Course Notes for
Petroleum Engineering 311
Reservoir Petrophysics
Authors:
1980 — Von Gonten, W.D. 1986 — McCain, W.D., Jr. 1990 — Wu, C.H.
PETROLEUM ENGINEERING 311
RESERVOIR PETROPHYSICS
CLASS NOTES (1992)
Instructor/Author:
Ching H. Wu
DEPARTMENT OF PETROLEUM ENGINEERING
TEXAS A&M UNIVERSITYCOLLEGE STATION, TEXAS
ii
TABLE OF CONTENTS
I. ROCK POROSITY I-1
I) Definition I-1II) Classification I-1
III) Range of values of porosity I-2VI) Factors affecting porosity I-3V) Measurement of porosity I-5
VI) Subsurface measurement of porosity I-13VII) Compressibility of porous rocks I-25
II. SINGLE PHASE FLOW IN POROUS ROCK II-1
I) Darcy's equation II-11II) Reservoir systems II-15
III. BOUNDARY TENSION AND CAPILLARY PRESSURE III-11
I) Boundary tension III-1II) Wettability III-3
III) Capillary pressure III-5IV) Relationship between capillary pressure and saturation III-13V) Relationship between capillary pressure and saturation history III-14
VI) Capillary pressure in reservoir rock III-17VII) Laboratory measurement of capillary pressure III-19
VIII) Converting laboratory data to reservoir conditions III-25IX) Determining water saturation in reservoir from capillary pressure data III-27X) Capillary pressure variation III-29
XI) Averaging capillary pressure data III-31
IV. FLUID SATURATIONS IV-1
I) Basic concepts of hydrocarbon accumulation IV-1II) Methods for determining fluid saturations IV-1
V. ELECTRICAL PROPERTIES OF ROCK-FLUID SYSTEMS V-1
I) Electrical conductivity of fluid saturated rock V-1II) Use of electrical Formation Resistivity Factor, Cementation Factor, and
Saturation Exponent V-8III) Laboratory measurement of electrical properties of rock V-9IV) Effect of clay on resistivity V-18
VI. MULTIPHASE FLOW IN POROUS ROCK VI-1
I) Effective permeability VI-1II) Relative permeability VI-2
III) Typical relative permeability curves VI-2IV) Permeability ratio (relative permeability ratio) VI-14V) Measurement of relative permeability VI-14
VI) Uses of relative permeability data VI-33
iii
VII. STATISTICAL MEASURES VII-1
I) Introduction VII-1II) Frequency Distributions VII-2
III) Histogram VII-3IV) Cumulative Frequency Distributions VII-6V) Normal Distribution VII-8
VI) Log Normal Distribution VII-9VII) Measures of Central Tendency VII-10
VIII) Measures of Variability (dispersion) VII-11IX) Normal Distribution VII-12X) Log Normal Distribution VII-16
I - 1
I. ROCK POROSITY
I) Definition
A measure of the pore space available for the storage of fluids inrock
In general form:
Porosity = φ = VpVb
= Vb - VmVb
where:φ is expressed in fraction
Vb = Vp + VmVb = bulk volume of reservoir rock, (L3)
Vp = pore volume, (L3)
Vm= matrix volume, (L3)
II) Classification
A. Primary (original) Porosity
Developed at time of deposition
B. Secondary Porosity
Developed as a result of geologic processoccurring after deposition
C. Total Porosity
φt = total pore spaceVb
= Vb - VmVb
D. Effective Porosity
φe = interconnected pore spaceVb
1. Clean sandstones: φe = φt2. Carbonate, cemented sandstones: φe < φt
I - 3
VI) Factors affecting porosity
A. Factors:1. Particle shape2. Particle arrangement3. Particle size distribution4. Cementation5. Vugs and fractures
B. Particle shape
Porosity increases as particle uniformity decreases.
C. Packing Arrangement
Porosity decreases as compaction increases
60005000400030002000100000
10
20
30
40
50
DEPTH OF BURIAL, ft
POR
OSI
TY
, %
EFFECT OF NATURAL COMPACTION ON POROSITY
(FROM KRUMBEIN AND SLOSS.)
SANDSTONES
SHALES
I - 4
D. Particle Size Distribution
Porosity decreases as the range of particle size increases
GRAIN SIZE DIAMETER, MM
INTERSTITIAL MATERIALSAND MUD FRAGMENTS
FRAMEWORK FRACTION
CLEAN SAND
SHALY SAND
SAND SILT CLAY100
0
1.0 0.1 0.01 0.001
WE
IGH
T %
E. Interstitial and Cementing Material
1. Porosity decreases as the amount of interstitial material increases
2. Porosity decreases as the amount of cementing material increases
3. Clean sand - little interstitial materialShaly sand - has more interstitial material
F. Vugs, Fractures
1. Contribute substantially to the volume of pore spaces
2. Highly variable in size and distribution
3. There could be two or more systems of pore openings - extremely complex
I - 5
V) Measurement of porosity
φ = Vb - VmVb
= VpVb
Table of matrix densities
Lithology ρm (g/cm3)___________ ___________
Quartz 2.65
Limestone 2.71
Dolomite 2.87
A. Laboratory measurement
1. Conventional core analysis
a. measure any two
1) bulk volume, Vb2) matrix volume, Vm3) pore volume, Vp
b. bulk volume
1) calculate from dimensions2) displacement method
a) volumetric (measure volume)
(1) drop into liquid and observe volume chargeof liquid
(2) must prevent test liquid from entering poresspace of sample
(a) coat with paraffin(b) presaturate sample with test liquid(c) use mercury as test liquid
b) gravimetric (measure mass)
(1) Change in weight of immersed sample-prevent test liquid from entering pore space
(2) Change in weight of container and test fluidwhen sample is introduced
I - 6
c. matrix volume
1) assume grain density
Vm = dry weightmatrix density
2) displacement method
Reduce sample to particle size, then
a) volumetric
b) gravimetric
3) Boyle's Law: P1V1 = P2V2
a)
P(1)
V(1)
VALVE CLOSED
b) Put core in second chamber, evacuate
c) Open valve
P(2)
VALVE OPEN
CORE
V2 = Volumetric of first chamber &volume of second chamber-matrixvolume or core ( calculated)
VT = Volume of first chamber +volume second chamber (known)
4) Vm =VT - V2
I - 7
d. pore volume
1) gravimetric
Vp = saturated weight - dry weightdensity of saturated fluid
2) Boyle's Law: P1V1 = P2V2
a)
P(1)
V(1)
VALVE CLOSED
CORE
b) Put core in Hassler sleeve, evacuate
c) Open valve
P(2)
V(1)
VALVE OPEN
CORE
V2 = Volume of first chamber + porevolume of core (calculated)
3) Vp = V2 - V1
I - 8
2. Application to reservoir rocks
a. intergranular porosity(sandstone, some carbonates)
1) use representative plugs from whole core in laboratory measurements
2) don't use sidewall cores
b. secondary porosity(most carbonates)
1) use whole core in laboratory measurements
2) calculate bulk volume from measurements
3) determine matrix or pore volume fromBoyle's Law procedure
I - 9
Example I-1
A core sample coated with paraffin was immersed in a Russell tube. The dry sample weighed 20.0gm. The dry sample coated with paraffin weighed 20.9 gm. The paraffin coated sample displaced10.9 cc of liquid. Assume the density of solid paraffin is 0.9 gm/cc. What is the bulk volume ofthe sample?
Solution:
Weight of paraffin coating = 20.9 gm - 20.0 gm = 0.9 gm
Volume of paraffin coating = 0.9 gm / (0.9 gm/cc) = 1.0 cc
Bulk volume of sample = 10.9 cc - 1.0 cc = 9.9 cc
Example I-2
The core sample of problem I-1 was stripped of the paraffin coat, crushed to grain size, andimmersed in a Russell tube. The volume of the grains was 7.7 cc. What was the porosity of thesample? Is this effective or total porosity.
Solution:
Bulk Volume = 9.9 cc
Matrix Volume = 7.7 cc
φ = Vb - Vm
Vb = 9.9 cc- 7.7 cc
9.9 cc = 0.22
It is total porosity.
I - 10
Example I-3
Calculate the porosity of a core sample when the following information is available:
Dry weight of sample = 427.3 gm
Weight of sample when saturated with water = 448.6 gm
Density of water = 1.0 gm/cm3
Weight of water saturated sample immersed inwater = 269.6 gm
Solution:
Vp = sat. core wt. in air - dry core wt. density of water
Vp = 448.6 gm - 427.3 gm
1 gm/cm3
Vp = 21.3 cm3
Vb = sat. core wt. in air - sat. core wt. in waterdensity of water
Vb = 448.6 gm - 269.6 gm
1 gm/cm3
Vb = 179.0 cm3
φ = VpVb
= 21.3 cm3179.0 cm3
= .119
φ = 11.9%
I - 11
What is the lithology of the sample?
Vm = Vb - Vp
Vm = 179.0 cm3 - 21.3 cm3 = 157.7 cm3
ρm = wt. of dry sample = 427.3 gm = 2.71 gm/(cm3)
matrix vol. 157.7 cm3
The lithology is limestone.
Is the porosity effective or total? Why?
Effective, because fluid was forced into the pore space.
I - 12
Example I-4
A carbonate whole core (3 inches by 6 inches, 695 cc) is placed in cell two of a BoylesLaw device. Each of the cells has a volume of 1,000 cc. Cell one is pressured to 50.0 psig. Celltwo is evacuated. The cells are connected and the resulting pressure is 28.1 psig. Calculate theporosity of the core.
Solution:
P1V
1= P
2V
2
V1
= 1,000 cc
P1
= 50 psig + 14.7 psia = 64.7 psia
P2
= 28.1 psig + 14.7 = 42.8 psia
V2
= (64.7 psia) (1,000 cc) / (42.8 psia)
V2
= 1,512 cc
Vm
= VT - V2
Vm
= 2,000 cc - 1,512 cc - 488 cc
φ =
VT - VmVT
= 695 cc - 488 cc695 cc
= .298 = 29.8%
I - 13
VI) Subsurface measurement of porosity
A. Types of logs from which porosity can be derived
1. Density log:
φd = ρm - ρLρm - ρf
2. Sonic log:
φs = ∆tL - ∆tm∆tf - ∆tm
3. Neutron log:
e-kφ = CNf
Table of Matrix Properties(Schlumberger, Log Interpretation Principles, Volume I)
Lithology ∆tm µsec/ft ρm gm/cc
Sandstone 55.6 2.65
Limestone 47.5 2.71
Dolomite 43.5 2.87
Anhydrite 50.0 2.96
Salt 67.0 2.17
Water 189.0 1.00
I - 14
B. Density Log
1. Measures bulk density of formation
FORMATION
GAMMA RAYSOURCE
SHORT SPACEDETECTOR
LONG SPACEDETECTOR
MUD CAKE
2. Gamma rays are stopped by electrons - the denser the rock the fewer gammarays reach the detector
3. Equation
ρL = ρm 1 - φ + ρf φ
φd = ρm - ρLρm - ρf
I - 15
FORMATION DENSITY LOG
4240
4220
4200
4180
4160
4140
4120
4100
20016012080400 3.02.82.62.42.22.0
ρ, gm/ccGR, API depth, ft
I - 16
Example I-5
Use the density log to calculate the porosity for the following intervals assuming ρmatrix = 2.68
gm/cc and ρfluid = 1.0 gm/cc.
Interval, ft ρL, gm/cc
φd ,%
__________ _________ ______
4143-4157 2.375 184170-4178 2.350 204178-4185 2.430 154185-4190 2.400 174197-4205 2.680 04210-4217 2.450 14
Example:
Interval 4,143 ft -4,157 ft :
ρL
= 2.375 gm/cc
φd = ρm - ρLρm - ρf
= 2.68 gm/cc - 2.375 gm/cc2.68 gm/cc - 1.0 gm/cc
= 0.18
I - 17
C. Sonic Log
1. Measures time required for compressional sound waves to travel throughone foot of formation
D
B
AT
R1
R2E
C
2. Sound travels more slowly in fluids than in solids. Pore space is filled withfluids. Travel time increases as porosity increases.
3. Equation
∆tL = ∆tm 1 - φ + ∆tf φ (Wylie Time Average Equation)
I - 18
SONIC LOG
4240
4220
4200
4180
4160
4140
4120
4100
2001000 140 120 100 80 60 40
GR, API ∆T, µ seconds/ftdepth, ft
I - 19
Example I-6
Use the Sonic log and assume sandstone lithology to calculate the porosity for the followingintervals.
Interval ∆tL φs ,%
(ft) µ second/ft
4,144-4,150 86.5 25
4,150-4,157 84.0 24
4,171-4,177 84.5 24
4,177-4,187 81.0 21
4,199-4,204 53.5 1
4,208-4,213 75.0 17
Example:
Interval 4144 ft - 4150 ft :
∆tL = 86.5 µ-sec/ft
φs = ∆tL - ∆tm∆tf - ∆tm
= 86.5 µ sec/ft- 51.6 µ sec/ft189.0 µ sec/ft- 51.6 µ sec/ft
= 0.25
I - 20
D. Neutron Log
1. Measures the amount of hydrogen in the formation (hydrogen index)
MaximumAverage EnergyNumber Loss/ Atomic Atomic
Element Collisions Collision, % Collision Number
Calcium 371 8 40.1 20Chlorine 316 10 35.5 17Silicone 261 12 28.1 14Oxygen 150 21 16.0 8Carbon 115 28 12.0 6Hydrogen 18 100 1.0 1
.1 1 10 102 103
O
104
Si
105
H
106 107
1
10
102
103
CLEAN SAND POROSITY = 15%
NEUTRON ENERGY IN ELECTRON VOLTS
REL
ATIV
E PR
OBA
BIL
ITY
FOR C
OLL
ISIO
N
.1 1 10 102 103 104 105
H
106
O
107
Si10-3
10-2
10-1
1
CLEAN SAND POROSITY = 15%
SLO
WIN
G D
OW
N P
OW
ER
NEUTRON ENERGY IN ELECTRON VOLTS
2. In clean, liquid filled formations, hydrogen index is directly proportional toporosity. Neutron log gives porosity directly.
3. If the log is not calibrated, it is not very reliable for determining porosity.Run density log to evaluate porosity, lithology, and gas content.
I - 21
NEUTRON DENSITY LOG
4240
4220
4200
4180
4160
4140
4120
4100
2000 30 -10
GR, API φ (CDL)depth, ft
I - 22
Example I-7
Use the neutron log to determine porosity for the following intervals.
Solution:
Interval φ n (ft) (%) .
4,143-4,149 23
4,149-4,160 20
4,170-4,184 21
4,198-4,204 9
4,208-4,214 19
I - 23
Example I-8
Calculate the porosity and lithology of the Polar No. 1 drilled in Lake Maracaibo. The depth ofinterest is 13,743 feet. A density log and a sonic log were run in the well in addition to thestandard Induction Electric Survey (IES) survey.
The readings at 13,743 feet are:
bulk density = 2.522 gm/cctravel time = 62.73 µ-sec/ft
Solution:
Assume fresh water in pores.
Assume sandstone:
ρm = 2.65 gm/cc
∆tm = 55.5 µ-sec/ft
φd = ρm - ρLρm - ρf
= 2.65 gm/cc - 2.522 gm/cc2.65 gm/cc - 1.0 gm/cc
= 7.76%
φs = ∆tL - ∆tm∆tf - ∆tm
= 62.73 µ sec/ft- 55.5 µ sec/ft189.0 µ sec/ft - 55.5 µ sec/ft
= 5.42%
Assume limestone:
ρm = 2.71 gm/cc
∆tm = 47.5 µ-sec/ft
φd = ρm - ρLρm - ρf
= 2.71 gm/cc - 2.522 gm/cc2.71 gm/cc - 1.0 gm/cc
= 10.99%
φs = ∆tL - ∆tm∆tf - ∆tm
= 62.73 µ sec/ft - 47.5 µ sec/ft189.0 µ sec/ft - 47.5 µ sec/ft
= 10.76%
I - 24
Assume dolomite:
ρm = 2.87 gm/cc
∆tm = 43.5 µ-sec/ft
φd = ρm - ρLρm - ρf
= 2.87 gm/cc - 2.522 gm/cc2.87 gm/cc - 1.0 gm/cc
= 18.619%
φs = ∆tL - ∆tm∆tf - ∆tm
= 62.73 µ sec/ft - 43.5 µ sec/ft189.0 µ sec/ft - 43.5 µ sec/ft
= 13.22%
φlimestone = 11%
Since both logs "read" nearly the same porosity when a limestone lithology was assumed then the hypothesis that the lithology is limestone is accepted.
