logging as
well as density of gas is less as compared to oil. In neutron logging, we bombarded neutrons from
radioactive methods and from hydrogen neutron slows down so neutrons becomes more slow in case of
oil as hydrogen content is less in oil as compared to gas.
AVERAGE SATURATION
If we have different wells in an area like, 92 wells in SUI then parameters (properties) of each well
varies. Porosity of a reservoir can vary due to different compaction from top at different places and
thickness of reservoir can also vary.
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 34
Now, to calculate saturation of whole reservoir, we calculate average saturation which is given by;
Where So is the oil saturation that varies as connate water saturation also varies in different wells. h is
the thickness and is the porosity that varies in different wells.
PROBLEM
Calculate average oil and water saturation for reservoir whose sample properties are given?
SAMPLE h So Sw
1 1 10% 75% 25%
2 1.5 12% 77% 23%
3 1 11% 79% 21%
4 2 13% 74% 26%
5 2.1 14% 78% 22%
6 1.1 10% 75% 25%
SOLUTION
SAMPLE h hSo hSw
1 0.1 0.075 0.025
2 0.18 0.1386 0.0414
3 0.11 0.0869 0.0231
4 0.26 0.1924 0.0676
5 0.294 0.2293 0.0647
6 0.11 0.0825 0.0275
1.054 0.8047 0.2493
Average saturation of oil will be;
So = 0.8047 / 1.054
So = 0.763 = 76.3%
Average saturation of water will be;
Sw = 0.2493 / 1.054
Sw = 0.237 = 23.7%
Average saturation of water can also be calculated by (1 So) as in the reservoir only oil and water is
present as the data suggests.
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 35
LABORATORY METHODS OF MEASURING SATURATION
In order to determine saturation of any fluid, we should know its volume in the pores of the core and
the total pore volume of the sample core. For pore volume we know the porosity of the core samples as
we know average porosity of the region from where cores are extracted (which can be determined by
either density or sonic or neutron log) and we can find the bulk volume of each core by measuring the
dimensions through Vernier caliper and Vp=Vb.
Now for oil volume (for oil saturation) and for water volume (for water saturation) we can use either
from the following two Methods or equipment in the laboratory. We cannot find gas volume because
gas no longer remain in pores and escape during extraction of core and we find gas saturation indirectly
from Sg=1-Sw-So).
In Retort (distillation of both oil and water takes place in it)
In Dean stark apparatus (extraction of only water takes place and oil volume is determined
through gravimetric method).
RETORT APPARATUS
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 36
THEORETICAL BACKGROUND
Retort means a tube used for distillation. Distillation is a method of separating mixture based on
difference in their volatility in boiling liquid mixture. It is a physical separation process and not a
chemical reaction.
Heating element converts electricity into heat energy through the process of joule heating. Joule heating
is a process by which the passage of an electric current through resistor creates heat (heat generates
due to resistivity of material). Heating element could be of Nickel or chromium.
Q I2R
Thermocouple is a temperature sensor and it works opposite to heating element (means in this heat
energy is converted into electric energy and it gives us the temperature on digital screen is shown). It is
put in between two metal plates, one at low temperature (known value) and the other is at temperature
of retort heater when the temperature difference is greater, and the voltage difference is greater and
hence current flows which gives us temperature with reference to the other metal.
Thermo controller acts as same as fuse in our houses such that whenever temperature gets increased
above the desired temperature or allowable temperature, the fuse gets melted and heating stops as
electrical circuit through heating element breaks (it may be a thermostat to control temperature by
using bimetallic strip).
Screens are present to restrict the solid contents so that they may not fall at bottom causing increase in
the fluid volume (inaccuracy comes in reading).
PROCEDURE
In retort we work in two stages:
Heating is done at first at 200oC at which all water in pores (connate water and free water) and also
the bond water (water in the crystals) vaporizes and after condensation collected in the centrifuge
tube (that is calibrated to give the extracted volume of the fluid).
In the second stage, heating is done at 600oC at which all oil is vaporize (except those of heavy
contents that forms carbon residue after thermal cracking instead of being vaporized, which stick to
the grains), after being vaporized, the oil is condensed and collected in the centrifuge tube above
the water collected.
DRAWBACKS
The water collected in the centrifuge tube is more than the water in the pores of the rocks (as
bound water is also vaporized along with the pore water and collected in the tube and we are only
concerned with pore water)
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 37
The volume of oil collected in the tube is smaller than that present in the pores (as heavy oil
contents form solid residue that remain stick to the grains and we are concerned with total volume
of oil in the pores whether heavy or light).
