Formation Resistivity
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Transcript of Formation Resistivity
A. SUMMARY
The topic investigated in this experiment was formation resistivity of a porous media (sand
pack). This is important because resistivity as an electrical property of reservoir rocks is a very
important property in the quantitative interpretation of well logs.
The objective of this lab was to set up a simple circuit to measure the resistivity of packed
sands in order to determine the formation factor, resistivity index (for different grades of sand)
and relate the electrical properties to porosity, saturation and pore sizes.
To carry out the experiment, a formation resistivity apparatus consisting of a formation
resistivity cell and formation resistivity circuit was used. Also a conductivity meter and a
vernier calliper were used. See figures 1-2 for the experimental set up.
From the experiment, it was found out that there is a correlation between formation resistivity
factor (F) and porosity, also between formation resistivity factor and cementation factor (m). In
addition, it was also found out that there is a relationship between the connate water saturation
and formation resistivity index. These correlations or relationships and other ones are discussed
in detail in the discussion section.
B. INTRODUCTION :
The objectives of this lab was to measure the electrical resistivity of porous media (sand pack)
under both full and partial brine saturation, and to link the electrical properties obtained to the
relevant petro-physical parameters.
This experiment was important because the electrical properties of a formation or reservoir
rocks play a very important role in the interpretation or analysis of well logs. Reservoir rocks
are capable of transmitting an electric current as a result of the adsorbed water (connate water)
they contain. The connate water contains dissolved salts that constitute some form of
electrolyte capable of conducting current through the formation.
To offset this conduction of electric current, there has to be some form of resistance. This
resistance is called resistivity. Thus the electrical resistivity of a fluid-saturated rock or
Page 1 of 16
formation is the ability of the formation to impede the flow of electric current through it. In
other words, it is the reciprocal of the conductivity.
C. THEORY:
Measuring the resistivity of a formation could help to determine water saturation, which could
then be used to calculate the volume of oil and /or gas in place. Thus resistivity could be said to
be a very viable tool for evaluating how much could be produced from a formation or
reservoir. This is because the principal objective of well log interpretation is the identification
of porous zones containing hydrocarbons and the determination of the water saturation. In line
with this objective, resistivity is a veritable parameter in well log analysis.
The resistivity of reservoir rocks is a function of salinity of formation water, effective porosity
and the quantity of hydrocarbons trapped in the pore space. An increase in porosity causes a
decrease in resistivity, while an increase in petroleum content increases the resistivity
(Donaldson and Tiab, 2004).
Since the solid matrix of rock sample is usually non-conducting, if the rock sample is then
saturated with a conducting fluid like brine (solution of sodium chloride), the resistivity of the
porous rock sample fully saturated with brine (Ro) can then be calculated. Also, the resistivity
of the brine (Rw) can be determined by using a conductivity meter and then expressing it a
reciprocal of the value shown on the conductivity meter. Alternatively, the resistivity of the
brine can be calculated by applying a voltage across a cell with length (L) and cross sectional
area (A) with brine in the cell, and then record the amount of flow of current. With these
parameters, the resistivity of the brine can then be calculated as:
Rw = rwAL
= EI w
AL (1)
Page 2 of 16
Where:
Rw is the resistivity of the brine (Ohm-m)
rw is the total resistance in ohms across the cell with brine (Ohms)
A is the cross sectional area of the cell with brine (m2)
L is the length of the cell with brine (m)
E is the voltage applied across the cell with brine (volts)
Iw is the amount of current flowing through the cell with brine (Amperes)
Note: It was the latter method that was used in this experiment because there was a problem
using the conductivity meter provided.
The resistivity a porous rock sample (in this case sand) can be determined by fully saturating
the sample with brine, and applying a voltage across the packed sand, thus allowing a current
to flow through it. From Ohms law, the resistivity can then be calculated as:
Ro=RAL
= EI o
AL (2)
Where:
Ro is the resistivity of the rock (in Ohm-m) fully saturated with brine of resistivity Rw
R is the total resistance in Ohms (Ω)
A is the core cross sectional area in meter square (m2)
L is the length of the core in meters (m)
E is the voltage across the packed sand (Volts)
Io is the current flowing through the sample (Ampere)
The resistivity of most sedimentary rocks can range from 0.2 to 2000 Ohm-m. The resistivity
of poorly consolidated sand can range from 0.20 Ohm-m for sands containing primarily
saltwater, to several Ohm-m for oil-bearing sands. For well consolidated sandstones, the
resistivity could range from 1 to 1,000 Ohm-m or more depending on the amount of shale
inter-bedding (Tiab and Donaldsn, 2004)
Page 3 of 16
The relationship between the resistivity of a porous rock medium (Ro) fully saturated with
brine and the resistivity of the brine is equal to a constant known as the formation resistivity
factor, this relationship was given by Archie as:
Ro
Rw
=F (3)
Where:
Ro is the resistivity of the rock fully saturated with brine (Ohm-m)
Rw is the resistivity of the brine (Ohm-m)
F is the formation resistivity factor (dimensionless)
It is pertinent to know here that Ro will be greater than Rw; hence ‘F’ will always be greater
than unity or one.
