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Model Solutions to Examination
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1. Complete the sections above but do not seal until the examination is finished.
2. Insert in box on right the numbers of the questions attempted.
3. Start each question on a new page.
4. Rough working should be confined to left hand pages.
5. This book must be handed in entire with the top corner sealed.
6. Additional books must bear the name of the candidate, be sealed and be affixed to the first book by means of a tag provided
Subject:
INSTRUCTIONS TO CANDIDATES
8 Pages
PLEASE READ EXAMINATION REGULATIONS ON BACK COVER
No. Mk.
NAME:REGISTRATION
NO.:
COURSE:YEAR:SIGNATURE:Complete this section but do not
seal until the examination
is finished
FORMATION EVALUATION
2A1A1A1A1A1.
a.a.a.a.a. Clays have a wide range of densities (2.2 - 2.65 g/cc). Presence of
clay in the pores of a sandstone could therefore result in
misinterpretation of the matrix density and therefore the porosity of
the sandstone.
b.b.b.b.b. The bound water and OH groups on clay minerals will result in an
overestimation of porosity when using the neutron log.
c.c.c.c.c. Bound water will also have an effect on resistivity measurements.
d.d.d.d.d. The electrostatic charges on the surface of clay minerals present in
the sandstone affects the conductivity of the sandstone and
therefore the resistivity. Smectites will have the greatest effect,
with Illite and finally Kaolinite having the lowest effect.
A2.A2.A2.A2.A2.
a.a.a.a.a. The principal controls on porosity of a formation depend on the type
of porosity: intergranular porosity and/or secondary porosity. The
intergranular porosity of a granular rock such as sandstone is a
function of stacking and sorting of the rock grains. The denser the
packing the lower the porosity. Stacking can result in porosities of
between 47.6% (for particles of the same size stacked on top of each
other to 25.9% for particles of the same size with the particles
sitting in troughs between layers. A variety in size (poorly sorted) and
shape of particles will result in a reduction of porosity.
Secondary porosity is caused by dissolving of limestone or dolomite
causing vugs and caverns. Fracturing also creates secondary porosity.
Model Solutions to Examination
3
b.b.b.b.b. Permeability is heavily dependant on fracture aperature and density.
The denser the fracture population the greater the permeability.
Porosity is rarely affected by fractures since the fractures generally
contributes less than 1% to the porosity.
A3.A3.A3.A3.A3.
a.a.a.a.a. The sources of Gamma radiation are: Potassium K40, Uranium U238,
and Thorium Th232
b.b.b.b.b. K40 is present in illitic shales and clays, feldspar and micas
U238 is present in phosphates and uranium salts
Th232 is present in phosphates and shales
A4A4A4A4A4.
a.a.a.a.a. Compressional wave velocities provide porosity information
b.b.b.b.b. Shear and compressional wave velocity waves are used to calculate the
mechanical properties of rocks such as Poissons ratio for sand control
and borehole stability in drilling
c.c.c.c.c. Stonley waves are used to predict permeability and the presence of
open fractures.
4A5.A5.A5.A5.A5.
a.a.a.a.a.D M S
SaltFresh MudFresh Mud
System
Resi
stiv
ity,
R
Salt MudSystem
Resi
stiv
ity,
R
WaterZone
R*
S M DR*
R0
Rwincreasing
RtSo
Rx0
Rx0
A6.A6.A6.A6.A6.
a.a.a.a.a. The density method:
= ma bma f
= .67 .31.67 .0
= 0.216
2 22 1
b.b.b.b.b. The Acoustic method:
= Logt mat
ft mat
= 84 55.56
185.2 55.56
= 0 219.
Model Solutions to Examination
5
The values from the two techniques are similar. The differences could
be due to errors in assumed fluid and matrix densities and travel
times. The difference could also be due to dispersed clays in the pore
space affecting the log readings.
c.c.c.c.c. The saturation of the rock is given by:
w mn w
t
S Ra
=
1R
Therefore:
wS = 1 37
0 2160 04 1
271 851 65
.
