PETROLEUM GEOLOGY AND BASIC ROCK PROPERTIES
Petroleum is not found in underground lakes or rivers, but it
exists within the void space of certain rocks.
Requirements for Commercial Oil Accumulations
Certain requirements must be fulfilled for a commercial
petroleum deposit to be present. These are
1. A source: material from which oil is formed
2. Porous and permeable beds (reservoir rocks) in which
the petroleum may migrate and accumulate after being
formed.
3. A trap: subsurface condition restricting further
movement of oil such that it may accumulate in
commercial quantities.
1
Source of Petroleum
Most geologists conclude that:
1.
Petroleum originates from organic material, primarily
vegetable, which has been altered by heat, bacterial
action, pressure and other agents over long periods
of time.
2.
Conditions favouring petroleum formation are found
only in sedimentary rocks.
3.
The principal sediments generally considered as
probable source rocks are shales and limestones that
were originally muds under saline water.
2
Porous and Permeable Beds (Reservoir Rocks)
After its formation, petroleum may migrate from the
source rock into porous and permeable beds where it
accumulates
and continues
its
migration
until finally
trapped. The forces causing this migration are
1. Compaction of sediments as depth of burial increases.
2. Diastrophism: crustal movements causing pressure
differentials
and
consequent
subsurface
fluid
movements.
3. Capillary forces causing oil to be expelled from fine
pores by the preferential entry of water.
4. Gravity which promotes fluid segregation according to
density differences.
3
4
Porosity
Porosity is a measure of the void space within a rock
expressed as a fraction (or percentage) of the bulk volume
of rock

Vb  Vs Vp

Vb
Vb
where
 = porosity
Vb = bulk volume of rock
Vs = net volume occupied by solid
Vp = pore volume
Rock porosity can be classified as
1.
Absolute porosity: total porosity of the rock,
regardless of whether or not the individual voids are
connected, and
2.
Effective porosity: only that porosity due to voids
which are interconnected.
It is the effective porosity which is of interest to the oil
industry.
5
Geological porosity has been classified in two types:
1.
Primary porosity (intergranular): Porosity formed at
the time sediment was deposited.
The voids
contributing to this type are the spaces between
individual grains of the sediment.
2.
Secondary porosity:
Voids formed after the
sediment was deposited.
Porosity of this type is
subdivided
classes
into
three
based
on
the
mechanism of formation.
i.
Solution porosity: voids formed by the solution
of the more soluble portions of the rock in
percolating surface and subsurface waters
containing carbonic and other organic acids.
Voids of this origin may range from small vugs
to cavernous openings.
ii.
Fractures, fissures and joints:
voids of this
type are common in many sedimentary rocks and
6
are formed by structural failure of the rock
under loads caused by various diastrophism such
as folding and faulting. This form of porosity is
extremely hard to evaluate quantitatively due
to its irregularity.
iii.
Dolomitization:
This is a process by which
limestone (CaCO3) is transformed into dolomite
Ca Mg(CO3)2.
2CaCO3 + MgCl2
Dolomite
is
CaMg(CO3)2 + CaCl2
normally
more
porous
than
limestone.
Typical Porosity Magnitude
Type of sedimentary rocks
Clean, consolidated and reasonably uniform
sand
Carboniferous rocks (limestone and dolomite)
porosity
20%
6 – 8%
7
Quantitative Use of Porosity Data
Let us assume that porosity has been measured and may be
used to determine the quantity of fluid which may be stored
within the rock.
Consider a bulk volume of rock with a surface area of 1 acre
and a thickness of 1 foot. This constitutes the basic rock
volume measurement used in oil field calculations, an acrefoot.
It is a standard practice to express all liquid volumes in
terms of barrels (bbl). Conversion factors used are:
1 acre = 43,560 ft2
1 acre-ft = 43,560 ft3
1 bbl = 42 gal = 5.61 ft3
1 acre-ft =
43,560
= 7758 bbl
5.61
Then the pore space within a rock,
Vp (bbl/acre-ft) = 7758 x 
where  is the porosity of the rock
8
Oil in Place = N =
where
7758  So 7758  (1 - Sw )

