M.Sc. in Petroleum Engineering
2002-2003
1.
Flow in Porous Media
Dr. R. W. Zimmerman
Section 1
Page 1
Diffusion Equation for Fluid Flow in Porous Rocks
1.1. Darcy’s law and the definition of permeability
The basic law governing the flow of fluids through porous
media is Darcy’s law, which was formulated by the French civil
engineer Henry Darcy in 1856 on the basis of his experiments on
vertical water filtration through sand beds. Darcy found that his
data could be described by
Q=
where:
CAΔ(P − ρgz)
L
(1)
P = pressure [Pa]
ρ = density [kg/m3]
g = gravitational acceleration [m/s2]
z = vertical coordinate (measured downwards) [m]
L = length of sample [m]
Q = volumetric flowrate [m3/s]
C = constant of proportionality [m2/Pa s]
A = cross-sectional area of sample [m2]
Any consistent set of units can be used in Darcy’s law, such
as SI units, cgs units, British engineering units, etc.
Unfortunately, in the oil industry it is common to use “oilfield
units”, which are not consistent.
Darcy’s law is mathematically analogous to other linear
phenomenological transport laws, such as Ohm’s law for electrical
conduction, Fick’s law for solute diffusion, or Fourier’s law for heat
conduction.
Department of Earth Science and Engineering
Imperial College
M.Sc. in Petroleum Engineering
2002-2003
Flow in Porous Media
Dr. R. W. Zimmerman
Section 1
Page 2
• Why does the term “ P − ρgz ” govern the flowrate?
Recall from fluid mechanics that Bernoulli’s equation (which
essentially embodies the principle of “conservation of energy”)
contains the terms
v 2 1 ⎜⎛
ρv 2 ⎟⎞
− gz +
= ⎜ P − ρgz +
,
ρ
2
ρ⎝
2 ⎟⎠
P
where
(2)
P/ρ is related to the enthalpy per unit mass,
gz is the gravitational energy per unit mass,
v2/2 is the kinetic energy per unit mass.
As fluid velocities in a reservoir are usually very small, the third
term is negligible, and we see that the combination “ P − ρgz ”
represents an energy-type term. It seems reasonable that the fluid
would flow from regions of higher to lower energy, and, therefore,
the driving force for flow should be the gradient (i.e., rate of
spatial change) of P − ρgz .
Subsequent to Darcy’s initial discovery, it has been found
that, all other factors being equal, Q is inversely proportional to
the fluid viscosity, μ [Pa ⋅ s] . It is therefore convenient to factor
out μ, and put C = k/μ, where k is known as the permeability, with
dimensions [m2].
It is also more convenient to work with the volumetric flow
per unit area, q = Q/A. Darcy’s law is therefore usually written as
q=
Q k Δ(P − ρgz)
=
,
A μ
L
(3)
where the flux q has dimensions of [m/s]. It is perhaps easier to
think of these units as [m3/m2s].
Department of Earth Science and Engineering
Imperial College
M.Sc. in Petroleum Engineering
2002-2003
Flow in Porous Media
Dr. R. W. Zimmerman
Section 1
Page 3
For transient processes in which the flux varies from pointto-point, we need a differential form of Darcy’s law. In the vertical
direction, this equation would take the form
qv =
Q −k d(P − ρgz)
=
.
A μ
dz
(4)
The minus sign is included because the fluid flows in the direction
from higher to lower potential.
The differential form of Darcy’s law for one-dimensional,
horizontal flow is
qH =
Q −k d(P − ρgz) −k dP
=
=
.
μ dx
A μ
dx
(5)
In most rocks the permeability kH in the horizontal plane is
different than the vertical permeability, kV; in most cases, kH > kV.
The permeabilities in any two orthogonal directions within the
horizontal plane may also differ. However, in this course we will
usually assume that kH = kV.
• The permeability is a function of rock type, and also varies with
stress, temperature, etc., but does not depend on the fluid; the
effect of the fluid on the flowrate is accounted for by the viscosity
term in eq. (4) or (5).
• Permeability has units of m2, but in petroleum engineering it is
conventional to use “Darcy” units, defined by
1 Darcy = 0.987 × 10
−12
2
m ≈ 10
−12
2
m .
(6)
The Darcy unit is defined such that a rock having a permeability of
1 Darcy would transmit 1 cc of water (with viscosity 1 cP) per
second, through a region of 1 sq. cm. cross-sectional area, if the
pressure drop along the direction of flow were 1 atm per cm.
Department of Earth Science and Engineering
Imperial College
M.Sc. in Petroleum Engineering
2002-2003
Flow in Porous Media
Dr. R. W. Zimmerman
Section 1
Page 4
Many soils and sands that civil engineers must deal with
have permeabilities on the order of a few Darcies. The original
purpose of the “Darcy” definition was therefore to avoid the need
for using small prefixes such as 10-12, etc. Fortunately, a Darcy is
nearly a round number in SI units, so conversion between the two
units is easy.
• The numerical value of k for a given rock depends on the
diameter of the pores in the rock, d, as well as on the degree of
interconnectivity of the void space. Very roughly speaking,
k ≈ d 2 /1000 . Typical values for intact (i.e., unfractured) rock are
given in the following table:
Rock Type
k (Darcies)
k (m2)
coarse gravel
103 - 104
10-9 - 10-8
sands, gravels
100 - 103
10-12 - 10-9
fine sand, silt
10-4 - 100
10-16 - 10-12
clay, shales
10-9 - 10-6
10-21 - 10-18
limestones
100 - 102
10-12 - 10-10
sandstones
10-5 - 101
10-17 - 10-11
weathered chalk
100 - 102
10-12 - 10-10
unweathered chalk
10-9 - 10-1
10-21 - 10-13
granite, gneiss
10-8 - 10-4
10-20 - 10-16
• The permeabilities of various rock and soil types vary over many
orders of magnitude. However, the permeabilities of petroleum
reservoir rocks tend to be in the range of 0.001-1.0 Darcies. It is
therefore convenient to refer to permeability of reservoir rocks in
units of “milliDarcies” (mD), which equal 0.001 Darcies.
Department of Earth Science and Engineering
Imperial College
M.Sc. in Petroleum Engineering
2002-2003
Flow in Porous Media
Dr. R. W. Zimmerman
Section 1
Page 5
• The values in the above table are for intact rock. In some
reservoirs, however, the permeability is due mainly to an
interconnected network of fractures. The permeabilities of
fractured rock masses tend to be in the range 1 mD to 10 Darcies.
In a fractured reservoir, the reservoir-scale permeability is not
closely related to the core scale permeability that one would
measure in the laboratory.
1.2. Datum levels and corrected pressure
If the fluid is in static equilibrium, then q = 0, and eq. (1.1.4)
yields
d(P − ρgz)
=0
dz
⇒
P − ρgz = constant .
(1)
If we take z = 0 to be at sea level, where the fluid pressure is
atmospheric, then
Pstatic ( z ) = Patm + ρgz .
(2)
As we always measure the pressure as “gauge pressure”
(i.e., the pressure above atmospheric), we can essentially neglect
Patm in eq. (2). We then see by comparing eq. (2) with eq. (1.1.4)
that only the pressure above and beyond the static pressure given
by eq. (2) plays a role in “driving” the flow. In a sense, then, the
term ρgz is superfluous, because it only contributes to the static
pressure, but does not contribute to the driving force for flow.
In order to remove this extraneous term, it is common to
define a corrected pressure, Pc , as
Pc = P − ρgz .
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M.Sc. in Petroleum Engineering
2002-2003
Flow in Porous Media
Dr. R. W. Zimmerman
Section 1
Page 6
In terms of the corrected pressure, Darcy’s law (for, say,
horizontal flow) can be written as
q=
Q −k dPc
=
.
A μ dx
(4)
Instead of using sea level (z = 0) as the “datum”, we often
use some depth zo such that equal amounts of initial oil-in-place
lie above and below zo. In this case,
Pc = P − ρg ( z − zo ) .
(5)
The choice of the datum level is immaterial, in the sense that it
only contributes a constant term to the corrected pressure, and so
does not contribute to the pressure gradient.
The corrected pressure defined in eq. (5) can be interpreted
as the pressure of a hypothetical fluid at depth zo that would be in
hydrostatic equilibrium with the fluid that exists at the actual
pressure at depth z.
1.3. Representative elementary volume
Darcy’s law is a macroscopic law that is intended to be
meaningful over regions that are much larger than the size of a
single pore. In other words, when we talk about the permeability
at a point “(x,y,z)” in the reservoir, we cannot be referring to the
permeability at a mathematically infinitesimal “point”, because a
given point may, for example, lie in a sand grain, not in the pore
space! The property of permeability is in fact only defined for a
porous medium, not for an individual pore. Hence, the
permeability is a property that is in some sense “averaged out”
over a certain region of space surrounded the mathematical point
Department of Earth Science and Engineering
Imperial College
M.Sc. in Petroleum Engineering
2002-2003
Flow in Porous Media
Dr. R. W. Zimmerman
Section 1
Page 7
(x,y,z). This region must be large enough to encompass a
statistically significant number of pores. Likewise, the “pressure”
that we use in Darcy’s law is actually an average pressure taken
over a small region of space.
For example, consider Fig. 1.3.1 below, which shows a few
pores in a sandstone. Two position vectors, R1 and R2, are
indicated in the figure. However, when we refer to the “pressure”
at a certain location in the reservoir, we do not distinguish
between two nearby points such as these. Instead, the entire
region shown in the figure would be represented by an average
pressure that is taken over the indicated circular region, which is
known as a “representative elementary volume” (REV). Similarly,
the permeability of a rock is only defined over the REV length
scale.
R1
R2
Fig. 1.3.1. Representative Elementary Volume (REV).
In practice, we rarely need to have a precise idea of the size
of the REV. Roughly, it must be at least one order of magnitude
larger than the pore size. Although it is important to be aware of
this concept, for most reservoir engineering purposes, no explicit
consideration of this issue is required.
Department of Earth Science and Engineering
Imperial College
M.Sc. in Petroleum Engineering
2002-2003
Flow in Porous Media
Dr. R. W. Zimmerman
Section 1
Page 8
1.4. Radial, steady-state flow to a well
Before we proceed to derive the general transient equation
that governs flow through porous media, we will examine a simple
(but illustrative) problem that can be solved using only Darcy’s
law: flow to a well in a circular reservoir that has a constant
pressure at its outer boundary.
Consider a reservoir of thickness H and horizontal
permeability k, that is fully penetrated by a vertical well of radius
Rw . Assume that at some radius Ro , the pressure remains at its
undisturbed value, Po . If we pump oil from this well at a rate Q,
what will be the steady-state pressure distribution in the
reservoir?
Q
P = Po
Ro
H
R
Fig. 1.4.1. Well in a bounded reservoir.
