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

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)

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)

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.

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

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

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

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 + sand (g) 318.5Grain Volume (cm3) 8.95


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)


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


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

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)

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

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

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:

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.

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

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.

of 16

http://www.onepetro.org/mslib/servlet/onepetropreview?id=00016790

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.

of 16