FORMULAE FOR CORRECTION OF VOLUME
For the correct volume of oil and water, we use correction method for which following formulae are to
be noted:
Co = Fraction of oil left in the rock with respect to the total oil recovered, mathematically:
Correct oil volume = (1+Co) x oil recovered
Cw = Amount of excessive water recovered due to dehydration with respect to dry mass,
mathematically:
Correct water volume = volume recovered (mass of dry core x Cw)
REASON OF DIVISION
Co and Cw is divided by volume of oil recovered and dry mass of core respectively on which they
depend.
PARAMETERS FOR CORRECTION VALUES
In the first stage we work for calculating the correction values for which we take random core plug from
the plugs achieved from large core piece and dry it by some method (but it will be containing bound
water), the sample (calibrating sample) will be crushed to inch sizes and we saturate it be pouring
some known amount of water first through pipette over the crushed sample, from the poured amount,
some quantity will be absorbed and some will flow down into the centrifuge tube from which the
amount of unabsorbed water can be seen and hence the difference between pored volume and
unabsorbed volume will give us the volume introduced for water and then we do the same for oil, the oil
will displace water (we saturate it first by water and then with oil as same thing happens in the reservoir
as the reservoir is initially water saturated) and hence the volume of introduced oil and water are
calculated, now we follow the procedure of retort experiment (first heating at 200oC and then at 600oC)
finally we observe the volume of oil and water recovered that will be containing errors and then we
calculate correction value given by formula in theoretical background.
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 38
Once the correction values are calculated, now we work on other core samples (such that test
specimen) and observe from them the in corrected oil and water volume and then by applying
correction we can get the right results from all the core samples and then calculate oil, water and gas
saturations as So=Vo/Vp, Sw=Vw/Vp and Sg=1-So-Sw. (Vw does not include bound water as we
required).
ILLUSTRATION OF FORMULAE FOR CALCULATING THE CORRECT VOLUME
Once we have calculated the correction factor from the calibrating sample then while using the formula
of the correction factor on test specimen, we know correction factor (same for all cores) and volume
recovered from that sample (different for different cores and can be observed in centrifuge tube after
extraction), now if we multiply the Co value and volume recovered we get (volume introduced-volume
recovered) or in general the unrecovered volume of oil so adding this unrecovered volume (Co x volume
recovered) to volume recovered then we will get total oil volume in pores or the correct volume such
that:
Correct volume = (Co x volume recovered) + volume recovered or (Co+1) x volume recovered.
Same is the case with correct water volume formula such that Once we have calculated the correction
factor from the calibrating sample then while using the formula of the correction factor on test
specimen, we know correction factor (same for all cores) and dry mass of that sample (different for
different cores and can be calculated by using weighing balance and putting the core on which the
experiment has been performed on it), now if we multiply the Co value and dry mass we get (volume
recovered-volume introduced) or in general the excess volume of water or bound water so subtracting
this (Cw x mass of dry sample) from volume of water recovered will give us the actual water volume in
the pores as:
Cw= volume of water recovered-(Cw x mass of dry sample)
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 39
EXTRACTION METHOD
In this method we also find water volume occupied in the pores and oil volume in the pores (through
gravimetric measurement as only the extraction of water is possible in this experiment by this apparatus
and not of oil), the heat provided to the pore water to vaporize is not provided directly (as we did in
retort) but in this case we have a solvent in the tub at the bottom (usually toluene), the solvent could be
any hydrocarbon but it should possess all following three properties:
Its B.P>B.P. of water.
It should be immiscible with water.
Must be lighter than water.
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 40
The heating element in the bottom region generates heat by Joule heating process and this heat is
transferred to toluene which raises its temperature until it B.P. is achieved, the toluene vapors thus
forming flow to upper region and strikes with the core sample and exchange their heat with water in the
pores of the core due to which water also vaporizes (that is why we have kept BP of solvent greater than
BP of water otherwise solvent will not be able to achieve the temperature required to boil water as if
Boiling temperature of solvent would be small as compared to that of water because at boiling point
the temperature becomes constant). The toluene vapors after exchanging some of their energy with
water flow along with water vapors into the condensing tube region (where temperature is kept low by
flowing cold water in surrounding tubes and heat always flow from hotter region to colder region), the
vapors of water and toluene after being condensed are accumulated in the graduated tube that gives us
the volume of water in the pores (we will note the reading when the increase in water volume in the
tube gets stopped). In this experiment, only the extraction of water and indeed the pore water (free and
connate water) has taken place as we have not kept a very high boiling temperature solvent and from
water volume we can calculate water saturation as Sw=Vw/Vp where pore volume can be calculated as
described earlier. Now for oil saturation we use gravimetric measurement as described below:
GRAVIMETRIC MEASUREMENT
Since, So=Vo/Vp or So= (mo/o) Vp
Or So = (mass of saturated core mass of dry core mass of water)/ oVp
In above formula, the mass of saturated core can be taken when the core plug is freshly prepared (large
core is extracted from reservoir and is cut into small plugs which are being used in the experiment and
we find their weight when they were initially saturated with water and hydrocarbons when taken from
reservoir before performing the experiment), now after the experiment has been performed, the core
will be containing oil in the pores, the oil can be removed by keeping the core in the sun or keep it
overnight as oil is volatile and evaporate and then we achieve dry core, we find its mass (dry mass) and
the mass of water can be determined through:
Mw=w x Vw; if w= 1 g/cc then mw=Vw.
