reservoir petrophysics[1]

Course Notes for

Petroleum Engineering 311

Reservoir Petrophysics

Authors:

1980 — Von Gonten, W.D. 1986 — McCain, W.D., Jr. 1990 — Wu, C.H.

PETROLEUM ENGINEERING 311

RESERVOIR PETROPHYSICS

CLASS NOTES (1992)

Instructor/Author:

Ching H. Wu

DEPARTMENT OF PETROLEUM ENGINEERING

TEXAS A&M UNIVERSITYCOLLEGE STATION, TEXAS

ii

TABLE OF CONTENTS

I. ROCK POROSITY I-1

I) Definition I-1II) Classification I-1

III) Range of values of porosity I-2VI) Factors affecting porosity I-3V) Measurement of porosity I-5

VI) Subsurface measurement of porosity I-13VII) Compressibility of porous rocks I-25

II. SINGLE PHASE FLOW IN POROUS ROCK II-1

I) Darcy's equation II-11II) Reservoir systems II-15

III. BOUNDARY TENSION AND CAPILLARY PRESSURE III-11

I) Boundary tension III-1II) Wettability III-3

III) Capillary pressure III-5IV) Relationship between capillary pressure and saturation III-13V) Relationship between capillary pressure and saturation history III-14

VI) Capillary pressure in reservoir rock III-17VII) Laboratory measurement of capillary pressure III-19

VIII) Converting laboratory data to reservoir conditions III-25IX) Determining water saturation in reservoir from capillary pressure data III-27X) Capillary pressure variation III-29

XI) Averaging capillary pressure data III-31

IV. FLUID SATURATIONS IV-1

I) Basic concepts of hydrocarbon accumulation IV-1II) Methods for determining fluid saturations IV-1

V. ELECTRICAL PROPERTIES OF ROCK-FLUID SYSTEMS V-1

I) Electrical conductivity of fluid saturated rock V-1II) Use of electrical Formation Resistivity Factor, Cementation Factor, and

Saturation Exponent V-8III) Laboratory measurement of electrical properties of rock V-9IV) Effect of clay on resistivity V-18

VI. MULTIPHASE FLOW IN POROUS ROCK VI-1

I) Effective permeability VI-1II) Relative permeability VI-2

III) Typical relative permeability curves VI-2IV) Permeability ratio (relative permeability ratio) VI-14V) Measurement of relative permeability VI-14

VI) Uses of relative permeability data VI-33

iii

VII. STATISTICAL MEASURES VII-1

I) Introduction VII-1II) Frequency Distributions VII-2

III) Histogram VII-3IV) Cumulative Frequency Distributions VII-6V) Normal Distribution VII-8

VI) Log Normal Distribution VII-9VII) Measures of Central Tendency VII-10

VIII) Measures of Variability (dispersion) VII-11IX) Normal Distribution VII-12X) Log Normal Distribution VII-16

I - 1

I. ROCK POROSITY

I) Definition

A measure of the pore space available for the storage of fluids inrock

In general form:

Porosity = φ = VpVb

= Vb - VmVb

where:φ is expressed in fraction

Vb = Vp + VmVb = 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 processoccurring after deposition

C. Total Porosity

φt = total pore spaceVb

= Vb - VmVb

D. Effective Porosity

φe = interconnected pore spaceVb

1. Clean sandstones: φe = φt2. Carbonate, cemented sandstones: φe < φt

I - 3

VI) Factors affecting porosity

A. Factors:1. Particle shape2. Particle arrangement3. Particle size distribution4. Cementation5. Vugs and fractures

B. Particle shape

Porosity increases as particle uniformity decreases.

C. Packing Arrangement

Porosity decreases as compaction increases

60005000400030002000100000

10

20

30

40

50

DEPTH OF BURIAL, ft

POR

OSI

TY

, %

EFFECT OF NATURAL COMPACTION ON POROSITY

(FROM KRUMBEIN AND SLOSS.)

SANDSTONES

SHALES

I - 4

D. Particle Size Distribution

Porosity decreases as the range of particle size increases

GRAIN SIZE DIAMETER, MM

INTERSTITIAL MATERIALSAND MUD FRAGMENTS

FRAMEWORK FRACTION

CLEAN SAND

SHALY SAND

SAND SILT CLAY100

0

1.0 0.1 0.01 0.001

WE

IGH

T %

E. 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 materialShaly sand - has more interstitial material

F. 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 - 5

V) Measurement of porosity

φ = Vb - VmVb

= VpVb

Table of matrix densities

Lithology ρm (g/cm3)___________ ___________

Quartz 2.65

Limestone 2.71

Dolomite 2.87

A. Laboratory measurement

1. Conventional core analysis

a. measure any two

1) bulk volume, Vb2) matrix volume, Vm3) pore volume, Vp

b. bulk volume

1) calculate from dimensions2) displacement method

a) volumetric (measure volume)

(1) drop into liquid and observe volume chargeof liquid

(2) must prevent test liquid from entering poresspace of sample

(a) coat with paraffin(b) presaturate sample with test liquid(c) use mercury as test liquid

b) gravimetric (measure mass)

(1) Change in weight of immersed sample-prevent test liquid from entering pore space

(2) Change in weight of container and test fluidwhen sample is introduced

I - 6

c. matrix volume

1) assume grain density

Vm = dry weightmatrix density

2) displacement method

Reduce sample to particle size, then

a) volumetric

b) gravimetric

3) Boyle's Law: P1V1 = P2V2

a)

P(1)

V(1)

VALVE CLOSED

b) Put core in second chamber, evacuate

c) Open valve

P(2)

VALVE OPEN

CORE

V2 = Volumetric of first chamber &volume of second chamber-matrixvolume or core ( calculated)

VT = Volume of first chamber +volume second chamber (known)

4) Vm =VT - V2

I - 7

d. pore volume

1) gravimetric

Vp = saturated weight - dry weightdensity of saturated fluid

2) Boyle's Law: P1V1 = P2V2

a)

P(1)

V(1)

VALVE CLOSED

CORE

b) Put core in Hassler sleeve, evacuate

c) Open valve

P(2)

V(1)

VALVE OPEN

CORE

V2 = Volume of first chamber + porevolume of core (calculated)

3) Vp = V2 - V1

I - 8

2. Application to reservoir rocks

a. intergranular porosity(sandstone, some carbonates)

1) use representative plugs from whole core in laboratory measurements

2) don't use sidewall cores

b. secondary porosity(most carbonates)

1) use whole core in laboratory measurements

2) calculate bulk volume from measurements

3) determine matrix or pore volume fromBoyle's Law procedure

I - 9

Example I-1

A core sample coated with paraffin was immersed in a Russell tube. The dry sample weighed 20.0gm. The dry sample coated with paraffin weighed 20.9 gm. The paraffin coated sample displaced10.9 cc of liquid. Assume the density of solid paraffin is 0.9 gm/cc. What is the bulk volume ofthe 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, andimmersed in a Russell tube. The volume of the grains was 7.7 cc. What was the porosity of thesample? Is this effective or total porosity.

Solution:

Bulk Volume = 9.9 cc

Matrix Volume = 7.7 cc

φ = Vb - Vm

Vb = 9.9 cc- 7.7 cc

9.9 cc = 0.22

It is total porosity.

I - 10

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 inwater = 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 waterdensity of water

Vb = 448.6 gm - 269.6 gm

1 gm/cm3

Vb = 179.0 cm3

φ = VpVb

= 21.3 cm3179.0 cm3

= .119

φ = 11.9%

I - 11

What is the lithology of the sample?

Vm = Vb - Vp

Vm = 179.0 cm3 - 21.3 cm3 = 157.7 cm3

ρm = wt. of dry sample = 427.3 gm = 2.71 gm/(cm3)

matrix vol. 157.7 cm3

The lithology is limestone.

Is the porosity effective or total? Why?

Effective, because fluid was forced into the pore space.

I - 12

Example I-4

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

Solution:

P1V

1= P

2V

2

V1

= 1,000 cc

P1

= 50 psig + 14.7 psia = 64.7 psia

P2

= 28.1 psig + 14.7 = 42.8 psia

V2

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

V2

= 1,512 cc

Vm

= VT - V2

Vm

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

φ =

VT - VmVT

= 695 cc - 488 cc695 cc

= .298 = 29.8%

I - 13

VI) Subsurface measurement of porosity

A. Types of logs from which porosity can be derived

1. Density log:

φd = ρm - ρLρm - ρf

2. Sonic log:

φs = ∆tL - ∆tm∆tf - ∆tm

3. Neutron log:

e-kφ = CNf

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

Lithology ∆tm µsec/ft ρm 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

Water 189.0 1.00

I - 14

B. Density Log

1. Measures bulk density of formation

FORMATION

GAMMA RAYSOURCE

SHORT SPACEDETECTOR

LONG SPACEDETECTOR

MUD CAKE

2. Gamma rays are stopped by electrons - the denser the rock the fewer gammarays reach the detector

3. Equation

ρL = ρm 1 - φ + ρf φ


I - 15

FORMATION DENSITY LOG

4240

4220

4200

4180

4160

4140

4120

4100

20016012080400 3.02.82.62.42.22.0

ρ, gm/ccGR, API depth, ft

I - 16

Example I-5

Use the density log to calculate the porosity for the following intervals assuming ρmatrix = 2.68

gm/cc and ρfluid = 1.0 gm/cc.

Interval, ft ρL, gm/cc

φd ,%

__________ _________ ______

4143-4157 2.375 184170-4178 2.350 204178-4185 2.430 154185-4190 2.400 174197-4205 2.680 04210-4217 2.450 14

Example:

Interval 4,143 ft -4,157 ft :

ρL

= 2.375 gm/cc


= 2.68 gm/cc - 2.375 gm/cc2.68 gm/cc - 1.0 gm/cc

= 0.18

I - 17

C. Sonic Log

1. Measures time required for compressional sound waves to travel throughone foot of formation

D

B

AT

R1

R2E

C

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

3. Equation

∆tL = ∆tm 1 - φ + ∆tf φ (Wylie Time Average Equation)

I - 18

SONIC LOG

4240

4220

4200

4180

4160

4140

4120

4100

2001000 140 120 100 80 60 40

GR, API ∆T, µ seconds/ftdepth, ft

I - 19

Example I-6

Use the Sonic log and assume sandstone lithology to calculate the porosity for the followingintervals.

Interval ∆tL φs ,%

(ft) µ second/ft

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 = 86.5 µ-sec/ft


= 86.5 µ sec/ft- 51.6 µ sec/ft189.0 µ sec/ft- 51.6 µ sec/ft

= 0.25

I - 20

D. Neutron Log

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

MaximumAverage EnergyNumber Loss/ Atomic Atomic

Element Collisions Collision, % Collision Number

Calcium 371 8 40.1 20Chlorine 316 10 35.5 17Silicone 261 12 28.1 14Oxygen 150 21 16.0 8Carbon 115 28 12.0 6Hydrogen 18 100 1.0 1

.1 1 10 102 103

O

104

Si

105

H

106 107

1

10

102

103

CLEAN SAND POROSITY = 15%

NEUTRON ENERGY IN ELECTRON VOLTS

REL

ATIV

E PR

OBA

BIL

ITY

FOR C

OLL

ISIO

N

.1 1 10 102 103 104 105

H

106

O

107

Si10-3

10-2

10-1

1

CLEAN SAND POROSITY = 15%

SLO

WIN

G D

OW

N P

OW

ER

NEUTRON ENERGY IN ELECTRON VOLTS

2. In clean, liquid filled formations, hydrogen index is directly proportional toporosity. 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 - 21

NEUTRON DENSITY LOG

4240

4220

4200

4180

4160

4140

4120

4100

2000 30 -10

GR, API φ (CDL)depth, ft

I - 22

Example I-7

Use the neutron log to determine porosity for the following intervals.

Solution:

Interval φ n (ft) (%) .

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 - 23

Example I-8

Calculate the porosity and lithology of the Polar No. 1 drilled in Lake Maracaibo. The depth ofinterest is 13,743 feet. A density log and a sonic log were run in the well in addition to thestandard Induction Electric Survey (IES) survey.

The readings at 13,743 feet are:

bulk density = 2.522 gm/cctravel time = 62.73 µ-sec/ft

Solution:

Assume fresh water in pores.

