formation resistivity

MSc. PETROLEUM ENGINEERRING

EAB_7_152 Petro Physics

Dr. Elsa Aristodemou

Permeability and Porosity

By

Muhammad Kamal

Student ID # 3325610

Submission Date: 30/11/2014

Muhammad Kamal (3325610)

FORMATION RESISTIVITY

ABSTRACT:

We can measure the Resistivity of three Sands which are Coarse sand, Medium sand and Fine sand for the determination of Resistivity index I and Formation Factor F. We can also measure the Porosity of each sand. When we measure Formation F and Porosity for each sand, then we plot a graph between Log(F) vs Log(Ø) to obtain Cementation Factor M. Likewise when we measure the resistivity index and water saturation then we plot a graph between Log(I) and Log(Sw) to obtain the value of n.

INTRODUCTION:

In this experiment we have to find Electrical Resistivity of porous media and measure the cementation factor and saturation exponent. We can find the electrical resistivity for both full and partial brine saturation.

We can find Formation Factor using formula

F=Ro/Rw

Where Ro=Resistivity of Rock fully saturated with Brine (100% brine saturation) And Rw=Resistivity of Brine.

We can find Resistivity Index using formula

I=Rt/Ro where Rt is resistivity of Brine at different saturations. Ro=Resistivity of brine at 100% saturation.

We use Archie equation for the determination of Cementation Factor.

F=a/Øm

Where Ø=Porosity, a=Tortuosity factor, m=Cementation factor

I=1/Swn

Where I=resistivity index, Sw=water saturation, n=Saturation Exponent

PROCEDURE:

First of all take the internal diameter and length of the cell using Vernier Calliper. Make sure that the brine we will use is 20,000ppm. Note the temperature of Brine using thermometer. Take the weight of the resistivity cell filled with brine and note the reading of voltage and

current on digital multimeter. After that we can find the resistivity of brine.


We always apply some gel on bottom and top of resistivity cell to avoid leakage. Now instead of brine apply coarse sand in resistivity cell and inject the brine into the cell

with the help of syringe so that resistivity cell is full with brine and sand i.e Sw=100% and note the the readings on DDM for voltage and current.

Now we eject some brine with syringe and note reading of V and I, now Sw is not 100%. We can note the readings accordingly.


RESULTS & DISCUSION:


Cell weight (g)

Cell+sand (g)

Grain volume Vg(cm3)

Pore volume Vp (cm3)

Bulk volume Vb (cm3)

Porosity % =Vp/Vb

Area (m2)

Length (m)

Rw (ohm m)

294.79 320.53 9.75 6.92 16.67 41.5116977 0.000404 0.0412 0.00175

Coarse Sand

V (volt)

A (amp)

R (ohm)

Ro (ohm-m)

Rt (ohm-m) Weight Sw Log Sw F I Log I m n

1.8 7.01 0.256776034 0.002517901 328.07 1 0 1.4388005 1 03.5 5.33 0.656660413 0.006439097 327.39 0.979567308 -0.008965718 2.55733 0.40778635

3.95 4.88 0.80942623 0.007937092 327.33 0.977764423 -0.009765769 3.15227 0.498622795.33 3.31 1.610271903 0.015790045 327.07 0.969951923 -0.013249792 6.27111 0.797344735.82 2.98 1.953020134 0.019150974 327.01 0.968149038 -0.014057782 7.60593 0.881152237.22 1.69 4.272189349 0.041892342 326.98 0.967247596 -0.014462341 16.6378 1.22109601

Resisitivity

1.864 3.782

We know that Diameter=22.7mm D=2.27cm D=0.0227 Length=41.2mm L=4.12 L=0.0412

Area=3.14*(0.0227)2/4 A=0.000404m2 A=4.045cm2

Bulk Volume=A*L Vb=4.045*4.12 Vb=16.67cm3

Rw=V/I * A/L V=1.35vollts I=7.55 put the values in above equation Rw=0.00175 ohm-m

Density=m/v Vg=m/density Density=2.64 g/cm3 Vg=320.53-294.79/2.64 Vg=9.75cm3


Vp=Vb-Vg Vp=6.92cm3

Ø=Vp/Vb Ø=42%

Calculations for Coarse Sand Resistivity : V=1.80 I=7.01 R=V/I R=0.256 ohm

Ro=RA/L By substituting the values R=0.00251 ohm (At 100% saturation)

F=Ro/Rw F=0.00251/0.00175 = 1.438

When saturation is not 100% (from reading 2-6), we use this formula Sw= (cell+sand weight)-(cell weight)/(cell+sand weight at 100% saturation)-(cell weight) Cell+Sand=327.39 Cell weight=294.79 Cell+Sand weight at 100% saturation =328.07 By putting all the values Sw=91% Similarly we can take rest of the readings.

