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PETROLEUM ENGINEERING 311 RESERVOIR PETROPHYSICS CLASS NOTES (1992)

Instructor/Author: Ching H. Wu

DEPARTMENT OF PETROLEUM ENGINEERING TEXAS A&M UNIVERSITY COLLEGE STATION, TEXAS

TABLE OF CONTENTS I.

ROCK POROSITY I) II) III) VI) V) VI) VII)

II.

I-1

Definition Classification Range of values of porosity Factors affecting porosity Measurement of porosity Subsurface measurement of porosity Compressibility of porous rocks

I-1 I-1 I-2 I-3 I-5 I-13 I-25

SINGLE PHASE FLOW IN POROUS ROCK

II-1

I) Darcy's equation II) Reservoir systems

II-11 II-15

III. BOUNDARY TENSION AND CAPILLARY PRESSURE I) II) III) IV) V) VI) VII) VIII) IX) X) XI)

Boundary tension Wettability Capillary pressure Relationship between capillary pressure and saturation Relationship between capillary pressure and saturation history Capillary pressure in reservoir rock Laboratory measurement of capillary pressure Converting laboratory data to reservoir conditions Determining water saturation in reservoir from capillary pressure data Capillary pressure variation Averaging capillary pressure data

IV. FLUID SATURATIONS

III-1 III-3 III-5 III-13 III-14 III-17 III-19 III-25 III-27 III-29 III-31 IV-1

I) Basic concepts of hydrocarbon accumulation II) Methods for determining fluid saturations V.

III-11

IV-1 IV-1

ELECTRICAL PROPERTIES OF ROCK-FLUID SYSTEMS

V-1

I) Electrical conductivity of fluid saturated rock II) Use of electrical Formation Resistivity Factor, Cementation Factor, and Saturation Exponent III) Laboratory measurement of electrical properties of rock IV) Effect of clay on resistivity

V-1

VI. MULTIPHASE FLOW IN POROUS ROCK I) II) III) IV) V) VI)

Effective permeability Relative permeability Typical relative permeability curves Permeability ratio (relative permeability ratio) Measurement of relative permeability Uses of relative permeability data

ii

V-8 V-9 V-18 VI-1 VI-1 VI-2 VI-2 VI-14 VI-14 VI-33

VII. STATISTICAL MEASURES I) II) III) IV) V) VI) VII) VIII) IX) X)

VII-1

Introduction Frequency Distributions Histogram Cumulative Frequency Distributions Normal Distribution Log Normal Distribution Measures of Central Tendency Measures of Variability (dispersion) Normal Distribution Log Normal Distribution

VII-1 VII-2 VII-3 VII-6 VII-8 VII-9 VII-10 VII-11 VII-12 VII-16

iii

I. ROCK POROSITY I)

Definition A measure of the pore space available for the storage of fluids in rock

In general form: Porosity = φ =

Vp Vb - Vm = Vb Vb

where: φ is expressed in fraction Vb = Vp + Vm Vb = bulk volume of reservoir rock, (L3) Vp = pore volume, (L3) Vm= matrix volume, (L3)

II)

Classification A.

Primary (original) Porosity Developed at time of deposition

B.

Secondary Porosity Developed as a result of geologic process occurring after deposition

C.

Total Porosity φt =

D.

total pore space Vb - Vm = Vb Vb

Effective Porosity φe = 1. 2.

interconnected pore space Vb Clean sandstones: φe = φt Carbonate, cemented sandstones: φe < φt

I-1

VI)

Factors affecting porosity A.

Factors: 1. Particle shape 2. Particle arrangement 3. Particle size distribution 4. Cementation 5. Vugs and fractures

B.

Particle shape Porosity increases as particle uniformity decreases.

C.

Packing Arrangement Porosity decreases as compaction increases

EFFECT OF NATURAL COMPACTION ON POROSITY (FROM KRUMBEIN AND SLOSS.) 50

40 SANDSTONES

POROSITY, %

30

20

SHALES 10

0 0

1000

2000

3000

4000

DEPTH OF BURIAL, ft

I-3

5000

6000

D.

Particle Size Distribution Porosity decreases as the range of particle size increases

SAND

SILT

CLAY

100

CLEAN SAND FRAMEWORK FRACTION

WEIGHT %

SHALY SAND

INTERSTITIAL MATERIALS AND MUD FRAGMENTS 0 1.0

0.1

0.01

0.001

GRAIN SIZE DIAMETER, MM

E.

F.

Interstitial and Cementing Material 1.

Porosity decreases as the amount of interstitial material increases

2.

Porosity decreases as the amount of cementing material increases

3.

Clean sand - little interstitial material Shaly sand - has more interstitial material

Vugs, Fractures 1.

Contribute substantially to the volume of pore spaces

2.

Highly variable in size and distribution

3.

There could be two or more systems of pore openings - extremely complex

I-4

V)

Measurement of porosity φ=

Vb - Vm Vp = Vb Vb

Table of matrix densities Lithology ρ m (g/cm3) ___________ ___________

A.

Quartz

2.65

Limestone

2.71

Dolomite

2.87

Laboratory measurement 1.

Conventional core analysis a.

measure any two 1) 2) 3)

b.

bulk volume, Vb matrix volume, Vm pore volume, Vp

bulk volume 1) 2)

calculate from dimensions displacement method a)

volumetric (measure volume) (1)

drop into liquid and observe volume charge of liquid

(2)

must prevent test liquid from entering pores space of sample (a) (b) (c)

b)

coat with paraffin presaturate sample with test liquid use mercury as test liquid

gravimetric (measure mass) (1)

Change in weight of immersed sampleprevent test liquid from entering pore space

(2)

Change in weight of container and test fluid when sample is introduced

I-5

c.

matrix volume 1)

assume grain density dry weight Vm = matrix density

2)

displacement method Reduce sample to particle size, then

3)

a)

volumetric

b)

gravimetric

Boyle's Law: P1V1 = P 2V2 a) P(1)

V(1)

VALVE CLOSED b)

Put core in second chamber, evacuate

c)

Open valve

P(2)

CORE

VALVE OPEN

4)

Vm

V2

=

Volumetric of first chamber & volume of second chamber-matrix volume or core ( calculated)

VT

=

Volume of first chamber + volume second chamber (known)

=V T - V2

I-6

d.

pore volume 1)

gravimetric Vp =

2)

saturated weight - dry weight density of saturated fluid

Boyle's Law: P1V1 = P 2V2 a) P(1)

V(1)

CORE

VALVE CLOSED

b)

Put core in Hassler sleeve, evacuate

c)

Open valve

P(2)

V(1)

CORE

VALVE OPEN

3)

V2

=

Vp

= V2 - V1

I-7

Volume of first chamber + pore volume of core (calculated)

2.

Application to reservoir rocks a.

b.

intergranular porosity (sandstone, some carbonates) 1)

use representative plugs from whole core in laboratory measurements

2)

don't use sidewall cores

secondary porosity (most carbonates) 1)

use whole core in laboratory measurements

2)

calculate bulk volume from measurements

3)

determine matrix or pore volume from Boyle's Law procedure

I-8

Example I-1 A core sample coated with paraffin was immersed in a Russell tube. The dry sample weighed 20.0 gm. The dry sample coated with paraffin weighed 20.9 gm. The paraffin coated sample displaced 10.9 cc of liquid. Assume the density of solid paraffin is 0.9 gm/cc. What is the bulk volume of the sample?

Solution: Weight of paraffin coating = 20.9 gm - 20.0 gm = 0.9 gm Volume of paraffin coating = 0.9 gm / (0.9 gm/cc) = 1.0 cc Bulk volume of sample = 10.9 cc - 1.0 cc = 9.9 cc

Example I-2 The core sample of problem I-1 was stripped of the paraffin coat, crushed to grain size, and immersed in a Russell tube. The volume of the grains was 7.7 cc. What was the porosity of the sample? Is this effective or total porosity.

Solution: Bulk Volume

=

9.9 cc

Matrix Volume

=

7.7 cc

V - Vm 9.9 cc- 7.7 cc φ= b = = 0.22 Vb 9.9 cc

It is total porosity.

I-9

Example I-3 Calculate the porosity of a core sample when the following information is available: Dry weight of sample = 427.3 gm Weight of sample when saturated with water = 448.6 gm Density of water = 1.0 gm/cm3 Weight of water saturated sample immersed in water = 269.6 gm

Solution: Vp

=

sat. core wt. in air - dry core wt. density of water

Vp

=

448.6 gm - 427.3 gm 1 gm/cm3

Vp

=

21.3 cm3

Vb

=

sat. core wt. in air - sat. core wt. in water density of water

Vb

=

448.6 gm - 269.6 gm 1 gm/cm3

Vb

=

179.0 cm3

φ

=

Vp = 21.3 cm3 = .119 Vb 179.0 cm3

φ

=

11.9%

I - 10

What is the lithology of the sample? Vm

=

Vb - Vp

Vm

=

179.0 cm3 - 21.3 cm3 = 157.7 cm 3

ρm

=

wt. of dry sample matrix vol.

= 2.71 gm/(cm3) 157.7 cm3

= 427.3 gm

The lithology is limestone. Is the porosity effective or total? Why? Effective, because fluid was forced into the pore space.

I - 11

Example I-4 A carbonate whole core (3 inches by 6 inches, 695 cc) is placed in cell two of a Boyles Law device. Each of the cells has a volume of 1,000 cc. Cell one is pressured to 50.0 psig. Cell two is evacuated. The cells are connected and the resulting pressure is 28.1 psig. Calculate the porosity of the core.

Solution: P V 1 1

=

P V 2 2

V 1

=

1,000 cc

P

=

50 psig + 14.7 psia = 64.7 psia

=

28.1 psig + 14.7 = 42.8 psia

V 2

=

(64.7 psia) (1,000 cc) / (42.8 psia)

V 2

=

1,512 cc

V m

=

VT - V2

V m

=

2,000 cc - 1,512 cc - 488 cc

φ

=

VT - Vm 695 cc - 488 cc = = .298 = 29.8% VT 695 cc

P

1 2

I - 12

VI)

Subsurface measurement of porosity A.

Types of logs from which porosity can be derived 1.

Density log: ρ -ρ φd = m L ρm - ρf

2.

Sonic log: φs =

3.

∆tL - ∆tm ∆tf - ∆tm

Neutron log: e-kφ = CNf

Table of Matrix Properties (Schlumberger, Log Interpretation Principles, Volume I) Lithology

∆tm µsec/ft

ρ m gm/cc

Sandstone

55.6

2.65

Limestone

47.5

2.71

Dolomite

43.5

2.87

Anhydrite

50.0

2.96

Salt

67.0

2.17

189.0

1.00

Water

I - 13

B.

Density Log 1.

Measures bulk density of formation

M UD CAKE

FORM ATION

GAM M A RAY SOURCE

SHORT SPACE DETECTOR

LONG SPACE DETECTOR

2.

Gamma rays are stopped by electrons - the denser the rock the fewer gamma rays reach the detector

3.

Equation ρL =

ρm 1 - φ + ρf φ

ρ -ρ φd = m L ρm - ρf

I - 14

FORMATION DENSITY LOG

GR, API

ρ, gm/cc

depth, ft 4100

4120

4140

4160

4180

4200

4220

4240 0

40

80

120

160

200

2.0

I - 15

2.2

2.4

2.6

2.8

3.0

Example I-5 Use the density log to calculate the porosity for the following intervals assuming ρ matrix = 2.68 gm/cc and ρ fluid = 1.0 gm/cc. Interval, ft

ρ

__________

L, gm/cc _________

4143-4157 4170-4178 4178-4185 4185-4190 4197-4205 4210-4217

2.375 2.350 2.430 2.400 2.680 2.450

φd ,% ______ 18 20 15 17 0 14

Example: Interval 4,143 ft -4,157 ft : ρ = 2.375 gm/cc L ρ -ρ 2.68 gm/cc - 2.375 gm/cc φd = m L = = 0.18 ρm - ρf 2.68 gm/cc - 1.0 gm/cc

I - 16

C.

Sonic Log 1.

Measures time required for compressional sound waves to travel through one foot of formation

T

A

B C

R1 D

E R2

2.

Sound travels more slowly in fluids than in solids. Pore space is filled with fluids. Travel time increases as porosity increases.

3.

Equation ∆tL = ∆tm 1 - φ + ∆tf φ

I - 17

(Wylie Time Average Equation)

SONIC LOG

GR, API

∆T, µ seconds/ft

depth, ft 4100

4120

4140

4160

4180

4200

4220

4240 0

100

200

140

I - 18

120

100

80

60

40

Example I-6 Use the Sonic log and assume sandstone lithology to calculate the porosity for the following intervals.

∆tL µ second/ft

Interval (ft)

φs ,%

4,144-4,150

86.5

25

4,150-4,157

84.0

24

4,171-4,177

84.5

24

4,177-4,187

81.0

21

4,199-4,204

53.5

1

4,208-4,213

75.0

17

Example: Interval 4144 ft - 4150 ft : ∆tL φs =

= 86.5 µ-sec/ft ∆tL - ∆tm 86.5 µ sec/ft- 51.6 µ sec/ft = = 0.25 ∆tf - ∆tm 189.0 µ sec/ft- 51.6 µ sec/ft

I - 19

D.

Neutron Log 1.

Measures the amount of hydrogen in the formation (hydrogen index)

Maximum Energy Loss/ Collision, %

Average Number Collisions

Element Calcium Chlorine Silicone Oxygen Carbon Hydrogen

371 316 261 150 115 18

Atomic Collision

8 10 12 21 28 100

40.1 35.5 28.1 16.0 12.0 1.0

103

20 17 14 8 6 1

CLEAN SAND POROSITY = 15%

O

102 Si 10

H

SLOWING DOWN POWER

RELATIVE PROBABILITY FOR COLLISION

CLEAN SAND POROSITY = 15%

Atomic Number

1

10-1

H

O

10-2

Si

10-3

1

.1

1

10

102 10 3 10 4 105

106 107

.1

NEUTRON ENERGY IN ELECTRON VOLTS

1

10

102 10 3 10 4 105

106 107

NEUTRON ENERGY IN ELECTRON VOLTS

2.

In clean, liquid filled formations, hydrogen index is directly proportional to porosity. Neutron log gives porosity directly.

3.

If the log is not calibrated, it is not very reliable for determining porosity. Run density log to evaluate porosity, lithology, and gas content.

I - 20

NEUTRON DENSITY LOG

GR, API

φ (CDL)

depth, ft 4100

4120

4140

4160

4180

4200

4220

4240 0

200

I - 21

30

-10

Example I-7 Use the neutron log to determine porosity for the following intervals.

Solution:

Interval (ft)

φ

n (%)

.