Are the tools measuring total or effective porosity? Why?
The density log measures total compressibility because is "sees" the entire rock volume,including all pores. The sonic log tends to measure the velocity of compressional waves that travel through interconnected pore structures as well as the rock matrix. The general consensus is that the sonic log measures effective porosity when we use the Wyllie "time-average" equation.
It is expected that the effective porosity is always less than ,or equal to,the total porosity.
I - 25
VII) Compressibility of porous rocks
Compressibility, c is the fractional change in volume per unit change in pressure:
c = - 1V
∂V∂P T
≅ -
∆VV T∆P
A. Normally pressured reservoirs
1. Downward force by the overburden must be balanced by upward force ofthe matrix and the fluid
Fm Ff
Fo
2. Thus,
Fo = Fm + Ff
it follows that
Po = pm + pf
3. Po ≅ 1.0 psi/ft
Pf ≅ 0.465 psi/ft
I - 26
4. As fluid is produced from a reservoir, the fluid pressure,Pf will usually decrease:
a. the force on the matrix increasesb. causing a decrease in bulk volumec. and a decrease in pore volume
B. Types of compressibility
1. Matrix Compressibility, cm
cm ≅ 0
2. Bulk Compressibility cb
used in subsidence studies
3. Formation Compressibility, cf - also called pore volume compressibility
a. important to reservoir engineers
1) depletion of fluid from pore spaces2) internal rock stress changes3) change in stress results in change in
Vp, Vm, Vb
4) by definition
cf = - 1Vp
∂Vp∂pm
b. since overburden pressure, Po, is constant
dPm = - dPf
I - 27
1) Thus,
cf = - 1Vp
∂Vp∂pm
2) where the subscript of f on cf means"formation" and the subscript of f on Pfmeans "fluid"
3) procedure
(a) measure volume of liquid expelled as afunction of "external" pressure
(b) "external" pressure may be taken torepresent overburden pressure, Po
(c) fluid pressure, pf, is essentially constant, thus,
dPo = dPm
(d) expelled volume increases as porevolume, vp, decreases, thus,
dVp = - dVexpelled
(e) from definition
cf = - 1Vp
∂Vp∂pm
it follows that
cf = + 1Vp
∆ Vp expelled
∆Po
I - 28
(f) plot
CU
MU
LATIV
E V
OLU
ME
EXPE
LLED
PO
RE V
OLU
ME
OVERBURDEN PRESSURE, psi
slope = cf.
I - 29
C. Measurement of compressibility
1) Laboratory core sample
a) apply variable internal and external pressures
b) internal rock volume changes
2) Equipment
InternalPressureGauge
HydraulicPump
OverburdenPressureGauge
HydraulicPump
Copper - JacketedCore
Mercury Sight Gauge
Apparatus for measuring pore volume compressibility (hydrostatic)
I - 30
Example I-9
Given the following lab data, calculate the pore volume compressibility for a sandstone sample at4,000 and 6,000 psi.
pore volume = 50.0 cc
pressure, psi vol. fluid expelled, cc
1000 0.244 2000 0.324 3000 0.392 4000 0.448 5000 0.500 6000 0.546 7000 0.596 8000 0.630
Solution:
from graph
@ 4,000 psi:
Slope = 0.0094000 psi
cf = 2.25 X 10-6 1psi
@ 6000 psi:
Slope = 0.0116000 psi
cf = 1.83 X 10-6 1psi
I - 31
10000800060004000200000.000
0.005
0.010
0.015
COMPACTION PRESSURE, psi
VO
LU
ME
EX
PEL
LE
D, c
cPO
RE
VO
LU
ME
, cc
I - 32
INITIAL POROSITY AT ZERO NET PRESSURE, %
3025201510501
10
100
PORE-VOLUME COMPRESSIBILITY AT 75 %LITHOSTATIC PRESSURE VS INITIAL SAMPLEPOROSITY FOR CONSOLIDATED SANDSTONES.
CONSOLIDATED SANDSTONES
HALL'S CORRELATION
PO
RE
VO
LU
ME
CO
MP
RE
SS
IBIL
ITY
X 1
0-6
psi-
1P
OR
E V
OL
UM
E C
OM
PR
ES
SIB
ILIT
Y X
10-
6 ps
i-1
3025201510501
10
100
PORE-VOLUME COMPRESSIBILITY AT 75 %LITHOSTATIC PRESSURE VS INITIAL SAMPLEPOROSITY FOR UNCONSOLIDATED SANDSTONES.
UNCONSOLIDATED SANDSTONES
HALL'S CORRELATION
INITIAL POROSITY AT ZERO NET PRESSURE, %
I - 33
E. Abnormally pressured reservoirs
"abnormal pressure": fluid pressures greater than or less than the hydrostatic fluidpressure expected from an assumed linear pressure gradient
PRESSURED
EPTH
NORMAL LINEAR
SUBNORMAL(LOWER)
SURNORMAL(GREATER)
I - 34
Compressibility/Porosity Problem No. 1
A limestone sample weighs 241.0 gm. The limestone sample coated with paraffin was found toweigh 249.5 gm. The coated sample when immersed in a partially filled graduated cylinderdisplaced 125.0 cc of water. The density of the paraffin is 0.90 gm/cc.
What is the porosity of the rock? Does the process measure total or effective porosity?
Solution:
Vm = wt. dryρls
= 241.0 gm2.71 gm/cc
= 88.9 cc
Vparaffin = wt. coated sample - st. uncoated sampleρ
Vparaffin = 249.5 gm - 241.0 gm
0.90 gm/cc = 9.4 cc
Vb = 125 cc - 9.4 cc = 115.6 cc
Vp = Vb - Vm
Vp = 115.6 cc - 88.9 cc - 26.7 cc
φ = VpVb
= 26.7 cc115.6 cc
= 0.231
φ = 23.1% (total porosity)
I - 35
Compressibility/Porosity Problem No. 2
You are furnished with the results of a sieve analysis of a core from Pete well #1. Previouslaboratory work indicates there is a correlation between grain size and porosity displayed by thoseparticular particles. The correlation is seen below:
gravel - 25% porosity
coarse sand - 38% porosity
fine sand - 41% porosity
What would be the minimum porosity of the mixture?What basic assumption must be made in order to work the problem?
Solution:
Begin calculation with a volume of 1 cu. ft.
remaining remaining pore matrix
component volume porosity volume (ft3) (%) (ft3)
___
void space 1.000 100.0 0.000
gravel 0.250 25.0 0.750
coarse sand 0.095 9.5 0.905
fine sand 0.039 3.9 0.961
Final porosity - 3.9%
(Complete mixing of the grains)
I - 36
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 2.40gm/cc.
The Highgrade #1 Well was drilled and cored through the reservoir. A rock sample was sent to thelaboratory and the following tests were run.
pressure cum. pore vol. change (psig) (-cc)_________ 1,000 0.122 2,000 0.162 3,000 0.196 4,000 0.224 5,000 0.250 6,000 0.273 7,000 0.298 8,000 0.315
The dry weight of the core sample was found to be 140 gm while the sample dimensions were1.575 inches long and 1.960 inches in diameter.
Assuming the compressibility at 4,500 psi is the average compressibility in the reservoir, howmuch subsidence occurs when the reservoir pressure declines from 5,500 psi to 3,500 psi?
Calculate:
A. Reservoir Porosity
B. Sample Pore Volume
C. Compressibility at 4,500 psi
D. Amount of Ground Subsidence.
Solution:
A. Reservoir Porosity
φ = ρm - ρLρm - ρf
= 2.65 - 2.402.65 - 1.00
= 15.22%
I - 37
B. Sample Pore Volume
L = (1.575 in) (2.54 cm/in) = 4.0 cm
D = (1.960 in) (2.54 cm/in) = 5.0 cm
Vb = bulk volume = πD2h4
= 3.14 5.0 2 4.0
4.0 = 78.5 cc
Vm = matrix volume = 140 gm cc2.65 gm
= 52.8 cc
Vp = Vb - Vm = 78.5 cc - 52.8 cc
Vp = 25.7 cc
C. Compressibility (see graph)
Vp = 25.7 cc
D. Subsidence
∆H = H cp φ ∆P
∆H = 85 ft 9.69x 10-7 psi-1 0.152 2,000 psi
∆H = 0.026 ft
∆H = 0.32 inches
Note: the pore volume (formation) compressibility is somewhat smaller than usuallyencountered. An experienced engineer would be wary of this small number. Also it wasassumed that the formation compressibility was exactly the same as the bulk volumecompressibility. Experience shows that this is not the case.
I - 38
8000600040002000000.0040
0.0060
0.0080
0.0100
0.0120
0.0140
POROSITY PROBLEM No. 3
PRESSURE, psig
SLOPE =.0118 - .0068
7000 - 2000
VO
LU
ME
EX
PEL
LE
D, c
cPO
RE
VO
LU
ME
, cc Cp = 9.96 x 10-7 psi -1
I - 39
Compressibility Problem
A 160-acre and 100 ft thick reservoir has a porosity of 11%. The pore compressibility is 5.0 x 10-6 (1/psi). If the pressure decreases 3,000 psi, what is the subsidence (ft)? Assume Cf = Cb
Solution:
A = 160 (43,560) = 6,969,600 ft2
Vb = 100 (6,969,600) = 696,960,000 ft3
Vp = Vb(f) = (696,960,000) (.11) = 76,665,600 ft3
Cp = - 1Vp
dVpdp
5 x 10-6 (1/psi) = -176,665,600 ft3
dVp
3,000 psi
dVp = 1.15 x 106 ft3
∆H = 1.15 x 106 ft3 x 16,969,600 ft2
= 0.165 ft
II - 1
II. SINGLE PHASE FLOW IN POROUS ROCK
I) Darcy's equation (1856)
A. Water flow through sand filters
A
Z
WATER
DARCY'S FOUNTAIN.
SAND
q
q
h1 - h2
h1
h2
q = kA(h1 - h2)µL
Length of sand pack,L = Z
1. constant of proportionality, k, characteristics of particular sand pack, notsample size
2. Darcy's work confined to sand packs that were 100% saturated with water
3. equation extended to include other liquids using viscosity
II - 2
q = kA(h1 - h2)µL
B. Generalized form of Darcy's equation
1. Equation
vs = -kµ
dPds
- ρg
1.0133 x 106 dzds
-1
+1 90o 180o 270o 360o
Θ
s
Vs
+X
+Y
-Z
+Z
θ
2. Nomenclature
vs = superficial velocity (volume flux along path s) - cm/sec
vs/φ = interstitial velocity - cm/sec
ρ = density of flowing fluid - gm/cm3
g = acceleration of gravity - 980 cm/sec2
dP = pressure gradient along s - atm/cmds
µ = viscosity - centipoise
k = permeability - darcies
II - 3
3. Conversion factors
dyne = gm cm/sec2 = a unit of force
atm = 1.0133 x 106 dyne/cm2
ρgh = dyne/cm2 = a unit of pressure
poise = gm/cm sec = dyne sec/cm2
4. The dimensions of permeability
L = length
m = mass
t = time
vs = L/t
µ = m/Lt
ρ = m/L3
p = m/Lt2
g = L/t2
vs = - kµ
dpds
- ρg
1.0133 x 106 dzds
Lt = - k
m/Lt m/Lt2
L -
m/L3 L/t2 LL
k = L2 = cross-sectional area
II - 4
5. Definition of Darcy units
a. conventional units would be:
1) feet squared in the English system
2) centimeter squared in the cgs system
b. both are too large for use in porous media
c. definition of darcy
A porous medium has a permeability of one darcy when a single-phase fluid of onecentipoise that completely fills the voids of the medium will flow through it under conditions ofviscous flow at a rate of one cubic centimeter per second per square centimeter cross-sectional areaunder a pressure or equivalent hydraulic gradient of one atmosphere per centimeter.
q = k A P1 - P2µL
II - 5
II) Reservoir systems
A. Flow of incompressible liquid
1. Horizontal, linear flow system
L
AqP1
qP2
a. Conditions
1) horizontal system, dzds
= 0
2) linear system, A = constant
3) incompressible liquid, q = constant
4) laminar flow, can use Darcy's equation
5) non-reactive fluid, k = constant
6) 100% saturated with one fluid
7) constant temperature, µ, q
II - 6
b. derivation of flow equation
vs = - kµ
dPds
- ρg
1.0133 x 106 dzds
vs = - kµ
dPds
= qA
q ds0
L = - kA
µ dP
p1
p2
q L - 0 = - kAµ
P2 - P1
q = kALµ
P2 - P1
Note: P1 acts at L = 0
P2 acts at L = L
q is + if flow is from L = 0 to L = L
II - 7
Example II-1
What is the flow rate of a horizontal rectangular system when the conditions are as follows:
permeability = k = 1 darcy
area = A = 6 ft2
viscosity = µ = 1.0 cp
length = L = 6 ft
inlet pressure = P1 = 5.0 atm
outlet pressure = P2 = 2.0 atm
Solution:
We must insure all the variables are in the correct units.
k = 1 darcy
A = 6 ft2 (144 in2/1 ft2) (6.45 cm2/1 in2) = 5572.8 cm2
L = 6 ft (12 in/1 ft) (2.54 cm/1 in) = 182.88 cm
P1 = 5.0 atm
P2 = 2.0 atm
q = kALµ
P2 - P1
q = (1) (5,572.8 ) (5.0 - 2.0) (1) (182.88)
q = 91.42 cm3 / sec
II - 8
2. Non-horizontal, linear system
Θ
-Z
P1
S
X
P2
a. Conditions
1) non-horizontal system, dzds
= sinθ = constant
2) linear system, A = constant
3) incompressible liquid, q = constant
4) laminar flow, use Darcy equation
5) non-reactive fluid, k = constant
6) 100% saturated with one fluid
7) constant temperature µ, q
II - 9
b. derivation of equation
vs = - kµ
dPds
- ρg
1.0133 x 106 dzds
vs = - qA
= - kµ
dPds
+ k ρg sin θ
µ 1.0133 x 106
q ds0
L = - kA
µ dp
P1
P2 + kA ρg sin θ
µ 1.0133 x 106 ds
0
L
q = - kAµL
P1 - P2 + ρgLsinθ
1.0133 x 106
II - 10
3. Vertical, upward flow, linear system
L
x
h
FLOW UNDERHEAD h
a. Conditions
1) vertical system, dzds = sinθ = constant
2) upward flow, q = 270°, sinθ = - 1
3) linear system, A = constant
4) incompressible liquid, q = constant
5) laminar flow, use Darcy equation
6) non-reactive fluid, k = constant
7) 100% saturated with one fluid
8) constant temperature, µ
II - 11
b. derivation of flow equation
vs = kµ
dPds
- ρg
1.0133 x 106 dzds
vs = qA
= - kµ
dPds
+ ρg
1.0133 x 106
q = kAµ
P1 - P2L
- ρg
1.0133 x 106
P1 = - ρg (h + x + L)
1.0133 x 106
P2 = ρg x
1.0133 x 106
P1 - P2L
= ρg h
1.0133 x 106 L +
ρg
1.0133 x 106
q = kAµ
ρg h
1.0133 x 106 L +
ρg
1.0133 x 106 -
ρg
1.0133 x 106
q = kAµL
ρg h
1.0133 x 106
II - 12
4. Horizontal, radial flow system
h
rw
re
Pe
Pw
re rw
a. Conditions
1) horizontal system, dzds = 0
2) radial system, A = 2πrh , ds = - dr, flow is inward
3) constant thickness, h = constant
4) incompressible liquid, q = constant
5) laminar flow, use Darcy equation
6) non-reactive fluid, k = constant
7) 100% saturated with liquid,
8) constant temperature, µ, q
II - 13
b. Derivation of flow Equation
vs = - kµ
dPds
- ρg
1.0133 x 106 dzds
vs = + kµ
dPdr
= qA
= q
2πrh
q2πh
drr
rw
re = k
µ dp
pw
pe
q2πh
1n(re) - 1n( rw) = kµ
Pe - Pw
q = 2πhkµ 1n (re/rw)
Pe - Pw
Note: if q is + , flow is from re to rw
B. Flow of gas (compressible fluid)
1. horizontal, linear flow system
L
Aq
P1
q
P2
a. Conditions
1) horizontal system, dzds = 0
2) linear system, A = constant
3) compressible gas flow, q = f(p)
4) laminar flow, use Darcy equation
5) non-reactive fluid, k= constant
6) 100% saturated with one fluid
7) constant temperature
II - 14
b. Assumptions
1) µ, Z = constant
2) Z(and µ ) can be determined at mean pressure
c. Derivation of equation for qsc
vs = - kµ
dPds
- ρg
1.0133 x 106 dzds
vs = - kµ
dPds
= qAds
but
q = Psc qscz T
PTsc
thus
Psc T qscTsc A
dso
L = - k PdP
µzp1
p2
Psc T qscTsc A
L -0 = - kµz
P2
2 - P12
2
qsc = kAµL
Tsc
Tz Psc
P12 - P2
2
2
Note: real gas equation of state
Pq = Z n R T
where q = volumetric flow/timen = mass flow/time
thus,Pq
Pscqsc = Z n R T
n R Tsc
q = Psc qscz T
Tsc 1P
where qsc is constant
Z is determined at P, T
II - 15
d. Derivation of equation for q
qsc = kAµL
Tsc
Tz Psc
P12 - P2
2
2
but
qsc = P q Tsc
Z Psc T = k A
µL
TscT z Psc
P12 - P22
2
q = k AµL
1P
P12 - P22
2
q = k AµL
2P1 + P2
P12 - P22
2
q = k AµL
P1 - P2
This equation is identical to the equation for horizontal, linear flow of incompressible liquid
thus
if gas flow rate is determined at mean pressure, P, the equation for incompressible liquidcan be used for compressible gas!
