PETROLEUM ENGINEERING 311 RESERVOIR PETROPHYSICS CLASS NOTES (1992)
Instructor/Author: Ching H. Wu
DEPARTMENT OF PETROLEUM ENGINEERING TEXAS A&M UNIVERSITY COLLEGE STATION, TEXAS
TABLE OF CONTENTS I.
ROCK POROSITY I) II) III) VI) V) VI) VII)
II.
I-1
Definition Classification Range of values of porosity Factors affecting porosity Measurement of porosity Subsurface measurement of porosity Compressibility of porous rocks
I-1 I-1 I-2 I-3 I-5 I-13 I-25
SINGLE PHASE FLOW IN POROUS ROCK
II-1
I) Darcy's equation II) Reservoir systems
II-11 II-15
III. BOUNDARY TENSION AND CAPILLARY PRESSURE I) II) III) IV) V) VI) VII) VIII) IX) X) XI)
Boundary tension Wettability Capillary pressure Relationship between capillary pressure and saturation Relationship between capillary pressure and saturation history Capillary pressure in reservoir rock Laboratory measurement of capillary pressure Converting laboratory data to reservoir conditions Determining water saturation in reservoir from capillary pressure data Capillary pressure variation Averaging capillary pressure data
IV. FLUID SATURATIONS
III-1 III-3 III-5 III-13 III-14 III-17 III-19 III-25 III-27 III-29 III-31 IV-1
I) Basic concepts of hydrocarbon accumulation II) Methods for determining fluid saturations V.
III-11
IV-1 IV-1
ELECTRICAL PROPERTIES OF ROCK-FLUID SYSTEMS
V-1
I) Electrical conductivity of fluid saturated rock II) Use of electrical Formation Resistivity Factor, Cementation Factor, and Saturation Exponent III) Laboratory measurement of electrical properties of rock IV) Effect of clay on resistivity
V-1
VI. MULTIPHASE FLOW IN POROUS ROCK I) II) III) IV) V) VI)
Effective permeability Relative permeability Typical relative permeability curves Permeability ratio (relative permeability ratio) Measurement of relative permeability Uses of relative permeability data
ii
V-8 V-9 V-18 VI-1 VI-1 VI-2 VI-2 VI-14 VI-14 VI-33
VII. STATISTICAL MEASURES I) II) III) IV) V) VI) VII) VIII) IX) X)
VII-1
Introduction Frequency Distributions Histogram Cumulative Frequency Distributions Normal Distribution Log Normal Distribution Measures of Central Tendency Measures of Variability (dispersion) Normal Distribution Log Normal Distribution
VII-1 VII-2 VII-3 VII-6 VII-8 VII-9 VII-10 VII-11 VII-12 VII-16
iii
I. ROCK POROSITY I)
Definition A measure of the pore space available for the storage of fluids in rock
In general form: Porosity = φ =
Vp Vb - Vm = Vb Vb
where: φ is expressed in fraction Vb = Vp + Vm Vb = bulk volume of reservoir rock, (L3) Vp = pore volume, (L3) Vm= matrix volume, (L3)
II)
Classification A.
Primary (original) Porosity Developed at time of deposition
B.
Secondary Porosity Developed as a result of geologic process occurring after deposition
C.
Total Porosity φt =
D.
total pore space Vb - Vm = Vb Vb
Effective Porosity φe = 1. 2.
interconnected pore space Vb Clean sandstones: φe = φt Carbonate, cemented sandstones: φe < φt
I-1
VI)
Factors affecting porosity A.
Factors: 1. Particle shape 2. Particle arrangement 3. Particle size distribution 4. Cementation 5. Vugs and fractures
B.
Particle shape Porosity increases as particle uniformity decreases.
C.
Packing Arrangement Porosity decreases as compaction increases
EFFECT OF NATURAL COMPACTION ON POROSITY (FROM KRUMBEIN AND SLOSS.) 50
40 SANDSTONES
POROSITY, %
30
20
SHALES 10
0 0
1000
2000
3000
4000
DEPTH OF BURIAL, ft
I-3
5000
6000
D.
Particle Size Distribution Porosity decreases as the range of particle size increases
SAND
SILT
CLAY
100
CLEAN SAND FRAMEWORK FRACTION
WEIGHT %
SHALY SAND
INTERSTITIAL MATERIALS AND MUD FRAGMENTS 0 1.0
0.1
0.01
0.001
GRAIN SIZE DIAMETER, MM
E.
F.
Interstitial and Cementing Material 1.
Porosity decreases as the amount of interstitial material increases
2.
Porosity decreases as the amount of cementing material increases
3.
Clean sand - little interstitial material Shaly sand - has more interstitial material
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-4
V)
Measurement of porosity φ=
Vb - Vm Vp = Vb Vb
Table of matrix densities Lithology ρ m (g/cm3) ___________ ___________
A.
Quartz
2.65
Limestone
2.71
Dolomite
2.87
Laboratory measurement 1.
Conventional core analysis a.
measure any two 1) 2) 3)
b.
bulk volume, Vb matrix volume, Vm pore volume, Vp
bulk volume 1) 2)
calculate from dimensions displacement method a)
volumetric (measure volume) (1)
drop into liquid and observe volume charge of liquid
(2)
must prevent test liquid from entering pores space of sample (a) (b) (c)
b)
coat with paraffin presaturate sample with test liquid use mercury as test liquid
gravimetric (measure mass) (1)
Change in weight of immersed sampleprevent test liquid from entering pore space
(2)
Change in weight of container and test fluid when sample is introduced
I-5
c.
matrix volume 1)
assume grain density dry weight Vm = matrix density
2)
displacement method Reduce sample to particle size, then
3)
a)
volumetric
b)
gravimetric
Boyle's Law: P1V1 = P 2V2 a) P(1)
V(1)
VALVE CLOSED b)
Put core in second chamber, evacuate
c)
Open valve
P(2)
CORE
VALVE OPEN
4)
Vm
V2
=
Volumetric of first chamber & volume of second chamber-matrix volume or core ( calculated)
VT
=
Volume of first chamber + volume second chamber (known)
=V T - V2
I-6
d.
pore volume 1)
gravimetric Vp =
2)
saturated weight - dry weight density of saturated fluid
Boyle's Law: P1V1 = P 2V2 a) P(1)
V(1)
CORE
VALVE CLOSED
b)
Put core in Hassler sleeve, evacuate
c)
Open valve
P(2)
V(1)
CORE
VALVE OPEN
3)
V2
=
Vp
= V2 - V1
I-7
Volume of first chamber + pore volume of core (calculated)
2.
Application to reservoir rocks a.
b.
intergranular porosity (sandstone, some carbonates) 1)
use representative plugs from whole core in laboratory measurements
2)
don't use sidewall cores
secondary porosity (most carbonates) 1)
use whole core in laboratory measurements
2)
calculate bulk volume from measurements
3)
determine matrix or pore volume from Boyle's Law procedure
I-8
Example I-1 A core sample coated with paraffin was immersed in a Russell tube. The dry sample weighed 20.0 gm. The dry sample coated with paraffin weighed 20.9 gm. The paraffin coated sample displaced 10.9 cc of liquid. Assume the density of solid paraffin is 0.9 gm/cc. What is the bulk volume of the sample?
Solution: Weight of paraffin coating = 20.9 gm - 20.0 gm = 0.9 gm Volume of paraffin coating = 0.9 gm / (0.9 gm/cc) = 1.0 cc Bulk volume of sample = 10.9 cc - 1.0 cc = 9.9 cc
Example I-2 The core sample of problem I-1 was stripped of the paraffin coat, crushed to grain size, and immersed in a Russell tube. The volume of the grains was 7.7 cc. What was the porosity of the sample? Is this effective or total porosity.
Solution: Bulk Volume
=
9.9 cc
Matrix Volume
=
7.7 cc
V - Vm 9.9 cc- 7.7 cc φ= b = = 0.22 Vb 9.9 cc
It is total porosity.
I-9
Example I-3 Calculate the porosity of a core sample when the following information is available: Dry weight of sample = 427.3 gm Weight of sample when saturated with water = 448.6 gm Density of water = 1.0 gm/cm3 Weight of water saturated sample immersed in water = 269.6 gm
Solution: Vp
=
sat. core wt. in air - dry core wt. density of water
Vp
=
448.6 gm - 427.3 gm 1 gm/cm3
Vp
=
21.3 cm3
Vb
=
sat. core wt. in air - sat. core wt. in water density of water
Vb
=
448.6 gm - 269.6 gm 1 gm/cm3
Vb
=
179.0 cm3
φ
=
Vp = 21.3 cm3 = .119 Vb 179.0 cm3
φ
=
11.9%
I - 10
What is the lithology of the sample? Vm
=
Vb - Vp
Vm
=
179.0 cm3 - 21.3 cm3 = 157.7 cm 3
ρm
=
wt. of dry sample matrix vol.
= 2.71 gm/(cm3) 157.7 cm3
= 427.3 gm
The lithology is limestone. Is the porosity effective or total? Why? Effective, because fluid was forced into the pore space.
I - 11
Example I-4 A carbonate whole core (3 inches by 6 inches, 695 cc) is placed in cell two of a Boyles Law device. Each of the cells has a volume of 1,000 cc. Cell one is pressured to 50.0 psig. Cell two is evacuated. The cells are connected and the resulting pressure is 28.1 psig. Calculate the porosity of the core.
Solution: P V 1 1
=
P V 2 2
V 1
=
1,000 cc
P
=
50 psig + 14.7 psia = 64.7 psia
=
28.1 psig + 14.7 = 42.8 psia
V 2
=
(64.7 psia) (1,000 cc) / (42.8 psia)
V 2
=
1,512 cc
V m
=
VT - V2
V m
=
2,000 cc - 1,512 cc - 488 cc
φ
=
VT - Vm 695 cc - 488 cc = = .298 = 29.8% VT 695 cc
P
1 2
I - 12
VI)
Subsurface measurement of porosity A.
Types of logs from which porosity can be derived 1.
Density log: ρ -ρ φd = m L ρm - ρf
2.
Sonic log: φs =
3.
∆tL - ∆tm ∆tf - ∆tm
Neutron log: e-kφ = CNf
Table of Matrix Properties (Schlumberger, Log Interpretation Principles, Volume I) Lithology
∆tm µsec/ft
ρ 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
189.0
1.00
Water
I - 13
B.
Density Log 1.
Measures bulk density of formation
M UD CAKE
FORM ATION
GAM M A RAY SOURCE
SHORT SPACE DETECTOR
LONG SPACE DETECTOR
2.
Gamma rays are stopped by electrons - the denser the rock the fewer gamma rays reach the detector
3.
Equation ρL =
ρm 1 - φ + ρf φ
ρ -ρ φd = m L ρm - ρf
I - 14
FORMATION DENSITY LOG
GR, API
ρ, gm/cc
depth, ft 4100
4120
4140
4160
4180
4200
4220
4240 0
40
80
120
160
200
2.0
I - 15
2.2
2.4
2.6
2.8
3.0
Example I-5 Use the density log to calculate the porosity for the following intervals assuming ρ matrix = 2.68 gm/cc and ρ fluid = 1.0 gm/cc. Interval, ft
ρ
__________
L, gm/cc _________
4143-4157 4170-4178 4178-4185 4185-4190 4197-4205 4210-4217
2.375 2.350 2.430 2.400 2.680 2.450
φd ,% ______ 18 20 15 17 0 14
Example: Interval 4,143 ft -4,157 ft : ρ = 2.375 gm/cc L ρ -ρ 2.68 gm/cc - 2.375 gm/cc φd = m L = = 0.18 ρm - ρf 2.68 gm/cc - 1.0 gm/cc
I - 16
C.
Sonic Log 1.
Measures time required for compressional sound waves to travel through one foot of formation
T
A
B C
R1 D
E R2
2.
Sound travels more slowly in fluids than in solids. Pore space is filled with fluids. Travel time increases as porosity increases.
3.
Equation ∆tL = ∆tm 1 - φ + ∆tf φ
I - 17
(Wylie Time Average Equation)
SONIC LOG
GR, API
∆T, µ seconds/ft
depth, ft 4100
4120
4140
4160
4180
4200
4220
4240 0
100
200
140
I - 18
120
100
80
60
40
Example I-6 Use the Sonic log and assume sandstone lithology to calculate the porosity for the following intervals.
∆tL µ second/ft
Interval (ft)
φs ,%
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 φs =
= 86.5 µ-sec/ft ∆tL - ∆tm 86.5 µ sec/ft- 51.6 µ sec/ft = = 0.25 ∆tf - ∆tm 189.0 µ sec/ft- 51.6 µ sec/ft
I - 19
D.
Neutron Log 1.
Measures the amount of hydrogen in the formation (hydrogen index)
Maximum Energy Loss/ Collision, %
Average Number Collisions
Element Calcium Chlorine Silicone Oxygen Carbon Hydrogen
371 316 261 150 115 18
Atomic Collision
8 10 12 21 28 100
40.1 35.5 28.1 16.0 12.0 1.0
103
20 17 14 8 6 1
CLEAN SAND POROSITY = 15%
O
102 Si 10
H
SLOWING DOWN POWER
RELATIVE PROBABILITY FOR COLLISION
CLEAN SAND POROSITY = 15%
Atomic Number
1
10-1
H
O
10-2
Si
10-3
1
.1
1
10
102 10 3 10 4 105
106 107
.1
NEUTRON ENERGY IN ELECTRON VOLTS
1
10
102 10 3 10 4 105
106 107
NEUTRON ENERGY IN ELECTRON VOLTS
2.
In clean, liquid filled formations, hydrogen index is directly proportional to porosity. 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 - 20
NEUTRON DENSITY LOG
GR, API
φ (CDL)
depth, ft 4100
4120
4140
4160
4180
4200
4220
4240 0
200
I - 21
30
-10
Example I-7 Use the neutron log to determine porosity for the following intervals.
Solution:
Interval (ft)
φ
n (%)
.