Assume sandstone:

ρm = 2.65 gm/cc

∆tm = 55.5 µ-sec/ft



= 7.76%


= 62.73 µ sec/ft- 55.5 µ sec/ft189.0 µ sec/ft - 55.5 µ sec/ft

= 5.42%

Assume limestone:

ρm = 2.71 gm/cc




= 10.99%


= 62.73 µ sec/ft - 47.5 µ sec/ft189.0 µ sec/ft - 47.5 µ sec/ft

= 10.76%

I - 24

Assume dolomite:

ρm = 2.87 gm/cc




= 18.619%


= 62.73 µ sec/ft - 43.5 µ sec/ft189.0 µ sec/ft - 43.5 µ sec/ft

= 13.22%

φlimestone = 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 - 25

VII) Compressibility of porous rocks

Compressibility, c is the fractional change in volume per unit change in pressure:

c = - 1V

∂V∂P T

≅ -

∆VV T∆P

A. Normally pressured reservoirs

1. Downward force by the overburden must be balanced by upward force ofthe matrix and the fluid

Fm Ff

Fo

2. Thus,

Fo = Fm + Ff

it follows that

Po = pm + pf

3. Po ≅ 1.0 psi/ft

Pf ≅ 0.465 psi/ft

I - 26

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

a. the force on the matrix increasesb. causing a decrease in bulk volumec. and a decrease in pore volume

B. Types of compressibility

1. Matrix Compressibility, cm

cm ≅ 0

2. Bulk Compressibility cb

used in subsidence studies

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

a. important to reservoir engineers

1) depletion of fluid from pore spaces2) internal rock stress changes3) change in stress results in change in

Vp, Vm, Vb

4) by definition

cf = - 1Vp

∂Vp∂pm

b. since overburden pressure, Po, is constant

dPm = - dPf

I - 27

1) Thus,

cf = - 1Vp

∂Vp∂pm

2) where the subscript of f on cf means"formation" and the subscript of f on Pfmeans "fluid"

3) procedure

(a) measure volume of liquid expelled as afunction of "external" pressure

(b) "external" pressure may be taken torepresent overburden pressure, Po

(c) fluid pressure, pf, is essentially constant, thus,

dPo = dPm

(d) expelled volume increases as porevolume, vp, decreases, thus,

dVp = - dVexpelled

(e) from definition

cf = - 1Vp

∂Vp∂pm

it follows that

cf = + 1Vp

∆ Vp expelled

∆Po

I - 28

(f) plot

CU

MU

LATIV

E V

OLU

ME

EXPE

LLED

PO

RE V

OLU

ME

OVERBURDEN PRESSURE, psi

slope = cf.

I - 29

C. Measurement of compressibility

1) Laboratory core sample

a) apply variable internal and external pressures

b) internal rock volume changes

2) Equipment

InternalPressureGauge

HydraulicPump

OverburdenPressureGauge

HydraulicPump

Copper - JacketedCore

Mercury Sight Gauge

Apparatus for measuring pore volume compressibility (hydrostatic)

I - 30

Example I-9

Given the following lab data, calculate the pore volume compressibility for a sandstone sample at4,000 and 6,000 psi.

pore volume = 50.0 cc

pressure, psi vol. fluid expelled, cc

1000 0.244 2000 0.324 3000 0.392 4000 0.448 5000 0.500 6000 0.546 7000 0.596 8000 0.630

Solution:

from graph

@ 4,000 psi:

Slope = 0.0094000 psi

cf = 2.25 X 10-6 1psi

@ 6000 psi:

Slope = 0.0116000 psi

cf = 1.83 X 10-6 1psi

I - 31

10000800060004000200000.000

0.005

0.010

0.015

COMPACTION PRESSURE, psi

VO

LU

ME

EX

PEL

LE

D, c

cPO

RE

VO

LU

ME

, cc

I - 32

INITIAL POROSITY AT ZERO NET PRESSURE, %

3025201510501

10

100

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

CONSOLIDATED SANDSTONES

HALL'S CORRELATION

PO

RE

VO

LU

ME

CO

MP

RE

SS

IBIL

ITY

X 1

0-6

psi-

1P

OR

E V

OL

UM

E C

OM

PR

ES

SIB

ILIT

Y X

10-

6 ps

i-1

3025201510501

10

100

PORE-VOLUME COMPRESSIBILITY AT 75 %LITHOSTATIC PRESSURE VS INITIAL SAMPLEPOROSITY FOR UNCONSOLIDATED SANDSTONES.

UNCONSOLIDATED SANDSTONES

HALL'S CORRELATION

INITIAL POROSITY AT ZERO NET PRESSURE, %

I - 33

E. Abnormally pressured reservoirs

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

PRESSURED

EPTH

NORMAL LINEAR

SUBNORMAL(LOWER)

SURNORMAL(GREATER)

I - 34

Compressibility/Porosity Problem No. 1

A limestone sample weighs 241.0 gm. The limestone sample coated with paraffin was found toweigh 249.5 gm. The coated sample when immersed in a partially filled graduated cylinderdisplaced 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. dryρls

= 241.0 gm2.71 gm/cc

= 88.9 cc

Vparaffin = wt. coated sample - st. uncoated sampleρ

Vparaffin = 249.5 gm - 241.0 gm

0.90 gm/cc = 9.4 cc

Vb = 125 cc - 9.4 cc = 115.6 cc

Vp = Vb - Vm

Vp = 115.6 cc - 88.9 cc - 26.7 cc

φ = VpVb

= 26.7 cc115.6 cc

= 0.231

φ = 23.1% (total porosity)

I - 35


You are furnished with the results of a sieve analysis of a core from Pete well #1. Previouslaboratory work indicates there is a correlation between grain size and porosity displayed by thoseparticular 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 remaining pore matrix

component volume porosity volume (ft3) (%) (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 - 36


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 2.40gm/cc.

The Highgrade #1 Well was drilled and cored through the reservoir. A rock sample was sent to thelaboratory and the following tests were run.

pressure cum. pore vol. change (psig) (-cc)_________ 1,000 0.122 2,000 0.162 3,000 0.196 4,000 0.224 5,000 0.250 6,000 0.273 7,000 0.298 8,000 0.315

The dry weight of the core sample was found to be 140 gm while the sample dimensions were1.575 inches long and 1.960 inches in diameter.

Assuming the compressibility at 4,500 psi is the average compressibility in the reservoir, howmuch 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

φ = ρm - ρLρm - ρf

= 2.65 - 2.402.65 - 1.00

= 15.22%

I - 37

B. 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 = bulk volume = πD2h4

= 3.14 5.0 2 4.0

4.0 = 78.5 cc

Vm = matrix volume = 140 gm cc2.65 gm

= 52.8 cc

Vp = Vb - Vm = 78.5 cc - 52.8 cc

Vp = 25.7 cc

C. Compressibility (see graph)

Vp = 25.7 cc

D. 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 usuallyencountered. An experienced engineer would be wary of this small number. Also it wasassumed that the formation compressibility was exactly the same as the bulk volumecompressibility. Experience shows that this is not the case.

I - 38

8000600040002000000.0040

0.0060

0.0080

0.0100

0.0120

0.0140

POROSITY PROBLEM No. 3

PRESSURE, psig

SLOPE =.0118 - .0068

7000 - 2000

VO

LU

ME

EX

PEL

LE

D, c

cPO

RE

VO

LU

ME

, cc Cp = 9.96 x 10-7 psi -1

I - 39

Compressibility Problem

A 160-acre and 100 ft thick reservoir has a porosity of 11%. The pore compressibility is 5.0 x 10-6 (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

Cp = - 1Vp

dVpdp

5 x 10-6 (1/psi) = -176,665,600 ft3

dVp

3,000 psi

dVp = 1.15 x 106 ft3

∆H = 1.15 x 106 ft3 x 16,969,600 ft2

= 0.165 ft

II - 1

II. SINGLE PHASE FLOW IN POROUS ROCK

I) Darcy's equation (1856)

A. Water flow through sand filters

A

Z

WATER

DARCY'S FOUNTAIN.

SAND

q

q

h1 - h2

h1

h2

q = kA(h1 - h2)µL

Length of sand pack,L = Z

1. constant of proportionality, k, characteristics of particular sand pack, notsample 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 - 2

q = kA(h1 - h2)µL

B. Generalized form of Darcy's equation

1. Equation

vs = -kµ

dPds

- ρg

1.0133 x 106 dzds

-1

+1 90o 180o 270o 360o

Θ

s

Vs

+X

+Y

-Z

+Z

θ

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 = pressure gradient along s - atm/cmds

µ = viscosity - centipoise

k = permeability - darcies

II - 3

3. 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

4. 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 = - kµ

dpds

- ρg

1.0133 x 106 dzds

Lt = - k

m/Lt m/Lt2

L -

m/L3 L/t2 LL

k = L2 = cross-sectional area

II - 4

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 onecentipoise that completely fills the voids of the medium will flow through it under conditions ofviscous flow at a rate of one cubic centimeter per second per square centimeter cross-sectional areaunder a pressure or equivalent hydraulic gradient of one atmosphere per centimeter.

q = k A P1 - P2µL

II - 5

II) Reservoir systems

A. Flow of incompressible liquid

1. Horizontal, linear flow system

L

AqP1

qP2

a. Conditions

1) horizontal system, dzds

= 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 - 6

b. derivation of flow equation

vs = - kµ

dPds

- ρg

1.0133 x 106 dzds

vs = - kµ

dPds

= qA

q ds0

L = - kA

µ dP

p1

p2

q L - 0 = - kAµ

P2 - P1

q = kALµ

P2 - P1

Note: P1 acts at L = 0

P2 acts at L = L

q is + if flow is from L = 0 to L = L

II - 7

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 cm2

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

P1 = 5.0 atm

P2 = 2.0 atm

q = kALµ

P2 - P1

q = (1) (5,572.8 ) (5.0 - 2.0) (1) (182.88)

q = 91.42 cm3 / sec

II - 8

2. Non-horizontal, linear system

Θ

-Z

P1

S

X

P2

a. Conditions

1) non-horizontal system, dzds

= sinθ = constant



4) laminar flow, use Darcy equation



7) constant temperature µ, q

II - 9

b. derivation of equation

vs = - kµ

dPds

- ρg

1.0133 x 106 dzds

vs = - qA

= - kµ

dPds

+ k ρg sin θ

µ 1.0133 x 106

q ds0

L = - kA

µ dp

P1

P2 + kA ρg sin θ

µ 1.0133 x 106 ds

0

L

q = - kAµL

P1 - P2 + ρgLsinθ

1.0133 x 106

II - 10

3. Vertical, upward flow, linear system

L

x

h

FLOW UNDERHEAD h

a. Conditions

1) vertical system, dzds = sinθ = constant

2) upward flow, q = 270°, sinθ = - 1






8) constant temperature, µ

II - 11

b. derivation of flow equation

vs = kµ

dPds

- ρg

1.0133 x 106 dzds

vs = qA

= - kµ

dPds

+ ρg

1.0133 x 106

q = kAµ

P1 - P2L

- ρg

1.0133 x 106

P1 = - ρg (h + x + L)

1.0133 x 106

P2 = ρg x

1.0133 x 106

P1 - P2L

= ρg h

1.0133 x 106 L +

ρg

1.0133 x 106

q = kAµ

ρg h

1.0133 x 106 L +

ρg

1.0133 x 106 -

ρg

1.0133 x 106

q = kAµL

ρg h

1.0133 x 106

II - 12

4. Horizontal, radial flow system

h

rw

re

Pe

Pw

re rw

a. Conditions

1) horizontal system, dzds = 0

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

3) constant thickness, h = constant




7) 100% saturated with liquid,

8) constant temperature, µ, q

II - 13

b. Derivation of flow Equation

vs = - kµ

dPds

- ρg

1.0133 x 106 dzds

vs = + kµ

dPdr

= qA

= q

2πrh

q2πh

drr

rw

re = k

µ dp

pw

pe

q2πh

1n(re) - 1n( rw) = kµ

Pe - Pw

q = 2πhkµ 1n (re/rw)

Pe - Pw

Note: if q is + , flow is from re to rw

B. Flow of gas (compressible fluid)

1. horizontal, linear flow system

L

Aq

P1

q

P2

a. Conditions

1) horizontal system, dzds = 0


3) compressible gas flow, q = f(p)


5) non-reactive fluid, k= constant


7) constant temperature

II - 14

b. Assumptions

1) µ, Z = constant

2) Z(and µ ) can be determined at mean pressure

c. Derivation of equation for qsc

vs = - kµ

dPds

- ρg

1.0133 x 106 dzds

vs = - kµ

dPds

= qAds

but

q = Psc qscz T

PTsc

thus

Psc T qscTsc A

dso

L = - k PdP

µzp1

p2

Psc T qscTsc A

L -0 = - kµz

P2

2 - P12

2

qsc = kAµL

Tsc

Tz Psc

P12 - P2

2

2

Note: real gas equation of state

Pq = Z n R T

where q = volumetric flow/timen = mass flow/time

thus,Pq

Pscqsc = Z n R T

n R Tsc

q = Psc qscz T

Tsc 1P

where qsc is constant

Z is determined at P, T

II - 15

d. Derivation of equation for q

qsc = kAµL

Tsc

Tz Psc

P12 - P2

2

2

but

qsc = P q Tsc

Z Psc T = k A

µL

TscT z Psc

P12 - P22

2

q = k AµL

1P

P12 - P22

2

q = k AµL

2P1 + P2

P12 - P22

2

q = k AµL

P1 - P2

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 liquidcan be used for compressible gas!

