PETROPHYSICS PRESENTED BY EMILE SOB N

PETROPHYSICS PRESENTED BY EMILE SOB N. PETROLEUM ENGINEER

INTRODUCTION PETROPHYSICS is the study of physical and chemical rock properties and their interactions with fluids Every hydrocarbon bearing reservoir is a valuable asset. To ensure the best possible return, it is important to understand as much as possible about the reservoir. This is a bit difficult because we cannot physically see the reservoir in question. Techniques such as Seismic data acquisition, Well logging, Core Analysis, PVT analysis, Well Testing, etc… produce valuable data which helps build the simulated reservoir model, and therefore improves our understanding of the underground substructure.

ROCK CLASSIFICATION Rocks are broadly classified into three groups - Igneous - Sedimentary - Metamorphic

ROCK CLASSIFICATION IGNEOUS ROCKS:They originate from the solidification of molten material emanating from below the earth surface. Volcanic igneous rocks are glassy in texture.

Plutonic igneous rocks are formed in slower cooling intrusive magmas which allow the time needed for the atoms to arrange themselves into a crystalline grain structure.

ROCK CLASSIFICATION SEDIMENTARY ROCKS:They are formed from the materials of older uplifted formations which have been broken down by erosion and transported by the elements to lower elevations where they are deposited.

Consolidation of sands, silts, pebbles and clays by pressure of many thounsand feets of overlying sediments, and cementation by precipitatesfrom percolating waters act to convert these materials into sandstones and conglomerates. Sedimentary rocks are classified into two groups - Clastic Sedimentary rocks(The rocks of detrital origin or debris from older rocks) such as sandstone, siltstone and shales

- Non-clastic sedimentary rocks(rocks of biochemical or chemical precipitate origin) such as limestone, dolomite and clays

ROCK CLASSIFICATION SEDIMENTARY ROCKS:They are formed from the materials of older uplifted formations which have been broken down by erosion and transported by the elements to lower elevations where they are deposited.

ROCK CLASSIFICATION SEDIMENTARY ROCKS:They are formed from the materials of older uplifted formations which have been broken down by erosion and transported by the elements to lower elevations where they are deposited.

ROCK CLASSIFICATION METAMORPHIC ROCKS:They are formed other sedimentary deposits by alteration under great heat and pressure. Examples of metamorphic rocks include: - Marble – metamorphosed limestone

- Hornfeld – converted from shale or tuff - Gneiss – Similar to granite but metamorphically consolidated Oil and gas are not usually found in igneous or metamorphic rocks as both are so non-porous that hydrocarbons cannot accumulate or be extracted from them.

THE RESERVOIR A reservoir is an underground geologic formation which contains hydrocarbons

To form a commercial reservoir of hydrocarbons, a geological formation must possess three essential characteristics: • Sufficient void space to contain hydrocarbons(Porosity)

• Adequate connectivity of these pore spaces to allow transportation over large distances(Permeability) • A capacity to trap sufficient quantities of hydrocarbon to prevent upward migration from the source beds.

THE RESERVOIR A reservoir is an underground geologic formation which contains hydrocarbons

THE RESERVOIR A reservoir is an underground geologic formation which contains hydrocarbons

RESERVOIR ROCK PROPERTIES - POROSITY POROSITY: Porosity is a measure of the void spaces within the rock

RESERVOIR ROCK PROPERTIES - POROSITY We can see that particles of carbonate materials that make up sandstones never fit together perfectly due to high degree of irregularity in shape. What factors affect porosity? Porosity can also be defined as the fraction of the bulk volume of the reservoir that is not occupied by the solid framework of the reservoir. This can be expressed in mathematical form as: Where Ø= porosity, Vb = bulk volume of the reservoir rock

Vgr = Grain volume Vp = Pore volume

What are typical values of porosity?