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 - 22
Example I-8 Calculate the porosity and lithology of the Polar No. 1 drilled in Lake Maracaibo. The depth of interest is 13,743 feet. A density log and a sonic log were run in the well in addition to the standard Induction Electric Survey (IES) survey. The readings at 13,743 feet are: bulk density travel time
= 2.522 gm/cc = 62.73 µ-sec/ft
Solution: Assume fresh water in pores. Assume sandstone: ρ m = 2.65 gm/cc ∆tm = 55.5 µ-sec/ft ρ -ρ 2.65 gm/cc - 2.522 gm/cc φd = m L = = 7.76% ρm - ρf 2.65 gm/cc - 1.0 gm/cc
φs =
∆tL - ∆tm 62.73 µ sec/ft- 55.5 µ sec/ft = = 5.42% ∆tf - ∆tm 189.0 µ sec/ft - 55.5 µ sec/ft
Assume limestone: ρm
= 2.71 gm/cc
∆tm
= 47.5 µ-sec/ft
ρ -ρ 2.71 gm/cc - 2.522 gm/cc φd = m L = = 10.99% ρm - ρf 2.71 gm/cc - 1.0 gm/cc φs =
∆tL - ∆tm 62.73 µ sec/ft - 47.5 µ sec/ft = = 10.76% ∆tf - ∆tm 189.0 µ sec/ft - 47.5 µ sec/ft
I - 23
Assume dolomite: ρm
= 2.87 gm/cc
∆tm
= 43.5 µ-sec/ft
φd
ρ -ρ 2.87 gm/cc - 2.522 gm/cc = m L= = 18.619% ρm - ρf 2.87 gm/cc - 1.0 gm/cc
φs
=
φlimestone
∆tL - ∆tm 62.73 µ sec/ft - 43.5 µ sec/ft = = 13.22% ∆tf - ∆tm 189.0 µ sec/ft - 43.5 µ sec/ft = 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 - 24
VII)
Compressibility of porous rocks Compressibility, c is the fractional change in volume per unit change in pressure: ∆V ∂V V T c=- 1 ≅ V ∂P T ∆P A.
Normally pressured reservoirs 1.
Downward force by the overburden must be balanced by upward force of the matrix and the fluid Fo
Ff
Fm
2.
Thus, Fo
= Fm + Ff
it follows that
3.
Po
=
pm + pf
Po
≅
1.0 psi/ft
Pf
≅
0.465 psi/ft
I - 25
4.
As fluid is produced from a reservoir, the fluid pressure, Pf will usually decrease: a. b. c.
B.
the force on the matrix increases causing a decrease in bulk volume and a decrease in pore volume
Types of compressibility 1.
Matrix Compressibility, cm cm
2.
≅ 0
Bulk Compressibility cb used in subsidence studies
3.
Formation Compressibility, cf - also called pore volume compressibility a.
important to reservoir engineers 1) 2) 3)
depletion of fluid from pore spaces internal rock stress changes change in stress results in change in Vp, Vm, Vb
4)
by definition ∂Vp cf = - 1 Vp ∂pm
b.
since overburden pressure, Po, is constant dPm
= - dP f
I - 26
1)
Thus, ∂Vp cf = - 1 Vp ∂pm
2)
where the subscript of f on cf means "formation" and the subscript of f on Pf means "fluid"
3)
procedure (a)
measure volume of liquid expelled as a function of "external" pressure
(b)
"external" pressure may be taken to represent overburden pressure, Po
(c)
fluid pressure, pf, is essentially constant, thus, dPo
(d)
expelled volume increases as pore volume, vp, decreases, thus, dVp
(e)
= dP m
= - dVexpelled
from definition ∂Vp cf = - 1 Vp ∂pm it follows that ∆ Vp expelled cf = + 1 Vp ∆Po
I - 27
plot
CUMULATIVE PORE VOLUME VOLUME EXPELLED
(f)
OVERBURDEN PRESSURE, psi
slope = cf.
I - 28
C.
Measurement of compressibility 1)
2)
Laboratory core sample a)
apply variable internal and external pressures
b)
internal rock volume changes
Equipment
Internal Pressure Gauge
Hydraulic Pump Mercury Sight Gauge Overburden Pressure Gauge
Hydraulic Pump Copper - Jacketed Core
Apparatus for measuring pore volume compressibility (hydrostatic)
I - 29
Example I-9 Given the following lab data, calculate the pore volume compressibility for a sandstone sample at 4,000 and 6,000 psi. pore volume
=
50.0 cc
pressure, psi
vol. fluid expelled, cc
1000 2000 3000 4000 5000 6000 7000 8000
0.244 0.324 0.392 0.448 0.500 0.546 0.596 0.630
Solution: from graph @ 4,000 psi: 0.009 4000 psi
Slope
=
cf
= 2.25 X 10-6
1 psi
@ 6000 psi: 0.011 6000 psi
Slope
=
cf
= 1.83 X 10-6
1 psi
I - 30
VOLUME EXPELLED, cc PORE VOLUME, cc
0.015
0.010
0.005
0.000 0
2000
4000
6000
8000
COMPACTION PRESSURE, psi
I - 31
10000
PORE VOLUME COMPRESSIBILITY X 10-6 psi-1
100
PORE-VOLUME COMPRESSIBILITY AT 75 % LITHOSTATIC PRESSURE VS INITIAL SAMPLE POROSITY FOR CONSOLIDATED SANDSTONES.
CONSOLIDATED SANDSTONES
10
HALL'S CORRELATION
1 0
5
10
15
20
25
30
PORE VOLUME COMPRESSIBILITY X 10-6 psi-1
INITIAL POROSITY AT ZERO NET PRESSURE, % PORE-VOLUME COMPRESSIBILITY AT 75 % LITHOSTATIC PRESSURE VS INITIAL SAMPLE POROSITY FOR UNCONSOLIDATED SANDSTONES.
100
UNCONSOLIDATED SANDSTONES
10
HALL'S CORRELATION
1 0
5
10
15
20
25
30
INITIAL POROSITY AT ZERO NET PRESSURE, %
I - 32
E.
Abnormally pressured reservoirs "abnormal pressure": fluid pressures greater than or less than the hydrostatic fluid pressure expected from an assumed linear pressure gradient
PRESSURE
DEPTH
NORMAL LINEAR
SUBNORMAL (LOWER)
SURNORMAL (GREATER)
I - 33
Compressibility/Porosity Problem No. 1 A limestone sample weighs 241.0 gm. The limestone sample coated with paraffin was found to weigh 249.5 gm. The coated sample when immersed in a partially filled graduated cylinder displaced 125.0 cc of water. The density of the paraffin is 0.90 gm/cc. What is the porosity of the rock? Does the process measure total or effective porosity?
Solution:
Vm =
wt. dry 241.0 gm = = 88.9 cc ρ ls 2.71 gm/cc
Vparaffin =
wt. coated sample - st. uncoated sample ρ
Vparaffin =
249.5 gm - 241.0 gm = 9.4 cc 0.90 gm/cc
Vb
= 125 cc - 9.4 cc = 115.6 cc
Vp
= Vb - Vm
Vp
= 115.6 cc - 88.9 cc - 26.7 cc
φ
=
φ
= 23.1%
Vp 26.7 cc = = 0.231 Vb 115.6 cc
(total porosity)
I - 34
Compressibility/Porosity Problem No. 2 You are furnished with the results of a sieve analysis of a core from Pete well #1. Previous laboratory work indicates there is a correlation between grain size and porosity displayed by those particular particles. The correlation is seen below: gravel
-
25% porosity
coarse sand
-
38% porosity
fine sand
-
41% porosity
What would be the minimum porosity of the mixture? What basic assumption must be made in order to work the problem? Solution: Begin calculation with a volume of 1 cu. ft. remaining pore volume (ft3)
component
porosity (%)
remaining matrix volume (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 - 35
Compressibility/Porosity Problem No. 3 A sandstone reservoir has an average thickness of 85 feet and a total volume of 7,650 acre-feet. Density log readings through the fresh water portion of the reservoir indicate a density of 2.40 gm/cc. The Highgrade #1 Well was drilled and cored through the reservoir. A rock sample was sent to the laboratory and the following tests were run. pressure (psig) 1,000 2,000 3,000 4,000 5,000 6,000 7,000 8,000
cum. pore vol. change (-cc)_________ 0.122 0.162 0.196 0.224 0.250 0.273 0.298 0.315
The dry weight of the core sample was found to be 140 gm while the sample dimensions were 1.575 inches long and 1.960 inches in diameter. Assuming the compressibility at 4,500 psi is the average compressibility in the reservoir, how much subsidence occurs when the reservoir pressure declines from 5,500 psi to 3,500 psi? Calculate: A.
Reservoir Porosity
B.
Sample Pore Volume
C.
Compressibility at 4,500 psi
D.
Amount of Ground Subsidence.
Solution: A.
Reservoir Porosity ρ -ρ φ = m L = 2.65 - 2.40 = 15.22% ρ m - ρ f 2.65 - 1.00
I - 36
B.
C.
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
2 3.14 5.0 2 4.0 = bulk volume = πD h = = 78.5 cc 4 4.0
Vm
= matrix volume = 140 gm
Vp
= Vb - Vm = 78.5 cc - 52.8 cc
Vp
= 25.7 cc
Compressibility (see graph) Vp
D.
cc = 52.8 cc 2.65 gm
= 25.7 cc
Subsidence ∆H
= H cp φ ∆P
∆H
= 85 ft 9.69x 10 -7 psi-1 0.152 2,000 psi
∆H
= 0.026 ft
∆H
= 0.32 inches
Note: the pore volume (formation) compressibility is somewhat smaller than usually encountered. An experienced engineer would be wary of this small number. Also it was assumed that the formation compressibility was exactly the same as the bulk volume compressibility. Experience shows that this is not the case.
I - 37
POROSITY PROBLEM No. 3 0.0140
SLOPE =
VOLUME EXPELLED, cc PORE VOLUME, cc
0.0120
.0118 - .0068 7000 - 2000
C p = 9.96 x 10-7 psi -1
0.0100
0.0080
0.0060
0.0040 0
2000
4000
6000
PRESSURE, psig
I - 38
8000
Compressibility Problem A 160-acre and 100 ft thick reservoir has a porosity of 11%. The pore compressibility is 5.0 x 106 (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 dVp Cp = - 1 Vp dp 5 x 10-6 (1/psi) =
dVp -1 76,665,600 ft3 3,000 psi
dVp = 1.15 x 106 ft3 ∆H = 1.15 x 106 ft3 x
1 = 0.165 ft 6,969,600 ft2
I - 39
II. SINGLE PHASE FLOW IN POROUS ROCK I)
Darcy's equation (1856) A.
Water flow through sand filters
A
q
h 1 - h2
h1
Z
h2
q
WATER
SAND
DARCY'S FOUNTAIN.
kA(h1 - h2) µL Length of sand pack,L = Z q=
1.
constant of proportionality, k, characteristics of particular sand pack, not sample 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 - 1
q=
B.
kA(h1 - h2) µL
Generalized form of Darcy's equation 1.
Equation ρg dz vs = -k dP µ ds 1.0133 x 10 6 ds
+Z
90o
+1
Vs
180o
270o
s Θ
+X
-Z -1
θ
+Y
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 ds
= pressure gradient along s - atm/cm
µ
= viscosity - centipoise
k
= permeability - darcies
II - 2
360o
3.
4.
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
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
ρg dp dz = - k µ ds 1.0133 x 10 6 ds
L t
m/Lt2 - m/L3 L/t2 L = - k m/Lt L L
k
= L2 = cross-sectional area
II - 3
5.
Definition of Darcy units a.
conventional units would be: 1)
feet squared in the English system
2)
centimeter squared in the cgs system
b.
both are too large for use in porous media
c.
definition of darcy
A porous medium has a permeability of one darcy when a single-phase fluid of one centipoise that completely fills the voids of the medium will flow through it under conditions of viscous flow at a rate of one cubic centimeter per second per square centimeter cross-sectional area under a pressure or equivalent hydraulic gradient of one atmosphere per centimeter.
q
=
k A P1 - P2 µL
II - 4
II)
Reservoir systems A.
Flow of incompressible liquid 1.
q P1
Horizontal, linear flow system
A
q P2
L
a.
Conditions 1)
dz horizontal system, ds = 0
2)
linear system, A = constant
3)
incompressible liquid, q = constant
4)
laminar flow, can use Darcy's equation
5)
non-reactive fluid, k = constant
6)
100% saturated with one fluid
7)
constant temperature, µ, q
II - 5
b.
derivation of flow equation ρg dP dz ds 1.0133 x 10 6 ds
vs
= - k µ
vs
q = - k dP = µ ds A L
q
ds 0
qL-0 q
p2 kA = dP µ p 1 = - kA P2 - P1 µ
= kA P2 - P1 Lµ
Note: P1 acts at L = 0 P2 acts at L = L q is + if flow is from L = 0 to L = L
II - 6
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 cm 2
L
= 6 ft (12 in/1 ft) (2.54 cm/1 in) = 182.88 cm
P1
= 5.0 atm
P2
= 2.0 atm q q
q
= kA P2 - P1 Lµ = (1) (5,572.8) (5.0 - 2.0) (1) (182.88) = 91.42 cm 3/sec
II - 7
2.
Non-horizontal, linear system
P2
-Z
S Θ X
P1
a.
Conditions 1)
dz = sinθ non-horizontal system, ds = 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 - 8
b.
derivation of equation vs
ρg dz = - k dP µ ds 1.0133 x 10 6 ds
vs
= -
q = - k A µ
ds
= - kA µ
L q 0
q
k ρg sin θ dP + ds µ 1.0133 x 10 6 P2 P1
kA ρg sin θ dp + µ 1.0133 x 10 6
ρgLsinθ = - kA P1 - P2 + µL 1.0133 x 10 6
II - 9
L ds 0
3.
Vertical, upward flow, linear system FLOW UNDER HEAD h
h
x
L
a.
