rock property -texas
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Course Notesfor
Petroleum Engineering 311
Reservoir Petrophysics
Authors:1980 Von Gonten, W.D.
1986 McCain, W.D., Jr.1990 Wu, C.H.
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" In The Name Of God "
N
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PETROLEUM ENGINEERING 311
RESERVOIR PETROPHYSICS
CLASS NOTES (1992)
Instructor/ Author:
Ching H. Wu
DEPARTMENT OF PETROLEUM ENGINEERING
TEXAS A&M UNIVERSITYCOLLEGE STATION, TEXAS
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ii
TABLE OF CONTENTS
I. ROCK POROSITY I-1
I) Definition I-1
II) Classification I-1III) Range of values of porosity I-2VI) Factors affecting porosity I-3V) Measurement of porosity I-5
VI) Subsurface measurement of porosity I-13VII) Compressibility of porous rocks I-25
II. SINGLE PHASE FLOW IN POROUS ROCK II-1
I) Darcy's equation II-11II) Reservoir systems II-15
III. BOUNDARY TENSION AND CAPILLARY PRESSURE III-11
I) Boundary tension III-1II) Wettability III-3
III) Capillary pressure III-5IV) Relationship between capillary pressure and saturation III-13V) Relationship between capillary pressure and saturation history III-14
VI) Capillary pressure in reservoir rock III-17VII) Laboratory measurement of capillary pressure III-19
VIII) Converting laboratory data to reservoir conditions III-25IX) Determining water saturation in reservoir from capillary pressure data III-27X) Capillary pressure variation III-29
XI) Averaging capillary pressure data III-31
IV. FLUID SATURATIONS IV-1
I) Basic concepts of hydrocarbon accumulation IV-1II) Methods for determining fluid saturations IV-1
V. ELECTRICAL PROPERTIES OF ROCK-FLUID SYSTEMS V-1
I) Electrical conductivity of fluid saturated rock V-1II) Use of electrical Formation Resistivity Factor, Cementation Factor, and
Saturation Exponent V-8III) Laboratory measurement of electrical properties of rock V-9IV) Effect of clay on resistivity V-18
VI. MULTIPHASE FLOW IN POROUS ROCK VI-1
I) Effective permeability VI-1II) Relative permeability VI-2
III) Typical relative permeability curves VI-2IV) Permeability ratio (relative permeability ratio) VI-14V) Measurement of relative permeability VI-14
VI) Uses of relative permeability data VI-33
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iii
VII. STATISTICAL MEASURES VII-1
I) Introduction VII-1II) Frequency Distributions VII-2
III) Histogram VII-3IV) Cumulative Frequency Distributions VII-6
V) Normal Distribution VII-8VI) 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
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I - 1
I. ROCK POROSITY
I) Definition
A measure of the pore space available for the storage of fluids in
rock
In general form:
Porosity = =Vp
Vb=
Vb- VmVb
where:
is expressed in fraction
Vb= Vp+ Vm
Vb= bulk volume of reservoir rock, (L3)
Vp= pore volume, (L3)
Vm= matrix volume, (L3)
II) Classification
A. Primary (original) Porosity
Developed at time of deposition
B. Secondary Porosity
Developed as a result of geologic processoccurring after deposition
C. Total Porosity
t=total pore space
Vb=
Vb- VmVb
D. Effective Porosity
e=interconnected pore space
Vb
1. Clean sandstones: e = t
2. Carbonate, cemented sandstones: e < t
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VI) Factors affecting porosity
A. Factors:1. Particle shape2. Particle arrangement
3. Particle size distribution4. Cementation5. Vugs and fractures
B. Particle shape
Porosity increases as particle uniformity decreases.
C. Packing Arrangement
Porosity decreases as compaction increases
6000500040003000200010000
0
10
20
30
40
50
DEPTH OF BURIAL, ft
POROSITY,%
EFFECT OF NATURAL COMPACTION ON POROSITY
(FROM KRUMBEIN AND SLOSS.)
SANDSTONES
SHALES
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D. Particle Size Distribution
Porosity decreases as the range of particle size increases
GRAIN SIZE DIAMETER, MM
INTERSTITIAL MATERIALS
AND MUD FRAGMENTS
FRAMEWORK
FRACTION
CLEAN SAND
SHALY SAND
SAND SILT CLAY100
0
1.0 0.1 0.01 0.001
WEIGHT%
E. Interstitial and Cementing Material
1. Porosity decreases as the amount of interstitial material increases
2. Porosity decreases as the amount of cementing material increases
3. Clean sand - little interstitial materialShaly sand - has more interstitial material
F. Vugs, Fractures
1. Contribute substantially to the volume of pore spaces
2. Highly variable in size and distribution
3. There could be two or more systems of pore openings - extremely complex
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V) Measurement of porosity
=Vb- Vm
Vb =
Vp
Vb
Table of matrix densities
Lithology m(g/cm3)
___________ ___________
Quartz 2.65
Limestone 2.71
Dolomite 2.87
A. Laboratory measurement
1. Conventional core analysis
a. measure any two
1) bulk volume, Vb2) matrix volume, Vm3) pore volume, Vp
b. bulk volume
1) calculate from dimensions2) displacement method
a) volumetric (measure volume)
(1) drop into liquid and observe volume chargeof liquid
(2) must prevent test liquid from entering poresspace of sample
(a) coat with paraffin(b) presaturate sample with test liquid
(c) use mercury as test liquidb) gravimetric (measure mass)
(1) Change in weight of immersed sample-prevent test liquid from entering pore space
(2) Change in weight of container and test fluidwhen sample is introduced
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c. matrix volume
1) assume grain density
Vm=dry weight
matrix density
2) displacement method
Reduce sample to particle size, then
a) volumetric
b) gravimetric
3) Boyle's Law: P1V1= P2V2
a)
P(1)
V(1)
VALVE CLOSED
b) Put core in second chamber, evacuate
c) Open valve
P(2)
VALVE OPEN
CORE
V2 = Volumetric of first chamber &volume of second chamber-matrixvolume or core ( calculated)
VT = Volume of first chamber +
volume second chamber (known)
4) Vm =VT- V2
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d. pore volume
1) gravimetric
Vp=saturated weight - dry weight
density of saturated fluid
2) Boyle's Law: P1V1= P2V2
a)
P(1)
V(1)
VALVE CLOSED
CORE
b) Put core in Hassler sleeve, evacuate
c) Open valve
P(2)
V(1)
VALVE OPEN
CORE
V2 = Volume of first chamber + pore
volume of core (calculated)
3) Vp
= V2- V
1
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2. Application to reservoir rocks
a. intergranular porosity(sandstone, some carbonates)
1) use representative plugs from whole core in
laboratory measurements
2) don't use sidewall cores
b. secondary porosity(most carbonates)
1) use whole core in laboratory measurements
2) calculate bulk volume from measurements
3) determine matrix or pore volume fromBoyle's Law procedure
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Example I-1
A core sample coated with paraffin was immersed in a Russell tube. The dry sample weighed 20.0gm. The dry sample coated with paraffin weighed 20.9 gm. The paraffin coated sample displaced10.9 cc of liquid. Assume the density of solid paraffin is 0.9 gm/cc. What is the bulk volume ofthe sample?
Solution:
Weight of paraffin coating = 20.9 gm - 20.0 gm = 0.9 gm
Volume of paraffin coating = 0.9 gm / (0.9 gm/cc) = 1.0 cc
Bulk volume of sample = 10.9 cc - 1.0 cc = 9.9 cc
Example I-2
The core sample of problem I-1 was stripped of the paraffin coat, crushed to grain size, andimmersed in a Russell tube. The volume of the grains was 7.7 cc. What was the porosity of thesample? Is this effective or total porosity.
Solution:
Bulk Volume = 9.9 cc
Matrix Volume = 7.7 cc
=Vb- Vm
Vb= 9.9 cc- 7.7 cc
9.9 cc= 0.22
It is total porosity.
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Example I-3
Calculate the porosity of a core sample when the following information is available:
Dry weight of sample = 427.3 gm
Weight of sample when saturated with water = 448.6 gm
Density of water = 1.0 gm/cm3
Weight of water saturated sample immersed inwater = 269.6 gm
Solution:
Vp
= sat. core wt. in air - dry core wt.
density of water
Vp = 448.6 gm - 427.3 gm
1 gm/cm3
Vp = 21.3 cm3
Vb = sat. core wt. in air - sat. core wt. in water
density of water
Vb = 448.6 gm - 269.6 gm 1 gm/cm3
Vb = 179.0 cm3
=Vp
Vb = 21.3 cm3
179.0 cm3= .119
= 11.9%
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What is the lithology of the sample?
