Introduction to Reservoir Petrophysics

INTRODUCTION TO RESERVOIR PETROPHYSICS

MUHAMMAD SHOAIB (ROLL # 22) HAFIZ MUHAMMAD BILAL (ROLL # 04)

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 2

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 3

Introduction to Reservoir Petrophysics (By Shoaib and Bilal)

Page 4

POROSITY It is the ratio of pore volume to the bulk volume Porosity = Ø = 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 5

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 6

 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 7

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 = πr2h



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 8



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  BOYLE’S LAW (Helium porosimeter) P1

closed

He

evacuated

Chamber 1       

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

Chamber 1     

Chamber 2

open

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 Boyle’s 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 Boyle’s law pressure should be in absolute so convert it by adding 14.7 (atmospheric pressure).

Introduction to Reservoir Petrophysics (By Shoaib and Bilal)

Page 9

   

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  BOYLE’S 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 Boyle’s 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 Boyle’s 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 10

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; ρ1 = M1/V1 where volume of liquid is volume of pycnometer ρ1 = 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/ρ1 (ρ1 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 11



Vgrains = 10.0 – 8.50 = 1.50cm3 Density of grains (matrix density) can be calculated by; ρ (grains) = Mgrains / Vgrains = 4.02/1.50 = 2.68gm/cm3

Introduction to Reservoir Petrophysics (By Shoaib and Bilal)

Page 12

Introduction to Reservoir Petrophysics (By Shoaib and Bilal)

Page 13

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. dρ/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. dρ/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 14

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 15

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 16

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 17

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 18

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 19

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 20

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 21

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 22

1) 2) 3) 4)

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. In ductile grains, the deformation of grains take place and some grains may elongate under pressure to block pore spaces. If grains are brittle, then large grains may break into smaller ones and reduce porosity. 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 23

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 24

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,

Hδ+ OHδWhile gas is non- polar compound so polar compound (water) and non-polar compound (gas) can’t 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 25

Introduction to Reservoir Petrophysics (By Shoaib and Bilal)

Page 26

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) can’t 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 27

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 28

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 29

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 Darcy’s force Capillary force Gravity forces are due to different phases as densities of fluids are different. Darcy’s 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 Darcy’s forces are equal.

But as we start production, pressure difference between the reservoir and well bore increases so, Darcy’s 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 30

MOVEABLE OIL SATURATION The oil saturation that could be produced by primary drive mechanism is called moveable oil saturation.

RESOURCES

DISCOVERED

PRODUCABLE

RESERVES

UNDISCOVERED

UNPRODUCABLE

CUMMULATIVE PRODUCTION

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 can’t be produced like we can’t produce tight gas as permeability is low. Techniques are required which are not economical in current situations.

Introduction to Reservoir Petrophysics (By Shoaib and Bilal)

Page 31

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 Φ as Vp = Φ 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 32

Now, we can calculate pore volume as Vp = ΦVb 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 33

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 Ø is the porosity that varies in different wells. PROBLEM Calculate average oil and water saturation for reservoir whose sample properties are given? SAMPLE 1 2 3 4 5 6 SOLUTION

h 1 1.5 1 2 2.1 1.1

SAMPLE 1 2 3 4 5 6 Σ

ø 10% 12% 11% 13% 14% 10%

øh 0.1 0.18 0.11 0.26 0.294 0.11 1.054

So 75% 77% 79% 74% 78% 75%

øhSo 0.075 0.1386 0.0869 0.1924 0.2293 0.0825 0.8047

Sw 25% 23% 21% 26% 22% 25%

øhSw 0.025 0.0414 0.0231 0.0676 0.0647 0.0275 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 34

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=øVb. 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 35

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 α 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 36

 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 ½ 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 37

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 38

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:   

It’s 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 39

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/ρo) Vp Or So = (mass of saturated core – mass of dry core – mass of water)/ ρoVp 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=ρw x Vw; if ρw= 1 g/cc then mw=Vw. IMPORTANCE OF SATURATION 1) 2) 3)

Amount of hydrocarbons can be calculated. Moveable and residual oil saturation can be calculated. 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 40

Introduction to Reservoir Petrophysics (By Shoaib and Bilal)

Page 41

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/μL ; 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 42

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 43

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………………………… K120

Oil

Water

60