4,143-4,149

23

4,149-4,160

20

4,170-4,184

21

4,198-4,204

9

4,208-4,214

19

I - 22

Example I-8 Calculate the porosity and lithology of the Polar No. 1 drilled in Lake Maracaibo. The depth of interest is 13,743 feet. A density log and a sonic log were run in the well in addition to the standard Induction Electric Survey (IES) survey. The readings at 13,743 feet are: bulk density travel time

= 2.522 gm/cc = 62.73 µ-sec/ft

Solution: Assume fresh water in pores. Assume sandstone: ρ m = 2.65 gm/cc ∆tm = 55.5 µ-sec/ft ρ -ρ 2.65 gm/cc - 2.522 gm/cc φd = m L = = 7.76% ρm - ρf 2.65 gm/cc - 1.0 gm/cc

φs =

∆tL - ∆tm 62.73 µ sec/ft- 55.5 µ sec/ft = = 5.42% ∆tf - ∆tm 189.0 µ sec/ft - 55.5 µ sec/ft

Assume limestone: ρm

= 2.71 gm/cc

∆tm

= 47.5 µ-sec/ft

ρ -ρ 2.71 gm/cc - 2.522 gm/cc φd = m L = = 10.99% ρm - ρf 2.71 gm/cc - 1.0 gm/cc φs =

∆tL - ∆tm 62.73 µ sec/ft - 47.5 µ sec/ft = = 10.76% ∆tf - ∆tm 189.0 µ sec/ft - 47.5 µ sec/ft

I - 23

Assume dolomite: ρm

= 2.87 gm/cc

∆tm

= 43.5 µ-sec/ft

φd

ρ -ρ 2.87 gm/cc - 2.522 gm/cc = m L= = 18.619% ρm - ρf 2.87 gm/cc - 1.0 gm/cc

φs

=

φlimestone

∆tL - ∆tm 62.73 µ sec/ft - 43.5 µ sec/ft = = 13.22% ∆tf - ∆tm 189.0 µ sec/ft - 43.5 µ sec/ft = 11%

Since both logs "read" nearly the same porosity when a limestone lithology was assumed then the hypothesis that the lithology is limestone is accepted. Are the tools measuring total or effective porosity? Why? The density log measures total compressibility because is "sees" the entire rock volume,including all pores. The sonic log tends to measure the velocity of compressional waves that travel through interconnected pore structures as well as the rock matrix. The general consensus is that the sonic log measures effective porosity when we use the Wyllie "time-average" equation. It is expected that the effective porosity is always less than ,or equal to,the total porosity.

I - 24

VII)

Compressibility of porous rocks Compressibility, c is the fractional change in volume per unit change in pressure: ∆V ∂V V T c=- 1 ≅ V ∂P T ∆P A.

Normally pressured reservoirs 1.

Downward force by the overburden must be balanced by upward force of the matrix and the fluid Fo

Ff

Fm

2.

Thus, Fo

= Fm + Ff

it follows that

3.

Po

=

pm + pf

Po

≅

1.0 psi/ft

Pf

≅

0.465 psi/ft

I - 25

4.

As fluid is produced from a reservoir, the fluid pressure, Pf will usually decrease: a. b. c.

B.

the force on the matrix increases causing a decrease in bulk volume and a decrease in pore volume

Types of compressibility 1.

Matrix Compressibility, cm cm

2.

≅ 0

Bulk Compressibility cb used in subsidence studies

3.

Formation Compressibility, cf - also called pore volume compressibility a.

important to reservoir engineers 1) 2) 3)

depletion of fluid from pore spaces internal rock stress changes change in stress results in change in Vp, Vm, Vb

4)

by definition ∂Vp cf = - 1 Vp ∂pm

b.

since overburden pressure, Po, is constant dPm

= - dP f

I - 26

1)

Thus, ∂Vp cf = - 1 Vp ∂pm

2)

where the subscript of f on cf means "formation" and the subscript of f on Pf means "fluid"

3)

procedure (a)

measure volume of liquid expelled as a function of "external" pressure

(b)

"external" pressure may be taken to represent overburden pressure, Po

(c)

fluid pressure, pf, is essentially constant, thus, dPo

(d)

expelled volume increases as pore volume, vp, decreases, thus, dVp

(e)

= dP m

= - dVexpelled

from definition ∂Vp cf = - 1 Vp ∂pm it follows that ∆ Vp expelled cf = + 1 Vp ∆Po

I - 27

plot

CUMULATIVE PORE VOLUME VOLUME EXPELLED

(f)

OVERBURDEN PRESSURE, psi

slope = cf.

I - 28

C.

Measurement of compressibility 1)

2)

Laboratory core sample a)

apply variable internal and external pressures

b)

internal rock volume changes

Equipment

Internal Pressure Gauge

Hydraulic Pump Mercury Sight Gauge Overburden Pressure Gauge

Hydraulic Pump Copper - Jacketed Core

Apparatus for measuring pore volume compressibility (hydrostatic)

I - 29

Example I-9 Given the following lab data, calculate the pore volume compressibility for a sandstone sample at 4,000 and 6,000 psi. pore volume

=

50.0 cc

pressure, psi

vol. fluid expelled, cc

1000 2000 3000 4000 5000 6000 7000 8000

0.244 0.324 0.392 0.448 0.500 0.546 0.596 0.630

Solution: from graph @ 4,000 psi: 0.009 4000 psi

Slope

=

cf

= 2.25 X 10-6

1 psi

@ 6000 psi: 0.011 6000 psi

Slope

=

cf

= 1.83 X 10-6

1 psi

I - 30

VOLUME EXPELLED, cc PORE VOLUME, cc

0.015

0.010

0.005

0.000 0

2000

4000

6000

8000

COMPACTION PRESSURE, psi

I - 31

10000

PORE VOLUME COMPRESSIBILITY X 10-6 psi-1

100

PORE-VOLUME COMPRESSIBILITY AT 75 % LITHOSTATIC PRESSURE VS INITIAL SAMPLE POROSITY FOR CONSOLIDATED SANDSTONES.

CONSOLIDATED SANDSTONES

10

HALL'S CORRELATION

1 0

5

10

15

20

25

30

PORE VOLUME COMPRESSIBILITY X 10-6 psi-1

INITIAL POROSITY AT ZERO NET PRESSURE, % PORE-VOLUME COMPRESSIBILITY AT 75 % LITHOSTATIC PRESSURE VS INITIAL SAMPLE POROSITY FOR UNCONSOLIDATED SANDSTONES.

100

UNCONSOLIDATED SANDSTONES

10

HALL'S CORRELATION

1 0

5

10

15

20

25

30

INITIAL POROSITY AT ZERO NET PRESSURE, %

I - 32

E.

Abnormally pressured reservoirs "abnormal pressure": fluid pressures greater than or less than the hydrostatic fluid pressure expected from an assumed linear pressure gradient

PRESSURE

DEPTH

NORMAL LINEAR

SUBNORMAL (LOWER)

SURNORMAL (GREATER)

I - 33

Compressibility/Porosity Problem No. 1 A limestone sample weighs 241.0 gm. The limestone sample coated with paraffin was found to weigh 249.5 gm. The coated sample when immersed in a partially filled graduated cylinder displaced 125.0 cc of water. The density of the paraffin is 0.90 gm/cc. What is the porosity of the rock? Does the process measure total or effective porosity?

Solution:

Vm =

wt. dry 241.0 gm = = 88.9 cc ρ ls 2.71 gm/cc

Vparaffin =

wt. coated sample - st. uncoated sample ρ

Vparaffin =

249.5 gm - 241.0 gm = 9.4 cc 0.90 gm/cc

Vb

= 125 cc - 9.4 cc = 115.6 cc

Vp

= Vb - Vm

Vp

= 115.6 cc - 88.9 cc - 26.7 cc

φ

=

φ

= 23.1%

Vp 26.7 cc = = 0.231 Vb 115.6 cc

(total porosity)

I - 34

Compressibility/Porosity Problem No. 2 You are furnished with the results of a sieve analysis of a core from Pete well #1. Previous laboratory work indicates there is a correlation between grain size and porosity displayed by those particular particles. The correlation is seen below: gravel

-

25% porosity

coarse sand

-

38% porosity

fine sand

-

41% porosity

What would be the minimum porosity of the mixture? What basic assumption must be made in order to work the problem? Solution: Begin calculation with a volume of 1 cu. ft. remaining pore volume (ft3)

component

porosity (%)

remaining matrix volume (ft3) ___

void space

1.000

100.0

0.000

gravel

0.250

25.0

0.750

coarse sand

0.095

9.5

0.905

fine sand

0.039

3.9

0.961

Final porosity - 3.9% (Complete mixing of the grains)

I - 35

Compressibility/Porosity Problem No. 3 A sandstone reservoir has an average thickness of 85 feet and a total volume of 7,650 acre-feet. Density log readings through the fresh water portion of the reservoir indicate a density of 2.40 gm/cc. The Highgrade #1 Well was drilled and cored through the reservoir. A rock sample was sent to the laboratory and the following tests were run. pressure (psig) 1,000 2,000 3,000 4,000 5,000 6,000 7,000 8,000

cum. pore vol. change (-cc)_________ 0.122 0.162 0.196 0.224 0.250 0.273 0.298 0.315

The dry weight of the core sample was found to be 140 gm while the sample dimensions were 1.575 inches long and 1.960 inches in diameter. Assuming the compressibility at 4,500 psi is the average compressibility in the reservoir, how much subsidence occurs when the reservoir pressure declines from 5,500 psi to 3,500 psi? Calculate: A.

Reservoir Porosity

B.

Sample Pore Volume

C.

Compressibility at 4,500 psi

D.

Amount of Ground Subsidence.

Solution: A.

Reservoir Porosity ρ -ρ φ = m L = 2.65 - 2.40 = 15.22% ρ m - ρ f 2.65 - 1.00

I - 36

B.

C.

Sample Pore Volume L

= (1.575 in) (2.54 cm/in) = 4.0 cm

D

= (1.960 in) (2.54 cm/in) = 5.0 cm

Vb

2 3.14 5.0 2 4.0 = bulk volume = πD h = = 78.5 cc 4 4.0

Vm

= matrix volume = 140 gm

Vp

= Vb - Vm = 78.5 cc - 52.8 cc

Vp

= 25.7 cc

Compressibility (see graph) Vp

D.

cc = 52.8 cc 2.65 gm

= 25.7 cc

Subsidence ∆H

= H cp φ ∆P

∆H

= 85 ft 9.69x 10 -7 psi-1 0.152 2,000 psi

∆H

= 0.026 ft

∆H

= 0.32 inches

Note: the pore volume (formation) compressibility is somewhat smaller than usually encountered. An experienced engineer would be wary of this small number. Also it was assumed that the formation compressibility was exactly the same as the bulk volume compressibility. Experience shows that this is not the case.

I - 37

POROSITY PROBLEM No. 3 0.0140

SLOPE =

VOLUME EXPELLED, cc PORE VOLUME, cc

0.0120

.0118 - .0068 7000 - 2000

C p = 9.96 x 10-7 psi -1

0.0100

0.0080

0.0060

0.0040 0

2000

4000

6000

PRESSURE, psig

I - 38

8000

Compressibility Problem A 160-acre and 100 ft thick reservoir has a porosity of 11%. The pore compressibility is 5.0 x 106 (1/psi). If the pressure decreases 3,000 psi, what is the subsidence (ft)? Assume Cf = Cb Solution: A

= 160 (43,560)

= 6,969,600 ft2

Vb

= 100 (6,969,600)

= 696,960,000 ft3

Vp

= Vb(f) = (696,960,000) (.11) = 76,665,600 ft3 dVp Cp = - 1 Vp dp 5 x 10-6 (1/psi) =

dVp -1 76,665,600 ft3 3,000 psi

dVp = 1.15 x 106 ft3 ∆H = 1.15 x 106 ft3 x

1 = 0.165 ft 6,969,600 ft2

I - 39

II. SINGLE PHASE FLOW IN POROUS ROCK I)

Darcy's equation (1856) A.

Water flow through sand filters

A

q

h 1 - h2

h1

Z

h2

q

WATER

SAND

DARCY'S FOUNTAIN.

kA(h1 - h2) µL Length of sand pack,L = Z q=

1.

constant of proportionality, k, characteristics of particular sand pack, not sample size

2.

Darcy's work confined to sand packs that were 100% saturated with water

3.

equation extended to include other liquids using viscosity

II - 1

q=

B.

kA(h1 - h2) µL

Generalized form of Darcy's equation 1.

Equation ρg dz vs = -k dP µ ds 1.0133 x 10 6 ds

+Z

90o

+1

Vs

180o

270o

s Θ

+X

-Z -1

θ

+Y

2.

Nomenclature vs

= superficial velocity (volume flux along path s) - cm/sec

vs/φ

= interstitial velocity - cm/sec

ρ

= density of flowing fluid - gm/cm3

g

= acceleration of gravity - 980 cm/sec2

dP ds

= pressure gradient along s - atm/cm

µ

= viscosity - centipoise

k

= permeability - darcies

II - 2

360o

3.

4.

Conversion factors dyne

= gm cm/sec2 = a unit of force

atm

= 1.0133 x 106 dyne/cm2

ρgh

= dyne/cm2 = a unit of pressure

poise

= gm/cm sec = dyne sec/cm2

The dimensions of permeability L

= length

m

= mass

t

= time

vs

= L/t

µ

= m/Lt

ρ

= m/L3

p

= m/Lt2

g

= L/t2

vs

ρg dp dz = - k µ ds 1.0133 x 10 6 ds

L t

m/Lt2 - m/L3 L/t2 L = - k m/Lt L L

k

= L2 = cross-sectional area

II - 3

5.

Definition of Darcy units a.

conventional units would be: 1)

feet squared in the English system

2)

centimeter squared in the cgs system

b.

both are too large for use in porous media

c.

definition of darcy

A porous medium has a permeability of one darcy when a single-phase fluid of one centipoise that completely fills the voids of the medium will flow through it under conditions of viscous flow at a rate of one cubic centimeter per second per square centimeter cross-sectional area under a pressure or equivalent hydraulic gradient of one atmosphere per centimeter.

q

=

k A P1 - P2 µL

II - 4

II)

Reservoir systems A.

Flow of incompressible liquid 1.

q P1

Horizontal, linear flow system

A

q P2

L

a.

Conditions 1)

dz horizontal system, ds = 0

2)

linear system, A = constant

3)

incompressible liquid, q = constant

4)

laminar flow, can use Darcy's equation

5)

non-reactive fluid, k = constant

6)

100% saturated with one fluid

7)

constant temperature, µ, q

II - 5

b.

derivation of flow equation ρg dP dz ds 1.0133 x 10 6 ds

vs

= - k µ

vs

q = - k dP = µ ds A L

q

ds 0

qL-0 q

p2 kA = dP µ p 1 = - kA P2 - P1 µ

= kA P2 - P1 Lµ

Note: P1 acts at L = 0 P2 acts at L = L q is + if flow is from L = 0 to L = L

II - 6

Example II-1 What is the flow rate of a horizontal rectangular system when the conditions are as follows: permeability = k = 1 darcy area = A = 6 ft2 viscosity = µ = 1.0 cp length = L = 6 ft inlet pressure = P1 = 5.0 atm outlet pressure = P2 = 2.0 atm

Solution: We must insure all the variables are in the correct units. k

= 1 darcy

A

= 6 ft2 (144 in2/1 ft2) (6.45 cm2/1 in2) = 5572.8 cm 2

L

= 6 ft (12 in/1 ft) (2.54 cm/1 in) = 182.88 cm

P1

= 5.0 atm

P2

= 2.0 atm q q

q

= kA P2 - P1 Lµ = (1) (5,572.8) (5.0 - 2.0) (1) (182.88) = 91.42 cm 3/sec

II - 7

2.

Non-horizontal, linear system

P2

-Z

S Θ X

P1

a.

Conditions 1)

dz = sinθ non-horizontal system, ds = constant

2)

linear system, A = constant

3)

incompressible liquid, q = constant

4)

laminar flow, use Darcy equation

5)

non-reactive fluid, k = constant

6)

100% saturated with one fluid

7)

constant temperature µ, q

II - 8

b.

derivation of equation vs

ρg dz = - k dP µ ds 1.0133 x 10 6 ds

vs

= -

q = - k A µ

ds

= - kA µ

L q 0

q

k ρg sin θ dP + ds µ 1.0133 x 10 6 P2 P1

kA ρg sin θ dp + µ 1.0133 x 10 6

ρgLsinθ = - kA P1 - P2 + µL 1.0133 x 10 6

II - 9

L ds 0

3.

Vertical, upward flow, linear system FLOW UNDER HEAD h

h

x

L

a.