Bo
Bo
N
= tank oil in place, bbl/acre-ft
So = Fraction of pore space occupied by
oil (the oil saturation)
Sw = The water saturation
Bo = The formation volume factor for the
oil at the reservoir pressure,
res. bbl/STB
The water within the pore is commonly called the connate
water.
The pore space is assumed to be occupied by oil and water
only and that no free gas is present. So the equation above
must be applied to the reservoir at or above the bubble
point and is generally used to compute the initial oil in place
(IOIP).
For the gas stored in a particular sand a similar expression
may be derived. The gas volume is commonly expressed in
terms of SCF or in MCF (thousands of standard cubic feet).
9
From the Gas Law
PsVs PVp
=
Ts
zT
where subscript, s, denotes standard conditions, zs = 1.0,
and is not shown. Then;
Vs = G = Vp ×
PTs
zTPs
where G is the standard gas volume contained in Vp at
conditions P, T, z.
But:
Vp = 43,560(1-Sw) ft3/acre-ft
Ts = 4600 + 600 = 5200F
Ps = 14.7 psia
Substituting of these values in the equation gives:
G = 43,560(1-Sw)x
520 p
×
14 .7 zT
Or
G
1540(1 - S w )P
MCF/acre - ft
zT
10
Reserve estimation
Any oil finding has to be interpretated in term of money or
in term of economic evaluation.
Initial Oil in Place (IOIP) has already being defined in term
of bbl/acre.ft. It can also be expressed as
IOIP 
7758  volume of reservoir  porosity  oil saturation
oil formation volume factor
IOIP 
Where
bbl
7758  Ah    So
Bo
A = area of reservoir in acre
h = height or thickness of reservoir in feet
 = porosity in fraction
So = oil saturation in fraction
Bo = Oil formation volume factor, res bbl/STB
Only a portion of the IOIP that can be recovered.
This
portion is known as the oil reserve.
11
Oil reserve is dependent on the recovery factor (RF).
Reserve = IOIP x RF
Permeability
Permeability is defined as a measure of a rock’s ability to
transmit fluids.
An empirical relationship was developed by a French
hydrologist Henry D’arcy who studied the flow of water
through unconsolidated sand.
This law in its differential form is:
v
where
k dP
 dL
(1)
v = apparent flow velocity
 = viscosity of the flowing fluid
dP/dL = pressure gradient in the direction of
the flow
k = permeability of the porous media
12
Consider the linear system of the figure below
dP
P1
P2
q2
q1
dL
L
The following assumptions are necessary to the development
of the basic flow equations:
1.
2.
3.
4.
5.
6.
Steady state flow conditions exists
The pore space of the rock is 100% saturated with
the flowing fluid. Under this restriction, k is the
absolute permeability.
The viscosity of the flowing fluid is constant.
Isothermal conditions prevail.
Flow is horizontal and linear.
Flow is laminar.
With these restrictions, let
v
where
q
A
(2)
q = volumetric rate of flow of fluid
A = cross-sectional area perpendicular to flow
direction
13
Case 1: Linear Incompressible Fluid Flow
Substitution of (2) into (1) gives
q
k dP

A
 dL
(3)
Separation of variables and insertion of the limits depicted
by the figure, gives
q L
k
dL  
A 0



q=
kA(P1 P2 )
μL
P2
P1
dP
(4)
or
k=
quL
AΔP
(5)
Unit for the above relationship is
If
then,
q = 1 cm3/s
A = 1 cm2
 = 1 centipoise
P/L = 1 atmosphere/cm
k = 1 darcy
14
A permeability of one darcy is much higher than that
commonly found in reservoir rocks.
Consequently, a more
common unit is the millidarcy, where
1 darcy = 1000 millidarcys
Case II: Linear Compressible Fluid Flow
Consider the same linear system(referring to the box
figure), but the flowing fluid is now compressible.
Assuming that Boyle’s law is valid (z = 1)
P1q1 = P2q2 = constant
P.q = -
q2

L
0
kA dP
×
P = P2q2
μ dL
dL  
kA 1
x
 P2

P2
P1
P dP
From which
kA P12  P22 1
q2 
x
x
L
2
P2
15
Expressing the above equation in term of qg, the rate of gas
flow at the average pressure in the system is
kA P12  P22 1
qg 
x
x
L
2
P
P P
P  1 2 , and
2
But
Therefore
qg 
P12  P22 (P1  P2 )(P1  P2 )

2
2
kA P
L
Which is exactly the same as equation (4).
An expression for the standard flow rate, qgas is obtained
from Charles’ Law:
Ps qgs
Ts

