02 General

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General

Symbols Used in Log InterpretationGen-1

(former Gen-3

1

dhHole

diameter

didj

h

rj

(Invasion diameters)

Adjacent bed

Zone oftransition

orannulus

Flushed

zone

Adjacent bed

(Bedthickness)

Mud

hmc

dh

Rm

Rs

Rs

Resistivity of the zone

Resistivity of the water in the zone

Water saturation in the zone

Rmc

Mudcake

Rmf

Sxo

Rxo

Rw

Sw

Rt

Ri

Uninvadedzone

PurposeThis diagram presents the symbols and their descriptions and rela-

tions as used in the charts. See Appendixes D and E for identifica-

tion of the symbols.

DescriptionThe wellbore is shown traversing adjacent beds above and below the

zone of interest. The symbols and descriptions provide a graphical

representation of the location of the various symbols within the well-

bore and formations.

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General

2

Estimation of Formation Temperature with Depth

PurposeThis chart has a twofold purpose. First, a geothermal gradient can

be assumed by entering the depth and a recorded temperature at

that depth. Second, for an assumed geothermal gradient, if the tem-

perature is known at one depth in the well, the temperature atanother depth in the well can be determined.

Description

Depth is on the y-axis and has the shallowest at the top and the

deepest at the bottom. Both feet and meters are used, on the left

and right axes, respectively. Temperature is plotted on the x-axis,

with Fahrenheit on the bottom and Celsius on the top of the chart.

The annual mean surface temperature is also presented in

Fahrenheit and Celsius.

ExampleGiven: Bottomhole depth = 11,000 ft and bottomhole tempera-

ture = 200F (annual mean surface temperature = 80F).

Find: Temperature at 8,000 ft.

Answer: The intersection of 11,000 ft on the y-axis and 200F

on the x-axis is a geothermal gradient of approximately

1.1F/100 ft (Point A on the chart).

Move upward along an imaginary line parallel to the con-

structed gradient lines until the depth line for 8,000 ft is

intersected. This is Point B, for which the temperature

on the x-axis is approximately 167F.

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General

3

General

Estimation of Formation Temperature with DepthGen-2

(former Gen-6

80 100 150 200 250 300 350

60 100 150 200 250 300 350

27 50 75 100 125 150 175

16 25 50 75 100 125 150 175

Temperature (F)

Temperature (C)

Temperature gradient conversions: 1F/100 ft = 1.823

C/100 m

1C/100 m = 0.5486F/100 ft

Depth

(thousands

of feet)

Depth(thousandsof meters)

Annual meansurface temperature

Annual meansurface temperature

5

10

15

20

25

1

2

3

4

5

6

7

8

0.6 0.8 1.0 1.2 1.4 1.6F/100 ft

1.09 1.46 1.82 2.19 2.55 2.92C/100 m

B

A

Geothermal gradient

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General

4

Estimation of Rmf and RmcFluid Properties

Gen-3(former Gen-7

Purpose

Direct measurements of filtrate and mudcake samples are preferred.

When these are not available, the mud filtrate resistivity (Rmf) and

mudcake resistivity (Rmc) can be estimated with the following

methods.

Description

Method 1: Lowe and Dunlap

For freshwater muds with measured values of mud resistivity (R m)

between 0.1 and 2.0 ohm-m at 75F [24C] and measured values of

mud density (m) (also called mud weight) in pounds per gallon:

Method 2: Overton and Lipson

For drilling muds with measured values of Rm between 0.1 and10.0 ohm-m at 75F [24C] and the coefficient of mud (Km) given

as a function of mud weight from the table:

ExampleGiven: Rm = 3.5 ohm-m at 75F and mud weight = 12 lbm/gal

[1,440 kg/m3].

Find: Estimated values of Rmfand Rmc.

Answer: From the table, Km = 0.584.Rmf= (0.584)(3.5)1.07 = 2.23 ohm-m at 75F.

Rmc = 0.69(2.23)(3.5/2.23)2.65 = 5.07 ohm-m at 75F.

log .= . .R

Rmf

mm

( )0 396 0 0475

R K R

R RR

R

mf m m

mc mf m

mf

= ( )

= ( )

1 07

2 65

0 69

.

.

. .

