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Kwame Nkrumah University of science and

technology, Kumasi

Petroleum engineering programme

Reservoir Petro physics handout

Lecturer; K. SARKODIE

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ii

TABLE OF CONTENTS

I. ROCK POROSITY I-1

I) Definition I-1

II) Classification I-1III) 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

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I. ROCK POROSITY

I) Definition

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

rock

In general form:

Porosity = =Vp

Vb=

Vb - VmVb

where:

is expressed in fraction

Vb = Vp + Vm

Vb = bulk volume of reservoir rock, (L3)

Vp = pore volume, (L3)

Vm= matrix volume, (L3)

II) Classification

A. Primary (original) Porosity

Developed at time of deposition

B. Secondary Porosity

Developed as a result of geologic processoccurring after deposition

C. Total Porosity

t =total pore space

Vb=

Vb - VmVb

D. Effective Porosity

e =interconnected pore space

Vb

1. Clean sandstones: e = t

2. Carbonate, cemented sandstones: e < t

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VI) Factors affecting porosity

A. Factors:1. Particle shape2. Particle arrangement

3. Particle size distribution4. Cementation5. Vugs and fractures

B. Particle shape

Porosity increases as particle uniformity decreases.

C. Packing Arrangement

Porosity decreases as compaction increases

6000500040003000200010000

0

10

20

30

40

50

DEPTH OF BURIAL, ft

POROSITY,%

EFFECT OF NATURAL COMPACTION ON POROSITY

(FROM KRUMBEIN AND SLOSS.)

SANDSTONES

SHALES

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D. Particle Size Distribution

Porosity decreases as the range of particle size increases

GRAIN SIZE DIAMETER, MM

INTERSTITIAL MATERIALS

AND MUD FRAGMENTS

FRAMEWORK

FRACTION

CLEAN SAND

SHALY SAND

SAND SILT CLAY100

0

1.0 0.1 0.01 0.001

WEIGHT%

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

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V) Measurement of porosity

=Vb - Vm

Vb=

Vp

Vb

Table of matrix densities

Lithology m (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 liquidb) 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

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c. matrix volume

1) assume grain density

Vm =dry weight

matrix 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

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d. pore volume

1) gravimetric

Vp =saturated weight - dry weight

density 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 + pore

volume of core (calculated)

3) Vp

= V2

- V1

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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

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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

=Vb - Vm

Vb= 9.9 cc- 7.7 cc

9.9 cc= 0.22

It is total porosity.

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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 water

density of water

Vb = 448.6 gm - 269.6 gm1 gm/cm3

Vb = 179.0 cm3

=Vp

Vb= 21.3 cm3

179.0 cm3= .119

= 11.9%

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What is the lithology of the sample?

Vm = Vb - Vp

Vm

= 179.0 cm3 - 21.3 cm3 = 157.7 cm3

m = 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.

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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 the

porosity of the core.

Solution:

P1V

1= P

2V

2

V1

= 1,000 cc

P

1

= 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

=

VT - VmVT

= 695 cc - 488 cc695 cc

= .298 = 29.8%

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VI) Subsurface measurement of porosity

A. Types of logs from which porosity can be derived

1. Density log:

d = m - Lm - f

2. Sonic log:

s =tL - tm

tf- tm

3. Neutron log:

e- k =CNf

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

Lithologytm sec/ft m gm/cc

Sandstone 55.6 2.65

Limestone 47.5 2.71

Dolomite 43.5 2.87

Anhydrite 50.0 2.96

Salt 67.0 2.17

Water 189.0 1.00

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B. Density Log

1. Measures bulk density of formation

FORMATION

GAMMA RAY

SOURCE

SHORT SPACE

DETECTOR

LONG SPACE

DETECTOR

MUD CAKE

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

3. Equation

L = m 1 - +f

d = m - Lm - f

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FORMATION DENSITY LOG

4240

4220

4200

4180

4160

4140

4120

4100

20016012080400 3.02.82.62.42.22.0

, gm/ccGR, API depth, ft

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Example I-5

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

gm/cc and fluid = 1.0 gm/cc.

Interval, ft L, gm/cc

d ,%

__________ _________ ______

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 :

L

= 2.375 gm/cc

d =m - Lm - f

=2.68 gm/cc - 2.375 gm/cc

2.68 gm/cc - 1.0 gm/cc= 0.18

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C. Sonic Log

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

D

B

AT

R1

R2

E

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 - +tf (Wylie Time Average Equation)

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SONIC LOG

4240

4220

4200

4180

4160

4140

4120

4100

2001000 140 120 100 80 60 40

GR, API T, seconds/ftdepth, ft

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Example I-6

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

Interval tLs ,%

(ft) 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 -sec/ft

s =tL - tm

tf-tm=

86.5 sec/ft- 51.6 sec/ft

189.0 sec/ft- 51.6 sec/ft = 0.25

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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

RELATIVEPROBABILITY

FORCOLLISION

.1 1 10 102 103 104 105

H

106

O

107

Si

10-3

10-2

10-1

1

CLEAN SAND POROSITY = 15%

SLOWINGDOWNPOWER

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.

