7. Interpretation

8/14/2019 7. Interpretation

1/25

1. Log Interpretation

Defining Log InterpretationControls on Fluid SaturationsWellbore EnvironmentVshaleLithologyPorosityEstimating RwWater SaturationExercisesThe Logging Operation

3. Definition of Well Log Interpretation Takes measured properties, singly or in combination, to interpret lithology, porosity and water saturation:

lithology - GR; NPHI-RHOB, PEF, Spectral GR porosity - RHOB; NPHI-RHOB; DT water saturation - Archie Equation

based on principle that water is the only conductive rock component if we know porosity (phit) and water resistivity (Rw: salinity), then water (Sw) and hydrocarbon (Shc)

saturation can be derived from whole rock resistivity (Rt)

Sw = ((a * Rw ) / ( phitm * Rt ))1/n

default values for a, m and n are 1, 2 and 2

4. FLUID SATURATION - fraction of the pore space occupied by a fluid phase

Sw is calculated by Archie equation using measurements from resistivity tool. The downhole environment is generallyconsidered as controlled and described by :

Grains coated by water (water wetted formation) Water fills part of measured pore space Measured as percent Defines So, Sg, Sw So = 1-Sw Depends on texture and column height Narrower pore throats restrict capillary entry pressure Capillary entry pressure low for well sorted, coarse grained Sandstones with large pores and pore throats,

hence low Sw Cap entry pressure important for drainage process


2/25

5. Controls on Sw Capillarity properties of Rock Position in a hydrocarbon column

Higher in the column you get higher buoyancy pressure so oil or gas can

overcome entry pressure into smaller pores

6. Interpretation and the Wellbore Environment

! " # $

% & % ' % (&)* " & +

$

$

7. Lithology Interpretation Gamma Ray , $ , -*+. /*+. 0

1 &" & , " $


3/25

8. Vshale A normalised measure of shale contentNumerous methods, most common is interpolation of GR response between cleansand and 100% shale

Vshale = GR - GRcleanGRshale - GRclean

baselines calibrated using cuttings, core, NPHI/RHOB etc.

Eg. GR at point A = 60GR clean = 20GR Shale = 140

Vshale = 60 - 20 = 33%140 - 20

This will not work in radioactive zones, eg. Feldspars, some dolomites.

0 2 3

4 .

25 /* 6 -(0 7 & )/ )/ 6 /(

894 2 !:

10.The Chart Shown provides Clay mineralogy estimate

using the NGT and PEF values. Because the porosity andcomposition of clays vary, the minerals on the plot aregeneral areas rather than unique points.

EG.THOR = 10.6 ppmU = 4.5 ppmK = 3.9%PEF = 3.2

Clay mineral is Illite


4/25

11. Porosity Density-Neutron method Best log derivation of porosity

Automatically corrects for shaliness/gas

- these have opposite effects on the RHOB and NPHI

Chartbook solution

Equation approximation

porosity = (( nphi2+ phit_d

2)/ 2)0.5

12. Lithology/Porosity InterpretationNeutron-Density Combination

;

9

! " ! #$ %&

0 8.2

8.2 >/ &)?


5/25

4 8& + # ! ' 8 0 ? 8 ' ! :340984 .B

ChartBook Pg. 56 62 for various borehole conditions and Neutron Tool Types


6/25

14. Exercise:

Determine Lithology and Porosity for the points shown(use QuickPetro, Chart CP1e)

Point NPHI RHOB GR Dens-NeutPorosity

CrossplotPorosity

VshalefromGR

Lithology

A .15 2.45 15

B .18 2.32 15

C .24 2.25 25

D .25 2.60 115

E .27 2.31 15

F .24 2.30 15

G .25 2.46 15

15. Lithology Interpretation PEF

.4

.4 C# .4 )?

-)>/ (*?> + *)( */ 8 .4

%09>"

5 (0 ( ' ,+For quantitative interpretation, must accountfor porosity -convert to Uma and RHOma

Eg 1.PEF = 3.65RHOB = 2.52PHIT =16%


7/25


8/25

17. Lithology Interpretation Cuttings andcores

1 ! &

; @ A ( -* "

$ '

Point NPHI RHOB GRDens-NeutPorosity

CrossplotPorosity

Vshale fromGR Lithology

A

B

C

18 .Example of Neutron-Density

Pnt NPHI RHOB GR PEF

Dens-NeutPoro.

XplotPoro

VshfromGR Lith

A 0.0 2.95 5 6.9

B .30 2.15 150 2.4

C .21 2.55 75 3.0

D .27 2.10 75 2.0

E .10 2.45 120 2.4

F .22 2.25 15 2.1

G .03 2.65 15 3.5


9/25

19. Summary of Lithology Interpretation

! " "

"

.