Note: real gas equation of state
Pq = Z n R T
thus
Psc qscP q
= n R Tsc z n R T
where
P = P1 + P2
2
P = volumetric flow rate at P, T
z is determined at P, T
qsc = P q Tscz Psc T
II - 16
2. Horizontal, radial flow system
h
rw
re
Pe
Pw
re rw
a. Conditions
1) horizontal system dzds
= 0
2) radial system, A = 2πrL, ds = - dr,inward flow
3) constant thickness, h = constant
4) compressible gas flow, q = f (P)
5) laminar flow, use Darcy equation
6) non-reactive fluid, k = constant
7) 100% saturated with one fluid
8) constant temperature
II - 17
b. Assumptions
µz = constant
z (and µ ) can be determined at mean pressure
c. derivation of equation for qsc
vs = - kµ
dPds
- ρg
1.0133 x 106 dzds
vs = - kµ
dPds
= qA
but
q = Psc qsc z T
PTsc
and
A = 2πrh and ds = - dr
thus
Psc T qsc2Tsc π h
drr
rw
re = k
ρdPµz
Pw
Pe
PscT qsc2 Tsc π h
1n rerw
= kµz
Pe2 - Pw2
2
qsc = 2 π h kµ 1n re/rw
Tsc
Psc zT
Pe2 - Pw2
2
II - 18
d. derivation of equation for q
qsc = 2 π h kµ 1n re/rw
Tsc
Psc zT
Pe2 - Pw2
2
but
q = P q Tscz Psc T
thus
P q Tscz Psc T
= 2 π h kµ 1n re/rw
Tsc
Psc zT
Pe2 - Pw2
2
q = 2 π h kµ 1n re/rw
1P
(Pe2 - Pw2 )
2
q = 2 π h kµ 1n re/rw
2Pe + Pw
(Pe2 - Pw2 )
2
q = 2 π h kµ 1n re/rw
Pe - Pw
Note: Equation for real gas is identical to equation for incompressible liquid whenvolumetric flow rate of gas, q, is measured at mean pressure.
II - 19
C. Conversion to Oilfield Units
Symbol Darcy units Oil field
q cc/sec bbl/d or cu ft/dk darcy mdA sq cm sq fth cm ftP atm psiaL cm ftµ cp cpr gm/cc lb/cu ft
Example:
q = hkA P1 - P2
µ L in Darcy's units
q ccsec = q bbl
d 5.615 cu ft
bbl
1,728 cu incu ft
16.39 cccu in
d24hr
hr3,600 sec
q ccsec = 1.841 q bbl
d
k darcy = k md
darcy1,000md
k darcy = 0.001 k md
A sq cm = 929.0 sq cm
sq ft A sq ft
A sq cm = 929.0 A sq ft
P1 - P2 atm = P1 - P2 psia atm14.696 psia
P1 - P2 atm = 0.06805 P1 - P2 psia
L cm = L ft 30.48 cmft
meter = 100 cm
1.841 q = 0.001 k 929.0 A .06805 P1 - P2
µ 30.48 L
q = 0.01127 k A P1 - P2
µ L in oilfield units
II - 20
D. Table of Equations
1. Darcy Units
System Fluid Equation
Horizontal,Linear
IncompressibleLiquid
q = kAµ L
P1 - P2
Dipping,Linear
IncompressibleLiquid q = kA
µ L P1 - P2 +
ρ g L sin θ1.0133 x 106
Horizontal,Radial
IncompressibleLiquid
q = 2 π k hµ ln (re/rw)
Pe - Pw
Horizontal,Linear
RealGas qsc = kA
µ L Tsc
Tz Psc
P12 - P2
2 2
q = kAµ L
P1 - P2
Horizontal,Radial
Real Gasqsc = π k h
µ ln (re/rw) Tsc
Tz Psc Pe
2 - Pw 2
q = 2 π k hµ ln (re/rw)
Pe - Pw
II - 21
2. Oilfield Units
System Fluid Equation
Horizontal,Linear
IncompressibleLiquid
q = 0.001127 kAµL
P1 - P2
q = res bbl/d
Dipping,Linear
IncompressibleLiquid
q = 0.001127 kAµL
P1 - P2
+ ρg L sinθ
1.0133 x 106
Horizontal,Radial
IncompressibleLiquid
q = .007082 khµ ln (re/rw)
Pe - Pw
Horizontal,Linear Real Gas
qsc = .1118 k Aµ L z T
P1 2 - P2
2
qsc = scf/d
q = .001127 kAµL
P1 - P2
q = res bbl/d
Horizontal,Radial Real Gas
qsc = .7032 k hµ ln (re/rw) Tz
Pe 2 - Pw 2
q = .007082 khµ ln (re/rw)
Pe - Pw
II - 22
Example II-2
What is the flow rate of a horizontal rectangular system when the conditions are as follows:
permeability = k = 1 darcyarea = A = 6 ft2
viscosity = µ = 1.0 cplength = L = 6 ftinlet pressure = P1 = 5.0 atm.outlet pressure = P2 = 2.0 atm.
Solutions:
We must insure that all the variables are in the correct units.
k = 1 darcy = 1,000 mdA = 6 ft2L = 6 ftP1 = (5.0 atm) (14.7 psi/atm) = 73.5 psiP2 = (2.0 atm) (14.7 psi/atm) = 29.4 psi
q = 1.1271 x 10-3 kAµL
P1 - P2
q = 1.1271 x 10-3 1,000 61 6
73.5 - 29.4
q = 49.7 bbl / day
II - 23
Example II-3
Determine the oil flow rate in a radial system with the following set of conditions:
K = 300 md re = 330 ft
h = 20 ft rw = 0.5 ft
Pe =2,500 psia re/rw = 660
Pw =1,740 psia ln (re/rw) = 6.492
µ = 1.3 cp
Solution:
q = 7.082 x 10-3 kH Pe - Pw
µ ln Re / Rw
q = 7.082 x 1--3 300 20 2,500 - 1,740
1.3 6.492
q = 3,826 res bbl/d
II - 24
E. Layered Systems
1. Horizontal, linear flow parallel to bedding
A
B
C
q q
P1 P2
L
W
qt = qA + qB + qC
h = hA + hB + hC
let k be "average" permeability,
then
qt = k wh P1 - P2µ L
and
qt = kA whAµ L
P1 - P2 + kB whBµ L
P1 - P2 + kC whCµ L
P1 - P2
then
k h = kA hA + kB hB + kC hC
k = ∑j = 1
n kj hj
h
II - 25
2. Horizontal, radial flow parallel to bedding
Pw
rw
re
qA
qB
qC
ht
hA
hB
hC
re
Pe
again
qt = qA + qB + qC
h = hA + hB + hC
qt = 2 π k hµ ln (re/rw)
Pe - Pw
and
qt = 2 π kA hAµ ln (re/rw)
Pe - Pw + 2 π kB hBµ ln (re/rw)
Pe - Pw
+ 2 π kc hcµ ln (re/rw)
Pe - Pw
then
k h = kA hA + kB hB + kC hC
and again
k = ∑j = 1
n kj hj
h
II - 26
3. Horizontal, linear flow perpendicular to bedding
L
h
P2P1
W
qkA∆PALA
A B C
kB∆PBLB
kC∆PCLC
q
qt = qA = qB = qC
p1 - p2 = ∆PA + ∆PB + ∆PC
L = LA + LB + LC
qt = k wh P1 - P2
µ L
and since P1 - P2 = ∆ PA + ∆PB + ∆PC
P1 - P2 = qt µ Lk wh
= qA µLA
kA wh +
qB µLBkB wh
+ qC µLCkC wh
since qt = qA = qB = qC
Lk
= LAkA
+ LBkB
+ LCkC
thusK = L
∑j = 1
n
Ljkj
II - 27
4. Horizontal, radial, flow perpendicular to bedding
h
rwrArBrC
q
PCPBPAPw
qt = qA = qB = qC
Pe - Pw = ∆PA + ∆PB + ∆PC
q = 2 π k h Pe - Pw µ ln (re/rw)
Pe - Pw = qt µ ln (re/rw)
2 π k h =
qA µ ln (rA/rw)2 π kAh
+ qB µ ln (rB/rA)
2 π kB h +
qC µ ln (rC/re)2 π kC h
then
k = ln re/rw
∑j = 1
n
ln(rj/rj-1)
kj
II - 28
Example II-4
Damaged zone near wellbore
k1 = 10 md r1 = 2 ft
k2 = 200 md r2 = 300 ft
rw = 0.25 ft
Solution:
k = ln (re/rw)
∑j = 1
n
ln (rj/rj-1 )
kj
k = ln 300
0.25
ln 2/0.25
10 +
ln 300/2200
k = 30.4 md
The permeability of the damaged zone near the wellbore influences the average permeability more
than the permeability of the undamaged formation.
II - 29
F. Flow through channels and fractures
1. Flow through constant diameter channel
L
A
a. Poiseuille's Equation for viscous flow through capillary tubes
q = πr48 µ L
P1 - P2
A = π r2, therefore
q = Ar28 µ L
P1 - P2
b. Darcy's law for linear flow of liquids
q = kAµ L
P1 - P2
assuming these flow equations have consistent units
Ar28 µ L
P1 - P2 = kAµ L
P1 - P2
thus
k = r28
= d2
32
where d = inches, k = 20 x 109 d2 md
II - 30
Example II-4
A. Determine the permeability of a rock composed of closely packed capillaries
0.0001 inch in diameter.
B. If only 25 percent of the rock is pore channels (f = 0.25), what will the
permeability be?
Solution:
A. k = 20 x 109 d2
k = 20 x 109 (0.0001 in)2
k = 200 md
B. k = 0.25 (200 md)
k = 50 md
II - 31
2. Flow through fractures
b
v = qA
= h212 µ L
(P1-P2)
q = b2 A 12 µ L
(P1 -P2)
setting this flow equation equal to Darcy's flow equation,
b2 A12 µ L
P1 - P2 = kAµ L
P1 - P2
solve for permeability of a fracture:
k = b2
12 in darcy units, or
k = 54 x 109 b2
where b = inchesk = md
II - 32
Example II-6
Consider a rock of very low matrix permeability, 0.01 md, which contains on the average afracture 0.005 inches wide and one foot in lateral extent per square foot of rock.
Assuming the fracture is in the direction of flow, determine the average permeability using theequation for parallel flow.
Solution:
k = ∑
kj Aj
A , similar to horizontal, linear flow parallel to fracture
k = matrix k matrix area + fracture k fracture area
total area
k = 0.01 12 in 2 + 12 in 0.005 in
144 in2 +
54 x 109 x 0.005 2 12 in x 0.005 in
144 in2
k = 1.439 + 81,000
144
k = 563 md
II - 33
III) Laboratory measurement of permeability
A. Procedure
1. Perm plug method
a. cut small, individual samples (perm plugs) from larger core
b. extract hydrocarbons in extractor
c. dry core in oven
d. flow fluid through core at several rates
TURBULENCE
SLOPE = k / m
P12 - P 2
2
2L
qsc PscA
qsc = kA P1
2 - P2 2
2 µ L Psc horizontal, linear, real gas flow with
T = Tsc and Z = 1.0
qsc PscA
= kµ
P1
2 - P2 2
2L
k = ( slope ) m
II - 34
2. Whole core method
a. prepare whole core in same manner as perm plugs
b. mount core in special holders and flow fluid through core as in permplug method
TO FLOWMETER
HIGH AIRPRESSURE
PIPE
RUBBERTUBING
CORE
LOW AIRPRESSURE(FLOW)
VERTICAL FLOW
c. the horizontal flow data must be adjusted due to complex flow path
d. whole core method gives better results for limestones
II - 35
B. Factors which affect permeability measurement
1. Fractures - rocks which contain fractures in situ frequently separate alongthe planes of natural weakness when cored. Thus laboratory measurementsgive "matrix" permeability which is lower than in situ permeability becausetypically only the unfractured parts of the sample are analyzed forpermeability.
2. Gas slippage
a. gas molecules "slip" along the grain surfaces
b. occurs when diameter of the capillary openings approaches the meanfree path of the gas molecules
c. Darcy's equation assumes laminar flow
d. gas flow path with slippage
e. called Klinkenberg effect
f. mean free path is function of size of molecule thus permeabilitymeasurements are a function of type of gas used in laboratorymeasurement.
II - 36
0
kCALCULATED
1P
H2
N2
CO2
g. mean free path is a function of pressure, thus Klinkenberg effect isgreater for measurements at low pressures - negligible at highpressures.
h. permeability is a function of size of capillary opening, thusKlinkenberg effect is greater for low permeability rocks.
i. effect of gas slippage can be eliminated by making measurements atseveral different mean pressures and extrapolating to high pressure(1/p => 0)
0
kMEASURED
1P
II - 37
Example II-7
Another core taken at 8815 feet from the Brazos County well was found to be very shaly. Therewas some question about what the true liquid permeability was, since nitrogen was used in thepermeameter.
Calculate the equivalent liquid permeability from the following data.
Mean MeasuredPressure Permeability ( atm ) ( md )
1.192 3.762.517 3.044.571 2.769.484 2.54
Solution:
Plot kmeasured vs. 1/pressure
Intercept is equivalent to liquid permeability
From graph:
kliq = 2.38 md
0
1
2
3
4
5
GA
S P
ERM
EA
BIL
ITY,
md
0.0 0.2 0.4 0.6 0.8 1.0
RECIPROCAL MEAN PRESSURE, atm-1
kgas = 2.38276 + 1.64632
Equivalent Liquid Permeability = 2.38 md
Pbar
II - 38
3. Reactive fluids
a. Formation water reacts with clays
1) lowers permeability to liquid
2) actual permeability to formation water is lower than lab permeabilityto gas
1000010001001011
10
100
1000
WA
TER P
ERM
EA
BIL
ITY,
md
AIR PERMEABILITY, md
Water concentration20,000 - 25,000 ppm Cl ion.
RELATIONSHIP OF PERMEABILITIES MEASUREDWITH AIR TO THOSE MEASURED WITH WATER
b. Injection water may,if its salinity is less than that of the formation water,reduce the permeability due to clay swelling.
II - 39
Effect of Water Salinity on Permeability of Natural Cores(Grains per gallon of chloride ion as shown).