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NEUTRON DENSITY LOG

4240

4220

4200

4180

4160

4140

4120

4100

2000 30 -10

GR, API (CDL)depth, ft

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Example I-7

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

Solution:

Interval 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

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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/cc

travel time = 62.73 -sec/ft

Solution:

Assume fresh water in pores.

Assume sandstone:

m = 2.65 gm/cc

tm = 55.5 -sec/ft

d =m - Lm-f

=2.65 gm/cc - 2.522 gm/cc

2.65 gm/cc - 1.0 gm/cc= 7.76%

s =tL - tm

tf-tm

=62.73 sec/ft- 55.5 sec/ft

189.0 sec/ft - 55.5 sec/ft

= 5.42%

Assume limestone:

m = 2.71 gm/cc

tm = 47.5 -sec/ft

d =m - L

m-f

=2.71 gm/cc - 2.522 gm/cc

2.71 gm/cc - 1.0 gm/cc

= 10.99%

s =tL - tm

tf-tm

=62.73 sec/ft - 47.5 sec/ft

189.0 sec/ft - 47.5 sec/ft= 10.76%

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Assume dolomite:

m = 2.87 gm/cc

tm = 43.5 -sec/ft

d =m - Lm -f

=2.87 gm/cc - 2.522 gm/cc

2.87 gm/cc - 1.0 gm/cc= 18.619%

s =tL - tm

tf- tm

=62.73 sec/ft - 43.5 sec/ft

189.0 sec/ft - 43.5 sec/ft= 13.22%

limestone = 11%

Since both logs "read" nearly the same porosity when a limestone lithology wasassumed 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 rockvolume,including all pores. The sonic log tends to measure the velocity ofcompressional waves that travel through interconnected pore structures as well as therock matrix. The general consensus is that the sonic log measures effective porositywhen we use the Wyllie "time-average" equation.

It is expected that the effective porosity is always less than ,or equal to,the totalporosity.

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VII) Compressibility of porous rocks

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

c = -1VV

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

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4. As fluid is produced from a reservoir, the fluid pressure,Pfwill 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

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1) Thus,

cf= -1

Vp

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= -1

Vp

Vp

pm

it follows that

cf= +1

Vp

Vp expelled

Po

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(f) plot

CUMU

LATIVEVOLUMEEXPELLED

POREVOLUME

OVERBURDEN PRESSURE, psi

slope = cf.

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C. Measurement of compressibility

1) Laboratory core sample

a) apply variable internal and external pressures

b) internal rock volume changes

2) Equipment

Internal

Pressure

Gauge

HydraulicPump

Overburden

Pressure

Gauge

Hydraulic

Pump

Copper - J acketed

Core

Mercury Sight Gauge

Apparatus for measuring pore volume compressibility (hydrostatic)

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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 0.2442000 0.3243000 0.3924000 0.4485000 0.5006000 0.5467000 0.5968000 0.630

Solution:

from graph

@ 4,000 psi:

Slope = 0.0094000 psi

cf = 2.25 X 10-6 1

psi

@ 6000 psi:

Slope = 0.0116000 psi

cf = 1.83 X 10-6 1

psi

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1000080006000400020000

0.000

0.005

0.010

0.015

COMPACTION PRESSURE, psi

VOLUMEEXPELLED,

cc

POREVOLUME,cc

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INITIAL POROSITY AT ZERO NET PRESSURE, %

302520151050

1

10

100

PORE-VOLUME COMPRESSIBILITY AT 75 %

LITHOSTATIC PRESSURE VS INITIAL SAMPLE

POROSITY FOR CONSOLIDATED SANDSTONES.

CONSOLIDATED SANDSTONES

HALL'S CORRELATION

POREVOLUMECOMPRESSIBILITYX10-6psi-1

PORE

VOLUMECOMPRESSIBILITYX10-6psi-1

302520151050

1

10

100

PORE-VOLUME COMPRESSIBILITY AT 75 %

LITHOSTATIC PRESSURE VS INITIAL SAMPLE

POROSITY FOR UNCONSOLIDATED SANDSTONES.

