Reservoir Petrophysics Class Notes

PETROLEUM ENGINEERING 311

RESERVOIR PETROPHYSICS

CLASS NOTES (1992)

Instructor/Author:

Ching H. Wu

TABLE OF CONTENTS

I. ROCK POROSITY I-1

I) Definition I-1

II) Classification I-1

III) Range of values of porosity I-2

VI) Factors affecting porosity I-3

V) Measurement of porosity I-5

VI) Subsurface measurement of porosity I-13

VII) Compressibility of porous rocks I-25

II. SINGLE PHASE FLOW IN POROUS ROCK II-1

I) Darcy's equation II-11

II) Reservoir systems II-15

III. BOUNDARY TENSION AND CAPILLARY PRESSURE III-11

I) Boundary tension III-1

II) Wettability III-3

III) Capillary pressure III-5

IV) Relationship between capillary pressure and saturation III-13 V) Relationship between capillary pressure and saturation history III-14 VI) Capillary pressure in reservoir rock III-17 VII) Laboratory measurement of capillary pressure III-19 VIII) Converting laboratory data to reservoir conditions III-25 IX) Determining water saturation in reservoir from capillary pressure data III-27

X) Capillary pressure variation III-29

XI) Averaging capillary pressure data III-31

IV. FLUID SATURATIONS IV-1

I) Basic concepts of hydrocarbon accumulation IV-1 II) Methods for determining fluid saturations IV-1 V. ELECTRICAL PROPERTIES OF ROCK-FLUID SYSTEMS V-1 I) Electrical conductivity of fluid saturated rock V-1 II) Use of electrical Formation Resistivity Factor, Cementation Factor, and

Saturation Exponent V-8

III) Laboratory measurement of electrical properties of rock V-9

IV) Effect of clay on resistivity V-18

VI. MULTIPHASE FLOW IN POROUS ROCK VI-1

I) Effective permeability VI-1

VII. STATISTICAL MEASURES VII-1

I) Introduction VII-1

II) Frequency Distributions VII-2

III) Histogram VII-3

IV) Cumulative Frequency Distributions VII-6

V) Normal Distribution VII-8

VI) Log Normal Distribution VII-9

VII) Measures of Central Tendency VII-10

VIII) Measures of Variability (dispersion) VII-11

IX) Normal Distribution VII-12

I. ROCK POROSITY

I) Definition

A measure of the pore space available for the storage of fluids in rock In general form: Porosity = φ = Vp Vb = Vb - VmVb where: φ is expressed in fraction Vb = Vp + Vm

Vb = bulk volume of reservoir rock, (L3) Vp = pore volume, (L3)

Vm= matrix volume, (L3)

II) Classification

A. Primary (original) Porosity Developed at time of deposition B. Secondary Porosity

Developed as a result of geologic process occurring after deposition

C. Total Porosity

φ_{t =} total pore space

Vb = Vb - VmVb

D. Effective Porosity

φ_{e =} interconnected pore space

Vb

VI) Factors affecting porosity A. Factors:

1. Particle shape 2. Particle arrangement 3. Particle size distribution 4. Cementation

5. Vugs and fractures B. Particle shape

Porosity increases as particle uniformity decreases. C. Packing Arrangement

Porosity decreases as compaction increases

6000 5000 4000 3000 2000 1000 0 0 10 20 30 40 50 POROSITY, %

EFFECT OF NATURAL COMPACTION ON POROSITY

(FROM KRUMBEIN AND SLOSS.)

SANDSTONES

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 CLAY

100

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 material

V) Measurement of porosity

φ = Vb - Vm

Vb = Vp Vb Table of matrix densities

Lithology ρ_{m (g/cm}3) ___________ ___________ Quartz 2.65 Limestone 2.71 Dolomite 2.87 A. Laboratory measurement

1. Conventional core analysis a. measure any two

1) _{bulk volume, Vb} 2) _{matrix volume, Vm} 3) _{pore volume, Vp} b. bulk volume

1) calculate from dimensions 2) displacement method

a) volumetric (measure volume)

(1) drop into liquid and observe volume charge of liquid

(2) must prevent test liquid from entering pores space of sample

(a) coat with paraffin

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

b) gravimetric (measure mass)

sample-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

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}

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 from

Example I-1

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

Solution:

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

Example I-2

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

Solution: Bulk Volume = 9.9 cc Matrix Volume = 7.7 cc φ = Vb - Vm Vb = 9.9 cc- 7.7 cc9.9 cc = 0.22 It is total porosity.

Example I-3

Calculate the porosity of a core sample when the following information is available: Dry weight of sample = 427.3 gm

Weight of sample when saturated with water = 448.6 gm Density of water = 1.0 gm/cm3

Weight of water saturated sample immersed in water = 269.6 gm

Solution:

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

Vp = 448.6 gm - 427.3 gm 1 gm/cm3

Vp = 21.3 cm3

Vb = sat. core wt. in air - sat. core wt. in water density of water Vb = 448.6 gm - 269.6 gm 1 gm/cm3 Vb = 179.0 cm3 φ = Vp Vb = 21.3 cm3179.0 cm3 = .119

What is the lithology of the sample?

Vm = _{Vb - Vp}

Vm = 179.0 cm3 - 21.3 cm3 = 157.7 cm3

ρ_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?