Mud Weight

lbm/gal kg/m3 Km

10 1,200 0.847

11 1,320 0.708

12 1,440 0.584

13 1,560 0.488

14 1,680 0.412

16 1,920 0.380

18 2,160 0.350

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10 20 50 100 200 500 1,000 2,000 5,000 10,000 20,000 50,000 100,000 300,000

2.0

1.5

1.0

0.5

0

0.5

2.0

1.0

0

Total solids concentration (ppm or mg/kg)

Multiplier

Multipliers that do not vary appreciably for low concentrations(less than about 10,000 ppm) are shown at the left margin of the chart

Li (2.5)

NH4 (1.9)

Na and CI (1.0)

NO3 (0.55)Br (0.44)

I (0.28)

OH (5.5)

Mg

Mg

K

K

Ca

Ca

CO3

CO3

SO4

SO4

HCO3

HCO3

General

5

Purpose

This chart is used to approximate the parts-per-million (ppm) con-

centration of a sodium chloride (NaCl) solution for which the total

solids concentration of the solution is known. Once the equivalent

concentration of the solution is known, the resistivity of the solution

for a given temperature can be estimated with Chart Gen-6.

DescriptionThe x-axis of the semilog chart is scaled in total solids concentration

and the y-axis is the weighting multiplier. The curve set represents

the various multipliers for the solids typically in formation water.

Example

Given: Formation water sample with solids concentrations

of calcium (Ca) = 460 ppm, sulfate (SO4) = 1,400 ppm,

and Na plus Cl = 19,000 ppm. Total solids concentration

= 460 + 1,400 + 19,000 = 20,860 ppm.

Find: Equivalent NaCl solution in ppm.

Answer: Enter the x-axis at 20,860 ppm and read the multiplier

value for each of the solids curves from the y-axis:

Ca = 0.81, SO4 = 0.45, and NaCl = 1.0. Multiply each

concentration by its multiplier:

(460 0.81) + (1,400 0.45) + (19,000 1.0) = 20,000 ppm.

Equivalent NaCl Salinity of SaltsGen-4

(former Gen-8

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General

6

Concentration of NaCl SolutionsGen-5

=

API141 5

131 5.

.sg at 60 F

g/L at

77F

ppm grains/gal

at 77F F/100 ft C/100 ft API

Oil GravityConcentrations of NaCl Solutions

Temperature Gradient

Conversion

Specific

gravity (sg) at 60F

Density of NaCl

solution at

77F [25C]

0.15 150

200

300

400

500

600

800

1,000

1,500

2,000

3,000

4,000

5,000

6,000

8,00010,000

15,000

20,000

30,000

40,000

60,000

80,000

100,000

150,000

200,000

250,000

101.00

1.005

1.01

1.02

1.03

1.04

1.05

1.061.071.081.091.10

1.121.141.161.181.20

1.0

1.5

2.0

2.5

3.0

0.60

0.62

0.64

0.66

0.68

0.70

0.72

0.74

0.76

0.78

0.80

0.82

0.84

0.860.88

0.900.920.940.960.981.001.021.041.061.08

3.5

0.6

0.7

0.8

0.9

1.0

1.1

1.2

1.3

1.4

1.5

1.6

1.7

100

90

80

70

60

50

40

30

20

10

0

1.8

1.9

2.0

12.5

15

20

25

30

40

50

60708090100

125

150

200

250

300

400

5006007008009001,000

1,250

1,500

2,000

2,500

3,000

4,000

5,000

6,0007,0008,0009,00010,00012,50015,00017,500

0.2

0.3

0.4

0.5

0.6

0.8

1.0

1.5

2

3

4

5

6

810

15

20

30

40

50

60

80

100

125

150

200

250300

1F/100 ft = 1.822C/100 m

1C/100 m = 0.5488F/100 ft

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General

7

Resistivity of NaCl Water Solutions

Purpose

This chart has a twofold purpose. The first is to determine the resis-

tivity of an equivalent NaCl concentration (from Chart Gen-4) at a

specific temperature. The second is to provide a transition of resis-

tivity at a specific temperature to another temperature. The solutionresistivity value and temperature at which the value was determined

are used to approximate the NaCl ppm concentration.

Description

The two-cycle log scale on the x-axis presents two temperature

scales for Fahrenheit and Celsius. Resistivity values are on the left

four-cycle log scale y-axis. The NaCl concentration in ppm and

grains/gal at 75F [24C] is on the right y-axis. The conversion

approximation equation for the temperature (T) effect on the

resistivity (R) value at the top of the chart is valid only for the

temperature range of 68 to 212F [20 to 100C].