Resistivity Index(I) = Rt/Ro I=0.0064/0.00251 I=2.549

For Cementation Factor m, we know Archie Equation

F=a/Øm

Log F=-mlog(Ø)+log a

Where m=cementation factor and a=Tortuosity

For Saturation Exponent n

I=1/(Sw)n

Log I =-nlogSw

Y=mx+c

We can calculate three values of n by ploting a graph between Log(I) And Log(Sw) for Coarse, Medium and Fine sand.


Results of Medium Sand


Cell weight (g)

Cell+sand (g)




Porosity % =Vp/Vb

Area (m2)

Length (m)

Rw (ohm m)

294.79 321.24 10.018 6.65 16.67 39.8920216 0.000404 0.0412 0.00175

Medium Sand

V(volt) A (amp) R(ohm) Ro(ohm-m) Rt(ohm-m) Weight Sw Log Sw F I Log I m n1.61 7.3 0.220547945 0.002162655 327.98 1 0 1.2358 1 01.69 7.21 0.234396671 0.00229845 327.57 0.987646882 -0.0053983 1.06279236 0.02642.25 6.65 0.338345865 0.00331776 327.26 0.978306719 -0.009525 1.53411479 0.18595.25 3.67 1.430517711 0.01402741 327.08 0.972883399 -0.0119392 6.48619832 0.8127.45 1.35 5.518518519 0.05411363 326.99 0.970171738 -0.0131514 25.0218542 1.39837.98 0.89 8.966292135 0.08792189 326.87 0.966556192 -0.0147729 40.6546165 1.6091

Resisitivity

1.864 1.835

Results of Fine Sand:


Cell weight (g)

Cell+sand (g)




Porosity % =Vp/Vb

Area (m2)

Length (m)

Rw (ohm m)

294.79 320.57 9.76 6.9 16.67 41.3917217 0.000404 0.0412 0.00175

Fine Sand

V (volts)

A (amps)

R (ohm)

Ro (ohm-m)

Rt (ohm-m) Weight Sw Log Sw F I Log I m n

1.85 6.91 0.267727931 0.002625293 327.91 1 0 1.50017 1 02.67 6.08 0.439144737 0.00430618 327.12 0.976147343 -0.0104846 1.64026494 0.21494.02 4.72 0.851694915 0.00835157 327.04 0.973731884 -0.0115606 3.1811956 0.50264.53 4.12 1.099514563 0.01078165 326.95 0.971014493 -0.0127743 4.10683548 0.61354.79 3.89 1.231362468 0.01207453 326.67 0.962560386 -0.016572 4.59930522 0.66275.64 3.1 1.819354839 0.01784028 326.55 0.958937198 -0.0182098 6.79553618 0.8322

1.864

Resisitivity

1.539

Once we have all the three values of porosity and formation factor, then we plot a graph of Log(F) vs Log(Ø) to find the value of cementation factor.


Sand F Porosity Log F Log (porosity)Coarse 1.43880052 0.415117 0.158000586 -0.381829481Medium 1.2358026 0.398983 0.091949105 -0.399045608Fine 1.5001674 0.414208 0.176139724 -0.382781517

Y=-1.864x+3.2062

Log F=-mlog(Ø) + Log a

-m=-1.864 ; m=1.864

Turtosity(a)= 3.20

CONCLUSION:

When we plot a graph between Current and Voltage, we can see that there is a inverse relationship between them. It means that when voltage increases, current decreases and when current increases, voltage decreases. We have large jump in readings because packing is uneven near the bottom of the cell and the amount of brine removed from resistivity cell.

Also whenever we insert the new sand sample in the resistivity cell, we have to apply gel on the upper and bottom side of the resistivity cell to avoid the leakage. Otherwise we problem with calculations.

From the tables of coarse sand, medium sand, fine sand, we can see that the pore space decreases and resistivity increases.


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