Chapter 4

Permeability
The concept of permeability is similar to that of cars on highways. Imagine that
there is a highway with only one lane and there are 300 cars that need to pass
through it; the flow of cars will be difficult as there is limited space to pass through
it. However, if the number of lanes increases to five, then the flow will become
much easier. In this case, permeability is a measure of the ease with which the
cars flow on the highway. Higher permeability means easier flow which, in this
case, will happen with the five-lane highway. Conversely, the one-lane highway
has lower permeability. Similarly, in rocks, permeability is a measure of the
ease with which the fluid flows in the porous medium. Figure 4.1 represents
a schematic showing the difference between low and high permeability rocks
in a reservoir. The permeability of a rock measured when it is 100% saturated
with a single phase (water, oil, or gas) is often called “single-phase permeability”,
“absolute permeability,” or just “permeability”. If there are two fluids flowing in a
rock, then it relates to another concept known as “relative permeability,” which
will be discussed in Chapter 9. Permeability is part of RCAL and is considered as a
flow or transport property that helps in understanding the flow in the reservoir.
The concept of permeability was first introduced by French civil engineer Henry
Darcy in 1856 when he performed an experiment on sand filtrates and analyzed
the concept of permeability. Darcy’s law for a single phase (liquid) is expressed
as:

kA
q=− dP (4.1)
µL

where q is the flow rate [m3/s], k is the permeability [m2], A is the core cross-
sectional area perpendicular to the flow [m2], L is the length of the core [m], dP is
the pressure difference across the core [Pa or N/m2], and µ is the viscosity of the
injected fluid [Pa.s or N/m2.s]

This equation is the linear form of Darcy’s law for incompressible fluid, which is
discussed further in the following sections.

55
 (a) (b)

Figure 4.1: Schematic showing the cross section at the micro-scale of (a) a lower permeability
rock and (b) a higher permeability rock. The fluid flow is much easier in rock (b) compared to
rock (a).

4.1 Applications of Permeability

Permeability is a parameter that describes the flow in porous media. From Darcy’s
law, we can estimate the production flow rate from the reservoir to the surface.
This can be done by determining the permeability of the reservoir through
laboratory experiments on core samples extracted from the same reservoir, as
well as determining all the parameters associated with Darcy’s law.

4.2 Validity of Darcy’s Law for Single-Phase Permeability

Darcy’s law for single-phase flow is valid under some conditions, which include:

1. The core sample used needs to be 100% saturated with a single

phase (water, oil, or gas). If the system consists of more than one
fluid, then we need to consider relative permeability, which will be
discussed in Chapter 9.

2. The flow has to be laminar. Flow can be characterized as either

laminar or turbulent. Laminar flow is defined as the “slow,” uniform flow
while turbulent flow is defined as the “fast,” chaotic flow (Figure 4.2).
In order to determine whether the flow is laminar or turbulent, a
dimensionless number is used, known as the Reynolds number. This
number is obtained from the following equation:

ρvD
Re = (4.2)
µ

where Re is the Reynolds number [dimensionless], ρ is the density of the

fluid [kg/m3], v is the velocity of the fluid [m/s], D is the pipe diameter
[m], and µ is the viscosity of the fluid [Pa.s]. A flow with Reynolds number of
2100 or less is considered laminar.

This equation is mainly used in pipes as they have a fixed diameter;

however, for a porous medium, an average grain diameter is used. Since

56
 it is difficult to compute Reynolds number in porous media, another
technique is used to determine whether the flow is laminar or turbulent.
This technique is used when measuring permeability in the laboratory
and will be discussed in the following sections. It is important to mention
that in a reservoir, the flow is generally laminar.

3. The flow has to be steady-state flow. Steady-state flow means that

whatever enters the system leaves the system, or that there is no
volumetric change over time. This is true for fluid flow in core samples if
the same amount of fluid enters and leaves the system over a certain
period. If the flow is unsteady-state, then Darcy’s law is invalid.

(a) (b)

Figure 4.2: Schematic showing the two different types of flow: (a) laminar flow (a flow that is
uniform, smooth, and occurs at low flow rates) and (b) turbulent flow (a flow that is chaotic and
occurs at high flow rates compared to laminar flow). The flow in porous media has to be laminar
in order for Darcy’s law to be valid.

4.3 Darcy’s Law Under Different Boundary Conditions

In this section, we will derive Darcy’s law for different boundary conditions by
starting from the differential form of Darcy’s law:

kA dP
q=− (4.3)
µ dx

where q is the flow rate [m3/s], k is the permeability [m2], A is the cross-sectional
area perpendicular to flow [m2], dx is the change in length [m], dP is the pressure
difference across the core [Pa], and µ is the viscosity of the injected fluid [Pa.s].

Fluids can be either compressible or incompressible. “Fluid” is a term that

refers to liquids and gases. Compressible fluids are fluids that change volume
due to a change in pressure, such as gases. Liquids, on the other hand, are
considered incompressible because their volumes change negligibly with a
change in pressure. An example of compressible fluids in real-life applications
is scuba tanks, where air is compressed inside the tank. The actual volume of
the compressed air in the tank, under atmospheric pressure, is much larger
than the actual volume of the tank itself. However, if we wish to fill the same
tank with water (which is an almost incompressible fluid), the volume of water
that fills the tank will approximately be the same inside and outside the tank.
When we consider flow in porous media, we need to consider whether the flow is
compressible or incompressible, as this will change the governing equations. In

57
addition, we need to consider whether the flow is linear in Cartesian coordinates
(most laboratory experiments) or radial coordinates (reservoir conditions).

4.3.1 Case 1: Linear solution of Darcy’s Law for incompressible fluid

Let us assume we have the system shown in Figure 4.3, and we inject it with an
incompressible fluid such as water. Based on the system, we know that the length
is L, the inlet pressure is P1, and the outlet pressure is P2. In order for the flow to
occur from the inlet to outlet, P1 needs to be greater than P2, because fluid flows
from high pressure to low pressure. This is analogous to the movement of most
of the other flows, such as the flow of electrical charges from high to low voltage,
and heat transfer from high to low temperature. Our homes also have a real-life
example of such flows. In a vacuum cleaner, for the particles to flow to the dust
collector, a pressure lower than the atmospheric pressure is required inside the
appliance. The vacuum cleaner achieves this by creating a partial vacuum inside
the machine for the particles or dust to flow towards it.

Darcy’s law is usually written with a negative sign. This is because dP = P2 – P1 and
since P2 is smaller than P1, a negative sign will make the overall equation positive.
For simplicity, we will use the following differential form of Darcy’s law in this
book:

kA dP
q= (4.4)
µ dx

where dP = P1 – P2 and P1 > P2.

Now, we can rearrange the equation to obtain:

kA
qdx = dP (4.5)
µ

Taking the integral with the boundary limits as shown in Figure 4.3:
 L  P1
kA
q dx = dP (4.6)
0 µ P2

Then, the equation becomes:

kA
q (L − 0) = (P1 − P2 ) (4.7)
µ

Finally, we divide both sides by L:

kA
q= (P1 − P2 ) (4.8)
µL

58
Equation 4.8 is the final form of Darcy’s law for an incompressible linear system.

q
A

P1 dP P2

0 dx L

Figure 4.3: Schematic showing a cylindrical core sample used for permeability measurement,
where a cross section is taken to display the boundary limits of the linear system.

4.3.2 Case 2: Radial solution of Darcy’s Law for incompressible fluid

In this case, the only difference is that the flow occurs in a radial manner as
opposed to linear. This flow is more representative of a reservoir. One of the
changes is that the area perpendicular to the flow is the circumference of the
circle (2πr) multiplied by the thickness of the reservoir (h) (in contrast to πr2 in
the linear flow for a cylindrical-shaped object). The second change is that instead
of the flow varying in the x-coordinate, it will vary in the r-coordinate and thus
instead of having dP/dx, we will have dP/dr. Figure 4.4 shows the radial system
of the reservoir.

We start with the differential form of Darcy’s law (Equation 4.4) and substitute
the area perpendicular to flow, which is the circumference of the circle, and
change the coordinates from linear to radial. The equation becomes:

2kπrh dP
q= (4.9)
µ dr

59
Then, we rearrange the equation to obtain:
dr 2πkh
q = dP (4.10)
r µ

Now, we take the integral with the boundary limits in agreement with Figure 4.4,
thus obtaining:
 re  Pe
1 2πkh
q dr = dP (4.11)
rw r µ Pwf

where re is the reservoir’s outer radius [ft], rw is the well radius [ft], Pe is the
reservoir’s outer pressure [psia], and Pwf is the wellbore flowing pressure [psia].