Vm = Vb- Vp
Vm
= 179.0 cm3- 21.3 cm3= 157.7 cm3
m = wt. of dry sample = 427.3 gm = 2.71 gm/(cm3)
matrix vol. 157.7 cm3
The lithology is limestone.
Is the porosity effective or total? Why?
Effective, because fluid was forced into the pore space.
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VI) Subsurface measurement of porosity
A. Types of logs from which porosity can be derived
1. Density log:
d = m- Lm- f
2. Sonic log:
s =tL- tm
tf- tm
3. Neutron log:
e- k=CNf
Table of Matrix Properties(Schlumberger, Log Interpretation Principles, Volume I)
Lithologytm sec/ft m gm/cc
Sandstone 55.6 2.65
Limestone 47.5 2.71
Dolomite 43.5 2.87
Anhydrite 50.0 2.96
Salt 67.0 2.17
Water 189.0 1.00
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B. Density Log
1. Measures bulk density of formation
FORMATION
GAMMA RAY
SOURCE
SHORT SPACE
DETECTOR
LONG SPACE
DETECTOR
MUD CAKE
2. Gamma rays are stopped by electrons - the denser the rock the fewer gammarays reach the detector
3. Equation
L = m 1- +f
d = m- Lm- f
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FORMATION DENSITY LOG
4240
4220
4200
4180
4160
4140
4120
4100
20016012080400 3.02.82.62.42.22.0
, gm/ccGR, API depth, ft
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Example I-5
Use the density log to calculate the porosity for the following intervals assuming matrix = 2.68
gm/cc and fluid = 1.0 gm/cc.
Interval, ft L, gm/cc
d,%
__________ _________ ______
4143-4157 2.375 184170-4178 2.350 204178-4185 2.430 154185-4190 2.400 174197-4205 2.680 04210-4217 2.450 14
Example:
Interval 4,143 ft -4,157 ft :
L
= 2.375 gm/cc
d =m- Lm- f
=2.68 gm/cc - 2.375 gm/cc
2.68 gm/cc - 1.0 gm/cc= 0.18
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C. Sonic Log
1. Measures time required for compressional sound waves to travel throughone foot of formation
D
B
AT
R1
R2
E
C
2. Sound travels more slowly in fluids than in solids. Pore space is filled withfluids. Travel time increases as porosity increases.
3. Equation
tL= tm 1 - +tf (Wylie Time Average Equation)
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SONIC LOG
4240
4220
4200
4180
4160
4140
4120
4100
2001000 140 120 100 80 60 40
GR, API T, seconds/ftdepth, ft
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Example I-6
Use the Sonic log and assume sandstone lithology to calculate the porosity for the followingintervals.
Interval tLs ,%
(ft) second/ft
4,144-4,150 86.5 25
4,150-4,157 84.0 24
4,171-4,177 84.5 24
4,177-4,187 81.0 21
4,199-4,204 53.5 1
4,208-4,213 75.0 17
Example:
Interval 4144 ft - 4150 ft :
tL = 86.5 -sec/ft
s =tL- tm
tf-tm=
86.5 sec/ft- 51.6 sec/ft
189.0 sec/ft- 51.6 sec/ft = 0.25
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D. Neutron Log
1. Measures the amount of hydrogen in the formation (hydrogen index)
MaximumAverage EnergyNumber Loss/ Atomic Atomic
Element Collisions Collision, % Collision Number
Calcium 371 8 40.1 20Chlorine 316 10 35.5 17Silicone 261 12 28.1 14Oxygen 150 21 16.0 8Carbon 115 28 12.0 6Hydrogen 18 100 1.0 1
.1 1 10 102 103
O
104
Si
105
H
106 107
1
10
102
103
CLEAN SAND POROSITY = 15%
NEUTRON ENERGY IN ELECTRON VOLTS
RELATIVEPROBABILITY
FORCOLLISION
.1 1 10 102 103 104 105
H
106
O
107
Si
10-3
10-2
10-1
1
CLEAN SAND POROSITY = 15%
SLOWINGDOWNPOWER
NEUTRON ENERGY IN ELECTRON VOLTS
2. In clean, liquid filled formations, hydrogen index is directly proportional toporosity. Neutron log gives porosity directly.
3. If the log is not calibrated, it is not very reliable for determining porosity.Run density log to evaluate porosity, lithology, and gas content.
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NEUTRON DENSITY LOG
4240
4220
4200
4180
4160
4140
4120
4100
2000 30 -10
GR, API (CDL)depth, ft
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Example I-7
Use the neutron log to determine porosity for the following intervals.
Solution:
Interval n (ft) (%) .
4,143-4,149 23
4,149-4,160 20
4,170-4,184 21
4,198-4,204 9
4,208-4,214 19
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Example I-8
Calculate the porosity and lithology of the Polar No. 1 drilled in Lake Maracaibo. The depth ofinterest is 13,743 feet. A density log and a sonic log were run in the well in addition to thestandard Induction Electric Survey (IES) survey.
The readings at 13,743 feet are:
bulk density = 2.522 gm/cc
travel time = 62.73 -sec/ft
Solution:
Assume fresh water in pores.
Assume sandstone:
m= 2.65 gm/cc
tm = 55.5 -sec/ft
d =m- Lm-f
=2.65 gm/cc - 2.522 gm/cc
2.65 gm/cc - 1.0 gm/cc= 7.76%
s =tL- tm
tf-tm
=62.73 sec/ft- 55.5 sec/ft
189.0 sec/ft - 55.5 sec/ft
= 5.42%
Assume limestone:
m = 2.71 gm/cc
tm = 47.5 -sec/ft
d =m- L
m-f
=2.71 gm/cc - 2.522 gm/cc
2.71 gm/cc - 1.0 gm/cc
= 10.99%
s =tL- tm
tf-tm
=62.73 sec/ft - 47.5 sec/ft
189.0 sec/ft - 47.5 sec/ft = 10.76%
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Assume dolomite:
m = 2.87 gm/cc
tm = 43.5 -sec/ft
d =m- Lm-f
=2.87 gm/cc - 2.522 gm/cc
2.87 gm/cc - 1.0 gm/cc= 18.619%
s =tL- tm
tf- tm
=62.73 sec/ft - 43.5 sec/ft
189.0 sec/ft - 43.5 sec/ft = 13.22%
limestone = 11%
Since both logs "read" nearly the same porosity when a limestone lithology wasassumed then the hypothesis that the lithology is limestone is accepted.
Are the tools measuring total or effective porosity? Why?
The density log measures total compressibility because is "sees" the entire rockvolume,including all pores. The sonic log tends to measure the velocity ofcompressional waves that travel through interconnected pore structures as well as therock matrix. The general consensus is that the sonic log measures effective porositywhen we use the Wyllie "time-average" equation.
It is expected that the effective porosity is always less than ,or equal to,the totalporosity.
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VII) Compressibility of porous rocks
Compressibility, c is the fractional change in volume per unit change in pressure:
c = -1VV
P T -
VV T
P
A. Normally pressured reservoirs
1. Downward force by the overburden must be balanced by upward force ofthe matrix and the fluid
Fm Ff
Fo
2. Thus,
Fo = Fm+ Ff
it follows that
Po = pm+ pf
3. Po 1.0 psi/ft
Pf 0.465 psi/ft
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4. As fluid is produced from a reservoir, the fluid pressure,Pfwill usually decrease:
a. the force on the matrix increasesb. causing a decrease in bulk volumec. and a decrease in pore volume
B. Types of compressibility
1. Matrix Compressibility, cm
cm 0
2. Bulk Compressibility cb
used in subsidence studies
3. Formation Compressibility, cf- also called pore volume compressibility
a. important to reservoir engineers
1) depletion of fluid from pore spaces2) internal rock stress changes3) change in stress results in change in
Vp, Vm, Vb
4) by definition
cf= - 1Vp
Vp
pm
b. since overburden pressure, Po, is constant
dPm = - dPf
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1) Thus,
cf= -1
Vp
Vp
pm
2) where the subscript of f on cfmeans
"formation" and the subscript of f on Pfmeans "fluid"
3) procedure
(a) measure volume of liquid expelled as afunction of "external" pressure
(b) "external" pressure may be taken torepresent overburden pressure, Po
(c) fluid pressure, pf, is essentially constant, thus,
dPo = dPm
(d) expelled volume increases as porevolume, vp, decreases, thus,
dVp = - dVexpelled
(e) from definition
cf= -1
Vp
Vp
pm
it follows that
cf= +1
Vp
Vp expelled
Po
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(f) plot
CUMU
LATIVEVOLUMEEXPELLED
POREVOLUME
OVERBURDEN PRESSURE, psi
slope = cf.