Note: real gas equation of state

Pq = Z n R T

thus

Psc qscP q

= n R Tsc z n R T

where

P = P1 + P2

2

P = volumetric flow rate at P, T

z is determined at P, T

qsc = P q Tscz Psc T

II - 16

2. Horizontal, radial flow system

h

rw

re

Pe

Pw

re rw

a. Conditions

1) horizontal system dzds

= 0

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

3) constant thickness, h = constant

4) compressible gas flow, q = f (P)




8) constant temperature

II - 17

b. Assumptions

µz = constant

z (and µ ) can be determined at mean pressure

c. derivation of equation for qsc

vs = - kµ

dPds

- ρg

1.0133 x 106 dzds

vs = - kµ

dPds

= qA

but

q = Psc qsc z T

PTsc

and

A = 2πrh and ds = - dr

thus

Psc T qsc2Tsc π h

drr

rw

re = k

ρdPµz

Pw

Pe

PscT qsc2 Tsc π h

1n rerw

= kµz

Pe2 - Pw2

2

qsc = 2 π h kµ 1n re/rw

Tsc

Psc zT

Pe2 - Pw2

2

II - 18

d. derivation of equation for q

qsc = 2 π h kµ 1n re/rw

Tsc

Psc zT

Pe2 - Pw2

2

but

q = P q Tscz Psc T

thus

P q Tscz Psc T

= 2 π h kµ 1n re/rw

Tsc

Psc zT

Pe2 - Pw2

2

q = 2 π h kµ 1n re/rw

1P

(Pe2 - Pw2 )

2


2Pe + Pw

(Pe2 - Pw2 )

2


Pe - Pw

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

II - 19

C. Conversion to Oilfield Units

Symbol Darcy units Oil field

q cc/sec bbl/d or cu ft/dk darcy mdA sq cm sq fth cm ftP atm psiaL cm ftµ cp cpr gm/cc lb/cu ft

Example:

q = hkA P1 - P2

µ L in Darcy's units

q ccsec = q bbl

d 5.615 cu ft

bbl

1,728 cu incu ft

16.39 cccu in

d24hr

hr3,600 sec

q ccsec = 1.841 q bbl

d

k darcy = k md

darcy1,000md

k darcy = 0.001 k md

A sq cm = 929.0 sq cm

sq ft A sq ft

A sq cm = 929.0 A sq ft

P1 - P2 atm = P1 - P2 psia atm14.696 psia

P1 - P2 atm = 0.06805 P1 - P2 psia

L cm = L ft 30.48 cmft

meter = 100 cm

1.841 q = 0.001 k 929.0 A .06805 P1 - P2

µ 30.48 L

q = 0.01127 k A P1 - P2

µ L in oilfield units

II - 20

D. Table of Equations

1. Darcy Units

System Fluid Equation

Horizontal,Linear

IncompressibleLiquid

q = kAµ L

P1 - P2

Dipping,Linear

IncompressibleLiquid q = kA

µ L P1 - P2 +

ρ g L sin θ1.0133 x 106

Horizontal,Radial


q = 2 π k hµ ln (re/rw)

Pe - Pw

Horizontal,Linear

RealGas qsc = kA

µ L Tsc

Tz Psc

P12 - P2

2 2

q = kAµ L

P1 - P2

Horizontal,Radial

Real Gasqsc = π k h

µ ln (re/rw) Tsc

Tz Psc Pe

2 - Pw 2

q = 2 π k hµ ln (re/rw)

Pe - Pw

II - 21

2. Oilfield Units

System Fluid Equation

Horizontal,Linear


q = 0.001127 kAµL

P1 - P2

q = res bbl/d

Dipping,Linear


q = 0.001127 kAµL

P1 - P2

+ ρg L sinθ

1.0133 x 106

Horizontal,Radial


q = .007082 khµ ln (re/rw)

Pe - Pw

Horizontal,Linear Real Gas

qsc = .1118 k Aµ L z T

P1 2 - P2

2

qsc = scf/d

q = .001127 kAµL

P1 - P2

q = res bbl/d

Horizontal,Radial Real Gas

qsc = .7032 k hµ ln (re/rw) Tz

Pe 2 - Pw 2

q = .007082 khµ ln (re/rw)

Pe - Pw

II - 22

Example II-2

What is the flow rate of a horizontal rectangular system when the conditions are as follows:

permeability = k = 1 darcyarea = A = 6 ft2

viscosity = µ = 1.0 cplength = L = 6 ftinlet 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 = 1 darcy = 1,000 mdA = 6 ft2L = 6 ftP1 = (5.0 atm) (14.7 psi/atm) = 73.5 psiP2 = (2.0 atm) (14.7 psi/atm) = 29.4 psi

q = 1.1271 x 10-3 kAµL

P1 - P2

q = 1.1271 x 10-3 1,000 61 6

73.5 - 29.4

q = 49.7 bbl / day

II - 23

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 = 7.082 x 10-3 kH Pe - Pw

µ ln Re / Rw

q = 7.082 x 1--3 300 20 2,500 - 1,740

1.3 6.492

q = 3,826 res bbl/d

II - 24

E. Layered Systems

1. Horizontal, linear flow parallel to bedding

A

B

C

q q

P1 P2

L

W

qt = qA + qB + qC

h = hA + hB + hC

let k be "average" permeability,

then

qt = k wh P1 - P2µ L

and

qt = kA whAµ L

P1 - P2 + kB whBµ L

P1 - P2 + kC whCµ L

P1 - P2

then

k h = kA hA + kB hB + kC hC

k = ∑j = 1

n kj hj

h

II - 25

2. Horizontal, radial flow parallel to bedding

Pw

rw

re

qA

qB

qC

ht

hA

hB

hC

re

Pe

again

qt = qA + qB + qC

h = hA + hB + hC

qt = 2 π k hµ ln (re/rw)

Pe - Pw

and

qt = 2 π kA hAµ ln (re/rw)

Pe - Pw + 2 π kB hBµ ln (re/rw)

Pe - Pw

+ 2 π kc hcµ ln (re/rw)

Pe - Pw

then

k h = kA hA + kB hB + kC hC

and again

k = ∑j = 1

n kj hj

h

II - 26

3. Horizontal, linear flow perpendicular to bedding

L

h

P2P1

W

qkA∆PALA

A B C

kB∆PBLB

kC∆PCLC

q

qt = qA = qB = qC

p1 - p2 = ∆PA + ∆PB + ∆PC

L = LA + LB + LC

qt = k wh P1 - P2

µ L

and since P1 - P2 = ∆ PA + ∆PB + ∆PC

P1 - P2 = qt µ Lk wh

= qA µLA

kA wh +

qB µLBkB wh

+ qC µLCkC wh

since qt = qA = qB = qC

Lk

= LAkA

+ LBkB

+ LCkC

thusK = L

∑j = 1

n

Ljkj

II - 27

4. Horizontal, radial, flow perpendicular to bedding

h

rwrArBrC

q

PCPBPAPw

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)

2 π k h =

qA µ ln (rA/rw)2 π kAh

+ qB µ ln (rB/rA)

2 π kB h +

qC µ ln (rC/re)2 π kC h

then

k = ln re/rw

∑j = 1

n

ln(rj/rj-1)

kj

II - 28

Example II-4

Damaged zone near wellbore

k1 = 10 md r1 = 2 ft

k2 = 200 md r2 = 300 ft

rw = 0.25 ft

Solution:

k = ln (re/rw)

∑j = 1

n

ln (rj/rj-1 )

kj

k = ln 300

0.25

ln 2/0.25

10 +

ln 300/2200

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 - 29

F. Flow through channels and fractures

1. Flow through constant diameter channel

L

A

a. Poiseuille's Equation for viscous flow through capillary tubes

q = πr48 µ L

P1 - P2

A = π r2, therefore

q = Ar28 µ L

P1 - P2

b. Darcy's law for linear flow of liquids

q = kAµ L

P1 - P2

assuming these flow equations have consistent units

Ar28 µ L

P1 - P2 = kAµ L

P1 - P2

thus

k = r28

= d2

32

where d = inches, k = 20 x 109 d2 md

II - 30

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?

Solution:

A. k = 20 x 109 d2

k = 20 x 109 (0.0001 in)2

k = 200 md

B. k = 0.25 (200 md)

k = 50 md

II - 31

2. Flow through fractures

b

v = qA

= h212 µ L

(P1-P2)

q = b2 A 12 µ L

(P1 -P2)

setting this flow equation equal to Darcy's flow equation,

b2 A12 µ L

P1 - P2 = kAµ L

P1 - P2

solve for permeability of a fracture:

k = b2

12 in darcy units, or

k = 54 x 109 b2

where b = inchesk = md

II - 32

Example II-6

Consider a rock of very low matrix permeability, 0.01 md, which contains on the average afracture 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 theequation for parallel flow.

Solution:

k = ∑

kj Aj

A , similar to horizontal, linear flow parallel to fracture

k = matrix k matrix area + fracture k fracture area

total area

k = 0.01 12 in 2 + 12 in 0.005 in

144 in2 +

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 - 33

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

SLOPE = k / m

P12 - P 2

2

2L

qsc PscA

qsc = kA P1

2 - P2 2

2 µ L Psc horizontal, linear, real gas flow with

T = Tsc and Z = 1.0

qsc PscA

= kµ

P1

2 - P2 2

2L

k = ( slope ) m

II - 34

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 permplug method

TO FLOWMETER

HIGH AIRPRESSURE

PIPE

RUBBERTUBING

CORE

LOW AIRPRESSURE(FLOW)

VERTICAL FLOW

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

d. whole core method gives better results for limestones

II - 35

B. Factors which affect permeability measurement

1. Fractures - rocks which contain fractures in situ frequently separate alongthe planes of natural weakness when cored. Thus laboratory measurementsgive "matrix" permeability which is lower than in situ permeability becausetypically only the unfractured parts of the sample are analyzed forpermeability.

2. Gas slippage

a. gas molecules "slip" along the grain surfaces

b. occurs when diameter of the capillary openings approaches the meanfree 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 permeabilitymeasurements are a function of type of gas used in laboratorymeasurement.

II - 36

0

kCALCULATED

1P

H2

N2

CO2

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

h. permeability is a function of size of capillary opening, thusKlinkenberg effect is greater for low permeability rocks.

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

0

kMEASURED

1P

II - 37

Example II-7

Another core taken at 8815 feet from the Brazos County well was found to be very shaly. Therewas some question about what the true liquid permeability was, since nitrogen was used in thepermeameter.

Calculate the equivalent liquid permeability from the following data.

Mean MeasuredPressure Permeability ( atm ) ( md )

1.192 3.762.517 3.044.571 2.769.484 2.54

Solution:

Plot kmeasured vs. 1/pressure

Intercept is equivalent to liquid permeability

From graph:

kliq = 2.38 md

0

1

2

3

4

5

GA

S P

ERM

EA

BIL

ITY,

md

0.0 0.2 0.4 0.6 0.8 1.0

RECIPROCAL MEAN PRESSURE, atm-1

kgas = 2.38276 + 1.64632

Equivalent Liquid Permeability = 2.38 md

Pbar

II - 38

3. Reactive fluids

a. Formation water reacts with clays

1) lowers permeability to liquid

2) actual permeability to formation water is lower than lab permeabilityto gas

1000010001001011

10

100

1000

WA

TER P

ERM

EA

BIL

ITY,

md

AIR PERMEABILITY, md

Water concentration20,000 - 25,000 ppm Cl ion.