(1)
The analogue of eq. (1.1.5) for flow in the R direction is
Q=
Department of Earth Science and Engineering
− kA dP
.
μ dR
(1)
Imperial College
M.Sc. in Petroleum Engineering
2002-2003
Flow in Porous Media
Dr. R. W. Zimmerman
Section 1
Page 9
(2) The cross-sectional area normal to the flow, at a radial
distance R from the centre of the well, is 2πRH (i.e., a cylindrical
surface of height H, and perimeter 2πR ), so
Q=
−2πkH
μ
R
dP
.
dR
(2)
(3) Separate the variables and integrate from R = Ro to some
generic location R:
dR −2πkH
=
dP
R
μQ
→
P 2πkH
dR
=−∫
dP
∫
R
μ
Q
Ro
Po
→
ln
→
P(R) = Po −
R
R −2πkH
=
(P − Po )
Ro
μQ
μQ ln⎜⎛ R ⎞⎟ .
2πkH ⎝ Ro ⎠
(3)
Eq. (3) is the famous Dupuit-Thiem equation. It shows that the
pressure varies logarithmically. Most of the pressure drawdown
occurs in the vicinity of the well, whereas far from the well, the
pressure varies slowly.
The pressure distribution for steady-state flow to a well
located at the centre of a circular reservoir is shown in Fig. 1.4.2:
Department of Earth Science and Engineering
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M.Sc. in Petroleum Engineering
2002-2003
Flow in Porous Media
Dr. R. W. Zimmerman
Section 1
Page 10
(P-Po )2πkH/μ Q
0.00
1.00
2.00
3.00
Thiem formula
4.00
5.00
0
0.2
0.4
0.6
R/Ro
0.8
1
Fig. 1.4.2. Steady-state flow to a well in a bounded reservoir.
We can make the following comments about eq. (3):
• If fluid is pumped from the well, then (mathematically) Q is
negative, and P(R) will be less than Po .
• The amount by which P(R) is less than Po is called the
pressure drawdown.
• The only reservoir parameter that affects the pressure
drawdown is the “permeability-thickness” product, kH.
• The pressure varies logarithmically as a function of radial
distance from the well. This same type of dependence also
occurs in transient problems.
• The pressure drawdown in the well is found by setting R = Rw
in eq. (3):
μ Q ⎛ Rw ⎞
⎟⎟ .
Pw = Po −
ln⎜⎜
(4)
2πkH ⎝ Ro ⎠
Department of Earth Science and Engineering
Imperial College
M.Sc. in Petroleum Engineering
2002-2003
Flow in Porous Media
Dr. R. W. Zimmerman
Section 1
Page 11
• Since we are usually interested in fluid flowing towards the well
(i.e., “production”), it is common to define Q > 0 for production,
in which case we write eq. (4) as
μQ ⎛ Rw ⎞
Pw = Po +
ln ⎜ ⎟ .
2πkH ⎜⎝ Ro ⎟⎠
(5)
1.5. Conservation of mass equation
Darcy’s law in itself does not contain sufficient information to
allow us to solve transient (i.e., time-dependent) problems
involving subsurface flow. In order to develop a complete
governing equation that applies to transient problems, we must
first derive a mathematical expression of the principle of
conservation of mass.
Consider flow through a one-dimensional tube of crosssectional area A; in particular, let’s focus on the region between
two locations x and x+Δx:
A
q(x+Δx)
q(x)
x
x+Δx
Fig. 1.5.1. Prismatic region used to derive an equation for
conservation of mass.
The main idea behind the application of the principle of
conservation of mass is
Flux in - Flux out = Increase in amount stored .
Department of Earth Science and Engineering
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M.Sc. in Petroleum Engineering
2002-2003
Flow in Porous Media
Dr. R. W. Zimmerman
Section 1
Page 12
Note that the property that is conserved is the mass of the fluid,
not the volume of the fluid.
Consider the period of time between time t and time t +Δt.
To be concrete, assume that the fluid is flowing from left to right
through the core. During this time increment, the mass flux into
this region of rock between will be
Mass flux in = A(x)ρ(x)q(x)Δt.
(2)
The mass flux out of this region of rock will be
Mass flux out = A(x+Δx)ρ(x+Δx)q(x+Δx)Δt.
(3)
The amount of fluid mass stored in the region is denoted by m, so
the conservation of mass equation takes the form
[A(x) ρ(x)q(x) − A(x + Δx) ρ(x + Δx)q(x + Δx )]Δt
= m (t + Δt ) − m( t ) .
(4)
For one-dimensional flow, such as through a cylindircal core, A(x)
= A = constant. So we can factor out A, divide both sides by Δt,
and let Δt → 0 :
m (t + Δt ) − m (t ) dm
=
,
Δt →0
Δt
dt
− A[ ρq ( x + Δx ) − ρq ( x )] = lim
(5)
where we temporarily treat ρq as a single entity.
But m = ρVp , where Vp is the pore volume of the rock
contained in the slab between x and x+Δx. So,
m = ρVp = ρφV = ρφAΔx ,
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Flow in Porous Media
Dr. R. W. Zimmerman
−A[ ρq ( x + Δx ) − ρq ( x )] =
Section 1
Page 13
d ( ρφ )
AΔx .
dt
(7)
Now divide both sides by AΔx, and let Δx → 0 :
−
d ( ρq ) d ( ρφ )
=
.
dx
dt
(8)
Eq. (8) is the basic equation of conservation of mass for 1-D
linear flow in a porous medium. It is exact, and applies to gases,
liquids, high or low flowrates, etc.
In its most general, three-dimensional form, the equation of
conservation of mass can be written as
d ( ρφ )
d( ρq ) d ( ρq ) d ( ρq )
+
+
=−
.
dx
dy
dz
dt
(9)
The mathematical operation on the left-hand side of eq. (9) is
known as the divergence of ρq; it represents the rate at which
fluid diverges from a given region, per unit volume.
1.6. Diffusion equation in Cartesian coordinates
Transient flow of a fluid through a porous medium is
governed by a certain type of partial differential equation known
as a diffusion equation. In order to derive this equation, we
combine Darcy’s law, the conservation of mass equation, and an
equation that describes the manner in which fluid is stored inside
a porous rock. (Strangely enough, this last aspect of flow through
porous media was only first understood many decades after
Darcy’s law was discovered!)
Department of Earth Science and Engineering
Imperial College
M.Sc. in Petroleum Engineering
2002-2003
Flow in Porous Media
Dr. R. W. Zimmerman
Section 1
Page 14
Let’s look more closely at the right-hand side of eq. (1.5.8),
and use the product rule (and chain rule) of differentiation:
d( ρφ )
dφ
dρ
=ρ
+φ
dt
dt
dt
=ρ
dφ dP
dρ dP
+φ
dP dt
dP dt
⎡⎛
⎞ ⎛
⎞⎤
= ρφ ⎢⎜ 1 dφ ⎟ + ⎜ 1 dρ ⎟ ⎥ dP
⎣⎝ φ dP ⎠ ⎝ ρ dP ⎠ ⎦ dt
= ρφ (cφ +cf ) dP ,
dt
where
(1)
cf is the compressibility of the fluid,
cφ is the compressibility of the rock formation.
Now look at the left-hand side of eq. (1.5.8). The flux q is
given by Darcy’s law, eq. (1.1.5):
⎡
⎤
d( ρq)
d ⎡ − ρk dP ⎤ k ⎢ d 2P dρ dP ⎥
⎢
⎥=
−
=−
ρ
+
dx
dx ⎢⎣ μ dx ⎥⎦ μ ⎢⎣ dx 2 dx dx ⎥⎦
⎡
⎤
k ⎢ d 2P dρ dP dP ⎥
=
ρ
+
μ ⎢⎣ dx 2 dP dx dx ⎦⎥
⎡
2⎤
⎣
⎦
ρk ⎢ d 2P ⎛ 1 dρ ⎞ ⎛ dP ⎞ ⎥
⎟⎜ ⎟
=
+⎜
μ ⎢⎢ dx 2 ⎜⎝ ρ dP ⎟⎠ ⎜⎝ dx ⎟⎠ ⎥⎥
Department of Earth Science and Engineering
Imperial College
M.Sc. in Petroleum Engineering
2002-2003
Flow in Porous Media
Dr. R. W. Zimmerman
⎡
Section 1
Page 15
2⎤
⎛ dP ⎞ ⎥
ρk ⎢ d 2 P
=
+ cf ⎜⎜ ⎟⎟ ⎥ .
μ ⎢⎢ dx 2
⎝ dx ⎠ ⎥
⎣
(2)
⎦
Now equate eqs. (1) and (2) to arrive at
⎛ dP ⎞ 2 φμ (cf + cφ ) dP
d 2P
+ c f ⎜⎜ ⎟⎟ =
.
k
dt
dx 2
⎝ dx ⎠
(3)
The second term on the left is usually negligible compared to the
first. To “prove” that this is the case, we can use eq. (1.4.5), and
ignore the difference between x and R, to find
⎛ dP ⎞ 2
⎛ μQ ⎞ 2
⎟⎟ ,
cf ⎜⎜ ⎟⎟ ≈ cf ⎜⎜
⎝ dx ⎠
⎝ 2πkHR ⎠
d 2P
dx
→
2
Ratio =
≈
μQ
2πkHR
2
(4)
,
(5)
cf μQ cf (Po − Pw )
=
.
2πkH ln(Ro /Rw )
(6)
Typical values of these parameters (for liquids) are
cf ≈ 10 −10 /Pa ,
7
Po − Pw ≈ 10 MPa = 10 Pa ,
4
ln(Ro / Rw ) ≈ ln(1000 m / 0.1m) = ln(10 ) ≈ 10 ,
→
10 −10 × 107
Ratio =
= 10 −4 << 1 .
10
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M.Sc. in Petroleum Engineering
2002-2003
Flow in Porous Media
Dr. R. W. Zimmerman
Section 1
Page 16
This example shows that, for liquids, the nonlinear term in eq. (3)
is small. In practice, it is usually neglected. For gases, however, it
cannot be neglected (see section 9).
The one-dimensional, linearised form of the diffusion
equation is therefore
dP
k d 2P
=
,
dt φμct dx 2
(8)
in which the total compressibility is given by
ct = cformation + cfluid = cφ + cf .
(9)
The product of the compressibility and the porosity, φc, is called
the storativity. Typical values of these compressibilities are shown
in the following table:
Rock (or Fluid) Type
c (1/Pa)
Clay
10-6 - 10-8
Sand
10-7 - 10-9
Gravel
10-8 - 10-10
Intact rock
10-9 - 10-11
Jointed rock
10-10 - 10-12
Water
5 x 10-10
Oil
10-9
For many rocks, the pore compressibility is negligible, and the
storage is due mainly to the fluid compressibility; for soils and
sands, the opposite is the case. In general, both contributions to
the total compressibility must be taken into account.