Conditions 1)

dz vertical system, ds = sinθ = constant

2)

upward flow, q = 270°, sinθ = - 1

3)

linear system, A = constant

4)

incompressible liquid, q = constant

5)

laminar flow, use Darcy equation

6)

non-reactive fluid, k = constant

7)

100% saturated with one fluid

8)

constant temperature, µ

II - 10

b.

derivation of flow equation ρg dP dz ds 1.0133 x 10 6 ds

vs

=

k µ

vs

=

q = - k A µ

q

=

kA µ

P1

= -

P2

=

ρg dP + ds 1.0133 x 10 6 ρg P1 - P2 L 1.0133 x 10 6

ρg (h + x + L) 1.0133 x 10 6 ρg x 1.0133 x 10 6 P1

L

P2

=

ρg h ρg + 1.0133 x 10 6 L 1.0133 x 10 6

ρg h ρg ρg q = kA + µ 1.0133 x 10 6 L 1.0133 x 10 6 1.0133 x 10 6

q

ρg h = kA µL 1.0133 x 10 6

II - 11

4.

Horizontal, radial flow system

re Pw

rw

re

Pe

rw

h

a.

Conditions 1)

dz horizontal system, ds = 0

2)

radial system, A = 2πrh , ds = - dr, flow is inward

3)

constant thickness, h = constant

4)

incompressible liquid, q = constant

5)

laminar flow, use Darcy equation

6)

non-reactive fluid, k = constant

7)

100% saturated with liquid,

8)

constant temperature, µ, q

II - 12

b.

Derivation of flow Equation ρg dP dz ds 1.0133 x 10 6 ds

vs

= - k µ

vs

q q = + k dP = = µ dr A 2πrh re

q 2πh

rw

q 2πh q

dr = k r µ

pe dp pw

1n(re) - 1n( rw) = k Pe - Pw µ =

2πhk P - Pw µ 1n (re/rw) e

Note: if q is + , flow is from r e to rw B.

Flow of gas (compressible fluid) 1.

q

horizontal, linear flow system

A

q

P1

P2

L a.

Conditions 1)

dz horizontal system, ds = 0

2)

linear system, A = constant

3)

compressible gas flow, q = f(p)

4)

laminar flow, use Darcy equation

5)

non-reactive fluid, k= constant

6)

100% saturated with one fluid

7)

constant temperature II - 13

b.

Assumptions

c.

1)

µ, Z

2)

Z(and µ ) can be determined at mean pressure

= constant

Derivation of equation for qsc vs

ρg dz = - k dP µ ds 1.0133 x 10 6 ds

vs

q = - k dP = µ ds Ads

but q

Psc qscz T PTsc

=

thus L

p2

Psc T qsc Tsc A

o

Psc T qsc Tsc A

P22 - P12 L -0 = - k µz 2

ds = - k p1

2 P21 - P2 qsc = kA Tsc µL Tz Psc 2 Note: real gas equation of state Pq

= ZnRT

where q n

= volumetric flow/time = mass flow/time

thus, Pq Pscqsc q

=

= ZnRT n R Tsc Psc qscz T 1 Tsc P

where qsc is constant Z is determined at P, T

II - 14

PdP µz

Derivation of equation for q

d.

qsc

=

2 P1 - P22 kA Tsc µL Tz Psc 2

but Tsc P12 - P22 = k A µL T z Psc 2

qsc =

P q Tsc Z Psc T

q =

k A 1 P12 - P22 µL P 2

q

=

P12 - P22 k A 2 µL P1 + P 2 2

q

=

k A P - P 1 2 µL

This equation is identical to the equation for horizontal, linear flow of incompressible liquid thus if gas flow rate is determined at mean pressure, P, the equation for incompressible liquid can be used for compressible gas! Note: real gas equation of state Pq

= ZnRT

thus Psc qsc n R Tsc = znRT Pq where P

=

P1 + P 2 2

P

=

volumetric flow rate at P, T

z is determined at P, T qsc

=

P q Tsc z Psc T

II - 15

2.

Horizontal, radial flow system

re Pw

rw

re

Pe

rw

h

a.

Conditions 1)

dz horizontal system ds = 0

2)

radial system, A = 2πrL, ds = - dr, inward flow

3)

constant thickness, h = constant

4)

compressible gas flow, q = f (P)

5)

laminar flow, use Darcy equation

6)

non-reactive fluid, k = constant

7)

100% saturated with one fluid

8)

constant temperature

II - 16

b.

Assumptions µz = constant z (and µ ) can be determined at mean pressure

c.

derivation of equation for qsc ρg dz vs = - k dP µ ds 1.0133 x 10 6 ds q vs = - k dP = µ ds A

but q

P q zT = sc sc PTsc

and A = 2πrh and ds = - dr thus Psc T qsc 2Tsc π h

re rw

dr = k r

Pe Pw

ρdP µz

PscT qsc r P2 - P2w 1n r e = k e µz 2 2 Tsc π h w qsc

=

Tsc P2e - P2w 2πhk µ 1n re/rw Psc zT 2

II - 17

d.

derivation of equation for q qsc

Tsc P2e - P2w 2πhk µ 1n re/rw Psc zT 2

=

but q

=

P q Tsc z Psc T

thus P q Tsc z Psc T

=

Tsc 2πhk µ 1n re/rw Psc zT

P2e - P2w 2

q

=

2 π h k 1 (P2e - P2w) µ 1n re/rw P 2

q

=

(P2e - P2w) 2πhk 2 µ 1n re/rw Pe + P w 2

q

=

2πhk P - Pw µ 1n re/rw e

Note: Equation for real gas is identical to equation for incompressible liquid when volumetric flow rate of gas, q, is measured at mean pressure.

II - 18

C.

Conversion to Oilfield Units

Symbol

Darcy units

q k A h P L µ r

Oil field

cc/sec darcy sq cm cm atm cm cp gm/cc

bbl/d or cu ft/d md sq ft ft psia ft cp lb/cu ft

Example: q=

hkA P1 - P2 µL in Darcy's units

cc = q bbl 5.615 cu ft q sec d bbl

1,728 cu in cu ft

16.39 cc cu in

cc = 1.841 q bbl q sec d darcy k darcy = k md 1,000md k darcy = 0.001 k md A sq cm =

929.0 sq cm A sq ft sq ft

A sq cm = 929.0 A sq ft P1 - P2

atm =

P1 - P2

P1 - P2

atm = 0.06805

psia

atm 14.696 psia

P1 - P2

psia

L cm = L ft 30.48 cm ft meter = 100 cm 1.841 q = q =

0.001 k

929.0 A .06805 µ 30.48 L

0.01127 k A P1 - P2 µL

P1 - P2

in oilfield units

II - 19

d 24hr

hr 3,600 sec

D.

Table of Equations 1.

System

Darcy Units

Fluid

Equation

Horizontal, Linear

Incompressible Liquid

q =

kA µL

Dipping, Linear

Incompressible Liquid

q =

kA µL

Horizontal, Radial

Incompressible Liquid

q =

Horizontal, Linear

Real Gas

Horizontal, Radial

Real Gas

P1 - P2

P1 - P2

2 π kh µ ln (re/rw)

q = kA µL

P1 - P2

qsc =

π kh ln (re/rw)

q =

2 π kh µ ln (re/rw)

II - 20

ρ g L sin θ 1.0133 x 10 6

Pe - Pw

Tsc qsc = kA µ L Tz Psc

µ

+

P21 - P22 2

Tsc Tz Psc

Pe - Pw

Pe 2 - Pw 2

2.

System Horizontal, Linear

Oilfield Units

Fluid

Equation

Incompressible Liquid

q = 0.001127 kA µL

P1 - P2

q = res bbl/d Dipping, Linear

Incompressible Liquid

q = 0.001127 kA µL

+

Horizontal, Radial

Incompressible Liquid

Horizontal, Linear

Real Gas

P1 - P2

ρg L sinθ 1.0133 x 10 6

q = .007082

kh µ ln (re/rw) kA µ LzT

qsc = .1118

Pe - Pw

P12 - P22

qsc = scf/d q = .001127 kA µL

P1 - P2

q = res bbl/d Horizontal, Radial

Real Gas

qsc = .7032

kh µ ln (re/rw) Tz

q = .007082

II - 21

kh µ ln (re/rw)

Pe 2 - Pw2

Pe - Pw

Example II-2 What is the flow rate of a horizontal rectangular system when the conditions are as follows: permeability = k = 1 darcy area = A = 6 ft2 viscosity = µ = 1.0 cp length = L = 6 ft inlet pressure = P1 = 5.0 atm. outlet pressure = P2 = 2.0 atm.

Solutions: We must insure that all the variables are in the correct units. k A L P1 P2

= = = = =

1 darcy = 1,000 md 2 6 ft 6 ft (5.0 atm) (14.7 psi/atm) = 73.5 psi (2.0 atm) (14.7 psi/atm) = 29.4 psi

q = 1.1271 x 10-3 kA P1 - P2 µL q = 1.1271 x 10-3

1,000 6 1 6

73.5 - 29.4

q = 49.7 bbl / day

II - 22

Example II-3 Determine the oil flow rate in a radial system with the following set of conditions: K

=

300 md

re

= 330 ft

h

=

20 ft

rw

= 0.5 ft

Pe =2,500 psia

re/rw

= 660

Pw =1,740 psia

ln (re/rw)

= 6.492

µ

=

1.3 cp

Solution:

q=

q=

7.082 x 10 -3 kH Pe - Pw µ ln Re / Rw 7.082 x 1--3

300 20 2,500 - 1,740 1.3 6.492

q = 3,826 res bbl/d

II - 23

E. Layered Systems 1.

Horizontal, linear flow parallel to bedding

P1

P2

q

q

A B C

W L qt = qA + qB + qC h = hA + hB + hC let k be "average" permeability, then

qt =

k wh P1 - P2 µL

and qt =

kA whA µL

P1 - P2

+

kB whB µL

P1 - P2

then k h = kA hA + kB hB + kC hC k

=

n kj hj ∑ h j=1

II - 24

+

kC whC µL

P1 - P2

2.

Horizontal, radial flow parallel to bedding

re re

hA

ht

rw

qA

hB

qB

hC

qC

Pe

Pw again qt = qA + qB + qC h = hA + hB + hC qt =

2πk h µ ln (re/rw)

qt =

2 π kA hA µ ln (re/rw)

Pe - Pw

and Pe - Pw

+

+

2 π kB hB µ ln (re/rw)

2 π kc hc µ ln (re/rw)

then k h = kA hA + kB hB + kC hC and again k =

n kj hj ∑ h j=1

II - 25

Pe - Pw

Pe - Pw

3.

Horizontal, linear flow perpendicular to bedding A

B

C

P1

P2

q

q

kA

kB

∆P A

∆P B

kC ∆P C

LA

LB

LC

h

W

L

qt = qA = qB = qC p1 - p2 = ∆PA + ∆PB + ∆PC L = LA + LB + LC qt = and since

k wh

P1 - P2 µL

P1 - P2 = ∆ PA + ∆PB + ∆PC P1 - P2 =

since

qt µ L q µLA q µLB q µLC = A + B + C kA wh kB wh kC wh k wh

qt = qA = qB = qC L = LA + LB + LC kA kB kC k

thus K =

L n Lj ∑ j = 1 kj

II - 26

4.

Horizontal, radial, flow perpendicular to bedding q

Pw

PA

PB

PC

h rw rA rB

rC

qt = qA = qB = qC Pe - Pw = ∆PA + ∆PB + ∆PC q =

2 π k h Pe - Pw µ ln (re/rw)

Pe - Pw =

qt µ ln (re/rw) 2π k h +

qA µ ln (rA/rw) 2 π kAh

qB µ ln (rB/rA) q µ ln (rC/re) + C 2 π kB h 2 π kC h

then k =

=

ln re/rw n ln(rj/rj-1) ∑ kj j=1

II - 27

Example II-4 Damaged zone near wellbore k1 = 10 md k2 = 200 md

r1 r2 rw

= 2 ft = 300 ft = 0.25 ft

Solution: k =

k =

ln (re/rw) n ln (rj/rj-1 ) ∑ kj j=1 ln 300 0.25 ln 2/0.25 ln 300/2 + 10 200

k = 30.4 md

The permeability of the damaged zone near the wellbore influences the average permeability more than the permeability of the undamaged formation.

II - 28

F.

Flow through channels and fractures 1.

Flow through constant diameter channel

A

L

a.

Poiseuille's Equation for viscous flow through capillary tubes q=

πr4 8µL

P1 - P2

A = π r2, therefore q=

b.

Ar2 8µL

P1 - P2

Darcy's law for linear flow of liquids q = kA µL

P1 - P2

assuming these flow equations have consistent units Ar2 8µL

P1 - P2

= kA µL

P1 - P2

thus 2 k = r2 = d 32 8 where d = inches, k = 20 x 10 9 d2 md

II - 29

Example II-4 A.

Determine the permeability of a rock composed of closely packed capillaries 0.0001 inch in diameter.

B.

If only 25 percent of the rock is pore channels (f = 0.25), what will the permeability be?

A.

k

= 20 x 109 d2

k

= 20 x 109 (0.0001 in) 2

k

= 200 md

k

= 0.25 (200 md)

k

= 50 md

Solution:

B.

II - 30

2.

Flow through fractures

b

v=

q h2 (P 1-P 2) = A 12 µ L

2 q = b A (P1 -P2) 12 µ L

setting this flow equation equal to Darcy's flow equation, b2 A P1 - P2 = kA P1 - P2 12 µ L µL solve for permeability of a fracture: 2 k = b in darcy units, or 12 k = 54 x 109 b2 where b k

= =

inches md

II - 31

Example II-6 Consider a rock of very low matrix permeability, 0.01 md, which contains on the average a fracture 0.005 inches wide and one foot in lateral extent per square foot of rock. Assuming the fracture is in the direction of flow, determine the average permeability using the equation for parallel flow. Solution:

∑ kj Aj k =

k = k =

A matrix k 0.01

, similar to horizontal, linear flow parallel to fracture

matrix area + fracture k total area 12 in 2 + 12 in 144 in2

fracture area

0.005 in

+

54 x 109 x 0.005 2 12 in x 0.005 in 144 in2 k =

1.439 + 81,000 144

k = 563 md

II - 32

III)

Laboratory measurement of permeability A.

Procedure 1.

Perm plug method a.

cut small, individual samples (perm plugs) from larger core

b.

extract hydrocarbons in extractor

c.

dry core in oven

d.

flow fluid through core at several rates

TURBULENCE

qsc P sc A

SLOPE = k / m

P 12 - P 22

2L

qsc =

kA P12 - P22 2 µ L Psc

horizontal, linear, real gas flow with T = Tsc and Z = 1.0

qsc Psc = k A µ

P12 - P22 2L

k = ( slope ) m

II - 33

2.

Whole core method a.

prepare whole core in same manner as perm plugs

b.

mount core in special holders and flow fluid through core as in perm plug method

VERTICAL FLOW LOW AIR PRESSURE (FLOW)

CORE RUBBER TUBING

HIGH AIR PRESSURE PIPE

TO FLOWMETER

c.

the horizontal flow data must be adjusted due to complex flow path

d.

whole core method gives better results for limestones

II - 34

B.

Factors which affect permeability measurement 1.

Fractures - rocks which contain fractures in situ frequently separate along the planes of natural weakness when cored. Thus laboratory measurements give "matrix" permeability which is lower than in situ permeability because typically only the unfractured parts of the sample are analyzed for permeability.

2.

Gas slippage a.

gas molecules "slip" along the grain surfaces

b.

occurs when diameter of the capillary openings approaches the mean free path of the gas molecules

c.

Darcy's equation assumes laminar flow

d.

gas flow path with slippage

e.

called Klinkenberg effect

f.

mean free path is function of size of molecule thus permeability measurements are a function of type of gas used in laboratory measurement.