RELATIONSHIP OF PERMEABILITIES MEASUREDWITH AIR TO THOSE MEASURED WITH WATER

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

II - 39

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 34 4080 1445 1380 1290 1190 885 17.2S 34 24800 11800 10600 10000 9000 7400 147.0S 34 40100 23000 18600 15300 13800 8200 270.0

S 34 4850 1910 1430 925 736 326 5.0S 34 22800 13600 6150 4010 3490 1970 19.5S 34 34800 23600 7800 5460 5220 3860 9.9

S 34 13600 5160 4640 4200 4150 2790 197.0S 34 7640 1788 1840 2010 2540 2020 119.0T 36 2630 2180 2140 2080 2150 2010 1960.0

T 36 3340 2820 2730 2700 2690 2490 2460.0T 36 2640 2040 1920 1860 1860 1860 1550.0T 36 3360 2500 2400 2340 2340 2280 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.

III- 1

III. BOUNDARY TENSION AND CAPILLARY PRESSURE

I) 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

LIQUID

SURFACE

III- 2

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 cm2 of surface (ergs/cm2) is termed"boundary energy"

2. also can be thought of as force in dynes acting along length of 1 cmrequired 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 "surfacetension"

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

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

I. Forces creating boundary tension

1. Forces

a. Law of Universal Gravitation applied between molecules

b. physical attraction (repulsion) between molecules

2. Liquid-Gas Boundary

attraction between molecules is directly proportional to their masses andinversely proportional to the square of the distance between them

3. Solid-Liquid Boundary

physical attraction between molecules of liquid and solid surface

III- 3

4. Liquid-Liquid Boundary

some of each

II) Wettability

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

Θ

σow

σos σws

OILWATER

OIL

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- 4

D. if the solid is "oil-wet"

σos ≥ σws

AT = -

cos θ = -

90° ≤ θ ≤ 180°

if θ = 180° - strongly oil-wet

θ = 830θ = 1580

θ = 350θ = 300

(A)

ISOOCTANE ISOOCTANE + 5.7% ISOQUINOLINE

ISOQUINOLINE NAPHTHENIC ACID

θ = 300 θ = 480 θ = 540 θ = 1060

(B)

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

III- 5

III) Capillary pressure

A. capillary pressure between air and water

ΘhAIR

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 thecircumference of the water-air-solid interface

= 2πr AT

b. total force down equals the weight of the column of waterconverted to force

= πr2 hgρw

c. thus when column of water comes to equilibrium

2πr AT = πr2 hgρw

d. units

cm dynecm = cm2 cm cm

sec2 gm

cm3

dyne = gm cm

sec2

dyne = force unite. adhesion tension

AT = 12

r hgρw dynecm

III- 6

2. liquid will rise in the tube until the vertical component of surface tensionequals the total force down

a. vertical component of surface tension is the surface tensionbetween air and water multiplied by the cosine of the contact angleacting along the water-air-solid interface

= 2πr σaw cosθ

b. total force down

= πr2 hgρw

c. thus when the column of water comes to equilibrium

2πr σaw cosθ = πr2 hgρw

d. units

cm dynecm = cm2 cm cm

sec2

gm

cm3

cm dynecm

= cm gm

sec2

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

h = 2 σaw cos θ

rg ρw

III- 7

4. capillary pressure (air-water system)

Θh

WATER

PaA' A

AIR

Pa

Pw

B'

B

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 waterbetween A & B

pw = pa - ρwgh

units: dyne

cm2 =

dyne

cm2 -

gm ⋅ cm

cm3 ⋅ sec2 cm

c. thus between B' and B there is a pressure difference

pa - pw = pa - (pa - ρwgh)

pa - pw = ρwgh

d. call this pressure difference between B' and B "capillary pressure"Pc = pa - pw = ρwgh

e. remember

h = 2 σgw cos θ

rg ρw

III- 8

f. thus

Pc = 2 σgw cos θ

r

B. capillary pressure between oil and water

Θh

WATER

OIL

1. liquid will rise in the tube until the vertical component of surface tensionequals the total force down

a. vertical component of surface tension equals the surface tensionbetween oil and water multiplied by the cosine of the contact angleacting 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 ispartially offset (bouyed) by the weight of the column of oil outsidethe capillary

c. thus, total force down equals the weight of the column of waterminus 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- 9

3) net weight per unit area acting to pull surface down

= ρwh - ρoh = h(ρw - ρo)

4) total force down

= πr2 gh (ρw - ρo)

d. thus when the column of water comes toequilibrium

2πr σow cosθ = πr2 gh (ρw - ρo)

2. thus the equilibrium for the height of the column of water

h = 2 σow cos θrg (ρw - ρo)

3. capillary pressure (oil-water system)

Θh

WATER

PoA

Po

Pw

B'

B

OIL

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 waterbetween A and B

Pwb = Pwa - ρwgh

c. pressure at B' equal to the pressure at A' minus the head of oilbetween A' and B'

III- 10

Pob = Poa - ρogh

d. thus capillary pressure, the difference between pressure at B' andthe pressure at B is

Pc = Pob - Pwb

Pc = (Poa - ρogh) - (Pwa - ρwgh)

since Poa = Pwa

Pc = (ρw - ρo)gh

e. remember

h = 2 σow cos θrg (ρw - ρo)

f. thus

Pc = 2 σow cos θ

r

4. same expression as for the air-solid system except for the boundarytension term

Pc = 2 σ cos θr

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- 11

ADHESION TENSION

AIR

WATER

AIR

HgWATER

AIR

1/radius of tube

D. an important result to remember

1. pwb < pob

2. thus, the pressure on the concave side of a curved surface is greater thanthe pressure on the convex side

3. or, pressure is greater in the non-wetting phase

E. capillary pressure-unconsolidated sand

1. the straight capillary previously discussed is useful for explaining basicconcepts - but it is a simple and ideal system

2. packing of uniform spheres

Pc = σ 1R1

+ 1R2

R1 and R2 are the principal radii of curvature for a liquid adhering to two spheres incontact with each other.

3. by analogy to capillary tube

1R1

+ 1R2

= 2 cos θr

where Pc = 2 σ cos θr

call it Rm(mean radius), i.e.

1Rm

= 2 cos θrm

= (∆ρ)gh

σ

III- 12

F. wettability-consolidated sand

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

2. Funicular distribution - wetting phase is continuous, completely coveringsurface of solid

(A) (B)

OIL OR GAS OIL OR GAS

SAND GRAIN SAND GRAIN

WATER WATER

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

III- 13

IV) Relationship between capillary pressure and saturation

A. remember that the height a liquid will rise in a tube depends on

1. adhesion2. fluid density3. variation of tube diameter with height

B. consider an experiment in which liquid is allowed to rise in a tube of varyingdiameter under atmospheric pressure. Pressure in the gas phase is increasedforcing the interface to a new equilibrium position.

R

ATMOSPHERIC PRESSURE

R

HIGHERPRESSURE

DEPENDENCE OF INTERFACIAL CURVATURE ON FLUID SATURATIONIN A NON-UNIFORM PORE

1. Capillary pressure is defined as the pressure difference across theinterface.

2. This illustrates:

a. Capillary pressure is greater for small radius of curvature than forlarge radius of curvature

b. An inverse relationship between capillary pressure and wetting-phase saturation

c. Lower wetting-phase saturation results in smaller radius ofcurvature which means that the wetting phase will occupy smallerpores in reservoir rock

III- 14

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 interfacebetween wetting fluid and non-wetting fluid reaches equilibrium(drainage)

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

Θ

SATURATION = 100%PC = LOW VALUE

Θ

SATURATION = 80%CAPILLARY PRESSURE = P C

R

LOW PC HIGHER P C

(A)

ΘSATURATION = 0%P C = HIGH VALUE

Θ

SATURATION = 10%CAPILLARY PRESSURE = P C

R

HIGHER P C LOW PC

(B)

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

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

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

III- 15

B. Leverett conducted a similar experiment with tubes filled with sand.

1008060402000.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

DATA FROM HEIGHT-SATURATION EXPERIMENTS ON CLEAN SANDS. (FROM LEVERETT)

∆ ρ

g h

σ(k

/ø)1

/2

Drainage

Imbibition

Drainage Imbibition

Sand I

Φ

Φ

Φ

Φ

Φ

Φ Sand II

WATER SATURATION, Sw %

1. capillary pressure is expressed in terms of a non-dimensional correlatingfunction ( 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- 16

PC

0 100WATER SATURATION, S W

IMBIBITION

DRAINAGE

III- 17

VI) Capillary pressure in reservoir rock

Pw = Po/w -ρwh

144Po = Po/w -

ρoh

144

Water Oil

Pw2 Po2

Po1 = Pw1

100% Water

Oil and Water

Pc = Po - Pw = h144

ρw - ρo

Where: Po = pressure in oil phase, psiaPw = pressure in water phase, psiah = distance above 100% water level, ftPo/w = pressure at oil-water contact, psia

ρw = density of water, lb/cf

ρo = density of oil, lb/cf

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

III- 18

HEIGHTABOVEO-W-C

PRESSURE

PCPO = PO/W -

144ρoH

Pw = PO/W -ρw H

144

III- 19

VII) Laboratory measurement of capillary pressure

A. Methods

1. porous diaphragm

2. mercury injection

3. centrifuge

4. dynamic method

B. Porous diaphragm

1. Start with core saturated with wetting fluid.

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

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

4. Repeat at successively higher pressures until no more wetting fluid willcome out.

5. Measure Sw periodically.

6. Results

7. Advantages

a. very accurateb. can use reservoir fluids

8. Disadvantages

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

III- 20

C. Mercury Injection Method

1. Force mercury into core - mercury is non-wetting phase - air (usuallyunder vacuum) is wetting phase

2. Measure pressure

3. Calculate mercury saturation

4. Advantages

a. fast-minutesb. reasonably accurate

5. Disadvantages

a. ruins coreb. difficult to relate data to oil-water systems

D. Centrifuge Method

CORE HOLDER BODY

WINDOW

TUBE BODY

1. Similar to porous disk method except centrifugal force (rather thanpressure) is applied to the fluids in the core

2. Pressure (force/unit area) is computed from centrifugal force (which isrelated to rotational speed)

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

4. Advantages

a. fastb. reasonably accuratec. use reservoir fluids

III- 21

E. Dynamic Method

CORE

OIL INLET

OIL BURETTE

∆Po

GAS OUTLET GAS INLET

∆Pg Pc

TO ATMOSPHERE

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. seems to simulate reservoir conditionsb. reservoir fluids can be used

5. Disadvantages

a. very tedious

III- 22

F. Comparison of methods

1. diaphragm method (restored state) is considered to be most accurate, thusused 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 injectionshould be five times greater than capillary pressure of air-watersystem by diaphragm method

b. capillary pressure scale for curves determined by mercury injectionis five times greater than scale for diaphragm air-water data

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

III- 23

Example VIII-1

Comparison of Mercury Injection Capillary Pressure Data with Porous Diaphragm Data

A. Calculate capillary pressure ratio,

PcAHgPcAW

, for the following data:

σAHg = 480 Dynes/cm σAW = 72 Dynes/cm

θ AHg = 140° θAW = 0°

B. Pore geometry is very complex. The curvature of the interface and pore radius arenot necessarily functions of contact angles. Calculate the ratio using therelationship.

PcAHgPcAW

= σAHgσAW

Solution:

(A) PcAHgPcAW

= σAHgcos θAHg

σAWcos θAW =

480 cos(140°)

72 cos (0°)

PcAHgPcAW

= 5.1

(B) PcAHgPcAW

@ σAHgσAW

= 48070

PcAHgPcAW

= 6.9

III- 24

Discussion:

A. Best way to determine the relationship between mercury and air-water datais to generate capillary pressure curves for each set of data and comparedirectly.

Mercury Injection and Porous Diaphragm Methods

B. For this given set of conditions, mercury injection method requires ahigher displacement pressure, must adjust ratio between scales until matchis obtained.

C. Minimum irreducible wetting phase saturations are the same.

D. Reduction in permeability results in a higher minimum irreducible wettingphase saturation. For both cases, mercury system still has higher requireddisplacement pressure.