Example I-4

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

Solution: P 1V1 = P2V2 V 1 = 1,000 cc P

1 = 50 psig + 14.7 psia = 64.7 psia P 2 = 28.1 psig + 14.7 = 42.8 psia V 2 = (64.7 psia) (1,000 cc) / (42.8 psia) V 2 = 1,512 cc V m = VT - V2 V m = 2,000 cc - 1,512 cc - 488 cc φ = VT - Vm_VT = 695 cc - 488 cc695 cc = .298 = 29.8%

VI) Subsurface measurement of porosity

A. Types of logs from which porosity can be derived 1. Density log: φ_{d =} ρm - ρL ρ_{m -} ρ_f 2. Sonic log: φ_{s =} ∆tL - ∆tm ∆_{tf -} ∆_tm 3. Neutron log: e-kφ = CNf

Table of Matrix Properties

(Schlumberger, Log Interpretation Principles, Volume I) Lithology ∆tmµsec/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

B. Density Log

1. Measures bulk density of formation

FORMATION GAM M A RAY SOURCE SHORT SPACE DETECTOR LONG SPACE DETECTOR M UD CAKE

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

FORMATION DENSITY LOG 4240 4220 4200 4180 4160 4140 4120 4100 200 160 120 80 40 0 2.0 2.2 2.4 2.6 2.8 3.0 ρ, gm/cc GR, API depth, ft

Example I-5

Use the density log to calculate the porosity for the following intervals assuming ρ_{matrix = 2.68} gm/cc and ρ_{fluid = 1.0 gm/cc.} Interval, ft ρ L, gm/cc φd ,% __________ _________ ______ 4143-4157 2.375 18 4170-4178 2.350 20 4178-4185 2.430 15 4185-4190 2.400 17 4197-4205 2.680 0 4210-4217 2.450 14 Example: Interval 4,143 ft -4,157 ft : ρ L = 2.375 gm/cc φ_{d =} ρm - ρL ρ_{m -} ρ_f = 2.68 gm/cc - 2.375 gm/cc2.68 gm/cc - 1.0 gm/cc = 0.18

C. Sonic Log

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

D B A T R1 R2 E C

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

3. Equation

SONIC LOG 4240 4220 4200 4180 4160 4140 4120 4100 200 100 0 140 120 100 80 60 40

Example I-6

Use the Sonic log and assume sandstone lithology to calculate the porosity for the following intervals. 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} φ_{s =} ∆tL - ∆tm ∆_{tf -} ∆_tm = 86.5 µ sec/ft- 51.6 µ sec/ft 189.0 µ sec/ft- 51.6 µ sec/ft = 0.25

D. Neutron Log

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

Maximum Average Energy

Number Loss/ Atomic Atomic

Element Collisions Collision, % Collision Number

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

RELATIVE PROBABILITY FOR COLLISION

.1 1 10 102 103 104 105 H 106 O 107 Si 10-3 10-2 10-1 1

CLEAN SAND POROSITY = 15%

SLOWING DOWN POWER

NEUTRON ENERGY IN ELECTRON VOLTS

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

NEUTRON DENSITY LOG 4240 4220 4200 4180 4160 4140 4120 4100 200 0 30 -10 GR, API depth, ft φ (CDL)

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

Example I-8

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

The readings at 13,743 feet are: bulk density = 2.522 gm/cc travel 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} φ_{d =} ρm - ρL ρ_{m -} ρ_f = 2.65 gm/cc - 2.522 gm/cc2.65 gm/cc - 1.0 gm/cc = 7.76% φ_{s =} ∆tL - ∆tm ∆_{tf -} ∆_tm = 62.73 µ sec/ft- 55.5 µ sec/ft 189.0 µ sec/ft - 55.5 µ sec/ft = 5.42% Assume limestone: ρm = 2.71 gm/cc ∆_{tm = 47.5 µ-sec/ft} φ_{d =} ρm - ρL ρ_{m -} ρ_f = 2.71 gm/cc - 2.522 gm/cc2.71 gm/cc - 1.0 gm/cc = 10.99% φ_{s =} ∆tL - ∆tm ∆_{tf -} ∆_tm = 62.73 µ sec/ft - 47.5 µ sec/ft 189.0 µ sec/ft - 47.5 µ sec/ft = 10.76%

Assume dolomite: ρm = 2.87 gm/cc ∆_{tm = 43.5 µ-sec/ft} φ_{d =} ρm - ρL ρ_{m -} ρ_f = 2.87 gm/cc - 2.522 gm/cc2.87 gm/cc - 1.0 gm/cc = 18.619% φ_{s =} ∆tL - ∆tm ∆_{tf -} ∆_tm = 62.73 µ sec/ft - 43.5 µ sec/ft 189.0 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.

