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    FORMATION EVALUATION

    PETE 663

    RESISTIVITY THEORY / SATURATION

    Summer 2010

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    IDEALIZEDLOG SET

    = 0.30

    = 0.35

    = 0.07

    R = 0.4

    R = 0.3

    R = 4

    R = 8

    Sand

    Shale

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    FOUR COMPONENTS OF SANDSTONE

    MATRIXFRAMEWORK

    (QUARTZ)

    FRAMEWORK

    (FELDSPAR)

    CEMENT

    PORE

    Note different use of matrix

    by geologists and engineers

    0.25 mm

    1. Framework

    2. Matrix3. Cement

    4. Pores

    Engineering

    matrix

    Geologists Classification

    Ayers, 2001

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    FLUID SATURATIONS

    Grain Water Gas Oil

    Initially, water fills pores and wets the rock surface

    Hydrocarbons migrate into the reservoir rock, displacing some water

    Hydrocarbon distribution determined by gravity and capillary forces,and by wettability

    Modified from NExT, 1999

    and matrix

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    RESISTIVITY OF ROCKS CONTAINING FLUID

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    RESISTIVITY DEFINITION OF THE

    OHM-METER

    From Halliburton (EL 1007)

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    RESISTIVITY

    Resistivity

    The voltage required to cause one amp to passthrough a cube having a face area of one

    square meter

    Units are ohm-m / m; usually ohm-m (.m)2

    tyConductivi

    1yResistivit =

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    RESISTIVITY MEASUREMENT

    Resistivity

    ( )( )

    ( )mL

    mAohms

    I

    V

    metersohmR

    2*)(

    =

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    RESISTIVITY OF EARTH MATERIALS

    tyConductivi

    1yResistivit =

    Incr

    easing

    Res

    istivity

    (1) Rock

    (2) Gas

    (3) Oil

    (4) Fresh Water

    (5) Salt Water

    Inc

    reasing

    Co

    nductivity

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    Resistivity of water Porosity of the formation,

    Pore geometry - tortuosity

    Lithology of the formation

    Degree of cementation, and

    Type and amount of clay in the rock

    FACTOR AFFECTING RESISTIVITY

    From J. Jensen, PETE 321 Lecture Notes

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    ELECTRICITY AND EARTH MATERIALS - 1

    Conduction is by ions in water

    Na+ and Cl- are very common

    Other monovalent ions K+ and OH-

    Common bivalent ions: Ca++, Mg++

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    Water resistivity

    controlled by

    Ion concentrations

    What type of ions

    What temperature

    Chart GEN-4 toconvert to NaCl

    equiv

    Chart GEN-5 fortemperature/resist

    for NaCl

    Electricity and Earth Materials - 2

    From Schlumberger

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    Resistivity

    of NaClSolutions

    ____

    GEN-5 H

    orGEN-9 S

    From Schlumberger

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    Ca = 460 ppm

    S04 = 1,400

    Na + Cl = 19,000

    TDS = 20,860

    TDS = 20,850 ppm

    0.81

    0.45

    (460)(0.81)+(1,400)(0.45)+(1)(19,000) = 20,000 ppm

    T = 75 deg. F

    H, GEN-4; SLB, GEN-8

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    75 deg. F

    SLB, GEN-9

    H, GEN 5

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    ARCHIES FIRST EQUATION

    Relates rock resistivity to Rw

    Ro

    = F Rw

    , where:

    Ro = Resistivity of a rock that is 100%saturated with formation water

    Rw

    = Resistivity of formation water

    F = Formation factor (formation resistivity factor)

    As the salt water content increases, the formationresistivity will decrease.

    A rock containing oil or gas will have a higherresistivity than the same rock completely saturatedwith salt water.

    As the shale content increases, the rock matrix willbecome more conductive.

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    Rt

    Ro

    Rw

    Cube of water

    having resistivity,

    Rw

    Non-shaly rock, 100% saturated

    with water having resistivity,

    Rw

    Rock containing pores saturated

    with water and hydrocarbons

    Inc

    reasing

    Re

    sistivity

    (1) Rock

    (2) Gas

    (3) Oil

    (4) Fresh Water

    (5) Salt Water

    Increasing

    Conductivity

    = 100%

    Sw = 100%

    = 20%

    Sw = 100%

    = 20%

    Sw = 20%

    SHC =80%

    mw

    o aR

    RF

    ==

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    The formation factor (F) depends on:

    Porosity of the formation;Pore geometry;

    Lithology of the formation;

    Degree of cementation; and

    Type and amount of clay in the rock.