III- 25

VIII) Converting laboratory data to reservoir conditions

PcL= 2σLcos θL

r

PcR= 2σRcos θR

r

setting r = r

r = 2σLcos θLPcL

= 2σRcos θRPcR

∴ PcR = σcos θ Rσcos θ L

PcL

where

PcR= reservoir capillary pressure, psi

PcL= 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- 26

Example III-2

Converting Laboratory Data to Reservoir Conditions

Express reservoir capillary pressure by using laboratory data.

lab data: σAW = 72 dynes

σAW = 0o

reservoir data: σOW = 24 dynes/cm

σOW = 20o

Solution:

PcR=

σcos θ Rσcos θ L

PcL

PcR=

24 cos20°

72 cos0° PcL

PcR=

0.333 PcL

III- 27

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 whichcapillary pressure is zero

Pc = (∆ρ)gh144 gc

in English units

∆ρ = ρw - ρO, lb/cu ft

g = 32 ft/sec2

gc = 32 lbm ftlbf sec2

h = ft

144 = (sq in)/(sq ft.)

thus

Pc = lbf/(sq in), psI

III- 28

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 watersaturation in the reservoir at a height of 40 ft. above the oil-water contact.

ρo = 0.85 gm/cm3 ρw = 1.0 gm/cm3

PCL

SW

00

10

20

50 100

8.38.3

Solution:

PcR= ρw − ρo h

144

PcR=

1.0 - 0.85 62.4 lbft3

40

144 = 2.6 psi

PcL=

PcR0.313

PcL= 2.6

0.313 = 8.3 psi

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

Sw = 50%

III -29

X) Capillary pressure variation

A. effect of permeability

1. displacement pressure increases as permeability decreases

2. minimum interstitial water saturation increases as permeability decreases

10090807060504030201000

20

40

60

80

100

120

140

160

180

200

RESERVOIR FLUID DISTRIBUTION CURVES

Sw %

(From Wright and Wooddy)

10

md

10

0 m

d

20

0 m

d

90

0 m

d

30

Heig

ht

above

zero

cap

illar

y pre

ssure

, ft

24

18

12

6

0

Oil

- W

ater

Cap

illar

y P

ress

ure

, psi

(re

serv

oir

condit

ions)

90

72

54

36

18

0A

ir -

Wate

r C

apill

ary

Pre

ssure

, psi

(l

abora

tory

dat

a)

III -30

B. Effect of grain size distribution

1008060402000

5

10

15

20

25

30 225.0

187.5

150.0

112.5

75.0

37.5

0

100 80 60 40 20 0

Sandstone Core

Porosity = 28.1%Permeability = 1.43 mdFactor = 7.5

Mer

cury

cap

illar

y pre

ssur

e, p

si

Wat

er/n

itro

gen

cap

illar

y pre

ssur

e, p

si

Hg

Water

1008060402000

10

20

30

40

50

60 348

290

232

174

116

58

0

100 80 60 40 20 0

Wat

er/n

itro

gen

cap

illar

y pre

ssur

e, p

si

Mer

cury

cap

illar

y pre

ssur

e, p

si

Water

Hg

Limestone CorePorosity = 23.0%Permeability = 3.36 mdFactor = 5.8

1. majority of grains same size, so most pores are same size - curve (a) (wellsorted)

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

III -31

XI) Averaging capillary pressure data

J-function

J Sw = Pcσcos θ

kφ

1/2

attempt to convert all capillary pressure data to a universal curve

universal curve impossible to generate due to wide range of differences existing inreservoirs

concept useful for given rock type from given reservoir

where

Pc = dyne/(sq cm) σ = dyne/cm k = (sq cm)

φ = fraction

or can use any units as long as you are consistent

III -32

10090807060504030201000.00.10.20.30.40.50.60.70.80.91.01.11.21.31.41.51.61.71.81.92.0

CAPILLARY RETENTION CURVES.

WATER SATURATION, Sw

CA

PIL

LA

RY P

RESSU

RE F

UN

CT

ION

, J

(From Rose and Bruce.)

LEVERETT

LEDUC

HAWKINS

KATIE ALUNDUM

EL ROBLE

KINSELLA

Reservoir Formation

Hawkins WoodbineEl Roble MorenoKinsella Viking

Katie DeeseLeduc Devonian

Alundum (consolidated)Leverett (unconsolidated)

III -33

Capillary Pressure Problem 1

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

Calculate:

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

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

2. The displacement pressure for a water saturated porcelain plate is 55 psi of air. What isthe diameter in inches of the largest pore in the porcelain plate? Assume 72 dynes/cmand 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 = 2σAWcos θrρg

= 2 72 cos0°

.01 1.0 980

h = 14.69 cm

(b) pressure drop in psi across interface

Pc = pa - pw = ρwgh = 1.0 980 14.69

Pc= 0.0142 atm

14.696 psiatm

Pc = 0.209 psi

III -34

(2)Pc

= 2σAWcos θr

Pc = 55 psi

Pc = 55 psi atm14.696 psi

1.0133 x 106 dynes/cm2

atm

= 3.792 x 106 dynes/cm2

r = 2σAW cos θPc

r = 2 72 cos0°3.792 x 106

= 3.797 x 10-5 cm in2.54 cm

r = 1.495 x 10-5 in

d = 2.99 x 10-5 in

III -35

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.

fluidslab

air-waterres

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

J = Pc k/φ 1/2

σ cos θ

III -36

10090807060504030201000.0

2.5

5.0

7.5

10.0

12.5

15.0

17.5

20.0

22.5

25.0

27.5

30.0

32.5

35.0

Sw %

PC

L, p

si

Solution:

(1) PcR

= σR cosθRσL cosθL

PcL

= 20 cos2560 cos0

PcL

PcR=

0.302 PcL

III -37

(2) PcR

= hR ρw - ρo144

= hR 1.1 - .863 62.4144

PcR= .103 hR

hR= 9.74 PcR

(3) J =Pc

σ cos θ k

φ1/2

=PcL

σAW cosθL k

φ L

1/2

=PcL

60 cos0° 37

.161/2

J = .253 PcL

Sw PcL PcR hR J % psi ps i ft assorted

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

III -38

1008060402000

2

4

6

8

10

Sw %

PcR

1008060402000

20

40

60

80

100

Sw %

h R

1008060402000

2

4

6

8

10

Sw %

J

IV - 1

IV. FLUID SATURATIONS

I) 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

II) 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 intoformation

Ph > Pres

2) for water base mud, filtrate displaces formation waterand oil from the area around the well (saturationslikely change)

3) for oil base mud, filtrate will be oil; saturations mayor may not change.

IV - 2

Example: Effects of flushing by mud filtrates

Coring with water base mud

Oil zone at minimum interstitial water saturation:

sat at surface flushing by bit trip to surface compared to res

Sw ↑ ↓ ? probably ↑

So ↓ ↓ ↓

Sg - ↑ ↑

Gas zone at minimum interstitial water saturation:


Sw ↑ ↓ ?

So - - -

Sg ↓ ↑ ?

Water zone:


Sw - ↓ ↓

So - - -

Sg - ↑ ↑

IV - 3

Coring With Oil Base Mud

Oil zone at minimum interstitial water saturation:


Sw - - -

So - ↓ ↓

Sg - ↑ ↑

Gas zone at minimum interstitial water saturation:


Sw - - -

So ↑ ↓ ↑

Sg ↓ ↑ ↓

Water zone:sat at surface

flushing by bit trip to surface compared to res

Sw ↓ ↓ ↓

So ↑ ↓ ↑

Sg - ↑ ↑

IV - 4

b. bringing core to surface

1) reduction in hydrostatic pressure causes gas to comeout of solution

2) gas displaces oil and water causing saturations tochange

2. laboratory methods

a. evaporation using retort distillation apparatus

HEATING ELEMENT

COOLING WATER IN

CONDENSERCOOLING WATER OUT

CORE

IV - 5

1) process

a) heat small sample of rock

b) oil and water vaporize, then condense ingraduated cylinder

c) record volumes of oil and water

d) correct quantity of oil

65605550454035302520150.9

1.0

1.1

1.2

1.3

1.4

Oil Gravity, °API at 60° F

Mult

iply

ing F

acto

r

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

IV - 6

e) determine saturations

Sw = VwVp

So = VoVp

Sg = 1 - So - Sw

where

Sw = water saturation, fractionSo = oil saturation, fractionSg = gas saturation, fractionVp = pore volume, ccVw = volume of water collected, ccVo = volume of oil collected, cc

2) disadvantages of retort process

a) must obtain temperature of 1000-1100oF tovaporize oil, water of crystallization fromclays also vaporizes causing increase in waterrecovery

WATERRECOVERED

TIME

00

PORE WATER

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

c) core sample ruined

IV - 7

3) advantages of retort process

a) short testing time required

b) acceptable results obtained

b. leaching using solvent extraction apparatus

GRADUATED TUBECORE

SOLVENT

HEATER

WATER IN

WATER OUT

1) process

a) weigh sample to be extracted

b) heat applied to system causes water from coreto vaporize

c) solvent leaches hydrocarbons from core

IV - 8

d) water condenses, collects in trap. Recordfinal water volume

e) reweigh core sample

f) determine volume of oil in sample

Vo = Wi - Wdry - Vw ρw

ρo

where:

Wi = weight of core sample after leaching

Wdry = weight of core sample after leaching

Sw = VwVp

So = VoVp

2) disadvantages of leaching

a) process is slow

b) volume of oil must be calculated

3) advantages of leaching

a) very accurate water saturation value obtained

b) heating does not remove water ofcrystallization

c) sample can be used for future analysis

IV - 9

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

GAS

OIL

WATER

SO0 50

So ≅ 0 in gas zone

So ≥ 15% in oil zone

0 ≤ So ≤ Sor in water zone

Sor = residual oil saturation

IV - 10

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 - 11

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 ρw = 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 - 12

ρo = 141.5131.5 + 35°API

= 0.85 gm/cc

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 - 13

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 ρw = 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 cc70 gm

100 gm = 1.014 cc

Vo = 4.5 cc80 gm

100 gm = 5.63 cc

Vw = 2.8 cc80 gm

100 gm = 3.50 cc

Vb = 45.7 cc105 gm

100 gm = 43.52 cc

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 - 14

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 - 15

Fluid Saturation Problem 1

Calculate porosity, water, oil, and gas saturations, and lithology from the following core analysisdata.

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 ρw = 1.0 gm/cc

oil gravity = 40° API

Solution: γo = 141.5131.5 + °API

γo = 141.5131.5 + 40°

= 0.825

ρo = 0.825 gm/cc

(1) φ =

VpVb

Vp = Vw + Vo + Vg

Wo = Wsat - Vwρw - Wdry

= 87.95 - 7.12(1.0) - 79.17

Wo = 1.66 gm

IV - 16

Vo = Woρo

Vo = 1.66 gm0.825 gm/cc

= 2.01 cc

Vw = Vwrec - Wsat - Wi / ρw

= 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 - Wimmρw

Vb = (87.95 - 48.95) gm1 gm/cc

= 39.0 cc

φ = 9.1339.0

= 23.4%

(2) Sw =

VwVp

Sw = 5.57 cc9.13 cc

= 61.0%

So = VoVp

So = 2.01 cc9.13 cc

= 22.0%

Sg =VgVp

Sg = 1.55 cc9.13 cc

= 17.0%

(3) Vm = Vb - Vp

IV - 17

Vm = 39 - 9.13 = 29.87 cc

ρm= Wdry

Vm

ρm=

79.17 gm/29.87 cc = 2.65 gmcc

. . lithology is sandstone

(4) water saturation at surface will probably be greater than reservoir watersaturation

oil saturation at surface will be less than reservoir oil saturation

gas saturation at surface will be greater than reservoir gas saturation

IV - 18

Fluid Saturation Problem 2

Calculate porosity, water saturation, oil saturation, gas saturation, and lithology from the followingcore 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 waterand 0.87 cc of oil

