INTERPRETACION DE REGISTROS

8/7/2019 INTERPRETACION DE REGISTROS

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INTERPRETACiÓN CUALITATIVA DE LOS REGISTROS ELECTRICO CONVECIONAL y MICROLOG

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rena de agua muy dulce Rw »Rmf, SP tiene desarrolloositivo, invasión moderada RlabR64>R16

rena de agua salada, Rw«Rmf, SP indica ligera arcillosidad

n el tope, invasión moderada a profunda, R16=25, R64=15, ésteon contrbución de la zona invadida mientras que RLat<1 :: Rt

ena de agua salada, Rw «Rmf, invasión somera,16> R64:: Rlat.SPno pareceestar afectadopor el espesor.

'ena delgada de agua salada, Rw «Rmf, invasión someraque R16»R64, Rlat no definida. SP afectado por espesor.

na gruesa de petróleo o gas, invasión muy somera ya quet = 38 R64=45:: Rt y R16>30con gran contribución de Rt.parece no estar afectado.

tita posiblemente derrumbada ya que R2"=R1"::Rm

ena de petróleo o gas Rlat=35(pico),R16=30 y R64=35 hacene sea difícil definir la profundidad de invasión dudosa.parece no estar afectado.

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2 INTERPRETACiÓN CUALITATIVA DE LOS REGIS-TROS ELECTRICO CONVECIONAL V MICROLOG

Arena de petróleo o gas, R64 = 25, posiblemente=Rt, Rlat no ofrece~ctura confiable por el espesor de la arena, diámetro de invasió,:" dudoso.SP parece no estar afectado.

Arena muy delgada, posiblemente petrolífera o gasífera.Didudoso. SP muy afectado

Arena gruesa productora de agua porque es una zona de transición conagua 100%en la zona inferior. Lainvasión es somera ya que en elacuífero Rlaty R64son bajos y R16::Rxoes más alto.-sP llega a SSP en el acuífero.

~rena (por el desarrollo de la curva de SP) saturada de agua salada.El MLno muestra separación debido posiblemente a derrumbes,ay que confirmar con el caliper.rena arcillosa (evidenciada en SP), posiblemente petrolífera. MLindica. rosidad. Invasión no definida.

jArena de agua salada situada debajo de una capa resistiva no permeable:::i I según el ML,el cual también muestra como arena la zona inferior dondeRlat y R64 son bajas porque Rw < Rmf indicando que es de agua salada.

La Invasión es somera por que R16 > Rlat=R64. SP::SSP en el acuífero.No debe cunfundir el cambio de resistividad como un contacto A/P

~Capas resistivas muy delgadas,el MLmuestra que no son porosas ni permeables

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Arena de agua salada Rw < Rmf, R64 y Rlat son:: 1 mietras R16:: 20indicando también que la invasión es somera. La base de la arena esligeramente arcillosa como lo indica la curva deSP.

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na capa delgada resistiva e impermeable encima de una arena de petróleo o gas debidoa la resistividad mostrada en la R64>1O=Rt.Lectura del Rlat no es confiable. La invasión1""."somera por que R16 posiblemente tiene contribución de la zona virgen.

na capa resistiva e impermeable encima de una arena de agua debido a la resistividadR64<5. La invasión es somera ya que R64 lee Ro. La Rlat no confiable.

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~~ Calizamasiva no porosa e impermeable.:i>~

~Lutita de muy alta resistividad, debe ser confirmado con el Rayos GmmaiC.; .

Caliza masiva.

~ La forma de la SP y algunas secciones del MLmuestran que tiene"- intercalaciones muy delgadas de capas porosas

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~Capa masiva de caliza, la redondez de la SP sugiere que puede habercierta permeabilidad muy baja, el MLno lo indica por la ausencia

_ del revoque-.--.

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Arena de petróleo o gas R16: R64=Rtsugiere que la invasión es muysomera.Existe un posible contacto A/Pa 3213'

.1 - na posiblemente de petróleo o gas penetrado parcialmente por el pozo



Ra '" Rt (Rat)

Most people, when dealing with the lateral, use the best apparent resistivity

(Ra) reading from the curve as Rt and do not correct for borehole, adjacent

bed and invasion effects. (These will be covered later in this chapter or

in the ¿ase of invasion, later in the book under departure curves.)

Rules

Conductive beds

Resistive beds

RULES

I Mid-PointI

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read the lowest resistivity in the lower half of the

bed.

Follow the'rules shown in Figure 2-12..

2/3 Thineak

A

! II~in

RmaxRt"'~(Rs)

Rml.n

e < ~ AO

e '" 1.5 AO

Figure 2-12 Ra approximates Rt Rules for the Lateral in Resistive Beds

Mid-Point Rule -- Find the point half way between the

top and bottom of the bed and then go down

one AO distance and read the resistivity

directly off the curve. This is Ra.

2/3 Rule -- From the top of the bed go down one AO

distance and read resistivity, at peak

value of resistivity read, Ra is 2/3 between

these two values.