Field Zone Ka K1000 K500 K300 K200 K100 Kw
S 34 4080 1445 1380 1290 1190 885 17.2S 34 24800 11800 10600 10000 9000 7400 147.0S 34 40100 23000 18600 15300 13800 8200 270.0
S 34 4850 1910 1430 925 736 326 5.0S 34 22800 13600 6150 4010 3490 1970 19.5S 34 34800 23600 7800 5460 5220 3860 9.9
S 34 13600 5160 4640 4200 4150 2790 197.0S 34 7640 1788 1840 2010 2540 2020 119.0T 36 2630 2180 2140 2080 2150 2010 1960.0
T 36 3340 2820 2730 2700 2690 2490 2460.0T 36 2640 2040 1920 1860 1860 1860 1550.0T 36 3360 2500 2400 2340 2340 2280 2060.0
Ka means permeability to air; K500 means permeability to 500 grains per gallon chloride solution;Kw means permeability to fresh water
4. Change in pore pressure
a. The removal of the core from the formation will likely result in a change in pore volume.This is likely to result in a change in permeability (+ or -).
b. The production of fluids,especially around the well,will result in a decrease in pore pressure and a reduction of in-situ permeability.
III- 1
III. BOUNDARY TENSION AND CAPILLARY PRESSURE
I) Boundary tension, σ
A. at the boundary between two phases there is an imbalance of molecular forces
B. the result is to contract the boundary to a minimum size
GAS
LIQUID
SURFACE
III- 2
C. the average molecule in the liquid is uniformly attracted in all directions
D. molecules at the surface attracted more strongly from below
E. creates concave or convex surface depending on force balance
F. creation of this surface requires work
1. work in ergs required to create 1 cm2 of surface (ergs/cm2) is termed"boundary energy"
2. also can be thought of as force in dynes acting along length of 1 cmrequired to prevent destruction of surface (dynes/cm) - this is called"boundary tension"
3. Boundary Energy = Boundary Tension x Length
G. Surface Tension - Boundary tension between gas and liquid is called "surfacetension"
H. Interfacial Tension - Boundary tension between two immiscible liquids or between a fluid and a solid is called "interfacial tension"
σgw = surface tension between gas and water
σgo = surface tension between gas and oil
σwo = interfacial tension between water and oil
σws = interfacial tension between water and solid
σos = interfacial tension between oil and solid
σgs = interfacial tension between gas and solid
I. Forces creating boundary tension
1. Forces
a. Law of Universal Gravitation applied between molecules
b. physical attraction (repulsion) between molecules
2. Liquid-Gas Boundary
attraction between molecules is directly proportional to their masses andinversely proportional to the square of the distance between them
3. Solid-Liquid Boundary
physical attraction between molecules of liquid and solid surface
III- 3
4. Liquid-Liquid Boundary
some of each
II) Wettability
A. forces at boundary of two liquids and a solid (or gas-liquid-solid)
Θ
σow
σos σws
OILWATER
OIL
SOLID
σws = σos + σow cos θ
B. Adhesion Tension, AT
AT = σws - σos = σow cos θ
C. if the solid is "water-wet"
σws ≥ σos
AT = +
cos θ = +
0° ≤ θ ≤ 90°
if θ = 0° - strongly water-wet
III- 4
D. if the solid is "oil-wet"
σos ≥ σws
AT = -
cos θ = -
90° ≤ θ ≤ 180°
if θ = 180° - strongly oil-wet
θ = 830θ = 1580
θ = 350θ = 300
(A)
ISOOCTANE ISOOCTANE + 5.7% ISOQUINOLINE
ISOQUINOLINE NAPHTHENIC ACID
θ = 300 θ = 480 θ = 540 θ = 1060
(B)
Interfacial contact angles. (A) Silica surface; (B) calcite surface
III- 5
III) Capillary pressure
A. capillary pressure between air and water
ΘhAIR
WATER
1. liquid will rise in the tube until total force up equals total force down
a. total force up equals adhesion tension acting along thecircumference of the water-air-solid interface
= 2πr AT
b. total force down equals the weight of the column of waterconverted to force
= πr2 hgρw
c. thus when column of water comes to equilibrium
2πr AT = πr2 hgρw
d. units
cm dynecm = cm2 cm cm
sec2 gm
cm3
dyne = gm cm
sec2
dyne = force unite. adhesion tension
AT = 12
r hgρw dynecm
III- 6
2. liquid will rise in the tube until the vertical component of surface tensionequals the total force down
a. vertical component of surface tension is the surface tensionbetween air and water multiplied by the cosine of the contact angleacting along the water-air-solid interface
= 2πr σaw cosθ
b. total force down
= πr2 hgρw
c. thus when the column of water comes to equilibrium
2πr σaw cosθ = πr2 hgρw
d. units
cm dynecm = cm2 cm cm
sec2
gm
cm3
cm dynecm
= cm gm
sec2
3. since AT = σaw cosθ, 1 and 2 above both result in
h = 2 σaw cos θ
rg ρw
III- 7
4. capillary pressure (air-water system)
Θh
WATER
PaA' A
AIR
Pa
Pw
B'
B
pressure relations in capillary tubes
a. pressure at A' is equal to pressure at A
Pa' = Pa
b. pressure at B is equal to the pressure at A minus the head of waterbetween A & B
pw = pa - ρwgh
units: dyne
cm2 =
dyne
cm2 -
gm ⋅ cm
cm3 ⋅ sec2 cm
c. thus between B' and B there is a pressure difference
pa - pw = pa - (pa - ρwgh)
pa - pw = ρwgh
d. call this pressure difference between B' and B "capillary pressure"Pc = pa - pw = ρwgh
e. remember
h = 2 σgw cos θ
rg ρw
III- 8
f. thus
Pc = 2 σgw cos θ
r
B. capillary pressure between oil and water
Θh
WATER
OIL
1. liquid will rise in the tube until the vertical component of surface tensionequals the total force down
a. vertical component of surface tension equals the surface tensionbetween oil and water multiplied by the cosine of the contact angleacting along the circumference of the water-oil-solid interface
= 2πr σow cosθ
b. the downward force caused by the weight of the column of water ispartially offset (bouyed) by the weight of the column of oil outsidethe capillary
c. thus, total force down equals the weight of the column of waterminus the weight of an equivalent column of oil converted to force
1) weight per unit area of water
= ρw h
2) weight per unit area of oil
= ρo h
III- 9
3) net weight per unit area acting to pull surface down
= ρwh - ρoh = h(ρw - ρo)
4) total force down
= πr2 gh (ρw - ρo)
d. thus when the column of water comes toequilibrium
2πr σow cosθ = πr2 gh (ρw - ρo)
2. thus the equilibrium for the height of the column of water
h = 2 σow cos θrg (ρw - ρo)
3. capillary pressure (oil-water system)
Θh
WATER
PoA
Po
Pw
B'
B
OIL
a. pressure at A' equals pressure at A
Poa = Pwa
b. pressure at B is equal to the pressure at A minus the head of waterbetween A and B
Pwb = Pwa - ρwgh
c. pressure at B' equal to the pressure at A' minus the head of oilbetween A' and B'
III- 10
Pob = Poa - ρogh
d. thus capillary pressure, the difference between pressure at B' andthe pressure at B is
Pc = Pob - Pwb
Pc = (Poa - ρogh) - (Pwa - ρwgh)
since Poa = Pwa
Pc = (ρw - ρo)gh
e. remember
h = 2 σow cos θrg (ρw - ρo)
f. thus
Pc = 2 σow cos θ
r
4. same expression as for the air-solid system except for the boundarytension term
Pc = 2 σ cos θr
C. remember adhesion tension is defined as
AT = σow cosθ,
and
Pc = 2 σow cos θ
r
thus
Pc = f (adhesion tension, 1/radius of tube)
III- 11
ADHESION TENSION
AIR
WATER
AIR
HgWATER
AIR
1/radius of tube
D. an important result to remember
1. pwb < pob
2. thus, the pressure on the concave side of a curved surface is greater thanthe pressure on the convex side
3. or, pressure is greater in the non-wetting phase
E. capillary pressure-unconsolidated sand
1. the straight capillary previously discussed is useful for explaining basicconcepts - but it is a simple and ideal system
2. packing of uniform spheres
Pc = σ 1R1
+ 1R2
R1 and R2 are the principal radii of curvature for a liquid adhering to two spheres incontact with each other.
3. by analogy to capillary tube
1R1
+ 1R2
= 2 cos θr
where Pc = 2 σ cos θr
call it Rm(mean radius), i.e.
1Rm
= 2 cos θrm
= (∆ρ)gh
σ
III- 12
F. wettability-consolidated sand
1. Pendular-ring distribution-wetting phase is not continuous, occupies thesmall interstices-non-wetting phase is in contact with some of the solid
2. Funicular distribution - wetting phase is continuous, completely coveringsurface of solid
(A) (B)
OIL OR GAS OIL OR GAS
SAND GRAIN SAND GRAIN
WATER WATER
Idealized representation of distribution of wetting and nonwetting fluidphase about intergrain contacts of spheres. (a) Pendular-ring distributions;(b) funicular distribution
III- 13
IV) Relationship between capillary pressure and saturation
A. remember that the height a liquid will rise in a tube depends on
1. adhesion2. fluid density3. variation of tube diameter with height
B. consider an experiment in which liquid is allowed to rise in a tube of varyingdiameter under atmospheric pressure. Pressure in the gas phase is increasedforcing the interface to a new equilibrium position.
R
ATMOSPHERIC PRESSURE
R
HIGHERPRESSURE
DEPENDENCE OF INTERFACIAL CURVATURE ON FLUID SATURATIONIN A NON-UNIFORM PORE
1. Capillary pressure is defined as the pressure difference across theinterface.
2. This illustrates:
a. Capillary pressure is greater for small radius of curvature than forlarge radius of curvature
b. An inverse relationship between capillary pressure and wetting-phase saturation
c. Lower wetting-phase saturation results in smaller radius ofcurvature which means that the wetting phase will occupy smallerpores in reservoir rock
III- 14
V) Relationship between capillary pressure and saturation history
A. consider an experiment using a non-uniform tube (pore in reservoir rock)
1. tube is filled with a wetting fluid and allowed to drain until the interfacebetween wetting fluid and non-wetting fluid reaches equilibrium(drainage)
2. tube is filled with non-wetting fluid and immersed in wetting fluidallowing wetting fluid to imbibe until the interface reaches equilibrium(imbibition)
Θ
SATURATION = 100%PC = LOW VALUE
Θ
SATURATION = 80%CAPILLARY PRESSURE = P C
R
LOW PC HIGHER P C
(A)
ΘSATURATION = 0%P C = HIGH VALUE
Θ
SATURATION = 10%CAPILLARY PRESSURE = P C
R
HIGHER P C LOW PC
(B)
Dependence of equilibrium fluid saturation upon the saturation history in anonuniform pore. (a) Fluid drains; (b) fluid imbibes. Same pore, samecontact angle, same capillary pressure, different saturation history
3. This is an oversimplified example, however it illustrates that therelationship between wetting-phase saturation and capillary pressure isdependent on the saturation process (saturation history)
a. for given capillary pressure a higher value of wetting-phasesaturation will be obtained from drainage than from imbibition
III- 15
B. Leverett conducted a similar experiment with tubes filled with sand.
1008060402000.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
DATA FROM HEIGHT-SATURATION EXPERIMENTS ON CLEAN SANDS. (FROM LEVERETT)
∆ ρ
g h
σ(k
/ø)1
/2
Drainage
Imbibition
Drainage Imbibition
Sand I
Φ
Φ
Φ
Φ
Φ
Φ Sand II
WATER SATURATION, Sw %
1. capillary pressure is expressed in terms of a non-dimensional correlatingfunction ( remember Pc = (∆ρ gh )
2. in general terms,
a. drainage means replacing a wetting fluid with a non-wetting fluid
b. imbibition means replacing a non-wetting fluid with a wetting fluid
III- 16
PC
0 100WATER SATURATION, S W
IMBIBITION
DRAINAGE
III- 17
VI) Capillary pressure in reservoir rock
Pw = Po/w -ρwh
144Po = Po/w -
ρoh
144
Water Oil
Pw2 Po2
Po1 = Pw1
100% Water
Oil and Water
Pc = Po - Pw = h144
ρw - ρo
Where: Po = pressure in oil phase, psiaPw = pressure in water phase, psiah = distance above 100% water level, ftPo/w = pressure at oil-water contact, psia
ρw = density of water, lb/cf
ρo = density of oil, lb/cf
At any point above the oil-water contact, po ≥ pw
III- 18
HEIGHTABOVEO-W-C
PRESSURE
PCPO = PO/W -
144ρoH
Pw = PO/W -ρw H
144
III- 19
VII) Laboratory measurement of capillary pressure
A. Methods
1. porous diaphragm
2. mercury injection
3. centrifuge
4. dynamic method
B. Porous diaphragm
1. Start with core saturated with wetting fluid.
2. Use pressure to force non-wetting fluid into core-displacing wetting fluidthrough the porous disk.
3. The pressure difference between the pressure in the non-wetting fluid andthe pressure in the wetting fluid is equal to Pc.
4. Repeat at successively higher pressures until no more wetting fluid willcome out.
5. Measure Sw periodically.
6. Results
7. Advantages
a. very accurateb. can use reservoir fluids
8. Disadvantages
a. very slow - up to 40 days for one coreb. pressure is limited by "displacement pressure" of porous disk
III- 20
C. Mercury Injection Method
1. Force mercury into core - mercury is non-wetting phase - air (usuallyunder vacuum) is wetting phase
2. Measure pressure
3. Calculate mercury saturation
4. Advantages
a. fast-minutesb. reasonably accurate
5. Disadvantages
a. ruins coreb. difficult to relate data to oil-water systems
D. Centrifuge Method
CORE HOLDER BODY
WINDOW
TUBE BODY
1. Similar to porous disk method except centrifugal force (rather thanpressure) is applied to the fluids in the core
2. Pressure (force/unit area) is computed from centrifugal force (which isrelated to rotational speed)
3. Saturation is computed from fluid removed (as shown in window)
4. Advantages
a. fastb. reasonably accuratec. use reservoir fluids
III- 21
E. Dynamic Method
CORE
OIL INLET
OIL BURETTE
∆Po
GAS OUTLET GAS INLET
∆Pg Pc
TO ATMOSPHERE
DYNAMIC CAPILLARY - PRESSURE APPARATUS(HASSLER'S PRINCIPLE)
1. establish simultaneous steady-state flow of two fluids through core
2. measure pressures of the two fluids in core (special wetted disks) -difference is capillary pressure
3. saturation varied by regulating quantity of each fluid entering core
4. advantages
a. seems to simulate reservoir conditionsb. reservoir fluids can be used
5. Disadvantages
a. very tedious
III- 22
F. Comparison of methods
1. diaphragm method (restored state) is considered to be most accurate, thusused as standard against which all other methods are compared
2. comparison of mercury injection data against diaphragm data
a. simple theory shows that capillary pressure by mercury injectionshould be five times greater than capillary pressure of air-watersystem by diaphragm method
b. capillary pressure scale for curves determined by mercury injectionis five times greater than scale for diaphragm air-water data
c. these comparisons plus more complex theory indicate that the ratiobetween mercury injection data and diaphragm data is about 6.9(other data indicate value between 5.8 and 7.5)
III- 23
Example VIII-1
Comparison of Mercury Injection Capillary Pressure Data with Porous Diaphragm Data
A. Calculate capillary pressure ratio,
PcAHgPcAW
, for the following data:
σAHg = 480 Dynes/cm σAW = 72 Dynes/cm
θ AHg = 140° θAW = 0°
B. Pore geometry is very complex. The curvature of the interface and pore radius arenot necessarily functions of contact angles. Calculate the ratio using therelationship.
PcAHgPcAW
= σAHgσAW
Solution:
(A) PcAHgPcAW
= σAHgcos θAHg
σAWcos θAW =
480 cos(140°)
72 cos (0°)
PcAHgPcAW
= 5.1
(B) PcAHgPcAW
@ σAHgσAW
= 48070
PcAHgPcAW
= 6.9
III- 24
Discussion:
A. Best way to determine the relationship between mercury and air-water datais to generate capillary pressure curves for each set of data and comparedirectly.
Mercury Injection and Porous Diaphragm Methods
B. For this given set of conditions, mercury injection method requires ahigher displacement pressure, must adjust ratio between scales until matchis obtained.
C. Minimum irreducible wetting phase saturations are the same.
D. Reduction in permeability results in a higher minimum irreducible wettingphase saturation. For both cases, mercury system still has higher requireddisplacement pressure.