Conditions 1)
dz vertical system, ds = sinθ = constant
2)
upward flow, q = 270°, sinθ = - 1
3)
linear system, A = constant
4)
incompressible liquid, q = constant
5)
laminar flow, use Darcy equation
6)
non-reactive fluid, k = constant
7)
100% saturated with one fluid
8)
constant temperature, µ
II - 10
b.
derivation of flow equation ρg dP dz ds 1.0133 x 10 6 ds
vs
=
k µ
vs
=
q = - k A µ
q
=
kA µ
P1
= -
P2
=
ρg dP + ds 1.0133 x 10 6 ρg P1 - P2 L 1.0133 x 10 6
ρg (h + x + L) 1.0133 x 10 6 ρg x 1.0133 x 10 6 P1
L
P2
=
ρg h ρg + 1.0133 x 10 6 L 1.0133 x 10 6
ρg h ρg ρg q = kA + µ 1.0133 x 10 6 L 1.0133 x 10 6 1.0133 x 10 6
q
ρg h = kA µL 1.0133 x 10 6
II - 11
4.
Horizontal, radial flow system
re Pw
rw
re
Pe
rw
h
a.
Conditions 1)
dz horizontal system, ds = 0
2)
radial system, A = 2πrh , ds = - dr, flow is inward
3)
constant thickness, h = constant
4)
incompressible liquid, q = constant
5)
laminar flow, use Darcy equation
6)
non-reactive fluid, k = constant
7)
100% saturated with liquid,
8)
constant temperature, µ, q
II - 12
b.
Derivation of flow Equation ρg dP dz ds 1.0133 x 10 6 ds
vs
= - k µ
vs
q q = + k dP = = µ dr A 2πrh re
q 2πh
rw
q 2πh q
dr = k r µ
pe dp pw
1n(re) - 1n( rw) = k Pe - Pw µ =
2πhk P - Pw µ 1n (re/rw) e
Note: if q is + , flow is from r e to rw B.
Flow of gas (compressible fluid) 1.
q
horizontal, linear flow system
A
q
P1
P2
L a.
Conditions 1)
dz horizontal system, ds = 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 - 13
b.
Assumptions
c.
1)
µ, Z
2)
Z(and µ ) can be determined at mean pressure
= constant
Derivation of equation for qsc vs
ρg dz = - k dP µ ds 1.0133 x 10 6 ds
vs
q = - k dP = µ ds Ads
but q
Psc qscz T PTsc
=
thus L
p2
Psc T qsc Tsc A
o
Psc T qsc Tsc A
P22 - P12 L -0 = - k µz 2
ds = - k p1
2 P21 - P2 qsc = kA Tsc µL Tz Psc 2 Note: real gas equation of state Pq
= ZnRT
where q n
= volumetric flow/time = mass flow/time
thus, Pq Pscqsc q
=
= ZnRT n R Tsc Psc qscz T 1 Tsc P
where qsc is constant Z is determined at P, T
II - 14
PdP µz
Derivation of equation for q
d.
qsc
=
2 P1 - P22 kA Tsc µL Tz Psc 2
but Tsc P12 - P22 = k A µL T z Psc 2
qsc =
P q Tsc Z Psc T
q =
k A 1 P12 - P22 µL P 2
q
=
P12 - P22 k A 2 µL P1 + P 2 2
q
=
k A P - P 1 2 µL
This equation is identical to the equation for horizontal, linear flow of incompressible liquid thus if gas flow rate is determined at mean pressure, P, the equation for incompressible liquid can be used for compressible gas! Note: real gas equation of state Pq
= ZnRT
thus Psc qsc n R Tsc = znRT Pq where P
=
P1 + P 2 2
P
=
volumetric flow rate at P, T
z is determined at P, T qsc
=
P q Tsc z Psc T
II - 15
2.
Horizontal, radial flow system
re Pw
rw
re
Pe
rw
h
a.
Conditions 1)
dz horizontal system ds = 0
2)
radial system, A = 2πrL, ds = - dr, inward flow
3)
constant thickness, h = constant
4)
compressible gas flow, q = f (P)
5)
laminar flow, use Darcy equation
6)
non-reactive fluid, k = constant
7)
100% saturated with one fluid
8)
constant temperature
II - 16
b.
Assumptions µz = constant z (and µ ) can be determined at mean pressure
c.
derivation of equation for qsc ρg dz vs = - k dP µ ds 1.0133 x 10 6 ds q vs = - k dP = µ ds A
but q
P q zT = sc sc PTsc
and A = 2πrh and ds = - dr thus Psc T qsc 2Tsc π h
re rw
dr = k r
Pe Pw
ρdP µz
PscT qsc r P2 - P2w 1n r e = k e µz 2 2 Tsc π h w qsc
=
Tsc P2e - P2w 2πhk µ 1n re/rw Psc zT 2
II - 17
d.
derivation of equation for q qsc
Tsc P2e - P2w 2πhk µ 1n re/rw Psc zT 2
=
but q
=
P q Tsc z Psc T
thus P q Tsc z Psc T
=
Tsc 2πhk µ 1n re/rw Psc zT
P2e - P2w 2
q
=
2 π h k 1 (P2e - P2w) µ 1n re/rw P 2
q
=
(P2e - P2w) 2πhk 2 µ 1n re/rw Pe + P w 2
q
=
2πhk P - Pw µ 1n re/rw e
Note: Equation for real gas is identical to equation for incompressible liquid when volumetric flow rate of gas, q, is measured at mean pressure.
II - 18
C.
Conversion to Oilfield Units
Symbol
Darcy units
q k A h P L µ r
Oil field
cc/sec darcy sq cm cm atm cm cp gm/cc
bbl/d or cu ft/d md sq ft ft psia ft cp lb/cu ft
Example: q=
hkA P1 - P2 µL in Darcy's units
cc = q bbl 5.615 cu ft q sec d bbl
1,728 cu in cu ft
16.39 cc cu in
cc = 1.841 q bbl q sec d darcy k darcy = k md 1,000md k darcy = 0.001 k md A sq cm =
929.0 sq cm A sq ft sq ft
A sq cm = 929.0 A sq ft P1 - P2
atm =
P1 - P2
P1 - P2
atm = 0.06805
psia
atm 14.696 psia
P1 - P2
psia
L cm = L ft 30.48 cm ft meter = 100 cm 1.841 q = q =
0.001 k
929.0 A .06805 µ 30.48 L
0.01127 k A P1 - P2 µL
P1 - P2
in oilfield units
II - 19
d 24hr
hr 3,600 sec
D.
Table of Equations 1.
System
Darcy Units
Fluid
Equation
Horizontal, Linear
Incompressible Liquid
q =
kA µL
Dipping, Linear
Incompressible Liquid
q =
kA µL
Horizontal, Radial
Incompressible Liquid
q =
Horizontal, Linear
Real Gas
Horizontal, Radial
Real Gas
P1 - P2
P1 - P2
2 π kh µ ln (re/rw)
q = kA µL
P1 - P2
qsc =
π kh ln (re/rw)
q =
2 π kh µ ln (re/rw)
II - 20
ρ g L sin θ 1.0133 x 10 6
Pe - Pw
Tsc qsc = kA µ L Tz Psc
µ
+
P21 - P22 2
Tsc Tz Psc
Pe - Pw
Pe 2 - Pw 2
2.
System Horizontal, Linear
Oilfield Units
Fluid
Equation
Incompressible Liquid
q = 0.001127 kA µL
P1 - P2
q = res bbl/d Dipping, Linear
Incompressible Liquid
q = 0.001127 kA µL
+
Horizontal, Radial
Incompressible Liquid
Horizontal, Linear
Real Gas
P1 - P2
ρg L sinθ 1.0133 x 10 6
q = .007082
kh µ ln (re/rw) kA µ LzT
qsc = .1118
Pe - Pw
P12 - P22
qsc = scf/d q = .001127 kA µL
P1 - P2
q = res bbl/d Horizontal, Radial
Real Gas
qsc = .7032
kh µ ln (re/rw) Tz
q = .007082
II - 21
kh µ ln (re/rw)
Pe 2 - Pw2
Pe - Pw
Example II-2 What is the flow rate of a horizontal rectangular system when the conditions are as follows: permeability = k = 1 darcy area = A = 6 ft2 viscosity = µ = 1.0 cp length = L = 6 ft inlet pressure = P1 = 5.0 atm. outlet pressure = P2 = 2.0 atm.
Solutions: We must insure that all the variables are in the correct units. k A L P1 P2
= = = = =
1 darcy = 1,000 md 2 6 ft 6 ft (5.0 atm) (14.7 psi/atm) = 73.5 psi (2.0 atm) (14.7 psi/atm) = 29.4 psi
q = 1.1271 x 10-3 kA P1 - P2 µL q = 1.1271 x 10-3
1,000 6 1 6
73.5 - 29.4
q = 49.7 bbl / day
II - 22
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=
q=
7.082 x 10 -3 kH Pe - Pw µ ln Re / Rw 7.082 x 1--3
300 20 2,500 - 1,740 1.3 6.492
q = 3,826 res bbl/d
II - 23
E. Layered Systems 1.
Horizontal, linear flow parallel to bedding
P1
P2
q
q
A B C
W L qt = qA + qB + qC h = hA + hB + hC let k be "average" permeability, then
qt =
k wh P1 - P2 µL
and qt =
kA whA µL
P1 - P2
+
kB whB µL
P1 - P2
then k h = kA hA + kB hB + kC hC k
=
n kj hj ∑ h j=1
II - 24
+
kC whC µL
P1 - P2
2.
Horizontal, radial flow parallel to bedding
re re
hA
ht
rw
qA
hB
qB
hC
qC
Pe
Pw again qt = qA + qB + qC h = hA + hB + hC qt =
2πk h µ ln (re/rw)
qt =
2 π kA hA µ ln (re/rw)
Pe - Pw
and Pe - Pw
+
+
2 π kB hB µ ln (re/rw)
2 π kc hc µ ln (re/rw)
then k h = kA hA + kB hB + kC hC and again k =
n kj hj ∑ h j=1
II - 25
Pe - Pw
Pe - Pw
3.
Horizontal, linear flow perpendicular to bedding A
B
C
P1
P2
q
q
kA
kB
∆P A
∆P B
kC ∆P C
LA
LB
LC
h
W
L
qt = qA = qB = qC p1 - p2 = ∆PA + ∆PB + ∆PC L = LA + LB + LC qt = and since
k wh
P1 - P2 µL
P1 - P2 = ∆ PA + ∆PB + ∆PC P1 - P2 =
since
qt µ L q µLA q µLB q µLC = A + B + C kA wh kB wh kC wh k wh
qt = qA = qB = qC L = LA + LB + LC kA kB kC k
thus K =
L n Lj ∑ j = 1 kj
II - 26
4.
Horizontal, radial, flow perpendicular to bedding q
Pw
PA
PB
PC
h rw rA rB
rC
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) 2π k h +
qA µ ln (rA/rw) 2 π kAh
qB µ ln (rB/rA) q µ ln (rC/re) + C 2 π kB h 2 π kC h
then k =
=
ln re/rw n ln(rj/rj-1) ∑ kj j=1
II - 27
Example II-4 Damaged zone near wellbore k1 = 10 md k2 = 200 md
r1 r2 rw
= 2 ft = 300 ft = 0.25 ft
Solution: k =
k =
ln (re/rw) n ln (rj/rj-1 ) ∑ kj j=1 ln 300 0.25 ln 2/0.25 ln 300/2 + 10 200
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.
II - 28
F.
Flow through channels and fractures 1.
Flow through constant diameter channel
A
L
a.
Poiseuille's Equation for viscous flow through capillary tubes q=
πr4 8µL
P1 - P2
A = π r2, therefore q=
b.
Ar2 8µL
P1 - P2
Darcy's law for linear flow of liquids q = kA µL
P1 - P2
assuming these flow equations have consistent units Ar2 8µL
P1 - P2
= kA µL
P1 - P2
thus 2 k = r2 = d 32 8 where d = inches, k = 20 x 10 9 d2 md
II - 29
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?
A.
k
= 20 x 109 d2
k
= 20 x 109 (0.0001 in) 2
k
= 200 md
k
= 0.25 (200 md)
k
= 50 md
Solution:
B.
II - 30
2.
Flow through fractures
b
v=
q h2 (P 1-P 2) = A 12 µ L
2 q = b A (P1 -P2) 12 µ L
setting this flow equation equal to Darcy's flow equation, b2 A P1 - P2 = kA P1 - P2 12 µ L µL solve for permeability of a fracture: 2 k = b in darcy units, or 12 k = 54 x 109 b2 where b k
= =
inches md
II - 31
Example II-6 Consider a rock of very low matrix permeability, 0.01 md, which contains on the average a fracture 0.005 inches wide and one foot in lateral extent per square foot of rock. Assuming the fracture is in the direction of flow, determine the average permeability using the equation for parallel flow. Solution:
∑ kj Aj k =
k = k =
A matrix k 0.01
, similar to horizontal, linear flow parallel to fracture
matrix area + fracture k total area 12 in 2 + 12 in 144 in2
fracture area
0.005 in
+
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 - 32
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
qsc P sc A
SLOPE = k / m
P 12 - P 22
2L
qsc =
kA P12 - P22 2 µ L Psc
horizontal, linear, real gas flow with T = Tsc and Z = 1.0
qsc Psc = k A µ
P12 - P22 2L
k = ( slope ) m
II - 33
2.
Whole core method a.
prepare whole core in same manner as perm plugs
b.
mount core in special holders and flow fluid through core as in perm plug method
VERTICAL FLOW LOW AIR PRESSURE (FLOW)
CORE RUBBER TUBING
HIGH AIR PRESSURE PIPE
TO FLOWMETER
c.
the horizontal flow data must be adjusted due to complex flow path
d.
whole core method gives better results for limestones
II - 34
B.
Factors which affect permeability measurement 1.
Fractures - rocks which contain fractures in situ frequently separate along the planes of natural weakness when cored. Thus laboratory measurements give "matrix" permeability which is lower than in situ permeability because typically only the unfractured parts of the sample are analyzed for permeability.
2.
Gas slippage a.
gas molecules "slip" along the grain surfaces
b.
occurs when diameter of the capillary openings approaches the mean free path of the gas molecules
c.
