Report for the laboratory part of Petrophysics

For the first two examples we had a task to measure and by that to calculate porosity

of saturated and dry rock samples plus calculating permeability of dry rock samples.

For the saturated samples we had to calculate porosity using the “Archimedes

Principle” and ration between volumes of pores versus volume of rock sample. While

for dry samples we had to calculate the porosity and permeability.

These are the measurements for saturated rock samples:

1) E 23/1/1: Limestone

Dry mass 27,873 g

Saturated mass: 27,913 g

Mass in water: 17,518 g

Length: 2,159 cm

Diameter: 2,412 cm

2) E10: Sandstone

Dry mass: 21,054 g

Saturated mass: 23,112 g

Mass in water: 12,957 g

Length: 2,140 cm

Diameter: 2,473 cm

3) I/B1-4: Sandstone

1Juraj Kesner

Dry mass: 26,720g

Saturated mass: 27,024 g

Mass in water: 16,660 g

Length: 2,160 cm

Diameter: 2,449 cm

Prior to calculating the porosity we measured the density and the volume of the samples.

ρ = mV

V=r ² π4

∗l

Sample Volume [cm3] Density [g/cm3]

E 23/1/1 9,884 2,82

E 10 10,270 2,05

I/B1-4 10,174 2,63

To calculate the porosity of the samples we had to use an equation that is directly dependent on the Archimedes principle.

ϕ = ms , a−md

ms ,a−ms ,w

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ms,a.................. saturated mass

md.................... dry mass

ms,w.................. mass in water

Results are:

Sample Φ-Archimedes [%]

E 23/1/1 0,37E 10 20,25

I/B1-4 2,93

The second porosity measurement is based on a relation between the volumes of the pore water and the whole volume of the sample.

ϕ = V p

V

Vp...................... volume of pore water

V……………….volume of the sample

Therefore we have to calculate the volume of pore water by using the following equation:

Vp = mpw

ρpw

mpw.................. mass of pore water

ρpw................... density of pore water

Where the mass of the pore water is equal to:

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m pw = m s,a - ma

And the final formula that we use to calculate the pore volume:

Vp = (ms , a−md)

ρpw

In this case the water that saturates the pores of the samples is fresh water so the table density is 1 g/cm3

SampleV-pore [g/cm³]

V-total [g/cm³] Φ-volume [%]

E 23/1/1 0,04 9,884 0,39E 10 2,058 10,270 20,04

I/B1-4 0,304 10,174 5,6

We can see that the Φ-Archimedes and the Φ-volume are very similar and that means that our calculations are good.

These are the measurements from dry sample rocks. We used the same formula to calculate the volume and bulk density.

Sample Length [cm]

Diameter [cm]

m dry [g]

Volume [cm³]

Bulk density [g/cm³]

I/B1-5 2,133 2,471 26,421 10,228 2,58E 11 2,090 2,410 21,164 9,533 2,22

E 27/1/2 2,139 2,472 27,973 10,265 2,72

We used He- pycnometer to measure grain density which we used to calculate the

porosity and we used permeameter to calculate the permeability only for the first two

samples (E 11 and E 27).

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ϕ=1−( ρbulk

ρgrain)

Sample Permeability [mD]

Grain density [g/cm³]

Φ [%]

E 11 5187,571 2,7358 18,84

E 27 0,06 2,7655 1,64

I/B1-5 2,7331 5,6

For the third measurement of the porosity we need the electrical properties of both

samples. Therefore we need to saturate our samples.

For the measurement of conductivity we wrapped the samples with an isolating tape.

After this, we place both of the samples after each other in a 4-point-light instrument,

which conducts electrical current through them to measure their resistance. The

intensity of the current is 10 000 nA for the first two samples and 1000 nA for the last

one. The next step is that we measure the currents voltage of both samples.

Sample U [mV]

I [nA]

I/B1-4 24,63 10 000

E 10 3,33 10 000

E 23 9,5 1000

After that we calculate the samples resistance and specific resistance with

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R = UI R0 = k * U

I

k....................... geometric factor (cross−section surface

length )

U...................... voltage

I........................ currents intensity

Sample U [mV]

I [nA]

R [Ω]

K [m]

R0

[Ωm]I/B1-4 24,63 10000 2463 0,0218 53,69

E 10 3,33 10000 333 0,0224 7,46

E 23 9,5 1000 9500 0,0212 201,4

In order to calculate the specific resistivity of the water we need the value of

conductivity which is 3,37 S/cm.

Rw = 1K =

13,37¿

¿ = 0,2967 [Ωm]

The final step is to calculate the porosity by using the formation factor of the samples.

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ϕ = m √ 1F

F....................... formation factor

m......................=2

F = R 0

Rw

Sample Ro [Ωm]

Rw [Ωm]

F [k]

Φ –Archie [%]

I/B1-4 53,29 0,2967 180,96 7,43

E 10 7,46 0,2967 25,14 19,94

E 23 201,4 0,2967 678,8 3,84

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