Then, we integrate the equation with the given boundary condition to obtain:

2πkh
q (ln re − ln rw ) = (Pe − Pwf ) (4.12)
µ
Finally, we rearrange the equation and use the natural logarithm properties to
obtain:
2πkh (Pe − Pwf )
q= (4.13)
µln rrwe
This is the final form of Darcy’s law for incompressible radial system.

Dealing with gases is different from liquids as gases are compressible fluids.
Hence, we need to account for the change in volume when deriving the equation.
We will discuss that in the later sections.

In addition, equations for Darcy’s law for different systems can also be modified
to account for the gravity term in vertical systems. Nevertheless, these equations
are rarely used in petroleum engineering, as the flow in the reservoir and in
laboratory experiments is mainly horizontal. Therefore, it is not discussed in
this book.

60
 re

q(r+dr)
dr

rw
Center of
the well
q(r)

Pwf dp Pe h

Pwf dr Pe h

rw
r
r + dr
re

(a) (b)
Figure 4.4: Schematic showing (a) a radial system where the flow occurs from the outer boundary
to the wellbore region and (b) a zoomed in section of one part of the reservoir to clearly explain
the system.

Unit Systems

When using Darcy’s law, three main unit systems can be used to find permeability.
These units are explained in Table 4.1. It is important to note that since
permeability values are very small in m2, a unit of Darcy [D] or milli Darcy [mD]
is used, named in honor of Henry Darcy. Moreover, the unit bbl/d in the table
indicates barrel (bbl) per day (d), which is a standard unit to quantify volume in
the oilfield units.

Table 4.1: Summary of different units used when dealing with permeability.

Parameter SI Units Darcy Units Oilfield Units

q (flow rate) [m3/s] [cm3/s] or [cc/s] [bbl/d]

L (length) [m] [cm] [ft]

A (area) [m2] [cm2] [ft2]

h (thickness) [m] [cm] [ft]

r (radius) [m] [cm] [ft]

P (pressure) [Pa] [atm] [psia]

k (permeability) [m2] [D] [mD]

µ (viscosity) [Pa.s] [cP] [cP]

61
By rearranging Equation 4.8, the following equation is obtained, which can be
used to find permeability in SI or Darcy units:

qµL
k= (4.14)
A(P1 − P2 )

However, for oilfield units, this equation is multiplied by a conversion factor for
linear flow:

qµL
k = 887.2 (4.15)
A(P1 − P2 )

and similarly, for radial flow:

qµln rrwe qµln rrwe

k = 887.2 = 141.2 (4.16)
2πh(Pe − Pwf ) h(Pe − Pwf )

Example 4.1
In an experiment designed to measure the permeability of a core sample
using brine, the cylindrical core was saturated with brine, and then water
was injected at a rate of 0.07 cm3/s. The pressure at the inlet was measured
to be 43 psig while the outlet was open to atmospheric pressure. Knowing
that the core length is 3.5 in, the core diameter is 1.48 in, and the brine
viscosity is 1 cP, calculate the permeability of the core.

Solution
The units must always be consistent when performing these calculations.
Therefore, the flow rate must be converted from cm3/s to bbl/day so that
all the variables are in oilfield units:

1 cm3/s = 0.54 bbl/d

0.07 cm3/s = 0.0378 bbl/d

Furthermore, the length and diameter of the core must also be converted
from inch to feet. This will be done in the calculation.

Equation 4.15 can be used to find permeability (in oilfield units):

62
 qµL
k = 887.2
A(P1 − P2 )
0.0378 × 1 × 3.5
k = 887.2
π( 0.74 2
12
= 19.04 mD
12 ) × (43 − 0)
Note that the answer when solving for permeability in oilfield units will
always be in mD. Also note that a conversion factor of 887.2 is used when
solving in oilfield units, but it is not used when solving for permeability in SI
or Darcy units.

Example 4.2
a) Derive the conversion of permeability from D to m2

b) Convert a permeability of 3.5 mD to m2

c) Convert a permeability of 7.3×10-12 m2 to D

Solution
a) For this conversion, the unit systems in Table 4.1 need to be analyzed.
Using Equation 4.14, we obtain:
qµL
k=
A(P1 − P2 )
We now need to reduce this equation to its base units for both the Darcy
unit system and the SI unit system. For the Darcy unit system, we obtain:

3
cm .cP. cm
2
cm .cP
k= 2 = =D
s. cm .atm s.atm

For the SI unit system, we obtain:

3
m .Pa.s.m
2
m .Pa.s
k= 2 = = m2
s.m .Pa s.Pa

It is apparent that a conversion factor is required to convert permeability

from D to m2. To find this conversion factor (we will call it β), the permeability
equations obtained from the two unit systems will be equated:

63
 2
D =β m
10
−4 m2 Pa 1 cP
2 2
cm .cP m .Pa.s β= × ×
=β m
2 101325 Pa 1000 cP
s.atm s.Pa
2
9.869 × 10−13
cm Pa cP 10
−4
β= 2 × × β= =
m atm Pa.s 101325 × 1000

Therefore, 1 D would be equal to 9.869 × 10-13 m2.

b) Using the conversion factor β:

2
1 D =β m

−3 −3 2
3.5 mD = 3.5 × 10 D = 3.5 × 10 ×β m

3.5 mD = 3.5 × 10
−3
× 9.869 × 10
−13
m
2
= 3.45 × 10−15 m2
c) Using the conversion factor β:
2
1 D =β m

2 D
1 m =
β
7.4 D
7.3 × 10
−12 7.3 × 10−12
−12 2
7.3 × 10 m = D = −13 D =
β 9.869 × 10

4.4 Laboratory Measurements of Absolute Permeability

The permeability of core samples is measured using either liquid or gas. The
procedure and the governing equation is different for each case.

4.4.1 Liquid Permeability

Before we start measuring the permeability of a core sample, we first need to

measure its dimensions (length and diameter) as length and area are part of
Darcy’s law. Secondly, we insert the core sample in a core holder as shown in
Figure 4.5. A common procedure is to vacuum the core using a vacuum pump
prior to injecting any liquid. This is to remove any air from the system and to
ensure the flow of only one phase. Then, a liquid (in this case water) is injected
at a specific rate using a pump. Before injecting the water, we need to ensure
that a confining pressure, similar to the overburden pressure that is squeezing
the core sample from all the sides, is applied. This is done in order to guarantee
that the water is only going through the core sample and not bypassing it as that
will generate errors in the measurement. A general rule of thumb is to apply a
confining pressure (the oil pressure in Figure 4.5) that is at least 1.5 times higher
than the water injection pressure. When water is injected at a constant flow rate,

64
a wait time is required in order to achieve a steady- state flow where the inlet and
outlet pressures become constant and do not fluctuate. After the steady-state
flow is achieved, we record the flow rate, inlet pressure, and outlet pressures. We
then move on to a new flow rate and follow the same procedure. After recording
a few data points, we can plot the data in order to find the permeability of the
core sample.

If we rearrange Darcy’s law for the liquid phase (Equation 4.8), it becomes:

q k dP
= (4.17)
A µ L

This equation is in the linear form y = mx+b, where y is the dependent variable
(q/A), x is the independent variable (dP/L), m is the slope (k/µ), and b is the
y-intercept. The y-intercept in this case is 0 since the curve goes through the
origin.

If we plot this figure with our data, a figure similar to Figure 4.6 will be generated.
The slope of that figure will be equivalent to the permeability divided by the
viscosity. In order to obtain the permeability, the slope needs to be multiplied
by the viscosity. When analyzing experimental data, make sure you follow
consistent units as shown in Table 4.1. Darcy’s units are the most commonly
used units when analyzing laboratory data.

0 00
30

Oil

0 00
30
0 00
30

Water
Core

Figure 4.5: Schematic showing the experimental set-up to measure the liquid permeability of
a core sample. In this system, we inject water through the core sample and use oil to apply a
confining pressure, which should be higher than the water injection pressure.

65
 q
A

k
Slope=
μ

k = (slope . μ)
(P1 - P2)
L

Figure 4.6: A plot analyzing laboratory measurements of liquid permeability. The y-axis is q/A and
the x-axis is dP/L while the slope is equivalent to k/µ.

4.4.2 Gas Permeability

Dealing with gas is different because gas is compressible as opposed to liquids.