IMPORTANCE OF SATURATION
1) Amount of hydrocarbons can be calculated.
2) Moveable and residual oil saturation can be calculated.
3) Important in all decision making factors like no. of wells required to drill and sizing of the well
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 41
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 42
PERMEABILITY
It is the property of the porous medium that measures the capacity and ability of formation to transmit
fluid
It is the measurement of ease with which a liquid can flow through a porous medium
Permeability is the property of rock when we are only concerned with absolute permeability and when
we talk about effective and relative permeability then we consider it as property of fluid and rock as
well.
TYPES OF PERMEABILTIES
ABSOLUTE PERMEABILITY
If the rock is saturated with one fluid then permeability is known as absolute permeability denoted by
K. It only depends on rock properties because we calculate absolute permeability when only a single
fluid exists in the pore and by the formula provided by Darcy as V = KdP/L ; since in this formula we
have already defined the parameter of viscosity such that nature of the fluid so no matter with which
fluid the core is 100% saturated, the absolute permeability will come to be same and that is why we do
not use o,w or g in suffix of K when used for absolute permeability as it is independent of nature of
fluids. There is no use of absolute permeability in our reservoirs as the oil and gas zones of our
reservoirs also contained connate water saturation that hinders the flow of gas or oil and decreases
their permeability (and the new permeability is the effective permeability), we only study the absolute
permeability in order to study the relative permeability. Absolute permeability remain constant for a
reservoir but effective and off course the relative permeabilities are changed during taking production
as saturation of oil and gas continue to change.
EFFECTIVE PERMEABILITY
The ability of one fluid to flow when other immiscible fluids are present in rock is known as effective
permeability.
When more than one immiscible fluids will be present, the other fluids will provide hindrance in the
path of the other particular fluid flow such that decreasing its flow area and hence the permeability of
that particular fluid will get decreased as compared to absolute permeability and the new permeability
obtained for that particular will be called as effective permeability which will always be less than
absolute permeability of rock. It depends on both rock and fluid properties such that in the presence of
more than one fluids, one fluid can travel much faster than the other due to low viscosity and hence its
permeability will be more (like gas as compared to oil) so it is depending on nature of fluids and also the
grains shape and size affect the permeability in general so it is depending on the rock type also.
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 43
Effective permeability is denoted by Ko, Kg and Kw for oil, gas and water. The effective permeability of a
fluid is a strong function of the saturation of that particular fluid in the pores and hence during
production the saturation changes so the effective permeabilities are also changed.
RELATIVE PERMEABILITY
It is the ratio of effective permeability to absolute permeability of a fluid.
Kro = Ko / K Krw = Kw / K Krg = Kg / K
We use the concept of relative permeability when more than one effective permeabilities exist in the
rock pores.
CONCLUSION AND EXAMPLE
The concept of absolute permeability comes when a single fluid exists in the pores which are not
possible in the reservoir condition due to presence of connate water so to study this core samples are
dried and then a single fluid is allowed to pass through the rock pores.
If more than one fluid exists in the pores (which are the case of our reservoir) then the concept of
effective and relative permeabilities come as permeability of each fluid will become of smaller value as
compared to the absolute permeability value of rock.
If only one effective permeability exists in the pores (such that one fluid is flowing and all other have
saturation less than critical saturation) then the concept of only effective permeability will exist and not
relative permeability because no use of it as no other effective permeability is existing so their relative
permabilities will also be zero and only one relative permeability will exist just like effective so no need
to use relative permeability concept. The concept of effective permeability exists because however
other fluids are not flowing but still they provide hindrance in the flow of the particular fluid (that is
flowing) by reducing its flow area and reducing its permeability below absolute value.
If more than one effective permeabilities exist (more than one fluid are flowing) then we use relative
permeability concepts which is the modification of effective permeability concept.
For example
Above bubble point pressure, no gas is free to move and all the gas dissolved in the oil so the
permeability of only oil exists and not of gas and water also, as connate water is not flow able but still
the permeability of oil is not equal to absolute permeability of rock because the saturation of oil is not
100% in the rock pores due to presence of connate water which reduces the flow area of oil in pores
reducing its permeability (here we use effective permeability concept of oil).
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 44
At bubble point, some gas molecules escape from oil which reduces the effective permeability of oil
further more (see effective permeability of a fluid does not remain same during production, as now both
connate water and gas molecules will be present in pores reducing the flow area much more but in
previous case there was only connate water in pores) but still the effective permeability of only oil exists
as gas has saturation below critical value and is not flowing (here we also use effective permeability
concept).