This ratio (F) is related to some important petro-physical parameters such as porosity and
connate water saturation. The relationship between the porosity and the formation resistivity
factor (F) is given as:
F= a
m (4)
Where:
φ is the fractional porosity of the rock sample
m is the cementation factor
a is a constant called tortuosity factor
The above formula is just an empirical formula; hence ‘a’ and ‘m’ are are usually taken as 0.81
and 2 respectively for chalky rocks and compacted formations or for clean sandstone. Although
ideally, ‘a’ and ‘m’ will depend on the types of rocks. For example, compact limestones, which
are very highly cemented, the value of m may be as high as 3. From equation (3), it can be seen
that there is a correlation between the porosity of a rock sample and the formation resistivity
factor (F), and also between the cementation factor (m) and the formation resistivity factor.
Page 4 of 16
The cementation factor is a function of the shape and distribution of pores in the rock.
Practically it is determined from a plot of the formation resistivity factor (F) against the
porosity on a log-log graph; with the slope equal to the cementation factor (m).
In a formation containing oil and/or gas, both of which are non-conductors of electricity, but
containing a certain amount of water, the resistivity of the formation will be a function of the
formation water saturation or brine saturation (Sw). Thus a rock that contains oil and /or gas
will have a higher resistivity than the same rock completely saturated with formation water or
brine. The relationship between the resistivity of a rock saturated with brine and oil (Rt) and
the resistivity of the same rock fully saturated with brine is given as:
R t
Ro
=I (5)
Where:
Rt is the resistivity of the rock saturated with brine and hydrocarbons (Ohm-m)
Ro is the resistivity of the same rock fully saturated with brine (Ohm-m)
I is the resistivity index (dimensionless)
Here Rt > Ro. If the formation is totally saturated with brine, then Rt = Ro, hence the resistivity
index will be equal to one. The resistivity index is a function of the brine saturation and can be
express as:
I = 1/Swn (6)
Where:
I is the resistivity index
Sw is the brine saturation (%)
n is the saturation exponent and is usually considered to be equal to 2. Although, according to
Donaldson and Siddiqui, this would also depend on the wettability of the rock (Donaldson and
Siddiqui, 1989). This saturation exponent is usually obtained from a log-log plot of the
resistivity index (I) against water saturation; with the slope been the saturation exponent (n).
D. EXPERIMENTAL EQUIPMENT
Page 5 of 16
In order to carry out the experiment, a formation resistivity apparatus consisting of a formation
resistivity cell and formation resistivity circuit was used. Also a conductivity meter for
measuring the conductivity of the brine was used, and a vernier calliper for measuring the
diameter and length of the core (packed sand) was used. See figure 1 – 2 below for schematic
diagrams of the experimental set up.
E. EXPERIMENTAL PROCEDURE AND OBSERVATIONS
The resistivity cell was disassembled, cleaned and then the length (L), diameter of the sand
pack compartment was determined using callipers. The cross section area (A) and the volume
were also calculated. After this, the resistivity of the provided (20,000 ppm) which is 2% was
measured by applying a voltage (E) across the brine and recording the amount of current (I)
flow. With these parameters and the dimensions of the cell, the brine resistivity was then
calculated from equation (1).
The circuit was connected, and made sure that the meters were set to the correct scale and that
the circuit connections were correct for measuring volts and amps, resistivity, before turning on
the power supply to the transformer (AC supply). The correct scales are at least 8 volts for the
voltmeter and smallest scale available on the ammeter. The circuit was tested by removing the
small jack plugs from the cell and connecting them together, it was then checked to see that the
values on the voltmeter and the ammeter reverse to make sure that the circuit is correct.
The complete cell assembly was weighted. And by removing the top of the cell by unfastening
the tow nuts and pack the square central section with one of the graded sands provided. Care
was taken to avoid getting sand on the ‘o’ ring to maintain the seal. When the cell was full, the
Page 6 of 16
top section was replaced. And again the ‘o’ ring was checked to obtain a good seal. The cell
was taped to settle the sand at the base to seal any void space. In some cases were there were
void spaces, the cell was then placed cell on its side and filled by one of the side orifice.