Darcy's equation assumes laminar flow
d.
gas flow path with slippage
e.
called Klinkenberg effect
f.
mean free path is function of size of molecule thus permeability measurements are a function of type of gas used in laboratory measurement.
II - 35
H2
N2 k CALCULATED
CO2
0 1 P g.
mean free path is a function of pressure, thus Klinkenberg effect is greater for measurements at low pressures - negligible at high pressures.
h.
permeability is a function of size of capillary opening, thus Klinkenberg effect is greater for low permeability rocks.
i.
effect of gas slippage can be eliminated by making measurements at several different mean pressures and extrapolating to high pressure (1/p => 0)
kMEASURED
0
1 P
II - 36
Example II-7 Another core taken at 8815 feet from the Brazos County well was found to be very shaly. There was some question about what the true liquid permeability was, since nitrogen was used in the permeameter. Calculate the equivalent liquid permeability from the following data. Mean Pressure ( atm ) 1.192 2.517 4.571 9.484
Measured Permeability ( md ) 3.76 3.04 2.76 2.54
Solution: Plot kmeasured vs. 1/pressure Intercept is equivalent to liquid permeability From graph: kliq = 2.38 md
GAS PERMEABILITY, md
5 4 3 2
k gas = 2.38276 + 1.64632 P bar
1
Equivalent Liquid Permeability = 2.38 md
0 0.0
0.2
0.4
0.6
0.8
RECIPROCAL MEAN PRESSURE, atm - 1
II - 37
1.0
3.
Reactive fluids a.
Formation water reacts with clays 1)
lowers permeability to liquid
2)
actual permeability to formation water is lower than lab permeability to gas
RELATIONSHIP OF PERMEABILITIES MEASURED WITH AIR TO THOSE MEASURED WITH WATER
WATER PERMEABILITY, md
1000
100
10
Water concentration 20,000 - 25,000 ppm Cl ion. 1 1
10
100
1000
10000
AIR PERMEABILITY, md
b. Injection water may,if its salinity is less than that of the formation water, reduce the permeability due to clay swelling.
II - 38
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 S S
34 34 34
4080 24800 40100
1445 11800 23000
1380 10600 18600
1290 10000 15300
1190 9000 13800
885 7400 8200
17.2 147.0 270.0
S S S
34 34 34
4850 22800 34800
1910 13600 23600
1430 6150 7800
925 4010 5460
736 3490 5220
326 1970 3860
5.0 19.5 9.9
S S T
34 34 36
13600 7640 2630
5160 1788 2180
4640 1840 2140
4200 2010 2080
4150 2540 2150
2790 2020 2010
197.0 119.0 1960.0
T T T
36 36 36
3340 2640 3360
2820 2040 2500
2730 1920 2400
2700 1860 2340
2690 1860 2340
2490 1860 2280
2460.0 1550.0 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.
II - 39
III. I)
BOUNDARY TENSION AND CAPILLARY PRESSURE 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
SURFACE
LIQUID
III- 1
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 cm 2 of surface (ergs/cm 2) is termed "boundary energy"
2.
also can be thought of as force in dynes acting along length of 1 cm required to prevent destruction of surface (dynes/cm) - this is called "boundary tension"
3.
Boundary Energy = Boundary Tension x Length
G.
Surface Tension - Boundary tension between gas and liquid is called "surface tension"
H.
Interfacial Tension - Boundary tension between two immiscible liquids or between a fluid and a solid is called "interfacial tension"
I.
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
Forces creating boundary tension 1.
2.
Forces a.
Law of Universal Gravitation applied between molecules
b.
physical attraction (repulsion) between molecules
Liquid-Gas Boundary attraction between molecules is directly proportional to their masses and inversely proportional to the square of the distance between them
3.
Solid-Liquid Boundary physical attraction between molecules of liquid and solid surface
III- 2
4.
Liquid-Liquid Boundary some of each
II)
Wettability A.
forces at boundary of two liquids and a solid (or gas-liquid-solid)
σow
OIL
OIL WATER Θ
σos
σw s
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- 3
D.
if the solid is "oil-wet" σos ≥ σws AT = cos θ = 90° ≤ θ ≤ 180° if θ = 180° - strongly oil-wet
θ=
300
θ=
θ = 1580
830
θ = 350 (A)
ISOOCTANE
θ = 300
ISOOCTANE + 5.7% ISOQUINOLINE
ISOQUINOLINE
θ = 480
θ = 540
NAPHTHENIC ACID
θ = 1060 (B)
Interfacial contact angles. (A) Silica surface; (B) calcite surface
III- 4
III)
Capillary pressure A.
capillary pressure between air and water
h
Θ
AIR
WATER
1.
liquid will rise in the tube until total force up equals total force down a.
total force up equals adhesion tension acting along the circumference of the water-air-solid interface = 2πr AT
b.
total force down equals the weight of the column of water converted to force = πr2 hgρ w
c.
thus when column of water comes to equilibrium 2πr A T = πr2 hgρ w
d.
units dyne gm cm cm = cm2 cm cm sec2 cm3 gm cm sec2 dyne = force unit adhesion tension dyne AT = 1 r hgρ w cm 2 dyne =
e.
III- 5
2.
liquid will rise in the tube until the vertical component of surface tension equals the total force down a.
vertical component of surface tension is the surface tension between air and water multiplied by the cosine of the contact angle acting along the water-air-solid interface = 2πr σaw cosθ
b.
total force down = πr2 hgρ w
c.
thus when the column of water comes to equilibrium 2πr σaw cosθ = πr 2 hgρ w
d.
units dyne gm cm cm = cm2 cm cm 2 sec cm3 gm cm dyne = cm cm sec2
3.
since AT = σaw cosθ, 1 and 2 above both result in h=
2 σaw cos θ rg ρw
III- 6
4.
capillary pressure (air-water system) Pa Pw
h A'
B' B
Θ
Pa A
AIR
WATER
pressure relations in capillary tubes a.
pressure at A' is equal to pressure at A Pa' = Pa
b.
pressure at B is equal to the pressure at A minus the head of water between A & B pw = pa - ρ wgh units:
c.
dyne dyne gm ⋅ cm = cm cm2 cm2 cm3 ⋅ sec2
thus between B' and B there is a pressure difference pa - pw = pa - (pa - ρ wgh) pa - pw = ρ wgh
d.
call this pressure difference between B' and B "capillary pressure" Pc = pa - pw = ρ wgh
e.
remember h=
2 σgw cos θ rg ρ w
III- 7
f.
thus Pc =
B.
2 σgw cos θ r
capillary pressure between oil and water
h
Θ
OIL
WATER
1.
liquid will rise in the tube until the vertical component of surface tension equals the total force down a.
vertical component of surface tension equals the surface tension between oil and water multiplied by the cosine of the contact angle acting along the circumference of the water-oil-solid interface = 2πr σow cosθ
b.
the downward force caused by the weight of the column of water is partially offset (bouyed) by the weight of the column of oil outside the capillary
c.
thus, total force down equals the weight of the column of water minus 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- 8
3)
net weight per unit area acting to pull surface down = ρ wh - ρ oh = h(ρ w - ρ o)
4)
total force down = πr2 gh (ρ w - ρ o) d.
thus when the column of water comes to equilibrium
2πr σow cosθ = πr 2 gh (ρ w - ρ o) 2.
thus the equilibrium for the height of the column of water h=
3.
2 σow cos θ rg (ρ w - ρo)
capillary pressure (oil-water system) Po h
Pw
B' B
Θ
Po A
OIL
WATER
a.
pressure at A' equals pressure at A Poa = Pwa
b.
pressure at B is equal to the pressure at A minus the head of water between A and B Pwb = Pwa - ρ wgh
c.
pressure at B' equal to the pressure at A' minus the head of oil between A' and B'
III- 9
Pob = Poa - ρ ogh d.
thus capillary pressure, the difference between pressure at B' and the pressure at B is Pc
= Pob - Pwb
Pc
= (Poa - ρ ogh) - (Pwa - ρ wgh)
since Poa
= (ρ w - ρ o)gh
Pc e.
remember h=
f.
2 σow cos θ rg (ρ w - ρo)
thus Pc =
4.
= Pwa
2 σow cos θ r
same expression as for the air-solid system except for the boundary tension term θ Pc = 2 σ cos r
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- 10
ADHESION TENSION
AIR
AIR
WATER
AIR
Hg
WATER
1/radius of tube
D.
E.
an important result to remember 1.
pwb < pob
2.
thus, the pressure on the concave side of a curved surface is greater than the pressure on the convex side
3.
or, pressure is greater in the non-wetting phase
capillary pressure-unconsolidated sand 1.
the straight capillary previously discussed is useful for explaining basic concepts - but it is a simple and ideal system
2.
packing of uniform spheres
Pc = σ 1 + 1 R1 R2 R1 and R2 are the principal radii of curvature for a liquid adhering to two spheres in contact with each other. 3.
by analogy to capillary tube 1 + 1 = 2 cos θ r R1 R2 θ where Pc = 2 σ cos r call it Rm(mean radius), i.e. 1 = 2 cos θ = (∆ρ)gh rm Rm σ
III- 11
F.
wettability-consolidated sand 1.
Pendular-ring distribution-wetting phase is not continuous, occupies the small interstices-non-wetting phase is in contact with some of the solid
2.
Funicular distribution - wetting phase is continuous, completely covering surface of solid
WATER
WATER
SAND GRAIN
SAND GRAIN
OIL OR GAS
OIL OR GAS
(A)
(B)
Idealized representation of distribution of wetting and nonwetting fluid phase about intergrain contacts of spheres. (a) Pendular-ring distributions; (b) funicular distribution
III- 12
IV)
Relationship between capillary pressure and saturation A.
remember that the height a liquid will rise in a tube depends on 1. 2. 3.
B.
adhesion fluid density variation of tube diameter with height
consider an experiment in which liquid is allowed to rise in a tube of varying diameter under atmospheric pressure. Pressure in the gas phase is increased forcing the interface to a new equilibrium position. ATMOSPHERIC PRESSURE
HIGHER PRESSURE
R R
DEPENDENCE OF INTERFACIAL CURVATURE ON FLUID SATURATION IN A NON-UNIFORM PORE 1.
Capillary pressure is defined as the pressure difference across the interface.
2.
This illustrates: a.
Capillary pressure is greater for small radius of curvature than for large radius of curvature
b.
An inverse relationship between capillary pressure and wettingphase saturation
c.
Lower wetting-phase saturation results in smaller radius of curvature which means that the wetting phase will occupy smaller pores in reservoir rock
III- 13
V)
Relationship between capillary pressure and saturation history A.
consider an experiment using a non-uniform tube (pore in reservoir rock) 1.
tube is filled with a wetting fluid and allowed to drain until the interface between wetting fluid and non-wetting fluid reaches equilibrium (drainage)
2.
tube is filled with non-wetting fluid and immersed in wetting fluid allowing wetting fluid to imbibe until the interface reaches equilibrium (imbibition)
LOW PC
HIGHER P C
HIGHER P C
LOW PC
R
Θ Θ R
Θ
Θ SATURATION = 100% PC = LOW VALUE
SATURATION = 80% CAPILLARY PRESSURE = P C
(A)
SATURATION = 0% P C = HIGH VALUE
SATURATION = 10% CAPILLARY PRESSURE = P C
(B)
Dependence of equilibrium fluid saturation upon the saturation history in a nonuniform pore. (a) Fluid drains; (b) fluid imbibes. Same pore, same contact angle, same capillary pressure, different saturation history
3.
This is an oversimplified example, however it illustrates that the relationship between wetting-phase saturation and capillary pressure is dependent on the saturation process (saturation history) a.
for given capillary pressure a higher value of wetting-phase saturation will be obtained from drainage than from imbibition
III- 14
B.
Leverett conducted a similar experiment with tubes filled with sand. DATA FROM HEIGHT-SATURATION EXPERIMENTS ON CLEAN SANDS. (FROM LEVERETT) 1.6 Φ
1.4
Drainage
Imbibition Sand I
1.2
Φ
Φ
Sand II
1.0 Φ
σ
∆ρgh
(k/ø)1/2
Φ
0.8 0.6
Drainage
Φ
0.4 0.2 Imbibition 0.0 0
20
40
60
80
100
WATER SATURATION, Sw %
1.
capillary pressure is expressed in terms of a non-dimensional correlating function ( remember Pc = (∆ρ gh )
2.
in general terms, a.
drainage means replacing a wetting fluid with a non-wetting fluid
b.
imbibition means replacing a non-wetting fluid with a wetting fluid
III- 15
DRAINAGE PC
IMBIBITION
0
WATER SATURATION, S W
III- 16
100
VI)
Capillary pressure in reservoir rock Water
Oil
ρw h P w = Po/w 144
P w2
P o2
P o = Po/w -
ρoh
144
Oil and Water P o1 = P w1 100% Water
Pc = P o - Pw = h 144
ρ w - ρo
Where:
Po Pw h Po/w ρw
= = = =
ρo
= density of oil, lb/cf
pressure in oil phase, psia pressure in water phase, psia distance above 100% water level, ft pressure at oil-water contact, psia
= density of water, lb/cf
At any point above the oil-water contact, po ≥ p w
III- 17
P O = PO/W -
PC HEIGHT ABOVE O-W-C
ρ H P w = PO/W - w 144
PRESSURE
III- 18
ρoH 144
VII) Laboratory measurement of capillary pressure A.
B.
Methods 1.
porous diaphragm
2.
mercury injection
3.
centrifuge
4.
dynamic method
Porous diaphragm 1.
Start with core saturated with wetting fluid.
2.
Use pressure to force non-wetting fluid into core-displacing wetting fluid through the porous disk.
3.
The pressure difference between the pressure in the non-wetting fluid and the pressure in the wetting fluid is equal to Pc.
4.
Repeat at successively higher pressures until no more wetting fluid will come out.
5.
Measure Sw periodically.
6.
Results
7.
Advantages a. b.
8.
very accurate can use reservoir fluids
Disadvantages a. b.
very slow - up to 40 days for one core pressure is limited by "displacement pressure" of porous disk
III- 19
C.
Mercury Injection Method 1.