After critical gas saturation has been achieved, the two effective permeabilities are now present (one of
oil and the other of gas) so now we use relative permeability concept and the relative permeability of oil
is decreased further more (quickly) due to increase in relative permeability of gas (as effective
permeability of gas will have more value as compared to oil, so gas travel faster than oil occupying more
pore spaces in less time and continue to leave less space for oil molecules to occupy due to which the
relative permeability or effective has decreased very quickly (but in previous two cases the matter was
not so severe as gas was not flowing and its effective permeability was zero).
CRITERIA OF RESERVOIR:
Poor K
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 45
3) Grain Shape:
The spherical grains have the greatest porosity while angular grains give small porosity value so the
permeability in rocks of grains of spherical shape is the greatest taking into account the shape of grains.
In case of four flat grains that are closely packed, permeability is less.
In case of four circular grains that are closely packed, permeability is high.
SECONDARY PERMEABILITY
Secondary permeability results from the alteration of rock matrix by;
Compaction (due to this large grains are converted into small ones and also shape of grains
change due to deformation)
Cementation
Fracturing
Solution
These are the four factors that can change the permeability such that cause secondary permeability.
Secondary permeability could be either greater or lesser than the original (primary permeability) such
that due to compaction and cementation increase, it always decreases but due to fracturing it may
increase or decrease. If the initial permeability is more, then due to fracturing secondary permeability
increases and vice versa if the initial permeability is less. Due to solution, secondary permeability
increases.
DIRTY SAND
Sand in which shale is present is called dirty sand. In sandstone formation (sandstone lithology) we have
shale presence in either of the following forms that decrease the permeability of sandstone as shale is
impermeable:
Laminar Shale
They are the thin beds of shale deposited between layers of clean sand.
Dispersed Shale
It is the shale that is adhering and coat on sand grains.
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 46
Structural Shale
They exist as grains of clay (shale) forming part of solid matrix along with sand grains (means just
like sand grains, the shale grains are also present but in lesser amount as the lithology is sandstone due
to its dominant presence).
If we arrange them in order of increasing permeability then it would be
Clean sand>Structural Shale> Laminar Shale > Dispersed Shale
Permeability Porosity Relationship
There is no relationship between porosity and permeability for example, shale is porous and
impermeable but if we talk specifically for a particular reservoir then we could say that as porosity
increases, permeability also increases
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 47
DRAINAGE PROCESS
It is generally assumed that reservoir rocks are deposited or formed in marine environment where water
is present initially and after that oil encroaches into pores until water saturation becomes connate water
saturation. When discovered, the reservoir pore spaces are filled with connate water saturation and oil
saturation. The laboratory procedure is to first saturate the core with water, then displace the water to
a residual or connate, water saturation with oil. This process is called drainage process. In the depletion
process, the non-wetting phase fluid is continuously increased, and the wetting phase fluid is
continuously decreased.
Drainage process is used to simulate the initial saturation distribution in the reservoir. Drainage process
occurs at the time of formation of reservoir.
IMBIBITION PROCESS
The reverse of the drainage process is imbibition process but it is not exactly reverse of drainage process
as in drainage process, 100% water is present initially but after imbibition process 100% water cant be
achieved. The imbibition process is performed in the laboratory by first saturating the core with the
water (wetting phase), then displacing the water to its irreducible (connate) saturation by injection oil.
This drainage procedure is designed to establish the original fluid saturations that are found when the
reservoir is discovered. The wetting phase (water) is reintroduced into the core and the water (wetting
phase) is continuously increased. This is the imbibition process and is intended to produce the relative
permeability data needed for water drive or water flooding calculations.
Some oil remains in the end and saturation of that oil is known as residual oil saturation.
It describes the residual oil saturation of the reservoir after the water flooding or natural water influx.
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 48
Sum of relative permeabilities is equal to or less than one.
Kro+ Krw + Krg 1
Consider only oil is present between the pores oil
K = 15 md (which is same for the rock)
Ko = 15 md
So, Kro= 1
If connate water is present with oil then oil
K = 15 md and Ko = 13 md
So, Kro< 1
And Kcw = 0 (as connate water does not flow) so Krcw =0
In this case, Kro+ Krw< 1
Hence,
Kro+ Krw 1
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 49
TWO-PHASE RELATIVE PERMEABILITY
When a wetting and a non-wetting phase flow together in a reservoir rock, each phase follows separate
and distinct paths. The distribution of the two phases according to their wetting characteristics results in
characteristic wetting and non-wetting phase relative permeabilities. Since the wetting phase occupies
the smaller pore openings at small saturations, and these pore openings do not contribute materially to
flow, it follows that the presence of small wetting phase saturation will affect the non-wetting phase
permeability only to a limited extent. Since the non-wetting phase occupies the central or larger pore
openings which contribute materially to fluid flow through the reservoir, however, small non-wetting
phase saturation will drastically reduce the wetting phase permeability. In water-oil system, water being
considered as the wetting phase.