VII) Compressibility of porous rocks

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

c = - 1 V ∂V ∂_{P T}≅ - ∆V V T ∆P A. Normally pressured reservoirs

1. Downward force by the overburden must be balanced by upward force of the 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

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

a. the force on the matrix increases b. causing a decrease in bulk volume c. 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 spaces 2) internal rock stress changes

3) change in stress results in change in Vp, Vm, Vb 4) by definition cf = - 1 Vp ∂_Vp ∂_pm

b. _{since overburden pressure, Po, is constant} dPm = - dPf

1) Thus, cf = - 1

Vp

∂_Vp ∂_pm

2) _{where the subscript of f on cf means} "formation" and the subscript of f on Pf means "fluid"

3) procedure

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

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

(c) _{fluid pressure, pf, is essentially constant, thus,} dPo = dPm

(d) expelled volume increases as pore volume, vp, decreases, thus, dVp = - dVexpelled (e) from definition

cf = - 1 Vp ∂_Vp ∂_pm it follows that cf = + 1

V

_p ∆_{Vp expelled} ∆_Po

(f) plot

CUMULATIVE VOLUME EXPELLED

PORE VOLUME

OVERBURDEN PRESSURE, psi

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 Hydraulic Pump Overburden Pressure Gauge Hydraulic Pump Copper - Jacketed Core

Mercury Sight Gauge

Example I-9

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

pore volume = 50.0 cc

pressure, psi vol. fluid expelled, cc

1000 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 = 0.009 4000 psi cf = 2.25 X 10-6 _psi1 @ 6000 psi: Slope = 0.011 6000 psi cf = 1.83 X 10-6 _psi1

10000 8000 6000 4000 2000 0 0.000 0.005 0.010 0.015

COMPACTION PRESSURE, psi

VOLUME EXPELLED, cc

INITIAL POROSITY AT ZERO NET PRESSURE, % 30 25 20 15 10 5 0 1 10 100 PORE-VOLUME COMPRESSIBILITY AT 75 % LITHOSTATIC PRESSURE VS INITIAL SAMPLE POROSITY FOR CONSOLIDATED SANDSTONES.

CONSOLIDATED SANDSTONES

HALL'S CORRELATION

PORE VOLUME COMPRESSIBILITY X 10

-6 psi -1 -6 psi -1 10 100 PORE-VOLUME COMPRESSIBILITY AT 75 % LITHOSTATIC PRESSURE VS INITIAL SAMPLE POROSITY FOR UNCONSOLIDATED SANDSTONES.

E. Abnormally pressured reservoirs

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

PRESSURE DEPTH NORMAL LINEAR SUBNORMAL (LOWER) SURNORMAL (GREATER)

Compressibility/Porosity Problem No. 1

A limestone sample weighs 241.0 gm. The limestone sample coated with paraffin was found to weigh 249.5 gm. The coated sample when immersed in a partially filled graduated cylinder displaced 125.0 cc of water. The density of the paraffin is 0.90 gm/cc.

What is the porosity of the rock? Does the process measure total or effective porosity?

Solution:

Vm = wt. dry_ρ

ls = 241.0 gm2.71 gm/cc = 88.9 cc

Vparaffin = wt. coated sample - st. uncoated sample_ρ

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 = Vp Vb = 26.7 cc115.6 cc = 0.231 φ _{= 23.1% (total porosity)}

Compressibility/Porosity Problem No. 2

You are furnished with the results of a sieve analysis of a core from Pete well #1. Previous laboratory work indicates there is a correlation between grain size and porosity displayed by those particular particles. The correlation is seen below:

gravel - 25% porosity coarse sand - 38% porosity fine sand - 41% porosity

What would be the minimum porosity of the mixture?

What basic assumption must be made in order to work the problem?

Solution:

Begin calculation with a volume of 1 cu. ft.

remaining 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%

Compressibility/Porosity Problem No. 3

A sandstone reservoir has an average thickness of 85 feet and a total volume of 7,650 acre-feet. Density log readings through the fresh water portion of the reservoir indicate a density of 2.40 gm/cc.

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

pressure 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 were 1.575 inches long and 1.960 inches in diameter.

Assuming the compressibility at 4,500 psi is the average compressibility in the reservoir, how much subsidence occurs when the reservoir pressure declines from 5,500 psi to 3,500 psi?

Calculate:

A. Reservoir Porosity B. Sample Pore Volume

C. Compressibility at 4,500 psi D. Amount of Ground Subsidence.

Solution:

A. Reservoir Porosity

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 _{2.65 gm}cc = 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 usually encountered. An experienced engineer would be wary of this small number. Also it was assumed that the formation compressibility was exactly the same as the bulk volume compressibility. Experience shows that this is not the case.

8000 6000 4000 2000 0 0 0.0040 0.0060 0.0080 0.0100 0.0120 0.0140

POROSITY PROBLEM No. 3

PRESSURE, psig SLOPE = .0118 - .0068 7000 - 2000 VOLUME EXPELLED, cc PORE VOLUME, cc Cp = 9.96 x 10-7 psi -1

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

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

q = kA(h1 - h2)

µL

B. Generalized form of Darcy's equation 1. Equation vs = -k_µ dP_ds - ρg 1.0133 x 106 dz ds -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/cm ds

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 = - k_µ dp ds - ρg 1.0133 x 106 dzds L t = - km/Lt m/Lt2L - m/L3 L/t2 L L k = L2 = cross-sectional area

5. Definition of Darcy units

a. conventional units would be:

1) feet squared in the English system 2) centimeter squared in the cgs system b. both are too large for use in porous media c. definition of darcy

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

q = k A P1 - P2

II) Reservoir systems

A. Flow of incompressible liquid

1. Horizontal, linear flow system

L A q P1 q P2 a. Conditions 1) horizontal system, dz_ds = 0 2) linear system, A = constant