Force mercury into core - mercury is non-wetting phase - air (usually under vacuum) is wetting phase
2.
Measure pressure
3.
Calculate mercury saturation
4.
Advantages a. b.
5.
Disadvantages a. b.
D.
fast-minutes reasonably accurate
ruins core difficult to relate data to oil-water systems
Centrifuge Method
CORE HOLDER BODY WINDOW
TUBE BODY 1.
Similar to porous disk method except centrifugal force (rather than pressure) is applied to the fluids in the core
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. b. c.
fast reasonably accurate use reservoir fluids
III- 20
E.
Dynamic Method
GAS OUTLET
GAS INLET ∆Po
Pc
∆P g
CORE
TO ATMOSPHERE
OIL INLET
OIL BURETTE 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. b.
5.
seems to simulate reservoir conditions reservoir fluids can be used
Disadvantages a.
very tedious
III- 21
F.
Comparison of methods 1.
diaphragm method (restored state) is considered to be most accurate, thus used as standard against which all other methods are compared
2.
comparison of mercury injection data against diaphragm data a.
simple theory shows that capillary pressure by mercury injection should be five times greater than capillary pressure of air-water system by diaphragm method
b.
capillary pressure scale for curves determined by mercury injection is five times greater than scale for diaphragm air-water data
c.
these comparisons plus more complex theory indicate that the ratio between mercury injection data and diaphragm data is about 6.9 (other data indicate value between 5.8 and 7.5)
III- 22
Example VIII-1 Comparison of Mercury Injection Capillary Pressure Data with Porous Diaphragm Data
A.
Calculate capillary pressure ratio, PcAH g , for the following data: PcAW σAHg = 480 Dynes/cm θ AH = 140° g
B.
σAW = 72 Dynes/cm θ AW = 0°
Pore geometry is very complex. The curvature of the interface and pore radius are not necessarily functions of contact angles. Calculate the ratio using the relationship. PcAH g σAHg = PcAW σAW
Solution:
(A)
(B)
PcAH σ cos θAHg 480 cos(140°) g = AHg = PcAW 72 cos (0°) σAWcos θ AW
PcAH g PcAW
=
PcAH g PcAW
@
PcAH g PcAW
=
5.1
σAHg = 480 σAW 70
6.9
III- 23
Discussion: A.
Best way to determine the relationship between mercury and air-water data is to generate capillary pressure curves for each set of data and compare directly.
Mercury Injection and Porous Diaphragm Methods B.
For this given set of conditions, mercury injection method requires a higher displacement pressure, must adjust ratio between scales until match is obtained.
C.
Minimum irreducible wetting phase saturations are the same.
D.
Reduction in permeability results in a higher minimum irreducible wetting phase saturation. For both cases, mercury system still has higher required displacement pressure.
III- 24
VIII) Converting laboratory data to reservoir conditions
PcL
=
2σLcos θ L r
PcR
=
2σRcos θ R r
setting r = r r=
∴
2σLcos θ L 2σRcos θ R = PcL PcR
σcos θ R PcR = PcL σcos θ L
where Pc R
= reservoir capillary pressure, psi
Pc L
= 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- 25
Example III-2 Converting Laboratory Data to Reservoir Conditions Express reservoir capillary pressure by using laboratory data. σAW = 72 dynes
lab data:
σAW = 0o reservoir data:
σOW = 24 dynes/cm σOW = 20o
Solution:
PcR
PcR PcR
=
=
σcos θ R PcL σcos θ L 24 cos20° 72 cos0°
=
P cL
0.333 PcL
III- 26
IX)
Determining water saturation in reservoir from capillary pressure data A.
convert laboratory capillary pressure data to reservoir conditions
B.
calculate capillary pressure in reservoir for various heights above height at which capillary pressure is zero (∆ρ)gh Pc = 144 gc in English units ∆ρ
= ρ w - ρ O, lb/cu ft
g
= 32 ft/sec2
gc
= 32 lbm ft lbf sec2
h
= ft
144
= (sq in)/(sq ft.)
Pc
= lbf/(sq in), psI
thus
III- 27
Example III-3 Determining Water Saturation From Capillary Pressure Curve Given the relationship, PcR = 0.313 P cL, use the laboratory capillary pressure curve to calculate the water saturation in the reservoir at a height of 40 ft. above the oil-water contact. ρo = 0.85 gm/cm3
ρ w = 1.0 gm/cm3
20
P CL 10
8.3
0 0
50
100
SW
Solution: =
ρw − ρ o h 144
PcR
=
1.0 - 0.85 62.4 lb 40 ft3 = 2.6 psi 144
PcL
=
PcR 0.313
PcL
=
2.6 = 8.3 psi 0.313
PcR
move to the right horizontally from PcL = 8.3 psi to the capillary pressure curve. Drop vertically to the x-axis, read Sw. Sw = 50%
III- 28
X)
Capillary pressure variation A.
effect of permeability 1.
displacement pressure increases as permeability decreases
2.
minimum interstitial water saturation increases as permeability decreases
RESERVOIR FLUID DISTRIBUTION CURVES
18
12
6
10 md
90
72
160 140
54
120 100 80
36
60 40
18
20 0
0
0 0
10
20
30
40
50 Sw %
III -29
60
70
80
90
100
Air - Water Capillary Pressure, psi (laboratory data)
24
Height above zero capillary pressure, ft
Oil - Water Capillary Pressure, psi (reservoir conditions)
180
100 md
200
900 md
30
200 md
(From Wright and Wooddy)
30
225.0
25
187.5
20
150.0 Sandstone Core
15
112.5
Porosity = 28.1% Permeability = 1.43 md Factor = 7.5
10
75.0
5
37.5
Mercury capillary pressure, psi
Water/nitrogen capillary pressure, psi
Effect of grain size distribution
0
0 0
20
40
100
80
60
60
80
100
40
20
0
Water
60
348
50
290
40
232
30
174
20
116 Limestone Core Porosity = 23.0% Permeability = 3.36 md Factor = 5.8
10
58
60
80
0 100
40
20
0
0
0
20
40
Mercury capillary pressure, psi
Hg
Water/nitrogen capillary pressure, psi
B.
Water
100
80
60 Hg
1.
majority of grains same size, so most pores are same size - curve (a) (well sorted)
2.
large range in grain and pore sizes - curve (b) (poorly sorted)
III -30
XI)
Averaging capillary pressure data J-function J Sw =
Pc k 1/2 σcos θ φ
attempt to convert all capillary pressure data to a universal curve universal curve impossible to generate due to wide range of differences existing in reservoirs concept useful for given rock type from given reservoir where Pc σ k φ
= = =
dyne/(sq cm) dyne/cm (sq cm)
=
fraction
or can use any units as long as you are consistent
III -31
CAPILLARY RETENTION CURVES.
CAPILLARY PRESSURE FUNCTION, J
(From Rose and Bruce.) 2.0 1.9 1.8 1.7 1.6 1.5 1.4 1.3 1.2 1.1 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0
LEDUC
LEVERETT
KATIE
ALUNDUM
EL ROBLE KINSELLA
HAWKINS
0
10
20
30
40
50
60
70
WATER SATURATION, Sw
Reservoir
Formation
Hawkins El Roble Kinsella Katie Leduc Alundum Leverett
Woodbine Moreno Viking Deese Devonian (consolidated) (unconsolidated)
III -32
80
90
100
Capillary Pressure Problem 1 1.
A glass tube is placed vertically in a beaker of water. The interfacial tension between the air and water is 72 dynes/cm and the contact angle is 0 degree. Calculate:
2.
a.
the capillary rise of water in the tube if the radius of the tube is 0.01 centimeters.
b.
what is the difference in pressure in psi across the air-water interface in the tube.
The displacement pressure for a water saturated porcelain plate is 55 psi of air. What is the diameter in inches of the largest pore in the porcelain plate? Assume 72 dynes/cm and 0 degrees.
Solution: (1)
σAW = 72 dynes/cm ρW
= 1 gm/cm3
g
= 980 dynes/gm
θ
= 0o
(a)
capillary rise of water if radius is .01 cm h=
2 72 cos0° 2σAWcos θ = rρg .01 1.0 980
h = 14.69 cm (b)
pressure drop in psi across interface Pc
Pc Pc
=
pa - pw = ρ wgh = 1.0 980 14.69
=
0.0142 atm
=
0.209 psi
14.696 psi atm
III -33
(2)
Pc
=
2σAWcos θ r
Pc
=
55 psi
Pc
=
55 psi
=
3.792 x 106 dynes/cm2
r
=
2σAW cos θ Pc
r
=
2 72 cos0° in = 3.797 x 10 -5 cm 6 2.54 cm 3.792 x 10
r
=
1.495 x 10-5 in
d
=
2.99 x 10-5 in
atm 14.696 psi
1.0133 x 10 6 dynes/cm2 atm
III -34
Capillary Pressure Problem 2 Given the information below and graph of PcL vs. wetting phase saturation Sw , construct the curves for PcR, h in reservoir, and J-function vs. Sw. Water is the wetting phase in both the laboratory and the reservoir.
fluids
lab air-water
res oil-water
θ
0°
25°
σ
60 dyne/cm
20 dyne/cm
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
Pc k/φ 1/2 J= σ cos θ
III -35
35.0 32.5 30.0 27.5 25.0
PCL, psi
22.5 20.0 17.5 15.0
12.5 10.0 7.5 5.0 2.5
0.0 0
10
20
30
40
50
60
Sw %
Solution:
(1)
=
σR cosθ R Pc σL cosθ L L
=
20 cos25 P 60 cos0 cL
PcR
PcR
=
0.302 P cL
III -36
70
80
90
100
(2) P cR
PcR hR (3)
J
J
=
hR ρw - ρo 144
=
hR 1.1 - .863 62.4 144
=
.103 h R
=
9.74 P cR
=
Pc k 1/2 σ cos θ φ
=
PcL k 1/2 σAW cosθ L φ L
=
PcL 37 1/2 60 cos0° .16
=
.253 P cL
Sw %
PcL psi
PcR psi
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
hR ft
III -37
J assorted
10 8
Pc
R
6 4 2 0 0
20
40
60
80
100
Sw % 100 80
h R
60 40 20 0 0
20
40 60 Sw %
80
100
0
20
40 60 Sw %
80
100
10 8
J
6 4 2 0
III -38
IV. FLUID SATURATIONS I)
II)
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
Methods for determining fluid saturations A.
Core analysis (direct method) 1.
factors affecting fluid saturations a.
flushing by mud filtrate 1)
differential pressure forces mud filtrate into formation Ph > Pres
2)
for water base mud, filtrate displaces formation water and oil from the area around the well (saturations likely change)
3)
for oil base mud, filtrate will be oil; saturations may or may not change.
IV - 1
Example:
Effects of flushing by mud filtrates
Coring with water base mud Oil zone at minimum interstitial water saturation: sat at surface compared to res
flushing by bit trip to surface Sw
↑
↓
? probably ↑
So
↓
↓
↓
Sg
-
↑
↑
Gas zone at minimum interstitial water saturation: sat at surface compared to res
flushing by bit trip to surface Sw
↑
↓
?
So
-
-
-
Sg
↓
↑
?
Water zone: sat at surface compared to res
flushing by bit trip to surface Sw
-
↓
↓
So
-
-
-
Sg
-
↑
↑
IV - 2
Coring With Oil Base Mud Oil zone at minimum interstitial water saturation: sat at surface compared to res
flushing by bit trip to surface Sw
-
-
-
So
-
↓
↓
Sg
-
↑
↑
Gas zone at minimum interstitial water saturation: sat at surface compared to res
flushing by bit trip to surface Sw
-
-
-
So
↑
↓
↑
Sg
↓
↑
↓
Water zone: sat at surface compared to res
flushing by bit trip to surface Sw
↓
↓
↓
So
↑
↓
↑
Sg
-
↑
↑
IV - 3
b.
2.
bringing core to surface 1)
reduction in hydrostatic pressure causes gas to come out of solution
2)
gas displaces oil and water causing saturations to change
laboratory methods a.
evaporation using retort distillation apparatus
HEATING ELEMENT
CORE COOLING WATER IN
CONDENSER COOLING WATER OUT
IV - 4
1)
process a)
heat small sample of rock
b)
oil and water vaporize, then condense in graduated cylinder
c)
record volumes of oil and water
d)
correct quantity of oil
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
1.4
Multiplying Factor
1.3
1.2
1.1
1.0
0.9 15
20
25
30
35
40
45
50
Oil Gravity, °API at 60° F
IV - 5
55
60
65
e)
determine saturations
V Sw = w Vp
V So = o Vp
Sg = 1 - S o - Sw where Sw So Sg Vp Vw Vo 2)
= = = = = =
water saturation, fraction oil saturation, fraction gas saturation, fraction pore volume, cc volume of water collected, cc volume of oil collected, cc
disadvantages of retort process a)
must obtain temperature of 1000-1100oF to vaporize oil, water of crystallization from clays also vaporizes causing increase in water recovery
WATER RECOVERED PORE WATER 0
0
TIME
b)
at high temperatures, oil will crack and coke. (change in hydrocarbon molecules) amount of recoverable liquid decreases.
c)
core sample ruined
IV - 6
3)
b.
advantages of retort process a)
short testing time required
b)
acceptable results obtained
leaching using solvent extraction apparatus WATER OUT
WATER IN
GRADUATED TUBE
CORE
SOLVENT
HEATER
1)
process a)
weigh sample to be extracted
b)
heat applied to system causes water from core to vaporize
c)
solvent leaches hydrocarbons from core
IV - 7
d)
water condenses, collects in trap. Record final water volume
e)
reweigh core sample
f)
determine volume of oil in sample Vo =
Wi - Wdry - Vw ρ w ρo
where: Wi = after leaching Wdry
weight of core sample
= weight of core sample after leaching
V Sw = w Vp 2)
3)
V So = o Vp
disadvantages of leaching a)
process is slow
b)
volume of oil must be calculated
advantages of leaching a)
very accurate water saturation value obtained
b)
heating does not remove water of crystallization
c)
sample can be used for future analysis
IV - 8
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
0
GAS
OIL
WATER
So ≅ 0 in gas zone So ≥ 15% in oil zone 0 ≤ So ≤ Sor in water zone Sor = residual oil saturation
IV - 9
SO
50
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 - 10
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 ρ w = 1 gm/cm3 Wt new = 219.7 gm the water with in the sample is removed and collected Vw removed = 13.0 cm3 the oil is extracted and the sample is dried Wt dry = 199.5 gm calculate: (1)
porosity
(2)
water saturation
(3)
oil saturation assuming 35o API
(4)
gas saturation
(5)
matrix density
(6)
lithology
Solution: gas vol.