II - 35

H2

N2 k CALCULATED

CO2

0 1 P g.

mean free path is a function of pressure, thus Klinkenberg effect is greater for measurements at low pressures - negligible at high pressures.

h.

permeability is a function of size of capillary opening, thus Klinkenberg effect is greater for low permeability rocks.

i.

effect of gas slippage can be eliminated by making measurements at several different mean pressures and extrapolating to high pressure (1/p => 0)

kMEASURED

0

1 P

II - 36

Example II-7 Another core taken at 8815 feet from the Brazos County well was found to be very shaly. There was some question about what the true liquid permeability was, since nitrogen was used in the permeameter. Calculate the equivalent liquid permeability from the following data. Mean Pressure ( atm ) 1.192 2.517 4.571 9.484

Measured Permeability ( md ) 3.76 3.04 2.76 2.54

Solution: Plot kmeasured vs. 1/pressure Intercept is equivalent to liquid permeability From graph: kliq = 2.38 md

GAS PERMEABILITY, md

5 4 3 2

k gas = 2.38276 + 1.64632 P bar

1

Equivalent Liquid Permeability = 2.38 md

0 0.0

0.2

0.4

0.6

0.8

RECIPROCAL MEAN PRESSURE, atm - 1

II - 37

1.0

3.

Reactive fluids a.

Formation water reacts with clays 1)

lowers permeability to liquid

2)

actual permeability to formation water is lower than lab permeability to gas

RELATIONSHIP OF PERMEABILITIES MEASURED WITH AIR TO THOSE MEASURED WITH WATER

WATER PERMEABILITY, md

1000

100

10

Water concentration 20,000 - 25,000 ppm Cl ion. 1 1

10

100

1000

10000

AIR PERMEABILITY, md

b. Injection water may,if its salinity is less than that of the formation water, reduce the permeability due to clay swelling.

II - 38

Effect of Water Salinity on Permeability of Natural Cores (Grains per gallon of chloride ion as shown). Field

Zone

Ka

K1000

K500

K300

K200

K100

Kw

S S S

34 34 34

4080 24800 40100

1445 11800 23000

1380 10600 18600

1290 10000 15300

1190 9000 13800

885 7400 8200

17.2 147.0 270.0

S S S

34 34 34

4850 22800 34800

1910 13600 23600

1430 6150 7800

925 4010 5460

736 3490 5220

326 1970 3860

5.0 19.5 9.9

S S T

34 34 36

13600 7640 2630

5160 1788 2180

4640 1840 2140

4200 2010 2080

4150 2540 2150

2790 2020 2010

197.0 119.0 1960.0

T T T

36 36 36

3340 2640 3360

2820 2040 2500

2730 1920 2400

2700 1860 2340

2690 1860 2340

2490 1860 2280

2460.0 1550.0 2060.0

Ka means permeability to air; K500 means permeability to 500 grains per gallon chloride solution; Kw means permeability to fresh water 4.

Change in pore pressure

a. The removal of the core from the formation will likely result in a change in pore volume.This is likely to result in a change in permeability (+ or -). b. The production of fluids,especially around the well,will result in a decrease in pore pressure and a reduction of in-situ permeability.

II - 39

III. I)

BOUNDARY TENSION AND CAPILLARY PRESSURE Boundary tension, σ A.

at the boundary between two phases there is an imbalance of molecular forces

B.

the result is to contract the boundary to a minimum size

GAS

SURFACE

LIQUID

III- 1

C.

the average molecule in the liquid is uniformly attracted in all directions

D.

molecules at the surface attracted more strongly from below

E.

creates concave or convex surface depending on force balance

F.

creation of this surface requires work 1.

work in ergs required to create 1 cm 2 of surface (ergs/cm 2) is termed "boundary energy"

2.

also can be thought of as force in dynes acting along length of 1 cm required to prevent destruction of surface (dynes/cm) - this is called "boundary tension"

3.

Boundary Energy = Boundary Tension x Length

G.

Surface Tension - Boundary tension between gas and liquid is called "surface tension"

H.

Interfacial Tension - Boundary tension between two immiscible liquids or between a fluid and a solid is called "interfacial tension"

I.

σgw

= surface tension between gas and water

σgo

= surface tension between gas and oil

σwo

= interfacial tension between water and oil

σws

= interfacial tension between water and solid

σos

= interfacial tension between oil and solid

σgs

= interfacial tension between gas and solid

Forces creating boundary tension 1.

2.

Forces a.

Law of Universal Gravitation applied between molecules

b.

physical attraction (repulsion) between molecules

Liquid-Gas Boundary attraction between molecules is directly proportional to their masses and inversely proportional to the square of the distance between them

3.

Solid-Liquid Boundary physical attraction between molecules of liquid and solid surface

III- 2

4.

Liquid-Liquid Boundary some of each

II)

Wettability A.

forces at boundary of two liquids and a solid (or gas-liquid-solid)

σow

OIL

OIL WATER Θ

σos

σw s

SOLID

σws = σos + σow cos θ B.

Adhesion Tension, AT AT = σws - σos = σow cos θ

C.

if the solid is "water-wet" σws ≥ σos AT = + cos θ = + 0° ≤ θ ≤ 90° if θ = 0° - strongly water-wet

III- 3

D.

if the solid is "oil-wet" σos ≥ σws AT = cos θ = 90° ≤ θ ≤ 180° if θ = 180° - strongly oil-wet

θ=

300

θ=

θ = 1580

830

θ = 350 (A)

ISOOCTANE

θ = 300

ISOOCTANE + 5.7% ISOQUINOLINE

ISOQUINOLINE

θ = 480

θ = 540

NAPHTHENIC ACID

θ = 1060 (B)

Interfacial contact angles. (A) Silica surface; (B) calcite surface

III- 4

III)

Capillary pressure A.

capillary pressure between air and water

h

Θ

AIR

WATER

1.

liquid will rise in the tube until total force up equals total force down a.

total force up equals adhesion tension acting along the circumference of the water-air-solid interface = 2πr AT

b.

total force down equals the weight of the column of water converted to force = πr2 hgρ w

c.

thus when column of water comes to equilibrium 2πr A T = πr2 hgρ w

d.

units dyne gm cm cm = cm2 cm cm sec2 cm3 gm cm sec2 dyne = force unit adhesion tension dyne AT = 1 r hgρ w cm 2 dyne =

e.

III- 5

2.

liquid will rise in the tube until the vertical component of surface tension equals the total force down a.

vertical component of surface tension is the surface tension between air and water multiplied by the cosine of the contact angle acting along the water-air-solid interface = 2πr σaw cosθ

b.

total force down = πr2 hgρ w

c.

thus when the column of water comes to equilibrium 2πr σaw cosθ = πr 2 hgρ w

d.

units dyne gm cm cm = cm2 cm cm 2 sec cm3 gm cm dyne = cm cm sec2

3.

since AT = σaw cosθ, 1 and 2 above both result in h=

2 σaw cos θ rg ρw

III- 6

4.

capillary pressure (air-water system) Pa Pw

h A'

B' B

Θ

Pa A

AIR

WATER

pressure relations in capillary tubes a.

pressure at A' is equal to pressure at A Pa' = Pa

b.

pressure at B is equal to the pressure at A minus the head of water between A & B pw = pa - ρ wgh units:

c.

dyne dyne gm ⋅ cm = cm cm2 cm2 cm3 ⋅ sec2

thus between B' and B there is a pressure difference pa - pw = pa - (pa - ρ wgh) pa - pw = ρ wgh

d.

call this pressure difference between B' and B "capillary pressure" Pc = pa - pw = ρ wgh

e.

remember h=

2 σgw cos θ rg ρ w

III- 7

f.

thus Pc =

B.

2 σgw cos θ r

capillary pressure between oil and water

h

Θ

OIL

WATER

1.

liquid will rise in the tube until the vertical component of surface tension equals the total force down a.

vertical component of surface tension equals the surface tension between oil and water multiplied by the cosine of the contact angle acting along the circumference of the water-oil-solid interface = 2πr σow cosθ

b.

the downward force caused by the weight of the column of water is partially offset (bouyed) by the weight of the column of oil outside the capillary

c.

thus, total force down equals the weight of the column of water minus the weight of an equivalent column of oil converted to force 1)

weight per unit area of water = ρw h

2)

weight per unit area of oil = ρo h

III- 8

3)

net weight per unit area acting to pull surface down = ρ wh - ρ oh = h(ρ w - ρ o)

4)

total force down = πr2 gh (ρ w - ρ o) d.

thus when the column of water comes to equilibrium

2πr σow cosθ = πr 2 gh (ρ w - ρ o) 2.

thus the equilibrium for the height of the column of water h=

3.

2 σow cos θ rg (ρ w - ρo)

capillary pressure (oil-water system) Po h

Pw

B' B

Θ

Po A

OIL

WATER

a.

pressure at A' equals pressure at A Poa = Pwa

b.

pressure at B is equal to the pressure at A minus the head of water between A and B Pwb = Pwa - ρ wgh

c.

pressure at B' equal to the pressure at A' minus the head of oil between A' and B'

III- 9

Pob = Poa - ρ ogh d.

thus capillary pressure, the difference between pressure at B' and the pressure at B is Pc

= Pob - Pwb

Pc

= (Poa - ρ ogh) - (Pwa - ρ wgh)

since Poa

= (ρ w - ρ o)gh

Pc e.

remember h=

f.

2 σow cos θ rg (ρ w - ρo)

thus Pc =

4.

= Pwa

2 σow cos θ r

same expression as for the air-solid system except for the boundary tension term θ Pc = 2 σ cos r

C.

remember adhesion tension is defined as AT = σow cosθ, and Pc =

2 σow cos θ r

thus Pc = f (adhesion tension, 1/radius of tube)

III- 10

ADHESION TENSION

AIR

AIR

WATER

AIR

Hg

WATER

1/radius of tube

D.

E.

an important result to remember 1.

pwb < pob

2.

thus, the pressure on the concave side of a curved surface is greater than the pressure on the convex side

3.

or, pressure is greater in the non-wetting phase

capillary pressure-unconsolidated sand 1.

the straight capillary previously discussed is useful for explaining basic concepts - but it is a simple and ideal system

2.

packing of uniform spheres

Pc = σ 1 + 1 R1 R2 R1 and R2 are the principal radii of curvature for a liquid adhering to two spheres in contact with each other. 3.

by analogy to capillary tube 1 + 1 = 2 cos θ r R1 R2 θ where Pc = 2 σ cos r call it Rm(mean radius), i.e. 1 = 2 cos θ = (∆ρ)gh rm Rm σ

III- 11

F.

wettability-consolidated sand 1.

Pendular-ring distribution-wetting phase is not continuous, occupies the small interstices-non-wetting phase is in contact with some of the solid

2.

Funicular distribution - wetting phase is continuous, completely covering surface of solid

WATER

WATER

SAND GRAIN

SAND GRAIN

OIL OR GAS

OIL OR GAS

(A)

(B)

Idealized representation of distribution of wetting and nonwetting fluid phase about intergrain contacts of spheres. (a) Pendular-ring distributions; (b) funicular distribution

III- 12

IV)

Relationship between capillary pressure and saturation A.

remember that the height a liquid will rise in a tube depends on 1. 2. 3.

B.

adhesion fluid density variation of tube diameter with height

consider an experiment in which liquid is allowed to rise in a tube of varying diameter under atmospheric pressure. Pressure in the gas phase is increased forcing the interface to a new equilibrium position. ATMOSPHERIC PRESSURE

HIGHER PRESSURE

R R

DEPENDENCE OF INTERFACIAL CURVATURE ON FLUID SATURATION IN A NON-UNIFORM PORE 1.

Capillary pressure is defined as the pressure difference across the interface.

2.

This illustrates: a.

Capillary pressure is greater for small radius of curvature than for large radius of curvature

b.

An inverse relationship between capillary pressure and wettingphase saturation

c.

Lower wetting-phase saturation results in smaller radius of curvature which means that the wetting phase will occupy smaller pores in reservoir rock

III- 13

V)

Relationship between capillary pressure and saturation history A.

consider an experiment using a non-uniform tube (pore in reservoir rock) 1.

tube is filled with a wetting fluid and allowed to drain until the interface between wetting fluid and non-wetting fluid reaches equilibrium (drainage)

2.

tube is filled with non-wetting fluid and immersed in wetting fluid allowing wetting fluid to imbibe until the interface reaches equilibrium (imbibition)

LOW PC

HIGHER P C

HIGHER P C

LOW PC

R

Θ Θ R

Θ

Θ SATURATION = 100% PC = LOW VALUE

SATURATION = 80% CAPILLARY PRESSURE = P C

(A)

SATURATION = 0% P C = HIGH VALUE

SATURATION = 10% CAPILLARY PRESSURE = P C

(B)

Dependence of equilibrium fluid saturation upon the saturation history in a nonuniform pore. (a) Fluid drains; (b) fluid imbibes. Same pore, same contact angle, same capillary pressure, different saturation history

3.

This is an oversimplified example, however it illustrates that the relationship between wetting-phase saturation and capillary pressure is dependent on the saturation process (saturation history) a.

for given capillary pressure a higher value of wetting-phase saturation will be obtained from drainage than from imbibition

III- 14

B.

Leverett conducted a similar experiment with tubes filled with sand. DATA FROM HEIGHT-SATURATION EXPERIMENTS ON CLEAN SANDS. (FROM LEVERETT) 1.6 Φ

1.4

Drainage

Imbibition Sand I

1.2

Φ

Φ

Sand II

1.0 Φ

σ

∆ρgh

(k/ø)1/2

Φ

0.8 0.6

Drainage

Φ

0.4 0.2 Imbibition 0.0 0

20

40

60

80

100

WATER SATURATION, Sw %

1.

capillary pressure is expressed in terms of a non-dimensional correlating function ( remember Pc = (∆ρ gh )

2.

in general terms, a.

drainage means replacing a wetting fluid with a non-wetting fluid

b.

imbibition means replacing a non-wetting fluid with a wetting fluid

III- 15

DRAINAGE PC

IMBIBITION

0

WATER SATURATION, S W

III- 16

100

VI)

Capillary pressure in reservoir rock Water

Oil

ρw h P w = Po/w 144

P w2

P o2

P o = Po/w -

ρoh

144

Oil and Water P o1 = P w1 100% Water

Pc = P o - Pw = h 144

ρ w - ρo

Where:

Po Pw h Po/w ρw

= = = =

ρo

= density of oil, lb/cf

pressure in oil phase, psia pressure in water phase, psia distance above 100% water level, ft pressure at oil-water contact, psia

= density of water, lb/cf

At any point above the oil-water contact, po ≥ p w

III- 17

P O = PO/W -

PC HEIGHT ABOVE O-W-C

ρ H P w = PO/W - w 144

PRESSURE

III- 18

ρoH 144

VII) Laboratory measurement of capillary pressure A.

B.

Methods 1.

porous diaphragm

2.

mercury injection

3.

centrifuge

4.

dynamic method

Porous diaphragm 1.

Start with core saturated with wetting fluid.

2.

Use pressure to force non-wetting fluid into core-displacing wetting fluid through the porous disk.

3.

The pressure difference between the pressure in the non-wetting fluid and the pressure in the wetting fluid is equal to Pc.

4.

Repeat at successively higher pressures until no more wetting fluid will come out.

5.

Measure Sw periodically.

6.

Results

7.

Advantages a. b.

8.

very accurate can use reservoir fluids

Disadvantages a. b.

very slow - up to 40 days for one core pressure is limited by "displacement pressure" of porous disk

III- 19

C.

Mercury Injection Method 1.

Force mercury into core - mercury is non-wetting phase - air (usually under vacuum) is wetting phase

2.