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C. Measurement of compressibility
1) Laboratory core sample
a) apply variable internal and external pressures
b) internal rock volume changes
2) Equipment
Internal
Pressure
Gauge
HydraulicPump
Overburden
Pressure
Gauge
Hydraulic
Pump
Copper - J acketed
Core
Mercury Sight Gauge
Apparatus for measuring pore volume compressibility (hydrostatic)
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Example I-9
Given the following lab data, calculate the pore volume compressibility for a sandstone sample at4,000 and 6,000 psi.
pore volume = 50.0 cc
pressure, psi vol. fluid expelled, cc
1000 0.244 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 1
psi
@ 6000 psi:
Slope = 0.0116000 psi
cf = 1.83 X 10-6 1
psi
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1000080006000400020000
0.000
0.005
0.010
0.015
COMPACTION PRESSURE, psi
VOLUMEEXPELLED,
cc
POREVOLUME,cc
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INITIAL POROSITY AT ZERO NET PRESSURE, %
302520151050
1
10
100
PORE-VOLUME COMPRESSIBILITY AT 75 %
LITHOSTATIC PRESSURE VS INITIAL SAMPLE
POROSITY FOR CONSOLIDATED SANDSTONES.
CONSOLIDATED SANDSTONES
HALL'S CORRELATION
POREVOLUMECOMPRESSIBILITYX10-6psi-1
PORE
VOLUMECOMPRESSIBILITYX10-6psi-1
302520151050
1
10
100
PORE-VOLUME COMPRESSIBILITY AT 75 %
LITHOSTATIC PRESSURE VS INITIAL SAMPLE
POROSITY FOR UNCONSOLIDATED SANDSTONES.
UNCONSOLIDATED SANDSTONES
HALL'S CORRELATION
INITIAL POROSITY AT ZERO NET PRESSURE, %
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E. Abnormally pressured reservoirs
"abnormal pressure": fluid pressures greater than or less than the hydrostatic fluidpressure expected from an assumed linear pressure gradient
PRESSURE
DEPTH
NORMAL LINEAR
SUBNORMAL
(LOWER)
SURNORMAL
(GREATER)
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Compressibility/Porosity Problem No. 1
A limestone sample weighs 241.0 gm. The limestone sample coated with paraffin was found toweigh 249.5 gm. The coated sample when immersed in a partially filled graduated cylinderdisplaced 125.0 cc of water. The density of the paraffin is 0.90 gm/cc.
What is the porosity of the rock? Does the process measure total or effective porosity?
Solution:
Vm=wt. dry
ls=
241.0 gm
2.71 gm/cc= 88.9 cc
Vparaffin=wt. coated sample - st. uncoated sample
Vparaffin=249.5 gm - 241.0 gm
0.90 gm/cc= 9.4 cc
Vb = 125 cc - 9.4 cc = 115.6 cc
Vp = Vb- Vm
Vp = 115.6 cc - 88.9 cc - 26.7 cc
=Vp
Vb= 26.7 cc
115.6 cc= 0.231
= 23.1% (total porosity)
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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 0.750coarse sand 0.095 9.5 0.905
fine sand 0.039 3.9 0.961
Final porosity - 3.9%
(Complete mixing of the grains)
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B. Sample Pore Volume
L = (1.575 in) (2.54 cm/in) = 4.0 cm
D = (1.960 in) (2.54 cm/in) = 5.0 cm
Vb = bulk volume =D2h
4=
3.14 5.0 2 4.0
4.0= 78.5 cc
Vm = matrix volume = 140 gmcc
2.65 gm= 52.8 cc
Vp = Vb- Vm= 78.5 cc - 52.8 cc
Vp = 25.7 cc
C. Compressibility (see graph)
Vp = 25.7 cc
D. Subsidence
H = H cp P
H = 85 ft 9.69x 10-7psi-1 0.152 2,000 psi
H = 0.026 ft
H = 0.32 inches
Note: the pore volume (formation) compressibility is somewhat smaller than usuallyencountered. An experienced engineer would be wary of this small number. Also it wasassumed that the formation compressibility was exactly the same as the bulk volumecompressibility. Experience shows that this is not the case.
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800060004000200000
0.0040
0.0060
0.0080
0.0100
0.0120
0.0140
POROSITY PROBLEM No. 3
PRESSURE, psig
SLOPE =.0118 - .0068
7000 - 2000
VOL
UMEEXPELLED,cc
P
OREVOLUME,cc Cp= 9.96 x 10
-7psi -1
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Compressibility Problem
A 160-acre and 100 ft thick reservoir has a porosity of 11%. The pore compressibility is 5.0 x 10-
6(1/psi). If the pressure decreases 3,000 psi, what is the subsidence (ft)? Assume Cf = Cb
Solution:
A = 160 (43,560) = 6,969,600 ft2
Vb = 100 (6,969,600) = 696,960,000 ft3
Vp = Vb(f) = (696,960,000) (.11) = 76,665,600 ft3
Cp = -1
Vp dVp
dp
5 x 10-6(1/psi) = -1
76,665,600 ft3
dVp
3,000 psi
dVp= 1.15 x 106ft3
H= 1.15 x 106 ft3 x 1
6,969,600 ft2= 0.165 ft
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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)
L
Length of sand pack,L = Z
1. constant of proportionality, k, characteristics of particular sand pack, notsample size
2. Darcy's work confined to sand packs that were 100% saturated with water
3. equation extended to include other liquids using viscosity
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II - 2
q =kA(h1- h2)
L
B. Generalized form of Darcy's equation
1. Equation
vs=-k
dPds
-g
1.0133 x 106dzds
-1
+1 90o 180o 270o 360o
s
Vs
+X
+Y
-Z
+Z
2. Nomenclature
vs = superficial velocity (volume flux along
path s) - cm/sec
vs/ = interstitial velocity - cm/sec
= density of flowing fluid - gm/cm3
g = acceleration of gravity - 980 cm/sec2
dP = pressure gradient along s - atm/cmds
= viscosity - centipoise
k = permeability - darcies
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II - 3
3. Conversion factors
dyne = gm cm/sec2= a unit of force
atm = 1.0133 x 106dyne/cm2
gh = dyne/cm2= a unit of pressure
poise = gm/cm sec = dyne sec/cm2
4. The dimensions of permeability
L = length
m = mass
t = time
vs = L/t
= m/Lt
= m/L3
p = m/Lt2
g = L/t2
vs = -k
dp
ds-
g
1.0133 x 106 dzds
Lt = - k
m/Lt m/Lt2
L-
m/L3 L/t2 L
L
k = L2 = cross-sectional area
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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- P2L
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II - 5
II) Reservoir systems
A. Flow of incompressible liquid
1. Horizontal, linear flow system
L
Aq
P1
q
P2
a. Conditions
1) horizontal system,dzds
= 0
2) linear system, A = constant
3) incompressible liquid, q = constant
4) laminar flow, can use Darcy's equation
5) non-reactive fluid, k = constant
6) 100% saturated with one fluid
7) constant temperature, , q
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II - 6
b. derivation of flow equation
vs = -k
dPds
-g
1.0133 x 106 dzds
vs = - k
dP
ds =
q
A
q ds
0
L
= - kA
dPp1
p2
q L - 0 = - kA
P2- P1
q = kAL
P2- P1
Note: P1 acts at L = 0
P2 acts at L = L
q is + if flow is from L = 0 to L = L
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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 = = 1.0 cp
length = L = 6 ft
inlet pressure = P1= 5.0 atm
outlet pressure = P2= 2.0 atm
Solution:
We must insure all the variables are in the correct units.