UNCONSOLIDATED SANDSTONES

HALL'S CORRELATION

INITIAL POROSITY AT ZERO NET PRESSURE, %

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E. Abnormally pressured reservoirs

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

PRESSURE

DEPTH

NORMAL LINEAR

SUBNORMAL

(LOWER)

SURNORMAL

(GREATER)

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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. dry

ls=

241.0 gm

2.71 gm/cc= 88.9 cc

Vparaffin =wt. coated sample - st. uncoated sample

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

=Vp

Vb= 26.7 cc

115.6 cc= 0.231

= 23.1% (total porosity)

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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 remainingpore matrix

component volume porosity volume

(ft3) (%) (ft3)___

void space 1.000 100.0 0.000

gravel 0.250 25.0 0.750coarse sand 0.095 9.5 0.905

fine sand 0.039 3.9 0.961

Final porosity - 3.9%

(Complete mixing of the grains)

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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 0.1222,000 0.1623,000 0.1964,000 0.2245,000 0.2506,000 0.2737,000 0.2988,000 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

=m - Lm -f

= 2.65 - 2.402.65 - 1.00

= 15.22%

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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 =D2h

4=

3.14 5.0 2 4.0

4.0= 78.5 cc

Vm = matrix volume = 140 gmcc

2.65 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 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.

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800060004000200000

0.0040

0.0060

0.0080

0.0100

0.0120

0.0140

POROSITY PROBLEM No. 3

PRESSURE, psig

SLOPE =.0118 - .0068

7000 - 2000

VOL

UMEEXPELLED,cc

P

OREVOLUME,cc Cp = 9.96 x 10

-7 psi -1

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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 = -1

Vp dVp

dp

5 x 10-6 (1/psi) = -1

76,665,600 ft3

dVp

3,000 psi

dVp = 1.15 x 106 ft3

H= 1.15 x 106 ft3 x 1

6,969,600 ft2= 0.165 ft

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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)

L

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

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q =kA(h1 - h2)

L

B. Generalized form of Darcy's equation

1. Equation

vs =-k

dPds

-g

1.0133 x 106dzds

-1

+190o 180o 270o 360o

s

Vs

+X

+Y

-Z

+Z

2. Nomenclature

vs = superficial velocity (volume flux along

path s) - cm/sec

vs/ = interstitial velocity - cm/sec

= density of flowing fluid - gm/cm3

g = acceleration of gravity - 980 cm/sec2

dP = pressure gradient along s - atm/cmds

= viscosity - centipoise

k = permeability - darcies

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3. Conversion factors

dyne = gm cm/sec2 = a unit of force

atm = 1.0133 x 106 dyne/cm2

gh = 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

= m/Lt

= m/L3

p = m/Lt2

g = L/t2

vs = -k

dp

ds-

g

1.0133 x 106 dzds

Lt

= - km/Lt

m/Lt2L

-m/L3 L/t2 L

L

k = L2 = cross-sectional area

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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 - P2L

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II) Reservoir systems

A. Flow of incompressible liquid

1. Horizontal, linear flow system

L

Aq

P1

q

P2

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, , q

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b. derivation of flow equation

vs = -k

dPds

-g

1.0133 x 106 dzds

vs = - k

dP

ds =

q

A

q ds

0

L

= - kA

dPp1

p2

q L - 0 = - kA

P2 - P1

q = kAL

P2 - P1

Note: P1

acts at L = 0

P2

acts at L = L

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

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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 = = 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 = kAL

P2 - P1

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

q = 91.42 cm3/sec

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2. Non-horizontal, linear system

-Z

P1

S

X

P2

a. Conditions

1) non-horizontal system,dzds

= sin= constant



4) laminar flow, use Darcy equation



7) constant temperature , q

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b. derivation of equation

vs = -k

dPds

-g

1.0133 x 106dzds

vs = - qA = - k dPds +kg sin

1.0133 x 106

q ds

0

L

= - kA

dp

P1

P2 +

kA g sin

1.0133 x 106ds

0

L

q = - kAL

P1 - P2 +gLsin

1.0133 x 106

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3. Vertical, upward flow, linear system

L

x

h

FLOW UNDER

HEAD h

a. Conditions

1) vertical system, dzds = sin = constant

2) upward flow, q = 270, sin = - 1






8) constant temperature,

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b. derivation of flow equation

vs =k

dPds

-g

1.0133 x 106dzds

vs =qA

= - k

dPds

+ g

1.0133 x 106

q =kA

P1 - P2

L-

g

1.0133 x 106

P1 = -g (h + x + L)

1.0133 x 106

P2 =g x

1.0133 x 106

P1 - P2L

=g h

1.0133 x 106 L+

g

1.0133 x 106

q = kA

g h

1.0133 x 106 L+

g

1.0133 x 106-

g

1.0133 x 106

q = kA

L

g h

1.0133 x 106

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4. Horizontal, radial flow system

h

rw

re

Pe

Pw

re rw

a. Conditions

1) horizontal system,dzds = 0

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

3) constant thickness, h = constant




7) 100% saturated with liquid,

8) constant temperature, , q

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b. Derivation of flow Equation

