1. Muhammad Kamal (3325610) Page 1
MSc. PETROLEUM ENGINEERRING
EAB_7_152 Petro Physics
Dr. Elsa Aristodemou
Permeability and Porosity
By
Muhammad Kamal
Student ID # 3325610
SubmissionDate: 30/11/2014

2. Muhammad Kamal (3325610) Page 2
FORMATION RESISTIVITY
ABSTRACT:
We can measure the Resistivityof three Sandswhichare Coarse sand,MediumsandandFine sand
for the determinationof Resistivityindex IandFormationFactorF. We can alsomeasure the
Porosityof eachsand.When we measure FormationF andPorosityforeachsand, thenwe plota
graph betweenLog(F) vsLog(Ø) toobtainCementationFactorM. Likewisewhenwe measure the
resistivityindexandwatersaturationthenwe plotagraph betweenLog(I) andLog(Sw) toobtainthe
value of n.
INTRODUCTION:
In thisexperimentwe have tofindElectrical Resistivityof porousmediaandmeasure the
cementationfactorandsaturationexponent.We canfindthe electrical resistivityforbothfull and
partial brine saturation.
We can findFormationFactorusing formula
F=Ro/Rw
Where Ro=Resistivityof RockfullysaturatedwithBrine (100% brine saturation)
And Rw=Resistivityof Brine.
We can find ResistivityIndex usingformula
I=Rt/Ro
where Rtis resistivityof Brine atdifferentsaturations.
Ro=Resistivityof brine at100% saturation.
We use Archie equationforthe determinationof CementationFactor.
F=a/Øm
Where Ø=Porosity, a=Tortuosityfactor, m=Cementationfactor
I=1/Swn
Where I=resistivityindex,Sw=watersaturation,n=SaturationExponent
PROCEDURE:
 Firstof all take the internal diameterandlengthof the cell usingVernierCalliper.
 Make sure that the brine we will use is20,000ppm.
 Note the temperature of Brine usingthermometer.
 Take the weightof the resistivitycell filledwithbrine andnote the readingof voltage and
currenton digital multimeter.
 Afterthat we can findthe resistivityof brine.

3. Muhammad Kamal (3325610) Page 3
 We alwaysapply some gel onbottomandtop of resistivitycell toavoidleakage.
 Nowinsteadof brine applycoarse sandinresistivitycellandinjectthe brine intothe cell
withthe helpof syringe sothat resistivitycell isfullwithbrine andsandi.e Sw=100% and
note the the readingsonDDM for voltage andcurrent.
 Nowwe ejectsome brine withsyringe andnote readingof V andI, now Sw isnot 100%.
 We can note the readingsaccordingly.

4. Muhammad Kamal (3325610) Page 4
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 LogSw F I LogI m n
1.8 7.01 0.256776034 0.002517901 328.07 1 0 1.4388005 1 0
3.5 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 0.49862279
5.33 3.31 1.610271903 0.015790045 327.07 0.969951923 -0.013249792 6.27111 0.79734473
5.82 2.98 1.953020134 0.019150974 327.01 0.968149038 -0.014057782 7.60593 0.88115223
7.22 1.69 4.272189349 0.041892342 326.98 0.967247596 -0.014462341 16.6378 1.22109601
Resisitivity
1.864 3.782

5. Muhammad Kamal (3325610) Page 5
We knowthat
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
BulkVolume=A*L
Vb=4.045*4.12
Vb=16.67cm3
Rw=V/I* A/L
V=1.35vollts I=7.55
put the valuesinabove equation
Rw=0.00175 ohm-m
Density=m/v
Vg=m/density
Density=2.64g/cm3
Vg=320.53-294.79/2.64
Vg=9.75cm3

6. Muhammad Kamal (3325610) Page 6
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 substitutingthe values
R=0.00251 ohm (At 100% saturation)
F=Ro/Rw
F=0.00251/0.00175 = 1.438
Whensaturation is not 100% (fromreading2-6),we use thisformula
Sw= (cell+sandweight)-(cell weight)/(cell+sandweightat100% saturation)-(cell weight)
Cell+Sand=327.39
Cell weight=294.79
Cell+Sandweightat100% saturation=328.07
By puttingall the values
Sw=91%
Similarlywe cantake restof the readings.
ResistivityIndex(I) =Rt/Ro
I=0.0064/0.00251
I=2.549
For CementationFactor m, we knowArchie Equation
F=a/Øm
Log F=-mlog(Ø)+log a
Where m=cementationfactoranda=Tortuosity
For Saturation Exponentn
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.

7. Muhammad Kamal (3325610) Page 7
Results of Medium Sand
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)
294.79 321.24 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 n
1.61 7.3 0.220547945 0.002162655 327.98 1 0 1.2358 1 0
1.69 7.21 0.234396671 0.00229845 327.57 0.987646882 -0.0053983 1.06279236 0.0264
2.25 6.65 0.338345865 0.00331776 327.26 0.978306719 -0.009525 1.53411479 0.1859
5.25 3.67 1.430517711 0.01402741 327.08 0.972883399 -0.0119392 6.48619832 0.812
7.45 1.35 5.518518519 0.05411363 326.99 0.970171738 -0.0131514 25.0218542 1.3983
7.98 0.89 8.966292135 0.08792189 326.87 0.966556192 -0.0147729 40.6546165 1.6091
Resisitivity
1.864 1.835

8. Muhammad Kamal (3325610) Page 8
Results of Fine Sand:
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)
294.79 320.57 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 LogSw F I LogI m n
1.85 6.91 0.267727931 0.002625293 327.91 1 0 1.50017 1 0
2.67 6.08 0.439144737 0.00430618 327.12 0.976147343 -0.0104846 1.64026494 0.2149
4.02 4.72 0.851694915 0.00835157 327.04 0.973731884 -0.0115606 3.1811956 0.5026
4.53 4.12 1.099514563 0.01078165 326.95 0.971014493 -0.0127743 4.10683548 0.6135
4.79 3.89 1.231362468 0.01207453 326.67 0.962560386 -0.016572 4.59930522 0.6627
5.64 3.1 1.819354839 0.01784028 326.55 0.958937198 -0.0182098 6.79553618 0.8322
1.864
Resisitivity
1.539

9. Muhammad Kamal (3325610) Page 9
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 -0.381829481
Medium 1.2358026 0.398983 0.091949105 -0.399045608
Fine 1.5001674 0.414208 0.176139724 -0.382781517

10. Muhammad Kamal (3325610) Page 10
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.

11. Muhammad Kamal (3325610) Page 11

12. Muhammad Kamal (3325610) Page 12

13. Muhammad Kamal (3325610) Page 13

14. Muhammad Kamal (3325610) Page 14

15. Muhammad Kamal (3325610) Page 15

16. Muhammad Kamal (3325610) Page 16

17. Muhammad Kamal (3325610) Page 17