.
.
.
.
Sw = 0.124Sw = 0.124Sw = 0.124Sw = 0.124Sw = 0.124
The saturation is 12.4%. This is less than the critical saturation of
45% and therefore the zone will be productive.
A7.A7.A7.A7.A7.
a.a.a.a.a. The Rw is determined from the following:
Since Rmfeq = Rmf x 0.85
= 0.55 x 0.85
= 0.468 ohm.m
SSP = -71 = (61 + 0.133 x 140) log(0.468/Rweq)
6Rweq = 0.468/1071/79.62 = 0.468/7.79 = 0.06 ohm.m
From SP-2:From SP-2:From SP-2:From SP-2:From SP-2:
Rweq = 0.06 => Rw = 0.075Rweq = 0.06 => Rw = 0.075Rweq = 0.06 => Rw = 0.075Rweq = 0.06 => Rw = 0.075Rweq = 0.06 => Rw = 0.075
b.b.b.b.b. The logging suite would be:
Spectral GR - for basic correlation
- identify anomalous high GR zones which are not shale
- aid lithology identification
Neutron Density - for lithology and porosity information
- also Pe log from density for lithology
Induction log When Rmf / Rw exceeds 2.5 and Rw is below 1 ohm.m
then an induction log should be used in place of a laterolog. Since Rmf /
Model Solutions to Examination
7
Rw is around 7.5 (See graphic above) we run the induction log for Rt
determination.
Sonic Log - Lithology identification
- help characterise porosity type
- An Array sonic can be used for fracture
identification. Vp and Vs data can be useful for
rock mechanics studies.
A8.A8.A8.A8.A8.
a.a.a.a.a. The T2 response is a function of the pore size distribution and can
therefore be correlated to permeability.
The NMR measures the fluid filled porosity. However the NMR can
resolve the bound or capillary trapped water saturation from the
moveable water saturation.
In an appraisal well a correlated permeability can be coupled with the
BVF to give likely fluid production and potential rates.
8B9.B9.B9.B9.B9.
a.a.a.a.a. The following zones can be seen on the log (See log):
Zone Depth
1 11400 - 11468 Limestone2 11468 - 11542 Shaley sandstone, possibly gas bearing
possibly oil bearing
3 11542 - 12115 Shaley sandstoneGas bearing down to GWC at 12115
4 11930 - 1215 Shaley sandstone
5 12115 - Shaley sandstone
B10.B10.B10.B10.B10.
a.a.a.a.a.
Depth b n t M N
1 11410 2.60 0.04 57 0.83 0.602 11510 2.18 0.23 91 0.83 0.653 11655 2.15 0.22 94 0.83 0.684 12165 2.20 0.235 85 0.87 0.64
Depth b - n M-N
1 11410 Low Sandstone Limestone2 11510 Sandstone Low Sandstone3 11655 Sandstone with gas High sandstone4 12165 Sandstone Ambiguous Sandstone but
with secondary porosity
Point
Point
Model Solutions to Examination
9
b.b.b.b.b. Ambiguities:
11655 - N-D plot shows gas indications but M-N does not. Gas is
supported by resistivity separation
12165 - N-D indicates sandstone, M-N plot is ambiguous.
It is possible that the shale and gas effect are affecting the
interpretation.
B11.B11.B11.B11.B11.
a.a.a.a.a. The Humble Equation is:
F = 0 622 15.
.
or,
F = 0 812.
At 12165 :
Neutron - Density cross plot gives a porosity of Neutron - Density cross plot gives a porosity of Neutron - Density cross plot gives a porosity of Neutron - Density cross plot gives a porosity of Neutron - Density cross plot gives a porosity of = 0.275 = 0.275 = 0.275 = 0.275 = 0.275
Hence,
F = 0 620 2752 15
.
.
.
= 9.95
10
or
F = 0 812.