Sample 3 weighed 50 gm and contained 0.40 cc of gas space

assumeρw = 1.0 gm/cc

oil gravity = 40o API

Solution: γo = 141.5131.5 + °API

γo = 141.5131.5 + 40°

= 0.825

ρo = 0.825 gm/cc

(1) φ =

VpVb

Vp = Vo + Vw +Vg

Vo = 0.87 cc86.71 gm

x 100 = 1.00 cc100 gm

Vw = 1.90 cc86.71 gm

x 100 = 2.19 cc100 gm

Vg = 0.40 cc50 gm

x 100 = 0.80 cc100 gm

Vp = (1.00 + 2.19 + 0.80) cc/100 gm

IV - 19

Vp = 3.99 cc/100 gm

Vb = 51.72 cc130 gm

x 100 = 39.78 cc100 gm

φ = 3.99/10039.78/100

x 100 = 10%

(2) Sw =

VwVp

x 100

= 2.19 cc3.99 cc

x 100

Sw = 54.8%

(3) So =

VoVp

x 100

= 1.0 cc3.99 cc

x 100

So = 25.1%

(4) Sg =

VgVp

x 100

= 0.80 cc3.99 cc

x 100

Sg = 20.1%

(5) Vm = Vb - Vp

= 39.78 - 3.99

Vm = 35.79 cc/100 gm

Wm = Wsat - ρoVo - ρwVw /Wsat

= 86.71 - 0.825 0.87 - 1.0 1.9086.71

Wm = 97 gm/100 gm

IV - 20

ρm = WmVm

ρm = 97 gm/100 gm35.79 cc/100 gm

ρm = 2.71 gm/cc

. . lithology - limestone

(6) 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

V - 1

V. ELECTRICAL PROPERTIES OF ROCK-FLUID SYSTEMS

I) Electrical conductivity of fluid saturated rock

A. Definition of resistivity

L

ELECTRICALCURRENT FLOW A

given a box of length (L) and cross-sectional area (A) completely filled withbrine of resistivity (Rw)

the resistance of the brine in the box to current flow may be expressed as

r = Rw LA

r = resistance - ohm

Rw = resistivity - ohm meters

L = length - meters

A = area - (meters)2

V - 2

B. Nonconductors of electricity

1. oil

2. gas

3. pure water

4. minerals

5. rock fragments

C. Conductors of electricity

1. water with dissolved salts conducts electricity (low resistance)

2. clay

D. Development of saturation equation (ignore clay)

A AP

ELECTRICALCURRENT FLOW

L

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

a. the area available for current flow is the cross-sectional areaof the pores.

Ap < A

b. the path through the pores is Lp.

Lp > L

V - 3

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

r = RwLp

Ap, water filled cube

r = RoL

A, porous media

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)

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

Ap = cross-sectional area available for current

flow, m2

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

A = cross-sectional area of porous media, m2

L = length of porous media, m

3. Since

ApA

≅ porosity, φ

and

LpL

≅ tortuosity, a measure of rock cementation.

then

Ro = RwALp

ApL

V - 4

becomes

Ro = f Rw, φ, tortuosity

E. Electrical formation resistivity factor, F

1. the equation for resistivity of a formation 100% saturated with abrine of resistivity of Rw

Ro = f Rw, φ, tortuosity

2. can be written as

Ro = F Rw

where F is the electrical formation resistivity factor

F = RoRw

3. cementation factor, m

a. it has been found experimentally that the equation for F takesthe form

F = C φ-m

where C is a constantm is the cementation factor

b. thus

log F = log C - m log φ

V - 5

F

φ

1

10

100

0.01 0.1 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) 1.3Very slightly cemented (Gulf Coast type of sand, except Wilcox) 1.4-1.5Slightly cemented (most sands with 20% porosity or more) 1.6-1.7Moderately cemented (highly consolidated sands of 15% porosity or less) 1.8-1.9Highly cemented (low-porosity sands, quartzite, limestone,dolomite) 2.0-2.2

V - 6

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 withthis 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 = 142.1

φ = 15.41%

Humble: F = 0.62/φ2.15

φ2.15 = 0.62/F

φ = 0.6242.1

2.15

φ = 14.06%

The Humble equation was developed for sandstone.

V - 7

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

1. definition of resistivity index

I = RtRo

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

Sw = I- 1

n = RtRo

- 1n

where n is the saturation exponent ≅ 2.0

3. rearrange

Sw-n = RtRo

-n log Sw = log RtRo

V - 8

1

10

100

.1 1.0

Rt

Ro

Sw

slope = -n, when n is the saturation exponent

NOTE:

slope = log y1 - log y2log x1 - log x2

II) Use of Electrical Formation Resistivity Factor, Cementation Factor, and SaturationExponent

A. obtain porosity, φ, from electric log or core analysis

B. F = C φ-m

(usually use Archie or Humble equation)

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

D. Ro = F Rw

E. convert Rt from electric log to water saturation

Sw = RoRt

1n

V - 9

III) Laboratory measurement of electrical properties of rock

A. Apparatus

CORE

VOLTMETER

1000 OHM STD. RESISTOR

AC SOURCE

B. Calculations

1. resistance of core

E = Ir

where: E = voltage, volts

I = current, amperes

r = resistance, ohms

∴ rcore = EI

V - 10

2. resistivity of core

Rcore = rcoreA

L

substituting r = E

I into the equation

Rcore = EAIL

C. Procedure

1. determine resistance of core

a. set desired current from AC source, low current preferredso core does not heat up.

b. record voltage from voltmeter

2. determine resistivity of core

a. for the first test completely saturate core with brine Sw =100%, Rcore = Ro

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

c. repeat tests until Sw = Swir

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

RtRo

= resistivity index = I

V - 11

D. Determine saturation exponent, n

1. rearrange saturation equation

Sw = RoRt

1/n

Swn= Ro

Rt

RoRt

= Sw-n

log RtRo

= -n log Sw

2. Plot log RtRo

vs log Sw or log I vs log Sw

1

10

100

.1 1.0

Rt

Ro

Sw

I =

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

V - 12

Example V-2

Given the following data, calculate the electrical formation resistivity factor and saturation exponentof the core.

Rw = 55 ohm-cm

I = 0.01 amp

D = 2.54 cm

L = 3.2 cm

ESw Voltage across

Water Saturation, % Core, volts 100.0 7.64

86.0 10.5074.0 14.3463.0 20.1654.0 27.5249.0 = Swir 34.67

Solution:

(1) electrical formation resistivity factor

F = RoRw

ro = EI = 7.64

.01 = 764 ohm

Ro = roAL

= 764 2.542π/4

3.2 = 1210 ohm cm

F = RoRw

= 121055

= 22

V - 13

saturation exponent

-n log Sw = log RtRo

Swrt = E

IRt =

rtAL

RtRo

% (ohm) (ohm-cm)

1.00 1.000.86 1050 1663 1.374.74 1434 2271 1.877.63 2016 3192 2.638.54 2452 4358 3.601.49 3467 5490 4.537

1.11

10

Sw

Rt/

Ro

(1.0,1.0)

(.334,10)

V - 14

-n = slope = log 10 - log 1log .334 - log 1

-n = 1 - 0-.4763 - 0

n = 2.10 = saturation exponent

NOTE:

RtRo

= EtEo

so RtRo

could have been calculated as the ratio of voltage at

Sw divided by the voltage at Sw = 1.0

V - 15

E. Determine cementation factor, m, and constant C for electrical formationresistivity 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

c.F =

RoRw

2. plot data according to form of equation for electrical formationresistivity factor

F = C φ-m

log F = log C - mlog φ

F

φ

1

10

100

0.01 0.1 1.0

slope = -m, m = cementation factor

intercept = C (intercept found at φ = 1.0)

V - 16

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 resistivityfactor equation.

Porosity Formation Resistivity Factor

φ F

0.152 400.168 320.184 260.199 220.213 190.224 17

Solution:

F = C φ-m

log F = log C - m log φ

plot log F vs log φ

1.11

10

100

ø

F

V - 17

slope = log 50 - log 10log 0.137 - log 0.284

= -2.21

-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 - 18

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

1RoA

= 1Rclay

+ 1Ro

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

1RoA

= 1Rclay

+ 1F 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 - 19

3. plot

1

ROA

(OHM - M) -1

1RW

(OHM - M) -1

1RoA

= 1Rclay

+ 1F

1Rw

where

1Rclay

= intercept

1F

= slope

V - 20

D. effect of clay

1. define FA =

RoARw , clays reduced the apparent formation resistivity

factor

CLEAN SAND

SHALY SAND

F

FA

RW

2. formation resistivity factor decreases more gradually when clay ispresent in the formation

CLEAN SAND

SHALY SAND

F

1

10

100

φ0.1 1.0

V - 21

3. saturation exponent n is not constant when clay is present information.

1

10

100

.1 1.0

Rt

Ro

Sw

I =

CLEAN SAND LOW R wn = 2

CLEAN SAND HIGH R wn = 1

SHALY SAND n =?

CLEAN SANDSwn-1 = I

VI - 1

VI. MULTIPHASE FLOW IN POROUS ROCK

I) Effective permeability

A. Permeability, k, previously discussed applies only to flow when pores are100% saturated with one fluid - sometimes called absolute permeability

q = kA∆ρ

µL

B. When pore space contains more than one fluid, the above equation becomes

qo = koA∆PοµoL

qw = kwA∆PwµwL

qg = kgA∆Pg

µgL

where qo = flow rate of oil, volume/time

qw = flow rate of water, volume/time

qg = flow rate of gas, volume/time

and 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 ofporous media to a particular fluid when the porous media is saturated withmore 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 - 2

E. Darcy equation for multiple fluids in linear flow, in oilfield units

qo = 1.1271 x 10-3 ko A P1 - P2µoL

qw = 1.1271 x 10-3 kw A P1 - P2 w

µwL

qg = 1.1271 x 10-3 kg A P1 - P2 g

µgL

when k = md

A = ft2

P = psia

L = ft

q = res bbl/day

II) Relative permeability

A. Defined as the ratio of the effective permeability to a fluid at a givensaturation to the effective permeability to that fluid at 100% saturated(absolute permeability)

kro = kok

krw = kwk

krg = kgk

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

III) Typical relative permeability curves

A. Use subscript wp to represent the "wetting phase"

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

VI - 3

00 100

1

Kr

SWP, %

NON-WETTING PHASE

WETTING PHASE

1

2

34

MINIMUM INTERSTITIAL SWP EQUILIBRIUM SNWP

1.krwp = 1, only at Swp = 100%

2. rapid decrease in krwp as Swp decreases

3.krwp = 0, at minimum interstitial Swp

4.krnwp = 0, at equilibrium Snwp

Note that krwp + krnwp < 1.0

VI - 4

B. Effect of saturation history

1. two types of relative permeability curves

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

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

2. the typical relative permeability curve shown below represents aprocess in which

a. process begins with porous rock 100% saturated withwetting phase (Swp = 100%)

b. wetting phase is displaced with non-wetting phase (drainage)until wetting phase ceases to flow (Swp = minimuminterstitial 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 - 5

00 100

1

Kr

SWP, %

DRAINAGE

IMBIBITION

Krnwp

Krwp

minimum interstitial residual non-wetting

wetting phase saturation phase saturation

VI - 6

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 inlaboratory must correspond to the process in the reservoir

a. initial distribution of fluids in reservoir was by drainage

b. at and behind a water front (flood or encroachment) theprocess is imbibition

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

Fluids Present Wetting Phase

Water & Oil WaterWater & Gas Water

Oil & Gas Oil

VI - 7

C. Three phase relative permeability

1. often three phases are present in petroleum reservoirs

2. tertiary (triangular) diagram is used to represent a three-phase system

100% OIL100% WATER

100% GAS

VI - 8

3. relative permeability to oil in a three phase system

1%

5

10

60

40

20

100% OIL100% WATER

100% GAS

Note, kro is shown in %

a. dependence of relative permeability to oil on saturations ofother phases is established as follows:

1) oil phase has a greater tendency than gas to wet thesolid

2) interfacial tension between water and oil is less thanthat between water and gas

3) oil occupies portions of pore adjacent to water

4) at lower water saturations the oil occupies more ofthe smaller pores. The extended flow path lengthaccounts for the change in relative permeability to oilat constant oil saturation and varying water saturation

VI - 9

4. Relative permeability to water in a three-phase system

100% OIL100% WATER

100% GAS

0

10%

20%

40%

60%

80%

Krw

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

b. thus, krw can be plotted on cartesian coordinates againstSw.