Department of Earth Science and Engineering
Imperial College
M.Sc. in Petroleum Engineering
2002-2003
Flow in Porous Media
Dr. R. W. Zimmerman
Section 1
Page 17
Much of the remainder of this module will be devoted to solving
the diffusion equation in various situations. For now, we make the
following general remarks about it:
• The parameter that governs the rate at which fluid pressure
diffuses through a rock mass is the hydraulic diffusivity [m2/s],
which is defined by
DH =
k
.
φμ ct
(10)
• Roughly, the distance λ over which a pressure disturbance will
travel during an elapsed time t is (we will prove this in section
2.5)
λ = 4D H t =
4kt
.
φμct
(11)
• Conversely, the time required for a pressure disturbance to
travel a distance λ is found by inverting eq. (11):
t=
φμct λ2
4k
.
(12)
• Pressure pulses obey a diffusion equation, not a wave
equation, as one might have thought. Rather than travelling at
a constant speed, they travel at a speed that continually
decreases with time. To prove this, differentiate eq. (11) with
respect to time, and observe that the “velocity” of the pulse,
dλ/dt, decays like 1/ t .
Department of Earth Science and Engineering
Imperial College
M.Sc. in Petroleum Engineering
2002-2003
Flow in Porous Media
Dr. R. W. Zimmerman
Section 1
Page 18
1.7. Diffusion equation in radial coordinates
In petroleum engineering we are often interested in fluid
flowing towards a well, in which case it is more convenient to use
cylindrical (radial) coordinates, rather than Cartesian coordinates.
To derive the proper form of the diffusion equation in radial
coordinates, consider fluid flowing radially towards (or away from)
a vertical well, in a radially-symmetric manner. Return to eq.
(1.5.4), replace x with R, and note that A(R) = 2πRH :
R
q(R)
ΔR
q(R+ΔR)
Fig. 1.7.1. Annular region used in deriving diffusion equation in
radial co-ordinates.
[2πRHρ(R)q(R) − 2π (R + ΔR)Hρ(R + ΔR)q(R + ΔR)]Δt
= m(t + Δt) − m(t) .
(1)
As before, divide by Δt, and let Δt → 0 :
2πH[Rρ(R)q(R) − (R+ΔR)ρ (R+ΔR)q(R+ΔR)] =
dm
.
dt
(2)
On the right-hand side:
m = ρφV = ρφ 2πHRΔR ,
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Flow in Porous Media
Dr. R. W. Zimmerman
Section 1
Page 19
d( ρφ )
dm d( ρφ 2πHRΔR)
=
= 2πHR
ΔR .
dt
dt
dt
(4)
Equate eqs. (2) and (4), divide by ΔR, and let ΔR → 0 :
−
d( ρφ )
d( ρqR)
=R
.
dt
dR
(5)
Eq. (5) is the radial-flow version of the continuity (i.e.,
conservation of mass) equation.
Now use Darcy’s law in the form of eq. (1.4.1) for q on the
left-hand side, and eq. (1.6.1) on the right-hand side:
dP ⎞
dP
k d ⎛
⎜⎜ ρR
⎟⎟ = ρφ (cf + cφ )R
.
μ dR ⎝ dR ⎠
dt
(6)
Follow the same procedure as that which led to eq. (1.6.3), to find
⎛ dP ⎞ 2 φμ (cf + cφ ) dP
1 d ⎛ dP ⎞
⎜⎜ R
⎟⎟ + cf ⎜⎜ ⎟⎟ =
.
k
R dR ⎝ dR ⎠
dt
⎝ dR ⎠
(7)
For liquids, we again neglect the term cf (dP / dR) 2 , to arrive at
dP
k 1 d ⎛ dP ⎞
⎟⎟ .
⎜R
=
dt φμct R dR ⎜⎝ dR ⎠
(8)
Eq. (8) is the governing equation for transient, radial flow of a
liquid through porous rock. It is the governing equation for flow
during primary production, and it is the starting point for all welltest analysis methods. We will develop and analyse solutions to
this equation in later parts of this module.
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2002-2003
Flow in Porous Media
Dr. R. W. Zimmerman
Section 1
Page 20
1.8. Governing equations for multi-phase flow
In all of the derivations given thus far, we have assumed that
the pores of the rock are filled with a single-component, singlephase fluid. Oil reservoirs are typically filled with at least two
components, oil and water, and often also contain some
hydrocarbons in the gaseous phase. We will now present the
governing flow equations for an oil/water system, in a fairly
general form.
Recall from the rock properties module that Darcy’s law can
be generalised for two-phase flow by including a relative
permeability factor for each phase:
qw =
− kkrw dPw
,
μw dx
(1)
qo =
− kkro dPo
,
μo dx
(2)
where the subscripts w and o denote oil and water, respectively.
The two relative permeability functions are assumed to be known
functions of the phase saturations. For an oil-water system, the
two saturations are necessarily related to each other by
Sw + So = 1.
(3)
In general, the pressures in the two phases at each “point” in
the reservoir will be different. If the reservoir is oil-wet, the two
pressures will be related by
Po − Pw = Pcap (So ) ,
(4)
where the capillary pressure is given by some rock-dependent
function of the oil saturation.
Department of Earth Science and Engineering
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Flow in Porous Media
Dr. R. W. Zimmerman
Section 1
Page 21
As the volume of, say, oil, in a given region is equal to the
total pore volume multiplied by the oil saturation, the conservation
of mass equations for the two phases can be taken directly from
eq. (1.5.8), by inserting a saturation factor in the storage term:
d( ρoqo ) d(φρoSo )
=
,
dx
dt
(5)
d( ρwqw ) d(φρwSw )
=
.
dx
dt
(6)
−
−
The densities of the two phases are related to their
respective phase pressures by an equation of state:
ρo = ρo (Po ) .
(7)
ρw = ρw (Pw ) .
(8)
where the right-hand sides of eqs. (7) and (8) are known functions
of the pressure, and (for our present purposes) the temperature is
assumed constant.
Finally, the porosity must be some function of the phase
pressures, Po and Pw . Currently, little is known about the manner
in which these two pressures independently affect the porosity.
Fortunately, the capillary pressure is usually small, and so
Po ≈ Pw , in which case we can use the pressure-porosity
relationship that would be obtained in a laboratory test performed
under single-phase conditions, i.e.,
φ = φ (Po ) .
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Dr. R. W. Zimmerman
Section 1
Page 22
Eqs. (1)-(9) give nine equations for the nine unknowns
(count them!). In many situations, the equations are simplified to
allow solutions to be obtained. For example, in the BuckleyLeverett problem of immiscible displacement (module 4.1), the
densities are assumed to be constant, and the capillary pressure
is assumed to be zero.
If the fluid is slightly compressible (or if the pressure
variations are small), the equations of state are written as
ρ(Po ) = ρoi [1+ co (Po − Poi )],
(10)
etc., where the subscript i denotes the initial state, and the
compressibility co is taken to be a constant.
______________________________________________
Tutorial Sheet 1:
(1) A well located in a 100 ft. thick reservoir having permeability
100 mD produces 100 barrels/day of oil from a 10 in. diameter
wellbore. The viscosity of the oil is 0.4 cP. The pressure at a
distance of 1000 feet from the wellbore is 3000 psi. What is the
pressure in the wellbore? Conversion factors are as follows:
1 barrel = 0.1589 m3
1 Poise = 0.1 Newton-seconds/m2
1 foot = 0.3048 m
1 psi = 6895 N/ m2 = 6895 Pa
(2) Carry out a derivation of the diffusion equation for sphericallysymmetric flow, in analogy to the derivation given in section 1.7
for radial flow. The result should be an equation similar to eq.
(1.7.8), but with a slightly different term on the right-hand side.
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Section 2
Page 23
2. Line Source Solution for a Vertical Well in an
Infinite Reservoir
One of the most basic and important problems in petroleum
reservoir engineering, and the cornerstone of well-test analysis, is
the problem of flow of a single-phase, slightly compressible fluid
to a vertical well that is located in an infinite reservoir. This
problem can be formulated precisely as follows:
• Geometry: a vertical well that fully penetrates a reservoir which
is of uniform thickness, H, and which extends infinitely far in all
horizontal directions.
• Reservoir Properties: the reservoir is assumed to be isotropic
and homogeneous, with constant properties (i.e., permeability,
etc.) that do not vary with pressure.
• Initial and Boundary Conditions: the reservoir is initially at
uniform pressure. Starting at t = 0, fluid is pumped out of the
wellbore at a constant rate, Q.
• Wellbore diameter: it is assumed that the diameter of the
wellbore is infinitely small; this leads to a much simpler problem
than the more realistic finite-diameter case, but with little loss of
applicability, as we will see later.
Problem: to determine the pressure at all points in the reservoir,
including in the wellbore, as a function of the elapsed time since
the start of production.
The conditions outlined above lead to the so-called line source
solution, also known as the Kelvin solution or Theis solution. It
was derived by Kelvin in the 1880s in the context of heat
conduction; Charles Theis was the first to use it in the context of
flow to a well, in 1935.
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Flow in Porous Media
Dr. R. W. Zimmerman
Section 2
Page 24
2.1. Development of the line-source solution
The basic governing equation for this problem is the diffusion
equation in radial coordinates, eq. (1.7.8):
1 d ⎛ dP ⎞
⎜R
⎟.
=
dt R dR ⎜⎝ dR ⎟⎠
φμct dP
k
(1)
The assumptions that are inherent in this equation are:
(1) The reservoir is homogeneous and isotropic - i.e., k, φ, etc.,
do not vary with position in the reservoir, and so they can be
assumed to be constant.
(2) The thickness of the reservoir is uniform - this implies that
the flow will be horizontal flow to the well, with no vertical
component.
(3) The well fully penetrates the entire thickness of the reservoir
(or else there would be a vertical flow component).
(4) The fluid is only slightly compressible - this is implicit in
treating the compressibility term ct = cf + cφ as a constant.
• If the reservoir is anisotropic, the equations can be modified by
a change of variables that is equivalent to stretching the x and
y coordinates (de Marsily, pp. 178-9).
• Inhomogeneity is difficult to treat, and methods for this situation
are still being developed. Variation in the thickness of the
reservoir is somewhat equivalent to spatial variation in k (i.e.,
inhomogeneity).
• If the well does not fully penetrate the reservoir, we must add
d 2P/dz2 to the RHS of eq. (1). The solution to this much more
difficult problem is discussed by de Marsily, pp. 179-190.
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Flow in Porous Media
Dr. R. W. Zimmerman
Section 2
Page 25
• The case of a highly compressible fluid, such as a gas, will be
discussed in detail in section 9.
To solve the line-source problem, (or to solve any partial
differential equation), we not only need a governing equation, but
we also need initial conditions and boundary conditions. In this
case they are as follows.