Point Rule -- When the bed is 1.3 times AO (for 18'8"

lateral bed must be 24~ feet thick), Ra is

the peak value.

Thin Rule -- \~en the bed is thinner than AO distance

a very approximate value of Ra may be deter-

mined. (see Figure 2-12 for equation)

The transition thicknesses between these rules can be ver~ difficult.

Figure 2-13 was developed to aid in picking values.' On Figure 2-13 all

rlistances are either relative to AO or are picked from the maximum peak

at the bottom of the resistive bed.

31

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Figure 6-3

,.

,.

SH1PLIFIED NICROLOG

CALIP~R

RESISTIVITY

~

tite (impermeable)

1 ~ permeable--===-

lO ~ tite

tj permeable

~~, tit~ (separation at

high resistivity pad

leakage)

shale

bad washout Microlog will read Rm

tite

permeable

tite

shale

permeable oil zone

permeable water zone

impermeable shale

permeable oil zone

(no invasion)

permeable water zone

(no invasion)

shale

R2" (mitronormal)

R1"Xl" (microinverse)

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106



invasion reduces the invasion zone resistivity 1ess than the

virgin zone. Under these conditions, curves that read deeper

into the formation must have a resistivity that changes from

c10se to Ri to one approaching Rt.

Fai1ure of the curves to stack is usua11y due to bed thicknesseffects or an unca1ibrated curve.

Bed thickness checks:

If bed thickness approaches or is 1ess than AO or AM

spacing, the reading is probab1y not usab1e.

If normal (AM) is in a bed where e (bed thickness) is 1ess

than 4 AM,the apparent resistivity may be too low.k

If bed thickness appears to be a prob1em, correct the Ra's

using appropriate charts in Chapters 2 & 3.

If the former does not work, try correcting the curves for

boreho1e (mud) inf1uences. This is particu1ar1y true if

mud resistivity is much 1ess than 1 ohm m.

Watch for dead zones, ref1ection peaks, peaks next to

craters that can distort the Ra.

EXAMPLES

In this section we wi11 look at some ES examp1es and show the kind of

ana1ysis that may be used. A11 the curves wi11 be used together to come

up with an ana1ysis. The first is a simu1ated log that is meant to

reinforce the simple curve shape and Ra = Rt concepts.

Figures 4~6a, 4-6b, 4~c and 4-6d are a schematic type log for Ok1ahoma.

1 consider this a good review log. It stresses curve shapes of the

norma1s and lateral s and what to look for when qua1itative1y ana1yzing

an ES. The commentary presented here is in addition to that a1ready on

the log. Cutting effect is used synonymous1y with decay zone. The

reference wi11 be the zone number on the log.

1 Fresh water sand 40 feet thick. The SP thickness is used in this

discussion. A c10se ana1ysis of the short normal wou1d increase the

thickness one foot in every case.

, RSN = 37 ~N = 47 ~ (use 2/3 ru1e) = 2/3 (100 - 15) + 15 = 72

curves stack proper1y

2 36 foot bed

RSN = 27 ~Nto invasion.

Rmf greater than Rw

10 ~ = 1 curves stack proper1y difference is due

Note trai1er for 19 feet be10w bed into sha1e.

320 foot bed RSN = 21 ~N = 2-3 ~ = 1Just 1ike zone #2 on1y thinner and with 1ess invasion effect on LN

67

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Figure 4-6a A Schematic ES Lag With a Mid-Cantinent Curve Arrangement

(drawn by C. K. Ruddick~ 1955 with Sch1umberger)

AM = 16" & 64" AO = 19' Rm = 1

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4 6 foot bed. Critical thickness for a long normal in a resistive bed.

RSN = 19 ~N = 4 ~ = 2 curve values stack properly

Conductive bed Rt < 2 ohm m. SP reduced due to bed thickness.

Actually SP is a little too reduced for just bed thickness.

5 70 foot bed. Thick resistive bed.

RSN = 45 ~N = 39 ~. (mid point rule) = 37 Rt = 37

625 foot bed. SP rounded at bed boundaries usually due to high resistivity.

RSN = 34 ~N = 28 ~ = 37 (e = 1.3 AO and thus use peak R)

resistivities do not stack.

Using Figure 3-10 to correct for bed thickness effect F)N corrected = 35Rt is thus around 37. Invasion from resistivity curves. is not evident.

..

7 20 foot bed. SP is rounded so should be resistive bed.

RSN = 25 ~N = 17 (which corrects to 22 using Figure 3-10)

~ should not be used as AO is about equal to bed thickness and the bed

is a resistive bed. Rt = RLN X ~N = 22 X 22/25 = 19 ohm m.