Example OneGiven: NaCl equivalent concentration = 20,000 ppm.

Temperature of concentration = 75F.

Find: Resistivity of the solution.

Answer: Enter the ppm concentration on the y-axis and the tem-

perature on the x-axis to locate their point of intersec-

tion on the chart. The value of this point on the left

y-axis is 0.3 ohm-m at 75F.

Example Two

Given: Solution resistivity = 0.3 ohm-m at 75F.

Find: Solution resistivity at 200F [93C].

Answer 1: Enter 0.3 ohm-m and 75F and find their intersectionon the 20,000-ppm concentration line. Follow the line to

the right to intersect the 200F vertical line (interpolate

between existing lines if necessary). The resistivity value

for this point on the left y-axis is 0.115 ohm-m.

Answer 2: Resistivity at 200F = resistivity at 75F [(75 + 6.77)/

(200 + 6.77)] = 0.3 (81.77/206.77) = 0.1186 ohm-m.

continued on next page

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General

8

Resistivity of NaCl Water SolutionsGen-6

(former Gen-9

F 50 75 100 125 150 200 250 300 350 400C 10 20 30 40 50 60 70 80 90 100 120 140 160 180 200

Temperature

Resistivity

of solution

(ohm-m)

ppm

10

8

6

5

4

3

2

1

0.8

0.6

0.5

0.4

0.3

0.2

0.1

0.08

0.06

0.05

0.04

0.03

0.02

0.01

200

300

400

500

600700

800

1,0001,2001,4001,7002,000

3,000

4,0005,0006,0

007,0008,00010,00012,00014,00017,00020,000

30,000

40,00050,00060,00070,00080,000100,000120,000140,000170,000200,000250,000280,000

Conversion approximatedby R2 = R1 [(T1+ 6.77)/(T2+ 6.77)]F or R2 = R1 [(T1+ 21.5)/(T2+ 21.5)]C

300,000

NaClconcentration

(ppm orgrains/gal)

grains/galat 75F

10

15

20

25

30

40

50

100

150

200

250

300

400

500

1,000

1,500

2,000

2,500

3,000

4,000

5,000

10,000

15,000

20,000

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General

9

Density of Water and Hydrogen Index of Water and HydrocarbonsGen-7

PurposeThese charts are for determination of the density (g/cm3) and hydro-

gen index of water for known values of temperature, pressure, and

salinity of the water. From a known hydrocarbon density of oil, a

determination of the hydrogen index of the oil can be obtained.

Description: Density of Water

To obtain the density of the water, enter the desired temperature (F

at the bottom x-axis or C at the top) and intersect the pressure and

salinity in the chart. From that point read the density on the y-axis.

Example: Density of WaterGiven: Temperature = 200F [93C], pressure = 7,000 psi, and

salinity = 250,000 ppm.

Answer: Density of water = 1.15 g/cm3.

Example: Hydrogen Index of Salt WaterGiven: Salinity of saltwater = 125,000 ppm.

Answer: Hydrogen index = 0.95.

Example: Hydrogen Index of Hydrocarbons

Given: Oil density = 0.60 g/cm3.

Answer: Hydrocarbon index = approximately 0.91.

1.20

Pressure NaCl

14.7 psi1,000 psi7,000 psi

25 50 100

Temperature (C)

Temperature (F)

Hydrogen Index of Salt Water

Hydrogen Index of Live Hydrocarbons and Gas

Salinity (kppm or g/kg)

Hydrocarbon density (g/cm3)

150 200

0.85 44040030020010040

Hydrocarbons

Water

0.90

0.95

1.00

1.05

1.10

1.15

1.05

1.2

1.2

1.0

1.00

0

0.2

0.2

0.4

0.4

0.6

0.6

0.8

0.8

1.00

0.95

0.90

0.85 0 50 100 150 200 250

Waterdensity(g/cm3)

Hydrogenindex

Hydrogenindex

250,000ppm

200,000ppm

150,000ppm

100,000ppm

50,000ppm

Distilledwater

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General

10

Purpose

This chart can be used to determine more than one characteristic

of natural gas under different conditions. The characteristics are

gas density (g), gas pressure, and hydrogen index (Hgas).

