introduction to reservoir petrophysics

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


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


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.


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.


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.


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


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).


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)


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


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


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.


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:


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


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.


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


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


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


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.


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:


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.


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


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.


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.


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.


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


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.


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


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


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.


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.


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


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)


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.


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)


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.


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


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.


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).


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


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.


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


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.


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


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.


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.


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)


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


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


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


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.


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,


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:


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


UNITS IN C.G.S SYSTEM

Now, for field units

Field units are;


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;


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:


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


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


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).


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;


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


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;


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.


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


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


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.


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.

introduction to reservoir petrophysics

Documents

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effective porosity

porosity of rock sample

low porosity

porosity of porous material

bilal page

reservoir rock porous

bulk volume vgr