Thus, the governing equations will be different as well to account for this change
in volume. Measuring the gas permeability has advantages over measuring liquid
permeability. The measurements for gas take less time than those for liquid,
and gas does not wet the core, which means the core can be reused for further
analyses. The disadvantage is that gas permeability requires correction as it
tends to be overestimated compared to liquid permeability. Figure 4.7 shows the
typical experimental set-up used to measure gas permeability. Since it is difficult
to control the gas flow rate, the core holder is attached to a gas cylinder, and a
metering valve is used to vary the flow rate. There is a flow meter at the end of
the core to measure the flow rate at atmospheric conditions. The simplest way
to derive the equation for the gas permeability is to take the average pressure
across the core:

P 1 + P2
P̄ = (4.18)
2
where P is the average pressure across the core [atm].

We use the average pressure because the flow rate varies across the core and
an average value would be more representative of the flow in the core. We can
use Boyle’s law as shown in Equation 2.11. We divide both volumes by time (t) in
order to convert it to flow rate. Thus, the equation becomes:

qa Pa = q̄ P̄ (4.19)

66
where qa is the atmospheric flow rate [cm3/s], Pa is the atmospheric pressure [1
atm], and q is the average flow rate across the core [cm3/s].

We compare the average flow rate across the core to the flow rate at atmospheric
conditions; since the atmospheric pressure is 1 atm, it can be eliminated from
the equation.

If we rearrange Equation 4.19, it becomes:

q a Pa
q̄ = (4.20)
P̄
Now, we substitute this equation in Equation 4.8:

kA qa Pa
q̄ = (P1 − P2 ) = (4.21)
µL P̄

Next, we can move to the other side of the equation and substitute Equation
4.18 in it to obtain:

kA (P1 + P2 )
qa = (P1 − P2 ) (4.22)
µPa L 2

Then, by rearranging the equation and factorizing it, we obtain:

kA  2 
qa = P1 − P22 (4.23)
2µPa L

For simplicity, Pa is substituted as 1 atm when using Darcy’s units and thus the
equation becomes:

kA  2 
qa = P1 − P22 (4.24)
2µL

Similar to liquid permeability, we can rearrange the equation to find the gas
permeability across the core after acquiring several data points (Figure 4.8).
When dealing with gases, it is more common to reach a higher flow rate than
liquids, because gases have lower viscosity that can result in turbulent flow,
making Darcy’s law invalid. This can be seen from the plot. A laminar flow, which
might also be referred to as “Darcy’s flow,” will follow the slope from the origin
and, once the slope deviates, enter the turbulent flow regime or “non-Darcy’s
flow”. The data that falls in the turbulent flow regime has to be omitted
from the analysis.

When comparing gas permeability to liquid permeability, gas permeability

tends to be higher than the latter. This is due to a gas slippage at the pore walls
known as the Klinkenberg effect. This gas slip makes the permeability higher

67
than what it should be; therefore, it is not representative of the actual value.
Fortunately, this can be corrected by computing the gas permeability (kg) at every
data point and then plotting it against the inverse of P, as shown in Figure 4.9.
The y-intercept of this line is the equivalent liquid permeability (kL). The x-axis
of the plot is the inverse of P, so a zero value of x represents infinite pressure.
At infinite pressure, gas can be considered to behave as a liquid. One factor
that affects this slippage is gas molecular weight. As the gas molecular weight
increases, the slippage decreases since gas becomes heavier and closer to liquid.
This effect is shown in Figure 4.10.

0 00
30

Oil

0 00 0 00
30 30

Gas
Core
Metering Valve
Flow Meter

Gas Cylinder

Figure 4.7: Schematic showing the experimental set-up for measuring the gas permeability of a
core sample.

Laminar Flow Turbulent Flow

qa
A

k
Slope=
μ

k = (slope . μ)
(P12 - P22)
2L

Figure 4.8: Analysis of laboratory measurements of gas permeability using Pa = 1 atm.

68
 Slippage

kg

Absolute or Liquid Permeability (kL)

1
P

Figure 4.9: A plot showing how to correct the gas permeability to liquid permeability through
plotting the gas permeability (kg) of each data point as a function of the inverse of the average
pressure. The intercept of this line is the corrected liquid permeability (kL).

Increasing gas molecular weight

He

N2

CO2
kg

He < N2 < CO2

1
P

Figure 4.10: A plot showing the effect of gas molecular weight on gas slippage. The higher the gas
molecular weight, the lower the slippage.

69
 Example 4.3
A core was mounted in a gas permeameter to measure permeability.
Determine the liquid permeability of the core using gas (in Darcy’s units).
The laboratory data are as follows:

- Diameter of the core = 3.78 cm

- Length of the core = 7.63 cm
- Gas viscosity = 0.0148 cP

q [cm3/s] P1 [psig] P2 [psig]

9.07 6.45 1.27

3.44 2.67 0.49

1.41 1.13 0.22

Solution
First, we will ensure that the units are consistent. We will use Darcy units,
so the dimensions will be in cm, the viscosity will be in cP, the flow rate will
be in cm3/s, and the pressures will be in atm. The cross-sectional area of the
core will first be calculated:

d2 3.782 2
A=π =π = 11.22 cm
4 4

Next, the pressures will be converted from psig to atm. The table will then
look like the following:

q [cm3/s] P1 [atm] P2 [atm]

9.07 1.439 1.086

3.44 1.182 1.033

1.41 1.077 1.015

Since we are dealing with gas permeability, Equation 4.24 will be rearranged
so that the permeability for each flow rate can be calculated:

2qµL
kg =
A(P12 − P22 )

70
 Furthermore, the average pressure for each flow rate will also be calculated:

P 1 + P2
P̄ =
2
The permeability and average pressure for each flow rate is calculated and
recorded in the table below, along with the inverse of the average pressure:

kg [D] P [atm] 1⁄P [atm-1]

0.204 1.262 0.792

0.213 1.107 0.903

0.219 1.046 0.956

The permeability is then plotted against the inverse of the average pressure,
and the y-intercept of the line of best fit through the data point gives the
corrected liquid permeability of the core:
0.3

0.25

0.2

kg [D] 0.15
y= 0.0923x+0.1305

0.1

0.05

0
0 0.2 0.4 0.6 0.8 1
1
[atm-1]
P

From the equation of the line, the corrected liquid permeability of the core
is 0.1305 D.

4.5 Pressure Profile

By understanding rock permeability, we will be able to analyze the pressure

profile across core samples. The pressure profile differs depending on the
injected phase, and so the governing equations are different for gas and liquid.

71
4.5.1 Pressure Profile: Liquid Flow

By knowing the permeability, we can estimate the pressure at any point in the
core. This can be done by rearranging Darcy’s law for liquid (Equation 4.8) and
by considering the outlet pressure P2 as the varying parameter P(x). The resulting
equation becomes:
qµ
P (x) = P1 − x (4.25)
kA
By analyzing this equation, we can expect the pressure profile to be linear.
Furthermore, by varying the distance across the core, we can find the pressure
across the core until we reach its end, which will represent P2. Figure 4.11 shows
the concept of pressure profile when using a liquid.

P1 P(x1) P2

P1

P(x1)
P

P2

0
0 x1 L
x

Figure 4.11: Pressure profile for a core sample when using a liquid.

4.5.2 Pressure Profile: Gas Flow

Since dealing with gases is different from dealing with liquids, we have to use
Equation 4.23 for this analysis. Again, we will replace P2 by P(x) and rearrange
the equation to obtain:

2qa µ
P 2 (x) = P12 − x (4.26)
kA

72
Then, we will take the square root for both sides to obtain:

2qa µ
P (x) = P12 − x (4.27)
kA
For this equation, the pressure profile would have a parabolic shape as shown
in Figure 4.12.

The concept of pressure profile can go beyond core samples and can be extended
to analyze reservoirs. It is very important to understand the pressure profile as it
can help in understanding fluid flow in reservoirs.

P1 P(x1) P2

P1

P(x1)

P2

0
0 x1 L
x

Figure 4.12: Pressure profile for a core sample when using a gas.

73
 Example 4.4
The pressure profile inside a core sample is given below, where nitrogen of
viscosity 0.0178 cP flows at a constant rate. Knowing that the permeability
of the rock is 12.4 mD, the length is 6 in, and the diameter is 1in,
what is the expected gas flow rate if the flow is in accordance with
Darcy’s law?
350

300

250

200
P [psig]
150

100

50

0
0 1 2 3 4 5 6 7
Distance from inlet [in]

Solution
It is important to ensure that all the units are consistent when solving such
problems. In this case, all the given units will be converted to Darcy units
in order to obtain the flow rate in cm3/s. This means that the length and
diameter need to be converted from inches to cm, and the pressure needs
to be converted from psig to atm.

Initially, a point x on the pressure profile will be selected, and its pressure
and distance will be recorded. At a distance of 6 in, the pressure is recorded
to be 75 psig from the pressure profile.