III- 25
VIII) Converting laboratory data to reservoir conditions
PcL= 2σLcos θL
r
PcR= 2σRcos θR
r
setting r = r
r = 2σLcos θLPcL
= 2σRcos θRPcR
∴ PcR = σcos θ Rσcos θ L
PcL
where
PcR= reservoir capillary pressure, psi
PcL= capillary pressure measured in laboratory, psi
σL = interfacial tension measured in laboratory, dynes/cm
σR = reservoir interfacial tension, dynes/cm
θR = reservoir contact angle, degrees
θL = laboratory contact angle, degrees
III- 26
Example III-2
Converting Laboratory Data to Reservoir Conditions
Express reservoir capillary pressure by using laboratory data.
lab data: σAW = 72 dynes
σAW = 0o
reservoir data: σOW = 24 dynes/cm
σOW = 20o
Solution:
PcR=
σcos θ Rσcos θ L
PcL
PcR=
24 cos20°
72 cos0° PcL
PcR=
0.333 PcL
III- 27
IX) Determining water saturation in reservoir from capillary pressure data
A. convert laboratory capillary pressure data to reservoir conditions
B. calculate capillary pressure in reservoir for various heights above height at whichcapillary pressure is zero
Pc = (∆ρ)gh144 gc
in English units
∆ρ = ρw - ρO, lb/cu ft
g = 32 ft/sec2
gc = 32 lbm ftlbf sec2
h = ft
144 = (sq in)/(sq ft.)
thus
Pc = lbf/(sq in), psI
III- 28
Example III-3
Determining Water Saturation From Capillary Pressure Curve
Given the relationship,
PcR = 0.313 P cL, use the laboratory capillary pressure curve to calculate the watersaturation in the reservoir at a height of 40 ft. above the oil-water contact.
ρo = 0.85 gm/cm3 ρw = 1.0 gm/cm3
PCL
SW
00
10
20
50 100
8.38.3
Solution:
PcR= ρw − ρo h
144
PcR=
1.0 - 0.85 62.4 lbft3
40
144 = 2.6 psi
PcL=
PcR0.313
PcL= 2.6
0.313 = 8.3 psi
move to the right horizontally from PcL = 8.3 psi to the capillary pressure curve. Drop verticallyto the x-axis, read Sw.
Sw = 50%
III -29
X) Capillary pressure variation
A. effect of permeability
1. displacement pressure increases as permeability decreases
2. minimum interstitial water saturation increases as permeability decreases
10090807060504030201000
20
40
60
80
100
120
140
160
180
200
RESERVOIR FLUID DISTRIBUTION CURVES
Sw %
(From Wright and Wooddy)
10
md
10
0 m
d
20
0 m
d
90
0 m
d
30
Heig
ht
above
zero
cap
illar
y pre
ssure
, ft
24
18
12
6
0
Oil
- W
ater
Cap
illar
y P
ress
ure
, psi
(re
serv
oir
condit
ions)
90
72
54
36
18
0A
ir -
Wate
r C
apill
ary
Pre
ssure
, psi
(l
abora
tory
dat
a)
III -30
B. Effect of grain size distribution
1008060402000
5
10
15
20
25
30 225.0
187.5
150.0
112.5
75.0
37.5
0
100 80 60 40 20 0
Sandstone Core
Porosity = 28.1%Permeability = 1.43 mdFactor = 7.5
Mer
cury
cap
illar
y pre
ssur
e, p
si
Wat
er/n
itro
gen
cap
illar
y pre
ssur
e, p
si
Hg
Water
1008060402000
10
20
30
40
50
60 348
290
232
174
116
58
0
100 80 60 40 20 0
Wat
er/n
itro
gen
cap
illar
y pre
ssur
e, p
si
Mer
cury
cap
illar
y pre
ssur
e, p
si
Water
Hg
Limestone CorePorosity = 23.0%Permeability = 3.36 mdFactor = 5.8
1. majority of grains same size, so most pores are same size - curve (a) (wellsorted)
2. large range in grain and pore sizes - curve (b) (poorly sorted)
III -31
XI) Averaging capillary pressure data
J-function
J Sw = Pcσcos θ
kφ
1/2
attempt to convert all capillary pressure data to a universal curve
universal curve impossible to generate due to wide range of differences existing inreservoirs
concept useful for given rock type from given reservoir
where
Pc = dyne/(sq cm) σ = dyne/cm k = (sq cm)
φ = fraction
or can use any units as long as you are consistent
III -32
10090807060504030201000.00.10.20.30.40.50.60.70.80.91.01.11.21.31.41.51.61.71.81.92.0
CAPILLARY RETENTION CURVES.
WATER SATURATION, Sw
CA
PIL
LA
RY P
RESSU
RE F
UN
CT
ION
, J
(From Rose and Bruce.)
LEVERETT
LEDUC
HAWKINS
KATIE ALUNDUM
EL ROBLE
KINSELLA
Reservoir Formation
Hawkins WoodbineEl Roble MorenoKinsella Viking
Katie DeeseLeduc Devonian
Alundum (consolidated)Leverett (unconsolidated)
III -33
Capillary Pressure Problem 1
1. A glass tube is placed vertically in a beaker of water. The interfacial tension between theair and water is 72 dynes/cm and the contact angle is 0 degree.
Calculate:
a. the capillary rise of water in the tube if the radius of the tube is 0.01centimeters.
b. what is the difference in pressure in psi across the air-water interface in thetube.
2. The displacement pressure for a water saturated porcelain plate is 55 psi of air. What isthe diameter in inches of the largest pore in the porcelain plate? Assume 72 dynes/cmand 0 degrees.
Solution:
(1) σAW = 72 dynes/cm
ρW = 1 gm/cm3
g = 980 dynes/gm
θ = 0o
(a) capillary rise of water if radius is .01 cm
h = 2σAWcos θrρg
= 2 72 cos0°
.01 1.0 980
h = 14.69 cm
(b) pressure drop in psi across interface
Pc = pa - pw = ρwgh = 1.0 980 14.69
Pc= 0.0142 atm
14.696 psiatm
Pc = 0.209 psi
III -34
(2)Pc
= 2σAWcos θr
Pc = 55 psi
Pc = 55 psi atm14.696 psi
1.0133 x 106 dynes/cm2
atm
= 3.792 x 106 dynes/cm2
r = 2σAW cos θPc
r = 2 72 cos0°3.792 x 106
= 3.797 x 10-5 cm in2.54 cm
r = 1.495 x 10-5 in
d = 2.99 x 10-5 in
III -35
Capillary Pressure Problem 2
Given the information below and graph of PcL vs. wetting phase saturation Sw , construct the
curves for PcR, h in reservoir, and J-function vs. Sw. Water is the wetting phase in both the
laboratory and the reservoir.
fluidslab
air-waterres
oil-water
θ 0° 25°
σ 60 dyne/cm 20 dyne/cm
ρwet 1.0 gm/cm3 1.1 gm/cm3
ρnon-wet 0 gm/cm3 0.863 gm/cm3
k 37 md variable
φ 16% variable
J = Pc k/φ 1/2
σ cos θ
III -36
10090807060504030201000.0
2.5
5.0
7.5
10.0
12.5
15.0
17.5
20.0
22.5
25.0
27.5
30.0
32.5
35.0
Sw %
PC
L, p
si
Solution:
(1) PcR
= σR cosθRσL cosθL
PcL
= 20 cos2560 cos0
PcL
PcR=
0.302 PcL
III -37
(2) PcR
= hR ρw - ρo144
= hR 1.1 - .863 62.4144
PcR= .103 hR
hR= 9.74 PcR
(3) J =Pc
σ cos θ k
φ1/2
=PcL
σAW cosθL k
φ L
1/2
=PcL
60 cos0° 37
.161/2
J = .253 PcL
Sw PcL PcR hR J % psi ps i ft assorted
15 32 9.7 94.1 8.1
20 19.5 5.9 57.4 4.9
25 15.6 4.7 45.9 3.9
30 13.2 4.0 38.8 3.3
40 9.9 3.0 29.1 2.5
50 7.8 2.4 22.9 2.0
60 6.0 1.8 17.6 1.5
70 4.7 1.4 13.8 1.2
80 3.7 1.1 10.9 0.9
90 2.8 0.8 8.2 0.7
100 2 0.6 5.9 0.5
III -38
1008060402000
2
4
6
8
10
Sw %
PcR
1008060402000
20
40
60
80
100
Sw %
h R
1008060402000
2
4
6
8
10
Sw %
J
IV - 1
IV. FLUID SATURATIONS
I) Basic concepts of hydrocarbon accumulation
A. Initially, water filled 100% of pore space
B. Hydrocarbons migrate up dip into traps
C. Hydrocarbons distributed by capillary forces and gravity
D. Connate water saturation remains in hydrocarbon zone
II) Methods for determining fluid saturations
A. Core analysis (direct method)
1. factors affecting fluid saturations
a. flushing by mud filtrate
1) differential pressure forces mud filtrate intoformation
Ph > Pres
2) for water base mud, filtrate displaces formation waterand oil from the area around the well (saturationslikely change)
3) for oil base mud, filtrate will be oil; saturations mayor may not change.
IV - 2
Example: Effects of flushing by mud filtrates
Coring with water base mud
Oil zone at minimum interstitial water saturation:
sat at surface flushing by bit trip to surface compared to res
Sw ↑ ↓ ? probably ↑
So ↓ ↓ ↓
Sg - ↑ ↑
Gas zone at minimum interstitial water saturation:
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 - ↓ ↓
So - - -
Sg - ↑ ↑
IV - 3
Coring With Oil Base Mud
Oil zone at minimum interstitial water saturation:
sat at surface flushing by bit trip to surface compared to res
Sw - - -
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 ↓ ↓ ↓
So ↑ ↓ ↑
Sg - ↑ ↑
IV - 4
b. bringing core to surface
1) reduction in hydrostatic pressure causes gas to comeout of solution
2) gas displaces oil and water causing saturations tochange
2. laboratory methods
a. evaporation using retort distillation apparatus
HEATING ELEMENT
COOLING WATER IN
CONDENSERCOOLING WATER OUT
CORE
IV - 5
1) process
a) heat small sample of rock
b) oil and water vaporize, then condense ingraduated cylinder
c) record volumes of oil and water
d) correct quantity of oil
65605550454035302520150.9
1.0
1.1
1.2
1.3
1.4
Oil Gravity, °API at 60° F
Mult
iply
ing F
acto
r
For converting distilled oil volume to oil volume originally in a sample, multiplyoil volume recovered by factor corresponding to gravity of oil in core
IV - 6
e) determine saturations
Sw = VwVp
So = VoVp
Sg = 1 - So - Sw
where
Sw = water saturation, fractionSo = oil saturation, fractionSg = gas saturation, fractionVp = pore volume, ccVw = volume of water collected, ccVo = volume of oil collected, cc
2) disadvantages of retort process
a) must obtain temperature of 1000-1100oF tovaporize oil, water of crystallization fromclays also vaporizes causing increase in waterrecovery
WATERRECOVERED
TIME
00
PORE WATER
b) at high temperatures, oil will crack and coke.(change in hydrocarbon molecules) amountof recoverable liquid decreases.
c) core sample ruined
IV - 7
3) advantages of retort process
a) short testing time required
b) acceptable results obtained
b. leaching using solvent extraction apparatus
GRADUATED TUBECORE
SOLVENT
HEATER
WATER IN
WATER OUT
1) process
a) weigh sample to be extracted
b) heat applied to system causes water from coreto vaporize
c) solvent leaches hydrocarbons from core
IV - 8
d) water condenses, collects in trap. Recordfinal water volume
e) reweigh core sample
f) determine volume of oil in sample
Vo = Wi - Wdry - Vw ρw
ρo
where:
Wi = weight of core sample after leaching
Wdry = weight of core sample after leaching
Sw = VwVp
So = VoVp
2) disadvantages of leaching
a) process is slow
b) volume of oil must be calculated
3) advantages of leaching
a) very accurate water saturation value obtained
b) heating does not remove water ofcrystallization
c) sample can be used for future analysis
IV - 9
3. uses of core determined fluid saturation
a. cores cut with water base mud
1) presence of oil in formation
2) determination of oil/water contact
3) determination of gas/oil contact
GAS
OIL
WATER
SO0 50
So ≅ 0 in gas zone
So ≥ 15% in oil zone
0 ≤ So ≤ Sor in water zone
Sor = residual oil saturation
IV - 10
b. cores cut with oil base mud ("natural state" cores)
1) minimum interstitial water saturation
2) hydrocarbon saturation
3) oil/water contact
B. Capillary pressure measurements (discussed in Chapter VIII)
C. Electric logs
IV - 11
Example IV-1
You want to analyze a core sample containing oil, water and gas.
Vb bulk volume = 95 cm3
Wt initial = 216.7 gm
the sample was evacuated and the gas space was saturated with water ρw = 1 gm/cm3
Wt new = 219.7 gm
the water with in the sample is removed and collected
Vw removed = 13.0 cm3
the oil is extracted and the sample is dried
Wt dry = 199.5 gm
calculate:
(1) porosity
(2) water saturation
(3) oil saturation assuming 35o API
(4) gas saturation
(5) matrix density
(6) lithology
Solution:
gas vol. = 219.7 - 216.7 ; Vg = 3 cc
water vol. = 13 - 3 ; Vw = 10 cc
Wt fluids = 219.7 - 199.5 = 20.2 gm
Wt oil = 20.2 - 10 - 3 = 7.2 gm
IV - 12
ρo = 141.5131.5 + 35°API
= 0.85 gm/cc
Vo = 7.2/0.85 = 8.49 cc
Vp = 8.49 + 3 + 10 = 21.47 cc
φ = 21.47/95 = 22.6%
Sw = 10/21.47 = 46.57%
So = 8.49/21.47 = 39.46%
Sg = 3/21.47 = 13.97%
ρm = 199.5/(95-21.47) = 2.71 gm/cc
lithology = limestone
IV - 13
Example IV-2
A core sample was brought into the laboratory for analysis. 70 gm of the core sample were placed
in a mercury pump and found to have 0.71 cc of gas volume. 80 gm of the core sample was
placed in a retort and found to contain 4.5 cc of oil and 2.8 cc of water. A piece of the original
sample weighing 105 gm was placed in a pycnometer and found to have a bulk volume of 45.7 cc.
(Assume ρw = 1.0 gm/cc and 35o API oil)
calculate:
(1) porosity
(2) water saturation
(3) oil saturation
(4) gas saturation
(5) lithology
Solution:
Vg = .71 cc70 gm
100 gm = 1.014 cc
Vo = 4.5 cc80 gm
100 gm = 5.63 cc
Vw = 2.8 cc80 gm
100 gm = 3.50 cc
Vb = 45.7 cc105 gm
100 gm = 43.52 cc
Wt matrix = 100 - 5.63(.85) - 3.5(1.0) = 91.71 gm
Vm = 43.52 - 1.014 - 5.63 - 3.50 = 33.37 cc
Vp = 1.014 + 5.63 + 3.50 = 10.14 cc
φ = 10.14/43.52 = 23.31%
IV - 14
Sw = 3.50/10.14 = 34.5%
So = 5.63/10.14 = 55.5%
Sg = 1.014/10.14 = 10%
ρm = (91.71/33.38) = 2.75
IV - 15
Fluid Saturation Problem 1
Calculate porosity, water, oil, and gas saturations, and lithology from the following core analysisdata.
How should the calculated saturations compare with the fluid saturations in the reservoir?
