Pete Val Mod Sol

Model Solutions to Examination

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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

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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

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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

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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

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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

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5

Oil

Water

GOC

OWC

16

Deliberately Left Blank

Model Solutions to Examination

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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