Measure pressure

3.

Calculate mercury saturation

4.

Advantages a. b.

5.

Disadvantages a. b.

D.

fast-minutes reasonably accurate

ruins core difficult to relate data to oil-water systems

Centrifuge Method

CORE HOLDER BODY WINDOW

TUBE BODY 1.

Similar to porous disk method except centrifugal force (rather than pressure) is applied to the fluids in the core

2.

Pressure (force/unit area) is computed from centrifugal force (which is related to rotational speed)

3.

Saturation is computed from fluid removed (as shown in window)

4.

Advantages a. b. c.

fast reasonably accurate use reservoir fluids

III- 20

E.

Dynamic Method

GAS OUTLET

GAS INLET ∆Po

Pc

∆P g

CORE

TO ATMOSPHERE

OIL INLET

OIL BURETTE DYNAMIC CAPILLARY - PRESSURE APPARATUS (HASSLER'S PRINCIPLE)

1.

establish simultaneous steady-state flow of two fluids through core

2.

measure pressures of the two fluids in core (special wetted disks) difference is capillary pressure

3.

saturation varied by regulating quantity of each fluid entering core

4.

advantages a. b.

5.

seems to simulate reservoir conditions reservoir fluids can be used

Disadvantages a.

very tedious

III- 21

F.

Comparison of methods 1.

diaphragm method (restored state) is considered to be most accurate, thus used as standard against which all other methods are compared

2.

comparison of mercury injection data against diaphragm data a.

simple theory shows that capillary pressure by mercury injection should be five times greater than capillary pressure of air-water system by diaphragm method

b.

capillary pressure scale for curves determined by mercury injection is five times greater than scale for diaphragm air-water data

c.

these comparisons plus more complex theory indicate that the ratio between mercury injection data and diaphragm data is about 6.9 (other data indicate value between 5.8 and 7.5)

III- 22

Example VIII-1 Comparison of Mercury Injection Capillary Pressure Data with Porous Diaphragm Data

A.

Calculate capillary pressure ratio, PcAH g , for the following data: PcAW σAHg = 480 Dynes/cm θ AH = 140° g

B.

σAW = 72 Dynes/cm θ AW = 0°

Pore geometry is very complex. The curvature of the interface and pore radius are not necessarily functions of contact angles. Calculate the ratio using the relationship. PcAH g σAHg = PcAW σAW

Solution:

(A)

(B)

PcAH σ cos θAHg 480 cos(140°) g = AHg = PcAW 72 cos (0°) σAWcos θ AW

PcAH g PcAW

=

PcAH g PcAW

@

PcAH g PcAW

=

5.1

σAHg = 480 σAW 70

6.9

III- 23

Discussion: A.

Best way to determine the relationship between mercury and air-water data is to generate capillary pressure curves for each set of data and compare directly.

Mercury Injection and Porous Diaphragm Methods B.

For this given set of conditions, mercury injection method requires a higher displacement pressure, must adjust ratio between scales until match is obtained.

C.

Minimum irreducible wetting phase saturations are the same.

D.

Reduction in permeability results in a higher minimum irreducible wetting phase saturation. For both cases, mercury system still has higher required displacement pressure.

III- 24

VIII) Converting laboratory data to reservoir conditions

PcL

=

2σLcos θ L r

PcR

=

2σRcos θ R r

setting r = r r=

∴

2σLcos θ L 2σRcos θ R = PcL PcR

σcos θ R PcR = PcL σcos θ L

where Pc R

= reservoir capillary pressure, psi

Pc L

= capillary pressure measured in laboratory, psi

σL

= interfacial tension measured in laboratory, dynes/cm

σR

= reservoir interfacial tension, dynes/cm

θR

= reservoir contact angle, degrees

θL

= laboratory contact angle, degrees

III- 25

Example III-2 Converting Laboratory Data to Reservoir Conditions Express reservoir capillary pressure by using laboratory data. σAW = 72 dynes

lab data:

σAW = 0o reservoir data:

σOW = 24 dynes/cm σOW = 20o

Solution:

PcR

PcR PcR

=

=

σcos θ R PcL σcos θ L 24 cos20° 72 cos0°

=

P cL

0.333 PcL

III- 26

IX)

Determining water saturation in reservoir from capillary pressure data A.

convert laboratory capillary pressure data to reservoir conditions

B.

calculate capillary pressure in reservoir for various heights above height at which capillary pressure is zero (∆ρ)gh Pc = 144 gc in English units ∆ρ

= ρ w - ρ O, lb/cu ft

g

= 32 ft/sec2

gc

= 32 lbm ft lbf sec2

h

= ft

144

= (sq in)/(sq ft.)

Pc

= lbf/(sq in), psI

thus

III- 27

Example III-3 Determining Water Saturation From Capillary Pressure Curve Given the relationship, PcR = 0.313 P cL, use the laboratory capillary pressure curve to calculate the water saturation in the reservoir at a height of 40 ft. above the oil-water contact. ρo = 0.85 gm/cm3

ρ w = 1.0 gm/cm3

20

P CL 10

8.3

0 0

50

100

SW

Solution: =

ρw − ρ o h 144

PcR

=

1.0 - 0.85 62.4 lb 40 ft3 = 2.6 psi 144

PcL

=

PcR 0.313

PcL

=

2.6 = 8.3 psi 0.313

PcR

move to the right horizontally from PcL = 8.3 psi to the capillary pressure curve. Drop vertically to the x-axis, read Sw. Sw = 50%

III- 28

X)

Capillary pressure variation A.

effect of permeability 1.

displacement pressure increases as permeability decreases

2.

minimum interstitial water saturation increases as permeability decreases

RESERVOIR FLUID DISTRIBUTION CURVES

18

12

6

10 md

90

72

160 140

54

120 100 80

36

60 40

18

20 0

0

0 0

10

20

30

40

50 Sw %

III -29

60

70

80

90

100

Air - Water Capillary Pressure, psi (laboratory data)

24

Height above zero capillary pressure, ft

Oil - Water Capillary Pressure, psi (reservoir conditions)

180

100 md

200

900 md

30

200 md

(From Wright and Wooddy)

30

225.0

25

187.5

20

150.0 Sandstone Core

15

112.5

Porosity = 28.1% Permeability = 1.43 md Factor = 7.5

10

75.0

5

37.5

Mercury capillary pressure, psi

Water/nitrogen capillary pressure, psi

Effect of grain size distribution

0

0 0

20

40

100

80

60

60

80

100

40

20

0

Water

60

348

50

290

40

232

30

174

20

116 Limestone Core Porosity = 23.0% Permeability = 3.36 md Factor = 5.8

10

58

60

80

0 100

40

20

0

0

0

20

40

Mercury capillary pressure, psi

Hg

Water/nitrogen capillary pressure, psi

B.

Water

100

80

60 Hg

1.

majority of grains same size, so most pores are same size - curve (a) (well sorted)

2.

large range in grain and pore sizes - curve (b) (poorly sorted)

III -30

XI)

Averaging capillary pressure data J-function J Sw =

Pc k 1/2 σcos θ φ

attempt to convert all capillary pressure data to a universal curve universal curve impossible to generate due to wide range of differences existing in reservoirs concept useful for given rock type from given reservoir where Pc σ k φ

= = =

dyne/(sq cm) dyne/cm (sq cm)

=

fraction

or can use any units as long as you are consistent

III -31

CAPILLARY RETENTION CURVES.

CAPILLARY PRESSURE FUNCTION, J

(From Rose and Bruce.) 2.0 1.9 1.8 1.7 1.6 1.5 1.4 1.3 1.2 1.1 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0

LEDUC

LEVERETT

KATIE

ALUNDUM

EL ROBLE KINSELLA

HAWKINS

0

10

20

30

40

50

60

70

WATER SATURATION, Sw

Reservoir

Formation

Hawkins El Roble Kinsella Katie Leduc Alundum Leverett

Woodbine Moreno Viking Deese Devonian (consolidated) (unconsolidated)

III -32

80

90

100

Capillary Pressure Problem 1 1.

A glass tube is placed vertically in a beaker of water. The interfacial tension between the air and water is 72 dynes/cm and the contact angle is 0 degree. Calculate:

2.

a.

the capillary rise of water in the tube if the radius of the tube is 0.01 centimeters.

b.

what is the difference in pressure in psi across the air-water interface in the tube.

The displacement pressure for a water saturated porcelain plate is 55 psi of air. What is the diameter in inches of the largest pore in the porcelain plate? Assume 72 dynes/cm and 0 degrees.

Solution: (1)

σAW = 72 dynes/cm ρW

= 1 gm/cm3

g

= 980 dynes/gm

θ

= 0o

(a)

capillary rise of water if radius is .01 cm h=

2 72 cos0° 2σAWcos θ = rρg .01 1.0 980

h = 14.69 cm (b)

pressure drop in psi across interface Pc

Pc Pc

=

pa - pw = ρ wgh = 1.0 980 14.69

=

0.0142 atm

=

0.209 psi

14.696 psi atm

III -33

(2)

Pc

=

2σAWcos θ r

Pc

=

55 psi

Pc

=

55 psi

=

3.792 x 106 dynes/cm2

r

=

2σAW cos θ Pc

r

=

2 72 cos0° in = 3.797 x 10 -5 cm 6 2.54 cm 3.792 x 10

r

=

1.495 x 10-5 in

d

=

2.99 x 10-5 in

atm 14.696 psi

1.0133 x 10 6 dynes/cm2 atm

III -34

Capillary Pressure Problem 2 Given the information below and graph of PcL vs. wetting phase saturation Sw , construct the curves for PcR, h in reservoir, and J-function vs. Sw. Water is the wetting phase in both the laboratory and the reservoir.

fluids

lab air-water

res oil-water

θ

0°

25°

σ

60 dyne/cm

20 dyne/cm

ρwet

1.0 gm/cm3

1.1 gm/cm3

ρnon-wet

0 gm/cm3

0.863 gm/cm3

k

37 md

variable

φ

16%

variable

Pc k/φ 1/2 J= σ cos θ

III -35

35.0 32.5 30.0 27.5 25.0

PCL, psi

22.5 20.0 17.5 15.0

12.5 10.0 7.5 5.0 2.5

0.0 0

10

20

30

40

50

60

Sw %

Solution:

(1)

=

σR cosθ R Pc σL cosθ L L

=

20 cos25 P 60 cos0 cL

PcR

PcR

=

0.302 P cL

III -36

70

80

90

100

(2) P cR

PcR hR (3)

J

J

=

hR ρw - ρo 144

=

hR 1.1 - .863 62.4 144

=

.103 h R

=

9.74 P cR

=

Pc k 1/2 σ cos θ φ

=

PcL k 1/2 σAW cosθ L φ L

=

PcL 37 1/2 60 cos0° .16

=

.253 P cL

Sw %

PcL psi

PcR psi

15

32

9.7

94.1

8.1

20

19.5

5.9

57.4

4.9

25

15.6

4.7

45.9

3.9

30

13.2

4.0

38.8

3.3

40

9.9

3.0

29.1

2.5

50

7.8

2.4

22.9

2.0

60

6.0

1.8

17.6

1.5

70

4.7

1.4

13.8

1.2

80

3.7

1.1

10.9

0.9

90

2.8

0.8

8.2

0.7

100

2

0.6

5.9

0.5

hR ft

III -37

J assorted

10 8

Pc

R

6 4 2 0 0

20

40

60

80

100

Sw % 100 80

h R

60 40 20 0 0

20

40 60 Sw %

80

100

0

20

40 60 Sw %

80

100

10 8

J

6 4 2 0

III -38

IV. FLUID SATURATIONS I)

II)

Basic concepts of hydrocarbon accumulation A.

Initially, water filled 100% of pore space

B.

Hydrocarbons migrate up dip into traps

C.

Hydrocarbons distributed by capillary forces and gravity

D.

Connate water saturation remains in hydrocarbon zone

Methods for determining fluid saturations A.

Core analysis (direct method) 1.

factors affecting fluid saturations a.

flushing by mud filtrate 1)

differential pressure forces mud filtrate into formation Ph > Pres

2)

for water base mud, filtrate displaces formation water and oil from the area around the well (saturations likely change)

3)

for oil base mud, filtrate will be oil; saturations may or may not change.

IV - 1

Example:

Effects of flushing by mud filtrates

Coring with water base mud Oil zone at minimum interstitial water saturation: sat at surface compared to res

flushing by bit trip to surface Sw

↑

↓

? probably ↑

So

↓

↓

↓

Sg

-

↑

↑

Gas zone at minimum interstitial water saturation: sat at surface compared to res

flushing by bit trip to surface Sw

↑

↓

?

So

-

-

-

Sg

↓

↑

?

Water zone: sat at surface compared to res

flushing by bit trip to surface Sw

-

↓

↓

So

-

-

-

Sg

-

↑

↑

IV - 2

Coring With Oil Base Mud Oil zone at minimum interstitial water saturation: sat at surface compared to res

flushing by bit trip to surface Sw

-

-

-

So

-

↓

↓

Sg

-

↑

↑

Gas zone at minimum interstitial water saturation: sat at surface compared to res

flushing by bit trip to surface Sw

-

-

-

So

↑

↓

↑

Sg

↓

↑

↓

Water zone: sat at surface compared to res

flushing by bit trip to surface Sw

↓

↓

↓

So

↑

↓

↑

Sg

-

↑

↑

IV - 3

b.

2.

bringing core to surface 1)

reduction in hydrostatic pressure causes gas to come out of solution

2)

gas displaces oil and water causing saturations to change

laboratory methods a.

evaporation using retort distillation apparatus

HEATING ELEMENT

CORE COOLING WATER IN

CONDENSER COOLING WATER OUT

IV - 4

1)

process a)

heat small sample of rock

b)

oil and water vaporize, then condense in graduated cylinder

c)

record volumes of oil and water

d)

correct quantity of oil

For converting distilled oil volume to oil volume originally in a sample, multiply oil volume recovered by factor corresponding to gravity of oil in core

1.4

Multiplying Factor

1.3

1.2

1.1

1.0

0.9 15

20

25

30

35

40

45

50

Oil Gravity, °API at 60° F

IV - 5

55

60

65

e)

determine saturations

V Sw = w Vp

V So = o Vp

Sg = 1 - S o - Sw where Sw So Sg Vp Vw Vo 2)

= = = = = =

water saturation, fraction oil saturation, fraction gas saturation, fraction pore volume, cc volume of water collected, cc volume of oil collected, cc

disadvantages of retort process a)

must obtain temperature of 1000-1100oF to vaporize oil, water of crystallization from clays also vaporizes causing increase in water recovery

WATER RECOVERED PORE WATER 0

0

TIME

b)

at high temperatures, oil will crack and coke. (change in hydrocarbon molecules) amount of recoverable liquid decreases.

c)

core sample ruined

IV - 6

3)

b.

advantages of retort process a)

short testing time required

b)

acceptable results obtained

leaching using solvent extraction apparatus WATER OUT

WATER IN

GRADUATED TUBE

CORE

SOLVENT

HEATER

1)

process a)

weigh sample to be extracted

b)

heat applied to system causes water from core to vaporize

c)

solvent leaches hydrocarbons from core

IV - 7

d)

water condenses, collects in trap. Record final water volume

e)

reweigh core sample

f)

determine volume of oil in sample Vo =

Wi - Wdry - Vw ρ w ρo

where: Wi = after leaching Wdry

weight of core sample

= weight of core sample after leaching

V Sw = w Vp 2)

3)

V So = o Vp

disadvantages of leaching a)

process is slow

b)

volume of oil must be calculated

advantages of leaching a)

very accurate water saturation value obtained

b)

heating does not remove water of crystallization

c)

sample can be used for future analysis

IV - 8

3.

uses of core determined fluid saturation a.

cores cut with water base mud 1)

presence of oil in formation

2)

determination of oil/water contact

3)

determination of gas/oil contact

0

GAS

OIL

WATER

So ≅ 0 in gas zone So ≥ 15% in oil zone 0 ≤ So ≤ Sor in water zone Sor = residual oil saturation

IV - 9

SO

50

b.

cores cut with oil base mud ("natural state" cores) 1)

minimum interstitial water saturation

2)

hydrocarbon saturation

3)

oil/water contact

B.

