Page 155 - Open-Hole Log Analysis and Formation Evaluation

Rock Matrix

be
In
Ry
Sp
7.4.
yN
Read
Read
|
=
value
=
which
Assume
&
chapter
The
porosity
cases
Calculate
recourse
for
@p F
Solution
QUESTION
6,
can
in
formation
in
FX
is
Ry
Ro/F
in =
R,,.
of
combined
we
be
the
=
#7.4
the
the

where
can
5420
to
factor,
made
this
oil/water
ft.
R,/R,.
Ry
be
used
have
wet
to
same
find
to
and
interval
a
0.62/¢7",
read
Rob”/a,
zone
'
contact
R,,,
find
cannot
in
value
can
equation
Archie’s
interval.
from
be
the
Water
be
this
the
|
5390-5400
R,, from
read
for R,.
ft.
read
well
dual
in
equation
the
requires
connate
is
a
from
at
logs

wet
water
the
written
5384
laterolog
in
ft.
zone
4
shown
: 4
wv
and
knowledge
the
in
of
density

of
resistivity.
log.
form
equal
a value
figures
Below this
both
Again
F,
7.3
These
(¢,)

level,
and
from

can

for
a
porosity,
the

of
3.0
*
8
04
, 05
1.0
20
‘
5
6
7
to
use
Use
90
100
200
Read
Read
berger
FIGURE
find
R,,
R,
With R,
Convert
this
Auseful
through
the
Compute
7.7
=
QUESTION
at
¢
S,,.
this
S,,
to
chart
chart
0.18
#7.5
at
to
5350
defined,
Thus,
for
0
porosity
an
ft.
find
Well Services.
at
5350
4
+
4
4
+
4
r
4 1 8 +4 5 Compute S,,. . +5 T° 7 ft. Read R, at 5297 a 3 Fr= pet 0.62 at Read Ry 4-8 | #7,3 5 v1 QUESTION 2 35-46 8 wot? 25 T } 20 -$ 20 +2 A 30 08 and Water Saturation. of. Porosity 15 40 | FIGURE 7.6 Definitions +3 06 60 Tt 10 + 89 +4 5 + 400 tT, % TY 600 { Porosity so T Oil
+
T
Ry.
: a 5350 ft. equation Saturation os 2 Tt? LOG ANALYSIS
ft.
(known
+5
+6
water
t*
+3
m?/m.
Nomogram for
Fo=FrRw
Archie’s
Fusing
as
a
Then
=
solving Archie’s
draw
saturation
a
0.62,
can
m
may
nomogram),
=
equation
be
; 8 : : ; 4 6 x 10 20 ° 40 80 80 100 200 300 400 600 yo 1,000 2,000 3,000 4,000 8.000 8,000 10,000 Q-m RA:
second
a
4
2
is
draw
2.15.
line
a
solved.
Determination.
shown
be.calculated
: 47 tL 5 +5 % Sw
frorn
first
in
Rp
ona
line
Courtesy
Sn
figure
=f

T
|
+
+1
7T8
712
+ 70
sto
tis
+ 16
,
+60
+ 90
+ 80
+50
20
a0
ta
25
from
7.7.
+100
through
Ry
Po
Schlun-

To

FR;
point-by-
R,,