vs = -k

dPds

-g

1.0133 x 106 dz

ds

vs = +

k

dP

dr =

q

A =

q

2rh

q

2h dr

rrw

re= k

dp

pw

pe

q

2h1n(re) - 1n( rw) =

k

Pe - Pw

q = 2hk 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

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b. Assumptions

1) , Z = constant

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

c. Derivation of equation for qsc

vs = -k

dPds

-g

1.0133 x 106 dzds

vs = -k

dPds

=q

Ads

but

q =

Psc qscz T

PTsc

thus

Psc T qscTsc A

ds

o

L

= - k PdPz

p1

p2

Psc T qscTsc A

L -0 = - kz

P2

2 - P12

2

qsc =kAL

Tsc

Tz Psc

P1 - P22

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 Tn R Tsc

q =Psc qscz T

Tsc 1P

where qsc is constant

Z is determined at P, T

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d. Derivation of equation for q

qsc =kAL

Tsc

Tz Psc

P1 - P22

but

qsc =P q Tsc

Z Psc T= k A

L

TscT z Psc

P1

2 - P22

2

q = k AL

1

P

P12 - P2

2

2

q = k AL

2P1 + P2

P1

2 - P22

2

q = k AL

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 qsc

P 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

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2. Horizontal, radial flow system

h

rw

re

Pe

Pw

re rw

a. Conditions

1) horizontal systemdzds

= 0

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

3) constant thickness, h = constant

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




8) constant temperature

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b. Assumptions

z = constant

z (and ) can be determined at mean pressure

c. derivation of equation for qsc

vs = -k

dPds

-g

1.0133 x 106dzds

vs = -k

dPds

=q

A

but

q =Psc qsc z T

PTsc

and

A = 2rh and ds = - dr

thus

Psc T qsc2Tsc h

drr

rw

re= k

dP

zPw

Pe

PscT qsc2 Tsch

1nrerw

= kz

Pe

2 - Pw2

2

qsc =2 h k

1n re/rw

TscPsc zT

Pe

2 - Pw2

2

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d. derivation of equation for q

qsc =2 h k

1n re/rw

TscPsc zT

Pe

2 - Pw2

2

but

q =P q Tscz Psc T

thus

P q Tscz Psc T

= 2 h k 1n re/rw

Tsc

Psc zT

Pe2 - Pw

2

2

q = 2 h k 1n re/rw

1P

(Pe2 - Pw2 )2

q = 2 h k 1n re/rw

2Pe + Pw

(Pe

2 - Pw2 )

2

q = 2 h k 1n re/rw

Pe - Pw

Note: Equation for real gas is identical to equation for incompressible liquid when

volumetric flow rate of gas,q, is measured at mean pressure.

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C. Conversion to Oilfield Units

Symbol Darcy units Oil field

q cc/sec bbl/d or cu ft/d

k darcy mdA sq cm sq fth cm ftP atm psiaL cm ft

cp cpr gm/cc lb/cu ft

Example:

q =hkA P1 - P2

L in Darcy's units

q ccsec

= q bbld

5.615 cu ftbbl

1,728 cu in

cu ft 16.39 cc

cu in d

24hr hr

3,600 sec

q ccsec

= 1.841 q bbld

k darcy = k md

darcy

1,000md

k darcy = 0.001 k md

A sq cm =929.0 sq cm

sq ftA sq ft

A sq cm = 929.0 A sq ft

P1 - P2 atm = P1 - P2 psiaatm

14.696 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

30.48 L

q =0.01127 k A P1 - P2

L in oilfield units

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D. Table of Equations

1. Darcy Units

System Fluid Equation

Horizontal,Linear

IncompressibleLiquid

q = kAL

P1 - P2

Dipping,Linear

IncompressibleLiquid q = kA

L P1 - P2 +

g L sin

1.0133 x 106

Horizontal,

Radial

Incompressible

Liquidq = 2k h

ln (re/rw) P

e- P

w

Horizontal,Linear

RealGas qsc =

kAL

Tsc

Tz Psc

P12 - P2

2

2

q = kA L

P1 - P2

Horizontal,Radial

Real Gasqsc =

k h ln (re/rw)

TscTz Psc

Pe2 - Pw 2

q = 2k hln (re/rw)

Pe - Pw

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2. Oilfield Units

System Fluid Equation

Horizontal,Linear


q = 0.001127kAL

P1 - P2

q = res bbl/d

Dipping,Linear


q = 0.001127 kAL

P1 - P2

+g L sin

1.0133 x 106

Horizontal,Radial


q = .007082 kh ln (re/rw)

Pe - Pw

Horizontal,Linear Real Gas

qsc = .1118k A

L z T P1

2 - P22

qsc = scf/d

q = .001127 kAL

P1 - P2

q = res bbl/d

Horizontal,Radial Real Gas

qsc = .7032k h

ln (re/rw) TzPe

2 - Pw2

q = .007082 kh ln (re/rw)