    FORMATION FACTOR

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    For a clean formation (no shale), the formation factor can

    usually be empirically correlated with porosity.

    a = constant 1.0 for most formations

    m = cementation factor 2 for most formations

    Common values

    F = 0.8/2 (Tixier) or 0.62/2.15 (Humble) for sandstones

    F = 0.8/2 for carbonates

    m

    aF

    =

    FORMATION FACTOR

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    FormationResistivity

    Factor

    VersusPorosity

    mw

    o a

    RRF

    ==

    As

    increases, more ofthe rock volume is

    comprised of pores

    filled with conductive

    fluid

    Therefore, Ro and Fdecrease

    RELATION OF FORMATION FACTOR

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    RELATION OF FORMATION FACTOR

    TO VOLUMETRIC POROSITY

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    RESISTIVITY FACTOR TO SHALINESS

    The formation factor (F) is constant for a clean

    sand; F decreases for shaly sand as value of Rwincreases.

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    SATURATION

    Amount of water per unit volume = Sw

    Amount of hydrocarbon per unit volume = (1 - Sw)

    Matrix

    Water

    Hydrocarbon(1-Sw)

    Sw

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    Relates Sw to Rt

    If Rt = R0, then the formation is 100% saturated

    with formation water. However, if Rt > Ro, thenthe formation contains oil or gas.

    General formula:

    t

    m

    w

    t

    w

    t

    on

    w

    R

    Ra

    R

    RF

    R

    RS

    **===

    For clean sands, n = 2 is common.

    Like a and m, n is measured in the lab.

    ARCHIES SECOND EQUATION

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    LAB EVALUATION OF N

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    Archies equation is based on the following

    relationships

    How Archies Equation Works

    Rock type 1

    Rock type 2

    1000

    100

    10

    1 .01 .1 1.0

    FR

    When water saturation is 100%

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    Determining Rw Directly: from measured water sample

    Indirectly: from openhole well logs

    SP logs

    Pickett plotsRwa technique

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    Rw Using the Rwa Technique

    t

    m

    waRR

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    Example Log

    GRC0 150

    SPCMV-160 40

    ACAL6 16

    ILDC0.2 200

    SNC0.2 200

    MLLCF0.2 200

    RHOC1.95 2.95

    CNLLC0.45 -0.15

    DTus/f150 50

    001) BONANZA 1

    10700

    10800

    10900

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    Rw Using Pickett Plots

    For water sands

    n1

    tm

    w

    w R

    Ra

    S

    /

    wwt SlognRaloglogmRlog

    wt aRloglogmRlog

    y InterceptxSlope

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    Well: Bonanza 1

    Zone: 10690.000 - 10980.000 ft

    Date: 27 Apr 1998 @ 11:31

    VCL < .3Deep Induction

    5000.01 0.1 1 10 100

    Porosity

    0.0

    1

    0.1

    1

    Ap

    parentRt

    0.2

    200

    0.6

    32456

    2

    6.3

    24555

    20

    63.2

    45553

    1.00

    Rw = 0.073

    a = 1

    m = 2

    n = 2

    Example Pickett Plot

    Pi k tt Pl t S l ti F S

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    Pickett Plot Solution For Sw

    1.00

    .8

    .6

    .4

    .2

    .1

    Well: Bonanza 1

    Zone: 10690.000 - 10980.000 ft

    Date: 27 Apr 1998 @ 11:32

    VCL < .3Deep induction

    5000.01 0.1 1 10 100

    Porosity

    0.0

    1

    0.1

    1

    Ap

    parentRt

    0.2

    200

    0.6

    32456

    2

    6.3

    24555

    20

    63.2

    45553

    Rw = 0.073

    a = 1

    m = 2

    n = 2

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    SUMMARY - RESISTIVITY

    Resistivity important property

    Depends on ions in water

    Water resistivity depends onConcentration

    Temperature Ion species

    Archies First Equation relates rock

    resistivity to Rw Archies Second Equation relates Sw to Rt