Point 1
Point 1 on the wetting phase relative permeability shows that a small saturation of the non-wetting
phase will drastically reduce the relative permeability of the wetting phase. The reason for this is that
the non-wetting phase occupies the larger pore spaces, and it is in these large pore spaces that flow
occurs with the least difficulty.
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 50
Point 2
Point 2 on the non-wetting phase relative permeability curve shows that the non-wetting phase begins
to flow at the relatively low saturation of the non-wetting phase and the saturation of the oil at this
point is called critical oil saturation Soc.
Point 3
Point 3 on the wetting phase relative permeability curve shows that the wetting phase will cease to flow
at a relatively large saturation and this is because the wetting phase preferentially occupies the smaller
pore spaces, where capillary forces are the greatest. The saturation of the water at this point is referred
to as the irreducible water saturation or connate water saturation (both terms are used
interchangeably).
Point 4
Point 4 on the non-wetting phase relative permeability curve shows that, at the lower saturations of the
wetting phase, changes in the wetting phase saturation have only a small effect on the magnitude of the
non-wetting phase relative permeability curve. The reason for the phenomenon at Point 4 is that at the
low saturations the wetting phase fluid occupies the small pore spaces which do not contribute
materially to flow, and therefore changing the saturation in these small pore spaces has a relatively
small effect on the flow of the non-wetting phase.
ILLUSTRATION
From point 4 to point 2
Initially saturation of oil is 100% and when saturation of oil is decreased and saturation of water is
increased initially, relative permeability of oil has not much effect as water occupies smaller pores which
do not contribute much to flow. As saturation of oil is decreased, relative permeability goes on
decreasing until it reaches critical oil saturation (saturation at which oil does not flow).
From point 1 to point 3
Initially saturation of water is 100% and when saturation of oil is increased, it occupies larger pore
spaces (due to its non-wetting behavior) and it is in these large pore spaces that flow occur with least
difficulty and relative permeability of water decreases rapidly and relative permeability of water
becomes zero at connate water saturation because at that saturation water stops flowing.
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 51
GAS OIL RELATIVE PERMEABILITY CURVE
In the reservoir along with oil and gas, connate water is present as well. Oil is present in smaller pores as
it is wetting phase in case of gas. Gas occupies larger pore spaces as it is non-wetting phase. The relative
permeability of connate water is zero so, relative permeability of oil and gas are drawn on graph.
Relative permeability of oil Kro
Initially liquid saturation is 100% and the relative permeability of oil is 1 (we have taken assumption that
relative permeability of oil is 1 as connate-water is present with oil at 100% liquid saturation which
provides hindrance so in actual case relative permeability cant be one). When the saturation of gas
increases then relative permeability of oil rapidly decreases as gas occupies larger pore spaces and in
these large pore spaces that flow occurs with least difficulty. Now, the relative permeability of oil value
becomes zero when saturation of oil reaches residual oil saturation (saturation at which oil cant flow)
and saturation of water is connate water saturation. Total saturation of the liquid, at which relative
permeability of oil is zero, is connate water plus residual oil saturation.
Relative permeability of gas Krg
Initially saturation of gas is 100% and relative permeability of gas is 1. When the saturation of liquid is
increased, relative permeability of gas has not much effect as liquid occupies smaller pores which do not
contribute much to flow. Now, the relative permeability of gas value becomes zero when saturation of
gas reaches residual gas saturation (saturation at which gas cant flow)
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 52
TYPES OF FLOW IN RESERVOIR
Linear flow
Radial flow
Spherical flow
LINEAR FLOW
Linear flow occurs when flow paths are parallel and fluid flows in single direction and cross-sectional
flow area is constant.
In inclined reservoir, flow can also be linear.
RADIAL FLOW
In a radial flow, fluid moves in all direction to the wells and converges at the well bore which means flow
occurs between two concentric cylinders, flow area changes at every location. Actually in reservoir,
radial flow is present. One larger cylinder is reservoir (diameter is drainage diameter or reservoir
diameter) and smaller cylinder is of well (diameter is well bore diameter). Drainage radius is the radius
up to which well can produce.
Flow occurs from reservoir to well bore due to difference in pressure.
Flow area is decreasing as radius is decreasing because A = 2rh
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 53
Darcy derived formula for linear flow and for radial flow, we need some modification. Darcy was
hydrologist and give law for purification of water. When petroleum engineers used it then with time
they require modifications.
DARCYS LAW
Apparent velocity of fluid is directly proportional to pressure gradient and inversely proportional to
viscosity of fluid
Negative sign shows that with length, pressure decreases.
This law is only applicable for some assumptions.