Initial Condition: At the start of production, the pressure in the
reservoir is assumed to be at some uniform value, Pi .
Boundary condition at infinity: Infinitely far from the well, the
pressure will always remain at its initial value, Pi .
Boundary condition at the wellbore: At the wellbore, which is
assumed to be infinitely small, the flux must be equal to Q (into
the well) at all times t > 0.
We can therefore formulate the problem in precise mathematical
terms as follows:
Governing PDE:
1 d ⎡ dP ⎤ φμc dP
⎢R
⎥=
,
k dt
R dR ⎢⎣ dR ⎥⎦
(1)
Initial condition:
P(R,t = 0) = Pi ,
(2)
BC at wellbore:
⎛ 2πkH dP ⎞
⎟⎟ = Q ,
R
lim ⎜⎜
dR
μ
⎝
⎠
R→0
(3)
BC at infinity:
lim P(R,t) = Pi .
(4)
R→∞
There are many ways to solve this equation, but we will
solve it using a method that avoids advanced techniques such as
Laplace transforms, etc.
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First define
transformation:
Flow in Porous Media
Dr. R. W. Zimmerman
a
new
Section 2
Page 26
variable
η=
φμcR 2
kt
using
the
Boltzmann
.
(5)
Now, rewrite eq. (1) in terms of the new variable, η. The
right-hand side transforms as follows:
dP dP dη 2φμcR dP φμcR 2 2 dP 2η dP
=
=
=
=
.
kt R dη R dη
dR dη dR
kt dη
(6)
So we see that differentiation with respect to R is equivalent to
differentiation with respect to η, followed by multiplication by 2η/R.
Hence,
1 d ⎡ dP ⎤ 1 2η d ⎛ dP ⎞ 4η d ⎛ dP ⎞
⎟.
⎢R
⎥=
⎜ 2η ⎟ =
⎜η
R dR ⎢⎣ dR ⎥⎦ R R dη ⎜⎝ dη ⎟⎠ R2 dη ⎜⎝ dη ⎟⎠
(7)
The left-hand side of eq. (1) transforms as follows:
dP dP dη
φμcR 2 dP −η dP
=
=−
=
,
2 dη
t
η
dt
d
dη dt
kt
→
φμc dP
k
−φμc η dP −φμ cR2 η dP −η 2 dP
=
=
= 2
. (8)
2 dη
kt
η
dt
k t dη
d
R
R
Using eqs. (7) and (8) in eq. (1) yields
d ⎛ dP ⎞
η dP
⎜⎜ η
⎟⎟ = −
.
4 dη
dη ⎝ dη ⎠
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Dr. R. W. Zimmerman
Section 2
Page 27
Eq. (9) is an ordinary differential equation for P as a function of η.
We must also transform the boundary/initial conditions so
that they apply to P(η) instead of to P(r,t). First note that both
limits, R → ∞ and t → 0 , correspond to the limit η → ∞ . Hence,
conditions (2,4) take the form
lim P(η) = Pi .
(10)
η →∞
Using eq. (6) in eq. (3) leads to a second BC:
⎛ 4π kH dP ⎞
⎜
η ⎟⎟ = Q ,
lim ⎜
dη ⎠
μ
η →0 ⎝
⎛ dP ⎞
μQ
→ lim ⎜⎜ η ⎟⎟ =
.
d
4
η
π
kH
⎝
⎠
η →0
(11)
The problem is now a two-point ODE boundary-value problem,
defined by eqs. (9-11).
To solve this problem, we first note that although eq. (9)
appears to be a 2nd-order equation, it is actually a first-order
equation for the function η(dP/dη). If we temporarily denote this
combination of terms by y, we can write eq. (9) as
y
dy
=− ,
4
dη
where
y =η
dP
,
dη
(12)
Now separate the variables, and integrate from η = 0 out to
an arbitrary value of η:
dη
dy
=−
4
y
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Flow in Porous Media
Dr. R. W. Zimmerman
⇒
y(η ) dy
∫
y (0)
y
Section 2
Page 28
η dη
= −∫
0
4
⇒
⎡ y(η) ⎤
η
⎢
⎥=−
ln
⎢⎣ y(0) ⎥⎦
4
⇒
y(η) = y (0)e
−η / 4
.
(13)
Now note that the boundary condition (11) is equivalent to
y(0) =
μQ
,
4π kH
(14)
which implies that eq. (13) can be written as
y(η) =
μQ −η / 4
e
.
4π kH
(15)
Recall that y = η(dP / dη) , and rewrite eq. (15) as
dP(η)
μ Q e −η / 4
=
.
dη
4π kH η
(16)
Eq. (16) can now be directly integrated to find P(η). We
cannot start the integral at η = 0, because we do not know the
pressure in the wellbore. We do, however, know from eq. (10) that
the pressure at η = ∞ must be equal to the initial pressure, Pi.
Therefore,
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Section 2
Page 29
μQ e −η / 4
dη
∫ dP = ∫
π
kH
η
4
Pi
∞
P(η )
⇒
η
μQ ∞e −η / 4
dη .
P(η) = Pi −
∫
4π kH η η
(17)
Now recall that η = φμcR / kt . We replace η with φμcR / kt
on the left-hand side of eq. (17), and also at the lower limit of
integration on the right, but not inside the integral, because inside
the integral, η is merely a dummy variable:
2
2
⎛
2⎞
∞
φμ
cR
μQ
e −η / 4
⎜
⎟
.
P⎜
⎟ = Pi − 4π kH ∫ 2 η dη
kt
⎝
⎠
φμcR
(18)
kt
Simplify the integrand by defining u = η / 4 , in which case
dη / η = du / u , and the lower limit of integration becomes
u = φμcR2 / 4kt :
⎛
⎞
∞
φμcR 2 ⎟
μQ
e −u
⎜
.
P⎜
⎟ = Pi − 4π kH ∫ 2 u du
4
kt
⎝
⎠
φμcR
(19)
4kt
The integral in eq. (19) is essentially the so-called “exponential
integral function”, which is defined by mathematicians as (see
Matthews and Russell, p. 131)
−Ei ( −x ) =
∞ −u
e
∫ u du .
x
(20)
Unfortunately, this function was defined long before it was first
used to solve the problem of a well in an infinite reservoir, and so
it contains extraneous minus signs that are inconvenient.
Department of Earth Sciences and Engineering
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Flow in Porous Media
Dr. R. W. Zimmerman
Section 2
Page 30
We can summarise the solution to this problem as follows:
P( R, t ) = Pi +
where
μQ
Ei ( −x ) ,
4π kH
∞ −u
e
∫ u du ,
x
(22)
x = φμcR2 / 4kt .
(23)
−Ei ( −x ) =
and
(21)
Numerical values for the pressure drawdown are found as
follows. Assume that we want to know the pressure at a certain
distance R from the centre of the well, at some time t.
(a)
Use these values of R and t to compute a value of x from eq.
(23).
(b)
Look up the value of -Ei(-x) from a table or graph of the
exponential integral function (see next page).
(c)
The pressure at (R,t) is then given by eq. (21).
In practice, the more common situation is that the pressure is
measured at some distance from the well, as a function of time,
and the data is used to infer the values for the reservoir
parameters, by fitting the data to the analytical solution. This
procedure will be demonstrated after first analysing the linesource solution in more detail.
Note: to find the pressure in the wellbore, we merely plug R = Rw
into eq. (21)! This is because R = Rw corresponds to the wellbore
wall, where the pressure must be the same as in the wellbore.
Tabular values of the Ei function are shown below. The table is
based on the one given by de Marsily, Quantitative Hydrogeology,
Academic Press, 1986:
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Flow in Porous Media
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Section 2
Page 31
Table 2.1.1. Exponential Integral Function, - Ei(-x)
x
1
2
3
4
x1
.219
.049
.013
x10-1
1.82
1.22
x10-2
4.04
x10-3
5
6
7
8
9
.0038 .0011 3.6e-4
1.2e-4
3.8e-5
1.2e-5
0.91
0.70
0.56
0.45
0.37
0.31
0.26
3.35
2.96
2.68
2.47
2.30
2.15
2.03
1.92
6.33
5.64
5.23
4.95
4.73
4.54
4.39
4.26
4.14
x10-4
8.63
7.94
7.53
7.25
7.02
6.84
6.69
6.55
6.44
x10-5
10.94 10.24 9.84
9.55
9.33
9.14
8.99
8.86
8.74
x10-6
13.24 12.55 12.14 11.85 11.63 11.45
11.29
11.16
11.04
x10-7
15.54 14.85 14.44 14.15 13.93 13.75
13.60
13.46
13.34
x10-8
17.84 17.15 16.74 16.46 16.23 16.05
15.90
15.76
15.65
x10-9
20.15 19.45 19.05 18.76 18.54 18.35
18.20
18.07
17.95
x10-10
22.45 21.76 21.35 21.06 20.84 20.66
20.50
20.37
20.25
x10-11
24.75 24.06 23.65 23.36 23.14 22.96
22.81
22.67
22.55
x10-12
27.05 26.36 25.96 25.67 25.44 25.26
25.11
24.97
24.86
x10-13
29.36 28.66 28.26 27.97 27.75 27.56
27.41
27.28
27.16
x10-14
31.66 30.97 30.56 30.27 30.05 29.87
29.71
29.58
29.46
x10-15
33.96 33.27 32.86 32.58 32.35 32.17
32.02
31.88
31.76
For example, if x = 5 x 10-7, then -Ei(-x) = 13.93.
2.2. Dimensionless pressure and time
Department of Earth Sciences and Engineering
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Flow in Porous Media
Dr. R. W. Zimmerman
Section 2
Page 32
Although the pressure drawdown seems to depend on many
variables and parameters, there are only two independent,
dimensionless mathematical variables in the line-source solution.
This can be proven from the pi-theorem of dimensional analysis,
or can be seen directly from eqs. (2.1.21-23).
Traditionally, these
dimensionless time,
variables
tD =
kt
φμ cR2
are
defined
as
,
the
(1)
and the dimensionless pressure drawdown,
ΔPD =
2πkH( Pi − P)
.
μQ
(2)
In terms of these dimensionless parameters, the line-source
solution takes the form
1
2
ΔPD = − Ei(−1/ 4t D ) .
(3)
The usefulness of dimensionless variables is that they allow
the pressure drawdown to be plotted and discussed in a form that
is applicable to all reservoirs, without being restricted to specific
values of the permeability, porosity, etc. These parameters are
accounted for by the definitions of the dimensionless variables.
Department of Earth Sciences and Engineering
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2002-2003
Flow in Porous Media
Dr. R. W. Zimmerman
Section 2
Page 33
Dimensionless Drawdown, ΔP D
Line Source Solution
00
-1
ΔPD = 2πkH(Pi-P)/μQ
-21
tD = kt/φμ cR2
-3
-42
-5
3
-7
44
0
200
400
600
800
1000
Dimensionless Time, t D
Fig. 2.2.1. Line-source solution in terms of dimensionless
variables.