The unwanted sand from the cell body was removed with an air line, and the cell re-weighted,
then the porosity of the packed sand was calculated.
The sand was then saturated with brine by using the syringe and a needle and injected
through the lower rubber bung. The saturated pack was weighted. The potential and current
were measured and the resistivity of the fully saturated pack (Ro) was calculated using
equation (2).
By using the syringe some air was introduced into the packed sand bed with the voltage and the
current been calculated. With these parameters and the dimensions of the cell, the resistivity in
a partially saturated state was also calculated using equation (2). Then the pack was re-
weighted. This was repeated for five more times by introducing more air in each case.
The above procedures were repeated for different grades of sand and then the formation factor,
the resistivity index, the cementation factor etc. were calculated in each case.
F. EXPERIMENTAL RESULTS AND CALCULATIONS:
The experimental results and calculated results are are shown in tables 1, 2, 3and 4.
Brine Concentration 20,000 ppm
Cell Dimensions:
Diameter (cm) 2.27 2.270E-02 m
Length (cm) 3.98 3.980E-02 m
Area (cm2 ) 4.05 4.048E-04 m2
Bulk Volume (cm3) 16.11 1.611E-05 m3
Voltage across cell with brine (Ampere) 0.78
Current flowing across cell with brine (Amperes) 8.13
Brine Resistance (Ohms) 9.594E-02
Page 7 of 16
Table 1: Experimental results for the dimensions of the resistivity cell
Brine Resistivity (Rw) 9.757E-04
Table 2: Experimental results obtained for sand 1 (Fine Sand)
Sand 1 (Fine Sand)
Cell weight (g) 294.75
cell + sand (g) 321.22Grain Volume (cm3) 10.03
Pore Volume (cm3) 6.08
Porosity (%) 37.76
Resistivity
V (Volts) A (Amps)R
(Ohm) Ro (Ohm-m) Rt (Ohm-m)Weight
(g)Sw (%) F I m n
1.42 7.33 0.194 1.970E-03 1.970E-03 327.47 1002.019 1.00 1.5 1.5
1.44 7.32 0.197 2.001E-03 327.25 99.328 1.015 1.5
1.46 7.29 0.200 2.037E-03 327.04 98.686 1.034 1.6
1.67 7.13 0.234 2.382E-03 326.81 97.983 1.209 1.6
2.15 6.65 0.323 3.288E-03 326.5 97.035 1.669 1.7
2.72 6.12 0.444 4.520E-03 326.17 96.027 2.294 1.8
Sand 2 (Medium Sand)
Cell weight (g) 294.88
cell + sand (g) 318.5Grain Volume (cm3) 8.95
Pore Volume (cm3) 7.16
Porosity (%) 44.46
Resistivity
V (Volts) A (Amps)R (Ohm) Ro (Ohm-m) Rt (Ohm-m)
Weight (g)
Sw (%) F I m n
1.13 7.56 0.149 1.520E-03 1.520E-03 325.67 1001.55
8 1.00 1.4 1.4
1.15 7.53 0.153 1.553E-03 325.43 99.221 1.02 1.4
1.17 7.52 0.156 1.582E-03 325.27 98.701 1.04 1.5
1.24 7.45 0.166 1.693E-03 325.02 97.889 1.11 1.7
1.34 7.36 0.182 1.852E-03 324.89 97.467 1.22 2.0
3.7 5.54 0.668 6.792E-03 324.74 96.980 4.47 2.3
Sand 3 (Coarse Sand)
Cell weight (g) 294.86
Page 8 of 16
Table 4: Experimental results obtained for sand 3 (Coarse Sand)
Table 3: Experimental results obtained for sand 2 (Fine Sand)
cell + sand (g) 319.8
Grain Volume (cm3) 9.45
Pore Volume (cm3) 6.66
Porosity (%) 41.36
Resistivity
V (Volts)A (Amps)
R (Ohm) Ro (Ohm-m) Rt (Ohm-m)
Weight (g)
Sw (%) F I m n
1.17 7.47 0.157 1.593E-03 1.593E-03 326.62 100 1.633 1.00 1.4 1.4
1.43 7.28 0.196 1.998E-03 325.94 97.859 1.25 2.2
3.52 5.24 0.672 6.832E-03 325.93 97.827 4.29 2.3
3.73 5.05 0.739 7.512E-03 325.91 97.764 4.72 2.3
3.80 4.97 0.765 7.776E-03 325.89 97.702 4.88 2.3
5.00 3.80 1.316 1.338E-02 325.88 97.670 8.40 3.2
0.01 0.1 11
10
100
Porosity (%)
For
mat
ion
Fac
tor
Figure 3: Log-log plot of formation resistivity factor versus porosity
Page 9 of 16
0.01 0.1 11
10
100
Brine Saturation (%)
Res
isti
vity
In
dex
Figure 4: Log-log plot of Resistivity index versus brine saturation
SAMPLE CALCULATIONS:
(a) Sample calculations for area of cell.