Pe - Pw

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Example II-2

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

permeability = k = 1 darcy

area = A = 6 ft2

viscosity = = 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 md

A = 6 ft2

L = 6 ftP1 = (5.0 atm) (14.7 psi/atm) = 73.5 psi

P2 = (2.0 atm) (14.7 psi/atm) = 29.4 psi

q = 1.1271 x 10-3 kAL

P1 - P2

q = 1.1271 x 10-31,000 6

1 6 73.5 - 29.4

q = 49.7 bbl / day

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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

= 1.3 cp

Solution:

q =7.082 x 10-3 kH Pe - Pw

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

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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 - P2

L

and

qt =kA whAL

P1 - P2 +kB whB L

P1 - P2 +kC whCL

P1 - P2

then

k h = kA hA + kB hB + kC hC

k = j = 1

nkj hj

h

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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 =2k h

ln (re/rw)

Pe - Pw

and

qt =2kA hAln (re/rw)

Pe - Pw +2kB hBln (re/rw)

Pe - Pw

+2 kc hcln (re/rw)

Pe - Pw

then

k h = kA hA + kB hB + kC hC

and again

k = j = 1

nkj hj

h

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3. Horizontal, linear flow perpendicular to bedding

L

h

P2

P1

W

q

kA

PA

LA

A B C

kB

PB

LB

kC

PC

LC

q

qt = qA = qB = qC

p1 - p2 = PA + PB + PC

L = LA + LB + LC

qt =

k wh P1 - P2

L

and since P1 - P2 = PA + PB + PC

P1 - P2 =qt L

k wh =

qA LA

kA wh +

qB LBkB wh

+qC LCkC wh

since qt = qA = qB = qC

Lk

= LAkA

+ LBkB

+ LCkC

thus

K = L

j = 1

n

Lj

kj

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4. Horizontal, radial, flow perpendicular to bedding

h

rw

rArB

rC

q

PCPBPAPw

qt = qA = qB = qC

Pe - Pw = PA + PB + PC

q =2 k h Pe - Pw

ln (re/rw)

Pe - Pw =qt ln (re/rw)

2 k h =

qA ln (rA/rw)

2 kAh

+qB ln (rB/rA)

2 kB h +

qC ln (rC/re)

2 kC h

then

k =ln re/rw

j = 1

n

ln(rj/rj-1)

kj

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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 300/2

200

k = 30.4 md

The permeability of the damaged zone near the wellbore influences the average permeability more

than the permeability of the undamaged formation.

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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 = r4

8L P1 - P2

A = r2, therefore

q = Ar2

8L P1 - P2

b. Darcy's law for linear flow of liquids

q = kA L

P1 - P2

assuming these flow equations have consistent units

Ar2

8L P1 - P2 =

kA L

P1 - P2

thus

k = r2

8

= d2

32

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

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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

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2. Flow through fractures

b

v =q

A

= h2

12

L

(P1-P2)

q = b2 A

12 L(P1 -P2)

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

b2 A12L

P1 - P2 = kA

L P1 - P2

solve for permeability of a fracture:

k = b2

12in darcy units, or

k = 54 x 109 b2

where b = inchesk = md

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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 the

equation 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

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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 - P22

2 L Psc horizontal, linear, real gas flow with

T = Tsc and Z = 1.0

qsc PscA

= k

P1

2 - P22

2L

k = ( slope ) m

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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 AIR

PRESSURE

PIPE

RUBBER

TUBING

CORE

LOW AIR

PRESSURE

(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

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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 for

permeability.

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.

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0

kCALCULATED

1

P

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

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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 k

measured

vs. 1/pressure

Intercept is equivalent to liquid permeability

From graph:

kliq = 2.38 md

0

1

2

3

4

5

G

ASPERMEABILITY,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

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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

100001000100101

1

10

100

1000

WATERPERMEABILITY,md

AIR PERMEABILITY, md

Water concentration

20,000 - 25,000 ppm Cl ion.

RELATIONSHIP OF PERMEABILITIES MEASURED

WITH 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.

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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 inpore 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 decreasein pore pressure and a reduction of in-situ permeability.

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III- 1

III. BOUNDARY TENSION AND CAPILLARY PRESSUREI) Boundary tension,

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

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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 orbetween a fluid and a solid is called "interfacial tension"

gw = surface tension between gas and water

go = surface tension between gas and oil

wo = interfacial tension between water and oil

ws = interfacial tension between water and solid

os = interfacial tension between oil and solid

gs = 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

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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)

ow

os ws

OIL

WATER

OIL

SOLID

ws = os + ow cos B. Adhesion Tension, AT

AT = ws - os = ow cos C. if the solid is "water-wet"

ws osAT = +

cos = +0 90if = 0 - strongly water-wet

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III- 4

D. if the solid is "oil-wet"

os wsAT = -

cos = -90 180if = 180 - strongly oil-wet

= 830 = 1580

= 350 = 300

(A)

ISOOCTANE ISOOCTANE + 5.7%

ISOQUINOLINE

ISOQUINOLINE NAPHTHENIC ACID

= 300 = 480 = 540 = 1060

(B)