=
219.7 - 216.7 ;
Vg
=
3 cc
water vol.
=
13 - 3
Vw
=
10 cc
Wt fluids
=
219.7 - 199.5 =
20.2 gm
Wt oil
=
20.2 - 10 - 3
7.2 gm
;
=
IV - 11
ρo
=
141.5 = 0.85 gm/cc 131.5 + 35°API
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 - 12
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 ρ w = 1.0 gm/cc and 35o API oil) calculate: (1)
porosity
(2)
water saturation
(3)
oil saturation
(4)
gas saturation
(5)
lithology
Solution: Vg
=
.71 cc 100 gm = 1.014 cc 70 gm
Vo
=
4.5 cc 100 gm = 5.63 cc 80 gm
Vw
=
2.8 cc 100 gm = 3.50 cc 80 gm
Vb
=
45.7 cc 100 gm = 43.52 cc 105 gm
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 - 13
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 - 14
Fluid Saturation Problem 1 Calculate porosity, water, oil, and gas saturations, and lithology from the following core analysis data. 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 ρ w = 1.0 gm/cc oil gravity = 40° API
Solution:
(1)
γo
=
141.5 131.5 + °API
γo
=
141.5 = 0.825 131.5 + 40°
ρo
=
0.825 gm/cc
φ
=
Vp Vb
Vp
=
Vw + Vo + Vg
Wo
=
Wsat - Vwρ w - Wdry
=
87.95 - 7.12(1.0) - 79.17
=
1.66 gm
Wo
IV - 15
(2)
(3)
Vo
=
Wo ρo
Vo
=
1.66 gm = 2.01 cc 0.825 gm/cc
Vw
=
Vwrec - Wsat - Wi / ρ 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 ρw
Vb
=
(87.95 - 48.95) gm = 39.0 cc 1 gm/cc
φ
=
9.13 = 23.4% 39.0 Vw Vp
Sw
=
Sw
=
5.57 cc = 61.0% 9.13 cc
So
=
Vo Vp
So
=
2.01 cc = 22.0% 9.13 cc
Sg
=
Sg
=
1.55 cc = 17.0% 9.13 cc
Vm
=
Vb - Vp
Vg Vp
IV - 16
Vm ρm ρm
=
39 - 9.13 = 29.87 cc
=
Wdry Vm
=
gm 79.17 gm/29.87 cc = 2.65 cc
. . lithology is sandstone (4)
water saturation at surface will probably be greater than reservoir water saturation oil saturation at surface will be less than reservoir oil saturation gas saturation at surface will be greater than reservoir gas saturation
IV - 17
Fluid Saturation Problem 2 Calculate porosity, water saturation, oil saturation, gas saturation, and lithology from the following core 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 water and 0.87 cc of oil Sample 3 weighed 50 gm and contained 0.40 cc of gas space assume ρ w = 1.0 gm/cc oil gravity = 40o API
Solution:
(1)
141.5 131.5 + °API
γo
=
γo
=
141.5 = 0.825 131.5 + 40°
ρo
=
0.825 gm/cc
φ
=
Vp Vb
Vp
=
Vo + Vw +Vg
Vo
=
0.87 cc x 100 = 1.00 cc 86.71 gm 100 gm
Vw
=
1.90 cc x 100 = 2.19 cc 86.71 gm 100 gm
Vg
=
0.40 cc x 100 = 0.80 cc 50 gm 100 gm
Vp
=
(1.00 + 2.19 + 0.80) cc/100 gm
IV - 18
(2)
Vp
=
3.99 cc/100 gm
Vb
=
51.72 cc x 100 = 39.78 cc 130 gm 100 gm
φ
=
3.99/100 x 100 = 10% 39.78/100
Sw
Sw
(3)
So
So
(4)
(5)
Sg
=
Vw x 100 Vp
=
2.19 cc x 100 3.99 cc
=
54.8%
=
Vo x 100 Vp
=
1.0 cc x 100 3.99 cc
=
25.1%
=
Vg x 100 Vp
=
0.80 cc x 100 3.99 cc
Sg
=
20.1%
Vm
=
Vb - Vp
=
39.78 - 3.99
Vm
=
35.79 cc/100 gm
Wm
=
Wsat - ρ oVo - ρ wVw /Wsat
=
86.71 - 0.825 0.87 - 1.0 1.90 86.71
=
97 gm/100 gm
Wm
IV - 19
(6)
ρm
=
Wm Vm
ρm
=
97 gm/100 gm 35.79 cc/100 gm
ρm
=
2.71 gm/cc
..
lithology - limestone
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
IV - 20
V. ELECTRICAL PROPERTIES OF ROCK-FLUID SYSTEMS I)
Electrical conductivity of fluid saturated rock A.
Definition of resistivity
ELECTRICAL CURRENT FLOW
A
L
given a box of length (L) and cross-sectional area (A) completely filled with brine of resistivity (Rw) the resistance of the brine in the box to current flow may be expressed as r = Rw L A r
=
resistance - ohm
Rw =
resistivity - ohm meters
L
=
length - meters
A
=
area - (meters)2
V-1
B.
C.
D.
Nonconductors of electricity 1.
oil
2.
gas
3.
pure water
4.
minerals
5.
rock fragments
Conductors of electricity 1.
water with dissolved salts conducts electricity (low resistance)
2.
clay
Development of saturation equation (ignore clay) AP
A
ELECTRICAL CURRENT FLOW
L
1.
the electrical current flows through the water (brine) a.
the area available for current flow is the cross-sectional area of the pores. Ap < A
b.
the path through the pores is Lp. Lp > L
V-2
2.
resistance to electrical flow through the porous media is equal to the resistance of a container of area Ap and length Lp filled with water (brine) r=
RwLp , water filled cube Ap
R L r = o , porous media A 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)
3.
Ro
= resistivity of formation 100% saturated with brine of resistivity, Rw, ohm-m
Ap
= cross-sectional area available for current flow, m 2
Lp
= actual path length ion (current) must travel through rock, m
A
= m2
cross-sectional area of porous media,
L
=
length of porous media, m
Since Ap ≅ porosity, φ A and Lp ≅ L tortuosity, a measure of rock cementation. then Ro =
RwALp ApL
V-3
becomes Ro = f Rw, φ, tortuosity E.
Electrical formation resistivity factor, F 1.
the equation for resistivity of a formation 100% saturated with a brine of resistivity of Rw Ro = f Rw, φ, tortuosity
2.
can be written as Ro = F R w where F is the electrical formation resistivity factor F=
3.
Ro Rw
cementation factor, m a.
it has been found experimentally that the equation for F takes the form F = C φ-m where C is a constant m is the cementation factor
b.
thus log F = log C - m log φ
V-4
100
F
10
1 0.01
0.1 φ
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) Very slightly cemented (Gulf Coast type of sand, except Wilcox) Slightly cemented (most sands with 20% porosity or more) Moderately cemented (highly consolidated sands of 15% porosity or less) Highly cemented (low-porosity sands, quartzite, limestone,dolomite)
1.3 1.4-1.5 1.6-1.7 1.8-1.9 2.0-2.2
V-5
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 with this 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=
1 42.1
φ = 15.41%
Humble:
F = 0.62/φ2.15 φ2.15 = 0.62/F φ = 2.15 0.62 42.1 φ = 14.06%
The Humble equation was developed for sandstone.
V-6
F.
Resistivity Index, I, and Saturation Exponent, n 1.
definition of resistivity index Rt Ro
I=
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 -1 R -1 Sw = I n = t n Ro where n is the saturation exponent ≅ 2.0
3.
rearrange Sw-n =
Rt Ro
-n log Sw = log
Rt Ro
V-7
100
Rt
10
Ro
1 .1
Sw
1.0
slope = -n, when n is the saturation exponent
NOTE: slope =
II)
log y1 - log y2 log x1 - log x2
Use of Electrical Formation Resistivity Factor, Cementation Factor, and Saturation Exponent A.
obtain porosity, φ, from electric log or core analysis
B.
F=Cφ
C.
obtain water resistivity, Rw, from water sample or electric log
D.
Ro = F R w
E.
convert Rt from electric log to water saturation
-m
(usually use Archie or Humble equation)
Ro 1n Sw = Rt
V-8
III)
Laboratory measurement of electrical properties of rock A.
Apparatus
AC SOURCE
1000 OHM STD. RESISTOR
CORE
VOLTMETER
B.
Calculations 1.
resistance of core E = Ir where: E
=
voltage, volts
I
=
current, amperes
r
=
resistance, ohms
∴ rcore = E I
V-9
2.
resistivity of core r A Rcore = core L substituting r = E into the equation I Rcore = EA IL
C.
Procedure 1.
2.
determine resistance of core a.
set desired current from AC source, low current preferred so core does not heat up.
b.
record voltage from voltmeter
determine resistivity of core a.
for the first test completely saturate core with brine Sw = 100%, R core = Ro
b.
for next test, desaturate core by 15-20%, until Sw < 100% Rcore = Rt
c.
repeat tests until Sw = S wir
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
Rt Ro
=
resistivity index = I
V - 10
D.
Determine saturation exponent, n 1.
rearrange saturation equation
Sw
Swn Ro Rt log
2.
=
Ro 1/n Rt
=
Ro Rt
=
Sw-n
Rt = -n log Sw Ro
Plot log
Rt vs log Sw or log I vs log Sw Ro
100
I=
Rt Ro
10
1 .1
3.
Sw
1.0
the slope of the plot is -n, where n is the saturation exponent
V - 11
Example V-2 Given the following data, calculate the electrical formation resistivity factor and saturation exponent of the core. Rw
=
55 ohm-cm
I
=
0.01 amp
D
=
2.54 cm
L
=
3.2 cm E Voltage across Core, volts 7.64 10.50 14.34 20.16 27.52 34.67
Sw Water Saturation, % 100.0 86.0 74.0 63.0 54.0 49.0 = Swir Solution: (1)
electrical formation resistivity factor F
=
Ro Rw
ro
=
E = 7.64 = 764 ohm I .01
Ro
=
roA 764 2.54 2π/4 = = 1210 ohm cm L 3.2
F
=
Ro 1210 = = 22 Rw 55
V - 12
saturation exponent
-n log Sw = log
Sw % 1.00 .86 .74 .63 .54 .49
Rt Ro
rt = E I (ohm)
rA Rt = t L (ohm-cm)
Rt Ro
1050 1434 2016 2452 3467
1663 2271 3192 4358 5490
1.000 1.374 1.877 2.638 3.601 4.537
(.334,10)
Rt/Ro
10
1 .1
1 Sw
V - 13
(1.0,1.0)
-n
=
slope
=
log 10 - log 1 log .334 - log 1
-n
=
1-0 -.4763 - 0
=
2.10
n
=
saturation exponent
NOTE: Rt Et = Ro Eo Rt could have been calculated as the ratio of voltage at Ro Sw divided by the voltage at Sw = 1.0 so
V - 14
E.
Determine cementation factor, m, and constant C for electrical formation resistivity 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 F=
c. 2.
Ro Rw
plot data according to form of equation for electrical formation resistivity factor F = C φ-m log F = log C - mlog φ
100
F
10
1 0.01
0.1 φ
slope = -m, m = cementation factor intercept = C (intercept found at φ = 1.0)
V - 15
1.0
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 resistivity factor equation. Porosity Formation Resistivity Factor φ
F
0.152 0.168 0.184 0.199 0.213 0.224
40 32 26 22 19 17
Solution: F = C φ-m log F = log C - m log φ plot log F vs log φ
F
100
10
1 .1
1 ø
V - 16
slope =
log 50 - log 10 = -2.21 log 0.137 - log 0.284
-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 - 17
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 1 = 1 + 1 RoA Rclay Ro 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
1 = 1 + 1 RoA Rclay F 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 - 18
3.
plot
1 ROA (OHM - M) -1
1 -1 RW (OHM - M)
1 = 1 + 1 1 RoA Rclay F Rw where 1 = intercept Rclay 1 = slope F
V - 19
D.
effect of clay
1.
define factor
FA =
RoA Rw , clays reduced the apparent formation resistivity
CLEAN SAND
F
SHALY SAND FA
RW 2.
formation resistivity factor decreases more gradually when clay is present in the formation
100 CLEAN SAND
F 10 SHALY SAND
1
0.1
φ
1.0
V - 20
3.
saturation exponent n is not constant when clay is present in formation.
100 CLEAN SAND LOW R w n=2
I=
Rt Ro
CLEAN SAND Swn-1 = I
10 SHALY SAND n =?
1 .1
CLEAN SAND HIGH R w n=1
Sw
V - 21
1.0
VI. MULTIPHASE FLOW IN POROUS ROCK I)
Effective permeability A.
Permeability, k, previously discussed applies only to flow when pores are 100% saturated with one fluid - sometimes called absolute permeability
q= B.
kA∆ρ µL
When pore space contains more than one fluid, the above equation becomes k A∆Pο qo = o µoL k A∆Pw qw = w µw L qg =
where
and
kgA∆Pg µgL qo
=
flow rate of oil, volume/time
qw
=
flow rate of water, volume/time
qg
=
flow rate of gas, volume/time
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 of porous media to a particular fluid when the porous media is saturated with more 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 - 1
E.
Darcy equation for multiple fluids in linear flow, in oilfield units k A P1 - P2 qo = 1.1271 x 10-3 o µoL k A P1 - P2 w qw = 1.1271 x 10-3 w µw L qg = 1.1271 x 10-3 when
II)
kg A P1 - P2 g µgL
k =
md
A =
ft2
P =
psia
L =
ft
q
res bbl/day
=
Relative permeability A.
