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8/7/2019 INTERPRETACION DE REGISTROS
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$P20.- .
LATERAl.ANO$l4OATNORMAL
AM."-O AQ'IS'.' 50 O
L~ NOIUIAL
A""'64' 50
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INTERPRETACiÓN CUALITATIVA DE LOS REGISTROS ELECTRICO CONVECIONAL y MICROLOG
R..' 1.25O...AT80"
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
I-J
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.
.1
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- 3INTERPRlETACIÓN CUALITATIVA DE LOSREGISTROSELECTRICO CONVECIONAL y MICROLOG
,,
;
ra--
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...-
5=t
~
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.
.,;g
~~ 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
"~
~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-.--.
.
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
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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
I
~..
~"<2e¡:{
¡
OO,..
Aa1
lill~- _RB-
1-----I
I
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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1J -f
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)
~ ;:.
IIJI)
1I,L/
((l
106
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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
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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
----. --
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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
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figure 4-6d Schematic Oklahoma Log (~ontinued)
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