VI - 10

5. Relative permeability to gas in a three-phase system

100% OIL100% WATER

100% GAS

50%

40

30

20

5

1

a. curves above indicate that krg is a function of saturations of

other phases present.

b. other research shows that krg is a unique function of gas

saturation

c. the other phases, oil and water, occupy the smaller poreopenings and wet the surface of the rock

d. therefore, krg should be dependent only on the total

saturation of the other two phases (i.e. 1-Sg) andindependent of how much of that total is composed of eitherphase

e. thus krg can be plotted on Cartesian coordinates against So +

Sw

VI - 11

1008060402000.0

0.2

0.4

0.6

0.8

1.0

krg

So + Sw

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

a. water

1) is located in smaller pore spaces and along sandgrains

2) therefore krw is a function of Sw only

3) thus plot krw against Sw on rectangular coordinates

b. gas

1) is located in center of larger pores

2) therefore krg is a function of Sg only

3) thus plot krg against Sg (or So + Sw) on rectangular

coordinate

VI - 12

c. oil

1) is located between water and gas in the pores and to acertain extent in the smaller pore spaces

2) therefore kro is a function of So, Sw, and Sg

3) thus plot kro against So, Sw, Sg on a triangulardiagram

4) if Sw can be considered to be constant (minimum

interstitial) kro can be plotted against So on arectangular diagram

1008060402000.0

0.2

0.4

0.6

0.8

1.0

kro

So, %

Minimum InterstitialWater Saturation

VI - 13

7. Flow in three-phase system

100% OIL100% WATER

100% GAS

5% oil

5% water

5% gas

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 - 14

IV) Permeability ratio (relative permeability ratio)

A. Definitions

1. When the permeability to water is zero (as at minimum interstitialwater saturation) it is sometimes convenient to use permeability ratioto represent the flow conductance of the rock to gas and oil as aratio.

permeability ratio = kgko

= krgkro

2. When the permeability to gas is zero (no gas or gas below "criticalgas saturation") it is sometimes convenient to use permeability ratioto represent the flow conductance of the rock to oil and water as aratio

permeability ratio = kokw

= krokrw

V) Measurement of relative permeability

A. 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

B. Laboratory Methods

1. Steady-state flow process

a. saturate core with wetting-phase fluid

b. inject wetting-phase fluid through core (this will determineabsolute permeability)

VI - 15

c. inject a mix of wetting-phase and non-wetting phase (startwith small fraction of non-wetting phase)

d. when inflow and outflow rates and portion of non-wettingphase equalize, record inlet pressure, outlet pressure andflow rates of each phase

e. measure fluid saturation in core (see below)

f. calculate relative permeability

ko = qoµoL

A∆p

kw = qwµwL

A∆p

g. repeat b through f with injection mixtures containingrelatively more non-wetting phase until irreducible wetting-phase saturation is reached

Sw, %0 1000

1

krkro krw

VI - 16

h. determination of fluid saturations

1) resistivity

Sw = RoRt

1n =

EoEt

1n

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 wettingphase less than 100%, volts

2) volumetric balance

3) gravimetric method - remove core and weigh it

Wf = Wt - Wd

where: Wf = weight of fluid in core, gm

Wt = weight of saturated core, gm

Wd = weight of dry core, gm

Wf = ρoVo + ρwVw

and

Vf = Vo + Vw

where: ρ = density, gm/cc

V = volume, cc

Sw = Vw/Vf

where: Sw = saturation of wetting phase

thus

VI - 17

Sw =

Wf/Vf - ρoρw - ρo

i. same procedure can be used starting with 100% saturation ofnon-wetting phase

1) injection ratio start with high ratio of non-wettingphase

2) procedure ends at residual non-wetting phasesaturation

3) then is a hysteresis effect of same type as discussedwith capillary pressure measurements

4) choice of starting saturation depends on reservoirprocess which is being simulated

j. end effects

1) causes of end effects

a) in the bulk of the core there is a wetting-phase saturation and a non-wetting phasesaturation, therefore there is a finite value ofcapillary pressure

b) thus there is a difference in pressure betweenthe wetting-phase and non-wetting phase

Pcap = Pnwp - Pwp

c) at the face of the core the pressures in thewetting-phase and the non-wetting phase areessentially equal

Pnwp = Pwp

thus capillary pressure is essentially zero

d) if capillary pressure is zero, the saturation ofthe wetting phase must be 100% (seecapillary pressure curve)

e) there must be a saturation gradient fromessential 100% wetting phase at the "end" tosome value of Swp less than 100% in thebulk of the core

VI - 18

25201510500

20

40

60

80

100

25201510500

20

40

60

80

100

Oil

Sat

urat

ion,

%

Distance from outflow face, cm

Inflow face

Theoretical saturationgradient

Oil

Sat

urat

ion,

%

Distance from outflow face, cm

Inflow face

Theoretical saturationgradient

2) elimination of end effects

a) install end pieces to contain end effects

b) flow at rapid rates to make end effectnegligible (pressure gradient > 2 psi/inch

EndSection

TestSection

MixingSection

Thermometer

Packing Nut Electrodes

CopperOrifice Plate

Inlet

InletHighly permeable disk

DifferentialPressure Taps

OutletBronze Screen

PENN STATE RELATIVE-PERMEABILITY APPARATUS

VI - 19

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 brine, 60,000 ppm

length = 2.30 cm oil, 40oAPIdiameter = 1.85 cm µw = 1.07 cp

area = 2.688 cm2 µo = 5.50 cpporosity = 25.5%

Oil Flow Water Flow Inlet Pressure Outlet Pressure Voltage Drop Electrical Currentcc/sec cc/sec psig psig volts amps

0.0000 1.1003 38.4 7.7 1.20 0.010.0105 0.8898 67.5 13.5 2.10 0.010.0354 0.7650 88.1 17.6 2.80 0.010.0794 0.3206 78.2 15.6 4.56 0.010.1771 0.1227 85.6 17.1 8.67 0.010.2998 0.0000 78.4 15.7 30.00 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 2.302.688 38.4 - 7.7 14.696

k = 0.482 darcy

VI - 20

2. Calculate effective permeabilities to oil and water

ko =

qoµoL

A∆P

ko =

0.0105 5.50 2.302.688 67.5 - 13.5 / 14.696

ko = 0.0134 darcy

kw =

qwµwL

A∆P

kw =

0.8898 1.07 2.302.688 67.5 - 13.5 / 14.696

kw = 0.2217 darcy

3. Calculate relative permeabilities

kro = kok

= .0134.482

= 0.028

krw = kwk

= .2217.482

= 0.460

4. Calculate water saturations

Sw = EoEt

1/2

Sw = 1.202.10

1/2 = .756

5. Results

Water Saturation Relative Permeability Relative Permeabilityto oil to water

Sw kro krw ko/kw

1.000 0.000 1.000 0.0000.756 0.028 0.460 0.0610.655 0.072 0.303 0.2380.513 0.182 0.143 1.2730.372 0.371 0.050 7.4190.200 0.686 0.000 -------

VI - 21

1008060402000.0

0.2

0.4

0.6

0.8

1.0

Kro

Krw

Sw, % pore space

Rel

ativ

e Pe

rmea

bilit

y

100806040200.01

.1

1

10

Sw, % of pore space

Per

mea

bili

ty R

atio

, ko/

k w

VI - 22

6. The data permit certain checks to be made

F = 0.62 φ-2.15

F = RoRw

Rw = 12 ohm-m for 60,000 ppm brine

Ro = EAIL

= 1.20 2.688

.01 2.3 = 140 ohm-m

F = 14012

= 11.7

Φ = .62F

12.15 = .62

11.7

12.15

Φ = .255

VI - 23

2. Displacement (unsteady-state)(Welge)

a. does not result in relative permeability only give permeabilityratio

b. procedure

1) mount core in holder

2) saturate with wetting phase (usually oil)

3) inject non-wetting phase (usually gas) at constantinlet and outlet pressures

4) measure

a) cumulative gas injected as a function of time

b) cumulative oil produced as a function of time

c. conditions

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

2) gas saturation can be described at mean pressure

Pm = Pi + Po

2

3) flow is horizontal and core is short so that effects ofgravity can be neglected

d. calculations

1) convert gas injected into pore volumes

Gipv = GipiLAφ pm

where Gi = cumulative gas injected (measured at pressure pi), cc

Gipv = cumulative gas injected in pore volume

pi = inlet pressure, psi

VI - 24

pm = pi + po2

, psi

LA φ = pore volume, cc

2) calculate average gas saturation, Sgav

Sgav

=Np

LAφ

where Np = cumulative oil produced, cc

LA φ = pore volume, cc

3) plot Sgav vs Gipv

Gipv00

Sgav

GAS BREAKTHROUGH

VI - 25

4) determine fractional flow of oil, fo

fo = d Sgavd Gipv

fo = slope of plot of Sgav vs qGipv

5) calculate permeability ratio, kg/ko

fo =

koA ∆pµoL

koA ∆pµoL

+ kgA ∆p

µgL

fo = ko/µo

ko/µo + kg/µg

kgko

= 1 - fofo µo/µg

where

kgko

= permeability ratio of gas to oil

fo = fractional flow of oil

6) Permeability ratio, kg/ko, calculated above appliesonly at the gas saturation of the outflow face, thusmust calculate Sgo

Sgo = Sgav - Gipvfo

where Sgo = gas saturation at outlet face of core

Gipv = cumulative gas

injected, pore volumes

fo = fractional flow of oil atoutlet face of core

VI - 26

e. advantages

1) minimum amount of equipment

2) rapid

f. disadvantages

1) results in kg/ko, not kro and krg

2) equations don't apply until gas breaks through, thusinitial value of gas saturation may be high, resulting

in incomplete kg/ko vs Sgo curve.

VI - 27

Example VI-2

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

Cumulative Gas Injection, Gi, cc Cumulative Oil Produced, Np, cc

14.0 14.650.2 19.5112.6 22.5202.3 25.5401.4 28.6546.9 30.4769.9 32.21226.5 33.43068.9 35.35946.6 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 cm2

Prepare to determine kg/ko by calculating Sgav and Gipv.

Solution:

1. Calculate Sgav

Sgav = Np

L A φ

Sgav = 14.6 cc14.3 cm 20.27 cm2 .210

Sgav = 0.24

2. Calculate Gipv

VI - 28

Gipv = GipiLAφ pm

Gipv = 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

SgavGipv pv

0.24 0.2640.32 0.9450.37 2.120.42 3.810.47 7.560.50 10.30.53 14.50.55 23.10.58 57.80.59 112.0

VI - 29

Example VI-3

A core sample initially saturated with oil is flooded with gas. The following data was obtained:

Sgav Gipv pv

0.24 0.2640.32 0.9450.37 2.120.42 3.810.47 7.560.50 10.30.53 14.50.55 23.10.58 57.80.59 112.0

µo = 2.25 cp

µg = 0.0185 cp

Calculate and construct a fg verses Sgo plot. Convert Sgavg to Sgo. Determine kg/ko for each of

the given saturations. Construct a graph of kg/ko versus Sgo.

Solution:

Plot Sgav vs. Gipv

The slope from this plot is fo.

Sgo = Sgav - fogipv

kg/ko = 1 - fo

fo µoµo

Sgav Gipv pv fo Sgo kg/ko 0.24 .264 .375 .141 .01370.32 .94 .075 .249 .1010.37 2.17 .0357 .294 .2220.42 3.81 .0214 .338 .3760.47 7.56 .0118 .381 .6890.50 10.3 .0092 .405 .8860.53 14.5 .0046 .463 1.780.55 23.1 .0013 .521 6.320.58 57.8 .0005 .550 16.40.59 112.0 .0001 .581 82.2

VI - 30

1201008060402000.2

0.3

0.4

0.5

0.6

Sgav

Gipv, pv

fraction

1.00.90.80.70.60.50.40.30.20.10.0.01

.1

1

10

100

kg/ko

Sg, %

VI - 31

C. Field determination of permeability ratios

1. equations

qgqo

=

kg A ∆p

µgL

ko A ∆pµoL

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

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

thus,

kgko

= qgqo

µgµo

replace qg/qo with

qgqo

= Bg Rp - Rs5.615 Bo

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

kgko

= µgBg Rp - Rsµo 5.615 Bo

2. procedure

a. producing gas-oil ratio, Rp, and physical properties, Bg,

Bo, Rs, µg, µo must be determined at some knownreservoir pressure

b. saturations in reservoir, Sg or So, must be calculated fromproduction data and material balance calculations

VI - 32

Example VI-4

Discovery pressure for your well was 4250 psia, temperature is 200oF, and initial producing gas-

oil 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 isnow 3000 psia. Producing gas-oil ratio is 18,100 SCF/STB. What is kg/ko in the reservoir at thistime.