Defined as the ratio of the effective permeability to a fluid at a given saturation to the effective permeability to that fluid at 100% saturated (absolute permeability) k kro = o k k krw = w k krg =
B.
III)
kg k
It is normally assumed that the effective permeability at 100% saturation is the same for all fluid in a particular rock. (not necessarily true in shaly sand)
Typical relative permeability curves A.
Use subscript wp to represent the "wetting phase" Use subscript nwp to represent the "non-wetting phase"
VI - 2
1
1
NON-WETTING PHASE
Kr
2
WETTING PHASE
3 0
4
0
SWP, % MINIMUM INTERSTITIAL S WP
100 EQUILIBRIUM S NWP
1.
krwp
2.
k rapid decrease in rwp as Swp decreases
3.
krwp
4.
krnwp
= 1, only at S wp = 100%
= 0, at minimum interstitial Swp = 0, at equilibrium Snwp
Note that krwp + krnwp < 1.0
VI - 3
B.
Effect of saturation history 1.
2.
two types of relative permeability curves a.
drainage curve - wetting phase is displaced by non-wetting phase, i.e., wetting phase saturation is decreasing
b.
imbibition curve - non-wetting phase is displaced by wetting phase, i.e., wetting phase saturation is increasing
the typical relative permeability curve shown below represents a process in which a.
process begins with porous rock 100% saturated with wetting phase (Swp = 100%)
b.
wetting phase is displaced with non-wetting phase (drainage) until wetting phase ceases to flow (Swp = minimum interstitial 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 - 4
1
Krnwp DRAINAGE
Kr
IMBIBITION
Krwp
0
0
SWP, %
100
minimum interstitial
residual non-wetting
wetting phase saturation
phase saturation
VI - 5
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 in laboratory must correspond to the process in the reservoir
5.
a.
initial distribution of fluids in reservoir was by drainage
b.
at and behind a water front (flood or encroachment) the process is imbibition
wetting preference for reservoir rocks is usually water first, then oil, finally gas
Fluids Present
Wetting Phase
Water & Oil Water & Gas Oil & Gas
Water Water Oil
VI - 6
C.
Three phase relative permeability 1.
often three phases are present in petroleum reservoirs
2.
tertiary (triangular) diagram is used to represent a threephase system
100% GAS
100% WATER
100% OIL
VI - 7
3.
relative permeability to oil in a three phase system
100% GAS
1%
5
10 20
40 60
100% WATER
100% OIL
Note, kro is shown in % a.
dependence of relative permeability to oil on saturations of other phases is established as follows: 1)
oil phase has a greater tendency than gas to wet the solid
2)
interfacial tension between water and oil is less than that between water and gas
3)
oil occupies portions of pore adjacent to water
4)
at lower water saturations the oil occupies more of the smaller pores. The extended flow path length accounts for the change in relative permeability to oil at constant oil saturation and varying water saturation
VI - 8
4.
Relative permeability to water in a three-phase system
100% GAS
0
Krw
10% 20% 40%
60% 80% 100% WATER
100% OIL
a.
straight lines indicate relative permeability to water is a function of water saturation only
b.
thus, krw can be plotted on cartesian coordinates against Sw.
VI - 9
5.
Relative permeability to gas in a three-phase system 100% GAS
50%
40
30
20 5
1
100% WATER
100% OIL
a.
k curves above indicate that rg is a function of saturations of other phases present.
b.
k other research shows that rg is a unique function of gas saturation
c.
the other phases, oil and water, occupy the smaller pore openings and wet the surface of the rock
d.
e.
k therefore, rg should be dependent only on the total saturation of the other two phases (i.e. 1-Sg) and independent of how much of that total is composed of either phase k thus rg can be plotted on Cartesian coordinates against So + Sw
VI - 10
1.0
0.8
0.6
krg 0.4
0.2
0.0 0
20
40
60
80
100
So + Sw 6.
Bottom line - for three-phase system in water wetted rock a.
b.
water 1)
is located in smaller pore spaces and along sand grains
2)
therefore krw is a function of Sw only
3)
thus plot krw against Sw on rectangular coordinates
gas 1)
is located in center of larger pores
2)
k therefore rg is a function of Sg only
3)
k thus plot rg against Sg (or So + Sw) on rectangular coordinate
VI - 11
c.
oil 1)
is located between water and gas in the pores and to a certain extent in the smaller pore spaces
2)
therefore kro is a function of So, S w, and S g
3)
thus plot kro against So, S w, S g on a triangular diagram
4)
if Sw can be considered to be constant (minimum interstitial) kro can be plotted against So on a rectangular diagram
1.0
0.8
0.6
kro 0.4
0.2
0.0 0
20
40
60
80
100
So, % Minimum Interstitial Water Saturation
VI - 12
7.
Flow in three-phase system
100% GAS
5% oil
5% gas
5% water
100% WATER
100% OIL
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 - 13
IV)
Permeability ratio (relative permeability ratio)
A.
Definitions 1.
When the permeability to water is zero (as at minimum interstitial water saturation) it is sometimes convenient to use permeability ratio to represent the flow conductance of the rock to gas and oil as a ratio.
permeability ratio =
2.
kg krg = ko kro
When the permeability to gas is zero (no gas or gas below "critical gas saturation") it is sometimes convenient to use permeability ratio to represent the flow conductance of the rock to oil and water as a ratio kr k permeability ratio = o = o kw krw
V)
Measurement of relative permeability A.
B.
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
Laboratory Methods 1.
Steady-state flow process a.
saturate core with wetting-phase fluid
b.
inject wetting-phase fluid through core (this will determine absolute permeability)
VI - 14
c.
inject a mix of wetting-phase and non-wetting phase (start with small fraction of non-wetting phase)
d.
when inflow and outflow rates and portion of non-wetting phase equalize, record inlet pressure, outlet pressure and flow rates of each phase
e.
measure fluid saturation in core (see below)
f.
calculate relative permeability q µ L ko = o o A∆p q µ L kw = w w A∆p
g.
repeat b through f with injection mixtures containing relatively more non-wetting phase until irreducible wettingphase saturation is reached
1
kro
kr
krw
0 0
Sw, %
VI - 15
100
h.
determination of fluid saturations 1)
resistivity
Sw =
Ro 1n Eo 1n = Rt Et
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 wetting phase less than 100%, volts
2) volumetric balance 3) gravimetric method - remove core and weigh it
where:
Wf
=
Wt - Wd
Wf
=
weight of fluid in core, gm
Wt
=
weight of saturated core, gm
Wd =
weight of dry core, gm
Wf
=
ρ oVo + ρ wVw
Vf
=
Vo + Vw
ρ
=
density, gm/cc
V
=
volume, cc
and
where:
where:
Sw =
Vw/Vf
Sw =
saturation of wetting phase
thus
VI - 16
Sw =
i.
j.
Wf/Vf - ρ o ρw - ρo
same procedure can be used starting with 100% saturation of non-wetting phase 1)
injection ratio start with high ratio of non-wetting phase
2)
procedure ends at residual non-wetting phase saturation
3)
then is a hysteresis effect of same type as discussed with capillary pressure measurements
4)
choice of starting saturation depends on reservoir process which is being simulated
end effects 1)
causes of end effects a)
in the bulk of the core there is a wettingphase saturation and a non-wetting phase saturation, therefore there is a finite value of capillary pressure
b)
thus there is a difference in pressure between the wetting-phase and non-wetting phase Pcap = P nwp - Pwp
c)
at the face of the core the pressures in the wetting-phase and the non-wetting phase are essentially equal Pnwp = P wp
thus capillary pressure is essentially zero d)
if capillary pressure is zero, the saturation of the wetting phase must be 100% (see capillary pressure curve)
e)
there must be a saturation gradient from essential 100% wetting phase at the "end" to some value of Swp less than 100% in the bulk of the core
VI - 17
100
100
Theoretical saturation gradient
80
Oil Saturation, %
Oil Saturation, %
80
Theoretical saturation gradient
60
Inflow face 40
60
Inflow face 40
20
20
0
0
0
5
10
15
20
0
25
5
Distance from outflow face, cm
2)
End Section
15
20
Distance from outflow face, cm
elimination of end effects a)
install end pieces to contain end effects
b)
flow at rapid rates to make end effect negligible (pressure gradient > 2 psi/inch
Thermometer Packing Nut
10
Electrodes
Test Section
Copper Orifice Plate
Inlet
Mixing Section
Differential Pressure Taps Bronze Screen
Highly permeable disk
Outlet
Inlet
PENN STATE RELATIVE-PERMEABILITY APPARATUS
VI - 18
25
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 length = 2.30 cm diameter = 1.85 cm area = 2.688 cm2
brine, 60,000 ppm oil, 40oAPI µw = 1.07 cp µo = 5.50 cp
porosity = 25.5% Oil Flow cc/sec
Water Flow cc/sec
Inlet Pressure psig
Outlet Pressure psig
Voltage Drop volts
Electrical Current amps
0.0000 0.0105 0.0354 0.0794 0.1771 0.2998
1.1003 0.8898 0.7650 0.3206 0.1227 0.0000
38.4 67.5 88.1 78.2 85.6 78.4
7.7 13.5 17.6 15.6 17.1 15.7
1.20 2.10 2.80 4.56 8.67 30.00
0.01 0.01 0.01 0.01 0.01 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 2.30 2.688 38.4 - 7.7 14.696
k
=
0.482 darcy
VI - 19
2.
Calculate effective permeabilities to oil and water qoµoL ko
=
A∆P
ko
=
0.0105 5.50 2.30 2.688 67.5 - 13.5 / 14.696
ko
=
0.0134 darcy qwµwL
kw
3.
=
A∆P
kw
=
0.8898 1.07 2.30 2.688 67.5 - 13.5 / 14.696
kw
=
0.2217 darcy
Calculate relative permeabilities k kro = o = .0134 = 0.028 k .482 k krw = w = .2217 = 0.460 k .482
4.
Calculate water saturations
5.
Sw
=
Eo 1/2 Et
Sw
=
1.20 1/2 = .756 2.10
Results
Water Saturation Sw
Relative Permeability to oil kro
Relative Permeability to water krw
ko/kw
1.000 0.756 0.655 0.513 0.372 0.200
0.000 0.028 0.072 0.182 0.371 0.686
1.000 0.460 0.303 0.143 0.050 0.000
0.000 0.061 0.238 1.273 7.419 -------
VI - 20
1.0
Relative Permeability
0.8
0.6
Kro Krw
0.4
0.2
0.0 0
20
40
60
80
Sw, % pore space
Permeability Ratio, ko/kw
10
1
.1
.01
0
20
40
60
Sw, % of pore space
VI - 21
80
100
100
6.
The data permit certain checks to be made
0.62 φ-2.15
F
=
F
=
Ro Rw
Rw
=
12 ohm-m for 60,000 ppm brine
Ro
=
EA = 1.20 2.688 = 140 ohm-m IL .01 2.3
F
=
140 = 11.7 12
Φ
=
1 1 .62 2.15 = .62 2.15 F 11.7
Φ
=
.255
VI - 22
2.
Displacement (unsteady-state)(Welge) a.
does not result in relative permeability only give permeability ratio
b.
procedure
c.
1)
mount core in holder
2)
saturate with wetting phase (usually oil)
3)
inject non-wetting phase (usually gas) at constant inlet and outlet pressures
4)
measure a)
cumulative gas injected as a function of time
b)
cumulative oil produced as a function of time
conditions 1)
pressure drop across core high enough to make end effects negligible,but not enough to cause turbulent (non-darcy) flow.
2)
gas saturation can be described at mean pressure P + Po Pm = i 2
3) d.
flow is horizontal and core is short so that effects of gravity can be neglected
calculations 1)
convert gas injected into pore volumes
Gipv =
where Gi
Gipi LAφ pm = cumulative gas injected (measured at pressure pi), cc
Gipv = cumulative gas injected in pore volume pi
VI - 23
=
inlet pressure, psi
pm
=
LA φ = 2)
pore volume, cc
calculate average gas saturation, Sgav =
3)
pi + po , psi 2
Np
Sgav
LAφ
where Np
=
cumulative oil produced, cc
LA φ
=
pore volume, cc
plot Sgav vs G ipv
Sgav GAS BREAKTHROUGH
0 0
Gipv
VI - 24
4)
determine fractional flow of oil, fo fo =
d Sgav d Gipv
fo = slope of plot of Sgav vs qGipv 5)
calculate permeability ratio, kg/ko koA ∆p µoL fo = koA ∆p kgA ∆p + µoL µgL fo =
ko/µo ko/µo + kg/µg
kg 1 - fo = ko fo µo/µg where
kg = permeability ratio of gas to oil ko
fo = fractional flow of oil 6)
Permeability ratio, kg/ko, calculated above applies only at the gas saturation of the outflow face, thus must calculate Sgo Sgo = S gav - Gipvfo where
Sgo Gipv fo
VI - 25
=
gas saturation at outlet face of core
= cumulative gas injected, pore volumes =
fractional flow of oil at outlet face of core
e.
f.
advantages 1)
minimum amount of equipment
2)
rapid
disadvantages 1) 2)
k results in kg/ko, not kro and rg equations don't apply until gas breaks through, thus initial value of gas saturation may be high, resulting in incomplete kg/ko vs Sgo curve.
VI - 26
Example VI-2 The data from an unsteady-state displacement of oil by gas in a 2 inch diameter by 5 5/8 inch long sandstone core are given below.
Cumulative Gas Injection, Gi, cc
Cumulative Oil Produced, Np, cc
14.0 50.2 112.6 202.3 401.4 546.9 769.9 1226.5 3068.9 5946.6
14.6 19.5 22.5 25.5 28.6 30.4 32.2 33.4 35.3 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 cm 2 G Prepare to determine kg/ko by calculating Sgav and ipv. Solution: 1.
Calculate Sgav Sgav = Sgav =
Np LAφ 14.6 cc 14.3 cm 20.27 cm2 .210
Sgav = 0.24 2.