Now, all the units will be converted to Darcy units:

Permeability = 12.4 mD = 0.0124 D

Length = 6 in = 15.24 cm
Diameter = 1 in = 2.54 cm
Distance, x = 6 in = 15.24 cm
P(x) = 75 psig = 5.1 + 1 = 6.1 atm
P1 = 300 psig = 20.41 + 1 = 21.41 atm

Note that for the pressure to be converted to atm, 1 atm needs to be added
since we have gauge pressure (psig) and not absolute pressure (psia).

74
 Now, the cross-sectional area of the sample will be calculated:
d2 2.542 2
A=π =π = 5.07 cm
4 4
Finally, Equation 4.26 can be rearranged to find the gas flow rate:
2qg µ
P (x)2 = P12 − x
kA

qg =
kA(P12 − P (x)2 )
2µx
=
0.0124 × 5.07 × (21.412 − 6.102 )
2 × 0.0178 × 15.24
= 48.8 cc/s

4.6 Flow in Layered Systems

The objective of understanding the flow in a layered system is to find the average
permeability across that system with beddings of different permeability. The
concept is similar to an electrical circuit; the average permeability will vary if the
flow is in parallel or series to the beddings in the system. In addition, it could also
vary if the flow is linear or radial. We will now examine each case individually.

4.6.1 Case 1: Linear Flow in Parallel

Let us assume a system similar to Figure 4.13 which has three beddings with
different permeabilities parallel to each other, and we try to find the average
permeability in that system. First, we can see a constant pressure difference
across the system; however, the flow rate will be different across each layer as
the flow rate is a function of the permeability of that bedding. Therefore, we can
say that the total flow rate (q) is equal to:

q = q1 + q2 + q3 (4.28)

We also know that the summation of the thickness of each layer is equal to the
total thickness of the system:

h = h1 + h2 + h3 (4.29)

The total flow rate of the system is equal to:

k̄W h (P1 − P2 )
q= (4.30)
µL

75
Then, we substitute for the flow rate from Darcy’s law in Equation 4.28:

k̄W h (P1 − P2 ) k1 W h1 (P1 − P2 ) k2 W h2 (P1 − P2 ) k3 W h3 (P1 − P2 )

q= = + +
µL µL µL µL
(4.31)

Now, we can factor the common parameters in the equation which will lead to:

k̄h = k1 h1 + k2 h2 + k3 h3 (4.32)

Finally, we can generalize the equation to obtain:

n
 k i hi
k̄ = (4.33)
i=1
h

This equation can be used to estimate the average permeability in a linear system
where the beddings are parallel to each other.

q1
P1 P2
q2
k1 h1

q
k2 h2 q
q3 h

k3 h3
W

Figure 4.13: Schematic showing a linear system with parallel beddings of different permeabilities.

4.6.2 Case 2: Linear Flow in Series

Let us assume a system shown in Figure 4.14 which has three beddings in series
with each other; each bedding has a different permeability. Here, the same flow
rate passes through all these rocks:

q = q 1 = q2 = q 3 (4.34)

However, the pressure across each rock is different, as shown in Figure 4.14 and
explained in Figure 4.15. It can be said that:

P1 − P2 = ∆P1 + ∆P2 + ∆P3 (4.35)

76
Furthermore, the total length is composed of the length of each rock:

L = L1 + L2 + L3 (4.36)

We know that the flow rate across the entire system is equal to:

k̄W h (P1 − P2 )
q= (4.37)
µL

Now, we will substitute Darcy’s law in Equation 4.35 by making ∆P of each

respective equation as the subject to obtain:

qµL qµL1 qµL2 qµL3

P1 − P2 = = + + (4.38)
k̄W h k1 W h k2 W h k3 W h
We can now remove the common parameters and the equation becomes:

L L1 L2 L3
= + + (4.39)
k̄ k1 k2 k3

We can also represent this equation in the following general form:

L
k̄ = n Li
(4.40)
i=1 ki

This equation can be used to estimate the average permeability in a linear

system, where the beddings are in series with each other.

P1 P2

k1 k2 k3
q q

P1 P2 P3
h
L1 L2 L3

Figure 4.14: Schematic showing a linear system with beddings of different permeabilities
in series.

77
 P1 P2
k1 > k2 > k3

k1 k2 k3

P1

k
P

P2

0
0 L
x

Figure 4.15: Schematic showing the pressure profile across a composite system with beddings in
series with each other.

Example 4.5
A rock consists of three layers of different permeabilities. The geometry
of the rock is given in the figure below, along with the permeability and
dimensions of each layer. Find the average permeability of the rock.

1.7 mD 0.5 in
q 32.5 mD
1 in
5.3 mD

3 in 5 in

Solution
The three layers given in this question can be numbered as follows: top-
left is layer 1, bottom-left is layer 2, and right is layer 3. To find the average
permeability of the entire rock, the combined permeability of layers 1 and 2
can first be found so that the combined section can be treated as one layer.

78
 This combined layer along with layer 3 can be used to find the overall
permeability.

Since the flow is horizontal, layers 1 and 2 are parallel to each other. The
combined parallel permeability of layers 1 and 2 can be calculated using
Equation 4.33:
n
 k i hi
k̄ =
h
i=1

k 1 h1 + k 2 h 2 1.7 × 0.5 + 5.3 × 0.5

k1,2 = = = 3.5 mD
h 0.5 + 0.5
Layers 1 and 2 combined are in series with layer 3. Therefore, the average
permeability can be calculated using Equation 4.40:
L
k̄ = n Li
i=1 ki

k̄ =
L1,2 + L3
L1,2 L3
= 3
3+5
5 =
+ 32.5
7.91 mD
k1,2 + k3 3.5

4.6.3 Case 3: Radial Flow in Parallel

The flow in parallel systems in radial orientation (Figure 4.16) is similar to that in
linear orientation, as will be proven below.

First, the overall flow rate is the summation of all the flow rates across the layers
(Equation 4.28), and the total thickness is the summation of all the thicknesses
across all the layers (Equation 4.29).

The total flow rate for this system is equivalent to:

2π k̄h (Pe − Pwf )

q=   (4.41)
µln rrwe

Now, if we substitute each flow rate in each layer in Equation 4.28, we will have:

2π k̄h (Pe − Pwf ) 2πk1 h1 (Pe − Pwf ) 2πk2 h2 (Pe − Pwf ) 2πk3 h3 (Pe − Pwf )
  =   +   +  
µln rrwe µln rrwe µln rrwe µln rrwe
(4.42)

We can now factor the common parameters to obtain:

k̄h = k1 h1 + k2 h2 + k3 h3 (4.43)

79
The form is indeed the same as the linear system and can also be represented in
the same general form:
n
 k i hi
k̄ = (4.44)
i=1
h

This equation can be used to estimate average permeability in a radial system

where the beddings are parallel to each other.

re
rW

k1 h1 q1

k2 h2 q2

q3
k3 h3

Figure 4.16: Schematic showing a radial system with beddings of different permeabilities in
parallel.

4.6.4 Case 4: Radial Flow in Series

The flow in series for a radial system is shown in Figure 4.17. We know that the
flow rate across the layers is the same, as shown in Equation 4.34; however, the
pressure difference in each layer is different as shown in Equation 4.35. The flow
rate in a radial system is expressed in Equation 4.41.

By substituting Darcy’s law in Equation 4.41 and making ∆P of each respective

equation as the subject, we obtain:
     
re r1 re
qµ ln rw qµ ln rw qµ ln r1
Pe − Pwf = = + (4.45)
2π k̄h 2πk1 h 2πk2 h

80
Then, we eliminate the common parameters and rearrange the equation to
obtain this generic form:
 
re
ln rw
k̄ = r  (4.46)
n ln
(i+1)
ri
i=1 ki

where r(i+1) represents the outer radius of layer i and ri represents the inner layer.

This equation can be used to estimate the average permeability in a radial

system where the beddings are in series with each other.

re

r1

rW

h k2 k1

Figure 4.17: Schematic showing a radial system with beddings of different permeabilities in
series.

Example 4.6
A radial system consists of three layers with the following properties:

- Layer 1 has an inner radius of 0.25 ft, an outer radius of 5 ft,

and a permeability of 10 mD
- Layer 2 has an inner radius of 5 ft, an outer radius of 150 ft,
and a permeability of 80 mD
- Layer 3 has an inner radius of 150 ft, an outer radius of 750 ft,
and a permeability of 150 mD

81
 Calculate the average permeability of this system if the flow is in series.

Solution
In this radial system, the flow is in series as it moves from one layer to the
other from the inside out, as shown in Figure 4.17. As such, the equation for
calculating average permeability for radial flow in series will be used.