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

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III- 5

III) Capillary pressure

A. capillary pressure between air and water

hAIR

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

= 2r AT

b. total force down equals the weight of the column of water

converted to force

= r2 hgw

c. thus when column of water comes to equilibrium

2r AT = r2 hgw

d. units

cmdynecm

= cm2 cm cm

sec2

gm

cm3

dyne =gm cm

sec2

dyne = force unit

e. adhesion tension

AT =12

r hgwdynecm

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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

= 2r aw cos

b. total force down

= r2 hgw

c. thus when the column of water comes to equilibrium

2r aw cos = r2 hgw

d. units

cmdynecm

= cm2 cm cm

sec2

gm

cm3

cmdynecm

= cmgm

sec2

3. since AT = aw cos, 1 and 2 above both result in

h =2 aw cos

rg w

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III- 7

4. capillary pressure (air-water system)

h

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 - wgh

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 - wgh)

pa - pw = wgh

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

e. remember

h =2 gw cos

rg w

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III- 8

f. thus

Pc =2 gw cos

r

B. capillary pressure between oil and water

h

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

= 2r ow cos

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

= w h

2) weight per unit area of oil

= o h

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III- 9

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

= wh - oh = h(w - o)

4) total force down

= r2 gh (w - o)

d. thus when the column of water comes toequilibrium

2r ow cos = r2 gh (w - o)

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

h =2 ow cos

rg (w - o)

3. capillary pressure (oil-water system)

h

WATER

Po

A

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 - wgh

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

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III- 10

Pob = Poa - ogh

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

Pc

= Pob

- Pwb

Pc = (Poa - ogh) - (Pwa - wgh)

since Poa = Pwa

Pc = (w - o)gh

e. remember

h = 2 ow cos rg (w - o)

f. thus

Pc =2 ow cos

r

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

Pc

= 2 cos

r

C. remember adhesion tension is defined as

AT = ow cos,

and

Pc =2 ow cos

r

thus

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

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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 = 1

R1+ 1

R2R1 and R2 are the principal radii of curvature for a liquid adhering to two spheres in

contact with each other.

3. by analogy to capillary tube

1R1

+ 1R2

= 2 cos r

where Pc = 2 cos r

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

1Rm

= 2 cos rm

=()gh

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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 covering

surface 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

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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 density

3. 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

HIGHER

PRESSURE

DEPENDENCE OF INTERFACIAL CURVATURE ON FLUID SATURATIOIN 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

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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)

SATURATION = 100%

PC = LOW VALUE

SATURATION = 80%

CAPILLARY PRESSURE = P C

R

LOW PC HIGHER P C

(A)

SATURATION = 0%

PC = HIGH VALUE

SATURATION = 10%

CAPILLARY PRESSURE = P

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

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III- 15

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

100806040200

0.0

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)

g

h

(k/)1/2

Drainage

Imbibition

Drainage Imbibition

Sand I

Sand II

WATER SATURATION, Sw %

1. capillary pressure is expressed in terms of a non-dimensional correlating

function ( remember Pc = ( 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

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III- 16

PC

0 100WATER SATURATION, S W

IMBIBITION

DRAINAGE

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III- 17

VI) Capillary pressure in reservoir rock

Pw = Po/ w -

wh

144Po = Po/ w -

oh

144

Water Oil

Pw2 Po2

Po1 = Pw1

100% Water

Oil and Water

Pc = Po - Pw =h

144 w - o

Where: Po = pressure in oil phase, psia

Pw = pressure in water phase, psia

h = distance above 100% water level, ftP

o/w= pressure at oil-water contact, psia

w = density of water, lb/cf

o = density of oil, lb/cf

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

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III- 18

HEIGHT

ABOVE

O-W-C

PRESSURE

PCPO = PO/ W -

144

oH

Pw = PO/W -wH

144

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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

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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 is

related to rotational speed)

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

4. Advantages

a. fastb. reasonably accuratec. use reservoir fluids

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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. Disadvantagesa. very tedious

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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)

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III- 23

Example VIII-1

Comparison of Mercury Injection Capillary Pressure Data with Porous Diaphragm Data

A. Calculate capillary pressure ratio,

PcAHg

PcAW

, for the following data

AHg= 480 Dynes/c

AW= 72 Dynes/c

AHg= 140 AW= 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.

PcAHg

PcAW

=AHgAW

Solution:

(A) PcAHgPcAW

=AHgcos AHg

AWcos AW =480 cos(140)

72 cos (0)

PcAHg

PcAW

= 5.1

(B) PcAHgPcAW

@

AHgAW = 48070

PcAHg

PcAW

= 6.9

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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.