Where
P2  q2 kA 2
1

(P1  P22 )
Tf
2L
Tf
Ts = 600F (5200R)
Ps = 1 atm
Tf = flowing temperature
Thus,
kA(P12  P22 ) Ts 1
qgs 
x
x
2L
Tf Ps
16
Case III: Radial Incompressible Fluid Flow
From the diffrential form of equation (1) with notation and
sign convention as applied to radial flow in the figure.
q k dp
 x
A  dr
q
Pe
re
Pw
q
q
rw
q
h
But radial flow
Where
A = 2rh
r = radius or distance from centre, cm
h = thickness of the bed, cm
Substitution of 2rh for A and separation of variables gives
q

re
rw
dr 2hk

r


Pe
Pw
dP
17
Which when integrated is
This is the basic expression for the steady state radial flow
of a liquid. The units are the same as previously defined.
Case IV: Radial Compressible Fluid Flow
The same manner as in case II, the radial equations for
gases may be obtained.
By Boyle’s Law
Where subscripts refer to position at which q is specified:
well, external boundary, etc.
18
Conversions to Practical Units
The standard units which define the darcy are useful in
laboratory calculations.
For computations pertaining to
field problems it is more convenient to convert to practical
units by use of appropriate constant.
For example, convert
to
where
q
2hk Pe  Pw 
 lnre / rw 
q
Chk Pe  Pw 
 lnre / rw 
h = ft, k = darcy, Pe, Pw = psia,  = cp, q = bbl/day
Conversion factors needed: 1 bbl = 159,000 cm3
1 ft = 30.48 cm
1 atm = 14.7 psi
Then:
159,000 cm3 / bbl
q bbl/day x
24 x 3,600 s / day
cm  
1 atm 
2 h ft x 30.48
k Pe  Pw psi x
ft  
14.7 psi 


 lnre / rw 
or
q
(24)(3600)(2)(30.48) hk(Pe  Pw )
x
(159,000)(14.7)
 ln(re / rw )
19
q
7.07hk (Pe  Pw )
 ln(re  rw )
Typical permeability magnitude
In general, rocks having a permeability of 100 md or greater
are considered fairly permeable, while rocks with less than
50 md are considered tight.
Many productive limestone and dolomite matrices have
permeability below 1 md but due to the associated solution
cavities and fractures which contribute the bulk flow of the
flow capacity.
Current stimulation techniques of acidizing and hydraulic
fracturing allow commercial production to be obtained from
reservoir rocks once considered too tight to be of interest.
The
oil
and
gas
reservoirs
in
Malaysia
are
having
permeability between 50 to 2000 millidarcy.
20
Petroleum Traps
In order for petroleum to accumulate in commercial
quantities , it must, in its migration process, encounter a
subsurface rock condition which halts further migration and
causes the accumulation to take place.
Numerous systems of trap classification exist, such as:
1.
Structural traps: those traps formed by
deformation of the earth’s crust by either
faulting or folding.
21
2.
Stratigraphic traps:
those traps formed by
changes in lithology, generally a disappearance of
the containing bed or porosity zone.
22
3.
Combination traps:
traps having both structural
and stratigraphic features.
23
Oil is found in flank sands, upper beds
or the caprock
A feature of all traps is the impermeable cap rock which
forms the top of the trap.
Subsurface Pressure
24
The elevated pressures encountered with depth are due to
one or both of the causes:
1.
Hydrostatic pressure imposed by the weight of fluid
(predominantly water) which fills the voids of the
rocks above and/ or contiguous with the reservoir in
question.
2.
Overburden pressure die to the weight of the rocks
and their fluid content existing above the reservoir.
It is more common to find subsurface pressures varying as
a linear function of depth with a gradient close to the
hydrostatic gradient of fresh to moderately saline water.
Departures from this behaviour, both higher or lower, are
considered abnormal.
The abnormally high pressures are more important as a
source of serious drilling and production hazards.
Magnitude of subsurface Pressure
25
Pressure-depth relationships are commonly spoken of in
terms of gradients.
The hydrostatic gradient in fresh
water is 0.433 psi/ft of depth which is the quotient of 62.4
lb/ft3 divided by 144 in2/ft2.
Since most subsurface waters are saline, it is common to
find the gradient to be more than 0.433 psi/ft.
Studies from 100 high pressure wells in Texas-Louisiana
Gulf Coast showed a pressure gradient of 1.0 psi/ft.
This figure is commonly used and may be obtained by using
an average water saturated rock specific gravity 0f 2.3.
Hence the overburden gradient is
2.3 x 0.433  1.0 psi/ft
Subsurface Temperature
26
The earth is assumed to contain a molten core, it is logical
to assume that temperature should increase with depth.
This temperature-depth relationship is commonly a linear
function of the form:
TD = Ta + D
Where
TD = temperature of the reservoir at any depth, D
Ta = average surface temperature
 = temperature gradient, degrees/100 ft
D = depth, hundreds of ft
A normal gradient seems to about 1.60F/100 ft, although it
should be noted considerable variations occur in various
areas.
Several devices for measuring subsurface temperature are
available and will be discussed under temperature logging.
27