Capillary pressure measurements (discussed in Chapter VIII)

C.

Electric logs

IV - 10

Example IV-1 You want to analyze a core sample containing oil, water and gas. Vb bulk volume = 95 cm3 Wt initial = 216.7 gm the sample was evacuated and the gas space was saturated with water ρ w = 1 gm/cm3 Wt new = 219.7 gm the water with in the sample is removed and collected Vw removed = 13.0 cm3 the oil is extracted and the sample is dried Wt dry = 199.5 gm calculate: (1)

porosity

(2)

water saturation

(3)

oil saturation assuming 35o API

(4)

gas saturation

(5)

matrix density

(6)

lithology

Solution: gas vol.

=

219.7 - 216.7 ;

Vg

=

3 cc

water vol.

=

13 - 3

Vw

=

10 cc

Wt fluids

=

219.7 - 199.5 =

20.2 gm

Wt oil

=

20.2 - 10 - 3

7.2 gm

;

=

IV - 11

ρo

=

141.5 = 0.85 gm/cc 131.5 + 35°API

Vo

=

7.2/0.85 = 8.49 cc

Vp

=

8.49 + 3 + 10 = 21.47 cc

φ

=

21.47/95 = 22.6%

Sw

=

10/21.47 = 46.57%

So

=

8.49/21.47 = 39.46%

Sg

=

3/21.47 = 13.97%

ρm

=

199.5/(95-21.47) = 2.71 gm/cc

lithology = limestone

IV - 12

Example IV-2

A core sample was brought into the laboratory for analysis. 70 gm of the core sample were placed in a mercury pump and found to have 0.71 cc of gas volume. 80 gm of the core sample was placed in a retort and found to contain 4.5 cc of oil and 2.8 cc of water. A piece of the original sample weighing 105 gm was placed in a pycnometer and found to have a bulk volume of 45.7 cc. (Assume ρ w = 1.0 gm/cc and 35o API oil) calculate: (1)

porosity

(2)

water saturation

(3)

oil saturation

(4)

gas saturation

(5)

lithology

Solution: Vg

=

.71 cc 100 gm = 1.014 cc 70 gm

Vo

=

4.5 cc 100 gm = 5.63 cc 80 gm

Vw

=

2.8 cc 100 gm = 3.50 cc 80 gm

Vb

=

45.7 cc 100 gm = 43.52 cc 105 gm

Wt matrix = 100 - 5.63(.85) - 3.5(1.0) = 91.71 gm Vm

=

43.52 - 1.014 - 5.63 - 3.50 = 33.37 cc

Vp

=

1.014 + 5.63 + 3.50 = 10.14 cc

φ

=

10.14/43.52 = 23.31%

IV - 13

Sw

=

3.50/10.14 = 34.5%

So

=

5.63/10.14 = 55.5%

Sg

=

1.014/10.14 = 10%

ρm

=

(91.71/33.38) = 2.75

IV - 14

Fluid Saturation Problem 1 Calculate porosity, water, oil, and gas saturations, and lithology from the following core analysis data. How should the calculated saturations compare with the fluid saturations in the reservoir?

Oil well core with water base mud initial weight of saturated core = 86.4 gm after gas space was saturated with water, weight of core = 87.95 gm weight of core immersed in water = 48.95 gm core was extracted with water recovery being 7.12 cc after drying core in oven, core weighed 79.17 gm assume ρ w = 1.0 gm/cc oil gravity = 40° API

Solution:

(1)

γo

=

141.5 131.5 + °API

γo

=

141.5 = 0.825 131.5 + 40°

ρo

=

0.825 gm/cc

φ

=

Vp Vb

Vp

=

Vw + Vo + Vg

Wo

=

Wsat - Vwρ w - Wdry

=

87.95 - 7.12(1.0) - 79.17

=

1.66 gm

Wo

IV - 15

(2)

(3)

Vo

=

Wo ρo

Vo

=

1.66 gm = 2.01 cc 0.825 gm/cc

Vw

=

Vwrec - Wsat - Wi / ρ w

=

7.12 - (87.95 - 86.4)/(1.0)

Vw

=

5.57 cc

Vg

=

1.55 cc

Vp

=

5.57 + 2.01 + 1.55

Vp

=

9.13 cc

Vb

=

Wsat - Wimm ρw

Vb

=

(87.95 - 48.95) gm = 39.0 cc 1 gm/cc

φ

=

9.13 = 23.4% 39.0 Vw Vp

Sw

=

Sw

=

5.57 cc = 61.0% 9.13 cc

So

=

Vo Vp

So

=

2.01 cc = 22.0% 9.13 cc

Sg

=

Sg

=

1.55 cc = 17.0% 9.13 cc

Vm

=

Vb - Vp

Vg Vp

IV - 16

Vm ρm ρm

=

39 - 9.13 = 29.87 cc

=

Wdry Vm

=

gm 79.17 gm/29.87 cc = 2.65 cc

. . lithology is sandstone (4)

water saturation at surface will probably be greater than reservoir water saturation oil saturation at surface will be less than reservoir oil saturation gas saturation at surface will be greater than reservoir gas saturation

IV - 17

Fluid Saturation Problem 2 Calculate porosity, water saturation, oil saturation, gas saturation, and lithology from the following core analysis data. How should the saturations you have calculated compare with the fluid saturations in the reservoir?

Oil well core cut with an oil base mud Sample 1 weighed 130 gm and was found to have a bulk volume of 51.72 cc Sample 2 weighed 86.71 gm, and from the retort method was found to contain 1.90 cc of water and 0.87 cc of oil Sample 3 weighed 50 gm and contained 0.40 cc of gas space assume ρ w = 1.0 gm/cc oil gravity = 40o API

Solution:

(1)

141.5 131.5 + °API

γo

=

γo

=

141.5 = 0.825 131.5 + 40°

ρo

=

0.825 gm/cc

φ

=

Vp Vb

Vp

=

Vo + Vw +Vg

Vo

=

0.87 cc x 100 = 1.00 cc 86.71 gm 100 gm

Vw

=

1.90 cc x 100 = 2.19 cc 86.71 gm 100 gm

Vg

=

0.40 cc x 100 = 0.80 cc 50 gm 100 gm

Vp

=

(1.00 + 2.19 + 0.80) cc/100 gm

IV - 18

(2)

Vp

=

3.99 cc/100 gm

Vb

=

51.72 cc x 100 = 39.78 cc 130 gm 100 gm

φ

=

3.99/100 x 100 = 10% 39.78/100

Sw

Sw

(3)

So

So

(4)

(5)

Sg

=

Vw x 100 Vp

=

2.19 cc x 100 3.99 cc

=

54.8%

=

Vo x 100 Vp

=

1.0 cc x 100 3.99 cc

=

25.1%

=

Vg x 100 Vp

=

0.80 cc x 100 3.99 cc

Sg

=

20.1%

Vm

=

Vb - Vp

=

39.78 - 3.99

Vm

=

35.79 cc/100 gm

Wm

=

Wsat - ρ oVo - ρ wVw /Wsat

=

86.71 - 0.825 0.87 - 1.0 1.90 86.71

=

97 gm/100 gm

Wm

IV - 19

(6)

ρm

=

Wm Vm

ρm

=

97 gm/100 gm 35.79 cc/100 gm

ρm

=

2.71 gm/cc

..

lithology - limestone

water saturations should be fairly close in value oil saturation will be less than reservoir oil saturation gas saturation will be greater than reservoir gas saturation

IV - 20

V. ELECTRICAL PROPERTIES OF ROCK-FLUID SYSTEMS I)

Electrical conductivity of fluid saturated rock A.

Definition of resistivity

ELECTRICAL CURRENT FLOW

A

L

given a box of length (L) and cross-sectional area (A) completely filled with brine of resistivity (Rw) the resistance of the brine in the box to current flow may be expressed as r = Rw L A r

=

resistance - ohm

Rw =

resistivity - ohm meters

L

=

length - meters

A

=

area - (meters)2

V-1

B.

C.

D.

Nonconductors of electricity 1.

oil

2.

gas

3.

pure water

4.

minerals

5.

rock fragments

Conductors of electricity 1.

water with dissolved salts conducts electricity (low resistance)

2.

clay

Development of saturation equation (ignore clay) AP

A

ELECTRICAL CURRENT FLOW

L

1.

the electrical current flows through the water (brine) a.

the area available for current flow is the cross-sectional area of the pores. Ap < A

b.

the path through the pores is Lp. Lp > L

V-2

2.

resistance to electrical flow through the porous media is equal to the resistance of a container of area Ap and length Lp filled with water (brine) r=

RwLp , water filled cube Ap

R L r = o , porous media A thus Ro =

RwALp ApL

where r = resistance of rock cube with pores filled with brine, ohm Rw = formation brine resistivity, ohm-m (from water sample or SP log)

3.

Ro

= resistivity of formation 100% saturated with brine of resistivity, Rw, ohm-m

Ap

= cross-sectional area available for current flow, m 2

Lp

= actual path length ion (current) must travel through rock, m

A

= m2

cross-sectional area of porous media,

L

=

length of porous media, m

Since Ap ≅ porosity, φ A and Lp ≅ L tortuosity, a measure of rock cementation. then Ro =

RwALp ApL

V-3

becomes Ro = f Rw, φ, tortuosity E.

Electrical formation resistivity factor, F 1.

the equation for resistivity of a formation 100% saturated with a brine of resistivity of Rw Ro = f Rw, φ, tortuosity

2.

can be written as Ro = F R w where F is the electrical formation resistivity factor F=

3.

Ro Rw

cementation factor, m a.

it has been found experimentally that the equation for F takes the form F = C φ-m where C is a constant m is the cementation factor

b.

thus log F = log C - m log φ

V-4

100

F

10

1 0.01

0.1 φ

1.0

when intercept = C slope = -m, the cementation factor 4.

commonly used equation for electrical formation resistivity factor a.

Archie's Equation F = φ-m

b.

Humble Equation F = 0.62 φ-2.15 (best suited for sandstones)

Cementation Factor (m) and Lithology Lithology

m values

Unconsolidated rocks (loose sands, oolitic limestones) Very slightly cemented (Gulf Coast type of sand, except Wilcox) Slightly cemented (most sands with 20% porosity or more) Moderately cemented (highly consolidated sands of 15% porosity or less) Highly cemented (low-porosity sands, quartzite, limestone,dolomite)

1.3 1.4-1.5 1.6-1.7 1.8-1.9 2.0-2.2

V-5

Example V-1 Determine the porosity for a sandstone using Archie's and Humble equation . The formation water's resisitivity was 0.5 ohm-meters. The formation rock 100% saturated with this water was 21.05 ohm-meters. Which of the two equations will give the most reasonable answer? Solution: F =21.05/0.5 = 42.1

Archie's:

F = φ-m m = 2.0 for sandstone φ2 = 1/F φ=

1 42.1

φ = 15.41%

Humble:

F = 0.62/φ2.15 φ2.15 = 0.62/F φ = 2.15 0.62 42.1 φ = 14.06%

The Humble equation was developed for sandstone.

V-6

F.

Resistivity Index, I, and Saturation Exponent, n 1.

definition of resistivity index Rt Ro

I=

where Ro = resistivity of formation 100% saturated with water (brine) of resistivity Rw, ohm-m Rt =resistivity of formation with water (brine) saturation less than 100%, ohm-m 2.

it has been found experimentally that -1 R -1 Sw = I n = t n Ro where n is the saturation exponent ≅ 2.0

3.

rearrange Sw-n =

Rt Ro

-n log Sw = log

Rt Ro

V-7

100

Rt

10

Ro

1 .1

Sw

1.0

slope = -n, when n is the saturation exponent

NOTE: slope =

II)

log y1 - log y2 log x1 - log x2

Use of Electrical Formation Resistivity Factor, Cementation Factor, and Saturation Exponent A.

obtain porosity, φ, from electric log or core analysis

B.

F=Cφ

C.

obtain water resistivity, Rw, from water sample or electric log

D.

Ro = F R w

E.

convert Rt from electric log to water saturation

-m

(usually use Archie or Humble equation)

Ro 1n Sw = Rt

V-8

III)

Laboratory measurement of electrical properties of rock A.

Apparatus

AC SOURCE

1000 OHM STD. RESISTOR

CORE

VOLTMETER

B.

Calculations 1.

resistance of core E = Ir where: E

=

voltage, volts

I

=

current, amperes

r

=

resistance, ohms

∴ rcore = E I

V-9

2.

resistivity of core r A Rcore = core L substituting r = E into the equation I Rcore = EA IL

C.

Procedure 1.

2.

determine resistance of core a.

set desired current from AC source, low current preferred so core does not heat up.

b.

record voltage from voltmeter

determine resistivity of core a.

for the first test completely saturate core with brine Sw = 100%, R core = Ro

b.

for next test, desaturate core by 15-20%, until Sw < 100% Rcore = Rt

c.

repeat tests until Sw = S wir

where Swir

= minimum interstitial brine saturation (irreducible), fraction

Ro

= resistivity of core 100% saturated with brine, ohm-m

Rt

= resistivity of core less than 100% saturated with brine of Rw, ohm-m

Rt Ro

=

resistivity index = I

V - 10

D.

Determine saturation exponent, n 1.

rearrange saturation equation

Sw

Swn Ro Rt log

2.

=

Ro 1/n Rt

=

Ro Rt

=

Sw-n

Rt = -n log Sw Ro

Plot log

Rt vs log Sw or log I vs log Sw Ro

100

I=

Rt Ro

10

1 .1

3.

Sw

1.0

the slope of the plot is -n, where n is the saturation exponent

V - 11

Example V-2 Given the following data, calculate the electrical formation resistivity factor and saturation exponent of the core. Rw

=

55 ohm-cm

I

=

0.01 amp

D

=

2.54 cm

L

=

3.2 cm E Voltage across Core, volts 7.64 10.50 14.34 20.16 27.52 34.67

Sw Water Saturation, % 100.0 86.0 74.0 63.0 54.0 49.0 = Swir Solution: (1)

electrical formation resistivity factor F

=

Ro Rw

ro

=

E = 7.64 = 764 ohm I .01

Ro

=

roA 764 2.54 2π/4 = = 1210 ohm cm L 3.2

F

=

Ro 1210 = = 22 Rw 55

V - 12

saturation exponent

-n log Sw = log

Sw % 1.00 .86 .74 .63 .54 .49

Rt Ro

rt = E I (ohm)

rA Rt = t L (ohm-cm)

Rt Ro

1050 1434 2016 2452 3467

1663 2271 3192 4358 5490

1.000 1.374 1.877 2.638 3.601 4.537

(.334,10)

Rt/Ro

10

1 .1

1 Sw

V - 13

(1.0,1.0)

-n

=

slope

=

log 10 - log 1 log .334 - log 1

-n

=

1-0 -.4763 - 0

=

2.10

n

=

saturation exponent

NOTE: Rt Et = Ro Eo Rt could have been calculated as the ratio of voltage at Ro Sw divided by the voltage at Sw = 1.0 so

V - 14

E.

Determine cementation factor, m, and constant C for electrical formation resistivity factor equation 1.

test several core samples from reservoir with formation brine a.

determine Ro and f for each sample

b.

determine Rw for formation brine F=

c. 2.