= 10.7
Since,
Rxo = 0.71
RLLS = 0.31
RLLD = 0.31
Hence,
RLLD /RLLS = 1 (implies no invasion correction)
RLLD/RXO = 0.44
From Rint-9bFrom Rint-9bFrom Rint-9bFrom Rint-9bFrom Rint-9b
Rt / RXO = 0.41
Therefore,
Rt = 0.291 (Approximately equal to RLLD)
Rwa = Rt /F
Therefore,
RRRRRwawawawawa = 0.291/9.95 = 0.030 = 0.291/9.95 = 0.030 = 0.291/9.95 = 0.030 = 0.291/9.95 = 0.030 = 0.291/9.95 = 0.030
Model Solutions to Examination
11
or
RRRRRwawawawawa = 0.291/10.7 = 0.027 = 0.291/10.7 = 0.027 = 0.291/10.7 = 0.027 = 0.291/10.7 = 0.027 = 0.291/10.7 = 0.027
These values are very close.
B12.B12.B12.B12.B12.
Since,
Rxo = 2.5
RLLS = 17
RLLD = 45
RLLD /RLLS = 2.65
RLLD/RXO = 18
From Rint-9bFrom Rint-9bFrom Rint-9bFrom Rint-9bFrom Rint-9b
Rt / RLLD = 1.35
Therefore,
RRRRRttttt = 60.75 ohm.m = 60.75 ohm.m = 60.75 ohm.m = 60.75 ohm.m = 60.75 ohm.m
di = 38 inchesdi = 38 inchesdi = 38 inchesdi = 38 inchesdi = 38 inches
38 inches
60
Resistivity
Rx0 = 2.5
Rt
12
B13.B13.B13.B13.B13.
a.a.a.a.a. It is difficult to identify a maximum shale value :
The biggest shale peak is at 11543. This is 112 GAPI
Hence this will be used :
GR @ 11890 = 80
GRsand = 52
GRshale = 112
GRsand
shale sand
GR
I = GR GR
GR GR
I = 80 52
112 52
= 0.53= 0.53= 0.53= 0.53= 0.53
Hence the volume of shale at 11890 using the Olser Rocks modelHence the volume of shale at 11890 using the Olser Rocks modelHence the volume of shale at 11890 using the Olser Rocks modelHence the volume of shale at 11890 using the Olser Rocks modelHence the volume of shale at 11890 using the Olser Rocks model
is approx. 38%.is approx. 38%.is approx. 38%.is approx. 38%.is approx. 38%.
Model Solutions to Examination
13
The whole interval below 11468 is very shaley. The m and a assumed
for the Rwa equation assumes a clean sand. The value of m will
decrease in a shaley sand due to the conductivity of the shale, and the
value of Rwa calculated represents a minimum
To correct for the shales, one of the shale saturation equations may
be needed.
14
3
Gas
Model Solutions to Examination
15
5
Oil
Water
GOC
OWC
16
Deliberately Left Blank
Model Solutions to Examination
17
1
18
Schlumberger
4-16
Porosity and Lithology Determination fromFormation Density Log and CNL* Compensated Neutron Log
0 10 20 30 40CNLcor, neutron porosity index (p.u.) (apparent limestone porosity)
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
b, bu
lk de
nsity
(g/cm
3 )
D, de
nsity
por
osity
(p.u.
) (m
a =
2.
71;
f = 1.
0)
45
40
35
30
25
20
15
10
5
0
5
10
15
Poros
ity
SulfurSalt
Approximategascorrection
0
Anhy
drite
PolyhaliteLangbeinite
Calcit
e (limes
tone)
Quartz
sands
tone
Dolom
ite
0
45
5
15
10
20
25
30
35
40
30
0
5
15
10
20
25
35
40
30
0
5
15
10
20
25
35
40
Fresh water, liquid-filled holes (f = 1.0)
*Mark of Schlumberger Schlumberger
1
24
3
23 41
For CNL logs before 1986, or labeled NPHI