3) incompressible liquid, q = constant 4) laminar flow, can use Darcy's equation 5) non-reactive fluid, k = constant 6) 100% saturated with one fluid 7) constant temperature, µ, q

b. derivation of flow equation vs = - _µk dP_ds - ρg 1.0133 x 106 dzds vs = - _µk dP_ds = q A q ds 0 L = - kA_µ dP p 1 p2 q L - 0 = - kA µ P2 - P1 q = kA Lµ P2 - P1 Note: P 1 acts at L = 0 P 2 acts at L = L q is + if flow is from L = 0 to L = L

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 = kA Lµ P2 - P1 q = (1) (5,572.8 ) (5.0 - 2.0) (1) (182.88) q = 91.42 cm3 / sec

2. Non-horizontal, linear system Θ -Z P1 S X P2 a. Conditions 1) non-horizontal system, dz ds = sinθ = constant 2) linear system, A = constant

3) incompressible liquid, q = constant 4) laminar flow, use Darcy equation 5) non-reactive fluid, k = constant 6) 100% saturated with one fluid 7) constant temperature µ, q

b. derivation of equation vs = - _µk dP_ds - ρg 1.0133 x 106 dzds vs = - _Aq = - k µ dPds + k ρg sin θ µ 1.0133 x 106 q ds 0 L = - kA µ dp P1 P2 + kA ρg sin θ µ 1.0133 x 106 ₀ ds L q = - kA µL P1 - P2 + ρgLsinθ 1.0133 x 106

3. Vertical, upward flow, linear system L x h FLOW UNDER HEAD h a. Conditions 1) vertical system, dz ds = sinθ = constant 2) upward flow, q = 270°, sinθ = - 1 3) linear system, A = constant

4) incompressible liquid, q = constant 5) laminar flow, use Darcy equation 6) non-reactive fluid, k = constant 7) 100% saturated with one fluid 8) constant temperature, µ

b. derivation of flow equation vs = _µk dP_ds - ρg 1.0133 x 106 dz ds vs = _Aq = - k µ dPds + ρg 1.0133 x 106 q = kA_µ P1 - P2_L - ρg 1.0133 x 106 P1 = - ρg (h + x + L) 1.0133 x 106 P2 = ρg x 1.0133 x 106 P1 - P2 L = ρg h 1.0133 x 106 L + ρg 1.0133 x 106 q = kA µ ρg h 1.0133 x 106 L + ρg 1.0133 x 106 - ρg 1.0133 x 106 q = kA µL ρg h 1.0133 x 106

4. Horizontal, radial flow system h rw re Pe Pw re rw a. Conditions 1) horizontal system, dz ds = 0

2) radial system, A = 2πrh , ds = - dr, flow is inward 3) constant thickness, h = constant

4) incompressible liquid, q = constant 5) laminar flow, use Darcy equation 6) non-reactive fluid, k = constant 7) 100% saturated with liquid, 8) constant temperature, µ, q

b. Derivation of flow Equation vs = - _µk dP_ds - ρg 1.0133 x 106 dzds vs = + _µk dP_dr = q A = q 2πrh q 2πh drr rw re = k µ dp pw pe q 2πh 1n(re) - 1n( rw) = k µ Pe - Pw q = 2πhk µ _{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 A q P1 q P2 a. Conditions 1) horizontal system, dz ds = 0

b. Assumptions

1) µ, Z = constant

2) Z(and µ ) can be determined at mean pressure c. _{Derivation of equation for qsc}

vs = - k_µ dP ds - ρg 1.0133 x 106 dzds vs = - k_µ dP ds = q Ads but q = Psc qscz T PTsc thus Psc T qsc Tsc A _o ds L = - k PdP µz p1 p2 Psc T qsc Tsc A L -0 = - kµz P2 2 - P₁2 2 qsc = kA_µL Tsc Tz Psc P12 - P22 2 Note: real gas equation of state

Pq = Z n R T

where q = volumetric flow/time n = mass flow/time thus, Pq Pscqsc = Z n R Tn R Tsc q = Psc qscz T Tsc 1P

d. Derivation of equation for

q

qsc = kA_µL Tsc Tz Psc P12 - P22 2 but qsc = P q Tsc Z Psc T = k AµL Tsc T z Psc P12 - P22 2 q = k A µL 1_P P12 - P22 2 q = k A µL 2 P1 + P2 P12 - P22 2 q = k A µL 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 liquid can 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

2. Horizontal, radial flow system h rw re Pe Pw re rw a. Conditions 1) horizontal system dz ds = 0 2) radial system, A = 2πrL, ds = - dr, inward flow

3) constant thickness, h = constant 4) compressible gas flow, q = f (P) 5) laminar flow, use Darcy equation 6) non-reactive fluid, k = constant 7) 100% saturated with one fluid 8) constant temperature

b. Assumptions

µz = constant

z (and µ ) can be determined at mean pressure c. _{derivation of equation for qsc}

vs = - _µk dP_ds - ρg 1.0133 x 106 dz ds vs = - _µk dP_ds = q A but q = Psc qsc z T PTsc and A = 2πrh and ds = - dr thus Psc T qsc 2Tsc π h dr r rw re = k ρdP µz Pw Pe PscT qsc 2 Tsc π h 1n re rw = kµz Pe2 - Pw2 2 qsc = 2 _µ π h k 1n re/rw Tsc Psc zT Pe2 - Pw2 2

d. derivation of equation for q qsc = 2 _µ π h k 1n re/rw Tsc Psc zT Pe2 - Pw2 2 but q = P q Tsc z Psc T thus P q Tsc z Psc T = 2 π h k µ _{1n re/rw Psc zT}Tsc Pe2 - Pw 2 2 q = 2 π h k µ _{1n re/rw} 1P (Pe2 - Pw2 ) 2 q = 2 π h k µ _{1n re/rw Pe + Pw}2 (Pe2 - Pw2 2 ) q = 2 π h k µ _{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.