RESERVOIR ROCK PROPERTIES - POROSITY Example

A clean and dry core sample weighing 425 g was 100% saturated with 1,07 specific gravity brine. The new weight is 453 g. The core sample is 12 cm in length and 4 cm in Diameter. Calculate the porosity of the rock sample. Hints: - Calculate the bulk volume of the core sample - Calculate the weight and therefore volume of the brine

- Calculate the porosity of the core as 0,173 or 17,3%

RESERVOIR ROCK PROPERTIES - POROSITY In actual rock, porosity can be classified as:

- Absolute porosity: this is the total porosity of a rock, regardless of whether or not the individual pores are interconnected - Effective porosity: this is the porosity due to interconnected pores

Which porosity is of interest to the petroleum engineer?

RESERVOIR ROCK PROPERTIES - POROSITY Porosity can also be classifioed into two types: - Primary Porosity: this is also called intergranular porosity, this is the porosity formed at the time the sediments were deposited.

- Secondary Porosity: This type of porosity is made essentially of voids which were formed after the sediment was deposited. Secondary porosity can be can be subdivided into three classes - Solution Porosity: These are voids formed by the solution of the more soluble portions of the rock by subsurface waters containing carbonic and organic acids. This is also called vugular porosity because it is made up of voids. - Fractures, Fissures, and joints: Voids of this type are common in many sedimentary rock are are formed by structural faillure of the rock under loads.

- Dolomitization: This is the process by which limestone(CaCO3) is transformed into Dolomite(CaMg(CO2)3) The chemical reaction explaining this change is:

2CaCO3 + MgCl 2

CaMg(CO3)2 + CaCl2

RESERVOIR ROCK PROPERTIES - POROSITY The chemical reaction explaining this change is: 2CaCO3 + MgCl 2

CaMg(CO3)2 + CaCl2

Since dolomite is more porous than limestone, this results in a more porous material.

What factors affect porosity? The principal factors affecting intergranular porosity include - Uniformity of grain size: the presence of small particles such as clay, silt which may fit into tiny voids will greatly reduce the porosity - Degree of cementation and consolidation: cementing materials deposited around grain junctions reduced porosity - Particle shape: irregular shapes will lead to porosity reduction

- Packing: The manner in which grains are arranged affect resulting porosity - Amount of compaction during and after deposition

RESERVOIR ROCK PROPERTIES - POROSITY

RESERVOIR ROCK PROPERTIES - POROSITY Packing: The manner in which grains are arranged affect resulting porosity For cubic packing, porosity value is 47,6% For hexagonal packing, it is 39.5%

For rhombohedral packing, it is 25,9%

Exercice: Show that for cubic, hexagonal and rhombohedral packing, porosity values are as indicated above

RESERVOIR ROCK PROPERTIES - POROSITY Quantitative use of porosity data Consider a bud volume of rock with a surface area of one acre and one foot thick, this constitues a volume of one acre-ft.

1 acre = 43560 ft², 1acre-ft = 43560 ft3, 1bbl = 42 gal = 5,61 cu ft, This implies that 1acre-ft occupies a volume of 7758 bbl. It is obvious that the pore space within a rock is equal Vp=7758x Ø bbl/acre-ft Where Ø is the porosity of the rock in question.

Continuing this derivation, we obtain the famous Volumetric Equation of Oil in place:

𝑁= Where N = tank oil in place in bbl/acre-ft Sw =water saturation

Bo = Formation Volume factor

7758∅𝐴ℎ 1 − 𝑆𝑤 𝐵𝑜

RESERVOIR ROCK PROPERTIES - POROSITY Exercice How much tank oil exists in the following oil field? Area of field = 60 ares

Average thickness = 20 ft Average Porosity, Ø= 20% Average water saturation Sw = 30% Formation Volume Factor, Bo= 1,20

This should yield an oil in place value of 11,6 x106 bbl

RESERVOIR ROCK PROPERTIES - POROSITY Exercice Given the following data for a cylindrical core sample, compute its porosity Clean, dry weight of sample = 311 gm

Weight of sample with pores completely filled with a 1,05 sp gravity brine = 331 gm Diameter of sample = 4,0 cm Length of sample = 10,0 cm Ans Porosity Ø = 15,1%

What gross storage space exists in this rock? Ans 1170 bbl/acre-ft

RESERVOIR ROCK PROPERTIES - POROSITY Laboratory Measurement of Porosity In the laboratory measurement of porosity, it is necessary to determine only two of the three basic parameters(Bulk volume, pore volume and grain volume).