2.3. Range of applicability of the “line source” solution
The line-source solution assumes that the wellbore radius is
“zero”, when in reality it is always non-zero. Does this cause a
problem in practice?
• Not really; we can use the line-source solution as soon as the
“radius of penetration” of the pressure pulse, as predicted by
the line-source solution, is greater than Rw, the wellbore radius.
• This seems reasonable, and can be proven rigorously by
examining the solution to the diffusion equation with finite
wellbore radius (which will be developed in section 6).
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2002-2003
Flow in Porous Media
Dr. R. W. Zimmerman
Section 2
Page 34
According to eq. (1.6.12), the time required for the pressure
pulse to travel at least a distance Rw (starting from the “infinitelysmall” hypothetical borehole at R = 0) is
t>
φμcRw2
4k
.
(1)
If we use “typical” values for the parameters, such as
φ = 0.20 (a typical reservoir value),
μ = 0.001 Pa s (reasonable for liquid hydrocarbons),
c = 10
−10
-1
Pa (reasonable for liquid hydrocarbons),
Rw = 0.1m ,
k = 10
−14
2
m (10 mD - somewhat low, but possible),
then eq. (1) predicts that the line-source approximation will
become valid after an elapsed time of only 0.005 s!
Note: In terms of the dimensionless time defined by eq. (2.2.1),
condition (1) is equivalent to tDw > 0.25, where the subscript w
denotes the fact that this is the dimensionless time relative to the
position R = R w.
Although one should always check the validity of the linesource approximation, using eq. (1), in practice it is almost always
valid.
2.4. Logarithmic approximation to line-source solution
The exponential integral function is unfamiliar to most
engineers, and is difficult to calculate. Fortunately, for sufficiently
small values of x, which is to say, large values of t, the
Department of Earth Sciences and Engineering
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Dr. R. W. Zimmerman
Section 2
Page 35
exponential integral essentially becomes a logarithmic function,
which makes it very easy to use. To derive this “long-time”
approximation, we proceed as follows:
a. For large times, x will be small, and we can break up the
integral into two parts:
∞ −u
1 −u
∞ −u
e
e
e
−Ei(−x) = ∫
du = ∫
du + ∫
du .
u
u
u
x
x
1
(1)
b. Use the Taylor series for exp(-u) in the first integrand:
1 e −u
∫
x
u
11−
du = ∫
u
1!
+
u2
2!
u
x
u3
−
3!
+L
du .
(2)
c. Break up integral into a series of integrals, and evaluate them
term-by-term:
1 −u
11
11
11
11 2
e
du = ∫ du − ∫ du + ∫ udu + ∫ u du − L
∫
u
1! x
2! x
3! x
x
xu
= lnu
]1x − u ]1x + 2!1 u2 ⎥⎥
= (ln1− lnx) − (1− x ) +
= −ln x + x −
1
2⎤
⎦x
−
1
3⎤
1u ⎥
+L
3! 3 ⎥⎦
x
1
1
2
3
(1− x ) −
(1− x ) + L
2!2
3!3
⎧
⎫
1 2
1 3
1
1
x +
x + L − ⎨1−
+
+ L⎬
⎪⎩ 2!2 3!3
⎪⎭
2!2
3!3
→ −Ei(−x) = −ln x − lnγ + x −
Department of Earth Sciences and Engineering
1 2
1 3
x +
x + L,
2!2
3!3
Imperial College
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where
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Dr. R. W. Zimmerman
Section 2
Page 36
⎧
⎫ ∞e −u
1
1
du .
ln γ = ⎨1−
+
+ L⎬ − ∫
⎪⎩ 2!2 3!3
⎪⎭ 1 u
(4)
Eq. (4) may look messy, but the important point is that the term
lnγ is merely a number, and does not depend on x, so we can
evaluate it numerically to find
ln γ = ln(1.781) = 0.5772 .
(5)
[Note: Both γ and lnγ are sometimes known as “Euler’s number”,
so be careful when reading the literature on this topic!]
We can further simplify eq. (3) if we can find conditions
under which the power series terms are negligible. If this is the
case, we are left with only a log term and a constant.
First, recall that large t implies small x, and so
x−
1 2
1 3
x +
x +L < x .
2!2
3!3
(6)
If we want the power series terms to be, say, two orders of
magnitude less than γ (which itself is ~1), then we need
x=
φμcR2
< 0.01 ⇒
4kt
kt
φμcR 2
> 25 .
(7)
So, if the dimensionless time is greater than about 25, we have
from eqs. (2.1.21) and (3):
P(R,t) = Pi +
μQ
[ln x + ln γ ]
4π kH
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M.Sc. in Petroleum Engineering
2002-2003
Flow in Porous Media
Dr. R. W. Zimmerman
⇒
P(R,t ) = Pi +
μQ
ln(xγ )
4π kH
⎛
⎞
⇒
μQ ⎜ φμ cR2γ ⎟
P(R,t ) = Pi +
ln
4π kH ⎜⎝ 4kt ⎟⎠
⇒
μQ ⎜⎛ 4kt ⎟⎞
P(R,t ) = Pi −
ln
4π kH ⎜⎝ φμ cR 2γ ⎟⎠
⇒ P(R,t) = Pi −
⇒
Section 2
Page 37
μQ ln⎜⎛ 2.246kt ⎞⎟
4π kH ⎝ φμcR2 ⎠
⎡
(8)
⎤
μQ ⎢ ⎜⎛ kt ⎟⎞
⎥.
+
0.80907
P(R,t ) = Pi −
ln⎜
⎟
⎥
4π kH ⎢⎣ ⎝ φμcR 2 ⎠
⎦
(9)
The form given in eq. (8) is used in groundwater hydrology, and is
called Jacob’s approximation; the equivalent form of eq. (9) is
used in petroleum reservoir engineering, where it is called the
logarithmic approximation.
By comparing eq. (9) with eqs. (2.2.1-3), we see that the
dimensionless form of the logarithmic approximation can be
written as
ΔPD =
[
]
1
ln(t D ) + 0.80907)
2
(10)
For values of the dimensionless time that are typically of
interest in the wellbore, the logarithmic approximation is very
accurate. For example, the Fig. 2.4.1 below shows that the range
of validity of the logarithmic approximation seems to be consistent
with the criterion given in eq. (7).
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Flow in Porous Media
Dr. R. W. Zimmerman
Section 2
Page 38
Dimensionless Drawdown, ΔP D
Line Source Solution
-1
ΔPD = 2πkH(Pi-P)/μQ
0
tD = kt/φμ cR2
1
2
Exponential Integral Solution
Logarithmic Approximation
3
4
0.1
1
10
10 2
10 3
Dimensionless Time, tD
Fig. 2.4.1. Range of validity of the logarithmic approximation.
Although smaller values of tD are usually not of interest in the
wellbore, they are needed if one wants to calculate the drawdown
at a location far from the well. For tD < 1, the logarithmic
approximation is very poor. Verify this for yourself for, say, tD =
0.25, by evaluating eq. (9), and comparing the result to that
obtained from Table 2.1.1.
2.5. Instantaneous pulse of injected fluid
Some insight can be gained into the diffusive nature of flow
through porous media by considering the problem of a finite
amount of fluid injected into a well over a very small period of
time. This problem also introduces the important concept of
superposition, which will be described more rigorously in section
3.1.
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Flow in Porous Media
Dr. R. W. Zimmerman
Section 2
Page 39
Note: The problem of “injecting” fluid is mathematically
equivalent to that of “producing” fluid, except for the sign, but it is
easier to visualise the pressure pulse propagating into the
reservoir in the case of injection.
If we start injecting fluid at a rate Q (m3/s) at time t = 0, then,
according to eq. (2.1.21), the pressure at a distance R into the
reservoir will be
μQ ∞ e −u
P(R,t) = Pi +
du, x = φμcR2 / 4kt ,
∫
4πkH x u
(1)
where t is the elapsed time since the start of injection. The + sign
is used because we are injecting rather than extracting fluid, so
the pressure in the reservoir will increase.
Now imagine that we stop injecting fluid after a small amount
of time δt. This is equivalent to injecting fluid at a rate Q starting
at t = 0, and then producing fluid at a rate Q (or, equivalently,
injecting at a rate -Q) starting at time δt. The pressure drawdown
in the reservoir due to this fictitious production would be given by
the same line-source solution, except that:
(1) For extraction of fluid we must use a - sign in front of
the integral;
(2) If t is the elapsed time since the start of injection, then tδt will be the elapsed time since the start of the fictitious extraction
of fluid (i.e., since the end of injection!)
Hence, the full expression for the pressure will be
μQ
P(R,t) = Pi +
4πkH
∞
x=
e −u
μQ
du −
∫
4πkH
φμcR2 u
4kt
Department of Earth Sciences and Engineering
∞
∫
x=
φμcR2
4k (t − δt )
e −u
du
u
Imperial College
M.Sc. in Petroleum Engineering
2002-2003
Flow in Porous Media
Dr. R. W. Zimmerman
μQ
= Pi +
4πkH
x=
Section 2
Page 40
φμcR2
4k (t − δt )
∫
x=
φμcR2
e −u
du .
u
(2)
4kt
But if δt is small, then the two limits of integration are close
together, and we can use the following approximation:
x2
∫ f (x )dx ≈ f (x1)[x 2 − x1] ,
(3)
x1
which in the present case gives
−
μQ
4kt
P(R,t) ≈ Pi +
⋅
⋅e
4πkH φμcR 2
φμcR2
μQ
4kt
≈ Pi +
⋅
⋅e
4πkH φμcR 2
4kt
−
⎡ φμcR 2
2⎤
φμ
cR
⎥
⋅⎢
−
⎢⎣ 4k(t − δt)
4kt ⎥⎦
φμcR2
4kt
μQ δt −
≈ Pi +
e
4πkHt
⎡ φμcR 2δt ⎤
⎢
⎥
⋅
2
⎥⎦
⎢⎣ 4kt
φμcR2
4kt
.
(4)
As Q is the rate of injection in [m3/s], and δt is the duration of
the injection, the total volume of injected fluid is Qδt, which we
can denote as Q*, with units of [m3].
Now imagine that we are monitoring the pressure at some
distance R from the borehole. From eq. (4), we se that the
pressure buildup at R will be the product of two terms:
(a) an exponential term that increases with t, then levels off to a
value of 1 as t → ∞ ;
(b)
a 1/t term that decays to zero as t → ∞ .