'&8 ' 0

- Density-Neutron method (previously discussed)B @A

D 0 E E

" "

'&8 $

6 %% "7 8'"&! "&9

F G F 1 "

20. Density method' $ 1 $

#$8 6 % 0 + & ! % 0 ( &

F 1 /7(5 /=)5 /?=5 "

F F )*(5 *?5"

#


10/25

21Sonic porosity method $ ; H&2&, H2," + H2, "

#$8 6 : % % '0 ! ' &&

! F *7/( *7= *7- " F 1 (( 5 $#I >J 5 " F

22. Calibrating Log and core porosities; K 1 1 L*)(" - ( M*/("

& % 1 ; ! 0

23. Estimating porosity from logs - guidelines# # $ " "

24. Formation Water Rw+ $ "

;

Why is it important? Rock matrix material and hydrocarbons are effectively non-conductive Formation water (brine) is conductive Given 2 zones of equivalent porosity, with the same formation water but with differing bulk resistivity (Rt).

Higher resistivity equals lower water saturation, and therefore higher oil/gas saturation. Archie equation relates porosity, Rt and Rw to Sw


11/25

25. Archie equation

Sw = water saturationRw = Formation water resistivityRt = true formation resistivityphit = total porositya = Tortuosity factor

m = Cementation factorn = saturation exponent (most commonly used value is 2)

26. What controls formation water resistivity ! 1 H **? /** /(* 0 H F **? 0 H **7( 2 N

Guideline - always quote formation water resistivities with the corresponding temperature. Analternative is to quote a salinity instead, and then calculate the appropriate Rw for the formationtemperature under consideration.


12/25

Exercise: A new well as just been drilled. The Produced water from an offset well is testedRw is reported as 0.05 @ 75FMaximum reading thermometers show the formation is at 203 F

What Rw should be used ?

27. Calculating Rw1. Water sample (least reliable)2. Reversed Archie Equation in a water sand3. Pickett Plot4. Resistivity ratio method in a water sand5. SP method

28.Reversed Archie + $ H 0 )** O & H + ; H"

H "+ F) FF/

,* 6 ": ,

H F . F H F

* + '( ' ( ;

29. The Pickett CrossPlot (Pickett, 1972)One of the simplest and most effective crossplot methods in use.

Can help determine: formation water resistivity (Rw), cementation factor (m) matrix parameters for porosity logs

(Dtma and RHOma). Water Saturation

Method is based on the observation that true resistivity (Rt) is afunction of:

1. porosity (1/ m )2. water saturation (Sw),3.

cementation factor(m).

A Pickett crossplot is developed by plotting porosity values withdeep resistivity (RIld or RLLd) values on two-by three cycle log-logpaper

Wet zones plot along a lineto the Southwest. (Ro line)

Slope of line = m (usually =2)


13/25

30. Pickett crossplot Example= 10%m = 2Locate where 10% porosity line (horizontal) intersects Ro line, follow point vertically down and read Resistivity, thisis Ro.Ro = 5.6 ohms

Rw = mRo = .056

31. Resistivity ratio method + 5 # $

53 $

H F H5H1"PH

;H F H F ::'5:'"H1 F # 30:"H F

32. Sw ExampleGiven:Resistivity of Mud Filtrate = 0.25 @ 75 FRFT Water Sample indicated 50,000 ppm chloridesPorosity = 30% throughoutWhat is Sw in interval A

Resistivity is usually presented in a logarithmic scale in 3-decade or 4-decade tracks

33. Archie Equation; ::' :'" $ + $

, + 0 )J-> )J7= EE2 + $ H" H" " $

3

& + $ " & "

Guideline % " # &


14/25

+ $ %*& 6 %/ : ,*& ! ,;

0 F F 1 /" F H F H F

"

/ 6 ! % 0&;

F F F !

34. Archie Equation inputs41

5 0+ ) /0' ' ) /' %#0+ (0& *7/ /)(' % 9994


15/25

=>$? @,- '

Matrix parameters - density & travel time Fluid parameters - density & travel time

Archie equation exponents: a, m and n Formation water resistivity Rw Porosity calculation method

! $ (

. H #

H


16/25

EXERCISE 1

Zone A: Determine Rw and Salinity in this wet zone. Temperature = 250 0F

Zones B and C: Complete the chart for these two subzones.Average Temperature of the interval is 300 0F

What is expected production from Zone B? Zone C?Given that the well is presently shut in as it was producing from a deeper horizon which haswatered out, would you re-complete this well? If so, where would you perforate?

Hint, RHOB scaled 1g/cc full scale or 0.05 g/cc per small divisionNPHI scaled 60 pu full scale or 0.03 v/v per small division

Use Schlumberger Chartbook Gen-1For salinity, converting Rw to differing temperatures.