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III- 25

VIII) Converting laboratory data to reservoir conditions

PcL= 2Lcos L

r

PcR= 2Rcos R

r

setting r = r

r =2LcosLPcL

=2RcosRPcR

PcR= cos R

cos L

PcL

where

PcR= reservoir capillary pressure, psi

PcL= capillary pressure measured in laboratory, psi

L = interfacial tension measured in laboratory, dynes/cm

R = reservoir interfacial tension, dynes/cm

R = reservoir contact angle, degrees

L = laboratory contact angle, degrees

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III- 26

Example III-2

Converting Laboratory Data to Reservoir Conditions

Express reservoir capillary pressure by using laboratory data.

lab data: AW = 72 dynes

AW = 0o

reservoir data: OW = 24 dynes/cm

OW = 20o

Solution:

PcR

=cos R

cos L PcL

PcR

=24 cos20

72 cos0 PcL

PcR=

0.333PcL

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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 =()gh144 gcin English units

= w - O, lb/cu ft

g = 32 ft/sec2

gc = 32 lbm ft

lbf sec2h = ft

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

thus

Pc = lbf/(sq in), psI

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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.

o= 0.85 gm/cm3 w= 1.0 gm/cm3

PCL

SW

0

0

10

20

50 100

8.38.3

Solution:

PcR=

w o h

144

PcR=

1.0 - 0.85 62.4 lb

ft340

144= 2.6 psi

PcL=

PcR0.313

PcL= 2.6

0.313= 8.3 psi

move to the right horizontally fromPcL = 8.3 psi to the capillary pressure curve. Drop vertically

to the x-axis, read Sw.

Sw = 50%

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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

1009080706050403020100

0

20

40

60

80

100

120

140

160

180

200

RESERVOIR FLUID DISTRIBUTION CURVES

Sw %

(From Wright and Wooddy)

10md

100md

200md

900md

30

Heightabovezerocapillarypressure,

ft

24

18

12

6

0

Oil-WaterCapillaryPressure,ps

i

(reservoirconditions)

90

72

54

36

18

0

A

ir-WaterCapillaryPressure,psi

(laboratorydata)

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III -30

B. Effect of grain size distribution

100806040200

0

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 md

Factor = 7.5

Mercurycapillarypressure,psi

Water/nitrogencapillarypressure,psi

Hg

Water

100806040200

0

10

20

30

40

50

60 348

290

232

174

116

58

0

100 80 60 40 20 0

Water/nitrogencapillary

pressure,psi

Mercurycapillarypr

essure,psi

Water

Hg

Limestone Core

Porosity = 23.0%

Permeability = 3.36 md

Factor = 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)

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III -31

XI) Averaging capillary pressure data

J-function

J Sw = Pccos

k

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)

= dyne/cm

k = (sq cm)

= fraction

or can use any units as long as you are consistent

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III -32

1009080706050403020100

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.1

1.2

1.3

1.4

1.5

1.6

1.71.8

1.9

2.0

CAPILLARY RETENTION CURVES.

WATER SATURATION, Sw

CAPILLARYPRESSUREFUNCTION,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)

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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) A = 72 dynes/cm

W = 1 gm/cm3

g = 980 dynes/gm

= 0o

(a) capillary rise of water if radius is .01 cm

h =2AWcos rg

= 2 72 cos0.01 1.0 980

h = 14.69 cm

(b) pressure drop in psi across interface

Pc = pa - pw =wgh = 1.0 980 14.69

Pc= 0.0142 atm

14.696 psi

atm

Pc

= 0.209 psi

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III -34

(2)Pc

=2AWcos

r

Pc = 55 psi

Pc = 55 psi

atm14.696 psi

1.0133 x 106 dynes/cm2

atm

= 3.792 x 106 dynes/cm2

r = 2AWcos Pc

r = 2 72 cos0

3.792 x 106= 3.797 x 10-5 cm in

2.54 cm

r = 1.495 x 10

-5

in

d = 2.99 x 10-5 in

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III -35

Capillary Pressure Problem 2

Given the information below and graph ofPcL vs. wetting phase saturation Sw , construct the

curves forPcR, h in reservoir, and J-function vs. Sw. Water is the wetting phase in both the

laboratory and the reservoir.

fluidslab

air-waterres

oil-wate

0 25

60 dyne/cm 20 dyne/cm

wet 1.0 gm/cm3 1.1 gm/cm3

non-wet 0 gm/cm3 0.863 gm/cm3

k 37 md variable

16% variable

J =Pc k/ 1/2cos

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III -36

1009080706050403020100

0.0

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 %

PCL,psi

Solution:

(1)PcR

= RcosR

LcosL

PcL

= 20 cos25

60 cos0PcL

PcR

=0.302 PcL

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III -37

(2)PcR

=hR w- o

144

= hR 1.1 - .863 62.4

144

PcR = .103 hR

hR= 9.74 PcR

(3) J =Pc

cos k1/2

=PcL

AWcosL

k L

1/2

=PcL

60 cos0 37

.16

1/2

J = .253 PcL

SwPcL

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

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III -38

100806040200

0

2

4

6

8

10

Sw %

PcR

100806040200

0

20

40

60

80

100

Sw %

hR

100806040200

0

2

4

6

8

10

Sw %

J

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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>Pres2) for water base mud, filtrate displaces formation water

and oil from the area around the well (saturationslikely change)

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

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Example: Effects of flushing by mud filtrates

Coring with water base mud

Oil zone at minimum interstitial water saturation:

sat at surfaceflushing by bit trip to surface compared to res

Sw ? probablySo

Sg-

Gas zone at minimum interstitial water saturation:


Sw ?