Description

For known values of gas density, pressure, and temperature, the value

of Hgas can be determined. If only the gas pressure and temperature

are known, then the gas density and Hgas can be determined. If the

gas density and temperature are known, then the gas pressure and

Hgas can be determined.

Example

Given: Gas density = 0.2 g/cm3 and temperature = 200F.

Find: Gas pressure and hydrogen index.

Answer: Gas pressure = approximately 5,200 psi and Hgas = 0.44.

Density and Hydrogen Index of Natural GasGen-8

0

0.1

0.2

0.3

0.4

0.5

100 200 400300

17,50015,00012,500

10,0007,500

5,000

2,500

Temperature (F)

Pressure (psi)

Gas gravity = 0.65

Gas

density

(g/cm3)

Gas pressure 1,000 (psia)

0

0.1

0.2

0.3

0.7

Hgas

Gastemperature

(F)

Gas gravity = 0.6(Air = 1.0)

0.6100

150

200250300350

0.5

0.4

0.3

0.2

0.1

0

0 2 4 6 8 10

Gas

density

(g/cm3)

14.7

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General

11

Purpose

This chart is used to determine the sound velocity (ft/s) and sound

slowness (s/ft) of gas in the formation. These values are helpful in

sonic and seismic interpretations.

Description

Enter the chart with the temperature (Celsius along the top x-axis

and Fahrenheit along the bottom) to intersect the formation

pore pressure.

General

Sound Velocity of HydrocarbonsGen-9

Gas gravity = 0.65

Natural Gas

50 100 150 200 250 300 3500

1,000

2,000

3,000

4,000

5,000 200

250

30017,500

15,000

12,500

10,000

7,500

5,000

2,500

14.7

400

500

1,000

2,000

Temperature (C)

Pressure (psi)

Temperature (F)

Soundvelocity

(ft/s)

Soundslowness

(s/ft)

0 50 100 150 200

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Gas Effect on Compressional Slowness

General

Gen-9a

12

PurposeThis chart illustrates the effect that gas in the formation has on the

slowness time of sound from the sonic tool to anticipate the slowness

of a formation that contains gas and liquid.

DescriptionEnter the chart with the compressional slowness time (tc) from thesonic log on the y-axis and the liquid saturation of the formation on

the x-axis. The curves are used to determine the gas effect on the

basis of which correlation (Woods law or Power law) is applied. The

slowing effect begins sooner for the Power law correlation. The

Woods law correlation slightly increases tcvalues as the formationliquid saturation increases whereas the Power law correlation

decreases tcvalues from about 20% liquid saturation.

1000

200

100

200 s/ft

110 s/ft

90 s/ft

70 s/ft

50

tc(s/ft)

Wetsand

Sandstone

80604020

Woods law (e = 5) Power law (e = 3)

Liquid saturation (%)

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General

Gas Effect on Acoustic VelocitySandstone and Limestone

Gen-9b

13

Limestone

Sandstone

0 10 20 30 40

25

00

5

10

10

15

20

20

25

30 40

20

15

10

5

0

Porosity (p.u.)

Porosity (p.u.)

Velocity(1,000 ft/s)

Velocity(1,000 ft/s)

Vp

Vs

Vs

Vp

No gasGas bearing

No gasGas bearing

PurposeThis chart is used to determine porosity from the compressional

wave or shear wave velocity (Vp and Vs, respectively).

DescriptionEnter Vp or Vs on the y-axis to intersect the appropriate curve. Read

the porosity for the sandstone or limestone formation on the x-axis.

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General

14

PurposeLongitudinal (Bulk) Relaxation Time of Pure Water

This chart provides an approximation of the bulk relaxation time(T1) of pure water depending on the temperature of the water.

Transverse (Bulk and Diffusion) Relaxation Time of Water

in the Formation

Determining the bulk and diffusion relaxation time (T2) from this

chart requires knowledge of both the formation temperature and

the echo spacing (TE) used to acquire the data. These data are pre-

sented graphically on the log and are the basis of the water or

hydrocarbon interpretation of the zone of interest.

DescriptionLongitudinal Relaxation Time

The chart relation is for pure waterthe additives in drilling fluidsreduce the relaxation time (T1) of water in the invaded zone. The

two major contributors to the reduction are surfactants added to the

drilling fluid and the molecular interactions of the mud filtrate con-

tained in the pore spaces and matrix minerals of the formation.