Bulk Volume Measurement: Although the bulk volume may be computed from measurement of the dimensions of a uniformly shaped sample, the usual procedure utilizes the observation of fluid displaced by the sample. This procedure is helpful in determining volume of irregularly shaped samples. The fluid displaced can be observed either volumetrically or gravimetrically. In measuring bulk volume in this manner, we have to prevent fluid from penetraing into the pores by: - Coating the rock with paraffin or a similar substance - Saturating the rock by the fluid into which it is to immersed

- Using mercury which by virtue of its surface tension and wetting characteristics, does not tend to enter the small pore spaces

RESERVOIR ROCK PROPERTIES - POROSITY Laboratory Measurement of Porosity Example: Coated Sample immersed in Water(Gravimetric method). Weight of dry sample in air: 20.0 gm

Weight of dry sample coated with paraffin = 20.9 gm (density of parrafin = 0.9 gm/cc) Weight of coated sample immersed in water at 40° F = 10.0 gm (water density =1.10 gm/cc) Weight of paraffin: 20.9 – 20.0 = 0.9 gm, volume of paraffin = 0.9/0.9 = 1 cc Weight of water displaced = 20.9 – 10.0 = 10.9gm

Volume of water displaced = 10.9/1.1 = 9.9 cc Bulk volume of rock = 9.9 cc

RESERVOIR ROCK PROPERTIES - POROSITY Laboratory Measurement of Porosity Exercice: Water saturated Sample immersed in Water(Gravimetric method). Weight of dry sample in air: 20.0 gm

Weight of sample saturated with water in air = 22.5 gm Weight of water saturated sample immersed in water at 40° F = 12.6 gm (water density =1.0 gm/cc) Weight of water displaced = Volume of water displaced =

Bulk volume of rock = Ans Vb = 9.9 cc

RESERVOIR ROCK PROPERTIES - POROSITY Laboratory Measurement of Porosity Homework: Dry sample immersed in mercury Pycnometer. Weight of dry sample in air: 20.0 gm

Weight of pycnometer filled with mercury at 20° C = 350.0 gm Weight of pycnometer filled with mercury and sample at 20° C = 235.9 gm (density of mercury = 13.546 gm/cc)

Find the bulk volume of the rock

RESERVOIR ROCK PROPERTIES - POROSITY Laboratory Measurement of Porosity Sand-Gran volume. The grain volume can be determined from the dry weight of the sample and the sand-grain density. The value of 2.65 gm/cc can be used as the sand-grain density.

For more rigorous determination, either the melcher-Nutting or Russell methods can be employed. In each case, the bulk volume of a sample is determined, then either that sample or an adjacent one is reduced to grain size and the grain volume determined.

RESERVOIR ROCK PROPERTIES - POROSITY Laboratory Measurement of Porosity Exercice: Porosity determination by the Saturation method A dry sample havin a bulk volume of 9.9 cc has a dry weight of sample in the air = 20gm, the weight of the saturated sample in the air is 22.5 gm, The density of the saturating fluid is 1gm/cc Calculate the porosity of the sample. Is the value obtained effective or Absolute porosity? What then is the minimum value of the absolute porosity of that rock sample. Ans: 25.3%

RESERVOIR ROCK PROPERTIES - POROSITY Exercice: Oil in Place Calculation Calculate the initial oil in place (N) of an oil reservoir if A = 1 600 Acres, h=32 ft, Ø = 22% Sw = 20% and Bo = 1.23 bbl/STB