Oil well core with water base mud
initial weight of saturated core = 86.4 gm
after gas space was saturated with water, weight of core = 87.95 gm
weight of core immersed in water = 48.95 gm
core was extracted with water recovery being 7.12 cc
after drying core in oven, core weighed 79.17 gm
assume ρw = 1.0 gm/cc
oil gravity = 40° API
Solution: γo = 141.5131.5 + °API
γo = 141.5131.5 + 40°
= 0.825
ρo = 0.825 gm/cc
(1) φ =
VpVb
Vp = Vw + Vo + Vg
Wo = Wsat - Vwρw - Wdry
= 87.95 - 7.12(1.0) - 79.17
Wo = 1.66 gm
IV - 16
Vo = Woρo
Vo = 1.66 gm0.825 gm/cc
= 2.01 cc
Vw = Vwrec - Wsat - Wi / ρw
= 7.12 - (87.95 - 86.4)/(1.0)
Vw = 5.57 cc
Vg = 1.55 cc
Vp = 5.57 + 2.01 + 1.55
Vp = 9.13 cc
Vb = Wsat - Wimmρw
Vb = (87.95 - 48.95) gm1 gm/cc
= 39.0 cc
φ = 9.1339.0
= 23.4%
(2) Sw =
VwVp
Sw = 5.57 cc9.13 cc
= 61.0%
So = VoVp
So = 2.01 cc9.13 cc
= 22.0%
Sg =VgVp
Sg = 1.55 cc9.13 cc
= 17.0%
(3) Vm = Vb - Vp
IV - 17
Vm = 39 - 9.13 = 29.87 cc
ρm= Wdry
Vm
ρm=
79.17 gm/29.87 cc = 2.65 gmcc
. . lithology is sandstone
(4) water saturation at surface will probably be greater than reservoir watersaturation
oil saturation at surface will be less than reservoir oil saturation
gas saturation at surface will be greater than reservoir gas saturation
IV - 18
Fluid Saturation Problem 2
Calculate porosity, water saturation, oil saturation, gas saturation, and lithology from the followingcore analysis data.
How should the saturations you have calculated compare with the fluid saturations in the reservoir?
Oil well core cut with an oil base mud
Sample 1 weighed 130 gm and was found to have a bulk volume of 51.72 cc
Sample 2 weighed 86.71 gm, and from the retort method was found to contain 1.90 cc of waterand 0.87 cc of oil
Sample 3 weighed 50 gm and contained 0.40 cc of gas space
assumeρw = 1.0 gm/cc
oil gravity = 40o API
Solution: γo = 141.5131.5 + °API
γo = 141.5131.5 + 40°
= 0.825
ρo = 0.825 gm/cc
(1) φ =
VpVb
Vp = Vo + Vw +Vg
Vo = 0.87 cc86.71 gm
x 100 = 1.00 cc100 gm
Vw = 1.90 cc86.71 gm
x 100 = 2.19 cc100 gm
Vg = 0.40 cc50 gm
x 100 = 0.80 cc100 gm
Vp = (1.00 + 2.19 + 0.80) cc/100 gm
IV - 19
Vp = 3.99 cc/100 gm
Vb = 51.72 cc130 gm
x 100 = 39.78 cc100 gm
φ = 3.99/10039.78/100
x 100 = 10%
(2) Sw =
VwVp
x 100
= 2.19 cc3.99 cc
x 100
Sw = 54.8%
(3) So =
VoVp
x 100
= 1.0 cc3.99 cc
x 100
So = 25.1%
(4) Sg =
VgVp
x 100
= 0.80 cc3.99 cc
x 100
Sg = 20.1%
(5) Vm = Vb - Vp
= 39.78 - 3.99
Vm = 35.79 cc/100 gm
Wm = Wsat - ρoVo - ρwVw /Wsat
= 86.71 - 0.825 0.87 - 1.0 1.9086.71
Wm = 97 gm/100 gm
IV - 20
ρm = WmVm
ρm = 97 gm/100 gm35.79 cc/100 gm
ρm = 2.71 gm/cc
. . lithology - limestone
(6) water saturations should be fairly close in value
oil saturation will be less than reservoir oil saturation
gas saturation will be greater than reservoir gas saturation
V - 1
V. ELECTRICAL PROPERTIES OF ROCK-FLUID SYSTEMS
I) Electrical conductivity of fluid saturated rock
A. Definition of resistivity
L
ELECTRICALCURRENT FLOW A
given a box of length (L) and cross-sectional area (A) completely filled withbrine of resistivity (Rw)
the resistance of the brine in the box to current flow may be expressed as
r = Rw LA
r = resistance - ohm
Rw = resistivity - ohm meters
L = length - meters
A = area - (meters)2
V - 2
B. Nonconductors of electricity
1. oil
2. gas
3. pure water
4. minerals
5. rock fragments
C. Conductors of electricity
1. water with dissolved salts conducts electricity (low resistance)
2. clay
D. Development of saturation equation (ignore clay)
A AP
ELECTRICALCURRENT FLOW
L
1. the electrical current flows through the water (brine)
a. the area available for current flow is the cross-sectional areaof the pores.
Ap < A
b. the path through the pores is Lp.
Lp > L
V - 3
2. resistance to electrical flow through the porous media is equal to theresistance of a container of area Ap and length Lp filled with water(brine)
r = RwLp
Ap, water filled cube
r = RoL
A, porous media
thus
Ro = RwALp
ApL
where r = resistance of rock cube with pores filled with brine, ohm
Rw = formation brine resistivity, ohm-m (from water sample or SP log)
Ro = resistivity of formation 100% saturated with brine of resistivity, Rw, ohm-m
Ap = cross-sectional area available for current
flow, m2
Lp = actual path length ion (current) must travel through rock, m
A = cross-sectional area of porous media, m2
L = length of porous media, m
3. Since
ApA
≅ porosity, φ
and
LpL
≅ tortuosity, a measure of rock cementation.
then
Ro = RwALp
ApL
V - 4
becomes
Ro = f Rw, φ, tortuosity
E. Electrical formation resistivity factor, F
1. the equation for resistivity of a formation 100% saturated with abrine of resistivity of Rw
Ro = f Rw, φ, tortuosity
2. can be written as
Ro = F Rw
where F is the electrical formation resistivity factor
F = RoRw
3. cementation factor, m
a. it has been found experimentally that the equation for F takesthe form
F = C φ-m
where C is a constantm is the cementation factor
b. thus
log F = log C - m log φ
V - 5
F
φ
1
10
100
0.01 0.1 1.0
when intercept = C
slope = -m, the cementation factor
4. commonly used equation for electrical formation resistivity factor
a. Archie's Equation
F = φ-m
b. Humble Equation
F = 0.62 φ-2.15
(best suited for sandstones)
Cementation Factor (m) and Lithology
Lithology m values
Unconsolidated rocks (loose sands, oolitic limestones) 1.3Very slightly cemented (Gulf Coast type of sand, except Wilcox) 1.4-1.5Slightly cemented (most sands with 20% porosity or more) 1.6-1.7Moderately cemented (highly consolidated sands of 15% porosity or less) 1.8-1.9Highly cemented (low-porosity sands, quartzite, limestone,dolomite) 2.0-2.2
V - 6
Example V-1
Determine the porosity for a sandstone using Archie's and Humble equation .The formation water's resisitivity was 0.5 ohm-meters. The formation rock 100% saturated withthis water was 21.05 ohm-meters.
Which of the two equations will give the most reasonable answer?
Solution:F =21.05/0.5 = 42.1
Archie's: F = φ-m
m = 2.0 for sandstone
φ2 = 1/F
φ = 142.1
φ = 15.41%
Humble: F = 0.62/φ2.15
φ2.15 = 0.62/F
φ = 0.6242.1
2.15
φ = 14.06%
The Humble equation was developed for sandstone.
V - 7
F. Resistivity Index, I, and Saturation Exponent, n
1. definition of resistivity index
I = RtRo
where Ro = resistivity of formation 100% saturated with water (brine) of
resistivity Rw, ohm-m
Rt =resistivity of formation with water (brine) saturation less than
100%, ohm-m
2. it has been found experimentally that
Sw = I- 1
n = RtRo
- 1n
where n is the saturation exponent ≅ 2.0
3. rearrange
Sw-n = RtRo
-n log Sw = log RtRo
V - 8
1
10
100
.1 1.0
Rt
Ro
Sw
slope = -n, when n is the saturation exponent
NOTE:
slope = log y1 - log y2log x1 - log x2
II) Use of Electrical Formation Resistivity Factor, Cementation Factor, and SaturationExponent
A. obtain porosity, φ, from electric log or core analysis
B. F = C φ-m
(usually use Archie or Humble equation)
C. obtain water resistivity, Rw, from water sample or electric log
D. Ro = F Rw
E. convert Rt from electric log to water saturation
Sw = RoRt
1n
V - 9
III) Laboratory measurement of electrical properties of rock
A. Apparatus
CORE
VOLTMETER
1000 OHM STD. RESISTOR
AC SOURCE
B. Calculations
1. resistance of core
E = Ir
where: E = voltage, volts
I = current, amperes
r = resistance, ohms
∴ rcore = EI
V - 10
2. resistivity of core
Rcore = rcoreA
L
substituting r = E
I into the equation
Rcore = EAIL
C. Procedure
1. determine resistance of core
a. set desired current from AC source, low current preferredso core does not heat up.
b. record voltage from voltmeter
2. determine resistivity of core
a. for the first test completely saturate core with brine Sw =100%, Rcore = Ro
b. for next test, desaturate core by 15-20%, until Sw < 100%Rcore = Rt
c. repeat tests until Sw = Swir
where
Swir = minimum interstitial brine saturation (irreducible), fraction
Ro = resistivity of core 100% saturated with brine, ohm-m
Rt = resistivity of core less than 100% saturated with brine of Rw, ohm-m
RtRo
= resistivity index = I
V - 11
D. Determine saturation exponent, n
1. rearrange saturation equation
Sw = RoRt
1/n
Swn= Ro
Rt
RoRt
= Sw-n
log RtRo
= -n log Sw
2. Plot log RtRo
vs log Sw or log I vs log Sw
1
10
100
.1 1.0
Rt
Ro
Sw
I =
3. the slope of the plot is -n, where n is the saturation exponent
V - 12
Example V-2
Given the following data, calculate the electrical formation resistivity factor and saturation exponentof the core.
Rw = 55 ohm-cm
I = 0.01 amp
D = 2.54 cm
L = 3.2 cm
ESw Voltage across
Water Saturation, % Core, volts 100.0 7.64
86.0 10.5074.0 14.3463.0 20.1654.0 27.5249.0 = Swir 34.67
Solution:
(1) electrical formation resistivity factor
F = RoRw
ro = EI = 7.64
.01 = 764 ohm
Ro = roAL
= 764 2.542π/4
3.2 = 1210 ohm cm
F = RoRw
= 121055
= 22
V - 13
saturation exponent
-n log Sw = log RtRo
Swrt = E
IRt =
rtAL
RtRo
% (ohm) (ohm-cm)
1.00 1.000.86 1050 1663 1.374.74 1434 2271 1.877.63 2016 3192 2.638.54 2452 4358 3.601.49 3467 5490 4.537
1.11
10
Sw
Rt/
Ro
(1.0,1.0)
(.334,10)
V - 14
-n = slope = log 10 - log 1log .334 - log 1
-n = 1 - 0-.4763 - 0
n = 2.10 = saturation exponent
NOTE:
RtRo
= EtEo
so RtRo
could have been calculated as the ratio of voltage at
Sw divided by the voltage at Sw = 1.0
V - 15
E. Determine cementation factor, m, and constant C for electrical formationresistivity factor equation
1. test several core samples from reservoir with formation brine
a. determine Ro and f for each sample
b. determine Rw for formation brine
c.F =
RoRw
2. plot data according to form of equation for electrical formationresistivity factor
F = C φ-m
log F = log C - mlog φ
F
φ
1
10
100
0.01 0.1 1.0
slope = -m, m = cementation factor
intercept = C (intercept found at φ = 1.0)
V - 16
Example V-3
The laboratory test of Example IV-2 has been repeated for several core samples from the reservoir.Data is given below. Calculate the cementation factor and intercept for the formation resistivityfactor equation.
Porosity Formation Resistivity Factor
φ F
0.152 400.168 320.184 260.199 220.213 190.224 17
Solution:
F = C φ-m
log F = log C - m log φ
plot log F vs log φ
1.11
10
100
ø
F
V - 17
slope = log 50 - log 10log 0.137 - log 0.284
= -2.21
-m = slope = -2.21
m = 2.21 = cementation factor
intercept
log F = log C -m log f
log 10 = log C -2.21 log 0.284
log C = -.2082
C = 062 = intercept
V - 18
IV) Effect of clay on resistivity
A. ideally, only water conducts a current in rock
B. if clay is present, portion of current conducted through the clay
BRINE
CLAY
1RoA
= 1Rclay
+ 1Ro
where RoA = resistivity measured on sample of reservoir rock with clay, 100% saturated with brine of
resistivity Rw, ohm-m
Rclay = component of measured resistivity due to clay, ohm-m
Ro = component of measured resistivity due to brine, ohm-m
1RoA
= 1Rclay
+ 1F Rw
C. to determine electrical formation resistivity factor
1. measure resistivity of core sample (containing clay) in usual manner,
this will be RoA
2. measure resistivity of brine, Rw, in usual manner
V - 19
3. plot
1
ROA
(OHM - M) -1
1RW
(OHM - M) -1
1RoA
= 1Rclay
+ 1F
1Rw
where
1Rclay
= intercept
1F
= slope
V - 20
D. effect of clay
1. define FA =
RoARw , clays reduced the apparent formation resistivity
factor
CLEAN SAND
SHALY SAND
F
FA
RW
2. formation resistivity factor decreases more gradually when clay ispresent in the formation
CLEAN SAND
SHALY SAND
F
1
10
100
φ0.1 1.0
V - 21
3. saturation exponent n is not constant when clay is present information.
1
10
100
.1 1.0
Rt
Ro
Sw
I =
CLEAN SAND LOW R wn = 2
CLEAN SAND HIGH R wn = 1
SHALY SAND n =?
CLEAN SANDSwn-1 = I
VI - 1
VI. MULTIPHASE FLOW IN POROUS ROCK
I) Effective permeability
A. Permeability, k, previously discussed applies only to flow when pores are100% saturated with one fluid - sometimes called absolute permeability
q = kA∆ρ
µL
B. When pore space contains more than one fluid, the above equation becomes
qo = koA∆PοµoL
qw = kwA∆PwµwL
qg = kgA∆Pg
µgL
where qo = flow rate of oil, volume/time
qw = flow rate of water, volume/time
qg = flow rate of gas, volume/time
and ko = effective permeability to oil, md
kw = effective permeability to water, md
kg = effective permeability to gas, md
C. Effective permeability is a measure of the fluid conductance capacity ofporous media to a particular fluid when the porous media is saturated withmore than one fluid
D. Effective permeability is a function of:
1. geometry of the pores of the rock
2. rock wetting characteristics
3. fluid saturations
VI - 2
E. Darcy equation for multiple fluids in linear flow, in oilfield units
qo = 1.1271 x 10-3 ko A P1 - P2µoL
qw = 1.1271 x 10-3 kw A P1 - P2 w
µwL
qg = 1.1271 x 10-3 kg A P1 - P2 g
µgL
when k = md
A = ft2
P = psia
L = ft
q = res bbl/day
II) Relative permeability
A. Defined as the ratio of the effective permeability to a fluid at a givensaturation to the effective permeability to that fluid at 100% saturated(absolute permeability)
kro = kok
krw = kwk
krg = kgk
B. It is normally assumed that the effective permeability at 100% saturation isthe same for all fluid in a particular rock. (not necessarily true in shaly sand)
III) Typical relative permeability curves
A. Use subscript wp to represent the "wetting phase"
Use subscript nwp to represent the "non-wetting phase"
VI - 3
00 100
1
Kr
SWP, %
NON-WETTING PHASE
WETTING PHASE
1
2
34
MINIMUM INTERSTITIAL SWP EQUILIBRIUM SNWP
1.krwp = 1, only at Swp = 100%
2. rapid decrease in krwp as Swp decreases
3.krwp = 0, at minimum interstitial Swp
4.krnwp = 0, at equilibrium Snwp
Note that krwp + krnwp < 1.0
VI - 4
B. Effect of saturation history
1. two types of relative permeability curves
a. drainage curve - wetting phase is displaced by non-wettingphase, i.e., wetting phase saturation is decreasing
b. imbibition curve - non-wetting phase is displaced by wettingphase, i.e., wetting phase saturation is increasing
2. the typical relative permeability curve shown below represents aprocess in which
a. process begins with porous rock 100% saturated withwetting phase (Swp = 100%)
b. wetting phase is displaced with non-wetting phase (drainage)until wetting phase ceases to flow (Swp = minimuminterstitial wetting phase saturation)
c. then non-wetting phase is displaced with wetting phase(imbibition) until non-wetting phase ceases to flow (Swp =equilibrium or residual non-wetting phase saturation)
VI - 5
00 100
1
Kr
SWP, %
DRAINAGE
IMBIBITION
Krnwp
Krwp
minimum interstitial residual non-wetting
wetting phase saturation phase saturation
VI - 6
3. the word "hysteresis" describes the process in which the results (kr)are different when measurements are made in different directions
4. the procedure (drainage or imbibition) used to obtain kr data inlaboratory must correspond to the process in the reservoir
a. initial distribution of fluids in reservoir was by drainage
b. at and behind a water front (flood or encroachment) theprocess is imbibition
5. wetting preference for reservoir rocks is usually water first, then oil,finally gas
Fluids Present Wetting Phase
Water & Oil WaterWater & Gas Water
Oil & Gas Oil
VI - 7
C. Three phase relative permeability
1. often three phases are present in petroleum reservoirs
2. tertiary (triangular) diagram is used to represent a three-phase system
100% OIL100% WATER
100% GAS
VI - 8
3. relative permeability to oil in a three phase system
1%
5
10
60
40
20
100% OIL100% WATER
100% GAS
Note, kro is shown in %
a. dependence of relative permeability to oil on saturations ofother phases is established as follows:
1) oil phase has a greater tendency than gas to wet thesolid
2) interfacial tension between water and oil is less thanthat between water and gas
3) oil occupies portions of pore adjacent to water
4) at lower water saturations the oil occupies more ofthe smaller pores. The extended flow path lengthaccounts for the change in relative permeability to oilat constant oil saturation and varying water saturation
VI - 9
4. Relative permeability to water in a three-phase system
100% OIL100% WATER
100% GAS
0
10%
20%
40%
60%
80%
Krw
a. straight lines indicate relative permeability to water is afunction of water saturation only
b. thus, krw can be plotted on cartesian coordinates againstSw.