Ro Rw

plot data according to form of equation for electrical formation resistivity factor F = C φ-m log F = log C - mlog φ

100

F

10

1 0.01

0.1 φ

slope = -m, m = cementation factor intercept = C (intercept found at φ = 1.0)

V - 15

1.0

Example V-3 The laboratory test of Example IV-2 has been repeated for several core samples from the reservoir. Data is given below. Calculate the cementation factor and intercept for the formation resistivity factor equation. Porosity Formation Resistivity Factor φ

F

0.152 0.168 0.184 0.199 0.213 0.224

40 32 26 22 19 17

Solution: F = C φ-m log F = log C - m log φ plot log F vs log φ

F

100

10

1 .1

1 ø

V - 16

slope =

log 50 - log 10 = -2.21 log 0.137 - log 0.284

-m

=

slope = -2.21

m

=

2.21 = cementation factor

intercept log F

=

log C -m log f

log 10

=

log C -2.21 log 0.284

log C

=

-.2082

C

=

062 = intercept

V - 17

IV)

Effect of clay on resistivity A.

ideally, only water conducts a current in rock

B.

if clay is present, portion of current conducted through the clay BRINE

CLAY 1 = 1 + 1 RoA Rclay Ro where RoA

= resistivity measured on sample of reservoir rock with clay, 100% saturated with brine of resistivity Rw, ohm-m

Rclay

= component of measured resistivity due to clay, ohm-m

Ro

= component of measured resistivity due to brine, ohm-m

1 = 1 + 1 RoA Rclay F Rw C.

to determine electrical formation resistivity factor 1.

measure resistivity of core sample (containing clay) in usual manner, this will be RoA

2.

measure resistivity of brine, Rw, in usual manner

V - 18

3.

plot

1 ROA (OHM - M) -1

1 -1 RW (OHM - M)

1 = 1 + 1 1 RoA Rclay F Rw where 1 = intercept Rclay 1 = slope F

V - 19

D.

effect of clay

1.

define factor

FA =

RoA Rw , clays reduced the apparent formation resistivity

CLEAN SAND

F

SHALY SAND FA

RW 2.

formation resistivity factor decreases more gradually when clay is present in the formation

100 CLEAN SAND

F 10 SHALY SAND

1

0.1

φ

1.0

V - 20

3.

saturation exponent n is not constant when clay is present in formation.

100 CLEAN SAND LOW R w n=2

I=

Rt Ro

CLEAN SAND Swn-1 = I

10 SHALY SAND n =?

1 .1

CLEAN SAND HIGH R w n=1

Sw

V - 21

1.0

VI. MULTIPHASE FLOW IN POROUS ROCK I)

Effective permeability A.

Permeability, k, previously discussed applies only to flow when pores are 100% saturated with one fluid - sometimes called absolute permeability

q= B.

kA∆ρ µL

When pore space contains more than one fluid, the above equation becomes k A∆Pο qo = o µoL k A∆Pw qw = w µw L qg =

where

and

kgA∆Pg µgL qo

=

flow rate of oil, volume/time

qw

=

flow rate of water, volume/time

qg

=

flow rate of gas, volume/time

ko

=

effective permeability to oil, md

kw

=

effective permeability to water, md

kg

=

effective permeability to gas, md

C.

Effective permeability is a measure of the fluid conductance capacity of porous media to a particular fluid when the porous media is saturated with more than one fluid

D.

Effective permeability is a function of: 1.

geometry of the pores of the rock

2.

rock wetting characteristics

3.

fluid saturations

VI - 1

E.

Darcy equation for multiple fluids in linear flow, in oilfield units k A P1 - P2 qo = 1.1271 x 10-3 o µoL k A P1 - P2 w qw = 1.1271 x 10-3 w µw L qg = 1.1271 x 10-3 when

II)

kg A P1 - P2 g µgL

k =

md

A =

ft2

P =

psia

L =

ft

q

res bbl/day

=

Relative permeability A.

Defined as the ratio of the effective permeability to a fluid at a given saturation to the effective permeability to that fluid at 100% saturated (absolute permeability) k kro = o k k krw = w k krg =

B.

III)

kg k

It is normally assumed that the effective permeability at 100% saturation is the same for all fluid in a particular rock. (not necessarily true in shaly sand)

Typical relative permeability curves A.

Use subscript wp to represent the "wetting phase" Use subscript nwp to represent the "non-wetting phase"

VI - 2

1

1

NON-WETTING PHASE

Kr

2

WETTING PHASE

3 0

4

0

SWP, % MINIMUM INTERSTITIAL S WP

100 EQUILIBRIUM S NWP

1.

krwp

2.

k rapid decrease in rwp as Swp decreases

3.

krwp

4.

krnwp

= 1, only at S wp = 100%

= 0, at minimum interstitial Swp = 0, at equilibrium Snwp

Note that krwp + krnwp < 1.0

VI - 3

B.

Effect of saturation history 1.

2.

two types of relative permeability curves a.

drainage curve - wetting phase is displaced by non-wetting phase, i.e., wetting phase saturation is decreasing

b.

imbibition curve - non-wetting phase is displaced by wetting phase, i.e., wetting phase saturation is increasing

the typical relative permeability curve shown below represents a process in which a.

process begins with porous rock 100% saturated with wetting phase (Swp = 100%)

b.

wetting phase is displaced with non-wetting phase (drainage) until wetting phase ceases to flow (Swp = minimum interstitial wetting phase saturation)

c.

then non-wetting phase is displaced with wetting phase (imbibition) until non-wetting phase ceases to flow (Swp = equilibrium or residual non-wetting phase saturation)

VI - 4

1

Krnwp DRAINAGE

Kr

IMBIBITION

Krwp

0

0

SWP, %

100

minimum interstitial

residual non-wetting

wetting phase saturation

phase saturation

VI - 5

3.

the word "hysteresis" describes the process in which the results (kr) are different when measurements are made in different directions

4.

the procedure (drainage or imbibition) used to obtain kr data in laboratory must correspond to the process in the reservoir

5.

a.

initial distribution of fluids in reservoir was by drainage

b.

at and behind a water front (flood or encroachment) the process is imbibition

wetting preference for reservoir rocks is usually water first, then oil, finally gas

Fluids Present

Wetting Phase

Water & Oil Water & Gas Oil & Gas

Water Water Oil

VI - 6

C.

Three phase relative permeability 1.

often three phases are present in petroleum reservoirs

2.

tertiary (triangular) diagram is used to represent a threephase system

100% GAS

100% WATER

100% OIL

VI - 7

3.

relative permeability to oil in a three phase system

100% GAS

1%

5

10 20

40 60

100% WATER

100% OIL

Note, kro is shown in % a.

dependence of relative permeability to oil on saturations of other phases is established as follows: 1)

oil phase has a greater tendency than gas to wet the solid

2)

interfacial tension between water and oil is less than that between water and gas

3)

oil occupies portions of pore adjacent to water

4)

at lower water saturations the oil occupies more of the smaller pores. The extended flow path length accounts for the change in relative permeability to oil at constant oil saturation and varying water saturation

VI - 8

4.

Relative permeability to water in a three-phase system

100% GAS

0

Krw

10% 20% 40%

60% 80% 100% WATER

100% OIL

a.

straight lines indicate relative permeability to water is a function of water saturation only

b.

thus, krw can be plotted on cartesian coordinates against Sw.

VI - 9

5.

Relative permeability to gas in a three-phase system 100% GAS

50%

40

30

20 5

1

100% WATER

100% OIL

a.

k curves above indicate that rg is a function of saturations of other phases present.

b.

k other research shows that rg is a unique function of gas saturation

c.

the other phases, oil and water, occupy the smaller pore openings and wet the surface of the rock

d.

e.

k therefore, rg should be dependent only on the total saturation of the other two phases (i.e. 1-Sg) and independent of how much of that total is composed of either phase k thus rg can be plotted on Cartesian coordinates against So + Sw

VI - 10

1.0

0.8

0.6

krg 0.4

0.2

0.0 0

20

40

60

80

100

So + Sw 6.

Bottom line - for three-phase system in water wetted rock a.

b.

water 1)

is located in smaller pore spaces and along sand grains

2)

therefore krw is a function of Sw only

3)

thus plot krw against Sw on rectangular coordinates

gas 1)

is located in center of larger pores

2)

k therefore rg is a function of Sg only

3)

k thus plot rg against Sg (or So + Sw) on rectangular coordinate

VI - 11

c.

oil 1)

is located between water and gas in the pores and to a certain extent in the smaller pore spaces

2)

therefore kro is a function of So, S w, and S g

3)

thus plot kro against So, S w, S g on a triangular diagram

4)

if Sw can be considered to be constant (minimum interstitial) kro can be plotted against So on a rectangular diagram

1.0

0.8

0.6

kro 0.4

0.2

0.0 0

20

40

60

80

100

So, % Minimum Interstitial Water Saturation

VI - 12

7.

Flow in three-phase system

100% GAS

5% oil

5% gas

5% water

100% WATER

100% OIL

Arrows point to increasing fraction of respective components in stream Region of three-phase flow in reservoir centers around 20% gas, 30% oil, 50% water

VI - 13

IV)

Permeability ratio (relative permeability ratio)

A.

Definitions 1.

When the permeability to water is zero (as at minimum interstitial water saturation) it is sometimes convenient to use permeability ratio to represent the flow conductance of the rock to gas and oil as a ratio.

permeability ratio =

2.

kg krg = ko kro

When the permeability to gas is zero (no gas or gas below "critical gas saturation") it is sometimes convenient to use permeability ratio to represent the flow conductance of the rock to oil and water as a ratio kr k permeability ratio = o = o kw krw

V)

Measurement of relative permeability A.

B.

Methods 1.

Laboratory - steady-state flow process

2.

Laboratory - displacement (unsteady-state process)

3.

Calculation from capillary pressure data (not covered here)

4.

Calculation from field performance data

Laboratory Methods 1.

Steady-state flow process a.

saturate core with wetting-phase fluid

b.

inject wetting-phase fluid through core (this will determine absolute permeability)

VI - 14

c.

inject a mix of wetting-phase and non-wetting phase (start with small fraction of non-wetting phase)

d.

when inflow and outflow rates and portion of non-wetting phase equalize, record inlet pressure, outlet pressure and flow rates of each phase

e.

measure fluid saturation in core (see below)

f.

calculate relative permeability q µ L ko = o o A∆p q µ L kw = w w A∆p

g.

repeat b through f with injection mixtures containing relatively more non-wetting phase until irreducible wettingphase saturation is reached

1

kro

kr

krw

0 0

Sw, %

VI - 15

100

h.

determination of fluid saturations 1)

resistivity

Sw =

Ro 1n Eo 1n = Rt Et

where: Ro = resistivity of core 100% saturated with wetting-phase, ohm-m Rt

= resistivity of core with saturation of wetting phase less than 100%, ohm-m

Eo

= voltage across core 100%, saturated with wetting phase, volts

Et

=

voltage across core with saturation of wetting phase less than 100%, volts

2) volumetric balance 3) gravimetric method - remove core and weigh it

where:

Wf

=

Wt - Wd

Wf

=

weight of fluid in core, gm

Wt

=

weight of saturated core, gm

Wd =

weight of dry core, gm

Wf

=

ρ oVo + ρ wVw

Vf

=

Vo + Vw

ρ

=

density, gm/cc

V

=

volume, cc

and

where:

where:

Sw =

Vw/Vf

Sw =

saturation of wetting phase

thus

VI - 16

Sw =

i.

j.

Wf/Vf - ρ o ρw - ρo

same procedure can be used starting with 100% saturation of non-wetting phase 1)

injection ratio start with high ratio of non-wetting phase

2)

procedure ends at residual non-wetting phase saturation

3)

then is a hysteresis effect of same type as discussed with capillary pressure measurements

4)

choice of starting saturation depends on reservoir process which is being simulated

end effects 1)

causes of end effects a)

in the bulk of the core there is a wettingphase saturation and a non-wetting phase saturation, therefore there is a finite value of capillary pressure

b)

thus there is a difference in pressure between the wetting-phase and non-wetting phase Pcap = P nwp - Pwp

c)

at the face of the core the pressures in the wetting-phase and the non-wetting phase are essentially equal Pnwp = P wp

thus capillary pressure is essentially zero d)

if capillary pressure is zero, the saturation of the wetting phase must be 100% (see capillary pressure curve)

e)

there must be a saturation gradient from essential 100% wetting phase at the "end" to some value of Swp less than 100% in the bulk of the core

VI - 17

100

100

Theoretical saturation gradient

80

Oil Saturation, %

Oil Saturation, %

80

Theoretical saturation gradient

60

Inflow face 40

60

Inflow face 40

20

20

0

0

0

5

10

15

20

0

25

5

Distance from outflow face, cm

2)

End Section

15

20

Distance from outflow face, cm

elimination of end effects a)

install end pieces to contain end effects

b)

flow at rapid rates to make end effect negligible (pressure gradient > 2 psi/inch

Thermometer Packing Nut

10

Electrodes

Test Section

Copper Orifice Plate

Inlet

Mixing Section

Differential Pressure Taps Bronze Screen

Highly permeable disk

Outlet

Inlet

PENN STATE RELATIVE-PERMEABILITY APPARATUS

VI - 18

25

Example VI-1 The relative permeability apparatus shown above was used in a steady-state flow process to obtain the data given below at a temperature of 70oF. See figure on previous page.

The Core

The Fluids

sandstone length = 2.30 cm diameter = 1.85 cm area = 2.688 cm2

brine, 60,000 ppm oil, 40oAPI µw = 1.07 cp µo = 5.50 cp

porosity = 25.5% Oil Flow cc/sec

Water Flow cc/sec

Inlet Pressure psig

Outlet Pressure psig

Voltage Drop volts

Electrical Current amps

0.0000 0.0105 0.0354 0.0794 0.1771 0.2998

1.1003 0.8898 0.7650 0.3206 0.1227 0.0000

38.4 67.5 88.1 78.2 85.6 78.4

7.7 13.5 17.6 15.6 17.1 15.7

1.20 2.10 2.80 4.56 8.67 30.00

0.01 0.01 0.01 0.01 0.01 0.01

Draw the relative permeability curve Solution: 1.

Calculate absolute permeability using data with core 100% saturated with water

k

=

qwµwL A∆p

k

=

1.1003 1.07 2.30 2.688 38.4 - 7.7 14.696

k

=

0.482 darcy

VI - 19

2.

Calculate effective permeabilities to oil and water qoµoL ko

=

A∆P

ko

=

0.0105 5.50 2.30 2.688 67.5 - 13.5 / 14.696

ko

=

0.0134 darcy qwµwL

kw

3.

=

A∆P

kw

=

0.8898 1.07 2.30 2.688 67.5 - 13.5 / 14.696

kw

=

0.2217 darcy

Calculate relative permeabilities k kro = o = .0134 = 0.028 k .482 k krw = w = .2217 = 0.460 k .482

4.

Calculate water saturations

5.

Sw

=

Eo 1/2 Et

Sw

=

1.20 1/2 = .756 2.10

Results

Water Saturation Sw

Relative Permeability to oil kro

Relative Permeability to water krw

ko/kw

1.000 0.756 0.655 0.513 0.372 0.200

0.000 0.028 0.072 0.182 0.371 0.686

1.000 0.460 0.303 0.143 0.050 0.000

0.000 0.061 0.238 1.273 7.419 -------

VI - 20

1.0

Relative Permeability

0.8

0.6

Kro Krw

0.4

0.2

0.0 0

20

40

60

80

Sw, % pore space

Permeability Ratio, ko/kw

10

1

.1

.01

0

20

40

60

Sw, % of pore space

VI - 21

80

100

100

6.