C. Conversion to Oilfield Units

Symbol Darcy units Oil field

q cc/sec bbl/d or cu ft/d k darcy md A sq cm sq ft h cm ft P atm psia L cm ft µ cp cp r gm/cc lb/cu ft Example: q = hkA P1 - P2 µ L _{in Darcy's units} q cc sec = q bbl_d 5.615 cu ft_bbl 1,728 cu in_{cu ft} 16.39 cc_{cu in} d_{24hr 3,600 sec}hr q cc sec = 1.841 q bbl_d k darcy = k md darcy 1,000md 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 _{14.696 psia}atm P1 - P2 atm = 0.06805 P1 - P2 psia

D. Table of Equations 1. Darcy Units

System Fluid Equation Horizontal, Linear Incompressible Liquid q = kAµ L P1 - P2 Dipping, Linear Incompressible Liquid q = kA µ L P1 - P2 + ρ g L sin θ 1.0133 x 106 Horizontal, Radial Incompressible Liquid q = 2 π k h µ _{ln (re/rw)} Pe - Pw Horizontal, Linear Real Gas _{qsc =} kA µ L Tsc_{Tz Psc} P1 2 - P₂2 2 q = kA µ L P1 - P2 Horizontal, Radial Real Gas qsc = _{µ ln (re/rw)}π k h Tsc Tz Psc Pe 2 - Pw 2 q = 2 π k h µ _{ln (re/rw)} Pe - Pw

2. Oilfield Units

System Fluid Equation Horizontal, Linear Incompressible Liquid q = 0.001127 kAµL P1 - P2 q = res bbl/d Dipping, Linear Incompressible Liquid q = 0.001127 kAµL P1 - P2 + ρg L sinθ 1.0133 x 106 Horizontal, Radial Incompressible Liquid q = .007082 kh µ _{ln (re/rw)} Pe - Pw Horizontal,

Linear Real Gas qsc = .1118 k A µ L z T P1 2 - P2 2 qsc = scf/d q = .001127 kA µL P1 - P2 q = res bbl/d Horizontal,

Radial Real Gas qsc = .7032

k h

Example II-2

What is the flow rate of a horizontal rectangular system when the conditions are as follows: permeability = k = 1 darcy

area = A = 6 ft2 viscosity = µ = 1.0 cp length = L = 6 ft

inlet pressure = P1 = 5.0 atm. outlet pressure = P2 = 2.0 atm.

Solutions:

We must insure that all the variables are in the correct units. k = 1 darcy = 1,000 md

A = 6 ft2 L = 6 ft

P1 = (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 kA

µL P1 - P2

q = 1.1271 x 10-3 1,000 6

1 6 73.5 - 29.4 q = 49.7 bbl / day

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

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_{µ L} and qt = kA whA_{µ L} P1 - P2 + kB whB_{µ L} P1 - P2 + kC whC_{µ L} P1 - P2 then k h = kA hA + kB hB + kC hC

2. Horizontal, radial flow parallel to bedding Pw r w re qA qB qC ht hA hB hC re Pe again qt = qA + qB + qC h = hA + hB + hC qt = _{µ ln (re/rw)}2 π k h Pe - Pw and qt = _µ2 _{ln (re/rw)}π kA hA Pe - Pw + _µ2 _{ln (re/rw)}π kB hB Pe - Pw + 2 π kc hc Pe - Pw

3. Horizontal, linear flow perpendicular to bedding L h P2 P 1 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 _{µ L} and since _{P1 - P2 =} ∆ _{PA +} ∆_{PB +} ∆_PC P1 - P2 = qt µ L k wh = qA µLA kA wh + qB µLB kB wh + qC µLC kC wh since _{qt = qA = qB = qC} L k = LA kA + LB kB + LC kC thus

4. Horizontal, radial, flow perpendicular to bedding h r w r A r B r C q PC PB PA Pw 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 µ _{ln (rB/rA)} µ _{ln (rC/re)}

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.

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 = πr4

8 µ L P1 - P2

A = π r2, therefore

q = Ar2

8 µ L P1 - P2

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

µ L P1 - P2

assuming these flow equations have consistent units Ar2

8 µ L P1 - P2 = kAµ L P1 - P2

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

2. Flow through fractures b v = q A = h212 µ L (P1-P2) q = b2 A 12 µ L (P1 -P2)

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

b2 A

12 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 = inches

Example II-6

Consider a rock of very low matrix permeability, 0.01 md, which contains on the average a fracture 0.005 inches wide and one foot in lateral extent per square foot of rock.

Assuming the fracture is in the direction of flow, determine the average permeability using the equation for parallel flow.