Ans: 56.8 x106 STB

The equation for calculating Initial gas in place is given as: 𝐺=

43 560 𝐴ℎØ 1 − 𝑆𝑤 𝐵𝑔𝑖 𝑍𝑖𝑇 ) 𝑃𝑖

Where the initial gas formation volume factor is 𝐵𝑔𝑖 = 0,02829(

RESERVOIR ROCK PROPERTIES - POROSITY Exercice: Gas in Place Calculation Calculate the initial gas in place (G) of an oil reservoir if A = 1 320 Acres, h=45 ft, Ø = 17.5% Sw = 23% and T= 200° F, Pi = 4000 psi; zi = 0,916

Ans: 81.539 x109 SCF

Exercice: Show that the Initial gas in place is : 𝐺=

43 560 𝐴ℎØ 1 − 𝑆𝑤 𝐵𝑔𝑖

RESERVOIR ROCK PROPERTIES - POROSITY Group Work: methods of Porosity determination Group 1 – Washburn Bunting method Group 2 – Stevens Method

Group 3 – Kobe Porosimeter Group 4 – Boyle’s Law Porosimeter Group 5 – Saturation Method Group 6 – Core Laboratories Wet Sample

Group 7 – Core Laboratory Dry Sample Group 8 – Sand Density Method Group 10 A summary of all the methods above For each method, specify the type of sampling, the sample preparation, the fuction measured, the manner of measurement, the potential sources of errors

RESERVOIR ROCK PROPERTIES - PERMEABILITY In additon to being porous, a reservoir rock must have the ability to allow petroleum fluids to flow through its interconnected pores. This rock’s ability to conduct fluids is known as permeability. This is a clear indication that nonporous rocks have no permeability. The permeability of a rock depends on its effective porosity, therefore permeability is affected by rock grain size, grain shape, grain size distribution,(sorting), grain packing, and the degree of consolidation and cementation. According to Henry Darcy, the velocity of flow of a fluid through a porous medium is directly proportional to the pressure gradient and inversely proportional to the fluid viscosity. This can be expressed mathematically as 𝑣 ∝

𝑑𝑝 𝑑𝑙

1 𝜇

( ) 𝑘 𝜇

Removing the proportionality constant, it becomes 𝑣 = − ∗ We know that v=q/A,

𝑑𝑝 𝑑𝑙

RESERVOIR ROCK PROPERTIES - PERMEABILITY We know that v=q/A, inserting that into the equation, we get Removing the proportionality constant, it becomes 𝑘

=

𝑞μ𝐿 ∆𝑝𝐴

K in darcy, fluid velocity v in cm/s, flowrate q in cm3/s, Cross sectional area A in cm², p in atmosphere, viscosity in centipoise

RESERVOIR ROCK PROPERTIES - PERMEABILITY In deriving this equation the following assumptions where made

𝑘=

𝑞μ𝐿 ∆𝑝𝐴

1- Steady state flow conditions exist

2 – The pore space within the rock is 100% saturated with the flowing fluid, so that k is the absolute permeability 3 – The viscosity of the flowing fluid is constant

4- Isothermal condistions prevail 5- Flow is horizontal and linear 6- Flow is laminar

RESERVOIR ROCK PROPERTIES - PERMEABILITY The permeability k as defined above is the « absolute » permeability if the rock is 100% sturated with a single fluid or phase such as oil, gas or water. In the presence of more than one fluid, permeability is called « effective » permeability to oil (ko), water(kw) or gas(kg)

Ko + kw + kg < k(absolute permeability) In the presence of more than one fluid in the rock, the ratio of the effective permeability of any phase to the absolute permeability in the rock is known as the « relative permeability » kr of that phase 𝑘𝑜 𝑘 𝑘 𝑘𝑟𝑤 = 𝑤 𝑘 𝑘 𝑘𝑟𝑔 = 𝑔 𝑘 𝑘𝑟𝑜 =