k = 1 darcy
A = 6 ft2(144 in2/1 ft2) (6.45 cm2/1 in2) = 5572.8 cm2
L = 6 ft (12 in/1 ft) (2.54 cm/1 in) = 182.88 cm
P1 = 5.0 atm
P2 = 2.0 atm
q = kAL
P2- P1
q = (1) (5,572.8 ) (5.0 - 2.0) (1) (182.88)
q = 91.42 cm3/sec
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II - 8
2. Non-horizontal, linear system
-Z
P1
S
X
P2
a. Conditions
1) non-horizontal system,dzds
= sin= 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 , q
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II - 9
b. derivation of equation
vs = -k
dPds
-g
1.0133 x 106dzds
vs = - qA = - k dPds +k g sin
1.0133 x 106
q ds
0
L
= - kA
dp
P1
P2 +
kA g sin
1.0133 x 106 ds
0
L
q = - kAL
P1- P2+gLsin
1.0133 x 106
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II - 10
3. Vertical, upward flow, linear system
L
x
h
FLOW UNDER
HEAD h
a. Conditions
1) vertical system, dzds= sin= constant
2) upward flow, q = 270, sin= - 1
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,
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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 = 2rh , 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, , q
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II - 13
b. Derivation of flow Equation
vs = -k
dPds
-g
1.0133 x 106 dz
ds
vs = +
k
dP
dr =
q
A =
q
2rh
q
2h dr
rrw
re = k
dp
pw
pe
q
2h 1n(re) - 1n( rw) =
k
Pe - Pw
q = 2hk1n (re/rw)
Pe - Pw
Note: if q is + , flow is from reto 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
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II - 14
b. Assumptions
1) , Z = constant
2) Z(and ) can be determined at mean pressure
c. Derivation of equation for qsc
vs = -k
dPds
-g
1.0133 x 106 dzds
vs = -k
dPds
=q
Ads
but
q =
Psc qscz T
PTsc
thus
Psc T qscTsc A
ds
o
L
= - k PdPz
p1
p2
Psc T qscTsc A
L -0 = - kz
P2
2 - P12
2
qsc =kAL
Tsc
Tz Psc
P1 - P22
Note: real gas equation of state
Pq = Z n R T
where q = volumetric flow/timen = mass flow/time
thus,Pq
Pscqsc =Z n R Tn R Tsc
q =Pscqscz T
Tsc 1P
where qscis constant
Z is determined at P, T
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II - 15
d. Derivation of equation for q
qsc =kAL
Tsc
Tz Psc
P1 - P22
but
qsc=P q Tsc
Z Psc T = k A
L
TscT z Psc
P1
2 - P22
2
q = k AL
1
P
P12 - P2
2
2
q = k AL
2P1+ P2
P1
2 - P22
2
q = k AL
P1 - P2
This equation is identical to the equation for horizontal, linear flow of incompressible liquid
thus
if gas flow rate is determined at mean pressure, P, the equation for incompressible liquidcan be used for compressible gas!
Note: real gas equation of state
Pq = Z n R T
thus
Psc qsc
P q =
n R Tsc
z n R T
where
P =P1 + P2
2
P = volumetric flow rate at P, T
z is determined at P, T
qsc =P q Tscz Psc T
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II - 16
2. Horizontal, radial flow system
h
rw
re
Pe
Pw
re rw
a. Conditions
1) horizontal systemdzds
= 0
2) radial system, A = 2rL, ds = - dr,inward flow
3) constant thickness, h = constant
4) compressible gas flow, q = f (P)
5) laminar flow, use Darcy equation
6) non-reactive fluid, k = constant
7) 100% saturated with one fluid
8) constant temperature
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II - 17
b. Assumptions
z = constant
z (and ) can be determined at mean pressure
c. derivation of equation for qsc
vs= -k
dPds
-g
1.0133 x 106dzds
vs= -k
dPds
=q
A
but
q =Pscqscz T
PTsc
and
A = 2rh and ds = - dr
thus
PscT qsc2Tsch
drr
rw
re = k
dP
zPw
Pe
PscT qsc2 Tsch
1nrerw
= kz
Pe
2 - Pw2
2
qsc =2 h k
1n re/rw
TscPsczT
Pe
2 - Pw2
2
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II - 18
d. derivation of equation for q
qsc =2 h k
1n re/rw
TscPsczT
Pe
2 - Pw2
2
but
q =P q Tscz Psc T
thus
P q Tscz PscT
= 2 h k1n re/rw
Tsc
PsczT
Pe2 - Pw
2
2
q = 2 h k1n re/rw
1P
(Pe2 - Pw2 )2
q = 2 h k1n re/rw
2Pe + Pw
(Pe
2 - Pw2 )
2
q = 2 h k1n re/rw
Pe - Pw
Note: Equation for real gas is identical to equation for incompressible liquid when
volumetric flow rate of gas,q, is measured at mean pressure.
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II - 19
C. Conversion to Oilfield Units
Symbol Darcy units Oil field
q cc/sec bbl/d or cu ft/d
k darcy mdA sq cm sq fth cm ftP atm psiaL cm ft
cp cpr gm/cc lb/cu ft
Example:
q =hkA P1- P2
L in Darcy's units
q ccsec
= q bbld
5.615 cu ftbbl
1,728 cu in
cu ft 16.39 cc
cu in d
24hr hr
3,600 sec
q ccsec
= 1.841 q bbld
k darcy = k md
darcy
1,000md
k darcy = 0.001 k md
A sq cm =929.0 sq cm
sq ft A sq ft
A sq cm = 929.0 A sq ft
P1- P2 atm = P1- P2 psiaatm
14.696 psia
P1- P2 atm = 0.06805 P1 - P2 psia
L cm = L ft 30.48 cmft
meter = 100 cm
1.841 q =0.001 k 929.0 A .06805 P1- P2
30.48 L
q =0.01127 k A P1- P2
L in oilfield units
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II - 20
D. Table of Equations
1. Darcy Units
System Fluid Equation
Horizontal,Linear
IncompressibleLiquid
q = kAL
P1- P2
Dipping,Linear
IncompressibleLiquid q = kA
L P1- P2 +
g L sin
1.0133 x 106
Horizontal,
Radial
Incompressible
Liquidq = 2k h
ln (re/rw) P
e - P
w
Horizontal,Linear
RealGas qsc =
kAL
Tsc
Tz Psc
P12 - P2
2
2
q = kAL
P1 - P2
Horizontal,Radial
Real Gasqsc =
k h ln (re/rw)
TscTz Psc
Pe2 - Pw 2
q = 2k hln (re/rw)
Pe - Pw
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II - 21
2. Oilfield Units
System Fluid Equation
Horizontal,Linear
IncompressibleLiquid
q = 0.001127kAL
P1 - P2
q = res bbl/d
Dipping,Linear
IncompressibleLiquid
q = 0.001127 kAL
P1 - P2
+g L sin
1.0133 x 106
Horizontal,Radial
IncompressibleLiquid
q = .007082 kh ln (re/rw)
Pe - Pw
Horizontal,Linear Real Gas
qsc = .1118k A
L z T P1
2 - P22
qsc= scf/d
q = .001127 kAL
P1- P2
q = res bbl/d
Horizontal,Radial Real Gas
qsc = .7032k h
ln (re/rw) TzPe
2 - Pw 2
q = .007082 khln (re/rw)
Pe- Pw
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II - 22
Example II-2
What is the flow rate of a horizontal rectangular system when the conditions are as follows:
permeability = k = 1 darcy
area = A = 6 ft2
viscosity = = 1.0 cplength = L = 6 ftinlet pressure = P1= 5.0 atm.
outlet pressure = P2 = 2.0 atm.
