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 t he 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). m). In 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.
INTRODUC INTRODUCTI TIO ON :
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. para meters. 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 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.
Page 1 of 16
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 (R o) can then be calculated. Also, the resistivity of the brine (R w) 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 =
(1)
Where:
Rw is the resistivity of the brine (Ohm-m) r w is the total resistance in ohms across the cell with brine (Ohms) 2
A is the cross sectional area of the cell with brine (m ) 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.
Page 2 of 16
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:
(2)
Where: is the resistivity of the rock (in Ohm-m) fully saturated with brine of resistivity R w
R is the total resistance in Ohms
()
A is the core cross sectional area in meter square
(m 2 )
L is the length of the core in meters
(m)
E is the voltage across the packed sand
(Volts)
I o 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) 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:
(3)
Where: 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 R o will be greater than R w; hence µF¶ will always be greater than unity or one. Page 3 of 16
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:
(4)
J
Where:
J 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 µa¶ and µm¶ are are usually taken as 0.81 and 2 respectively for chalky rocks and compacted formations or for clean sandstone. Although ideally, µa¶ and µm¶ will depend on the t ypes 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. 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 (S w). 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:
(5)
Where:
R t is the resistivity of the rock saturated with brine and hydrocarbons
(Ohm-m)
R o is the resistivity of the same rock fully saturated with brine
(Ohm-m)
I is the resistivity index
(dimensionless) Page 4 of 16
Here
R t > R o. If the formation is totally saturated with brine, then R t = R o, hence the resistivity
index will be equal to one. The resistivity index is a function of the brine saturation and can be express as: n
I = 1/Sw
(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.
EXPER IMENTAL EQUIPMENT
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.
Page 5 of 16
E.
EXPER IMENTAL PROCEDUR E 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 µo¶ ring to maintain the seal. When the cell was full, the top section was replaced. And again the µo¶ 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 reweighted. 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.
Page 6 of 16
F.
EXPER IMENTAL R ESULTS AND CALCULATIONS:
The experimental results and calculated results are are shown in tables 1, 2, 3and 4. Table 1: Experimental results f or the dimensions of the resistivity cell
Brine Concentration
20,000 ppm Cell Dimensions:
Diameter (cm)
2.27
2.270E-02 m
Length (cm)
3.98
3.980E-02 m
4.05
4.048E-04 m
2
Bulk Volume (cm )
16.11
1.611E-05 m
3
Voltage across cell with brine (Ampere)
0.78
Current flowing across cell with brine (Amperes)
8.13
2
Area (cm ) 3
Brine Resistance (Ohms)
9.594E-02
Brine Resistivity (Rw)
9.757E-04
Table 2: Experimental results obtained f or sand 1 (Fine Sand) Sand 1 (Fine Sand)
Cell weight (g)
294.75
cell + sand (g) Grain Volume (cm3)
321.22 10.03 3
Pore Volume (cm )
6.08
Porosity (%)
37.76 R esistivity
V (Volts)
A (Amps)
R (Ohm)
1.42
7.33
0.194
1.44
7.32
1.46
R o (O hm-m)
1.970E-03
R t (Ohm-m)
Weig ht (g)
Sw (%)
F
2.019
I
1.00
m
1.5
n
1.970E-03
327.47
100
0.197
2.001E-03
327.25
99.328
1.015
1.5
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
Page 7 of 16
1.5
Table 3: Experimental results obtained f or sand 2 (Fine Sand) Sand 2 (Medi um Sand)
Cell weight (g)
294.88
cell + sand (g) Grain Volume 3 (cm )
318.5 8.95 3
Pore Volume (cm )
7.16
Porosity (%)
44.46 R esistivity
V (Volts)
A (Amps)
R (Ohm)
1.13
7.56
0.149
1.15
7.53
1.17
R o (O hm-m)
1.520E-03
R t (Ohm-m)
Weig ht (g)
Sw (%)
F
I
1.558
1.00
m
1.4
n
1.520E-03
325.67
100
1.4
0.153
1.553E-03
325.43
99.221
1.02
1.4
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
Table 4: Ex erimental results obtained f or sand 3 Coarse Sand Sand 3 (Coarse Sand)
Cell weight (g)
294.86
cell + sand (g)
319.8 3
9.45
Pore Volume (cm )
6.66
Grain Volume (cm ) 3
Porosity (%)
V (Volts)
41.36
A (Amps)
R (Ohm)
1.17
7.47
0.157
1.43
7.28
3.52
R o (O hm-m)
1.593E-03
R esistivity Weig ht R t (Ohm-m) (g)
Sw (%)
F
I
1.633
1.00
m
1.4
n
1.593E-03
326.62
100
0.196
1.998E-03
325.94
97.859
1.25
2.2
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
Page 8 of 16
1.4
100
r o t
c
a
F n
o i t
10
a
m r o
F
1 0.01
0.1
1
Porosity (%)
Figure
3: Log-log plot of formation resistivity factor versus porosity
100
x e d n I y t
i v i t s i s
10
e
R
1 0.01
0.1
1
Brine Saturation (%)
Figure
4: Log-log plot of Resistivity index versus brine saturation Page 9 of 16
SAMPLE CALCULATIONS:
(a)
Sample calculations f or area of cell.