Transverse Relaxation Time

The relaxation time (T2) determination is based on the formation

temperature and echo spacing used to acquire the measurement.

The TE value is listed in the parameter section of the log. Using

the T2 measurement from a known water sand or based on local

experience further aids in determining whether a zone of

interest contains hydrocarbons, water, or both.

Nuclear Magnetic Resonance Relaxation Times of WaterGen-10

Longitudinal (Bulk) Relaxation

Time of Water

Temperature (C)

Relaxationtime (s)

100

10

1.0

0.1

0.0120 60 100 140 180

T1

Transverse (Bulk and Diffusion)

Relaxation Time of Water

Temperature (C)

Relaxationtime (s)

100

10

1.0

0.1

0.0120 60 100 140 180

T2(TE = 0.2 ms)

T2(TE = 0.32 ms)

T2(TE = 1 ms)

T2(TE = 2 ms)

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General

15

Nuclear Magnetic Resonance Relaxation Times of HydrocarbonsGen-11a

Purpose

Longitudinal (Bulk) Relaxation Time ofCrude Oil

This chart is used to predict the T1 of crude oils with various viscosi-

ties and densities or specific gravities to assist in interpretation of

the fluid content of the formation of interest.

Transverse (Bulk and Diffusion) Relaxation Time

Known values of T2 and TE can be used to approximate the viscosity

by using this chart.

DiffusionCoefficients for Hydrocarbon and Water

These charts are used to predict the diffusion coefficient of hydro-

carbon as a function of formation temperature and viscosity and

of water as a function of formation temperature.

Description

Longitudinal (Bulk) Relaxation Time

This chart is divided into three distinct sections based on the compo-

sition of the oil measured. The type of oil contained in the formation

can be determined from the measured T1 and viscosity determined

from the transverse relaxation time chart.

Transverse (Bulk and Diffusion) Relaxation Time

The viscosity can be determined with values of the measured T2 and

TE for input to the longitudinal relaxation time chart to identify the

type of oil in the formation.

Transverse (Bulk and Diffusion) RelaxationTime of Crude Oil

T1 (s)

Longitudinal (Bulk) RelaxationTime of Crude Oil

Viscosity (cp)

0.1 1 10 100 1,000 10,000 100,000

0.001

0.01

0.1

1

10

0.0001

T2 (s)

Viscosity (cp)

0.1 1 10 100 1,000 10,000 100,000

0.001

0.01

0.1

1

10

0.0001

Diffusion(cm2/s)

Hydrocarbon Diffusion Coefficient103

104

105

106

107

1500 50 100 200

Temperature (C)

Oil (9 at 20C)

Oil (40 at 20C)

Diffusion(105 cm2/s)

Water Diffusion Coefficient

Temperature (C)

0 50 100 150 200

0

5

10

15

20

Light oil: 4560

API

0.650.75 g/cm3

Medium oil: 2540 API0.750.85 g/cm3

Heavy oil: 1020 API0.850.95 g/cm3

TE = 0.2 ms

TE = 0.32 msTE = 1 msTE = 2 ms

T1

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General

Nuclear Magnetic Resonance Relaxation Times of HydrocarbonsGen-11b

16

0 0.2 0.4 0.6 0.8 1.0 1.2

0

0.2

0.4

0.6

0.8

1.0

1.2

Hydrocarbon density (g/cm3)

Hydrogenindex


0

2

4

6

0 6,000 9,000

25C75C125C175C

Longitudinal (Bulk) RelaxationTime of Methane

Pressure (psi)

3,000 12,000

10

8

Diffusion (cm2/s)

1020.001

Transverse (Bulk and Diffusion)Relaxation Time of Methane

TE = 2 ms

100

104

0.01

0.1

1

10

103

T2 (s)

TE =1 ms

TE = 0.32 ms

TE = 0.2 ms

T1 (s)

10

15

20

25

30

35

50 100 150 200

Methane Diffusion Coefficient

Diffusion(104cm2/s)

Temperature (C)

1,600 psi

3,000

00

5

8,300

15,500

22,800

3,900

4,500

0

0

General

Purpose

Methane DiffusionCoefficient

This chart is used to determine the diffusion coefficient of methane

at a known formation temperature and pressure.