RSN

84 foot bed. RSN = 15 ~N rever sed - crater due to resistive bed thinner

than AM spacing. Lateral shows bed, dead zone below and 19 feet below

bed is reflection peak. SP reduced due to thin resistive bed.

guess at Rt = Rmax X RRm

' sl.n

12 X 5 = 20

39 50 foot bed.

upper 20 foot section RSN = 30 ~N = 17 ~ = 5 SP suppressed over

upper section, looks like hydrocarbons when commpared to lower 20

lower 20 foot section RSN = 24 ~N = 4 ~ = 1

total section looks like 20 feet of water at bottom, 10 feet of tran-

sition zone and 20 feet of hydrocarbons at the topo,

10 This is a unique situation that was noticed in Oklahoma and up into the

Rocky Mountains which ha~ be en described as colloidal. Apparently, some-

thing like illite is in pores. The SP is often larger than normal and

the resistivity is lower than normal. Sometimes there is no apparent

invasion. Porosity can, on a density log, appear to be over 20% and

the formation have no permeability. These very low resistivity zones

can produce hydrocarbons and no water.

11 Permeable, Rt about 6 ohm m. Shaly.

12 Probably tite and nonproductive.

69



Figure 4-6b Schematic Oklahoma Log (continued)

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13 Tite lime stringer on top of permeab1e bed. Note SP and resistivity.

LN is rever sed and short normal shows bed as resistive. Lateral

sees stringer as thin resistive bed. Lateral has dead zone for 19 feet

be10w stringer and then ref1ection peak.

Permeab1e zone 24 feet thick. RSN = 13 ~N = 3 Upper 20 feet of

permeab1e stringer is in dead zone of lateral and thus not usab1e.

~ at bottom of the bed 1 ohm m. Rt around 1 ohm m. Note trai1erbe10w permeab1e bed indicating Rt 1ess than Rs.

14 Typica1 thin tite stringer. Cou1d be lime or coa~stringer. Note

dead zone and ref1ection peak.

15 25 foot zone. Note ref1ection peak on lateral at top of bed due to

bed 19 feet above. RSN = 19 ~N = 2 ~ = 2

16 16 foot zone. Lateral not usab1e as dead zone covers complete bed.

R = 22 ~ = 8 Rt = 8 X 8 = 3SN -LN 22

17 Same as #16 on1y RSN = 16 ~N = 3 Lateral again not usab1e.Rt = 1 or 1ess.

Ref1ection peak sma11er due to 10w bed resistivity. Note trai1er on

lateral. Rt 1ess than Rs.

18 Tite thick zone.

19 SP shows some permeab1e streaks (marked as porous). If permeab1e

streaks were conductive they wou1d show up so zones are most probab1y

around the same resistivity as the tite zones. See Figure 4-7 be10w

Figure of Sch1umberger

O

Normal I~A

- - -

_ ~Lateral lA

___ ... O......

_ --'t"-~=-~-----

........

....

....

....

....

....

....

....

...

....

....

....

'-

--------

72

----. --



20 70 foot high resistivity bed. Curve on lateral looks normal.

Rt shou1d be at 1east 1500 ohm m. Norma1s peak near top and then

apparent resistivity reduces. In this cas~ the normal changes from

an apparent two e1ectrode device to a three e1ectrode device due

to the c10se proximity of N. For a modern Sch1umberger short normal,

this peak shou1d be 20 feet from the top of the bed (AN = 20 feet)

and for the long normal, it shou1d be 70 feet from the top of the

bed (AN = 70 feet). Curves for zone #20 may be good for old 4

conductor cable (rag 1ine). See Figure 4-7 for another examp1e

of norma1s turning into 3 e1ectrode curves.

The

andmaximum resistivity at peak is a function of bor~ho1e diameter

mud resistivity. The maximum a short normal can record is:

Rm _8RmAMANax - 2 2

d - dh s

for: Rm = 1 AM = 1.33 (16 inches)

AN = 20 feet

feet (8 inch boreho1e)

feet (3.5 inch sonde)

Rmax = 585 ohm m

for a 9 inch ho1e Rmax = 445 ohm m and for a 12~ inch ho1e

Rmax = 220 ohm m for an Rm = 1 ohm m.

If the short normal reaches this maximum 1eve1,it means that the

bed is essentia11y infinite1y resistive.

21 Lateral curve is hanging over from zone in #20. There is no lateral

trai1er be10w bed #21 and thus Rt is equa1 to or greater than Rs of 5.

RSN = 9 at top and reduces to 5. ~N has same va1ues. These are

thus good estimates of Rt.

22 In these older days it was common practice to dri11 a few feet into

the target zone and log. Care must be taken when eva1uating the

curves under these conditions. If you go to ideal curve shapes,

they wi11 he1p. The schematics be10w wi11 diagnose #22.

l'NORMAL S LATERAL S

CONDUCTIVE BED

RESISTIVE BED

~NDUCTIVE BEDRESISTIVE BED

for lateral in boreho1e

bottomed in hígh R bed

partia11y penetrated

BOREHOLE

BED

SP

IMPERMEABLE BED"_" -. ......

PERHEABLE BED

73



figure 4-6d Schematic Oklahoma Log (~ontinued)

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INTERPRETACION DE REGISTROS

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