Unidade 05 - Prop. Das Rochas

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    Fundamentos daIndstria Petrolfera

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    Unidade 5

    Propriedades das Rochas

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    Rock Properties

    Basic Rock Parameters Used in Reservoir Engineering

    (Porosity, Permeability, Saturations, Compressibility)

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    Scanning Electron MicrographNorphlet Formation, Offshore Alabama, USA

    Pores Provide theVolume to Contain

    Hydrocarbon Fluids

    Pore Throats RestrictFluid Flow

    PoreThroat

    Porosityin a Sandstone

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    Sandstone Composition Framework

    Grains

    Norphlet Sandstone, Offshore Alabama, USAGrains are About =< 0.25 mm in Diameter/Length

    PRF KF

    P

    KF = PotassiumFeldspar

    PRF = Rock Fragment

    P = Pore

    Potassium Feldspar isStained Yellow With aChemical Dye

    Pores are ImpregnatedWith Blue-Dyed Epoxy

    CEMENT

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    What is Porosity? And why is it

    important?

    Total Porosity

    Total Porosity isthe ratio of thevoid space to the

    total rock volume

    Effective Porosity

    Effective Porosityis the ratio of the

    interconnectedvoid space to thetotal rock volume

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    Porosity = 48%

    Cubic Packing of Spheres

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    Porosity = 27 %

    Rhombic Packing of Spheres

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    Porosity = 14%

    Packing of Two Sizes of Spheres

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    Inter-granular Porosity

    rombohedrally packed

    spheres: = 26%

    grain sorting, silt, clay and

    cementation effect porosity

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    Permeability (k)

    Permeability is the measure of capacity of rock to transmitfluid.

    Symbol : k

    Units : Darcy or milliDarcy (D or mD) in SI units m21 mD = 10**(-15) m2

    Source : Well tests, core analysis

    Range : 0.001 mD - 10,000 mD

    pA

    Lqk

    L

    pkAq

    or

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    Definition of 1 Darcy

    Darcy is a unit of

    Area

    1 mm2=

    1,013,250 Darcy

    1 Darcy =

    0.986923 m2

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    Effect of Grain Size in Permeability

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    Shale

    Sand

    h3

    h2

    h1

    hnet= h1+ h2+ h3

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    Saturations (Sw, So, Sg)H2O

    During deposition of the

    sedimentary rock the pore space issaturated with water of varyiingsalinity (20,000 to 200,000 ppm)The HC migrate into these porespaces and displace most of this

    water down to the irreducible

    water saturation.Sandstones are usually water wet.This means that the sand grainsare covered with a film of water(10-30%), and the oil or gasoccupies the more central part of

    the pores. These central positionsare also more beneficial for flow.Carbonates are usually oil or ofmixed wettability and hence theoil occupies a more difficultflowpath.

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    Open Hole Logs to determine Saturation

    Resistivity log

    Induction tools

    Laterolog tools

    Porosity log Neutron tools

    Density tools

    Sonic tools

    Magnetic resonance tools

    We run tools into the freshly

    drilled reservoir section whichmeasure physical rock propertieswhich we in turn use to calculatethe properties that we really areafter. Resistivitiy tools are used tocalculate saturations (why do think

    this could be used ????).The neutron measures the capacityof Hydrogen absorption and hencegives an indication of thehydrocarbon charge. The densitylog measures the density and

    hence is an indicator of porosity.From the sonic travel timesporosities can also bebackcalculated. Magneticresonance logs give a measure ofpermeability.

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    Archies Equation

    where:

    Sw: water saturation

    a: constant

    : porositym: cementation factor

    Rw: formation waterresistivity

    Rt: true resistivity (measuredby deep resistivitymeasurement)

    t

    w

    2

    2w

    t

    w

    15.2

    2w

    t

    w

    m

    nw

    R

    R1S

    )estone(lim

    R

    R62.0S

    )sandstone(

    R

    RaS

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    Exercise

    Calculate oil saturation in the following case

    Rock type: sandstone

    Rw: 0.0912 (at bottom hole temperature)

    : 0.2 Rt: 20 ohm meter

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    7.0S3.0S

    20

    0912.0

    2.0

    62.0S

    20

    0912.0

    2.0

    62.0S

    o

    w

    15.2w

    15.2

    2

    w

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    Total Compressibility (Ct)

    ggwwooft cScScScc Typically,

    Cw: 3E-6

    Co: 3E-6(Black oil)

    Cg: 1/pressure

    Cf: 4E-6

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    Exercise

    Calculate total compressibility in the following case

    cf: 3.6E-06

    co: 1.158E-05 So: 0.83

    cw: 2.277E-06 Sw: 0.17cg: 6.023E-05 Sg: 0

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    Solution

    Ct = 1.36 x 10-5 psi-1

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    Exercise

    Calculate total compressibility in the following case

    cf: 3.6E-06

    co: 1.158E-05 So: 0.73

    cw: 2.277E-06 Sw: 0.17cg: 6.023E-05 Sg: 0.1

    Compare solution with previous Example

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    Additional Reading Requirements

    Please read the chapters 26,27 and 28 of the PetroleumEngineering Handbook !