Using Equation 4.46:

 
re  750 
ln
k̄ =
n ln
rw
r
(i+1)
 = 5
ln( 0.25 )
ln 0.25
ln( 150
5 )
ln( 750
150 )
= 22.7 mD
ri
10 + 80 + 150
i=1 ki

4.7 Flow in Channels and Fractures

The concept of permeability can go beyond core samples and beddings, and can
be useful in estimating permeability in channels and fractures. Channels and
fractures can be superficially induced in the reservoir in order to increase the
permeability. However, some reservoirs can be naturally fractured.

4.7.1 Flow in Channels

Channels are created in reservoirs by injecting acids to dissolve the rock in

order to increase the permeability in the near-wellbore region. Increasing the
permeability will increase the flow rate of the hydrocarbons in the well, and will
hence increase the productivity. Channels can have different shapes; however,
the simplest one is the capillary tube shape (Figure 4.18), which also represents
the wormhole effect caused by acidizing.

The flow in a capillary tube is described by Poiseuille in the following equation:

πr4
q= (P1 − P2 ) (4.47)
8µL

We know that the area of the channel is equal to πr2; thus, we can write:

Ar2
q= (P1 − P2 ) (4.48)
8µL

82
Now, if we compare this equation with Darcy’s linear flow of liquids, we obtain:

kA Ar2
(P1 − P2 ) = (P1 − P2 ) (4.49)
µL 8µL

We can now eliminate all the common parameters to obtain:

r2
k= (4.50)
8
This represents a measure of permeability in channels. However, you need
to bear in mind that the channel’s radius would have a unit other than Darcy
(usually inches), and therefore unit conversion is required. If the porous medium
has n number of identical channels, then Equation 4.50 can be modified to:

r2
k=n (4.51)
8

P2

P1
l
ne
an
Ch

Matrix L

q
Figure 4.18: Capillary tube-shaped channels.

4.7.2 Flow in Fractures

Fractures can be either natural or induced, and the simplest model assumes
a slab of constant thickness as shown in Figure 4.19. The flow in this slab is
described by Buckingham in the following equation:

Ah2
q= (P1 − P2 ) (4.52)
12µL

Now, if we compare this equation with Darcy’s linear flow of liquids, we obtain:

kA Ah2
(P1 − P2 ) = q = (P1 − P2 ) (4.53)
µL 12µL

83
We can now eliminate all the common parameters to obtain:

h2
k= (4.54)
12
This should represent the permeability of fractures with a simplified shape.
Nevertheless, we still need to be careful when dealing with units.

W
L
Figure 4.19: Schematic showing a porous medium with a fracture.

Example 4.7
a) Find the fracture permeability (in D) of a cubic rock with a fracture of
width 0.5 inch passing completely through the rock.

b) Given that the rock has sides of 1.5 ft, a matrix permeability of 2.5 mD
and the flow is normal to the direction of the fracture, find the average
permeability of the rock (in mD).

Solution
a) Since the fracture permeability formula is in terms of SI units, the units
given in the question have to be converted to SI. Since 1 inch = 0.0254 m,
the fracture height, h, is:

h = 0.5 in = 0.5 × 0.0254 = 0.0127 m

Using the fracture permeability formula (Equation 4.54):

84
 h2 0.01272 −5 2
k= = = 1.344 × 10 m
12 12
Using the conversion of Darcy to m2 derived in Example 4.2, the fracture
permeability will be 1.36 × 107 D.

b) The fracture that passes through the length of the rock essentially divides
it into three sections. There is a matrix section above and below the fracture
with permeabilities of 2.5 mD, and then there is the fracture section itself
with a permeability of 1.36 × 107 D or 1.36 × 1010 mD. The exact position of
the fracture on the cube is not given, so it can be assumed that the fracture
is in the exact middle of the cube. The exact position of the fracture is
irrelevant, since the matrix will occupy the same area and hence the average
permeability would be the same regardless of where the fracture actually
lies. Since we assume that the fracture is in the middle, and the fracture has
a height of 0.5 in or 0.5/12 = 0.04 ft, then the length of each of the matrix
sections on either side of the fracture becomes:

1.5 − 0.04
L= = 0.73 ft
2
Since the flow is normal to the fracture, this becomes a case of linear
flow with three sections in series to one another. Therefore, the average
permeability of the rock can be found using Equation 4.40:

L
k̄ = n Li
i=1 ki

k̄ =
Lm + Lf + Lm
Lm Lf Lm
=
0.73 + 0.04 + 0.73
0.73 0.04
+ 1.36×10 0.73 = 2.57 mD
+ + 2.5 10 + 2.5
km kf km

As can be observed from this question, the average permeability of the rock
increases due to the fracture, since the fracture is basically empty space
through which a fluid can pass.

85
4.7.3 Average Permeability with Channels and Fractures

In the case where there are channels and/or fractures in a rock and the average
permeability needs to be found, area ratios can be used, as shown in the equation
below:
n
 k i Ai
k̄ = (4.55)
i=1
A

where Ai is the cross-sectional area of the channel, fracture, or surface [ft2] and A
is the total area [ft2].

As evident from this equation, the very high permeability of the channel or
fracture will cause the average permeability of the rock to increase significantly
even though they occupy a small area on the rock.

Example 4.8
A cube-shaped rock with sides of length 2.5 ft has a matrix permeability of
5 mD and a 0.3 inch diameter cylindrical channel that traverses the rock.

a) Calculate the permeability of the channel (in D).

b) Calculate the average permeability of the rock (in mD) when flow
is linear, and in the direction of the channels.

Solution
a) Since the channel permeability formula is in terms of SI units, the units
given in the question have to be converted to SI. Since 1 in = 0.0254 m, the
channel radius, r, is:

r = 0.15 in = 0.15 × 0.0254 = 0.00381 m

Using the channel permeability formula (Equation 4.50):

r2 0.003812 −6 2
k= = = 1.815 × 10 m
8 8
Using the conversion of Darcy to m2 derived in Example 4.2, the channel
permeability is 1.84 x 106 D.

86
 b) To find the average permeability, Equation 4.55 can be used. However,
first, the area of the matrix and the channel needs to be calculated.

The area of the channel is:

d2 0.32 −4 2
Ac = π =π = 4.909 × 10 ft
4 4

Note that the diameter is converted from inches to feet in the equation.

Now, the area of the matrix is:

Am = At − Ac = 2.52 − 4.909 × 10−4 = 6.2495 ft

2

Note that to find the surface area of the matrix, the surface area of the
channel has to be subtracted from the surface area of the cube.

Now, Equation 4.55 can be used to find the average permeability:

n
 k i Ai
k̄ =
A
i=1

which is equal to:

5 × 6.2495 + 1.84 × 109 × 4.909 × 10−4

= 1.45 × 105 mD
k m Am + k c Ac
=
Am + Ac 6.2495 + 4.909 × 10−4

Note that the channel permeability is converted from D to mD to be used

in this equation.

4.8 Summary

Permeability is part of RCAL and is a measure of the ability of a porous rock

to allow fluids to pass through it, and it can therefore help determine the rate
at which oil and gas will flow from the reservoir to the surface. Darcy’s law of
single-phase permeability is the equation used to determine permeability when
only one phase (water, oil, or gas) is present, and it is only valid when the core
sample is 100% saturated with the single phase and the flow is laminar and
steady-state. Darcy’s law for incompressible fluids has a solution for both linear
and radial systems. The three main unit systems used in Darcy’s equation are
SI units, Darcy’s units, and Oilfield units, where the unit for permeability is m2,
D (Darcy), and mD (milli Darcy), respectively. It is imperative that the units are
consistent when solving for permeability. The permeability of core samples
can be measured experimentally using either liquid or gas. However, when

87
measuring permeability using gas, a correction is required due to gas slippage
at the pore walls known as the Klinkenberg effect, which makes the permeability
higher than it should be. The pressure profile across a core sample when using
liquid is linear, whereas the pressure profile when using gas is parabolic. Average
permeability can be found for a system with layers of different dimensions and
permeabilities in series or parallel with one another depending on the direction
of fluid flow, where the orientation of the layers can be either linear or radial.
Permeability can also be calculated for channels and fractures using different
equations.

A summary of permeability is presented in Table 4.2.

Table 4.2: Definition of permeability and its importance to the petroleum industry.

Parameter Symbol Definition Importance

Permeability k A measure of the Permeability is

ease with which important in
fluid flows through determining the
a porous rock. rate at which oil
and gas will flow
from the reservoir.