Solutions:
We must insure that all the variables are in the correct units.
k = 1 darcy = 1,000 md
A = 6 ft2
L = 6 ftP1 = (5.0 atm) (14.7 psi/atm) = 73.5 psi
P2 = (2.0 atm) (14.7 psi/atm) = 29.4 psi
q = 1.1271 x 10-3 kAL
P1- P2
q = 1.1271 x 10-31,000 6
1 6 73.5 - 29.4
q = 49.7 bbl / day
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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
= 1.3 cp
Solution:
q =7.082 x 10-3kH Pe- Pw
ln Re/ Rw
q =7.082 x 1--3 300 20 2,500 - 1,740
1.3 6.492
q = 3,826 res bbl/d
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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 - P2
L
and
qt =kAwhAL
P1- P2 +kBwhBL
P1- P2 +kCwhCL
P1- P2
then
k h = kAhA+ kB hB + kC hC
k = j = 1
nkjhj
h
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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 =2k h
ln (re/rw)
Pe - Pw
and
qt =2kAhAln (re/rw)
Pe- Pw +2kB hBln (re/rw)
Pe - Pw
+2 kchcln (re/rw)
Pe- Pw
then
k h = kAhA + kBhB+ kChC
and again
k = j = 1
nkj hj
h
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II - 26
3. Horizontal, linear flow perpendicular to bedding
L
h
P2
P1
W
q
kA
PA
LA
A B C
kB
PB
LB
kC
PC
LC
q
qt = qA = qB = qC
p1 - p2 = PA + PB + PC
L = LA + LB + LC
qt =
k wh P1 - P2
L
and since P1 - P2 = PA+ PB + PC
P1 - P2 =qt L
k wh =
qA LA
kA wh +
qB LBkB wh
+qC LCkCwh
since qt = qA = qB = qC
Lk
= LAkA
+ LBkB
+ LCkC
thus
K = L
j = 1
n
Lj
kj
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II - 28
Example II-4
Damaged zone near wellbore
k1 = 10 md r
1= 2 ft
k2 = 200 md r2 = 300 ft
rw = 0.25 ft
Solution:
k =ln (re/rw)
j = 1
n
ln (rj/rj-1 )
kj
k =ln 300
0.25
ln 2/0.25
10 +
ln 300/2
200
k = 30.4 md
The permeability of the damaged zone near the wellbore influences the average permeability more
than the permeability of the undamaged formation.
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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 = r4
8L P1 - P2
A = r2, therefore
q = Ar2
8L P1 - P2
b. Darcy's law for linear flow of liquids
q = kAL
P1 - P2
assuming these flow equations have consistent units
Ar2
8L P1 - P2 =
kAL
P1- P2
thus
k = r2
8
= d2
32
where d = inches, k = 20 x 109d2md
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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 109d2
k = 20 x 109(0.0001 in)2
k = 200 md
B. k = 0.25 (200 md)
k = 50 md
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II - 31
2. Flow through fractures
b
v =q
A
= h2
12
L
(P1-P2)
q = b2A
12 L(P1-P2)
setting this flow equation equal to Darcy's flow equation,
b2A12L
P1 - P2 = kA
L P1 - P2
solve for permeability of a fracture:
k = b2
12 in darcy units, or
k = 54 x 109b2
where b = inchesk = md
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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 the
equation for parallel flow.
Solution:
k =
kjAj
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 109x 0.005 2 12 in x 0.005 in
144 in2
k =1.439 + 81,000
144
k = 563 md
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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
qscPscA
qsc =kA P1
2 - P22
2L Psc horizontal, linear, real gas flow with
T = Tscand Z = 1.0
qscPscA
= k
P1
2 - P22
2L
k = ( slope ) m
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B. Factors which affect permeability measurement
1. Fractures - rocks which contain fractures in situ frequently separate alongthe planes of natural weakness when cored. Thus laboratory measurementsgive "matrix" permeability which is lower than in situ permeability becausetypically only the unfractured parts of the sample are analyzed for
permeability.
2. Gas slippage
a. gas molecules "slip" along the grain surfaces
b. occurs when diameter of the capillary openings approaches the meanfree path of the gas molecules
c. Darcy's equation assumes laminar flow
d. gas flow path with slippage
e. called Klinkenberg effect
f. mean free path is function of size of molecule thus permeabilitymeasurements are a function of type of gas used in laboratorymeasurement.
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II - 36
0
kCALCULATED
1
P
H2
N2
CO2
g. mean free path is a function of pressure, thus Klinkenberg effect isgreater for measurements at low pressures - negligible at highpressures.
h. permeability is a function of size of capillary opening, thusKlinkenberg effect is greater for low permeability rocks.
i. effect of gas slippage can be eliminated by making measurements atseveral different mean pressures and extrapolating to high pressure(1/p => 0)
0
kMEASURED
1P
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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 k
measured
vs. 1/pressure
Intercept is equivalent to liquid permeability
From graph:
kliq= 2.38 md
0
1
2
3
4
5
G
ASPERMEABILITY,md
0.0 0.2 0.4 0.6 0.8 1.0
RECIPROCAL MEAN PRESSURE, atm-1
kgas= 2.38276 + 1.64632
Equivalent Liquid Permeability = 2.38 md
Pbar
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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
100001000100101
1
10
100
1000
WATERPERMEABILITY,md
AIR PERMEABILITY, md
Water concentration
20,000 - 25,000 ppm Cl ion.
RELATIONSHIP OF PERMEABILITIES MEASURED
WITH AIR TO THOSE MEASURED WITH WATER
b. Injection water may,if its salinity is less than that of the formation water,reduce the permeability due to clay swelling.
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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
Kameans permeability to air; K500means permeability to 500 grains per gallon chloride solution;
Kwmeans permeability to fresh water
4. Change in pore pressure
a. The removal of the core from the formation will likely result in a change inpore volume.This is likely to result in a change in permeability (+ or -).
b. The production of fluids,especially around the well,will result in a decreasein pore pressure and a reduction of in-situ permeability.
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III- 1
III. BOUNDARY TENSION AND CAPILLARY PRESSURE
I) Boundary tension,
A. at the boundary between two phases there is an imbalance of molecular forces
B. the result is to contract the boundary to a minimum size
GAS
LIQUID
SURFACE
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III- 2
C. the average molecule in the liquid is uniformly attracted in all directions
D. molecules at the surface attracted more strongly from below
E. creates concave or convex surface depending on force balance
F. creation of this surface requires work
1. work in ergs required to create 1 cm2of surface (ergs/cm2) is termed"boundary energy"
2. also can be thought of as force in dynes acting along length of 1 cmrequired to prevent destruction of surface (dynes/cm) - this is called"boundary tension"
3. Boundary Energy = Boundary Tension x Length
G. Surface Tension - Boundary tension between gas and liquid is called "surfacetension"
H. Interfacial Tension - Boundary tension between two immiscible liquids orbetween a fluid and a solid is called "interfacial tension"
gw = surface tension between gas and water
go = surface tension between gas and oil
wo = interfacial tension between water and oil
ws = interfacial tension between water and solid
os = interfacial tension between oil and solid
gs = interfacial tension between gas and solid
I. Forces creating boundary tension
1. Forces
a. Law of Universal Gravitation applied between molecules
b. physical attraction (repulsion) between molecules
2. Liquid-Gas Boundary
attraction between molecules is directly proportional to their masses andinversely proportional to the square of the distance between them
3. Solid-Liquid Boundary
physical attraction between molecules of liquid and solid surface
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III- 3
4. Liquid-Liquid Boundary
some of each
II) Wettability
A. forces at boundary of two liquids and a solid (or gas-liquid-solid)
ow
os ws
OIL
WATER
OIL
SOLID
ws = os + ow cos B. Adhesion Tension, AT
AT = ws - os = ow cos C. if the solid is "water-wet"
ws osAT = +
cos = +0 90if = 0 - strongly water-wet
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III- 4
D. if the solid is "oil-wet"
os wsAT = -
cos = -90 180if = 180 - strongly oil-wet
= 830 = 1580
= 350 = 300
(A)
ISOOCTANE ISOOCTANE + 5.7%
ISOQUINOLINE
ISOQUINOLINE NAPHTHENIC ACID
= 300 = 480 = 540 = 1060
(B)
Interfacial contact angles. (A) Silica surface; (B) calcite surface
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III- 5
III) Capillary pressure
A. capillary pressure between air and water
hAIR
WATER
1. liquid will rise in the tube until total force up equals total force down
