reserve estimation notes

8/12/2019 Reserve Estimation Notes

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Chapter 3 Reserve Estimation

Lecture notes for PET 370

Spring 2012Prepared by: Thomas W. Engler,

Ph.D., P.E.


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ParametersArea = 640 acresBoi = 1.2 rb/stb

Calculated from logh = 20 ft

f ave = 10%Sw ave = 30%

What is the OIP in MSTB?

OOIP = 5,793 MSTB

One well

H=20 f

Reserve Estimation OIP


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Calculated from log

What is the HCPV?

One well

H=10 f


H=20 1

2

Zone 1 Zone 2

H, ft 10 20

F , % 10 15

Sw, % 30 50

What is the OIP?

2.2 ft

9,102 MSTB

What is the averageporosity? 13.3%

What is the ave. Sw? 45.0%

What is the OIP usingThe ave values?


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Reserve Estimation

Porosity thickness weighted average

Averaging

h1

h2

h3

f 1

f 2

3 n

i ih

n

i i

hi

1

1f

f

n

i

n

iw

S

1 ih

i

1 i

hiwi

S

Water saturation volume weighted average


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Calculated from logh = 20 ftSw ave = 30%

two wells

H=20 f


f = 10% f = 20%

What is the averageporosity?

OOIP = 8,689 MSTB if A1 = A2


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Example of a HCPV map


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3D view of a multilayered reservoir, colors show oil saturation


Graduate students only.what is the OIP for the system below?


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Reserve Estimation Reservoir Volume

1*

13 n A

n

A

n

A

n

Ah

b

V

12 n A

n A

hb

V

or


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t

ww R

R F *S

Measured Water Sample

Published Water Tables

R w = R o/F

SP log

Depth merge & correlate

Sonic

Density

Neutron

others

Lithology

Gas/Liquid

m

a

f F

Induction log

Laterologs

So = 1 - S w

n

1iwiii

oi)S(1hB

7758A N

place,inoilVolumetric

f

Reserves,R = N * RF

Flowchart for Well Log Interpretation

VshSP

GR

Porosity logs


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Reserve Estimation

1. Shale content (Vsh) eliminate the portion of theformation which contains large quantities of shale.

Vshcutoff 20 to 30 %

(Note: shale plays up to 40 to 50%) GROSS SAND

Cutoff Values


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ross Sand

50 ft

15 ft

50 ft

otal=115ft

Reserve Estimation How much gross sand in thiswell?


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Reserve Estimation

1. Shale content (Vsh) eliminate the portion of theformation which contains large quantities of shale.

Vshcutoff 20 to 30 %

(Note: shale plays up to 40 to 50%) GROSS SAND

2. Porosity eliminate the portion of the formationwhich is low porosity (and low permeability) andtherefore would be non-productive.Sandstones

f cutoff 5% to 15%

consolidated friable, unconsolidated Carbonates

f cutoff 4% (Note: shale plays ~ 3 to 4%) NET SAND

Cutoff Values


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et Sand

50 ft

50 ft

otal=100ft

Reserve Estimation How much net sand in thiswell?


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Reserve Estimation

1. Shale content (Vsh) eliminate the portion of theformation which contains large quantities of shale.Vshcutoff 20 to 30 % (Note: shale plays up to 40 to 50%) GROSS SAND

2. Porosity eliminate the portion of the formation which is

low porosity (and low permeability) and therefore wouldbe non-productive.Sandstones

f cutoff 5% to 15% consolidated friable, unconsolidated

Carbonates

f cutoff 4% (Note: shale plays ~ 3 to 4%) NET SAND

3. Water saturation eliminate the portion of the formationwhich contains large volumes of water in the pore space.Sandstones

Swcutoff 60% Carbonates

Swcutoff 50%

NET PAY

Cutoff Values


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et Sand

50 ft

50 ft

otal=100ft

Reserve Estimation How much net pay in thiswell?

Net pay20 ft


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Electrical Properties of Rocks



f ave = 30%Sw ave = 30%

Example

OOIP = 869 Mstb


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Reserve Estimation

Recovery Factor can be estimated by:A. displacement efficiency studies

B. correlations based on statistical studies of particulartypes of reservoir mechanisms

C. All of the above

D. None of the above

E. I dont know I slept through Reservoir Engineering class

Recovery Factor


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Reserve Estimation

Recovery Factor can be estimated by:A. displacement efficiency studies

B. correlations based on statistical studies of particulartypes of reservoir mechanisms

C. All of the above

D. None of the above

E. I dont know I slept through ReservoirEngineering class

Recovery Factor


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Reserve Estimation Recovery Factor

rw re

S a t u r a t i o n

Sw

SoiSor

Som = Soi - Sor

Difference betweeninitial oil saturation, S oi and the residual oilsaturation, S or, thatremains after theformation is invaded bywater.

rw re

S a t u r a

t i o n

Sw

SoiSor

Som = Sxo - Sw

or

....for depletion drive, use rule of thumb of 1/2 of (RF) wd

Infer mud filtrate invasionas an efficientdisplacement mechanism,recovery factor is:

w

S

wS

xoS

wd r E

1

rw re

S a t u r a t i o n

Sw

SoiFrom displacement ofreservoir fluids (invasion)using well logs


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Reserve Estimation Recovery Factor

Driveminimum

Sandstonesaverage maximum minimum

Carbonatesaverage maximum

Water drive 27.8 51.1 86.7 6.3 43.6 80.5Solution gasdrive withoutsupplementaldrives

9.5 21.3 46.0 15.5 17.6 20.7

Solution gasdrive withsupplementaldrives

13.1 28.4 57.9 9.0 21.8 48.1

Gas cap drive 15.8 32.5 67.0 Combined with sandstoneData not availableGravity drainage 16.0 57.2 63.8

Gas depletion 75.0 85.0 95.0Gas water drive 50.0 70.0 80.0

Solution Gas Drive Reservoirs (Arps, 1962)

Statistical Performance

Recovery factor for different drive mechanisms

Soln Gor Oilgravity maximum

Sandstonesaverage minimum maximum

Carbonatesaverage minimum

60 153050

12.821.334.2

8.615.224.8

2.68.7

16.9

28.032.839.0

4.49.9

18.6

0.62.98.0

200 153050

13.322.237.4

8.815.226.4

3.38.4

17.6

27.532.339.8

4.59.8

19.3

0.92.67.4

600 15

3050

18.0

24.335.6

11.3

15.123.0

6.0

8.413.8

26.6

30.036.1

6.9

9.615.1

1.9

2.54.31000 15

3050

-34.433.7

-21.220.2

-12.611.6

-32.631.8

-13.212.0

-4.03.1

2000 153050

--

40.7

--

24.8

--

15.6

--

32.8

--

14.5

--

5.0


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f ave = 30%Sw ave = 30%

Example

OOIP = 869 Mstb

Assume Sxo = 70 % What is the R.F.?

Recovery FactorAssume sandstone reservoir, water drive.What is the R.F.?

RF = 51.1%

R = 444 MstbReserves (R) = ?

Reserves (R) = ?

RF = 57 %

R = 496 Mstb


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Chapter 11, Sec 11.4-11.7 , Bassiouni, Z: Theory,Measurement, and Interpretation of Well Logs, SPETextbook Series, Vol. 4, (1994)

Corelab, Fundamentals of Core Analysis, Houston, TX(1983), Chapter 7

References

reserve estimation notes

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