ASSUMPTIONS OF DARCY LAW
Flow is linear.
Rock is 100% saturated with flowing fluid.
Isothermal condition prevails.
Porous rock is homogeneous.
Flow is laminar.
Gravity forces are negligible.
Steady state condition exists.
Incompressible flow.
Isothermal condition means temperature remains constant because with temperature increase, velocity
also increases. Gravity forces are negligible because due to gravity, flow restricts. Incompressible flow
means with change in pressure, volume remains constant i.e. dV/dP=0. Homogeneous means all the
properties are same i.e. if porosity is 20% at one point then it is same for a reservoir and if the connate
water saturation is 10% then it will remain 10%.
Laminar flow and Turbulent flow
In laminar flow, the motion of the particles of fluid is very orderly with all particles moving in a straight
line parallel to pipe wall. It occurs at low flow rate. In turbulent flow, particles move haphazardly and
velocity is high so molecules collide with greater velocity due to which friction is increased and we get
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 54
more pressure drop. Darcy law is only valid for laminar flow and in case of turbulent flow; velocity is
directly proportional to square of pressure gradient.
Steady, unsteady and semi-steady state flow
If there is no change in pressure of fluid with respect to time then it is called steady state flow i.e.
dP/dt=0. If the change in pressure w.r.t time is constant then it is called semi-steady flow and if there is
no relation between pressure and time then it is called unsteady state flow.
Oil reservoirs having water aquifer is a steady state reservoir because pressure does not drop there.
DARCYS LAW FOR LINEAR FLOW
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 55
As much pressure gradient is present, more will be the flow. If viscosity is high, flow will be less. K is
absolute permeability and it is independent of fluid flowing i.e. it is constant for any formation. For same
pressure gradient, whether oil or gas is present, flow will be same. q is apparent flow rate here as Darcy
considered total cross-sectional area.
Darcy considered flow from whole core (piece) but actually in reservoir, fluid only flows from pores so if
we multiply the total area with porosity which is dimensionless quantity then we get pore area i.e.
Pore area = A
Now, this pore area will give actual flow rate rather than apparent flow rate which is considered by
Darcy.
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 56
UNITS IN C.G.S SYSTEM
FIELD UNITS
Now,
As the units we get in field are different as compared to the units that are applicable for formula i.e. CGS
units but we can convert these CGS units into field units.
It is because formula can only be used for CGS units.
Now,
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 57
Field units are;
PROBLEM
Brine is used to measure the absolute permeability of a core plug, the core sample is 4cm long and 3
cm2 in cross-section, the brine has a viscosity of 1 cp and flowing at constant rate of 0.5 cm3/sec under
2 atm pressure difference. Calculate absolute permeability?
Solution:
PROBLEM
Rework the above example assuming that an oil of 2 cp is used to measure the permeability under the
same differential pressure. Assuming flow rate is 0.25 cm3/sec.
Solution:
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 58
Conclusion:
The above two numerical shows that absolute permeability is independent of fluid flowing. With any
fluid the rock is 100% saturated, absolute permeability comes to be the same.
DARCYS LAW FOR RADIAL FLOW
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 59
UNITS IN C.G.S SYSTEM
Now, for field units
Field units are;
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 60
AVERAGING ABSOLUTE PERMEABILITY
In Darcys law, it is assumed that reservoir is homogeneous i.e. same properties throughout. It means
absolute permeability remains constant but if our reservoir is heterogeneous then its properties like
absolute permeability changes. Change in absolute permeability may be due to change in compaction or
may be due to shale that comes in between pore spaces.
In many cases, the reservoir contains distinct layers, blocks, or concentric rings of varying permeabilities.
Also, because smaller-scale heterogeneities always exist, core permeabilities must be averaged to
represent the flow characteristics of the entire reservoir or individual reservoir layers (units). The proper
way of averaging the permeability data depends on how permeabilities were distributed as the rock was
deposited. There are three simple permeability-averaging techniques that are commonly used to
determine an appropriate average permeability to represent an equivalent homogeneous system. These
are:
Weighted-average permeability
Harmonic-average permeability
Geometric-average permeability
WEIGHTED-AVERAGE PERMEABILITY
This averaging method is used to determine the average permeability of layered-parallel beds with
different permeabilities.
LAYERED RESERVOIR WITH SAME WIDTH
Consider the case where the flow system is comprised of three parallel layers that are separated from
one another by thin impermeable barriers, i.e., no cross flow, as shown in figure. All the layers have the
same width W. Vertical changes in permeability takes place in case of layered reservoir.
It is similar to parallel resistances electrical circuit;
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 61
Qt is the total flow rate which will is equal to the sum of flow rates through each layer. Flow will be more
where permeability is high.
Qt = Q1 + Q2 + Q3 ----- (1)
Where; Q1 is the flow rate through K1, Q2 is the flow rate through K2 and Q3 is the flow rate through K3.