a. total force up equals adhesion tension acting along thecircumference of the water-air-solid interface
= 2r AT
b. total force down equals the weight of the column of water
converted to force
= r2hgw
c. thus when column of water comes to equilibrium
2r AT = r2hgw
d. units
cmdynecm
= cm2cm cm
sec2
gm
cm3
dyne =gm cm
sec2
dyne = force unit
e. adhesion tension
AT=12
r hgwdynecm
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III- 6
2. liquid will rise in the tube until the vertical component of surface tensionequals the total force down
a. vertical component of surface tension is the surface tensionbetween air and water multiplied by the cosine of the contact angleacting along the water-air-solid interface
= 2r awcos
b. total force down
= r2hgw
c. thus when the column of water comes to equilibrium
2r awcos= r2hgw
d. units
cmdynecm
= cm2cm cm
sec2
gm
cm3
cmdynecm
= cmgm
sec2
3. since AT= awcos, 1 and 2 above both result in
h =2 awcos
rg w
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III- 7
4. capillary pressure (air-water system)
h
WATER
PaA' A
AIR
Pa
Pw
B'
B
pressure relations in capillary tubes
a. pressure at A' is equal to pressure at A
Pa' = Pa
b. pressure at B is equal to the pressure at A minus the head of waterbetween A & B
pw= pa- wgh
units:dyne
cm2 =
dyne
cm2-
gm cm
cm3 sec2 cm
c. thus between B' and B there is a pressure difference
pa- pw= pa- (pa- wgh)
pa- pw= wgh
d. call this pressure difference between B' and B "capillary pressure"Pc= pa- pw= wgh
e. remember
h =2 gwcos
rg w
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III- 8
f. thus
Pc=2 gwcos
r
B. capillary pressure between oil and water
h
WATER
OIL
1. liquid will rise in the tube until the vertical component of surface tensionequals the total force down
a. vertical component of surface tension equals the surface tensionbetween oil and water multiplied by the cosine of the contact angleacting along the circumference of the water-oil-solid interface
= 2r owcos
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
= wh
2) weight per unit area of oil
= oh
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III- 9
3) net weight per unit area acting to pull surface down
= wh - oh = h(w- o)
4) total force down
= r2gh (w-o)
d. thus when the column of water comes toequilibrium
2r owcos = r2gh (w-o)
2. thus the equilibrium for the height of the column of water
h =2 owcos
rg (w- o)
3. capillary pressure (oil-water system)
h
WATER
Po
A
Po
Pw
B'
B
OIL
a. pressure at A' equals pressure at A
Poa
= Pwa
b. pressure at B is equal to the pressure at A minus the head of waterbetween A and B
Pwb= Pwa- wgh
c. pressure at B' equal to the pressure at A' minus the head of oilbetween A' and B'
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III- 10
Pob= Poa- ogh
d. thus capillary pressure, the difference between pressure at B' andthe pressure at B is
Pc
= Pob
- Pwb
Pc = (Poa- ogh) - (Pwa- wgh)
since Poa = Pwa
Pc = (w- o)gh
e. remember
h = 2 owcos rg (w- o)
f. thus
Pc=2 owcos
r
4. same expression as for the air-solid system except for the boundarytension term
Pc= 2 cos
r
C. remember adhesion tension is defined as
AT = owcos,
and
Pc=2 owcos
r
thus
Pc= f (adhesion tension, 1/radius of tube)
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III- 11
ADHESION TENSION
AIR
WATER
AIR
HgWATER
AIR
1/ radius of tube
D. an important result to remember
1. pwb < pob
2. thus, the pressure on the concave side of a curved surface is greater thanthe pressure on the convex side
3. or, pressure is greater in the non-wetting phase
E. capillary pressure-unconsolidated sand
1. the straight capillary previously discussed is useful for explaining basicconcepts - but it is a simple and ideal system
2. packing of uniform spheres
Pc= 1
R1+ 1
R2R1 and R2 are the principal radii of curvature for a liquid adhering to two spheres in
contact with each other.
3. by analogy to capillary tube
1R1
+ 1R2
= 2 cos r
where Pc= 2 cos r
call it Rm(mean radius), i.e.
1Rm
= 2 cos rm
=()gh
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III- 12
F. wettability-consolidated sand
1. Pendular-ring distribution-wetting phase is not continuous, occupies thesmall interstices-non-wetting phase is in contact with some of the solid
2. Funicular distribution - wetting phase is continuous, completely covering
surface of solid
(A) (B)
OIL OR GAS OIL OR GAS
SAND GRAIN SAND GRAIN
WATER WATER
Idealized representation of distribution of wetting and nonwetting fluidphase about intergrain contacts of spheres. (a) Pendular-ring distributions;(b) funicular distribution
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III- 13
IV) Relationship between capillary pressure and saturation
A. remember that the height a liquid will rise in a tube depends on
1. adhesion2. fluid density
3. variation of tube diameter with height
B. consider an experiment in which liquid is allowed to rise in a tube of varyingdiameter under atmospheric pressure. Pressure in the gas phase is increasedforcing the interface to a new equilibrium position.
R
ATMOSPHERIC
PRESSURE
R
HIGHER
PRESSURE
DEPENDENCE OF INTERFACIAL CURVATURE ON FLUID SATURATIOIN A NON-UNIFORM PORE
1. Capillary pressure is defined as the pressure difference across theinterface.
2. This illustrates:
a. Capillary pressure is greater for small radius of curvature than forlarge radius of curvature
b. An inverse relationship between capillary pressure and wetting-phase saturation
c. Lower wetting-phase saturation results in smaller radius ofcurvature which means that the wetting phase will occupy smallerpores in reservoir rock
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III- 14
V) Relationship between capillary pressure and saturation history
A. consider an experiment using a non-uniform tube (pore in reservoir rock)
1. tube is filled with a wetting fluid and allowed to drain until the interfacebetween wetting fluid and non-wetting fluid reaches equilibrium
(drainage)
2. tube is filled with non-wetting fluid and immersed in wetting fluidallowing wetting fluid to imbibe until the interface reaches equilibrium(imbibition)
SATURATION = 100%
PC= LOW VALUE
SATURATION = 80%
CAPILLARY PRESSURE = P C
R
LOW PC HIGHER P C
(A)
SATURATION = 0%
PC= HIGH VALUE
SATURATION = 10%
CAPILLARY PRESSURE = P
R
HIGHER P C LOW PC
(B)
Dependence of equilibrium fluid saturation upon the saturation history in anonuniform pore. (a) Fluid drains; (b) fluid imbibes. Same pore, samecontact angle, same capillary pressure, different saturation history
3. This is an oversimplified example, however it illustrates that therelationship between wetting-phase saturation and capillary pressure isdependent on the saturation process (saturation history)
a. for given capillary pressure a higher value of wetting-phasesaturation will be obtained from drainage than from imbibition
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III- 15
B. Leverett conducted a similar experiment with tubes filled with sand.
100806040200
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
DATA FROM HEIGHT-SATURATION EXPERIMENTS
ON CLEAN SANDS. (FROM LEVERETT)
g
h
(k/)1/2
Drainage
Imbibition
Drainage Imbibition
Sand I
Sand II
WATER SATURATION, Sw %
1. capillary pressure is expressed in terms of a non-dimensional correlating
function ( remember Pc= (gh )
2. in general terms,
a. drainage means replacing a wetting fluid with a non-wetting fluid
b. imbibition means replacing a non-wetting fluid with a wetting fluid
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III- 16
PC
0 100WATER SATURATION, S W
IMBIBITION
DRAINAGE
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III- 17
VI) Capillary pressure in reservoir rock
Pw= Po/ w-
wh
144Po= Po/ w-
oh
144
Water Oil
Pw2 Po2
Po1= Pw1
100% Water
Oil and Water
Pc = Po - Pw =h
144 w - o
Where: Po = pressure in oil phase, psia
Pw = pressure in water phase, psia
h = distance above 100% water level, ftP
o/w= pressure at oil-water contact, psia
w = density of water, lb/cf
o = density of oil, lb/cf
At any point above the oil-water contact, popw
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III- 18
HEIGHT
ABOVE
O-W-C
PRESSURE
PCPO= PO/ W-
144
oH
Pw= PO/W-wH
144
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III- 19
VII) Laboratory measurement of capillary pressure
A. Methods
1. porous diaphragm
2. mercury injection
3. centrifuge
4. dynamic method
B. Porous diaphragm
1. Start with core saturated with wetting fluid.
2. Use pressure to force non-wetting fluid into core-displacing wetting fluidthrough the porous disk.
3. The pressure difference between the pressure in the non-wetting fluid andthe pressure in the wetting fluid is equal to Pc.