So - - -

Sg ?

Water zone:


Sw -

So - --

Sg -

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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 surfaceflushing by bit trip to surface compared to res

Sw

So

Sg-

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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

CONDENSER

COOLING WATER OUT

CORE

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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

6560555045403530252015

0.9

1.0

1.1

1.2

1.3

1.4

Oil Gravity, API at 60 F

Multiplying

Factor

For converting distilled oil volume to oil volume originally in a sample, multiply

oil volume recovered by factor corresponding to gravity of oil in core

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e) determine saturations

Sw =VwVp

So =VoVp

Sg = 1 - So - Sw

where

Sw = water saturation, fraction

So = oil saturation, fraction

Sg = gas saturation, fraction

Vp = pore volume, cc

Vw = volume of water collected, cc

Vo = 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

WATER

RECOVERED

TIME

0

0

PORE WATER

b) at high temperatures, oil will crack and coke.

(change in hydrocarbon molecules) amountof recoverable liquid decreases.

c) core sample ruined

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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

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

e) reweigh core sample

f) determine volume of oil in sample

Vo = Wi - Wdry - Vw wowhere:

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

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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

So0 in gas zoneSo 15% in oil zone0SoSorin water zoneSor = residual oil saturation

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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

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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 w= 1 gm/cm3

Wt new = 219.7 gm

the water with in the sample is removed and collected

Vw removed = 13.0 cm3

the oil is extracted and the sample is dried

Wt dry = 199.5 gm

calculate:

(1) porosity

(2) water saturation

(3) oil saturation assuming 35o API

(4) gas saturation

(5) matrix density

(6) lithology

Solution:

gas vol. = 219.7 - 216.7 ; Vg = 3 cc

water vol. = 13 - 3 ; Vw = 10 cc

Wt fluids = 219.7 - 199.5 = 20.2 gm

Wt oil = 20.2 - 10 - 3 = 7.2 gm

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o = 141.5

131.5 + 35API= 0.85 gm/cc

Vo = 7.2/0.85 = 8.49 cc

Vp = 8.49 + 3 + 10 = 21.47 cc

= 21.47/95 = 22.6%

Sw = 10/21.47 = 46.57%

So = 8.49/21.47 = 39.46%

Sg = 3/21.47 = 13.97%

m

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

lithology = limestone

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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 w = 1.0 gm/cc and 35o API oil)

calculate:

(1) porosity

(2) water saturation

(3) oil saturation

(4) gas saturation

(5) lithology

Solution:

Vg

= .71 cc70 gm 100 gm = 1.014 ccVo = 4.5 cc

80 gm100 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

= 10.14/43.52 = 23.31%

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Sw = 3.50/10.14 = 34.5%

So = 5.63/10.14 = 55.5%

Sg = 1.014/10.14 = 10%

m = (91.71/33.38) = 2.75

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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 w= 1.0 gm/cc

oil gravity = 40API

Solution: o = 141.5131.5 + API

o = 141.5131.5 + 40

= 0.825o = 0.825 gm/cc

(1) =

Vp

Vb

Vp = Vw + Vo + Vg

Wo = Wsat - Vww- Wdry

= 87.95 - 7.12(1.0) - 79.17

Wo = 1.66 gm

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Vo =Woo

Vo = 1.66 gm

0.825 gm/cc= 2.01 cc

Vw = Vwrec- Wsat - Wi/w

= 7.12 - (87.95 - 86.4)/(1.0)

Vw = 5.57 cc

Vg = 1.55 cc

Vp = 5.57 + 2.01 + 1.55

Vp = 9.13 cc

Vb =Wsat - Wimm

w

Vb =(87.95 - 48.95) gm

1 gm/cc= 39.0 cc

= 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 =

Vg

Vp

Sg = 1.55 cc9.13 cc

= 17.0%

(3) Vm = Vb - Vp

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Vm = 39 - 9.13 = 29.87 cc

m=

Wdry

Vm

m=

79.17 gm/29.87 cc = 2.65gmcc

. . 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

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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

assumew= 1.0 gm/cc

oil gravity = 40o API

Solution: o = 141.5131.5 +API

o = 141.5

131.5 + 40= 0.825

o = 0.825 gm/cc

(1) =

Vp

Vb

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

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