Similar to parallel circuits in which current distributes in three resistances and current will pass more in
the resistor where resistance is low.
Pressure difference in different layers is same similar to voltage difference that remains same in all three
resistances in parallel circuit.
Width is same so W1 = W2 = W3 = W
The flow from each layer can be calculated by applying Darcys equation in a linear form
Where;
Qt= total flow rate
Kaverage= average permeability for the entire model
W = width of the formation
ht= total thickness
Put the values in (1)
The average absolute permeability for a parallel-layered system can be expressed in the following form:
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 62
LAYERED RESERVOIR WITH VARIABLE AREA
Consider the case where the flow system is comprised of three parallel layers that are separated from
one another by thin impermeable barriers, i.e., no cross flow, as shown in figure. All the layers have
different width and thickness but pressure difference in layers is same.
Qt is the total flow rate which will is equal to the sum of flow rates through each layer.
Qt = Q1 + Q2 + Q3 ----- (1)
Where; Q1 is the flow rate through K1, Q2 is the flow rate through K2 and Q3 is the flow rate through K3.
The flow from each layer can be calculated by applying Darcys equation in a linear form
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 63
Put the values in (1)
PROBLEM
Given the following permeability data from a core analysis report, calculate the average permeability
of reservoir?
DEPTH, ft PERMEABILITY, md
3998 200
4002 130
4004 170
4006 180
4008 140
SOLUTION
THICKNESS hi, ft Ki hi ki
4 200 800
2 130 260
2 170 340
2 180 360
2 140 280
hi =12 hiKi =2040
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 64
HARMONIC-AVERAGE PERMEABILITY
FOR LINEAR FLOW
Permeability variations can occur laterally in a reservoir
as well as in the vicinity of a well bore. Consider figure
which shows an illustration of fluid flow through a series
combination of beds with different permeabilities. For a
steady-state flow, the flow rate is constant and the total
pressure drop P is equal to the sum of the pressure
drops across each bed, or
P= P1+ P2+ P3
Substituting for the pressure drop by applying Darcys
equation;
By simplifying, we get;
The above expression can be expressed in more general form as;
If permeability is more, pressure drop will be less and flow rate remains constant as
q = (KA P)/ L
It is similar to series electrical circuit in which current remains same in all resistances (equivalent to flow
rate that is same for each bed) and total voltage drop will be equal to sum of voltage drop across each
resistance (equivalent to pressure drop across each bed).
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 65
PROBLEM
A hydrocarbon reservoir is characterized by five distinct formation segments that are connected in
series. The length and permeability of each section of the five bed reservoirs are given below;
LENGTH, ft PERMEABILITY, md
150 80
200 50
300 30
500 20
200 10
SOLUTION
Li, ft Ki (L/K)i
150 80 1.875
200 50 4.000
300 30 10.000
500 20 25.000
200 10 20.000
1350 60.875
FOR RADIAL FLOW
Radial flow is a flow between two concentric cylinders. Radial
flow is also similar to series electric circuit as in it lateral changes
in permeability occur. Pressure changes in lateral direction with
change in permeability but flow rate remains same.
P = P1 + P2 + P3---- (1)
Darcys law for radial flow;
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 66
From (1) we get
Flow rate will be same although flow area decreases as we come toward well bore but at the same time,
velocity increases so, Q = AV (flow rate will be same).
Since, A1> A2 but V2> V1 therefore, flow rate remains same i.e.
Q=A1 V1 = A2 V2
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 67
PERMEABILITY DETERMINATION METHOD
Permeability is calculated through Darcys law. Absolute permeability is the property of rock and it is
independent of fluid flowing.
From Darcys law;
K = qL
AP
For determining absolute permeability, apparatus called Hassler Sleeve Core Holder is used.
Total area of core sample and length of core sample can be known easily. We create pressure difference
for the fluid to flow (P1 P2) and viscosity of fluid is known as well. Now,
K q
For flow rate, liquid is allowed to pass through core and collected in particular time in graduated
cylinder.
Q = V / t, in cm3/sec
Where V is volume of fluid collected in particular time t.
Absolute permeability can be calculated now as;
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 68
CALCULATION OF K BY USING GAS
Permeability measured with air or gas as the flowing fluid show different result from permeability
measured with liquid as a flowing fluid. The permeability of a core sample measured by flowing air is
always greater than permeability measured with liquid as a flowing fluid. This is because due to the
phenomenon called slippage because gases do not have a zero velocity at the contact of solid.
One of the conditions for the validity of Darcys law is the requirement of laminar flow. At low gas
pressure, this condition is broken. At low pressure, gas molecules are so far apart that they slip through
the pore channel almost without any interaction. At low pressure, gas molecules are moving randomly
so laminar flow does not exist.