4. Repeat at successively higher pressures until no more wetting fluid willcome out.
5. Measure Swperiodically.
6. Results
7. Advantages
a. very accurateb. can use reservoir fluids
8. Disadvantages
a. very slow - up to 40 days for one coreb. pressure is limited by "displacement pressure" of porous disk
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III- 20
C. Mercury Injection Method
1. Force mercury into core - mercury is non-wetting phase - air (usuallyunder vacuum) is wetting phase
2. Measure pressure
3. Calculate mercury saturation
4. Advantages
a. fast-minutesb. reasonably accurate
5. Disadvantages
a. ruins coreb. difficult to relate data to oil-water systems
D. Centrifuge Method
CORE HOLDER BODY
WINDOW
TUBE BODY
1. Similar to porous disk method except centrifugal force (rather thanpressure) is applied to the fluids in the core
2. Pressure (force/unit area) is computed from centrifugal force (which is
related to rotational speed)
3. Saturation is computed from fluid removed (as shown in window)
4. Advantages
a. fastb. reasonably accuratec. use reservoir fluids
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III- 21
E. Dynamic Method
CORE
OIL INLET
OIL BURETTE
Po
GAS OUTLET GAS INLET
Pg Pc
TO ATMOSPHERE
DYNAMIC CAPILLARY - PRESSURE APPARATUS
(HASSLER'S PRINCIPLE)
1. establish simultaneous steady-state flow of two fluids through core
2. measure pressures of the two fluids in core (special wetted disks) -difference is capillary pressure
3. saturation varied by regulating quantity of each fluid entering core
4. advantages
a. seems to simulate reservoir conditionsb. reservoir fluids can be used
5. Disadvantagesa. very tedious
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III- 22
F. Comparison of methods
1. diaphragm method (restored state) is considered to be most accurate, thusused as standard against which all other methods are compared
2. comparison of mercury injection data against diaphragm data
a. simple theory shows that capillary pressure by mercury injectionshould be five times greater than capillary pressure of air-watersystem by diaphragm method
b. capillary pressure scale for curves determined by mercury injectionis five times greater than scale for diaphragm air-water data
c. these comparisons plus more complex theory indicate that the ratiobetween mercury injection data and diaphragm data is about 6.9(other data indicate value between 5.8 and 7.5)
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III- 23
Example VIII-1
Comparison of Mercury Injection Capillary Pressure Data with Porous Diaphragm Data
A. Calculate capillary pressure ratio,
PcAHg
PcAW
, for the following data
AHg= 480 Dynes/c
AW= 72 Dynes/c
AHg= 140 AW= 0
B. Pore geometry is very complex. The curvature of the interface and pore radius arenot necessarily functions of contact angles. Calculate the ratio using therelationship.
PcAHg
PcAW
=AHgAW
Solution:
(A) PcAHgPcAW
=AHgcos AHg
AWcos AW =480 cos(140)
72 cos (0)
PcAHg
PcAW
= 5.1
(B) PcAHgPcAW
@
AHgAW = 48070
PcAHg
PcAW
= 6.9
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III- 24
Discussion:
A. Best way to determine the relationship between mercury and air-water datais to generate capillary pressure curves for each set of data and comparedirectly.
Mercury Injection and Porous Diaphragm Methods
B. For this given set of conditions, mercury injection method requires ahigher displacement pressure, must adjust ratio between scales until matchis obtained.
C. Minimum irreducible wetting phase saturations are the same.
D. Reduction in permeability results in a higher minimum irreducible wettingphase saturation. For both cases, mercury system still has higher requireddisplacement pressure.
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III- 25
VIII) Converting laboratory data to reservoir conditions
PcL= 2Lcos L
r
PcR= 2Rcos R
r
setting r = r
r =2LcosLPcL
=2RcosRPcR
PcR= cos R
cos L
PcL
where
PcR= reservoir capillary pressure, psi
PcL= capillary pressure measured in laboratory, psi
L = interfacial tension measured in laboratory, dynes/cm
R = reservoir interfacial tension, dynes/cm
R = reservoir contact angle, degrees
L = laboratory contact angle, degrees
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III- 26
Example III-2
Converting Laboratory Data to Reservoir Conditions
Express reservoir capillary pressure by using laboratory data.
lab data: AW = 72 dynes
AW= 0o
reservoir data: OW= 24 dynes/cm
OW = 20o
Solution:
PcR
=cos R
cos L PcL
PcR
=24 cos20
72 cos0 PcL
PcR=
0.333PcL
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III- 27
IX) Determining water saturation in reservoir from capillary pressure data
A. convert laboratory capillary pressure data to reservoir conditions
B. calculate capillary pressure in reservoir for various heights above height at whichcapillary pressure is zero
Pc=()gh144 gcin English units
= w- O, lb/cu ft
g = 32 ft/sec2
gc = 32 lbm ft
lbf sec2h = ft
144 = (sq in)/(sq ft.)
thus
Pc = lbf/(sq in), psI
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III- 28
Example III-3
Determining Water Saturation From Capillary Pressure Curve
Given the relationship,
PcR= 0.313 P cL, use the laboratory capillary pressure curve to calculate the watersaturation in the reservoir at a height of 40 ft. above the oil-water contact.
o= 0.85 gm/cm3 w= 1.0 gm/cm3
PCL
SW
0
0
10
20
50 100
8.38.3
Solution:
PcR=
w o h
144
PcR=
1.0 - 0.85 62.4 lb
ft340
144= 2.6 psi
PcL=
PcR0.313
PcL= 2.6
0.313= 8.3 psi
move to the right horizontally fromPcL = 8.3 psi to the capillary pressure curve. Drop vertically
to the x-axis, read Sw.
Sw= 50%
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III -29
X) Capillary pressure variation
A. effect of permeability
1. displacement pressure increases as permeability decreases
2. minimum interstitial water saturation increases as permeability decreases
1009080706050403020100
0
20
40
60
80
100
120
140
160
180
200
RESERVOIR FLUID DISTRIBUTION CURVES
Sw %
(From Wright and Wooddy)
10md
100md
200md
900md
30
Heightabovezerocapillarypressure,
ft
24
18
12
6
0
Oil-WaterCapillaryPressure,ps
i
(reservoirconditions)
90
72
54
36
18
0
A
ir-WaterCapillaryPressure,psi
(laboratorydata)
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III -30
B. Effect of grain size distribution
100806040200
0
5
10
15
20
25
30 225.0
187.5
150.0
112.5
75.0
37.5
0
100 80 60 40 20 0
Sandstone Core
Porosity = 28.1%
Permeability = 1.43 md
Factor = 7.5
Mercurycapillarypressure,psi
Water/nitrogencapillarypressure,psi
Hg
Water
100806040200
0
10
20
30
40
50
60 348
290
232
174
116
58
0
100 80 60 40 20 0
Water/nitrogencapillary
pressure,psi
Mercurycapillarypr
essure,psi
Water
Hg
Limestone Core
Porosity = 23.0%
Permeability = 3.36 md
Factor = 5.8
1. majority of grains same size, so most pores are same size - curve (a) (wellsorted)
2. large range in grain and pore sizes - curve (b) (poorly sorted)
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III -31
XI) Averaging capillary pressure data
J-function
J Sw = Pccos
k
1/2
attempt to convert all capillary pressure data to a universal curve
universal curve impossible to generate due to wide range of differences existing inreservoirs
concept useful for given rock type from given reservoir
where
Pc = dyne/(sq cm)
= dyne/cm
k = (sq cm)
= fraction
or can use any units as long as you are consistent
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III -32
1009080706050403020100
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.71.8
1.9
2.0
CAPILLARY RETENTION CURVES.
WATER SATURATION, Sw
CAPILLARYPRESSUREFUNCTION,J
(From Rose and Bruce.)
LEVERETT
LEDUC
HAWKINS
KATIE ALUNDUM
EL ROBLE
KINSELLA
Reservoir Formation
Hawkins WoodbineEl Roble MorenoKinsella Viking
Katie DeeseLeduc Devonian
Alundum (consolidated)Leverett (unconsolidated)
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III -33
Capillary Pressure Problem 1
1. A glass tube is placed vertically in a beaker of water. The interfacial tension between theair and water is 72 dynes/cm and the contact angle is 0 degree.
Calculate:
a. the capillary rise of water in the tube if the radius of the tube is 0.01centimeters.
b. what is the difference in pressure in psi across the air-water interface in thetube.
2. The displacement pressure for a water saturated porcelain plate is 55 psi of air. What isthe diameter in inches of the largest pore in the porcelain plate? Assume 72 dynes/cmand 0 degrees.