RESERVOIR ROCK PROPERTIES - PERMEABILITY Example: A 10cm long cylindrical core sample was subjected to a laboratory linear flow test under a pressure differential of 3,4 atm using a fluid of viscosity 2.5 cp. The diameter of the core is 4 cm. A flow rate of 0.35 cm3/s was obtained. Calculate the permeability of the core sample. Ans : k = 0,204 Darcy = 204 mD

RESERVOIR ROCK PROPERTIES - PERMEABILITY Homework: Show that in oil field units, the darcy equation can be written as

𝑞=

0,001127𝑘𝐴 𝑝1−𝑝2 𝜇𝐿

Where q is in bbl/day, k in md, p in psia, viscosity in centipoise, distance L in ft, cross-sectional area in ft² Exercice: An incompressible fluid flows in a linear porous media with the following properties:

L = 2000 ft, h = 20 ft, width = 300 ft, K = 100 md, Porosity = 15%, μ = 2 cp, p1 = 2000 psi, p2 = 1990 psi. Calculate:

- The flow rate in bbl/day Ans: 1.6905 bbl/day - The apparent fluid velocity in ft/day Ans: 0,0016 ft/day - The actual fluid velocity in ft/day Ans: 0,0105 ft/day

RESERVOIR ROCK PROPERTIES - PERMEABILITY Exercice: Average permeability in parallel beds What is the equivalent linear permeability of four parallel beds having equal widths and lengths under the following conditions?

Ans: 200 md

𝑞

RESERVOIR ROCK PROPERTIES - PERMEABILITY Homework: Calculate the arithmetic average and thickness-weighted average porosity from the following measurements:

Arithmetic Average Porosity = 11.67% Thickness Weighted Average Porosity = 12.11%

RESERVOIR ROCK PROPERTIES - PERMEABILITY Exercice: Average permeability in beds in series What would be the equivalent linear permeability if the beds are in series?

Ans: 80 md

𝑞

RESERVOIR ROCK PROPERTIES - SATURATION

Saturation is defined as that fraction, or percent of the pore volume occupied by a particular fuild (Oil, Gas or Water).

𝑇𝑜𝑡𝑎𝑙 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐹𝑙𝑢𝑖𝑑 𝐹𝑙𝑢𝑖𝑑 𝑆𝑎𝑡𝑢𝑟𝑎𝑡𝑖𝑜𝑛 = 𝑃𝑜𝑟𝑒 𝑉𝑜𝑙𝑢𝑚𝑒

RESERVOIR ROCK PROPERTIES - SATURATION

Applying this concept to each reservoir fluid gives 𝑇𝑜𝑡𝑎𝑙 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑂𝑖𝑙 𝑆𝑜 = 𝑃𝑜𝑟𝑒 𝑉𝑜𝑙𝑢𝑚𝑒 𝑇𝑜𝑡𝑎𝑙 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐺𝑎𝑠 𝑆𝑔 = 𝑃𝑜𝑟𝑒 𝑉𝑜𝑙𝑢𝑚𝑒 𝑇𝑜𝑡𝑎𝑙 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑊𝑎𝑡𝑒𝑟 𝑆𝑤 = 𝑃𝑜𝑟𝑒 𝑉𝑜𝑙𝑢𝑚𝑒 By definition, the sum of the saturations is 100%, therfore So+

Sg + Sw = 1

RESERVOIR ROCK PROPERTIES - SATURATION

Critical Oil Saturation, Soc For the oil phase to flow, the saturation of the oil must exceed a certain value termed critical oil saturation. Below this value and no matter how high the pressure, oil will not flow, it will remain in the pores. Residual oil Saturation, Sor

During the displacment process by water or gas injection, not all the oil is produced, some remaining oil will be left whose saturation is a bit larger than the critical oil saturation. This is the residual oil saturation Sor. The term Residual saturation is usually associated with the non wetting phase when it is being displaced by a wetting phase.

Of course, Critical Water (Swc)and Critical gas(Sgc) saturations also exist

RESERVOIR ROCK PROPERTIES - SATURATION

RESERVOIR ROCK PROPERTIES - SATURATION

Average Saturation Proper averaging of saturation data requires that the saturations values be weighted by both the interval thickness hi and the interval porosity Ø.