Longitudinal and Transverse Relaxation Times of Methane

These charts are used to determine the longitudinal relaxation time

(T1) of methane by using the formation temperature and pressure

(see Reference 48) and the transverse relaxation time (T2) of

methane by using the diffusion and echo spacing (TE), respectively.


This chart is used to determine the hydrogen index from the hydro-

carbon density.

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General

Capture Cross Section of NaCl Water Solutions

17

PurposeThe sigma value (w) of a saltwater solution can be determined fromthis chart. The sigma water value is used to calculate the water satu-

ration of a formation.

Description

Charts Gen-12 and Gen-13 define sigma water for pressure condi-

tions of ambient through 20,000 psi [138 MPa] and temperatures

from 68 to 500F [20 to 260C]. Enter the appropriate chart for

the pressure value with the known water salinity on the y-axis and

move horizontally to intersect the formation temperature. The sigma

of the formation water for the intersection point is on the x-axis.

ExampleGiven: Water salinity = 125,000 ppm, temperature = 68F at

ambient pressure, and formation temperature = 190F

at 5,000 psi.

Find: wat ambient conditions and wof the formation.Answer: w= 69 c.u. and wof the formation = 67 c.u.

If the sigma water apparent (wa) is known from a clean watersand, then the salinity of the formation can be determined by enter-

ing the chart from the sigma water value on the x-axis to intersect

the pressure and temperature values.

continued on next page

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General

18

Ambie

nt

200

F[93C]

68F[20

C]

400

F[205

C]

300

F[15

0C]

200

F[9

3C]

68F

[20

C]

400

F[205

C]

300

F[15

0C]

200

F[93

C]

68F

[20

C]

1,000

psi[6

.9MPa]

5,000

psi[34

MPa]

300

275

250

225

200

175

150

125

100

75

50

25

0

100

75

50

25

0

100

75

50

25

0

300

275

250

225

200

300

275

250

225

200

300

275

250

225

200

175

150

125

100

75

50

25

0

0 10 20 30 40 50 60 70 80 90 100 110 120 130 140

125

150

175175

150

125

Equivalent water salinity(1,000 ppm NaCl)

Capture Cross Section of NaCl Water SolutionsGen-12

(former Tcor-2a

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General

Capture Cross Section of NaCl Water SolutionsGen-13

(former Tcor-2b

19

10,000p

si[69

MPa]

500F[

260C]

400F[

205C]

300F[

150C]

200F[

93C

]

68F

[20C]

400F[

205C]

300F[

150C]

200F[

93C

]

68F

[20C]

400F[

205C]

300

F[150

C]

200F[

93C

]

68F[2

0C]

15,000p

si[10

3MPa]

20,000p

si[138

MPa]

300

275

250

225

200

175

150

125

100

75

50

25

0

100

75

50

25

0

100

75

50

25

0

300

275

250

225

200

300

275

250

225

200

300

275

250

225

200

175

150

125

100

75

50

25

0

0 10 20 30 40 50 60 70 80 90 100 110 120 130 140

Equivalent water salinity(1,000 ppm NaCl)

125

150

175175

150

125

Purpose

Chart Gen-13 continues Chart Gen-12 at higher pressure values for

the determination ofwof a saltwater solution.

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Purpose

Sigma hydrocarbon (h) for gas or oil can be determined by usingthis chart. Sigma hydrocarbon is used to calculate the water satura-

tion of a formation.

DescriptionOne set of charts is for measurement in metric units and the other

is for measurements in customary oilfield units.

For gas, enter the background chart of a chart set with the reser-

voir pressure and temperature. At that intersection point move left

to the y-axis and read the sigma of methane gas.

For oil, use the foreground chart and enter the solution gas/oil

ratio (GOR) of the oil on the x-axis. Move upward to intersect the

appropriate API gravity curve for the oil. From this intersection

point, move horizontally left and read the sigma of the oil on

the y-axis.

Example

Given: Reservoir pressure = 8,000 psi, reservoir temperature =

300F, gravity of reservoir oil = 30API, and solution

GOR = 200.

Find: Sigma gas and sigma oil.

Answer: Sigma gas = 10 c.u. and sigma oil = 21.6 c.u.

