Jul 10, 2008 ... Evidence for a variable Archie porosity exponent “m” and impact on saturation
calculations for Mesaverde tight gas sandstones: Piceance ...
Evidence for a variable Archie porosity exponent “m” and impact on saturation calculations for Mesaverde tight gas sandstones: Piceance, Uinta, Green River, Wind River, and Powder River basins Robert M Cluff, The Discovery Group Inc. Alan P. Byrnes, Kansas Geological Survey1 1presently
with Chesapeake Energy
Stefani Whittaker, The Discovery Group Inc. Dan Krygowski, The Discovery Group Inc. AAPG Rocky Mountain Section meeting, Denver, Colorado 10 July 2008
US DOE Project Summary z
DOE Contract # DE-FC26-05NT42660 z
z
completion date 30 June 2008
Organizations z z z
University of Kansas Center for Research, Inc. Kansas Geological Survey, Lawrence, KS The Discovery Group Inc., Denver, CO
z
Principal Investigators: Alan P. Byrnes, KGS; Bob Cluff, Discovery Group
z
project website is http://www.kgs.ku.edu/mesaverde
Objectives of this task z
Characterize Mesaverde electrical properties as a function of porosity and salinity z Archie
porosity (cementation) exponent “m” z Investigate behavior at low porosity end (100 mD 1000
Klinkenberg Permeability (4,000 psi, mD)
z
100 10 1 0.1
Green River Piceance Powder River Uintah Washakie Wind River logK=0.3Phi-3.7 logK=0.3Phi-5.7
0.01 0.001 0.0001 0.00001 0.000001
0.0000001 0
2
4
6
8
10
12
14
In situ calc Porosity (%)
16
18
20
22
24
Archie’s equation
Sw = (a / φ n
z z
* (Rw / Rt )
completely empirical – no theoretical basis “m” is the porosity or cementation exponent z
z
m)
loosely related to tortuosity of the current flow path, better thought of as electrical efficiency of the path
“n” is the saturation exponent z
related to change in conductivity path with changing saturation
Archie porosity exponent z
for a simple bundle of capillary tubes oriented parallel to current flow direction: m→1 z
insensitive to cross section shape, so fractures act like capillary tubes
as porosity increases and less of it participates in the conductive path, m ↑ z for an “average” sandstone comprised of spherical grains, m → 2 z
Resistivity of a simple rock model
0
∞ For straight capillary tubes:
Porosity (φ) Resistivity (Ro)
F = Ro/Rw = 1/φ
For rock with tortuous pores:
F = Ro/Rw = 1/φ m (after Doveton, 2005)
1 Rw
Capillary tube model for m m 1.0 >1 ~2 >2
m=1
after Herrick & Kennedy, 1993, SPWLA Paper HH
A
Φ
Rw L
high -P core holder electric insulator
then we measure R0. F = R0/Rw.
brine in
Resistance Reference Cell Frequency Generator
brine of known Rw & φ
Core Plug
Start with core plug saturated with
high-P fluid
Micropipette
Core measurement of the formation factor, F
Plated electrodes
When F and φ are plotted log-log 1000
m= 2
m= 3
100 m= 1
F 10
1 0.01
0.1
φ
1
log F = -m log φ
Observed porosity dependence of “m”
R2 = 0.63 (RMA)
40Kppm dataset is largest and used for base case cap m at 1.95 40K ppm brine data 2.2
2.2 2.1 2.0
In situ Archie Cementaiton Exponent
Empirical: m = 0.676 log φ + 1.22 each salinity is different
In situ Archie m
y = 0.5377x + 1.3313 R2 = 0.6331
1.9 1.8 1.7 1.6 1.5 1.4 1.3 1.2 1.1 1.0 -0.2 0.0
0.2
0.4
0.6
0.8
1.0
1.2
log In situ Porosity (%)
1.4
2.0 1.8 1.6 1.4 1.2 1.0 0
4
8
12
16
In situ Porosity (%)
20
24
Dual porosity model z
m = log[(φ-φ2)m1 + φ2m2]/log φ z
φ expressed as V/V
φ2 = 0.0035, m1=2, m2=1; SE both = 0.11 z rock behaves like a mixture of matrix porosity and cracks, with cracks dominating low porosity end cap at m = 1.95 (φ ~ 16%) 40K ppm brine data both models fit data 2.2
z z
φ = bulk porosity φ2 = fracture porosity m1 = matrix cementation exponent m2 = fracture cementation exponent
In situ Archie Cementaiton Exponent
z
2.0 1.8 1.6 1.4 1.2 1.0 0
4
8
12
16
In situ Porosity (%)
20
24
And a third way to look at it.... z z
z z
z