RESERVOIR ROCK PROPERTIES - SATURATION

Exercice: Average Saturation Calculate the average oil and connate water saturation from the following measurements: You can do it on Microsoft Excel

RESERVOIR ROCK PROPERTIES - SATURATION

Solution: Average Saturation

RESERVOIR ROCK PROPERTIES - SATURATION

Solution: Average Saturation

RESERVOIR ROCK PROPERTIES - SATURATION

What interpretation do you give to this graph? Define drainage and imbibition

RESERVOIR ROCK PROPERTIES - WETTABILITY Wettability is defined as the tendency of onr fluid to spread on or adhere to a solid surface in the presence of other immiscible fluids. Wettability can also be expressed by measuring the contact angle at the liquidsolid surface.

RESERVOIR ROCK PROPERTIES - WETTABILITY Complete wettability is evidenced by a zero contact angle Complete nonwetting is evidenced by a contact angle of 180°

RESERVOIR ROCK PROPERTIES - WETTABILITY The tendency of one fluid to displace another from a solid surface is determined by the relative wettability of the fluids to the solid.

RESERVOIR ROCK PROPERTIES – SURFACE AND INTERFACIAL TENSION When dealing with multi phase systems, it is important to consider the effect of the forces at the interface when two imiscible fluids are in contact - When there two fluids are liquid and gas, the force at the interface is called surface tension

- When the interface is between two liquids, the acting forces are called interfacial tension

RESERVOIR ROCK PROPERTIES – CAPILLARY PRESSURE Capillary pressure can be defined as the pressure of the nonwetting phase minus the pressure of the wetting phase

There are three types of capillary pressures

RESERVOIR ROCK PROPERTIES – ROCK COMPRESSIBILITY Three different types of compressibility must be distinguished in rocks

- Rock matrix compressibility, cr: this is the fractional change in volume of solid rock material(grains) with a unit change in pressure. - Rock-Bulk Combressibility, cB: This is defined as the fractional change in volume of the bulk volume of the tock with a unit change in pressure. - Pore Compressibility, cp: This is the fractional change in pore volume of rock with a unit change in pressure.

RESERVOIR ROCK PROPERTIES – ROCK COMPRESSIBILITY For most petroleum reservoir, the rock and bulk compressibility are considered small in comparison with the pore compressibility cp. Therefore the term formation compressibility cf, is the term commonly used to describe the total compressibility of the formation.

Typical values of formation compressibility range from 3x10-6 to 25x10-6 psi-1

RESERVOIR ROCK PROPERTIES – ROCK COMPRESSIBILITY Exercice: Calculate the reduction in pore volume of a reservoir due to a pressure drop of 10psi. The reservoir original pore volume is 1 million barrels with an estimated formation compressibility of 10x10-6 psi-1.

Ans: ∆Vp = 100 bbl

RESERVOIR ROCK PROPERTIES – ROCK COMPRESSIBILITY The reduction in the pore volume due to pressure decline can also be expressed in term of changes in reservoir porosity. The porosity change can be expressed as Ø= Where Øo is the original porosity Po the original pressure, in psi P is the current pressure in psi Ø Is the porosity at pressure p

Øo[ 1+cf (p-po)]

RESERVOIR ROCK PROPERTIES – ROCK COMPRESSIBILITY Exercice: Given the following data, calculate the porosity at 4 500 psi Øo = 18% Po = 5 000 psi Cf=10x10-6 psi-1

Ø= Øo[ 1+cf (p-po)] Ans: 0,179

RESERVOIR ROCK PROPERTIES – ROCK COMPRESSIBILITY Note: the total comprsssibility commonly used in transient flow equations and the material balance equation is given by:

For undersaturated reservoirs, the reservoir pressure is above the bubble point pressure and therefore there is no gas cap. the equation becomes

RESERVOIR ROCK PROPERTIES – NET PAY THICKNESS Reservoir performance prediction depends on the knowledge of the volume of oil originally in the reservoir. The reservoir is confined within certain geologic and fluid boundaries: GOC, WOC, GWC.