Area = π D2
4 = π ¿¿ = 4.048E-04 m2 = 4.048 cm2
(b) Sample calculations for bulk volume.
Volume = π r2l = area * length = 4.048E-04 * 3.980E-02 = 1.611E-05 m3 = 16.11 cm3
(c) Sample calculations for grain volume.
Grain volume = (mass of cell+sand−mass of dry cell )
density of sand =
(321.22−294.75 ) g
2.64 gcm−3 = 10.03 cm3
(d) Sample calculations for pore volume.
Grain volume = Bulk volume – Grain volume = (16.11 - 10.03 ) cm3 = 6.08 cm3
(e) Porosity sample calculation for sand 1 (Fine Sand)
Page 10 of 16
Porosity = Pore volumeBulk volume
* 100 % = 6.0816.11
* 100 % = 0.3776 *100 % = 37.76 %
(f) Brine resistivity sample calculation
Brine resistivity (Rw) = rwAL
= EI w
AL
Where:
rw is the total resistance in ohms across the cell with brine (Ohms)
A is the cross sectional area of the cell with brine (m2)
L is the length of the cell with brine (m)
E is the voltage applied across the cell with brine (volts)
Iw is the amount of current flowing through the cell with brine (Amperes)
Thus, Brine resistivity (Rw) = 0.788.13
4.048∗10−4
3.980∗10−2 = 9.757E-04 Ohm-m
(g) Core sample (sand) resistivity sample calculation
Core sample (sand) resistivity (Ro) = RAL
= EI o
AL
Where:
R is the total resistance in ohms across the cell with core sample fully saturated with brine of
resistivity Rw
A is the cross sectional area of the cell with core sample (m2)
L is the length of the cell with core sample (m)
E is the voltage applied across the sand medium fully saturated with brine (volts)
Io is the amount of current flowing through the sand medium fully saturated with brine
(Amperes)
Thus, core sample (sand) resistivity (Ro) = 1.427.33
4.048∗10−4
3.980∗10−2 = 1.970E-03 Ohm-m
Note: From the calculation above, it shows that the resistivity of the rock fully saturated with
brine (Ro) is greater than the resistivity of the brine (Rw). That is Ro > Rw. This is
theoretically correct and is true in all cases.
(h) Formation resistivity (F) sample calculation
Using equation (2): Ro
Rw
=F = 1.970E-039.757E-04
= 2.019
Page 11 of 16
(i) Formation resistivity index (I) sample calculation
Using equation (4): R t
Ro
=I = 1.970E-031.970E-03
=¿ 1.00
Note: From the calculation above, it can be seen that when the rock sample was fully saturated
with brine, the resistivity index resulted in unity (that is 1).
(j) Connate water saturation (Sw) calculation
Sw# = (Weight# - Weight of dry cell) / (Weight1 – Weight of dry cell) * 100%
= (327.25- 294.75) / (327.47– 294.75) * 100% = 99.328 %
Note: Weight1 is the weight of the previous cell with the core sample saturated with brine.
Weight# represents the weight of each of the preceding cell with the core sample saturated with brine.
(k) Cementation factor (m) sample calculation:
Cementation factor (m) = log [ [ Rw / Rt ] ]
Where φ = (0.81/F)0.5 = (0.81/2.019)0.5 = 0.6333
Thus, m = log [ [ 9.757E-04 /1.970E-03 ]0.6333 ] = 1.5
(G) DISCUSSION OF RESULTS:
From figure 3, it can be seen that there is a relationship between porosity and formation
resistivity as measured by formation resistivity factor. Resistivity decreases with increasing
porosity. This is because even though reservoir rocks conducts electricity due to the salinity of
water contained in their pores, porosity also plays an important role in the flow of electric
current in a medium (Butler, 2005). The more porous a rock is, the more the electric
conductance of the rock and hence the less the resistivity. This relationship can also be seen
from equation (3).
The above relationship where the resistivity of the rock was measured by the formation
resistivity factor is true for a rock fully saturated with brine, but in a formation containing oil
Page 12 of 16
and/or gas both of which are non-conductors of electricity with a certain amount of brine, the
resistivity will be a function of the brine saturation. The parameter or factor in this case is
called the resistivity index.