In case of a liquid due to adhesive forces, velocity of liquid at solid contact is zero as it wets the solid
surface so; the molecules of liquid at solid contact do not contribute in flow rate.
At high pressure, gas is almost incompressible so it behaves as liquid and shape of curve is same and the
velocity is very low (not zero) at solid contacts.
At low pressure, when the gas molecules strike solid grains, it does not wet the solid surface so velocity
is not zero and energy is transferred back to the gas molecules at center which is used for the
movement of gas molecules. This is called slippage so the flow rate of gas is greater than flow rate of
liquid.
This effect is called Klinkenberg effect.
With increase in pressure, viscosity of gas increases so; absolute permeability in this case is dependent
on q whereas, L/AP is constant. In case of high pressure, viscosity of gas is constant as gas behaves
like liquid and then absolute permeability only depends upon flow rate.
Correction of Klinkenberg effect
Gas permeability and liquid permeability comes different due to effect called Klinkenberg effect.
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 69
Pm = (P1 + P2)/2
At Pm and 1/Pm =0, gas permeability and liquid permeability becomes equal as shown in graph. With
increase in pressure i.e. decrease in 1/Pm error decreases. As in graph, error at 1 is less than error at 2
and error at 2 is less than error at 3. At 1/Pm = 0, error becomes zero.
Pm and pressure difference are independent of each order which can be explained by following example;
Pressure difference in both cases are same i.e. 200 100 = 10000 9900 but Pm in both cases are
different.
Pm in (1) = (100 +200)/2 = 150
Pm in (2) = (9900+10000)/2 = 9950
Pm in (2) is higher so it contains less error.
Correction of Klinkenberg effect is;
Kg = KL + c (1/Pm)
Which is obtained by slope equation y = c + mx
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 70
HORIZONTAL AND VERTICAL PERMEABILITY
A permeability which is in horizontal plane is known as horizontal permeability (KH) and the permeability
which is in vertical plane is known as vertical permeability (kV).
Generally, KH is greater than KV. If KH is greater than KV so it is better to drill vertical well.
If KV is greater than KH then it is better to deviate the well
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 71
ISOTROPIC PROPERTIES
Same properties in all direction are called isotropic properties. In case of isentropic reservoir;
KH = KV
On the contrary, properties are different in all direction in case of anisotropic reservoir so,
KH KV
Difference between homogeneous and isotropic
Homogeneous means same properties throughout while isotropic means same properties in all
direction. Consider reservoir is divided into two axes X and Y. It shows porosity of reservoir in X and Y
direction.
If the porosity of reservoir is same in X and Y direction say 20% then it is called isotropic and
homogeneous reservoir and if the porosity is different in two directions but same along one direction
say 20% along X axis and 10% along Y axis then it is called homogeneous and anisotropic reservoir.
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 72
-
Introduction to Reservoir Petrophysics (By Shoaib and Bilal) Page 73
FORMATION RESISTIVITY AND WATER SATURATION
Current is the flow of charges. For the flow of charges, free electrons are necessary that is why current
cant pass through wood.
Sedimentary formations are capable of transmitting an electric current only by means of the interstitial
and adsorbed water they contain. They would be nonconductive if they were entirely dry. The
interstitial or connate water containing dissolved salts constitutes an electrolyte capable of conducting
current, as these salts dissociate into positively charged cations, such as Na+ and Ca2+, and negatively
charged anions, such as Cl- and SO42-.These ions move under the influence of an electrical field and carry
an electrical current through the solution. The greater the salt concentration, the greater the
conductivity of connate water is. Freshwater, for example, has only a small amount of dissolved salts
and is, therefore, a poor conductor of an electric current. Oil and gas are nonconductors.
The electrical resistivity (reciprocal of conductivity) of a fluid-saturated rock is its ability to impede the
flow of electric current through that rock. Dry rocks exhibit infinite resistivity. The resistivity of reservoir
rocks is a function of salinity of formation water, effective porosity, and quantity of hydrocarbons
trapped in the pore space. Relationships among these quantities indicate that the resistivity decreases
with increasing porosity and increases with increasing petroleum content. Resistivity measurements are
also dependent upon pore geometry, formation stress, and composition of rock, interstitial fluids, and
temperature. Resistivity is, therefore, a valuable tool for evaluating the producibility of a formation.
More the salinity is present; more the amount of current can pass through and as the amount of current
is more, less will be the resistance as R=V/I. If resistance is less than resistivity will be less as well
because = (RA)/L. Conductivity can now easily measure by taking the reciprocal of resistivity i.e.
Conductivity = 1/ Resistivity
Or k = 1/
Dry rocks exhibit infinite resistivity. A rock that contains oil and gas will have more resistivity as
compared to rock that is completely saturated with formation water.
Oil bearing rock has some conductivity which is due to presence of connate water in it.The greater the
connate water saturation, the lower the formation resistivity.