Solution:
(1) A = 72 dynes/cm
W = 1 gm/cm3
g = 980 dynes/gm
= 0o
(a) capillary rise of water if radius is .01 cm
h =2AWcos rg
= 2 72 cos0.01 1.0 980
h = 14.69 cm
(b) pressure drop in psi across interface
Pc = pa- pw=wgh = 1.0 980 14.69
Pc= 0.0142 atm
14.696 psi
atm
Pc
= 0.209 psi
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III -34
(2)Pc
=2AWcos
r
Pc = 55 psi
Pc = 55 psi
atm14.696 psi
1.0133 x 106dynes/cm2
atm
= 3.792 x 106dynes/cm2
r = 2AWcos Pc
r = 2 72 cos0
3.792 x 106= 3.797 x 10-5cm in
2.54 cm
r = 1.495 x 10
-5
in
d = 2.99 x 10-5in
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III -35
Capillary Pressure Problem 2
Given the information below and graph ofPcL vs. wetting phase saturation Sw , construct the
curves forPcR, h in reservoir, and J-function vs. Sw. Water is the wetting phase in both the
laboratory and the reservoir.
fluidslab
air-waterres
oil-wate
0 25
60 dyne/cm 20 dyne/cm
wet 1.0 gm/cm3 1.1 gm/cm3
non-wet 0 gm/cm3 0.863 gm/cm3
k 37 md variable
16% variable
J =Pc k/ 1/2cos
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III -36
1009080706050403020100
0.0
2.5
5.0
7.5
10.0
12.5
15.0
17.5
20.0
22.5
25.0
27.5
30.0
32.5
35.0
Sw %
PCL,psi
Solution:
(1)PcR
= RcosR
LcosL
PcL
= 20 cos25
60 cos0PcL
PcR
=0.302 PcL
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III -37
(2)PcR
=hR w-o
144
= hR 1.1 - .863 62.4
144
PcR = .103 hR
hR= 9.74 PcR
(3) J =Pc
cos k1/2
=PcL
AWcosL
k L
1/2
=PcL
60 cos0 37
.16
1/2
J = .253 PcL
SwPcL
PcR hR J % psi ps i ft assorted
15 32 9.7 94.1 8.1
20 19.5 5.9 57.4 4.9
25 15.6 4.7 45.9 3.9
30 13.2 4.0 38.8 3.3
40 9.9 3.0 29.1 2.5
50 7.8 2.4 22.9 2.0
60 6.0 1.8 17.6 1.5
70 4.7 1.4 13.8 1.2
80 3.7 1.1 10.9 0.9
90 2.8 0.8 8.2 0.7
100 2 0.6 5.9 0.5
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III -38
100806040200
0
2
4
6
8
10
Sw %
PcR
100806040200
0
20
40
60
80
100
Sw %
hR
100806040200
0
2
4
6
8
10
Sw %
J
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IV - 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>Pres2) for water base mud, filtrate displaces formation water
and oil from the area around the well (saturationslikely change)
3) for oil base mud, filtrate will be oil; saturations mayor may not change.
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IV - 2
Example: Effects of flushing by mud filtrates
Coring with water base mud
Oil zone at minimum interstitial water saturation:
sat at surfaceflushing by bit trip to surface compared to res
Sw ? probablySo
Sg-
Gas zone at minimum interstitial water saturation:
sat at surfaceflushing by bit trip to surface compared to res
Sw ?
So - - -
Sg ?
Water zone:
sat at surfaceflushing by bit trip to surface compared to res
Sw -
So - - -
Sg -
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IV - 3
Coring With Oil Base Mud
Oil zone at minimum interstitial water saturation:
sat at surfaceflushing by bit trip to surface compared to res
Sw - - -
So -
Sg -
Gas zone at minimum interstitial water saturation:
sat at surfaceflushing by bit trip to surface compared to res
Sw - - -
So
Sg
Water zone: sat at surfaceflushing by bit trip to surface compared to res
Sw
So
Sg-
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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
CONDENSER
COOLING WATER OUT
CORE
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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
6560555045403530252015
0.9
1.0
1.1
1.2
1.3
1.4
Oil Gravity, API at 60 F
Multiplying
Factor
For converting distilled oil volume to oil volume originally in a sample, multiply
oil volume recovered by factor corresponding to gravity of oil in core
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IV - 6
e) determine saturations
Sw=VwVp
So=VoVp
Sg= 1 - So- Sw
where
Sw = water saturation, fraction
So = oil saturation, fraction
Sg = gas saturation, fraction
Vp = pore volume, cc
Vw = volume of water collected, cc
Vo = volume of oil collected, cc
2) disadvantages of retort process
a) must obtain temperature of 1000-1100oF tovaporize oil, water of crystallization fromclays also vaporizes causing increase in waterrecovery
WATER
RECOVERED
TIME
0
0
PORE WATER
b) at high temperatures, oil will crack and coke.
(change in hydrocarbon molecules) amountof recoverable liquid decreases.
c) core sample ruined
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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
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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 wowhere:
Wi = weight of core sample
after leaching
Wdry = weight of core sample
after leaching
Sw=VwVp
So=VoVp
2) disadvantages of leaching
a) process is slow
b) volume of oil must be calculated
3) advantages of leaching
a) very accurate water saturation value obtained
b) heating does not remove water ofcrystallization
c) sample can be used for future analysis
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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
So0 in gas zoneSo 15% in oil zone0SoSorin water zoneSor= residual oil saturation
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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
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IV - 11
Example IV-1
You want to analyze a core sample containing oil, water and gas.
Vb bulk volume = 95 cm3
Wtinitial = 216.7 gm
the sample was evacuated and the gas space was saturated with water w= 1 gm/cm3
Wt new = 219.7 gm
the water with in the sample is removed and collected
Vw removed = 13.0 cm3
the oil is extracted and the sample is dried
Wt dry = 199.5 gm
calculate:
(1) porosity
(2) water saturation
(3) oil saturation assuming 35oAPI
(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
Wtfluids = 219.7 - 199.5 = 20.2 gm
Wtoil = 20.2 - 10 - 3 = 7.2 gm
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IV - 12
o = 141.5
131.5 + 35API= 0.85 gm/cc
Vo = 7.2/0.85 = 8.49 cc
Vp = 8.49 + 3 + 10 = 21.47 cc
= 21.47/95 = 22.6%
Sw = 10/21.47 = 46.57%
So = 8.49/21.47 = 39.46%
Sg = 3/21.47 = 13.97%
m
= 199.5/(95-21.47) = 2.71 gm/cc
lithology = limestone
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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 w= 1.0 gm/cc and 35oAPI oil)
calculate:
(1) porosity
(2) water saturation
(3) oil saturation
(4) gas saturation
(5) lithology
Solution:
Vg
= .71 cc70 gm 100 gm = 1.014 ccVo = 4.5 cc
80 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
= 10.14/43.52 = 23.31%
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IV - 14
Sw = 3.50/10.14 = 34.5%
So = 5.63/10.14 = 55.5%
Sg = 1.014/10.14 = 10%
m = (91.71/33.38) = 2.75
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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 w= 1.0 gm/cc
oil gravity = 40API
Solution: o = 141.5131.5 + API
o = 141.5131.5 + 40
= 0.825o = 0.825 gm/cc
(1) =
Vp
Vb
Vp = Vw+ Vo+ Vg
Wo = Wsat- Vww- Wdry
= 87.95 - 7.12(1.0) - 79.17
Wo = 1.66 gm
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IV - 16
Vo =Woo
Vo = 1.66 gm
0.825 gm/cc= 2.01 cc
Vw = Vwrec- Wsat- Wi/w
= 7.12 - (87.95 - 86.4)/(1.0)
Vw = 5.57 cc
Vg = 1.55 cc
Vp = 5.57 + 2.01 + 1.55
Vp = 9.13 cc
Vb =Wsat- Wimm
w
Vb =(87.95 - 48.95) gm
1 gm/cc= 39.0 cc
= 9.1339.0
= 23.4%
(2) Sw =
VwVp
Sw = 5.57 cc9.13 cc
= 61.0%
So =VoVp
So = 2.01 cc9.13 cc
= 22.0%Sg =
Vg
Vp
Sg = 1.55 cc9.13 cc
= 17.0%
(3) Vm = Vb- Vp
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IV - 17
Vm = 39 - 9.13 = 29.87 cc
m=
Wdry
Vm
m=
79.17 gm/29.87 cc = 2.65gmcc
. . lithology is sandstone
(4) water saturation at surface will probably be greater than reservoir watersaturation
oil saturation at surface will be less than reservoir oil saturation
gas saturation at surface will be greater than reservoir gas saturation
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