RESERVOIR ROCK PROPERTIES – NET PAY THICKNESS Oil is contained in what is commonly reffered to as gross pay. Net pay is that part of the reservoir which contributes to oil recovery. Net pay is defined by imposing the following criteria: • Lower limits of porosity • Lower limits of permeability • Upper limits of water saturation

RESERVOIR ROCK PROPERTIES – RESERVOIR HETEROGENEITIES Reservoir heterogeneity can be defined as a variation in reservoir properties as a function of space. If the reservoir were homogeneous, measuring a resevoir property at any location in the reservoir will allow us to fully describe the reservoir.

Reservoir description is an heterogeneous reservoir need to take into account the variation of reservoir properties as a function of spacial location. There are essentially two types of heterogeneities: • Vertical heterogeneities • Areal heterogeneities

RESERVOIR ROCK PROPERTIES – FINAL EXERCICES/PROBLEMS Exercice 1 : An oil reservoir exists at its bubble point pressure of 3000 psia and temperature of 160° F. The oil has an API gravity of 42° and a gas-oil-ratio(Rs) of 600 SCF/STB. The specific gravity of the solution gas is 0.65. The following additional date are avalaible: • Reservoir area = 640 acres • Average thickness = 10 ft • Connate water saturation = 0.25 • Effective Porosity = 15%

1- Determine the specific gravity of the stock tank oil using the Api Gravity Ans: 0.8156 2- Calculate the oil formation volume factor Bo by applying Standing’s Equation. Ans: 1.396 bbl/STB 3- Calculate the initial oil in place. Ans: 4 276 998 STB

RESERVOIR ROCK PROPERTIES – FINAL EXERCICES/PROBLEMS Exercice 2 : Given Pi = 3500

Pb = 3500

T=160° F

A=1000 acres

h = 25ft

Swi = 30%

Ø= 12%

API Density = 45°

Rsb=750 scf/STB

γg = 0,7

Calculate • The initial oil in place in STB, also express the answer in cubic meters and in barrels • The volume of gas originally disolved in the oil

RESERVOIR ROCK PROPERTIES – FINAL EXERCICES/PROBLEMS Exercice 3: The folowing measrurements on pay zone are availlable:

Calculate

• The average porosity • The average oil and water saturations (assuming no gas)

RESERVOIR ROCK PROPERTIES – FINAL EXERCICES/PROBLEMS Exercice 4 : Assuming a steady-state laminar flow, calculate the permeability from the following measurements made on a core sample by using air: Flow rate = 2 cm3/sec

T = 65° F

Upstream pressure = 2 atm

A = 2 cm²

L = 3 cm

Viscosity = 0,018 cp

RESERVOIR ROCK PROPERTIES – FINAL EXERCICES/PROBLEMS Exercice 5: Calculate average permeability from the following core analysis data:

RESERVOIR ROCK PROPERTIES – FINAL EXERCICES/PROBLEMS Exercice 6 : Calculate the average permeability of a formation that consists of four beds in series, assuming • A linear system • A radial system with rw = 0,3 ft and re = 1,450 ft:

RESERVOIR ROCK PROPERTIES – FINAL EXERCICES/PROBLEMS Exercice 7: Calculate the porosity of the sample described below:

Is this the effective or total porosity of the sample? What is most probably the lithology of the sample? Explain why.

RESERVOIR ROCK PROPERTIES – FINAL EXERCICES/PROBLEMS Exercice 8 : A core 2.54 cm in length and 2.54 cm in diameter has a porosity of 22%. It is saturated with oil and water. The oil content is 1,5 cm3. What is the pore volume of the core? What are the oil and water saturations inside the core?

Divide the class in 10 groups. Each group should work on all the 8 problems separately and submit during our next class.