Department of Earth Sciences and Engineering
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2002-2003
Flow in Porous Media
Dr. R. W. Zimmerman
Section 2
Page 41
Their product gives a function that first increases, and then
decreases with time, as shown below in Fig. 2.5.1:
Response to Injection Pulse
2πH(P-Pi)μφ cR2 /μQ*
0.4
0.3
0.2
0.1
0
-0.1
0.01
0.1
10 1
1
Dimensionless Time, t
D
= kt/φμ cR 2
Fig. 2.5.1. Pressure buildup due to an injected pulse of fluid.
It seems reasonable to identify the time at which the
pressure buildup reaches its maximum value as the time “at which
the pressure pulse has arrived at location R”. This time is found
by setting dP/dt = 0 in eq. (4):
∂P
∂t R
⎡ −1 φμcR2 ⎤ − φμcR2
⎢ +
⎥e 4kt ,
=
2
3
4πkH ⎢⎣ t
4kt ⎥⎦
μQ∗
∂P
=0
∂t R
⇒
t =
φμcR 2
4k
,
(6)
which is equivalent to eq. (1.6.12).
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Flow in Porous Media
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Section 2
Page 42
2.6. Estimating permeability and storativity from a drawdown
test
In sections 2.1-4 we derived the line-source solution, and
showed how to use it to calculate the pressure drawdown, based
on assumed knowledge of the reservoir properties. However, the
most common use of this solution, and the other solutions that we
will derive shortly, is for the inverse problem:
• We use measured wellbore pressures, in conjunction with the
mathematical solutions, to estimate reservoir properties such
as permeability, porosity, etc.
• This process, known as well-test analysis, is the subject of a
subsequent module of the course. For now, we will do one
simple example to see how to calculate the reservoir
permeability from a drawdown test.
Recall from eq. (2.4.8) that, in the long-time regime,
μ Q ⎜⎛ 2.246kt ⎟⎞
P(R,t) = Pi −
ln
.
4π kH ⎜⎝ φμ cR 2 ⎟⎠
(1)
At the wellbore wall,
⎡
⎤
⎛ 2.246k ⎞
μQ ⎢
⎟⎥.
P(Rw ,t) = Pi −
lnt + ln⎜⎜
2⎟
4π kH ⎢⎣
⎝ φμcRw ⎠ ⎥⎦
(2)
In general, we will not know the values of any of the parameters
on the RHS of eq. (2); we will only know Q, and the wellbore
pressure as a function of time, P(Rw ,t ) ≡ Pw (t ) . In particular, we
don’t know k or φμc, so we can’t use criterion (2.4.7) to find out
when our data falls in the late-time regime.
Department of Earth Sciences and Engineering
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Dr. R. W. Zimmerman
Section 2
Page 43
However, the second logarithmic term, although unknown, is a
constant. Hence, if we plot Pw (t ) vs. lnt, the data will (eventually,
at large enough values of t) fall on a straight line!
The slope of this line on a semi-log plot gives kH, i.e.,
dPw
ΔPw
μQ
=
≡m=
,
d lnt
Δ lnt
4π kH
→
kH =
(3)
μQ
.
4π m
(4)
• Q is known, and μ can be measured, so the semi-log slope
gives the permeability-thickness product, kH.
• This method is unable to distinguish separately between the
effects of permeability and thickness; i.e., a thick impermeable
reservoir can give the same drawdown as a thin, permeable
reservoir.
Now, let’s do a simple example to learn how to calculate the
permeability of a reservoir from a drawdown test.
Example: A well with 4 in. radius produces oil with a viscosity of
0.3 cP, at a rate of 200 barrels/day, from a reservoir that is 15 ft.
thick. The pressure in the wellbore as a function of time is:
t (mins)
1
5
10
20
30
60
Pw (psi)
4740
4667
4633
4596
4573
35
Use the “semi-log straight line” method to estimate the
permeability, k.
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Flow in Porous Media
Dr. R. W. Zimmerman
Section 2
Page 44
Plot wellbore pressure against the logarithm of time:
Wellbore Pressure (psi)
4800
Wellbore Pressure (data)
Straight line (fitted)
4750
4700
4650
4600
4550
4500
1
(2)
10
t (mins)
100
Look for a straight line at long times, and find its slope:
m=
ΔP
4760 − 4510
=
= 54.3 psi
Δ lnt
2 × 2.303
= 54.3psi ×
6895 Pa
= 374,400 Pa .
psi
Note: Δ lnt has the same value regardless of which units are used
for t!
(3)
Calculate k from eq. (4). First convert all data to SI units:
μ = 0.3 cP ×
0.001 Pa ⋅ s
= 0.0003 Pa ⋅ s
cP
3
hr
bbl 0.1589 m3 day
−4 m
Q = 200
×
×
×
= 3.68 × 10
24hr 3600 s
day
bbl
s
Department of Earth Sciences and Engineering
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2002-2003
Flow in Porous Media
Dr. R. W. Zimmerman
H = 15 ft ×
→
Section 2
Page 45
0.3048 m
= 4.572 m
ft
μQ
(0.0003Pa s)(3.68 × 10-4 m 3/s)
k=
=
4πmH
4π (4.572m)(374,400Pa)
= 5.13 × 10 -15 m2 ×
1mD
0.987 × 10
-15
2
= 5.1mD .
m
We can also use a semi-log plot to estimate the storativity term,
φc. To see how this works, first note that on a semi-log plot of Pw
vs. lnt, there are two asymptotic straight lines:
(a) At early times, Pw = Pi (plots as a horizontal line).
(b) At late times, Pw slopes downward as a function of lnt,
according to eq. (4).
Q:
At what time t* do these straight lines intersect?
A:
When Pw (early t asymptote) = Pw (late t asymptote) :
a
μQ ⎜⎛ 2.246kt * ⎟⎞
Pi = Pi −
ln
4π kH ⎜⎝ φμ cRw2 ⎟⎠
⎛ 2.246kt * ⎞
⎟ =0
ln ⎜
⎜ φμ cR2 ⎟
⎝
w ⎠
a
a
→
2.246kt *
φμcRw2
φc =
=1
2.246kt *
μRw2
,
(5)
where t* is the intersection time; see the well test analysis module
for details.
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2002-2003
Flow in Porous Media
Dr. R. W. Zimmerman
Section 2
Page 46
______________________________________________
Tutorial Sheet 2:
(1) A well with 3 in. radius is located in a 40 ft. thick reservoir
having permeability 30 mD and porosity 0.20. The total
compressibility of the oil/rock system is 3x10-5/psi. The initial
pressure in the reservoir is 2800 psi. The well produces 448
barrels/day of oil having a viscosity of 0.4 cP.
(a) How long will it take in order for the line-source solution to be
applicable at the wellbore wall?
(b) What is the pressure in the wellbore after six days of
production, according to the line-source solution?
(c) How long will it take in order for Jacob’s logarithmic
approximation to be valid at the wellbore?
(d) What is the pressure in the wellbore after six days of
production, according to the logarithmic approximation?
(e) Answer questions (b)-(d) for a location that is 800 ft.
(horizontally) away from the wellbore.
Conversion factors can be found at end of Tutorial Sheet 1.
(2) A well with radius 0.3 ft. produces 200 barrels/day of oil, with
viscosity 0.6 cP, from a 20 ft. thick reservoir. The wellbore
pressures are as follows:
t (mins)
0
5
10
20
60
120
480
1440
2880
5760
Pw (psi)
4000
3943
3938
3933
3926
3921
3911
3904
3899
3894
Estimate the permeability and the storativity of the reservoir, using
the semi-log method presented in section 2.6.
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3.
Flow in Porous Media
Dr. R. W. Zimmerman
Section 3
Page 47
Superposition and Buildup Tests
3.1. Linearity and the principle of superposition
A basic property of the diffusion equation that governs flow
of a single-phase compressible liquid through a porous medium is
its linearity. Linearity is the most important property that any
differential equation can have, because it implies that the principle
of superposition can be used to construct solutions to the
equation. Most of the analytical methods that have been
developed to solve differential equations, such as Laplace
transforms, Green’s functions, separation of variables, etc., can
be used only on linear differential equations. These analytical
methods will be discussed in later sections of this module. In this
section we will discuss a simple form of the principle of
superposition that will allow us to solve many important reservoir
engineering problems, such as pressure buildup tests.
The diffusion equation is a linear PDE because both of the
differential operators that appear in it are linear operators. In
general, a differential operator M that operates on a function F is
linear if it has the following two properties:
M(F1 + F2) = M(F1) +M(F2 ) ,
(1)
M(cF1) = cM(F1) ,
(2)
where F1 and F2 are any two differentiable functions, and c is any
constant.
The process of partial differentiation is a linear operation,
since
[
]
dP (R,t ) dP2 (R,t )
d
P1(R,t) + P2 (R,t) = 1
+
,
dt
dt
dt
Department of Earth Science and Engineering
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(3)
M.Sc. in Petroleum Engineering
2003-2004
Flow in Porous Media
Dr. R. W. Zimmerman
[
Section 3
Page 48
]
dP (R,t)
d
cP1(R,t) = c 1
.
dt
dt
(4)
By this definition, we see that eq. (1.7.8) is a linear PDE.
This is will be the case if the coefficients that appear in the
diffusivity, such as φ, c, μ and k, are constants. If these
coefficients were instead functions of position or time, the
governing equations (see section 1.8) would still be linear, albeit
more difficult to solve. However, if any of the coefficients were
functions of pressure, the equation would no longer be linear. This
is the case, for example, with gas flow, for which the
compressibility varies with pressure (see section 9). It is also the
case for “stress-sensitive” reservoirs in which the permeability
varies with pressure.
A simple rule-of-thumb is that a differential equation will be
nonlinear if it contains any term in which the dependent variable
(in our case, P) or any of its derivatives appear to a power higher
than one, or are multiplied by one another. For example, M =
P(dP/dt) is a nonlinear operator, because it violates condition (1):
M {P1 + P2}= (P1 + P2)
d(P1 + P2 )
dt
⎡ dP dP ⎤
= (P1 + P2 ) ⎢ 1 + 2 ⎥
⎢⎣ dt
dt ⎦⎥
= P1
dP1
dP
dP
dP
+ P2 1 + P1 2 + P2 2
dt
dt
dt
dt
= M {P1}+ M {P2}+ P2
dP1
dP
+ P1 2
dt
dt
⇒ L{P1 + P2}≠ L{P1}+ L{P2}!
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Flow in Porous Media
Dr. R. W. Zimmerman
Section 3
Page 49
The importance of linearity is that it allows us to create new
solutions to the diffusion equation by adding together previously
known solutions. Care must be taken, however, with the initial
conditions and boundary conditions. For example, if P1 and P2 are
two solutions that each satisfy the diffusion equation (2.1.1) and
the initial condition (2.1.2), then the sum of P1 and P2 will also
satisfy the diffusion equation, but will not satisfy the correct initial
conditions, because
P1(R, t = 0) + P2 (R, t = 0) = Pi + Pi = 2Pi (!)