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

kgko

= µgBg Rp - Rsµo 5.615 Bo

kgko

= 0.0192 5.34 x 10-3 18100 - 5600.75 5.615 1.314

kgko

= 0.325

VI - 33

VI) Uses of relative permeability data

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

Sw, %0 1000

h, ft100% Water Production

100 % Sw

SP Log RT Log

Log Response Diagram

Sw, %0 1000

1

kr

VI - 34

B. Determination of height of 100% oil production

Sw, %0 1000

h, ft100% Water Production

100 % Sw

SP Log RT Log

Log Response Diagram

100% Oil Production

Sw, %0 1000

1

kr

VI - 35

C. Effect of permeability on thickness of transition zone

Sw, %0 1000

h, ft

Sw, %0 1000

1

krLow K

High K

Low K

High K

h = height of zone of interest

VI - 36

D. Fractional flow of water as a function of height

fw = qwqtot

= qwqo + qw

=

kw A ∆ Pµw L

ko A ∆ Pµo L

+ kw A ∆ Pµo L

= 1

1 + kokw

µwµo

fw

1

00 100Sw

00 100Sw

h

100

00

h

100

fw1

VI - 37

1008060402000

20

40

60

80

100

120

140

160

Fraction of water in produced fluid, %

Hei

ght a

bove

fre

e w

ater

leve

l, ft

10 md

50 md

100 md

200 md

This figure indicates that lower permeabilities result in longer transition zones

E. Determination of residual fluid saturations

1

00 100Sw

kr

Oil

Water

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 - 38

F. Interpretation of fractional flow curve

fw

1

00 100Sw

2 3

1

4

1. fw at water breakthrough

2. Sw at well at water breakthrough

3. Swav in reservoir between wells at water breakthrough

4.1

slope = pore volume of water injected

VI - 39

fw

1

0

0 100Oil Rec - % Oil in Place

1

2

3

0

Water inputPore vols.

VII - 1

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 - 2

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 - 3

III) Histogram

A histogram is a graphical representation of a frequency distribution.

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

Fre

quen

cy o

f O

ccur

ence

# of

Sam

ples

0

2

4

6

8

Porosity, %12 16 20 24

# of

Sam

ples

0

2

4

6

8

20 60 100 140

Permeability, md

VII - 4

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

Frequency Relative FrequencyRange of (No. of wells (No. of wells having thickness Relativethickness, having thickness values in each range, Frequencyft. values in the range) fraction of total wells) as percentage.

50-80 4 0.20 20%81-110 7 0.35 35%111-140 5 0.25 25%141-170 3 0.15 15%171-200 1 0.05 5%

20 1.00 100%

0 50 80 110 140 170 200

Random variable: net pay thickness, ft

Fre

quen

cy

0

2

4

6

8

10

VII - 5

Sometimes the relative frequency is plotted on a histogram

0 50 80 110 140 170 200


Rel

ativ

e Fr

eque

ncy

0

.1

.2

.3

.4

.5

VII - 6

IV) Cumulative Frequency Distributions

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

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

0 50 80 110 140 170 200


0

20%

40%

60%

80%

100%

Cum

ulat

ive

% le

ss th

an o

r eq

ual t

o

0

.2

.4

.6

Cum

ulat

ive

freq

uenc

y le

ss th

an f

or e

qual

to

.8

1.0

VII - 7

Occasionally a "cumulative frequency greater than or equal to" distribution is

plotted. Relative frequencies are summed from the highest class interval and plotted

at the lower ends of the intervals

0 50 80 110 140 170 200


0

20%

40%

60%

80%

100%

Cum

ulat

ive

% g

reat

er th

an o

r eq

ual t

o

0

.2

.4

.6

Cum

ulat

ive

freq

uenc

y gr

eate

r th

an f

or e

qual

to

.8

1.0

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 - 8

V) Normal Distribution

The normal distribution is continuous probability distribution having a symmetrical

shape similar to a bell, sometimes called a Gaussian distribution.

a a

µ − a µ + a

Random variable x

f(x)

Inflection pointof curve

This distribution is completely and uniquely defined by two values - the mean, m,

and standard deviation, σ.

VII - 9

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.

Random variable x

f(x)

Mode

Median (geometric mean)

Mean (arithmetic mean)

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 - 10

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

µ = xiΣ

i =1

n

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 . . . xn1/n

µg = xiπi=1

n 1/n

where µg = the geometric mean

VII - 11

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.

σ2 = xi - µ 2Σ

i=1

n

n-1

where σ2 is the variance

σ is the standard deviation

B. Mean deviation - another measure of the dispersion about the centraltendency

MD = xi - µΣ

i=1

n

n

For classified data

σ2 = fiΣ

j xi - µ 2

fiΣj

where fi = frequency for each class

xj = class mark

or

σ2 = frjΣj

xj - µ 2

where frj = relative frequency for each class

VII - 12

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

µ = xiΣ

i=1

n

n

where i refers to each individual data point and, for classified data

µ = fjΣ

j xj

fjΣj

where j refers to each class interval

fj is the frequency of the class

xj is the class mark

orµ = frjΣ

j xj

where frj is the relative frequency of the class.

xj is the class mark

VII - 13

Cum

ulat

ive

Freq

uenc

y

Cum

ulat

ive

Freq

uenc

y

Random variable x Random variable x

Cumulative frequency plottedon coordinate graph paper

Cumulative frequency plottedon normal probability paper

.001 .001

.999 .999

.001

.999

Cumulative % <

.001

.999

Cumulative % <

Ran

dom

var

iabl

e x,

dist

ribu

ted

norm

ally

Ran

dom

var

iabl

e x,

dist

ribu

ted

norm

ally

Normal probability graph paper

50%

µ

50%

µ

µ + a

84.1%

σ

VII - 14

Porosity and permeability data from a well in the Denver-Julesburg Basin

Porosity IntervalInterval Frequency Midpoint

i Percent fi xi fixi (xi - µ) (xi - µ)2 fi(xi - µ)2

1 7.0<x<10.0 1 8.5 8.5 -9.2 84.64 84.642 10.0<x<12.0 0 11.0 0.0 -6.7 44.89 0.003 12.0<x<14.0 1 13.0 13.0 -4.7 22.09 22.094 14.0<x<16.0 10 15.0 150.0 -2.7 7.29 72.905 16.0<x<18.0 12 17.0 204.0 -0.7 0.49 5.886 18.0<x<20.0 8 19.0 152.0 +1.3 1.69 13.527 20.0<x<22.0 7 21.0 147.0 +3.3 10.89 76.238 22.0<x<25.0 3 23.5 70.5 +5.8 33.64 100.92

42 745.0 376.18

µ = Σ fi xiΣ fi

= 745.042

= 17.7%

σ2 = fiΣ

jxi - µ 2

fiΣj

= 376.1842

= 8.96

σ = 8.96 = 2.99%

VII - 15

Cumulative

Frequency Cumulative

Than or Equal to Frequency

Porosity Upper Limit of Expressed as

Interval, % Frequency Interval Percentage

7.0<x<10.0 1 1 2.4%

10.0<x<12.0 0 1 2.4%

12.0<x<14.0 1 2 4.8%

14.0<x<16.0 10 12 28.6%

16.0<x<18.0 12 24 57.1%

18.0<x<20.0 8 32 76.2%

20.0<x<22.0 7 39 92.9%

22.0<x<25.0 3 42 100.0%

42

26

24

22

20

18

16

14

12

10

82 10 20 30 40 50 60 70 80 90 96

Cumulative % less than or equal to given values of porosity

Cor

e po

rosi

ty, %

at 50th percentile f = 17.7%

at 84th percentile f + σ = 20.7%

σ = 20.7 - 17.7 = 3%

VII - 16

X) Log Normal Distribution

Permeability data is usually assumed to have a log normal distribution.

For log normal distribution the median, mode, and mean have different numerical

values. The median has been chosen as the value of central tendency which best

represents the data.

The median of a log normal distribution is equal to the geometric mean.

Thus, for unclassified data

µ = xiπi=1

n 1/n

or

log (m) = 1n log xiΣ

i=1

n

VII - 17

CumulativeFrequency Less Cumulative

Permeability Than or Equal to FrequencyInterval Upper Limit of Expressed As

(millidarcies) Frequency Interval Percentage

0-50 2 2 4.8%51-100 2 4 9.5%

101-150 4 8 19.0%151-200 4 12 28.6%201-250 4 16 38.1%251-300 8 24 57.1%301-350 4 28 66.7%351-400 2 30 71.4%401-450 4 34 81.0%451-500 1 35 83.3%501-700 5 40 95.2%

701-1000 2 42 100.0%42

2 10 20 30 40 50 60 70 80 90 96

Cumulative % less than or equal to given values of permeability

10

100

1000

Cor

e pe

rmea

bili

ty, m

d

Lognormal probability graph paper

1

PETE 306 HANDOUT 3/5/92

Calculation of Permeability using Capillary Pressure Data

Purcell Approach(ABW: pages 167-172)

Three basic considerations:

1. Capillary pressure in a capillary,

Pc = 2 σ cosθr

2. Capillary flow: Poiseuille's law

qi = π ri

4 ∆p

8 µ L

3. Darcy's equation,

qt = kA ∆p µ L

Pc = dynes cm2 k = cm2

σ = dynes cm A = cm2

r = cm L = cm

q = cm2

secµ = poise = dynes-sec

cm2

∆p = dynecm2

2

Let V i = πr i2L,

then the flow rate in a capillary is

qi = Vr i

2∆p

8µL2, V = cm3

Sincer i = 2σ cosθ

Pci

qi =

σ cosθ 2

Pci 2 Vi

2 µ L2 ∆p

For a bundle of n capillary tubes,

qt = σ cosθ 2 ∆p

2 µ L2Σ

i = 1

i = n Vi

Pci2

Σi=1

i=n

Since,

qt = k A ∆pµL

k = σ cosθ 2

2 A L Vi

Pci2

Σi=n

i=n

3

Define the fractional volume of ith capillary

s i = V i

VT , s i = fraction

and

φ = VTA L

, φ = fraction

k = σ cosθ 2

2 φ Si

Pci2Σ

i=1

i=n

Introducing a lithology factor λ for deviation of the actual porespace,

k = σ cosθ 2

2 φλ Si

Pci2Σ

i=1

i=n

In integral form,

k = σ cosθ 2

2 φλ dS

Pc2

S=0

S=1

1

PETE 306 HANDOUT 4/16/92

Calculation of Relative Permeabilities usingCapillary Pressure Data

Purcell and Burdine Approach(ABW: pages 196-199)

Purcell approach:

The absolute permeability may be expressed as

k = σ cosθ 2

2 φλ dS

Pc2

S=0

S=1

The effective permeability of the wetting phase may be expressed as

kwt = σ cosθ 2

2 φλ dS

Pc2

S=0

S=Swt

The relative permeability of the wetting phase is the ratio of the wettingphase effective permeability to the absolute permeability

krwt = kwt

k =

dSPc

2S=0

S=Swt

dSPc

2S=0

S=1

2

Similarly, the effective permeability of the nonwetting phase may beexpressed as

knwt = σ cosθ 2

2 φλ dS

Pc2

S=Swt

S=1

The relative permeability of the nonwetting phase is the ratio of thenonwetting phase effective permeability to the absolute permeability

krnwt = knwt

k =

dSPc

2S=Swt

S=1

dSPc

2S=0

S=1

3

Burdine Approach:

Burdine considered the tortuosity factors for one-phase and multiphasesystems and modified the Purcell equations for the effectivepermeabilities.

λrwti = λiλwti

The relative permeability of the wetting phase is the ratio of the wettingphase effective permeability to the absolute permeability

krwt = kwt

k = λrwt

2

dSPc

2S=0

S=Swt

dSPc

2S=0

S=1

The tortuosity ratio is related to the minimum wetting-phase saturationSm, as

λrwt = Swt - Sm1 - Sm

4

Similarly, the relative permeability of the nonwetting phase is the ratioof the nonwetting phase effective permeability to the absolutepermeability

krnwt = knwt

k = λrnwt

2

dSPc

2S=Swt

S=1

dSPc

2S=0

S=1

The tortuosity ratio for the nonwetting phase is related to the minimumwetting-phase saturation, Sm, and the equilibrium saturation to thenonwetting phase, Se, as

λrnwt = Snwt - Se1 - Sm - Se

reservoir petrophysics[1]

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