This can be seen in figure 4 which shows a log-log plot of resistivity index versus brine
saturation. This can also be seen from tables 2-4 where a decrease in brine saturation (Sw)
caused an increase in the resistivity index (I). Thus as the brine saturation decreases, the
resistivity index increases; hence the formation resistivity would increase because of the
available volume for the flow of current.
It was also found from the experiment that for the same porosity, the values of the true
resistivity (Rt) of a given formation (i.e. the various sand media) when the formation was not
fully saturated were larger than the resistivity of the formation fully saturated with brine. The
reason for this was that there was less available volume for the flow of electric current in the
latter. The ratio between the former and the latter equals to the resistivity index (as shown in
equation 4).
The slope of the plot in figure 3 gave rise to an important parameter known as the cementation
factor or exponent. This factor is usually taken to be 2 for clean sandstones and can also be as
high as 3 for compact limestones which are highly cemented (Donaldson and Tiab, 2004). In
this experiment, this was found to be approximately 2 as well (exactly 1.5). This cementation
factor can be said to be a function of the shape and distribution of pores. Furthermore, the
degree of cementation of rock particles (in this case sand particles) will depend on the nature,
amount, and distribution of various cementing materials.
Thus, the less cemented sands will have higher porosities; hence lower resistivity factor
(in this experiment, it was sand 2). Therefore as the sand becomes more cemented, the porosity
decreased, and the formation resistivity factor increases.
The slope of the plot in figure 4 gave rise to an important parameter known as the saturation
exponent. This factor is usually taken to be 2 although this could also be affected by various
factors such as the brine saturation, wettability, overburden pressure, nature and microscopic
distribution of the reservoir fluids (Donaldson and Siddiqui, 1989)).
H. ERROR ANALYSIS:
Page 13 of 16
As it is with every experiment, it must be noted that some errors could have arose in this
experiment, this maybe in terms of systematic, random or human error. Some factors that
could be considered as possible sources of error are:
(1) If the base of the cell sand is not tapped properly, there could be void spaces left in the cell
and this could ultimately have an effect on the experimental results obtained like the
porosity of the sand media. This could in turn affect the resistivity formation factor.
(2) Another possible source of error could be in the circuit connections. If the connections are
not correct for measuring volts and amps respectively, this could also have an effect on the
results of this experiment.
(3) Other possible source of errors could be in the reading of the vernier calliper.
I. CONCLUSION:
At the end of this experiment, the following conclusions can be drawn:
(i) For a rock fully saturated with brine, the resistivity is measured by the resistivity
factor (F).
(ii) For a rock partially saturated with brine, the resistivity is measured by the resistivity
index (I).
(iii) A rock that contains oil/gas together brine will have a higher resistivity than the
same rock completely saturated with brine or formation water.
(iv) For a partially saturated rock (a rock saturated with brine and oil), the resistivity
index is a function of brine saturation. As brine saturation reduces, resistivity index
increases.
(v) For same porosity, the true resistivity (Rt) of a formation is larger than the
resistivity of the formation fully saturated with brine (Ro).
(vi) Porosity has an effect on the resistivity of a formation.
(vii) Even though the values of the cementation factor (m) and the saturation exponent
(n) are usually considered constant and equal to 2, ideally they vary from rock to
rock, and are also affected by other factors.
Page 14 of 16
Nomenclature
Symbols Units Meaning
A cm2 Cross sectional area of the cell
a Tortuosity factor
D cm Diameter of cell
E V Voltage applied across the cell
F Formation resistivity factor
Page 15 of 16
I Formation resistivity index
L cm Length of the core
m Cementation factor
n Saturation exponent
Io A
amount of current flowing through the sand medium fully
saturated with brine
Iw Aamount of current flowing through the cell with brine
R Ω
total resistance in ohms across the cell with core sample fully saturated with brine of resistivity Rw
Rw Ω-m Brine resistivity
Ro Ω-mResistivity of core (sand)
sample
Rt Ω-mResistivity of the core (sand) saturated with brine and oil
rw Ωtotal resistance across the cell with brine (Ohms)
REFERENCES
Donaldson, E. C. and Siddiqui, T. K. (1989) Relationship between the Archie Saturation
Exponent and Wettability: SPE Formation Evaluation. September, 4 (3), pp. 359-362. [Online]
Available from: http://www.onepetro.org/mslib/servlet/onepetropreview?id=00016790
[accessed 4 January 2011].
Butler, D.K. (2005) Introduction to Near-Surface Geophysics. Tulsa, Oklahoma: Society of
Exploration Geophysicists. pp.47- 48.
Page 16 of 16