6)
This difficulty can be circumvented by working with the
drawdown instead of the actual pressure. Linearity of the diffusion
equation implies that the drawdown ΔP = Pi − P(R, t ) satisfies the
same diffusion equation as does P(R,t) itself, since, for example,
dP(R, t )
d[ Pi − P(R, t )] dPi dP(R, t )
=
−
=−
, etc.
dt
dt
dt
dt
(7)
As the drawdown satisfies zero initial condition, by definition, eq.
(6) shows that the sum of two drawdown functions will also satisfy
the correct initial condition (i.e., that the drawdown must be zero
when t = 0.)
Likewise, the drawdown is also zero infinitely far from the
well, so if two drawdown functions satisfy the boundary condition
at R = ∞ , their sum will also satisfy this outer boundary condition.
3.2. Pressure buildup tests
In a pressure buildup test, a well that has been producing
fluid at a constant rate Q for some time t is then “shut in” - i.e.,
production is stopped. After this occurs, fluid will continue to flow
towards the well, because of the pressure gradient, but will not be
Department of Earth Science and Engineering
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Dr. R. W. Zimmerman
Section 3
Page 50
able to exit at the wellhead. Consequently, the pressure within the
well will rise back towards it initial value, Pi. The rate of this
pressure recovery in the well can be used to estimate both the
transmissivity, kH, and the initial pressure, Pi, of the reservoir.
The analysis of a pressure buildup test is based on the
principle of superposition that was discussed in the previous
section, and proceeds as follows:
First imagine that we produce at a rate Q, starting at t = 0, in
which case the pressure drawdown due to this production will be
⎛ −φμcR 2 ⎞
μQ
⎜
⎟.
ΔP1 = Pi − P1(R,t) = −
Ei ⎜
4πkH ⎝ 4kt ⎟⎠
(1)
Now consider the following fictitious problem, in which, at
some time t1, we begin to inject fluid into the reservoir at a rate Q.
The drawdown due to this injection would be given by the same
line-source solution, except that:
(a) The “elapsed time” used in the solution must be measured
from the start of injection, i.e., the variable must be t - t1;
(b) Since we are injecting rather than producing, we must use “Q” in the solution.
Therefore, the pressure drawdown due to this fictitious
injection is
⎛ −φμcR 2 ⎞
μQ
⎟.
ΔP2 = Pi − P2(R,t) =
Ei ⎜⎜
4πkH ⎝ 4k(t − t 1) ⎟⎠
(2)
NOTE: it is implicitly understood in all equations such as (2) that
the Ei function is taken to be zero when the term in brackets is
positive, which is to say when t < t1.
Department of Earth Science and Engineering
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Flow in Porous Media
Dr. R. W. Zimmerman
Section 3
Page 51
We now superimpose these two solutions (for the
drawdown!), putting ΔP = ΔP1 + ΔP2 . In light of the discussion
given in the previous section, this composite function is also a
solution to the diffusion equation. However, we note that
(i) For t < t1, the injection has not started, and so the composite
solution corresponds to production at rate Q.
(ii) For t > t1, the composite solution corresponds to production
at rate Q, and injection at rate Q, which is to say, a well that is
neither producing nor injecting, i.e., “shut in”!
(iii) Therefore, this superposition of solutions solves the problem
of production for a time t1, followed by shut-in:
ΔP(R,t) = ΔP1(R,t ) + ΔP2 (R,t)
⎛ −φμcR 2 ⎞
⎛ −φμcR2 ⎞
μQ
μQ
⎜
⎟
⎜
⎟
=−
Ei ⎜
Ei ⎜
+
⎟
4πkH ⎝ 4kt ⎠ 4πkH ⎝ 4k(t − t1) ⎟⎠
⎛ −φμcR 2 ⎞ ⎫⎪
μQ ⎧⎪ ⎜⎛ −φμcR2 ⎟⎞
⎟ ⎬.
⎨Ei ⎜
=−
− Ei ⎜⎜
⎟
⎟
4πkH ⎪⎩ ⎝ 4kt ⎠
⎝ 4k(t − t1) ⎠ ⎪⎭
Now recall that ΔP(R,t) = Pi − P(R,t) ,
P(R,t) = Pi − ΔP(R,t) , and so
in
(3)
which
case
⎛ −φμcR 2 ⎞ ⎫⎪
μQ ⎧⎪ ⎜⎛ −φμcR 2 ⎟⎞
⎟ ⎬.
⎨Ei ⎜
P(R,t) = Pi +
− Ei ⎜⎜
⎟
⎟
4πkH ⎪⎩ ⎝ 4kt ⎠
⎝ 4k(t − t 1) ⎠ ⎪⎭
(4)
If t is sufficiently large that we can use the logarithmic
approximation for both terms in eq. (4), then
Department of Earth Science and Engineering
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Section 3
Page 52
⎧
⎫
⎛ 2.246k ⎞
⎛ 2.246k ⎞ ⎪
μQ ⎪
⎟ − ln(t − t 1) − ln⎜
⎟⎬
⎨lnt + ln⎜⎜
P(R,t) = Pi −
⎟
⎜
⎟
2
2
4πkH ⎪⎩
⎝ φμcR ⎠
⎝ φμcR ⎠ ⎪⎭
→
P(R,t) = Pi −
→
μQ
{lnt − ln(t − t1)}
4πkH
P(R,t) = Pi −
t
μQ
ln
.
4πkH (t − t1)
(5)
This is the equation for the pressure in the wellbore during a
buildup test. However, the usual nomenclature used in buildup
tests is the following:
• The duration of production period is denoted by t, not t1.
• The duration of the shut-in period is denoted by Δt, not t-t1.
Using this notation, the shut-in pressure is given by
μQ ⎛ t + Δt ⎞
⎟.
Pw (t ) = Pi −
ln⎜
4πkH ⎜⎝ Δt ⎟⎠
(6)
This equation is used for graphical analysis of the data from a
buildup test, allowing estimation of the kH product and the initial
reservoir pressure. The ratio of times that appears in the
logarithmic term of eq. (6) is known as the Horner time, tH. Note
that it has the following peculiar properties:
(a) it doesn’t have units of time, but is dimensionless, and
(b) it becomes numerically smaller as the duration of the shut-in
period increases!
Be aware of these facts when using the Horner time.
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Section 3
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The wellbore pressure that would be measured during a
shut-in test is shown schematically in Fig. 3.2.1.
Pw
Δt
t
time
Fig. 3.2.1. Wellbore pressure during a buildup test.
3.3. Multi-rate flow tests
The superposition principle that was used to solve the
problem of a buildup test can also be used in the more general
situation in which the production rate is changed by discrete
amounts at various time intervals. For example, imagine that the
production rate is given by
Q = Q0
Q = Q1
for
for
0 < t < t1,
(1)
t > t1 .
(2)
To find the drawdown, we superpose the solution for
production at rate Q0 starting at time t = 0, plus a solution starting
at t1 that corresponds to the increment in the production rate,
Q1 − Q0 :
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μQ0 ⎜⎛ −φμcR 2 ⎟⎞ μ (Q1 − Q0 ) ⎜⎛ −φμ cR2 ⎟⎞
ΔP(R,t) = −
Ei
Ei ⎜
−
⎟ . (3)
4πkH
4πkH ⎜⎝ 4kt ⎟⎠
4k(t
−
t
)
1 ⎠
⎝
To verify that it is correct to use the flowrate increment, note
that for t > t1, the first Ei function corresponds to a flowrate of Q0 ,
and the second corresponds to a rate of Q1 − Q0 , so the total
flowrate is
Q(t > t1) = Q0 + (Q1 − Q0 ) = Q1.
(4)
The drawdown in the general case in which flowrate Qi
commences at time ti can therefore be represented by
− μQ0 ⎜⎛ −φμcR 2 ⎟⎞
μ (Qi − Qi −1 ) ⎜⎛ −φμ cR2 ⎟⎞
ΔP(R,t) =
Ei
Ei ⎜
−∑
⎟.
4πkH ⎜⎝ 4kt ⎟⎠ i =1 4πkH
4k(t
−
t
)
i
⎝
⎠
(5)
We can simplify the notation by defining the pressure
drawdown per unit of flowrate, with production starting at t = 0, as
ΔPQ(t). For the line-source in an infinite reservoir, this definition
takes the form
ΔPQ (R,t) ≡
ΔP(R,t;Q)
Q
⎛ −φμ cR2 ⎞
−μ
⎟,
⎜
≡
Ei ⎜
4πkH ⎝ 4kt ⎟⎠
(6)
with the understanding that ΔPQ(t) = 0 when t < 0.
Using definition (6), the drawdown in a multi-rate test can be
written as
ΔP(R,t) = Q0 ΔPQ (R,t) + ∑ (Qi − Qi −1) ΔPQ (R,t − t i ) .
i =1
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3.4. Convolution and variable-rate flow tests
The superposition formula (3.3.7) can be generalised further,
to the case where the flowrate at the well is some arbitrary (but
continuous) function of time, Q(t). We first note that an arbitrary
production schedule can always be approximated by a discrete
number of time periods during which the flowrate is constant, as
shown in Fig. 3.4.1:
Q
Q4
Q0
t0
t1
t2
t3
t4
t5
t
Fig. 3.4.1. Variable-rate production approximated by multi-rate
production.
Now recall that the time derivative of the flowrate can be
approximated as
dQ
ΔQ Qi − Qi −1
≈
=
.
dt t
Δt
t i − t i −1
(1)
i
Conversely, the flowrate increment can be approximated by
Qi − Qi −1 ≈
dQ(t i )
[t i − t i −1 ] .
dt
Using this approximation in eq. (3.3.7) gives
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dQ(t i )
ΔPQ (R,t − t i )[t i − t i −1 ] . (3)
dt
i =1
ΔP(R,t) = Q0 ΔPQ (R,t) + ∑
We now simplify the notation by rewriting eq. (3) in the following
equivalent form:
ΔP(R,t) = Q0 ΔPQ (R,t) + ∑ Q ′(t i ) ΔPQ (R,t − t i )Δt i .
(4)
i =1
As we make each time increment smaller, approximation (1)
becomes more accurate, and the “step-function” approximation to
Q(t) also becomes more accurate. In the limit as each time
increment goes to zero, the errors due to these approximations
will vanish. But as the time increments get smaller, the series in
eq. (4) becomes an integral with respect to ti:
ti = t
ΔP(R,t) = Q0 ΔPQ (R,t) + ∫ Q ′(t i )ΔPQ (R,t − t i )dt i .
(5)
ti = 0
The integral in eq. (5) ends at ti = t, because when ti > t, the
function ΔPQ(t-ti) is zero, by definition. Physically, this reflects the
fact that a change in flowrate that occurs at a time later than t
cannot possibly have an effect on the drawdown at time t.