G Calculate ipv
VI - 27
Gipv = Gipv =
Gipi LAφ pm 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 Sgav 0.24 0.32 0.37 0.42 0.47 0.50 0.53 0.55 0.58 0.59
Gipv pv 0.264 0.945 2.12 3.81 7.56 10.3 14.5 23.1 57.8 112.0
VI - 28
Example VI-3 A core sample initially saturated with oil is flooded with gas. The following data was obtained: Gipv pv
Sgav 0.24 0.32 0.37 0.42 0.47 0.50 0.53 0.55 0.58 0.59
0.264 0.945 2.12 3.81 7.56 10.3 14.5 23.1 57.8 112.0
µo = 2.25 cp µg = 0.0185 cp S Calculate and construct a fg verses Sgo plot. Convert gavg to Sgo. Determine kg/ko for each of the given saturations. Construct a graph of kg/ko versus S go. Solution: Plot Sgav vs. Gipv The slope from this plot is fo. Sgo = S gav - fogipv kg/ko =
1 - fo µ fo o µo
Sgav
Gipv pv
0.24 0.32 0.37 0.42 0.47 0.50 0.53 0.55 0.58 0.59
.264 .94 2.17 3.81 7.56 10.3 14.5 23.1 57.8 112.0
fo .375 .075 .0357 .0214 .0118 .0092 .0046 .0013 .0005 .0001
VI - 29
Sgo .141 .249 .294 .338 .381 .405 .463 .521 .550 .581
kg/ko .0137 .101 .222 .376 .689 .886 1.78 6.32 16.4 82.2
0.6
0.5
Sgav
0.4
fraction
0.3
0.2 0
20
40
60
80
100
120
Gipv, pv 100
10
kg/ko
1
.1
.01 0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
Sg, %
VI - 30
0.8
0.9
1.0
C.
Field determination of permeability ratios 1.
equations kg A ∆p µgL
qg qo = k A ∆p o µoL where qg qo
= gas flow rate measured at reservoir conditions, vol/time = oil flow rate measured at reservoir conditions, vol/time
thus, kg qg µg = ko qo µo replace qg/qo with qg Bg Rp - Rs qo = 5.615 B o 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 kg µgBg Rp - Rs = ko µo 5.615 Bo 2.
procedure a.
producing gas-oil ratio, Rp, and physical properties, Bg, Bo, R s, µg, µo must be determined at some known reservoir pressure
b.
saturations in reservoir, Sg or So, must be calculated from production data and material balance calculations
VI - 31
Example VI-4
Discovery pressure for your well was 4250 psia, temperature is 200oF, and initial producing gasoil 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 is now 3000 psia. Producing gas-oil ratio is 18,100 SCF/STB. What is kg/ko in the reservoir at this time. 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
kg ko
=
µgBg Rp - Rs µo 5.615 Bo
kg ko
=
0.0192 5.34 x 10 -3 18100 - 560 0.75 5.615 1.314
kg ko
=
0.325
VI - 32
VI)
Uses of relative permeability data A.
Determination of free water surface in reservoir (100% water production)
1
kr
Log Response Diagram
0 0
Sw, %
100
SP Log
h, ft 100% Water Production
100 % Sw
0 0
Sw, %
100
VI - 33
RT Log
B.
Determination of height of 100% oil production
1
kr
0 0
Sw, %
Log Response Diagram
100
SP Log
100% Oil Production
100% Water Production h, ft
100 % Sw
0 0
Sw, %
100
VI - 34
RT Log
C.
Effect of permeability on thickness of transition zone 1
Low K
kr
High K
0 0
Sw, %
100
h = height of zone of interest
h, ft
Low K High K
0 0
Sw, %
VI - 35
100
D.
Fractional flow of water as a function of height
kw A ∆ P µw L
q q 1 fw = q w = q +wq = = tot o w ko A ∆ P kw A ∆ P ko µw 1+ + kw µo µo L µo L 1
fw
100
0 0
Sw
100 h
100
0
h
0 0
Sw
100
VI - 36
0 fw
1
160
10 md 140
Height above free water level, ft
120 100 80
50 md 60 40
100 md 20
200 md
0 0
20
40
60
80
100
Fraction of water in produced fluid, %
This figure indicates that lower permeabilities result in longer transition zones E.
Determination of residual fluid saturations 1
Oil Water
kr
0 0
100
Sw 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 - 37
F.
Interpretation of fractional flow curve
1
1 4
fw
2
0 0
Sw
3
100
1.
fw at water breakthrough
2.
Sw at well at water breakthrough
3.
Swav in reservoir between wells at water breakthrough
4.
1 = pore volume of water injected slope
VI - 38
1
3
2 Water input Pore vols.
fw
1
0
0 0
100 Oil Rec - % Oil in Place
VI - 39
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 - 1
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 - 2
Histogram A histogram is a graphical representation of a frequency distribution.
Frequency of Occurence
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
8
# of Samples
6
4
2
0
12
16
20
24
Porosity, %
8
6
# of Samples
III)
4
2
0 20
60
100
Permeability, md
VII - 3
140
Net pay thickness data from 20 wells summarized as relative frequency data
Frequency (No. of wells having thickness values in the range)
Range of thickness, ft. 50-80 81-110 111-140 141-170 171-200
Relative Frequency (No. of wells having thickness values in each range, fraction of total wells)
4 7 5 3 1 20
0.20 0.35 0.25 0.15 0.05 1.00
10
Frequency
8 6 4 2
0 0
50
80
110
140
170
200
Random variable: net pay thickness, ft
VII - 4
Relative Frequency as percentage. 20% 35% 25% 15% 5% 100%
Sometimes the relative frequency is plotted on a histogram
Relative Frequency
.5 .4 .3 .2 .1 0 0
50
80
110
140
170
200
Random variable: net pay thickness, ft
VII - 5
IV)
Cumulative Frequency Distributions Relative frequencies are summed and plotted at the higher ends of the class intervals
1.0 .8 .6 .4 .2 0
Cumulative % less than or equal to
Cumulative frequency less than for equal to
to create a "cumulative frequency less than or equal to" distribution
100% 80% 60% 40% 20% 0 0
50
80
110
140
170
200
Random variable: net pay thickness, ft
VII - 6
Occasionally a "cumulative frequency greater than or equal to" distribution is plotted. Relative frequencies are summed from the highest class interval and plotted
1.0 .8 .6 .4 .2 0
Cumulative % greater than or equal to
Cumulative frequency greater than for equal to
at the lower ends of the intervals
100% 80% 60% 40% 20% 0 0
50
80
110
140
170
200
Random variable: net pay thickness, ft
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 - 7
V)
Normal Distribution The normal distribution is continuous probability distribution having a symmetrical shape similar to a bell, sometimes called a Gaussian distribution.
f(x) Inflection point of curve a
µ −a
a
µ +a
Random variable x
This distribution is completely and uniquely defined by two values - the mean, m, and standard deviation, σ.
VII - 8
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.
Mode f(x)
Median (geometric mean) Mean (arithmetic mean)
Random variable x
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 - 9
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 n Σ xi µ = i =1 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 . . . x n 1/n n 1/n µg = π xi i=1 where µg = the geometric mean
VII - 10
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. n Σ xi - µ 2 i=1 2 σ = n-1 where σ2 is the variance σ is the standard deviation
B.
Mean deviation - another measure of the dispersion about the central tendency n Σ xi - µ MD = i=1 n For classified data σ2 =
where
Σ fi xi - µ 2 j Σ fi j fi
=
frequency for each class
xj
=
class mark
or σ2 = Σ frj xj - µ 2 j f where rj = relative frequency for each class
VII - 11
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 n Σ xi µ = i=1n where i refers to each individual data point and, for classified data
µ=
Σ fj xj j Σ fj j
where j refers to each class interval fj is the frequency of the class xj is the class mark or µ = Σ frj xj j f where rj is the relative frequency of the class. xj is the class mark
VII - 12
.999 Cumulative Frequency
Cumulative Frequency
.999
.001
.001 Random variable x
Random variable x
Cumulative frequency plotted on coordinate graph paper
Cumulative frequency plotted on normal probability paper
Normal probability graph paper .999 Random variable x, distributed normally
Random variable x, distributed normally
.999
µ
.001
µ+a σ µ
.001 50%
50%
Cumulative % <
84.1%
Cumulative % <
VII - 13
Porosity and permeability data from a well in the Denver-Julesburg Basin
i 1 2 3 4 5 6 7 8
Porosity Interval Percent
Frequency fi
7.0
µ=
1 0 1 10 12 8 7 3 42
Σ fi xi
σ2 =
Σ fi
Interval Midpoint xi 8.5 11.0 13.0 15.0 17.0 19.0 21.0 23.5
fixi
(xi - µ)
(xi - µ)2
fi(xi - µ)2
8.5 0.0 13.0 150.0 204.0 152.0 147.0 70.5 745.0
-9.2 -6.7 -4.7 -2.7 -0.7 +1.3 +3.3 +5.8
84.64 44.89 22.09 7.29 0.49 1.69 10.89 33.64
84.64 0.00 22.09 72.90 5.88 13.52 76.23 100.92 376.18
= 745.0 = 17.7% 42
Σ fi xi - µ 2 j Σ fi j
= 376.18 = 8.96 42
σ = 8.96 = 2.99%
VII - 14
Porosity Interval, % 7.0
Frequency
Cumulative Frequency Than or Equal to Upper Limit of Interval
Cumulative Frequency Expressed as Percentage
1 0 1 10 12 8 7 3
1 1 2 12 24 32 39 42
2.4% 2.4% 4.8% 28.6% 57.1% 76.2% 92.9% 100.0%
42 26
24
22
Core porosity, %
20
18 16
14
12
10
8
2
10
20 30 40 50 60
70 80
90
96
Cumulative % less than or equal to given values of porosity
at 50th percentile at 84th percentile
f = 17.7% f + σ = 20.7% σ = 20.7 - 17.7 = 3%
VII - 15
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
n 1/n µ = π xi i=1
or n log (m) = 1 Σ log xi n i=1
VII - 16
Permeability Interval (millidarcies)
Frequency
0-50 51-100 101-150 151-200 201-250 251-300 301-350 351-400 401-450 451-500 501-700 701-1000
Cumulative Frequency Less Than or Equal to Upper Limit of Interval
Cumulative Frequency Expressed As Percentage
2 4 8 12 16 24 28 30 34 35 40 42
4.8% 9.5% 19.0% 28.6% 38.1% 57.1% 66.7% 71.4% 81.0% 83.3% 95.2% 100.0%
2 2 4 4 4 8 4 2 4 1 5 2 42 Lognormal probability graph paper
Core permeability, md
1000
100
10 2
10 20 30 40 50 60 70 80
90
96
Cumulative % less than or equal to given values of permeability
VII - 17
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, P c = 2 σ cosθ r
2.
Capillary flow: Poiseuille's law qi =
3.
π r4 i ∆p 8µL
Darcy's equation, qt = kA ∆p µL P c = dynes cm2
k
= cm2
σ = dynes cm
A
= cm2
r
L
= cm
=
2 q = cm sec
cm
µ = poise = dynes-sec cm2
∆p = dyne cm2 1
2 Let V i = πr i L,
then the flow rate in a capillary is 2
qi = Vr i ∆p, V = cm3 2 8µL Since r i = 2σ cosθ Pci σ cosθ 2 Vi qi = ∆p Pci 2 2 µ L 2 For a bundle of n capillary tubes, i = n σ cosθ 2 ∆p i=n V i qt = Σ Σ 2 i=1 2µL i=1 Pci 2 Since, qt
k
= k A ∆p µL
i=n σ cosθ 2 = Σ Vi 2 2 A L i=n Pci
2
Define the fractional volume of ith capillary
si =
Vi , VT
s i = fraction
and φ = VT , φ = fraction AL
k=
σ cosθ 2 2
φ
i=n
Σ i=1
Si Pci 2
Introducing a lithology factor λ for deviation of the actual pore space, k=
σ cosθ 2 2
φλ
i=n
Σ i=1
Si Pci 2
In integral form, S=1
σ cosθ 2 dS k= φλ 2 2 S=0 Pc
3
PETE 306 HANDOUT 4/16/92 Calculation of Relative Permeabilities using Capillary Pressure Data Purcell and Burdine Approach (ABW: pages 196-199) Purcell approach: The absolute permeability may be expressed as S=1
σ cosθ 2 dS k= φλ 2 2 S=0 Pc The effective permeability of the wetting phase may be expressed as S=Swt
kwt
σ cosθ 2 = φλ 2 S=0
dS P2c
The relative permeability of the wetting phase is the ratio of the wetting phase effective permeability to the absolute permeability S=Swt
krwt
= kwt = k
S=0
dS P2c
S=1
dS 2 S=0 Pc
1
Similarly, the effective permeability of the nonwetting phase may be expressed as S=1
knwt
σ cosθ 2 dS = φλ 2 2 S=Swt Pc
The relative permeability of the nonwetting phase is the ratio of the nonwetting phase effective permeability to the absolute permeability S=1
krnwt = knwt = k
dS 2 S=Swt Pc S=1
dS 2 S=0 Pc
2
Burdine Approach: Burdine considered the tortuosity factors for one-phase and multiphase systems and modified the Purcell equations for the effective permeabilities.
λrwti = λi λwti The relative permeability of the wetting phase is the ratio of the wetting phase effective permeability to the absolute permeability S=Swt
krwt
= kwt = λrwt 2 k
S=0
dS P2c
S=1
dS 2 S=0 Pc The tortuosity ratio is related to the minimum wetting-phase saturation Sm, as
λrwt = Swt - S m 1 - Sm
3
Similarly, the relative permeability of the nonwetting phase is the ratio of the nonwetting phase effective permeability to the absolute permeability S=1
krnwt = knwt = λrnwt 2 k
dS 2 S=Swt Pc S=1
dS 2 S=0 Pc The tortuosity ratio for the nonwetting phase is related to the minimum wetting-phase saturation, Sm, and the equilibrium saturation to the nonwetting phase, Se, as
λrnwt =
Snwt - S e 1 - Sm - Se
4