The data permit certain checks to be made

0.62 φ-2.15

F

=

F

=

Ro Rw

Rw

=

12 ohm-m for 60,000 ppm brine

Ro

=

EA = 1.20 2.688 = 140 ohm-m IL .01 2.3

F

=

140 = 11.7 12

Φ

=

1 1 .62 2.15 = .62 2.15 F 11.7

Φ

=

.255

VI - 22

2.

Displacement (unsteady-state)(Welge) a.

does not result in relative permeability only give permeability ratio

b.

procedure

c.

1)

mount core in holder

2)

saturate with wetting phase (usually oil)

3)

inject non-wetting phase (usually gas) at constant inlet and outlet pressures

4)

measure a)

cumulative gas injected as a function of time

b)

cumulative oil produced as a function of time

conditions 1)

pressure drop across core high enough to make end effects negligible,but not enough to cause turbulent (non-darcy) flow.

2)

gas saturation can be described at mean pressure P + Po Pm = i 2

3) d.

flow is horizontal and core is short so that effects of gravity can be neglected

calculations 1)

convert gas injected into pore volumes

Gipv =

where Gi

Gipi LAφ pm = cumulative gas injected (measured at pressure pi), cc

Gipv = cumulative gas injected in pore volume pi

VI - 23

=

inlet pressure, psi

pm

=

LA φ = 2)

pore volume, cc

calculate average gas saturation, Sgav =

3)

pi + po , psi 2

Np

Sgav

LAφ

where Np

=

cumulative oil produced, cc

LA φ

=

pore volume, cc

plot Sgav vs G ipv

Sgav GAS BREAKTHROUGH

0 0

Gipv

VI - 24

4)

determine fractional flow of oil, fo fo =

d Sgav d Gipv

fo = slope of plot of Sgav vs qGipv 5)

calculate permeability ratio, kg/ko koA ∆p µoL fo = koA ∆p kgA ∆p + µoL µgL fo =

ko/µo ko/µo + kg/µg

kg 1 - fo = ko fo µo/µg where

kg = permeability ratio of gas to oil ko

fo = fractional flow of oil 6)

Permeability ratio, kg/ko, calculated above applies only at the gas saturation of the outflow face, thus must calculate Sgo Sgo = S gav - Gipvfo where

Sgo Gipv fo

VI - 25

=

gas saturation at outlet face of core

= cumulative gas injected, pore volumes =

fractional flow of oil at outlet face of core

e.

f.

advantages 1)

minimum amount of equipment

2)

rapid

disadvantages 1) 2)

k results in kg/ko, not kro and rg equations don't apply until gas breaks through, thus initial value of gas saturation may be high, resulting in incomplete kg/ko vs Sgo curve.

VI - 26

Example VI-2 The data from an unsteady-state displacement of oil by gas in a 2 inch diameter by 5 5/8 inch long sandstone core are given below.

Cumulative Gas Injection, Gi, cc

Cumulative Oil Produced, Np, cc

14.0 50.2 112.6 202.3 401.4 546.9 769.9 1226.5 3068.9 5946.6

14.6 19.5 22.5 25.5 28.6 30.4 32.2 33.4 35.3 35.9

Other data T = 70oF,

µo = 2.25 cp,

µg = .0185 cp

φ = .210, p inlet = 5.0 psig, p out = 0.0 psig L = 5 5/8 x 2.54 = 14.3 cm A = p (2.54)2 = 20.27 cm 2 G Prepare to determine kg/ko by calculating Sgav and ipv. Solution: 1.

Calculate Sgav Sgav = Sgav =

Np LAφ 14.6 cc 14.3 cm 20.27 cm2 .210

Sgav = 0.24 2.

G Calculate ipv

VI - 27

Gipv = Gipv =

Gipi LAφ pm 14.0 cc 19.7 psia 14.3 cm 20.27 cm2 .210 19.7 psia + 14.7 psia /2

Gipv = 0.264 pv 3.

Results Sgav 0.24 0.32 0.37 0.42 0.47 0.50 0.53 0.55 0.58 0.59

Gipv pv 0.264 0.945 2.12 3.81 7.56 10.3 14.5 23.1 57.8 112.0

VI - 28

Example VI-3 A core sample initially saturated with oil is flooded with gas. The following data was obtained: Gipv pv

Sgav 0.24 0.32 0.37 0.42 0.47 0.50 0.53 0.55 0.58 0.59

0.264 0.945 2.12 3.81 7.56 10.3 14.5 23.1 57.8 112.0

µo = 2.25 cp µg = 0.0185 cp S Calculate and construct a fg verses Sgo plot. Convert gavg to Sgo. Determine kg/ko for each of the given saturations. Construct a graph of kg/ko versus S go. Solution: Plot Sgav vs. Gipv The slope from this plot is fo. Sgo = S gav - fogipv kg/ko =

1 - fo µ fo o µo

Sgav

Gipv pv

0.24 0.32 0.37 0.42 0.47 0.50 0.53 0.55 0.58 0.59

.264 .94 2.17 3.81 7.56 10.3 14.5 23.1 57.8 112.0

fo .375 .075 .0357 .0214 .0118 .0092 .0046 .0013 .0005 .0001

VI - 29

Sgo .141 .249 .294 .338 .381 .405 .463 .521 .550 .581

kg/ko .0137 .101 .222 .376 .689 .886 1.78 6.32 16.4 82.2

0.6

0.5

Sgav

0.4

fraction

0.3

0.2 0

20

40

60

80

100

120

Gipv, pv 100

10

kg/ko

1

.1

.01 0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Sg, %

VI - 30

0.8

0.9

1.0

C.

Field determination of permeability ratios 1.

equations kg A ∆p µgL

qg qo = k A ∆p o µoL where qg qo

= gas flow rate measured at reservoir conditions, vol/time = oil flow rate measured at reservoir conditions, vol/time

thus, kg qg µg = ko qo µo replace qg/qo with qg Bg Rp - Rs qo = 5.615 B o where Bg

=

formation volume factor of gas, res cu ft/scf

Bo

=

formation volume factor of oil, res bbl/STB

Rp

= producing gas-oil ratio, scf/STB must include both separator gas and stock tank gas)

thus kg µgBg Rp - Rs = ko µo 5.615 Bo 2.

procedure a.

producing gas-oil ratio, Rp, and physical properties, Bg, Bo, R s, µg, µo must be determined at some known reservoir pressure

b.

saturations in reservoir, Sg or So, must be calculated from production data and material balance calculations

VI - 31

Example VI-4

Discovery pressure for your well was 4250 psia, temperature is 200oF, and initial producing gasoil ratio was 740 SCF/STB. Stock tank oil gravity is 30oAPI and surface gas gravity is 0.7. Production history and correlations indicate the bubble point at 3500 psia. Reservoir pressure is now 3000 psia. Producing gas-oil ratio is 18,100 SCF/STB. What is kg/ko in the reservoir at this time. Solution: Correlations covered in the fluid properties portion of this course yield the following value of the physical properties of the gas and oil at 3000 psia and 200o F. Rs

=

560 SCF/STB

Bo

=

1.314 res. BBL/STB

Tpc of gas

= 390 oR

Ppc of gas

= 665 psia

z

=

0.86

Bg

=

0.0282 z T/p = 5.34 x 103 res cu ft/SCF

µg

=

0.0192 cp

µo

=

0.75 cp

kg ko

=

µgBg Rp - Rs µo 5.615 Bo

kg ko

=

0.0192 5.34 x 10 -3 18100 - 560 0.75 5.615 1.314

kg ko

=

0.325

VI - 32

VI)

Uses of relative permeability data A.

Determination of free water surface in reservoir (100% water production)

1

kr

Log Response Diagram

0 0

Sw, %

100

SP Log

h, ft 100% Water Production

100 % Sw

0 0

Sw, %

100

VI - 33

RT Log

B.

Determination of height of 100% oil production

1

kr

0 0

Sw, %

Log Response Diagram

100

SP Log

100% Oil Production

100% Water Production h, ft

100 % Sw

0 0

Sw, %

100

VI - 34

RT Log

C.

Effect of permeability on thickness of transition zone 1

Low K

kr

High K

0 0

Sw, %

100

h = height of zone of interest

h, ft

Low K High K

0 0

Sw, %

VI - 35

100

D.

Fractional flow of water as a function of height

kw A ∆ P µw L

q q 1 fw = q w = q +wq = = tot o w ko A ∆ P kw A ∆ P ko µw 1+ + kw µo µo L µo L 1

fw

100

0 0

Sw

100 h

100

0

h

0 0

Sw

100

VI - 36

0 fw

1

160

10 md 140

Height above free water level, ft

120 100 80

50 md 60 40

100 md 20

200 md

0 0

20

40

60

80

100

Fraction of water in produced fluid, %

This figure indicates that lower permeabilities result in longer transition zones E.

Determination of residual fluid saturations 1

Oil Water

kr

0 0

100

Sw Residual Oil Saturation

1.

Imbibition curve used in water flood calculations

2.

Maximum oil recovery = area (acre) x h(ft) x f x 7758 BBL/acre ft x ∆Sw

VI - 37

F.

Interpretation of fractional flow curve

1

1 4

fw

2

0 0

Sw

3

100

1.

fw at water breakthrough

2.

Sw at well at water breakthrough

3.

Swav in reservoir between wells at water breakthrough

4.

1 = pore volume of water injected slope

VI - 38

1

3

2 Water input Pore vols.

fw

1

0

0 0

100 Oil Rec - % Oil in Place

VI - 39

VII. STATISTICAL MEASURES I)

Introduction Usually we can not examine an entire "population" (i.e. we can not dig up an entire reservoir, cut it into plugs, and measure the porosity of every plug). We can only "sample" the population and use the properties of the sample to represent the properties of the population. Often we seek a single number (porosity or permeability) to represent the population (reservoir) for use in reservoir engineering calculations.

If the sample is representative of the population, we have a statistical basis for estimating properties of the population.

The sample data is said to be unclassified or classified depending on whether it is arranged or grouped in a particular order. Unclassified data is randomly arranged. The classification of data for a large number of samples will often provide additional information to help describe the physical properties of the population.

VII - 1

II)

Frequency Distributions

It is often useful to distribute data into classes. The number of individuals belonging to each class is called the class frequency. A tabular arrangement of these data according to class is called a frequency distribution or frequency table. Sometimes classified data is called grouped data.

The division of unclassified data into classified data is accomplished by allocating all data to respective class intervals. The midpoint of each class interval is called the class mark. Rules for forming frequency distributions

A.

Determine largest and smallest numbers in the raw data.

B.

Divide the range of numbers into a convenient number of equal sized class intervals. The number of class intervals depends on the data but is usually taken between 5 and 20 in number.

C.

The number of observations for each class interval is the class frequency.

D.

The relative frequency of a class is the frequency of the class divided by the total frequency of all the classes.

VII - 2

Histogram A histogram is a graphical representation of a frequency distribution.

Frequency of Occurence

The vertical scale is the number of data points - the class frequency - in each class. The width of the rectangle corresponds to the class interval.

Mean

Magnitude of Variable

8

# of Samples

6

4

2

0

12

16

20

24

Porosity, %

8

6

# of Samples

III)

4

2

0 20

60

100

Permeability, md

VII - 3

140

Net pay thickness data from 20 wells summarized as relative frequency data

Frequency (No. of wells having thickness values in the range)

Range of thickness, ft. 50-80 81-110 111-140 141-170 171-200

Relative Frequency (No. of wells having thickness values in each range, fraction of total wells)

4 7 5 3 1 20

0.20 0.35 0.25 0.15 0.05 1.00

10

Frequency

8 6 4 2

0 0

50

80

110

140

170

200

Random variable: net pay thickness, ft

VII - 4

Relative Frequency as percentage. 20% 35% 25% 15% 5% 100%

Sometimes the relative frequency is plotted on a histogram

Relative Frequency

.5 .4 .3 .2 .1 0 0

50

80

110

140

170

200

Random variable: net pay thickness, ft

VII - 5

IV)

Cumulative Frequency Distributions Relative frequencies are summed and plotted at the higher ends of the class intervals

1.0 .8 .6 .4 .2 0

Cumulative % less than or equal to

Cumulative frequency less than for equal to

to create a "cumulative frequency less than or equal to" distribution

100% 80% 60% 40% 20% 0 0

50

80

110

140

170

200

Random variable: net pay thickness, ft

VII - 6

Occasionally a "cumulative frequency greater than or equal to" distribution is plotted. Relative frequencies are summed from the highest class interval and plotted

1.0 .8 .6 .4 .2 0

Cumulative % greater than or equal to

Cumulative frequency greater than for equal to

at the lower ends of the intervals

100% 80% 60% 40% 20% 0 0

50

80

110

140

170

200

Random variable: net pay thickness, ft

Probability graph paper has been constructed so that data from certain probability distributions plot as a straight line. Different probability paper is used for data with different distributions

VII - 7

V)

Normal Distribution The normal distribution is continuous probability distribution having a symmetrical shape similar to a bell, sometimes called a Gaussian distribution.

f(x) Inflection point of curve a

µ −a

a

µ +a

Random variable x

This distribution is completely and uniquely defined by two values - the mean, m, and standard deviation, σ.

VII - 8

VI)

Log Normal Distribution The log normal distribution is a continuous probability distribution that appears similar to a normal distribution except that it is skewed to one side. It is also called an exponential distribution.

Mode f(x)

Median (geometric mean) Mean (arithmetic mean)

Random variable x

This distribution can also be completely and uniquely defined by the mean, m, and the standard deviation, σ. If random variable xi are log normally distributed then the variables log xi are normally distributed.

VII - 9

VII)

Measures of Central Tendency An average is a value which is typical or representative of a set of data. When a set of data is arranged according to magnitude the average value tends to lie in the center of these data. These averages are called measure of central tendency. mean - the arithmetic average value of the samples n Σ xi µ = i =1 n where xi = values of the variable of interest for each sample nµ = number of samples median - the value equalled or exceeded by exactly one-half of the samples. mode - the value which occurs with the greatest frequency geometric mean - the nth root of the product of n numbers µg = x1⋅ x2⋅ x3 . . . x n 1/n n 1/n µg = π xi i=1 where µg = the geometric mean

VII - 10

VIII) Measures of Variability (dispersion) A measure of central tendency is the "average" or expected value of a set of variables, however it does not show the spread or variability of the variables on either side of the central tendency. A.

Standard deviation - The square root of the mean of the squared deviations about µ, where deviation is defined as the distance of the variable from µ. n Σ xi - µ 2 i=1 2 σ = n-1 where σ2 is the variance σ is the standard deviation

B.

Mean deviation - another measure of the dispersion about the central tendency n Σ xi - µ MD = i=1 n For classified data σ2 =

where

Σ fi xi - µ 2 j Σ fi j fi

=

frequency for each class

xj

=

class mark

or σ2 = Σ frj xj - µ 2 j f where rj = relative frequency for each class

VII - 11

IX)

Normal Distribution Porosity data is usually assumed to have a normal distribution. For the normal distribution the mean, median and mode have the same numerical values. They are identical measures of central tendency. Thus, for unclassified data n Σ xi µ = i=1n where i refers to each individual data point and, for classified data

µ=

Σ fj xj j Σ fj j

where j refers to each class interval fj is the frequency of the class xj is the class mark or µ = Σ frj xj j f where rj is the relative frequency of the class. xj is the class mark

VII - 12

.999 Cumulative Frequency

Cumulative Frequency

.999

.001

.001 Random variable x

Random variable x

Cumulative frequency plotted on coordinate graph paper

Cumulative frequency plotted on normal probability paper

Normal probability graph paper .999 Random variable x, distributed normally

Random variable x, distributed normally

.999

µ

.001

µ+a σ µ

.001 50%

50%

Cumulative %