End of Chapter Questions

Question 4.1

A core of length 5 cm and diameter 3 cm is saturated with brine of viscosity 1

cP. Water is injected into this core at a rate of 0.13 cm3/s. The gauge pressure at
the inlet was measured to be 180 kPa while the outlet was open to atmospheric
pressure. Calculate the permeability of this core.

a) In m2

b) In mD

88
Question 4.2

A steady-state liquid permeability test on a core sample from a reservoir of

interest was conducted. The laboratory measured data are given below:

- Plug diameter: 2.52 cm

- Plug length: 2 cm

- Test oil viscosity: 1.82 cP

- Outlet pressure: 1 atm

- Inlet pressure: 2 atm

- Flow rate: 0.275 cm3/s

Calculate the permeability of the core.

Question 4.3

Measure the brine absolute permeability for a core sample (L= 76.38 mm, D=
37.85 mm). The rock is cylindrical, and the brine viscosity is 0.001048 Pa.s. The
rest of the required data are given in the table below.

q [m3/s] P1 [psig] P2 [psig]

9.97 × 10-9
5.7 0.8

3.53 × 10-8
16 1.1

8.33 × 10-8 35.5 1.1

89
Question 4.4

A core was mounted in a gas permeameter to measure permeability, and the

following laboratory data are obtained from the experiment:

- Diameter of the core = 2.59 cm

- Length of the core = 6.03 cm

- Gas viscosity = 0.0148 cP

q [cm3/s] P1 [kPa] P2 [kPa]

13.09 65.4 13.5

7.91 44.6 11.2

4.12 25.3 6.4

3.23 19.4 4.1

Determine the gas permeability and equivalent liquid permeability of the core
using the above data (in Darcy units).

Question 4.5

A rock sample has a length of 20.32 cm and a diameter of 5 cm. The pressure
profile inside the core sample is given below, where nitrogen of viscosity 0.0178
cP is flowing at a constant rate of 55 cm3/s. What is the permeability of the sample
if the flow is in accordance with Darcy’s law?

450
400
350
300
250
P [psig]
200
150
100
50
0
0 1 2 3 4 5 6 7 8 9
Distance from inlet [in]

90
Question 4.6

Nitrogen of 0.0178 cP viscosity is injected at an inlet pressure of 120 psig with

the outlet pressure set at 72 psig. The rock has a 1 in diameter, 4.5 in length, and
a permeability of 2.3 mD. Provide an accurate plot of the pressure profile inside
the rock.

Question 4.7

Rock samples 1, 2, 3 and 4 have similar cylindrical diameters of 3.5 cm. The
pressure profile inside the core samples is given below, where water of viscosity
1 cP is flowing at a constant rate of 1.2 cm3/s. Find the permeabilities of all the
four rock samples.

120
Rock 1

100 Rock 2

Rock 3
80
Rock 4

P [psia] 60

40

20

0
0 5 10 15 20 25 30

Distance from inlet [cm]

Question 4.8

Consider a layered reservoir consisting of three alternating layers 1 m thick and

1 m long, with k1= 1000 mD, k2 = 100 mD, and k3= 10 mD.

a) What is the average permeability of this reservoir if the fluid is flowing

parallel to the layering (kh)?

b) What is the average permeability of this reservoir if the fluid is flowing

perpendicular to the layering (kv)?

c) What is the kv/kh?

91
Question 4.9

A rock consists of six layers of different permeabilities. The geometry of the rock
is given in the figure below. The permeability and dimensions of each layer are
given in the table below.

a) Find the average permeability of this rock for horizontal flow (kh).

b) Find the average permeability of this rock for vertical flow (kv).

c) What is the kv/kh?

qv

qh 3
2
4

5 6

Layer Permeability [mD] Thickness [in] Length [in] Width [in]

1 50 3 30 10

2 60 3.5 9 10

3 30 1.7 21 10

4 40 1.8 21 10

5 20 2.5 18 10

6 10 2.5 12 10

92
Question 4.10

A radial system consists of four layers with the following properties:

Layer Inner Radius [m] Outer Radius [m] Permeability [D]

1 0.6 4.5 0.35

2 4.5 26.0 0.21

3 26.0 95.5 0.49

4 95.5 164.5 0.24

Calculate the average permeability of this system if the flow is in series.

Question 4.11

Consider the radial flow problem shown in top view (right), where the shaded
area represents a damaged zone due to the drilling operations. Assume there
are three horizontal layers of equal heights (thickness), and the permeability of
the undamaged formation is 45 mD in layer 1, 25 mD in layer 2, and 30 mD in
layer 3. Further assume that the radius of damage is 2.2 ft and the permeability
of the damaged zone is 1.5 mD in all the three layers. The radius of the wellbore
is 0.6 ft, and the radius of the reservoir is 450 ft.

a) What is the average permeability in each horizontal layer?

b) What is the average permeability of the three layers?

rd
rw

re

(Top View)

93
Question 4.12

A cube of reservoir rock with sides of 2.5 ft has a single vertical fracture of width
0.35 that passes completely through the rock. The matrix permeability is 1 mD.

a) What is the permeability of the fracture (in D)?

b) What is the average permeability of the cube if the flow is in the

direction parallel to the fracture (in mD)?

c) What is the average permeability of the cube if the flow is in the

direction normal to the fracture (in mD)?

Question 4.13

A cube-shaped rock with sides of length 1.2 m has a matrix permeability of 0.8
mD and four 0.65 cm diameter cylindrical channels that traverse the rock as
shown below. Calculate the average permeability of the rock (in mD) when flow
is linear and in the direction of the channels.

Question 4.14

The rock sample shown below has a channel that partially traverses the length
of the sample. The rock sample had a permeability of 8 mD before the channel
was drilled into it.

3 in

Channel diameter
1 in
1 mm

6 in

a) Calculate the permeability of the channel (in D).

b) Find the average permeability of the rock sample if the flow is linear
and in the direction of the channel (in mD).

94
95
96
 Chapter 5

Fluid
Saturation
The concept of fluid saturation can be explained through a glass filled with
different fluids (Figure 5.1). Figure 5.1a shows a glass filled entirely with water,
meaning that the entire pore volume (glass volume) is occupied by water and
thus the water saturation is 100%. Fluid saturation is the volume of a particular
fluid in a rock sample divided by the pore volume. In Figure 5.1b, an identical
glass is occupied by both water and oil. Here, the water saturation cannot be
100% as the oil is sharing some space with the water. Finally, Figure 5.1c has
water, oil, and gas in the same glass, and thus the saturation of each fluid will be
less than 100%.

20% Air
50%
Oil
30% Oil
100%
Water

50% 50%
Water Water

(a) (b) (c)

Figure 5.1: Schematic showing identical glasses filled with different fluids: (a) the glass is
filled with 100% water, representing Sw = 1, (b) the glass is filled with 50% water and 50% oil,
representing Sw = 0.5 and So = 0.5, and (c) the glass is filled with 50% water, 30% oil, and 20% air
(gas), representing Sw = 0.5, So = 0.3, and Sg = 0.2.

Similarly, rocks are filled with one or more fluids. Fluid saturation helps us
quantify the amount of hydrocarbons or water in the rock. We can classify the
saturation into three categories: water, oil, or gas. Water saturation, Sw, is the
volume of water in a rock divided by the pore volume:

Vw
Sw = (5.1)
Vp

97
where Sw is the water saturation [dimensionless], Vw is the volume of water in the
pore spaces [cm3], and Vp is the pore volume [cm3].

Similarly, oil saturation, So, is the oil volume divided by the pore volume:

Vo
So = (5.2)
Vp

where So is the oil saturation [dimensionless], Vo is the volume of oil in the pore
spaces [cm3], and Vp is the pore volume [cm3].

Finally, gas saturation, Sg, is the gas volume in a rock divided by the pore volume:

Vg
Sg = (5.3)
Vp

where Sg is the gas saturation [dimensionless], Vg is the volume of gas in the pore
spaces [cm3], and Vp is the pore volume [cm3].

The summation of saturations of all the fluids in a reservoir has to be 1 as the

pores have to be filled with at least one fluid. If a reservoir contains water, oil,
and gas, then the equation becomes:

Vw + V o + Vg
= Sw + So + Sg = 1 (5.4)
Vp

However, if a reservoir only contains oil and water, then Equation 5.4 will reduce
to Sw + So = 1. Although Equation 5.4 is very simple, it is helpful in finding an
unknown fluid saturation mathematically.

Similar to porosity, fluid saturation is either presented as a fraction or a

percentage; however, bear in mind that it should always be used as a fraction in
calculations.

Figure 5.2 shows an example of a microscopic rock slice illustrating water, oil,
and gas saturations in the pore spaces. Similar to porosity, fluid saturation is
important to estimate the amount of hydrocarbons in a reservoir. However, it is
important to distinguish between porosity and fluid saturation. Porosity tells us
the maximum storage capacity of a medium, while fluid saturation depicts the
exact amount of fluid occupying the pore spaces of the same medium.