Solution:

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

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 22 2L qsc Psc A qsc = kA P₂1 2 - P_µ 2 2

L Psc horizontal, linear, real gas flow with

T = Tsc and Z = 1.0

2. Whole core method

a. prepare whole core in same manner as perm plugs

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

B. Factors which affect permeability measurement

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

2. Gas slippage

a. gas molecules "slip" along the grain surfaces

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

c. Darcy's equation assumes laminar flow

d. gas flow path with slippage

e. called Klinkenberg effect

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

0 kCALCULATED 1 P H2 N2 CO2

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

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

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

Example II-7

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

Calculate the equivalent liquid permeability from the following data.

Mean Measured Pressure Permeability ( atm ) ( md ) 1.192 3.76 2.517 3.04 4.571 2.76 9.484 2.54 Solution:

Plot kmeasured vs. 1/pressure

Intercept is equivalent to liquid permeability

From graph: kliq = 2.38 md 2 3 4 5 kgas = 2.38276 + 1.64632 Pbar

3. Reactive fluids

a. Formation water reacts with clays 1) lowers permeability to liquid

2) actual permeability to formation water is lower than lab permeability to gas 10000 1000 100 10 1 1 10 100 1000 WATER PERMEABILITY, 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.

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 17.2 S 34 24800 11800 10600 10000 9000 7400 147.0 S 34 40100 23000 18600 15300 13800 8200 270.0 S 34 4850 1910 1430 925 736 326 5.0 S 34 22800 13600 6150 4010 3490 1970 19.5 S 34 34800 23600 7800 5460 5220 3860 9.9 S 34 13600 5160 4640 4200 4150 2790 197.0 S 34 7640 1788 1840 2010 2540 2020 119.0 T 36 2630 2180 2140 2080 2150 2010 1960.0 T 36 3340 2820 2730 2700 2690 2490 2460.0 T 36 2640 2040 1920 1860 1860 1860 1550.0 T 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. 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

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 cm required to prevent destruction of surface (dynes/cm) - this is called "boundary tension"

3. Boundary Energy = Boundary Tension x Length

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

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

σ_gw = surface tension between gas and water

σ_go = surface tension between gas and oil

σ_wo = interfacial tension between water and oil

σ_ws = interfacial tension between water and solid

σ_os = interfacial tension between oil and solid

σ_gs = interfacial tension between gas and solid

I. Forces creating boundary tension 1. Forces

4. Liquid-Liquid Boundary some of each

II) Wettability

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

Θ σ_ow σ_os σ_ws OIL WATER OIL SOLID σ_{ws =} σ_{os +} σ_{ow cos} θ B. _{Adhesion Tension, AT} AT = σws - σos = σow cos θ C. if the solid is "water-wet"

σ_ws ≥ σ_os

AT = + cos θ = + 0° ≤ θ ≤ 90°

D. if the solid is "oil-wet" σ_os ≥ σ_ws AT = -cos θ = -90° ≤ θ ≤ 180° if θ = 180° - strongly oil-wet θ = 830 θ = 1580 _{θ = 35}0 θ = 300 (A) ISOOCTANE ISOOCTANE + 5.7%

ISOQUINOLINE ISOQUINOLINE NAPHTHENIC ACID

θ = 300 _{θ = 48}0 _{θ = 540 θ = 106}0

(B)

III) Capillary pressure

A. capillary pressure between air and water

Θ

h

AIR

WATER

1. liquid will rise in the tube until total force up equals total force down a. total force up equals adhesion tension acting along the

circumference of the water-air-solid interface = 2π_{r AT}

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

= πr2 hgρ_w

c. thus when column of water comes to equilibrium 2π_{r AT =} πr2 hgρ_w d. units cm dyne cm = cm2 cm cm_sec2 gm cm3 dyne = gm cm sec2

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

a. vertical component of surface tension is the surface tension

between air and water multiplied by the cosine of the contact angle acting along the water-air-solid interface

= 2πr σ_{aw cos}θ b. total force down

= πr2 hgρ_w

c. thus when the column of water comes to equilibrium 2πr σ_{aw cos}θ = πr2 hgρ_w d. units cm dyne cm = cm2 cm cm_sec2 gm cm3 cm dyne cm = cm gm sec2

3. _{since AT =} σ_{aw cos}θ, 1 and 2 above both result in

h = 2 σaw cos θ rg ρ_w

4. capillary pressure (air-water system) Θ h WATER Pa A' _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 water between 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

f. thus

Pc = 2 σgw cos _r θ

B. capillary pressure between oil and water

Θ

h

WATER

OIL

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

a. vertical component of surface tension equals the surface tension between oil and water multiplied by the cosine of the contact angle acting along the circumference of the water-oil-solid interface

= 2πr σ_{ow cos}θ

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

3) net weight per unit area acting to pull surface down = ρ_{wh -} ρ_{oh = h(}ρ_{w -} ρ_o)

4) total force down

= πr2 gh (ρ_{w -} ρ_o)

d. thus when the column of water comes to equilibrium

2πr σ_{ow cos}θ = πr2 gh (ρ_{w -} ρ_o)

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

h = 2 σow cos θ rg (ρ_{w -} ρ_o)

3. capillary pressure (oil-water system)

Θ 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 water between A and B

Pob = Poa - ρogh

d. thus capillary pressure, the difference between pressure at B' and the 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 θ rg (ρ_{w -} ρ_o) f. thus Pc = 2 σow cos _r θ

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

Pc = 2 σ cos _r θ C. remember adhesion tension is defined as

AT = σow cosθ, and

ADHESION TENSION AIR WATER AIR Hg WATER 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 than the 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 basic concepts - but it is a simple and ideal system

2. packing of uniform spheres

Pc = σ 1 R1 + 1R2

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

3. by analogy to capillary tube 1

R1 + 1R2 = 2 cos

θ

r where Pc = 2 σ cos _r θ call it Rm(mean radius), i.e.