VI - 10
5. Relative permeability to gas in a three-phase system
100% OIL100% WATER
100% GAS
50%
40
30
20
5
1
a. curves above indicate that krg is a function of saturations of
other phases present.
b. other research shows that krg is a unique function of gas
saturation
c. the other phases, oil and water, occupy the smaller poreopenings and wet the surface of the rock
d. therefore, krg should be dependent only on the total
saturation of the other two phases (i.e. 1-Sg) andindependent of how much of that total is composed of eitherphase
e. thus krg can be plotted on Cartesian coordinates against So +
Sw
VI - 11
1008060402000.0
0.2
0.4
0.6
0.8
1.0
krg
So + Sw
6. Bottom line - for three-phase system in water wetted rock
a. water
1) is located in smaller pore spaces and along sandgrains
2) therefore krw is a function of Sw only
3) thus plot krw against Sw on rectangular coordinates
b. gas
1) is located in center of larger pores
2) therefore krg is a function of Sg only
3) thus plot krg against Sg (or So + Sw) on rectangular
coordinate
VI - 12
c. oil
1) is located between water and gas in the pores and to acertain extent in the smaller pore spaces
2) therefore kro is a function of So, Sw, and Sg
3) thus plot kro against So, Sw, Sg on a triangulardiagram
4) if Sw can be considered to be constant (minimum
interstitial) kro can be plotted against So on arectangular diagram
1008060402000.0
0.2
0.4
0.6
0.8
1.0
kro
So, %
Minimum InterstitialWater Saturation
VI - 13
7. Flow in three-phase system
100% OIL100% WATER
100% GAS
5% oil
5% water
5% gas
Arrows point to increasing fraction of respective components in stream
Region of three-phase flow in reservoir centers around 20% gas, 30% oil,50% water
VI - 14
IV) Permeability ratio (relative permeability ratio)
A. Definitions
1. When the permeability to water is zero (as at minimum interstitialwater saturation) it is sometimes convenient to use permeability ratioto represent the flow conductance of the rock to gas and oil as aratio.
permeability ratio = kgko
= krgkro
2. When the permeability to gas is zero (no gas or gas below "criticalgas saturation") it is sometimes convenient to use permeability ratioto represent the flow conductance of the rock to oil and water as aratio
permeability ratio = kokw
= krokrw
V) Measurement of relative permeability
A. Methods
1. Laboratory - steady-state flow process
2. Laboratory - displacement (unsteady-state process)
3. Calculation from capillary pressure data (not covered here)
4. Calculation from field performance data
B. Laboratory Methods
1. Steady-state flow process
a. saturate core with wetting-phase fluid
b. inject wetting-phase fluid through core (this will determineabsolute permeability)
VI - 15
c. inject a mix of wetting-phase and non-wetting phase (startwith small fraction of non-wetting phase)
d. when inflow and outflow rates and portion of non-wettingphase equalize, record inlet pressure, outlet pressure andflow rates of each phase
e. measure fluid saturation in core (see below)
f. calculate relative permeability
ko = qoµoL
A∆p
kw = qwµwL
A∆p
g. repeat b through f with injection mixtures containingrelatively more non-wetting phase until irreducible wetting-phase saturation is reached
Sw, %0 1000
1
krkro krw
VI - 16
h. determination of fluid saturations
1) resistivity
Sw = RoRt
1n =
EoEt
1n
where: Ro = resistivity of core 100% saturated with wetting-phase, ohm-m
Rt = resistivity of core with saturation of wetting phase less than 100%, ohm-m
Eo = voltage across core 100%, saturated with wetting phase, volts
Et = voltage across core with saturation of wettingphase less than 100%, volts
2) volumetric balance
3) gravimetric method - remove core and weigh it
Wf = Wt - Wd
where: Wf = weight of fluid in core, gm
Wt = weight of saturated core, gm
Wd = weight of dry core, gm
Wf = ρoVo + ρwVw
and
Vf = Vo + Vw
where: ρ = density, gm/cc
V = volume, cc
Sw = Vw/Vf
where: Sw = saturation of wetting phase
thus
VI - 17
Sw =
Wf/Vf - ρoρw - ρo
i. same procedure can be used starting with 100% saturation ofnon-wetting phase
1) injection ratio start with high ratio of non-wettingphase
2) procedure ends at residual non-wetting phasesaturation
3) then is a hysteresis effect of same type as discussedwith capillary pressure measurements
4) choice of starting saturation depends on reservoirprocess which is being simulated
j. end effects
1) causes of end effects
a) in the bulk of the core there is a wetting-phase saturation and a non-wetting phasesaturation, therefore there is a finite value ofcapillary pressure
b) thus there is a difference in pressure betweenthe wetting-phase and non-wetting phase
Pcap = Pnwp - Pwp
c) at the face of the core the pressures in thewetting-phase and the non-wetting phase areessentially equal
Pnwp = Pwp
thus capillary pressure is essentially zero
d) if capillary pressure is zero, the saturation ofthe wetting phase must be 100% (seecapillary pressure curve)
e) there must be a saturation gradient fromessential 100% wetting phase at the "end" tosome value of Swp less than 100% in thebulk of the core
VI - 18
25201510500
20
40
60
80
100
25201510500
20
40
60
80
100
Oil
Sat
urat
ion,
%
Distance from outflow face, cm
Inflow face
Theoretical saturationgradient
Oil
Sat
urat
ion,
%
Distance from outflow face, cm
Inflow face
Theoretical saturationgradient
2) elimination of end effects
a) install end pieces to contain end effects
b) flow at rapid rates to make end effectnegligible (pressure gradient > 2 psi/inch
EndSection
TestSection
MixingSection
Thermometer
Packing Nut Electrodes
CopperOrifice Plate
Inlet
InletHighly permeable disk
DifferentialPressure Taps
OutletBronze Screen
PENN STATE RELATIVE-PERMEABILITY APPARATUS
VI - 19
Example VI-1
The relative permeability apparatus shown above was used in a steady-state flow process to obtain
the data given below at a temperature of 70oF. See figure on previous page.
The Core The Fluids
sandstone brine, 60,000 ppm
length = 2.30 cm oil, 40oAPIdiameter = 1.85 cm µw = 1.07 cp
area = 2.688 cm2 µo = 5.50 cpporosity = 25.5%
Oil Flow Water Flow Inlet Pressure Outlet Pressure Voltage Drop Electrical Currentcc/sec cc/sec psig psig volts amps
0.0000 1.1003 38.4 7.7 1.20 0.010.0105 0.8898 67.5 13.5 2.10 0.010.0354 0.7650 88.1 17.6 2.80 0.010.0794 0.3206 78.2 15.6 4.56 0.010.1771 0.1227 85.6 17.1 8.67 0.010.2998 0.0000 78.4 15.7 30.00 0.01
Draw the relative permeability curve
Solution:
1. Calculate absolute permeability using data with core 100% saturated with water
k =qwµwL
A∆p
k = 1.1003 1.07 2.302.688 38.4 - 7.7 14.696
k = 0.482 darcy
VI - 20
2. Calculate effective permeabilities to oil and water
ko =
qoµoL
A∆P
ko =
0.0105 5.50 2.302.688 67.5 - 13.5 / 14.696
ko = 0.0134 darcy
kw =
qwµwL
A∆P
kw =
0.8898 1.07 2.302.688 67.5 - 13.5 / 14.696
kw = 0.2217 darcy
3. Calculate relative permeabilities
kro = kok
= .0134.482
= 0.028
krw = kwk
= .2217.482
= 0.460
4. Calculate water saturations
Sw = EoEt
1/2
Sw = 1.202.10
1/2 = .756
5. Results
Water Saturation Relative Permeability Relative Permeabilityto oil to water
Sw kro krw ko/kw
1.000 0.000 1.000 0.0000.756 0.028 0.460 0.0610.655 0.072 0.303 0.2380.513 0.182 0.143 1.2730.372 0.371 0.050 7.4190.200 0.686 0.000 -------
VI - 21
1008060402000.0
0.2
0.4
0.6
0.8
1.0
Kro
Krw
Sw, % pore space
Rel
ativ
e Pe
rmea
bilit
y
100806040200.01
.1
1
10
Sw, % of pore space
Per
mea
bili
ty R
atio
, ko/
k w
VI - 22
6. The data permit certain checks to be made
F = 0.62 φ-2.15
F = RoRw
Rw = 12 ohm-m for 60,000 ppm brine
Ro = EAIL
= 1.20 2.688
.01 2.3 = 140 ohm-m
F = 14012
= 11.7
Φ = .62F
12.15 = .62
11.7
12.15
Φ = .255
VI - 23
2. Displacement (unsteady-state)(Welge)
a. does not result in relative permeability only give permeabilityratio
b. procedure
1) mount core in holder
2) saturate with wetting phase (usually oil)
3) inject non-wetting phase (usually gas) at constantinlet and outlet pressures
4) measure
a) cumulative gas injected as a function of time
b) cumulative oil produced as a function of time
c. conditions
1) pressure drop across core high enough to make endeffects negligible,but not enough to cause turbulent(non-darcy) flow.
2) gas saturation can be described at mean pressure
Pm = Pi + Po
2
3) flow is horizontal and core is short so that effects ofgravity can be neglected
d. calculations
1) convert gas injected into pore volumes
Gipv = GipiLAφ pm
where Gi = cumulative gas injected (measured at pressure pi), cc
Gipv = cumulative gas injected in pore volume
pi = inlet pressure, psi
VI - 24
pm = pi + po2
, psi
LA φ = pore volume, cc
2) calculate average gas saturation, Sgav
Sgav
=Np
LAφ
where Np = cumulative oil produced, cc
LA φ = pore volume, cc
3) plot Sgav vs Gipv
Gipv00
Sgav
GAS BREAKTHROUGH
VI - 25
4) determine fractional flow of oil, fo
fo = d Sgavd Gipv
fo = slope of plot of Sgav vs qGipv
5) calculate permeability ratio, kg/ko
fo =
koA ∆pµoL
koA ∆pµoL
+ kgA ∆p
µgL
fo = ko/µo
ko/µo + kg/µg
kgko
= 1 - fofo µo/µg
where
kgko
= permeability ratio of gas to oil
fo = fractional flow of oil
6) Permeability ratio, kg/ko, calculated above appliesonly at the gas saturation of the outflow face, thusmust calculate Sgo
Sgo = Sgav - Gipvfo
where Sgo = gas saturation at outlet face of core
Gipv = cumulative gas
injected, pore volumes
fo = fractional flow of oil atoutlet face of core
VI - 26
e. advantages
1) minimum amount of equipment
2) rapid
f. disadvantages
1) results in kg/ko, not kro and krg
2) equations don't apply until gas breaks through, thusinitial value of gas saturation may be high, resulting
in incomplete kg/ko vs Sgo curve.
VI - 27
Example VI-2
The data from an unsteady-state displacement of oil by gas in a 2 inch diameter by 5 5/8 inch longsandstone core are given below.
Cumulative Gas Injection, Gi, cc Cumulative Oil Produced, Np, cc
14.0 14.650.2 19.5112.6 22.5202.3 25.5401.4 28.6546.9 30.4769.9 32.21226.5 33.43068.9 35.35946.6 35.9
Other data
T = 70oF, µo = 2.25 cp, µg = .0185 cp
φ = .210, p inlet = 5.0 psig, p out = 0.0 psig
L = 5 5/8 x 2.54 = 14.3 cm A = p (2.54)2 = 20.27 cm2
Prepare to determine kg/ko by calculating Sgav and Gipv.
Solution:
1. Calculate Sgav
Sgav = Np
L A φ
Sgav = 14.6 cc14.3 cm 20.27 cm2 .210
Sgav = 0.24
2. Calculate Gipv
VI - 28
Gipv = GipiLAφ pm
Gipv = 14.0 cc 19.7 psia
14.3 cm 20.27 cm2 .210 19.7 psia + 14.7 psia /2
Gipv = 0.264 pv
3. Results
SgavGipv pv
0.24 0.2640.32 0.9450.37 2.120.42 3.810.47 7.560.50 10.30.53 14.50.55 23.10.58 57.80.59 112.0
VI - 29
Example VI-3
A core sample initially saturated with oil is flooded with gas. The following data was obtained:
Sgav Gipv pv
0.24 0.2640.32 0.9450.37 2.120.42 3.810.47 7.560.50 10.30.53 14.50.55 23.10.58 57.80.59 112.0
µo = 2.25 cp
µg = 0.0185 cp
Calculate and construct a fg verses Sgo plot. Convert Sgavg to Sgo. Determine kg/ko for each of
the given saturations. Construct a graph of kg/ko versus Sgo.
Solution:
Plot Sgav vs. Gipv
The slope from this plot is fo.
Sgo = Sgav - fogipv
kg/ko = 1 - fo
fo µoµo
Sgav Gipv pv fo Sgo kg/ko 0.24 .264 .375 .141 .01370.32 .94 .075 .249 .1010.37 2.17 .0357 .294 .2220.42 3.81 .0214 .338 .3760.47 7.56 .0118 .381 .6890.50 10.3 .0092 .405 .8860.53 14.5 .0046 .463 1.780.55 23.1 .0013 .521 6.320.58 57.8 .0005 .550 16.40.59 112.0 .0001 .581 82.2
VI - 30
1201008060402000.2
0.3
0.4
0.5
0.6
Sgav
Gipv, pv
fraction
1.00.90.80.70.60.50.40.30.20.10.0.01
.1
1
10
100
kg/ko
Sg, %
VI - 31
C. Field determination of permeability ratios
1. equations
qgqo
=
kg A ∆p
µgL
ko A ∆pµoL
where qg = gas flow rate measured at reservoir conditions, vol/time
qo = oil flow rate measured at reservoir conditions, vol/time
thus,
kgko
= qgqo
µgµo
replace qg/qo with
qgqo
= Bg Rp - Rs5.615 Bo
where Bg = formation volume factor of gas, res cu ft/scf
Bo = formation volume factor of oil, res bbl/STB
Rp = producing gas-oil ratio, scf/STB must include both separator gas and stock tank gas)
thus
kgko
= µgBg Rp - Rsµo 5.615 Bo
2. procedure
a. producing gas-oil ratio, Rp, and physical properties, Bg,
Bo, Rs, µg, µo must be determined at some knownreservoir pressure
b. saturations in reservoir, Sg or So, must be calculated fromproduction data and material balance calculations
VI - 32
Example VI-4
Discovery pressure for your well was 4250 psia, temperature is 200oF, and initial producing gas-
oil ratio was 740 SCF/STB. Stock tank oil gravity is 30oAPI and surface gas gravity is 0.7.Production history and correlations indicate the bubble point at 3500 psia. Reservoir pressure isnow 3000 psia. Producing gas-oil ratio is 18,100 SCF/STB. What is kg/ko in the reservoir at thistime.