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 0.5 mm 0.5 mm 0.5 mm

Rock Water Rock Water Oil Rock Water Oil Gas

(a) (b) (c)

Figure 5.2: Schematic showing a cross section of a rock at the microscopic scale: (a) all the pore
spaces are filled with water (Sw = 1), (b) the pore spaces are filled with water and oil (Sw + So = 1),
and (c) the pore spaces are filled with water, oil, and gas (Sw + So + Sg = 1).

Example 5.1
A core sample with a pore volume of 15 cm3 contains water, oil, and gas. The
water volume within the sample is 6.3 cm3, while the oil volume is 5.4 cm3.
What is the gas saturation?

Solution
The water saturation can be found using Equation 5.1:

Vw 6.3
Sw = = = 0.42
Vp 15

The oil saturation can be found using Equation 5.2:

Vo 5.4
So = = = 0.36
Vp 15

The gas saturation can then be found using Equation 5.4:

S +S +S =1
w o g
S
g = 1 − Sw − So
Sg = 1 − 0.42 − 0.36 = 0.22

5.1 Measuring Fluid Saturation

Fluid saturation measurements can be classified into two types: direct and
indirect. Direct measurements include conventional core analysis techniques
such as extraction methods (retort distillation and Dean-Stark method), while
indirect measurements include electrical properties and capillary pressure,
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which will be discussed in Chapters 6 and 8, respectively. In this chapter, we will
focus on the direct methods, which are part of RCAL.

5.1.1 Extraction Method: Retort Distillation

For the retort distillation method (Figure 5.3), a core sample is placed in a
chamber and heated to around 1100 °F (≈593 °C). This is to evaporate all the
fluids in the system (oil and water). The vapors will rise and reach a condensing
tube where cold water is being circulated. The vaporized liquids will condense
back to liquid form and will be collected in the graduated cylinder after passing
through the condensing tube. Once we have the volumes of oil and water
from this method and by knowing the pore volume of the core sample, we can
calculate the water and oil saturations using Equations 5.1 and 5.2, respectively.
The advantages of the retort distillation method are that it can directly measure
oil and water saturations, and is a relatively fast method (usually takes less than
one hour). The main disadvantage of this method is that subjecting the core to
very high temperatures can damage it, thereby preventing the core from being
used for additional experimentation and analysis.

Cooling Water In

Core Sample

Condenser

Heating Element
1000 - 1100 ºF Graduated
Cylinder
Cooling Water
Out
Figure 5.3: Schematic showing the experimental set-up for the retort distillation extraction
method.

5.1.2 Extraction Method: Dean-Stark

This method is also known as Soxhlet extraction or solvent extraction.

Chemists Dean and Stark first designed this experimental set-up in 1920. For
this experiment, a core sample is placed at the top of a solvent flask, as shown
in Figure 5.4. The solvents used are usually toluene (hydrocarbon solvent) or
a mixture of toluene and methanol. Methanol can be used in the presence of
salty water. The solvent is heated to around 230 °F (110 °C, the boiling point of
toluene), so that the water present in both the core and the solvent evaporates
when the temperature in the system exceeds the boiling point of these fluids.
The toluene vapor will strip the oil from the core and travel upward as toluene is
miscible with oil. Once the vapor goes upward, it will reach the condensing tube
with circulating cooling water. Both fluids (water and solvent) will drop down
in the graduated cylinder. Since water is denser than the solvent, it will settle
at the bottom of the graduated tube, while the condensed solvent (being less
dense) will accumulate on top of the water until it drops down to the solvent
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flask. This method can measure the volume of water directly, and the water
saturation can be calculated using Equation 5.1 if the pore volume of the core
sample is known. Oil volume needs further calculations using material balance,
which will be discussed in the following section. We cannot measure the oil
volume/saturation from the Dean-Stark method directly, as the solvent mixes
with oil to form a new fluid that has different fluid properties than the original oil.
Therefore, collecting both fluids will be impractical. However, we can calculate the
oil saturation mathematically since the summation of all saturations is equal to 1
and compare the value with the one obtained from the material balance method.
The advantage of the Dean-Stark method is that it does not damage the core,
and the core sample can be used for future analysis. The disadvantages are that
the experiment is time-consuming as it usually takes about 48 hours, and that
time required for the experiment is also a function of the permeability; the lower
the permeability, the more time is required. Furthermore, in this experiment, we
can only measure saturation of one fluid directly, which is water, unlike the retort
distillation method which measures both water and oil saturations.

To drain

Condenser

Cooling Water entering

Graduate tube

Core Sample

Mesh

Solvent

Electric Heater

Figure 5.4: Schematic showing the experimental set-up for the Dean-Stark extraction method.

Material Balance

As mentioned earlier, the Dean-Stark method can only measure the water
volume. In order to find the oil volume and saturation, we need to use
material/mass balance. The concept is the same as discussed in Chapter 2.

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However, now we need to assume that there are two fluids in the system
(assuming no gas occupying the pores). It is important to note that we need to
record the weight of the core sample prior to the extraction process. First, we
know that the weight of the saturated core (assuming it contains oil and water)
prior to extraction is equal to:
Ws = Wd + Ww + Wo (5.5)
where Ws is the saturated weight [g], Wd is the dry weight of the sample that can
be obtained after the extraction process [g], Ww is the weight of the water in the
core sample [g], and Wo is the weight of the oil in the core sample [g].

We can say that the weights of water and oil in the core sample are equivalent to
their densities multiplied by their volumes, as shown below:

Ws = Wd + ρw Vw + ρo Vo (5.6)

where ρw is the density of water [g/cm3], Vw is the volume of water in the core
[cm3], ρo is the density of oil [g/cm3], and Vo is the volume of oil in the core [cm3].

We know that:

Vw + Vo + Vg = Vp (5.7)

where Vg is the volume of gas in the core [cm3], and Vp is the pore volume of the
core [cm3].

Since we only have oil and water, Equation 5.7 can be reduced to:

Vw + Vo = Vp (5.8)

For simplicity, let us assume:

Vo = x (5.9)

Then, we can rewrite Equation 5.8 to obtain:

Vw = Vp − x (5.10)

Then, we can substitute Equations 5.9 and 5.10 in Equation 5.6 to obtain:

Ws = Wd + ρw (Vp − x) + ρo x (5.11)

After that, we can rearrange the equation to obtain:

Ws − Wd − ρw Vp = x(ρo − ρw ) (5.12)

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Now, we can solve for x and replace it with Vo:

Ws − Wd − ρw Vp
Vo = (5.13)
ρo − ρw

Finally, we can divide both sides by the pore volume to find the oil saturation:

Ws − Wd − ρw Vp
So = (5.14)
Vp (ρo − ρw )

We can cross-check the water saturation value obtained from the Dean-Stark
method using the following simple term:

Sw = 1 − So (5.15)

where Sw is the water saturation [dimensionless] and So is the oil saturation

[dimensionless].

It is important to mention that the use of material balance to find the fluid
saturations can go beyond the Dean-Stark method to measure fluid saturations
obtained from different experiments, as we will explore in the following chapters.

Example 5.2
A core sample containing only water (ρw = 1 g/cm3) and oil (ρo = 0.87 g/cm3)
has a 13.6% porosity, 3 inch length, and 1.5 inch diameter. Its saturated
weight was measured to be 144.3 g, and its dry weight was measured to be
133.2 g. Calculate the water and oil saturations.

Solution
First, the dimensions of the sample need to be converted to cm so that the
units are consistent. Since 1 inch = 2.54 cm:

D = 1.5 in = 1.5 × 2.54 = 3.81 cm

L = 3 in = 3 × 2.54 = 7.62 cm
Then, the bulk volume of this core needs to be calculated. Since this is a
cylinder:

Vb = πr2 L
 2
3.81
Vb = π × × 7.62 = 86.87 cm3
2

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 We now find the pore volume:

Vp = φVb

Vp = 0.136 × 86.87 = 11.81 cm3
Since this rock contains only oil and water, the oil saturation can now be
found using Equation 5.14:

Ws − W d − ρ w V p 144.3 − 133.2 − 1 × 11.81

So = = = 0.462
V (ρ
p o − ρ w ) 11.81(0.87 − 1)

The water saturation can then be found using Equation 5.15:

Sw = 1 − So

Sw = 1 − 0.462 = 0.538

Example 5.3
A Dean-Stark apparatus was used to determine fluid saturations of a
sandstone cylindrical core plug, which measures 2 cm in diameter and 4.5
cm in length. The initial weight of the core plug prior to extraction was 76.63
g. The volume of water recovered from the core plug was 1.42 cm3. The
weight of the core plug after it was dried was 73.94 g. The porosity of the
core plug was determined to be 23.1%, and the densities of reservoir water
and oil were 1.04 g/cm3 and 0.82 g/cm3, respectively. Determine the initial
fluid saturations in the core plug.