F. wettability-consolidated sand

1. Pendular-ring distribution-wetting phase is not continuous, occupies the small interstices-non-wetting phase is in contact with some of the solid 2. Funicular distribution - wetting phase is continuous, completely covering

surface of solid

(A)

(B)

OIL OR GAS OIL OR GAS

SAND GRAIN SAND GRAIN

WATER WATER

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

IV) Relationship between capillary pressure and saturation

A. remember that the height a liquid will rise in a tube depends on 1. adhesion

2. fluid density

3. variation of tube diameter with height

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

R ATMOSPHERIC PRESSURE R HIGHER PRESSURE

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

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

2. This illustrates:

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

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

c. Lower wetting-phase saturation results in smaller radius of

V) Relationship between capillary pressure and saturation history

A. consider an experiment using a non-uniform tube (pore in reservoir rock) 1. tube is filled with a wetting fluid and allowed to drain until the interface

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

2. tube is filled with non-wetting fluid and immersed in wetting fluid allowing wetting fluid to imbibe until the interface reaches equilibrium (imbibition) Θ SATURATION = 100% PC = LOW VALUE Θ SATURATION = 80% CAPILLARY PRESSURE = P C R LOW PC HIGHER P C (A) Θ SATURATION = 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 a nonuniform pore. (a) Fluid drains; (b) fluid imbibes. Same pore, same contact angle, same capillary pressure, different saturation history

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

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

B. Leverett conducted a similar experiment with tubes filled with sand. 100 80 60 40 20 0 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)

∆ 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 correlating function ( remember Pc = (∆ρ gh )

2. in general terms,

a. drainage means replacing a wetting fluid with a non-wetting fluid b. imbibition means replacing a non-wetting fluid with a wetting fluid

PC

0

100

WATER SATURATION, S W

IMBIBITION

VI) Capillary pressure in reservoir rock Pw = Po/w - ρ₁₄₄w h Po = Po/w - ρ₁₄₄oh Water Oil Pw2 Po2 Po1 = Pw1 100% Water

Oil and Water

Pc = Po - Pw = ₁₄₄h ρw - ρo

Where: _Po = pressure in oil phase, psia Pw = pressure in water phase, psia

h = distance above 100% water level, ft Po/w = pressure at oil-water contact, psia

ρ_w = density of water, lb/cf

ρ_o = density of oil, lb/cf At any point above the oil-water contact, po ≥ pw

HEIGHT

ABOVE

O-W-C

PRESSURE

PC

PO = PO/W -

144

ρ

_oH

Pw = PO/W -

ρ

₁₄₄

w H

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 fluid through the porous disk.

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

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

5. _{Measure Sw periodically.} 6. Results

7. Advantages

a. very accurate

b. can use reservoir fluids 8. Disadvantages

a. very slow - up to 40 days for one core

C. Mercury Injection Method

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

2. Measure pressure

3. Calculate mercury saturation 4. Advantages

a. fast-minutes

b. reasonably accurate 5. Disadvantages

a. ruins core

b. 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 than pressure) is applied to the fluids in the core

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 conditions b. reservoir fluids can be used

5. Disadvantages a. very tedious

F. Comparison of methods

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

2. comparison of mercury injection data against diaphragm data

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

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

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

Example VIII-1

Comparison of Mercury Injection Capillary Pressure Data with Porous Diaphragm Data

A. Calculate capillary pressure ratio,

PcAH_g

PcAW, for the following data:

σ_AHg= 480 Dynes/cm _σ

AW= 72 Dynes/cm

θ _AHg = 140° _θ

AW= 0°

B. Pore geometry is very complex. The curvature of the interface and pore radius are not necessarily functions of contact angles. Calculate the ratio using the

relationship. PcAH_g PcAW = σ_AHg σ_AW Solution: (A) PcAH_g PcAW = σ_AHgcos θ_AHg σ_AWcos θ_AW = 480 cos(140°) 72 cos (0°) PcAH_g PcAW = 5.1 (B) PcAH_g PcAW @ σ_AHg σ_AW = 480₇₀ PcAH

Discussion:

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

Mercury Injection and Porous Diaphragm Methods

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

C. Minimum irreducible wetting phase saturations are the same.

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

VIII) Converting laboratory data to reservoir conditions PcL = 2σLcos r θL PcR = 2σRcos r θR setting r = r r = 2σLcos θL PcL = 2 σ_Rcos θ_R PcR ∴ _{PcR =} σcos θ R σcos θ _L PcL where

Pc_R = reservoir capillary pressure, psi

Pc_L = capillary pressure measured in laboratory, psi

σ_{L = interfacial tension measured in laboratory, dynes/cm} σ_{R = reservoir interfacial tension, dynes/cm}

θ_{R = reservoir contact angle, degrees} θ_{L = laboratory contact angle, degrees}

Example III-2

Converting Laboratory Data to Reservoir Conditions

Express reservoir capillary pressure by using laboratory data. lab data: σ_{AW = 72 dynes}