Area = (b)
=
2
3
= area * length = 4.048E-04 * 3.980E-02 = 1.611E-05 m = 16.11 cm3
Sample calculations f or grain volume.
Grain volume = (d)
2
= 4.048E-04 m = 4.048 cm
Sample calculations f or bulk volu me.
Volume = (c)
= = 10.03 cm
3
Sample calculations f or pore volume. 3
3
Grain volume = Bulk volume ± Grain volume = ( 16.11 - 10.03 ) cm = 6.08 cm (e)
Porosity sample calculation f or sand 1 (Fine Sand)
Porosity =
(f)
* 100 % = * 100 % = 0.3776 *100 % = 37.76 %
Brine resistivity sample calculation
Brine resistivity (Rw) =
Where:
r w is the total resistance in ohms across the cell with brine (Ohms) A is the cross sectional area of the cell with brine (m 2 ) 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 thr ough the cell with brine (Amperes) Thus, Brine resistivity (Rw) =
(g)
= 9.757E-04 Ohm-m
Core sample (sand) resistivity sample calculation
Core sample (sand) resistivity (R o) =
Where:
R is the total resistance in ohms across the cell with core sample fully saturated with brine of resistivity Rw
Page 10 of 16
A is the cross sectional area of the cell with core sample (m 2) 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 (R o) =
= 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 a ll cases.
(h)
Formation
Using
(i)
equation (2): Formation
Using
equation (4):
resistivity (F) sample calculation
= = 2.01
9
resistivity index (I) sample calculation
= 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) calculatio n
Sw# = (Weight# - Weight of dry cell) / (Weight 1 ± 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 f actor (m) sample calculation:
Cementation f actor (m) =
J
Where J = (0.81/F)0.5 = (0.81/2.019)0.5 = 0.6333 Thus, m =
= 1.5 Page 11 of 16
(G)
DISCUSSION OF R ESULTS:
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 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
ementation
c
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. Page 12 of 16
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 c emented, 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 Si ddiqui, 1989)).
ERROR ANALYSIS:
H.
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 t he 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 a mps respectively, this could also have a n 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.
Page 13 of 16
(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
A
cm
2
a
Meaning
Cross sectional area of the cell Tortuosity factor
D
cm
Diameter of cell
E
V
Voltage applied across the cell
F
Formation resistivity factor
I
Formation resistivity index
L
cm
Length of the core
m
Cementation factor
n
Saturation exponent
Io
A
Iw
A
amount of current flowing through the sand medium fully saturated with brine amount of current flowing through the cell with brine
R
total resistance in ohms across the cell with core sample fully saturated with brine of resistivity Rw
Rw
-m
Brine resistivity
-m
Resistivity of core (sand) sample
-m
Resistivity of the core (sand) saturated with brine and oil
Ro Rt r w
total resistance across the cell with brine (Ohms)
Page 15 of 16
R EFER ENCES
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. Donaldson, E.C. and Tiab, D. (2004) Petrophysics: theory and practice of measuring reservoir rock and fluid transport properties. 2nd ed. Oxford: Gulf Professional Publishing. Chapter 3.
Page 16 of 16