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20

Capture Cross Section of Hydrocarbons

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General

21

Capture Cross Section of HydrocarbonsGen-14

(former Tcor-1

20.0

17.5

15.0

12.5

10.0

7.5

5.0

2.5

0

68

125

200

300

400

500

Reservoir pressure (psia)

0 4,000 8,000 12,000 16,000 20,000

Methane

Temperature (F)

h (c.u.)

Customary

10 100 1,000 10,000

Solution GOR (ft3/bbl)

Liquid hydrocarbons

Condensate

30, 40, and 50API

20 and 60API

16

18

20

22

h (c.u.)

Metric

20.0

17.5

15.0

12.5

10.0

7.5

5.0

2.5

0

20

52

93

150

260

Reservoir pressure (mPa)

0 14 28 41 55 69 83 97 110 124 138

Methane

Temperature (C)

h (c.u.)

2 10 100 1,000 2,000

Solution GOR (m3/m3)

205

Condensate

Liquid hydrocarbons

0.78 to 0.88 mg/m3

0.74 or 0.94 mg/m3

16

18

20

22

h (c.u.)

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General

EPT* Propagation Time of NaCl Water SolutionsGen-15

(former EPTcor-1

22

PurposeThis chart is designed to determine the propagation time (t pw) of

saltwater solutions. The value of tpwof a water zone is used to deter-

mine the temperature variation of the salinity of the formation water.

DescriptionEnter the chart with the known salinity of the zone of interest and

move upward to the formation temperature curve. From that inter-

section point move horizontally left and read the propagation time

of the water in the formation on the y-axis. Conversely, enter the

chart with a known value of tpw from the EPT Electromagnetic

Propagation Tool log to intersect the formation temperature curve

and read the water salinity at the bottom of the chart.

90

80

70

60

50

40

30

20

0 50 100 150 200 250

Equivalent water salinity (1,000 ppm or g/kg NaCl)

tpw (ns/m)

120C250F

100C200F

80C175F

40C100F

150F60C

75F20C

125F

*Mark of Schlumberger Schlumberger

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General

23

EPT* Attenuation of NaCl Water SolutionsGen-16

(former EPTcor-2

Purpose

This chart is designed to estimate the attenuation of saltwater solu-

tions. The attenuation (Aw) value of a water zone is used in conjunc-

tion with the spreading loss determined from the EPT propagation

time measurement (tpl) to determine the saturation of the flushed

zone by using Chart SatOH-8.

Description

Enter the chart with the known salinity of the zone of interest and

move upward to the formation temperature curve. From that intersec-

tion point move horizontally left and read the attenuation of the water

in the formation on the y-axis. Conversely, enter the chart with a known

EATT attenuation value of Aw from the EPT Electromagnetic

Propagation Tool log to intersect the formation temperature curve

and read the water salinity at the bottom of the chart.

0 5 10 15 20 25 30

Uncorrected tpl (ns/m)

0 50 100 150 200 250

Equivalent water salinity (kppm or g/kg NaCl)

Correctionto EATT

(dB/m)

40

60

80

100

120

140

160

180

200

120C

250F100C200F80C

175F

40C100F

150F60C

75F20C

125F

5,000

4,000

3,000

2,000

1,000

0

Attenuation,Aw

(dB/m)

EPT-D Spreading Loss


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General

EPT* Propagation TimeAttenuation CrossplotSandstone Formation at 150F [60C]

Gen-16a

PurposeThis chart is used to determine the apparent resistivity of the mud

filtrate (Rmfa) from measurements from the EPT Electromagnetic

Propagation Tool. The porosity of the formation (EPT) can also beestimated. Porosity and mud filtrate resistivity values are used in

determining the water saturation.

Description

Enter the chart with the known attenuation and propagation time(tpl). The intersection of those values identifies Rmfaand EPT fromthe two sets of curves. This chart is characterized for a sandstone

formation at a temperature of 150F [60C].

ExampleGiven: Attenuation = 300 dB/m and tpl = 13 ns/m.

Find: Apparent resistivity of the mud filtrate and EPT porosity.

Answer: Rmfa= 0.1 ohm-m and EPT = 20 p.u.

1,000

900

800

700

600

500

400

300

200

100

07 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

1.0

40 50302010

2.0

5.010.0

50.0

Attenuation(dB/m)

tpl (ns/m)

Sandstone at 150F [60C]

Rmfa from EPT log (ohm-m)

EPT

poro

sit

y(

EPT)

0.02 0.05

0.1

0.2

0.5


Documents

02 General