Why is the minimum m ~ 1.2, instead of 1? A – for a distribution of cracks of different crosssectional area, the largest (widest) cracks will dominate the conductivity The high tail of the distribution determines the bulk conductivity, while the rest of the cracks act like “excess” porosity that do not participate (significantly) in the conductivity. Therefore m ↑
And are the “cracks” all fractures? z z
z
Probably not.............. Slot-like pores oriented preferentially parallel to bedding also act like conductive cracks
Thin parallel laminae of slightly coarser, more permeable sand will be crack-like
Salinity dependence of “m”
tested plugs with 20K, 40K, 80K, and 200K ppm brines Nearly all cores exhibit some salinity dependence 2.3
n=335
0.9
Core Conductivity (mho/m)
In situ Archie Cementation Exponent, (m, A=1)
1.0
0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0
2.2 2.1 2.0 1.9 1.8 1.7 1.6 1.5 1.4 1.3 1.2 1.1 1.0
0
2
4
6
8
10
12
14
16
18
Brine Conductivity (mho/m)
20
22
0.01
0.1
Brine Resistivity (ohm-m)
1
All data, all salinities Archie Cementaiton Exponent (m, a=1)
2.40 2.20 2.00 1.80 1.60 1.40 1.20
200K 80K
1.00
40K 20K
0.80 0
2
4
6
8
10 12 14 16 18 20 22
In situ Porosity (% )
Salinity dependence of “m” z
20K ppm y = 0.2267Ln(x) + 2.2979 2
R = 0.6619
z
1.50 Series1 Log. (Series1)
1.00
40K ppm
z
3.00
0.50
y = 0.2328Ln(x) + 2.409 2
R = 0.6547
2.50 0.050
0.100
0.150
0.200
0.250
2.00
insitu porosity (%)
Series1
1.50
Log. (Series1)
80K ppm
1.00
0.50
3.00
0.00 0.000
2.50
y = 0.2149Ln(x) + 2.4354 2
0.050
0.100
0.150
0.200
R = 0.5132
0.250
insitu porosity (%)
2.00 Axis Title
0.00 0.000
Axis Title
Axis Title
2.00
200K ppm
Series1
1.50
Log. (Series1) 1.00
3.00
0.50
2.50
y = 0.1621Ln(x) + 2.3222
0.00 0.000
2
R = 0.3633
2.00 0.050
0.100
0.150
insitu porosity (%)
0.200
0.250
Axis Title
2.50
m = a log φ + b intercept b drops with decreasing salinity slope is ~ constant
Series1
1.50
Log. (Series1) 1.00
0.50
0.00 0.000
0.050
0.100
0.150
insitu porosity (%)
0.200
0.250
Simple procedure to compute Sw z
determine Rw @ Tf conventionally Pickett plots – focus on the lower porosity, wetter sandstones z produced waters z your best guess....... z
z
convert Rw to 75°F by chart lookup or Arps equation
Pickett Plot example Rw = 0.306
pick m at low porosity end, where BVWirr ~ BVW
Williams PA 424-34 Piceance basin Kmv above “top gas”
Pickett plot Rw 0.306 ohmm @ 160°F = 0.7 @ 75°F (9K ppm)
Our new procedure z
compute m at 40K ppm from RMA regression: m40k = 0.676 log φ + 1.22 e.g. for 10% φ : m = 0.676 + 1.22 = 1.896
z
correct m for salinity effect by m = m40k + ((0.0118 φ – 0.355) * (log Rw + 0.758))
z
e.g. for 10% φ, Rw = 0.7 @ 75°F m = 1.896 + ((0.0118 * 10 – 0.355) * (log 0.7 + 0.758)) m = 1.896 + (-0.237 * 0.603) = 1.753
z
cap m at 1.95 (~12% porosity)
Practical impact Nominally, most of us use an m close to 2, but usually slightly less, for tight gas sand evaluations (e.g. 1.85, 1.90) z Variable m that DECREASES with decreasing porosity leads to lower Sw’s z Therefore, there is more gas in the tight rocks than we thought. z Above 10% porosity there is very little difference z
Example: Low porosity, wet zone
Moderate porosity, wet
“High” porosity gas zone
m is HIGHER than base case, so Sw is higher
20Kppm example, Natural Buttes
30K ppm example, Wamsutter
Summary & Conclusions 335 Kmv samples run at multiple salinities z Archie porosity exponent m varies with z
porosity z salinity z
z
m ↓ as porosity ↓ m ↓ as salinity ↓
behavior is consistent with increasing electrical efficiency with decreasing porosity, whatever the pore scale architecture
variable m model can be implemented with a simple equation relating m to porosity and formation water salinity z m is constant above ~12% porosity at 1.95 z lowering m at 5-12% φ increases GIP z see no impact below ~5% porosity z
BVWirr is typically 3-5% z no longer calculate Sw’s >> 1 z Sw = 1 at low φ validates Rw z
Visit our project website http://www.kgs.ku.edu/mesaverde
Questions?