Solution
First, the bulk volume of this core needs to be calculated. Since this is a
cylinder:
Vb = πr2 L
 2
2
Vb = π × × 4.5 = 14.14 cm3
2

Now, we find the pore volume:

Vp = φVb

Vp = 0.231 × 14.14 = 3.27 cm3

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 Since the volume of water is given, the water saturation can be found using
Equation 5.1:

Vw 1.42
Sw = = = 0.43
Vp 3.27

To find the oil saturation, we need to calculate the volume of oil.

First, we calculate the weight of water:

Ww = Vw ρw = 1.42 × 1.04 = 1.48 g

We can use this to find the weight of oil using Equation 5.5:
Ws = Wd + Ww + Wo

W o = Ws − W d − W w

Wo = 76.63 − 73.94 − 1.48 = 1.21 g
Note that the assumption made here is that the weight of gas is insignificant.

The volume of oil can then be found:

Wo 1.21
Vo = = = 1.48 cm3
ρo 0.82
The oil saturation can be found using Equation 5.2:

Vo 1.48
So = = = 0.45
Vp 3.27

Thereafter, the gas saturation can simply be found using Equation 5.4:
S w + So + S g = 1

Sg = 1 − Sw − So

Sg = 1 − 0.43 − 0.45 = 0.12

5.2 Limitations of Using Extraction Methods to Evaluate Reservoir’s

Saturation

It is difficult to evaluate the reservoir’s saturation using the conventional core

analysis (extraction methods) because of two main reasons: drilling muds and
fluid properties.

105
5.2.1 Drilling Muds

When drilling a reservoir, two types of muds are usually used: water-based mud
(WBM) and oil-based mud (OBM). When either mud is used, part of the mud leaks
into the reservoir, as the reservoir is both porous and permeable. For instance,
if we use a WBM, the water will enter the permeable reservoir and push the
fluids inside the reservoir. This means that the water saturation in the reservoir
will increase in the region near the well where the mud has entered. In drilling
terminology, this region is called the near-wellbore region. The extraction of core
samples using coring tools usually occurs in the near-wellbore region where
the mud has flushed the original reservoir fluids in that region. This makes the
samples less representative of the reservoir’s initial condition. The same thing
can be said when OBM is used, as the oil saturation will be greater than the
actual oil saturation.

5.2.2 Fluid Properties

The physical state of fluids is a function of temperature and pressure. Liquids

become gases at high temperatures, and gases become liquids at low
temperatures. Similarly, liquids become gases at low pressures and gases
become liquids at high pressures. When the core is extracted from the reservoir,
it goes through a journey from the reservoir at elevated pressures (thousands
of psia) and temperatures (around 177 °F) to the surface at ambient conditions
(atmospheric pressure and temperatures of around 77 °F, depending on the
location). The decrease in temperature is usually too insignificant to affect the
state of the fluid, as the temperature difference between the reservoir and the
surface is not substantial enough to cause a significant phase change. However,
the pressure drops significantly and reaches a threshold point where the gas in
the oil escapes. This threshold pressure is known as the bubble point pressure,
and occurs when the first bubble of gas leaves the oil. At pressures lower than
the bubble point pressure, more gas will evaporate and leave the oil, leading to
a higher gas saturation. Consequently, performing the extraction methods on
these core samples will lead to significant errors in estimating the reservoir’s
original saturations.

What we deal with in a reservoir is a combined effect of drilling mud and the fluid
properties. Figure 5.5 explains the effect of drilling mud and fluid properties
on fluid saturation from the reservoir to the surface. Initially, we have high oil
saturation in the reservoir. Then, after drilling with a WBM, the water will invade
the pore spaces to reduce the oil saturation. Then, when we extract the core,
the core undergoes pressure and temperature changes that will change its
fluid properties. Reducing the pressure from thousands of psia to atmospheric
pressure, when the core reaches the surface, will release plenty of gas dissolved
in oil, which will reduce the oil saturation even further. The purpose of this
section is to understand why using the extraction method directly on reservoir
rocks might not produce the most accurate results. In the following chapters, we
will discuss alternative methods to estimate fluid saturations in the reservoir.

106
Despite the associated errors, these extraction methods are still in use, as they
provide a good, fast, and cheap estimate of fluid saturations within a reservoir.
This can be useful in making decisions related to the amount of hydrocarbons
present in the reservoir.

Fluid Contents
At Surface
outside of
arrow
4 e
Oil Gas Water

c
15 40 45

rfa
At Su
Gas
Expansion
Journey to
Surface 3

Expansion
Shrinkage

Expulsion
2 In reservoir
After Flushing
with water-based mud
20 ― 80

1 In reservoir Original Fluids 70 ― 30

Figure 5.5: Schematic showing the change in fluid saturation in a core sample from the initial
reservoir condition until it reaches the surface. The numbers displayed are arbitrary numbers
that are only used to explain the concept.

5.3 Summary

Fluid saturation is a percentage that indicates how much fluid the pore space
inside a rock contains, which is defined as the volume of fluid in a rock divided
by its pore volume. In reservoir rocks, the fluids are usually hydrocarbons or
water. Some extraction methods used to measure fluid saturation include retort
distillation and the Dean-Stark method. In both of these methods, the fluid is
extracted from the rock sample and then measured. The Dean-Stark method
can only measure water saturation, unlike retort distillation that can measure
both oil and water saturations. Therefore, material balance analysis is used to
supplement the Dean-Stark method to calculate the oil saturation as well. Drilling
muds make it difficult to evaluate the reservoir’s saturation using extraction
methods because they interfere with the saturation of the extracted samples.
Similarly, extreme temperatures and pressures while extracting the core samples
also cause changes within them, due to which the extraction methods cannot
accurately determine the saturation within the reservoir.

A summary of fluid saturation is presented in Table 5.1.

107
 Table 5.1: Definition of fluid saturation and its importance to the petroleum industry.

Parameter Symbol Definition Importance

Fluid Si (where i The fraction of We use fluid

saturation can be water, pore volume saturation to
oil, or gas) occupied by quantify the volume
the fluid. of oil and/or gas in
the reservoirs.

End of Chapter Questions

Question 5.1

If a core sample with a pore volume of 22.5 cm3 contains an oil volume of 9.8 cm3
and has a gas saturation of 0.15, what is its water saturation?

Question 5.2

A cylindrical core has a diameter of 1 in, length of 1.8 in, and a porosity of 19.5%.
Using solvent extraction, 1.94 cm3 of water were collected from the core and the
volume of oil was found to be 1.72 cm3. Calculate water, oil, and gas saturations.

Question 5.3

The dry weight of a rock is measured to be 211.6 g, and its saturated weight is
measured to be 219.4 g. The rock has a length of 6.5 cm, a diameter of 3 cm, and
a porosity of 18.5%. Given that the rock contains only water (ρw = 1.02 g/cm3) and
oil (ρo = 0.84 g/cm3), calculate the water and oil saturations.

108
Question 5.4

Solvent extraction was used in order to measure the fluid saturation in a core.
The following data are given for the core:

- Length: 3.5 inches

- Diameter: 2 inches

- Porosity: 22.9%

- Initial weight of the core: 487.6 g

- Final dry weight of the core: 455.9 g

- Water volume measured: 14.23 cm3

- Water density: 1.04 g/cm3

- Oil density: 0.80 g/cm3

Calculate the fluid saturations.

Question 5.5

A core contains only oil and water within it. Find the water and oil saturations by
using the following data:

- Dry weight: 168.3 g

- Saturated weight: 203.2 g

- Weight of the core after water removal: 184.1 g

- Diameter: 3.62 cm

- Length: 6.92 cm

- Water density: 1.03 g/cm3

- Oil density: 0.79 g/cm3

109
Question 5.6

A Dean-Stark apparatus was used to determine the fluid saturations in a

cylindrical core of diameter 3 cm, length 5.8 cm, and porosity 15.2%. The initial
weight of the core prior to extraction was measured to be 151.6 g. The weight
of the core plug after it was extracted and dried was 146.3 g. The volume of
water recovered from the core in the experiment was 2.94 cm3. The densities of
reservoir water and oil are 1 g/cm3 and 0.86 g/cm3, respectively. Determine the
initial fluid saturations in the core.

110