17/25

A 888888888888888888888888888888888888888888888

; F G G G G G G G G G G G G G G' ; F G G G G G G G

B .0 3 *)/->(7=?J

)*)))/)-)>)()7)=)?)J/*/)

///-/>/(/7/=

/?/J

2 G G G G G2 G G G G G G

+ . G G G G G G+ ; 0 ." G G G G G G


18/25

37. Interpretation bits and pieces '

+

H ' 0 " & 41 01 ' ; 3

38 Getting More out of a GR/RES LWD suite9 1 & 3;' $

.

% EE ; 3;' $ 1

) * * +

Gross section describes the overall succession

The volume of shale (Vsh) provides a reference for lithology and input intoeffective porosity, shaley sand, Sw and permeability calculations normallycomputed from GR with ND. Net sand is commonly quantified using a Vsh

cut-off along with other log parameter cutoffs - can also use porosityqualifier net pay is net sand containing hydrocarbon. Reserve calculations are

highly sensitive to NtG

40. Choose Vsh% cut off todetermine net sand using

integrated log suite.

Net = A + B + C + D

Net-to-Gross = NET / Gross


19/25

41. Difference between minimum

and maximum cases results in:

So Which is Right?

42. Net Pay can be defined as that part of the reservoir section which will contribute to oil or gas production,

and that depends on the recovery mechanism:

For example: Water drive will recover more oil but may access less of the reservoir volume than straightforward

depletion As gas is so much more mobile than oil much more of the reservoir will contribute to production

(And the real fun starts with naturally and hydraulic

fractured reservoirs - small zone of inflow to the well, butlarge reservoir volume drained)

So always aim to base a net pay cutoff on permeability ifpossible, and make the cutoff a team decision

!" #$ %& $ ' $$( $!!$

$ ) $ * "' +)("& $, * '' $ -&, "! ' ./ "

0 12 "


20/25

44. Net Sand CalibrationExample from Angola

Three Facies examined in Nine cored wells.

Note that each facies has is own cutoff value

established by comparing the Net from core vs thatcalculated from logs.

' 4 R

' 6 "41 #8

. R R0 8 F 8 R

1 >

0 0 ( >+ (0 )

)" ! : - ; 5 5

/" ! 6 : R"

-" ' : 5 5 RR">" + R"(" 9 5 R7" D=" 3 R

Core vs Log net sand

0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0

9.0

0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0

Core Net (m)

LogNet(m)

PFAC 1

PFAC 2

PFAC 3

PFAC 33

PFAC333

0

500

1000

1500

2000

2500

3000

3500

20 22 24 26 28 30 32 34

Phi @1600psi

P

erm

1600

si Corelab hot

Anglolab hot

Anglolab cool

Anglolab flow

Corelab cool

Corelab flow


21/25

2 #' >

C ' 38

6 '

& 1

& B H +3 8

& ! .

+ D

0

#' > ! % 6

+ 41

$ $ K +

3 .

. B

: ,"

3 :

0 A S T & ST

0


22/25

5 -

A

;

+ " + 1 " + "

$

+ 1 K ; 3

" ,' 0 ' (

H $ $

3 . "

1 1 )** &

" + '

K . 41 "

4

K $ $ $ $ 1

: 30: 0.


23/25

#3& .


24/25

QUIZ:

1a. Given Rmf

= 2.5 -m @ 10oC, find Rmf

@ 52oC, using Chart Gen-9 (Figure B2).

Rmf

=

b. What is NaCl concentration of the mud filtrate in ppm?

2a. Given a solution salinity of 80 000 ppm, find the solution resistivity @ 121oC

Rm= @121oC

b. Given a solution salinity of 10 000 ppm @ 20oC, find the solution resistivity @ 50oC

Rm= @50oC

3. Using the log example on the next page and given that porosity is about 25%, determine:

a. Rt-(ILD uncorrected) at A and B

b. In which case is the mud filtrate more saline than formation fluids?

c. In which case is the mud filtrate fresher than formation fluids?

d. Which is water bearing? Hydrocarbon Bearing?

e. What is the Rw?

Bonus Questions:


25/25

1/240

1 09-JUN-1992 14:09

INPUT FILE(S) CREATION DATE

CP 32.6 FILE 8 09-JUN-1992 14:42

.20000 2000.0

SFL(OHMM)

.20000 2000.0

ILD(OHMM)

.20000 2000.0

ILM(OHMM)

-150.0 0.0

SP(MV )

0.0 150.00

GR(GAPI)

ILM(OHMM)

1800

---GR

---SP

---ILM

---ILD

---SFL

1700

---SP

---ILM

---ILD

SFL---

1725

Figure B32

4

4

4

44

44

44

4 4 44444

454

7. Interpretation

Documents