Solution:
Correlations covered in the fluid properties portion of this course yield the following value
of the physical properties of the gas and oil at 3000 psia and 200o F.
Rs = 560 SCF/STB
Bo = 1.314 res. BBL/STB
Tpc of gas = 390 oR
Ppc of gas = 665 psia
z = 0.86
Bg = 0.0282 z T/p = 5.34 x 103 res cu ft/SCF
µg = 0.0192 cp
µo = 0.75 cp
kgko
= µgBg Rp - Rsµo 5.615 Bo
kgko
= 0.0192 5.34 x 10-3 18100 - 5600.75 5.615 1.314
kgko
= 0.325
VI - 33
VI) Uses of relative permeability data
A. Determination of free water surface in reservoir (100% water production)
Sw, %0 1000
h, ft100% Water Production
100 % Sw
SP Log RT Log
Log Response Diagram
Sw, %0 1000
1
kr
VI - 34
B. Determination of height of 100% oil production
Sw, %0 1000
h, ft100% Water Production
100 % Sw
SP Log RT Log
Log Response Diagram
100% Oil Production
Sw, %0 1000
1
kr
VI - 35
C. Effect of permeability on thickness of transition zone
Sw, %0 1000
h, ft
Sw, %0 1000
1
krLow K
High K
Low K
High K
h = height of zone of interest
VI - 36
D. Fractional flow of water as a function of height
fw = qwqtot
= qwqo + qw
=
kw A ∆ Pµw L
ko A ∆ Pµo L
+ kw A ∆ Pµo L
= 1
1 + kokw
µwµo
fw
1
00 100Sw
00 100Sw
h
100
00
h
100
fw1
VI - 37
1008060402000
20
40
60
80
100
120
140
160
Fraction of water in produced fluid, %
Hei
ght a
bove
fre
e w
ater
leve
l, ft
10 md
50 md
100 md
200 md
This figure indicates that lower permeabilities result in longer transition zones
E. Determination of residual fluid saturations
1
00 100Sw
kr
Oil
Water
Residual Oil Saturation
1. Imbibition curve used in water flood calculations
2. Maximum oil recovery = area (acre) x h(ft) x f x 7758 BBL/acre ft x∆Sw
VI - 38
F. Interpretation of fractional flow curve
fw
1
00 100Sw
2 3
1
4
1. fw at water breakthrough
2. Sw at well at water breakthrough
3. Swav in reservoir between wells at water breakthrough
4.1
slope = pore volume of water injected
VI - 39
fw
1
0
0 100Oil Rec - % Oil in Place
1
2
3
0
Water inputPore vols.
VII - 1
VII. STATISTICAL MEASURES
I) Introduction
Usually we can not examine an entire "population" (i.e. we can not dig up an entire
reservoir, cut it into plugs, and measure the porosity of every plug). We can only
"sample" the population and use the properties of the sample to represent the
properties of the population. Often we seek a single number (porosity or
permeability) to represent the population (reservoir) for use in reservoir engineering
calculations.
If the sample is representative of the population, we have a statistical basis for
estimating properties of the population.
The sample data is said to be unclassified or classified depending on whether it is
arranged or grouped in a particular order. Unclassified data is randomly arranged.
The classification of data for a large number of samples will often provide
additional information to help describe the physical properties of the population.
VII - 2
II) Frequency Distributions
It is often useful to distribute data into classes . The number of individuals
belonging to each class is called the class frequency . A tabular arrangement of
these data according to class is called a frequency distribution or frequency table .
Sometimes classified data is called grouped data .
The division of unclassified data into classified data is accomplished by allocating
all data to respective class intervals . The midpoint of each class interval is called the
class mark .
Rules for forming frequency distributions
A. Determine largest and smallest numbers in the raw data.
B. Divide the range of numbers into a convenient number of equal sized class
intervals. The number of class intervals depends on the data but is usually
taken between 5 and 20 in number.
C. The number of observations for each class interval is the class frequency .
D. The relative frequency of a class is the frequency of the class divided by the
total frequency of all the classes.
VII - 3
III) Histogram
A histogram is a graphical representation of a frequency distribution.
The vertical scale is the number of data points - the class frequency - in each class.
The width of the rectangle corresponds to the class interval.
Mean
Magnitude of Variable
Fre
quen
cy o
f O
ccur
ence
# of
Sam
ples
0
2
4
6
8
Porosity, %12 16 20 24
# of
Sam
ples
0
2
4
6
8
20 60 100 140
Permeability, md
VII - 4
Net pay thickness data from 20 wells summarized as relative frequency data
Frequency Relative FrequencyRange of (No. of wells (No. of wells having thickness Relativethickness, having thickness values in each range, Frequencyft. values in the range) fraction of total wells) as percentage.
50-80 4 0.20 20%81-110 7 0.35 35%111-140 5 0.25 25%141-170 3 0.15 15%171-200 1 0.05 5%
20 1.00 100%
0 50 80 110 140 170 200
Random variable: net pay thickness, ft
Fre
quen
cy
0
2
4
6
8
10
VII - 5
Sometimes the relative frequency is plotted on a histogram
0 50 80 110 140 170 200
Random variable: net pay thickness, ft
Rel
ativ
e Fr
eque
ncy
0
.1
.2
.3
.4
.5
VII - 6
IV) Cumulative Frequency Distributions
Relative frequencies are summed and plotted at the higher ends of the class intervals
to create a "cumulative frequency less than or equal to" distribution
0 50 80 110 140 170 200
Random variable: net pay thickness, ft
0
20%
40%
60%
80%
100%
Cum
ulat
ive
% le
ss th
an o
r eq
ual t
o
0
.2
.4
.6
Cum
ulat
ive
freq
uenc
y le
ss th
an f
or e
qual
to
.8
1.0
VII - 7
Occasionally a "cumulative frequency greater than or equal to" distribution is
plotted. Relative frequencies are summed from the highest class interval and plotted
at the lower ends of the intervals
0 50 80 110 140 170 200
Random variable: net pay thickness, ft
0
20%
40%
60%
80%
100%
Cum
ulat
ive
% g
reat
er th
an o
r eq
ual t
o
0
.2
.4
.6
Cum
ulat
ive
freq
uenc
y gr
eate
r th
an f
or e
qual
to
.8
1.0
Probability graph paper has been constructed so that data from certain probability
distributions plot as a straight line. Different probability paper is used for data with
different distributions
VII - 8
V) Normal Distribution
The normal distribution is continuous probability distribution having a symmetrical
shape similar to a bell, sometimes called a Gaussian distribution.
a a
µ − a µ + a
Random variable x
f(x)
Inflection pointof curve
This distribution is completely and uniquely defined by two values - the mean, m,
and standard deviation, σ.
VII - 9
VI) Log Normal Distribution
The log normal distribution is a continuous probability distribution that appears
similar to a normal distribution except that it is skewed to one side. It is also called
an exponential distribution.
Random variable x
f(x)
Mode
Median (geometric mean)
Mean (arithmetic mean)
This distribution can also be completely and uniquely defined by the mean, m, and
the standard deviation, σ.
If random variable xi are log normally distributed then the variables log xi are
normally distributed.
VII - 10
VII) Measures of Central Tendency
An average is a value which is typical or representative of a set of data. When a set
of data is arranged according to magnitude the average value tends to lie in the
center of these data. These averages are called measure of central tendency .
mean - the arithmetic average value of the samples
µ = xiΣ
i =1
n
n
where xi = values of the variable of interest for each sample
nµ = number of samples
median - the value equalled or exceeded by exactly one-half of the samples.
mode - the value which occurs with the greatest frequency
geometric mean - the nth root of the product of n numbers
µg = x1⋅ x2⋅ x3 . . . xn1/n
µg = xiπi=1
n 1/n
where µg = the geometric mean
VII - 11
VIII) Measures of Variability (dispersion)
A measure of central tendency is the "average" or expected value of a set of
variables, however it does not show the spread or variability of the variables on
either side of the central tendency.
A. Standard deviation - The square root of the mean of the squared deviations
about µ, where deviation is defined as the distance of the variable from µ.
σ2 = xi - µ 2Σ
i=1
n
n-1
where σ2 is the variance
σ is the standard deviation
B. Mean deviation - another measure of the dispersion about the centraltendency
MD = xi - µΣ
i=1
n
n
For classified data
σ2 = fiΣ
j xi - µ 2
fiΣj
where fi = frequency for each class
xj = class mark
or
σ2 = frjΣj
xj - µ 2
where frj = relative frequency for each class
VII - 12
IX) Normal Distribution
Porosity data is usually assumed to have a normal distribution.
For the normal distribution the mean, median and mode have the same numerical
values. They are identical measures of central tendency.
Thus, for unclassified data
µ = xiΣ
i=1
n
n
where i refers to each individual data point and, for classified data
µ = fjΣ
j xj
fjΣj
where j refers to each class interval
fj is the frequency of the class
xj is the class mark
orµ = frjΣ
j xj
where frj is the relative frequency of the class.
xj is the class mark
VII - 13
Cum
ulat
ive
Freq
uenc
y
Cum
ulat
ive
Freq
uenc
y
Random variable x Random variable x
Cumulative frequency plottedon coordinate graph paper
Cumulative frequency plottedon normal probability paper
.001 .001
.999 .999
.001
.999
Cumulative % <
.001
.999
Cumulative % <
Ran
dom
var
iabl
e x,
dist
ribu
ted
norm
ally
Ran
dom
var
iabl
e x,
dist
ribu
ted
norm
ally
Normal probability graph paper
50%
µ
50%
µ
µ + a
84.1%
σ
VII - 14
Porosity and permeability data from a well in the Denver-Julesburg Basin
Porosity IntervalInterval Frequency Midpoint
i Percent fi xi fixi (xi - µ) (xi - µ)2 fi(xi - µ)2
1 7.0<x<10.0 1 8.5 8.5 -9.2 84.64 84.642 10.0<x<12.0 0 11.0 0.0 -6.7 44.89 0.003 12.0<x<14.0 1 13.0 13.0 -4.7 22.09 22.094 14.0<x<16.0 10 15.0 150.0 -2.7 7.29 72.905 16.0<x<18.0 12 17.0 204.0 -0.7 0.49 5.886 18.0<x<20.0 8 19.0 152.0 +1.3 1.69 13.527 20.0<x<22.0 7 21.0 147.0 +3.3 10.89 76.238 22.0<x<25.0 3 23.5 70.5 +5.8 33.64 100.92
42 745.0 376.18
µ = Σ fi xiΣ fi
= 745.042
= 17.7%
σ2 = fiΣ
jxi - µ 2
fiΣj
= 376.1842
= 8.96
σ = 8.96 = 2.99%
VII - 15
Cumulative
Frequency Cumulative
Than or Equal to Frequency
Porosity Upper Limit of Expressed as
Interval, % Frequency Interval Percentage
7.0<x<10.0 1 1 2.4%
10.0<x<12.0 0 1 2.4%
12.0<x<14.0 1 2 4.8%
14.0<x<16.0 10 12 28.6%
16.0<x<18.0 12 24 57.1%
18.0<x<20.0 8 32 76.2%
20.0<x<22.0 7 39 92.9%
22.0<x<25.0 3 42 100.0%
42
26
24
22
20
18
16
14
12
10
82 10 20 30 40 50 60 70 80 90 96
Cumulative % less than or equal to given values of porosity
Cor
e po
rosi
ty, %
at 50th percentile f = 17.7%
at 84th percentile f + σ = 20.7%
σ = 20.7 - 17.7 = 3%
VII - 16
X) Log Normal Distribution
Permeability data is usually assumed to have a log normal distribution.
For log normal distribution the median, mode, and mean have different numerical
values. The median has been chosen as the value of central tendency which best
represents the data.
The median of a log normal distribution is equal to the geometric mean.
Thus, for unclassified data
µ = xiπi=1
n 1/n
or
log (m) = 1n log xiΣ
i=1
n
VII - 17
CumulativeFrequency Less Cumulative
Permeability Than or Equal to FrequencyInterval Upper Limit of Expressed As
(millidarcies) Frequency Interval Percentage
0-50 2 2 4.8%51-100 2 4 9.5%
101-150 4 8 19.0%151-200 4 12 28.6%201-250 4 16 38.1%251-300 8 24 57.1%301-350 4 28 66.7%351-400 2 30 71.4%401-450 4 34 81.0%451-500 1 35 83.3%501-700 5 40 95.2%
701-1000 2 42 100.0%42
2 10 20 30 40 50 60 70 80 90 96
Cumulative % less than or equal to given values of permeability
10
100
1000
Cor
e pe
rmea
bili
ty, m
d
Lognormal probability graph paper
1
PETE 306 HANDOUT 3/5/92
Calculation of Permeability using Capillary Pressure Data
Purcell Approach(ABW: pages 167-172)
Three basic considerations:
1. Capillary pressure in a capillary,
Pc = 2 σ cosθr
2. Capillary flow: Poiseuille's law
qi = π ri
4 ∆p
8 µ L
3. Darcy's equation,
qt = kA ∆p µ L
Pc = dynes cm2 k = cm2
σ = dynes cm A = cm2
r = cm L = cm
q = cm2
secµ = poise = dynes-sec
cm2
∆p = dynecm2
2
Let V i = πr i2L,
then the flow rate in a capillary is
qi = Vr i
2∆p
8µL2, V = cm3
Sincer i = 2σ cosθ
Pci
qi =
σ cosθ 2
Pci 2 Vi
2 µ L2 ∆p
For a bundle of n capillary tubes,
qt = σ cosθ 2 ∆p
2 µ L2Σ
i = 1
i = n Vi
Pci2
Σi=1
i=n
Since,
qt = k A ∆pµL
k = σ cosθ 2
2 A L Vi
Pci2
Σi=n
i=n
3
Define the fractional volume of ith capillary
s i = V i
VT , s i = fraction
and
φ = VTA L
, φ = fraction
k = σ cosθ 2
2 φ Si
Pci2Σ
i=1
i=n
Introducing a lithology factor λ for deviation of the actual porespace,
k = σ cosθ 2
2 φλ Si
Pci2Σ
i=1
i=n
In integral form,
k = σ cosθ 2
2 φλ dS
Pc2
S=0
S=1
1
PETE 306 HANDOUT 4/16/92
Calculation of Relative Permeabilities usingCapillary Pressure Data
Purcell and Burdine Approach(ABW: pages 196-199)
Purcell approach:
The absolute permeability may be expressed as
k = σ cosθ 2
2 φλ dS
Pc2
S=0
S=1
The effective permeability of the wetting phase may be expressed as
kwt = σ cosθ 2
2 φλ dS
Pc2
S=0
S=Swt
The relative permeability of the wetting phase is the ratio of the wettingphase effective permeability to the absolute permeability
krwt = kwt
k =
dSPc
2S=0
S=Swt
dSPc
2S=0
S=1
2
Similarly, the effective permeability of the nonwetting phase may beexpressed as
knwt = σ cosθ 2
2 φλ dS
Pc2
S=Swt
S=1
The relative permeability of the nonwetting phase is the ratio of thenonwetting phase effective permeability to the absolute permeability
krnwt = knwt
k =
dSPc
2S=Swt
S=1
dSPc
2S=0
S=1
3
Burdine Approach:
Burdine considered the tortuosity factors for one-phase and multiphasesystems and modified the Purcell equations for the effectivepermeabilities.
λrwti = λiλwti
The relative permeability of the wetting phase is the ratio of the wettingphase effective permeability to the absolute permeability
krwt = kwt
k = λrwt
2
dSPc
2S=0
S=Swt
dSPc
2S=0
S=1
The tortuosity ratio is related to the minimum wetting-phase saturationSm, as
λrwt = Swt - Sm1 - Sm
4
Similarly, the relative permeability of the nonwetting phase is the ratioof the nonwetting phase effective permeability to the absolutepermeability
krnwt = knwt
k = λrnwt
2
dSPc
2S=Swt
S=1
dSPc
2S=0
S=1
The tortuosity ratio for the nonwetting phase is related to the minimumwetting-phase saturation, Sm, and the equilibrium saturation to thenonwetting phase, Se, as
λrnwt = Snwt - Se1 - Sm - Se