σ_{AW = 0}o

reservoir data: σ_{OW = 24 dynes/cm}

σ_{OW = 20}o Solution: PcR = σcos θ _R σcos θ _L PcL PcR = 24 cos20° 72 cos0° PcL PcR = 0.333 PcL

IX) Determining water saturation in reservoir from capillary pressure data A. convert laboratory capillary pressure data to reservoir conditions

B. calculate capillary pressure in reservoir for various heights above height at which capillary pressure is zero

Pc = (∆ρ)gh 144 gc in English units ∆ρ = ρ_{w -} ρ_{O, lb/cu ft} g = 32 ft/sec2 gc = _{32 lbm ft} lbf sec2 h = ft 144 = (sq in)/(sq ft.) thus Pc = lbf/(sq in), psI

Example III-3 Determining Water Saturation From Capillary Pressure Curve Given the relationship,

PcR = 0.313 P cL, use the laboratory capillary pressure curve to calculate the water saturation in the reservoir at a height of 40 ft. above the oil-water contact.

ρ_{o = 0.85 gm/cm3} ρ_{w = 1.0 gm/cm3} PCL SW 0 0 10 20 50 100 8.3 8.3 Solution: PcR = ρw − ρ144o h PcR = 1.0 - 0.85 62.4 lb ft3 40 144 = 2.6 psi

X) Capillary pressure variation A. effect of permeability

1. displacement pressure increases as permeability decreases

2. minimum interstitial water saturation increases as permeability decreases

100 90 80 70 60 50 40 30 20 10 0 0 20 40 60 80 100 120 140 160 180 200

RESERVOIR FLUID DISTRIBUTION CURVES

Sw %

(From Wright and Wooddy)

10 md

100 md

200 md

900 md

30

Height above zero capillary pressure, ft

24

18

12

6

0

Oil - Water Capillary Pressure, psi (reservoir conditions)

90 72 54 36 18 0

B. Effect of grain size distribution 100 80 60 40 20 0 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

Mercury capillary pressure, psi

Water/nitrogen capillary pressure, psi

Hg Water 100 80 60 40 20 0 0 10 20 30 40 50 60 348 290 232 174 116 58 0

Water/nitrogen capillary pressure, psi

Mercury capillary pressure, psi

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

XI) Averaging capillary pressure data J-function J Sw = Pc σcos θ k φ 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 in reservoirs

concept useful for given rock type from given reservoir where

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

100 90 80 70 60 50 40 30 20 10 0 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 1.7 1.8 1.9 2.0

CAPILLARY RETENTION CURVES.

WATER SATURATION, Sw

CAPILLARY PRESSURE FUNCTION, J

(From Rose and Bruce.)

LEVERETT LEDUC HAWKINS KATIE ALUNDUM EL ROBLE KINSELLA Reservoir Formation Hawkins Woodbine El Roble Moreno Kinsella Viking Katie Deese Leduc Devonian Alundum (consolidated) Leverett (unconsolidated)

Capillary Pressure Problem 1

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

Calculate:

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

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

2. The displacement pressure for a water saturated porcelain plate is 55 psi of air. What is the diameter in inches of the largest pore in the porcelain plate? Assume 72 dynes/cm and 0 degrees. Solution: (1) σAW = 72 dynes/cm ρ_{W = 1 gm/cm3} g = 980 dynes/gm θ = 0o

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

h = 2σAWcos θ rρg =

2 72 cos0° .01 1.0 980 h = 14.69 cm

(b) pressure drop in psi across interface

Pc = _{pa - pw =} ρ_{wgh = 1.0 980 14.69}

Pc = 0.0142 atm

14.696 psi atm Pc = 0.209 psi

(2) Pc = 2σAWcos r θ Pc = 55 psi P_{c = 55 psi} atm 14.696 psi 1.0133 x 10 6 dynes/cm2 atm = 3.792 x 106 dynes/cm2 r = 2σAW cos θ 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

Capillary Pressure Problem 2

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

fluids lab air-water res oil-water θ 0° 25° σ 60 dyne/cm 20 dyne/cm ρ_{wet 1.0 gm/cm3 1.1 gm/cm3} ρ_{non-wet 0 gm/cm3 0.863 gm/cm3} k 37 md variable φ _{16% variable} J = Pc k/φ 1/2 σ cos θ

100 90 80 70 60 50 40 30 20 10 0 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 = σ_Rcosθ_R σ_Lcosθ_L PcL = 20 cos25 60 cos0 PcL PcR = 0.302 PcL

(2) PcR = hR ρ144w - ρo = hR 1.1 - .863 62.4 144 PcR = .103 hR hR = 9.74 PcR (3) J ₌ Pc σ cos θ kφ 1/2 = PcL σ_AWcosθ_L k φ _L 1/2 = PcL 60 cos0° 37.16 1/2 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 80 60 40 20 0 0 2 4 6 8 10 Sw % Pc R 100 80 60 40 20 0 0 20 40 60 80 100 Sw % h R 100 80 60 40 20 0 0 2 4 6 8 10 J

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 into formation

Ph > Pres

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

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

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:

sat at surface

flushing by bit trip to surface compared to res

Sw ↑ ↓ ? So - - -Sg ↓ ↑ _? Water zone: sat at surface

flushing by bit trip to surface compared to res

Sw - ↓ ↓