Finally, we note that in the limit as each of the time
increments become infinitely small, the finite number of times that
we were denoting by ti evolve into a continuous variable, which
we will denote by τ. The drawdown can then be written as
dQ(τ )
ΔPQ (R,t − τ )dτ .
d
τ
0
t
ΔP(R,t) = Q0 ΔPQ (R,t) + ∫
(6)
The integral in eq. (6) is known as a convolution integral. Formula
(6) is also known as Duhamel’s principle, after the French
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physicist who first derived this equation in 1833 (in the context of
solving heat conduction problems).
The importance of eq. (6) is that it allows us to find the
drawdown for any production schedule, by merely performing a
single integral utilising the “constant-flowrate” solution. The
principle also works for other boundary-value problems that are
governed by a linear PDE. For example, if we have a reservoir
with, say, a closed outer boundary (see section 6), then we need
only to find the solution for the case of constant flowrate in a
reservoir with a closed outer boundary; the solution for a variable
flowrate then follows from eq. (6), with the constant-flowrate
solution playing the role of the function ΔPQ.
Another form of the convolution integral that is sometimes
more convenient to use can be derived by applying integration-byparts to the integral in (6). First, recall that the general expression
for integration by parts is
t
t
dg(τ )
df (τ )
dτ = f (τ )g(τ ) 0 − ∫ g( τ )
dτ .
∫ f (τ )
d
d
τ
τ
0
0
]
t
(7)
We now put f (τ ) = ΔPQ (R, t − τ ) and g(τ ) = Q(τ ) , in which case (6)
becomes
ΔP(R,t) = Q0 ΔPQ (R,t) + Q( τ )ΔPQ (R,t − τ )]
t
0
t
− ∫ Q(τ )
0
dΔPQ (R,t − τ )
dτ
dτ
= Q0 ΔPQ (R,t) + Q(t) ΔPQ (R,0) − Q(0)ΔPQ (R,t)
t
− ∫ Q(τ )
0
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dτ
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But Q(0) = Q0 , by definition, so the first and third terms on the
right cancel out. And ΔPQ (R,0) is the drawdown at time zero,
which must be zero, so the second term drops out.
Using the chain rule we can see that
dΔPQ (R,t − τ ) −dΔPQ (R,t − τ )
=
,
dτ
dt
in which case (8) can finally be written as
t
ΔP(R,t) = ∫ Q(τ )
0
dΔPQ (R,t − τ )
dτ .
dt
(9)
(10)
______________________________________________
Tutorial Sheet 3:
(1) Which, if any, of the following differential equations are
linear, and why (or why not)?
(a)
d2y
dy
+y
+ y = 0.
2
dx
dx
(b)
d2y
dy
+x
+ y = 0.
2
dx
dx
(c)
d2y
dy
+x
+ xy = 0 .
dx
dx 2
(2) What would be the wellbore pressure in a well in an infinite
reservoir, if the production rate increases linearly as a function of
time according to Q(t) = Q t / t , where Q and t are constants?
*
*
*
*
Use convolution, in the form of either (3.4.6) or (3.4.10), and recall
that ΔPQ(R,t) for a well in an infinite reservoir is given by eq.
(3.3.6).
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Effects of Faults and Linear Boundaries
4.1. Superposition of solutions in space
In section 3.1 we saw that if we add together two different
line-source solutions that “start” at different times, we still have a
legitimate solution to the diffusion equation. We can also add
together solutions that represent line-sources that are located at
different places in space, and the sum will represent a solution to
the diffusion equation.
The most simple and straightforward example of the use of
spatial superposition is for the problem of two wells located in an
infinite reservoir, as shown in Fig. 4.1.1. Well 1 is located at point
A, and the distance from A to a generic point in the reservoir is
denoted by R1. Well 2 is located at point B, and the distance from
B to a generic point in the reservoir is denoted by R2. Well 1
starts producing at rate Q1 at time t1, and well 2 starts producing
at rate Q2, starting at time t2.
R2
R1
Well 2
Well 1
k,φ,μ,c,H
Fig. 4.1.1. Two wells producing from an infinite, homogeneous
reservoir.
We now claim that the drawdown at any point in the reservoir
at time t is given by the sum of the two line-source solutions:
μQ1 ⎜⎛ −φμcR12 ⎟⎞ μQ2 ⎜⎛ −φμcR22 ⎟⎞
+
P(R,t) = Pi +
Ei
Ei
.
4π kH ⎜⎝ 4k(t − t 1) ⎟⎠ 4π kH ⎜⎝ 4k(t − t 2 ) ⎟⎠
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To prove that this is indeed the proper solution to this
problem, we note that:
(1) The line-source solution for well 1 satisfies the diffusion
equation at all points in the reservoir (except at R1 = 0, where it
gives an “infinite” drawdown), and at all times. Similarly for the
second line-source solution. Hence, their sum satisfies the
diffusion equation at all times, and at all points except R1 = 0 and
R2 = 0. But these two points are not actually located in the
reservoir, so we don’t care that the diffusion equation is not
satisfied there!
(2) Both line-source solutions satisfy the initial condition of
zero drawdown, so their sum also satisfies this initial condition.
(3) Both line-source solutions satisfy the far-field boundary
condition that the drawdown is zero “at infinity”, so their sum also
satisfies this boundary condition, i.e.,
P(R → ∞,t) = Pi +
μQ
μQ
Ei (−∞ )+
Ei (−∞)= Pi ,
4π kH
4π kH
(2)
because, as can be seen from Table 2.1.1, Ei(-x) goes to zero as
x becomes very large.
Hence, it seems that the drawdown given by eq. (1) is the
correct solution for the two-well problem. Obviously, spatial
superposition can be used for any number of wells.
The principle of superposition can also be used in
conjunction with the concept of “fictitious” wells, or “image” wells,
to solve problems such as finding the effect of a nearby
impermeable fault, as explained in the next section.
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4.2. Effect of an impermeable vertical fault
Hydrocarbon reservoirs are often transected by numerous
faults, many of which are nearly vertical. Due to mineral
deposition on the fault surfaces, accumulation of fault gouge, and
other geological processes, faults are often impermeable to flow.
In order to properly interpret the results of well tests, it is
important to understand the effect that an impermeable linear
boundary will have on a drawdown test.
It is therefore desirable to find the solution for the problem of
constant-rate production from a well located in an infinite,
homogeneous reservoir that is bounded on one side by a vertical
fault of infinite extent that is impermeable to flow. This solution
can easily be found from the Theis solution for a well in an infinite
(unbounded) reservoir, by using the principle of superposition.
Consider a well located at a perpendicular distance (i.e.,
nearest distance) d from an impermeable fault, which appears in
planview as a straight line that extends infinitely far in both
directions, as in Fig. 4.2.1. This well produces fluid at a constant
rate Q, starting at t = 0.
Now imagine a fictitious “image well” that is situated as the
“mirror image” of the first well (i.e., at a distance d on the other
side of the fault) which also produces at rate Q starting at t = 0, as
in Fig. 4.2.1.
By the symmetry of this situation, no fluid will cross the plane of
the fault. Hence, this plane will effectively act as a no-flow
boundary. We therefore have found the solution for a well near an
impermeable fault!
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R2
R1
d
Actual Well
k,φ,μ,c,H
d
Image Well
Impermeable fault
Fig. 4.2.1. Use of an image well to solve problem of a well near an
impermeable fault.
The pressure drawdown in this situation is the superposition
of the drawdown due to the actual well, plus the drawdown due to
the fictitious image well:
⎛ −φμcR2 ⎞
⎛ −φμcR 2 ⎞
μQ
μ
Q
1
2 ⎟.
⎟+
P(R,t) = Pi +
Ei ⎜⎜
Ei ⎜⎜
⎟
4π kH ⎝ 4kt ⎠ 4π kH ⎝ 4kt ⎟⎠
(1)
To find the pressure in the wellbore, we put
R1 = Rw , R2 = 2d − Rw ≈ 2d ,
(2)
in which case eq. (1) becomes
⎛ −φμcR 2 ⎞
⎛ −φμc(2d) 2 ⎞
μQ
μ
Q
w
⎟+
⎟ . (3)
Pw (t ) = Pi +
Ei ⎜⎜
Ei ⎜⎜
⎟
⎟
4π kH ⎝ 4kt ⎠ 4π kH ⎝
4kt
⎠
There are several time regimes to consider:
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(i) There is an early time regime, during which the logarithmic
approximation is not yet valid for the actual well solution (nor for
the image-well solution). From eq. (2.4.7) this regime is defined
by
25φμcR w2
t<
,
k
or
t Dw < 25 .
(4)
•
In this regime, we would have to use the full infinite series
expansion for the exponential integral function.
•
However, we saw in Section 2.3 that the duration of this
regime is usually very short, so there is no need for us to
study it further.
(ii) There is another regime in which the logarithmic
approximation can be used for the well solution, but the
drawdown in the wellbore due to the image solution is still
negligible. The start of this regime is defined by eq. (4), i.e.,
25φμcR w2
t>
,
k
or
t Dw > 25 .
(5)
This regime ends when the front edge of the pressure pulse from
the image well reaches the actual well (which is a distance 2d
away). We can consider that this occurs when the second Ei
function in eq. (3) reaches, say, 0.01.
According to Table 2.1.1, this occurs when the argument of the Ei
function reaches about 3.3. Therefore, the upper limit of this time
regime is defined by
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0.3φμcd 2
t<
,
k
or
Section 4
Page 64
t Dw < 0.3(d / Rw )2 .
(6)
• In this regime, the pressure in the well is given by
μQ ⎜⎛ 2.246kt ⎟⎞
Pw (t ) = Pi −
ln
.
4π kH ⎜⎝ φμcRw2 ⎟⎠
(7)
i.e., this is the drawdown that would occur in the absence of the
fault, because in this time regime the actual pressure pulse has
not yet had time to travel to the fault and reflect back to the
wellbore (where it can be detected).
• In particular, during this regime the slope of the wellbore
pressure curve on a semi-log plot will be
dPw ΔPw
− μQ
=
=
.
d lnt Δ lnt 4π kH
(8)
(iii) There is a late time regime during which the logarithmic
approximation is valid for both the actual well solution, and the
image well solution.
• This regime is defined by
25φμc(2d)2
t>
,
k
or
t Dw > 100(d /Rw ) 2 .
• In this regime, the pressure in the well is
μQ ⎜⎛ 2.246kt ⎟⎞
μQ ⎜⎛ 2.246kt ⎟⎞
−
Pw (t ) = Pi −
ln
ln
4π kH ⎜⎝ φμcRw2 ⎟⎠ 4π kH ⎜⎝ φμc(2d) 2 ⎟⎠
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Section 4
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⎡
⎤
⎛ 2.246k ⎞
⎛ 2.246k ⎞
μQ ⎢
⎜
⎟
⎜
⎟⎥
+ lnt + ln⎜
= Pi −
lnt + ln ⎜
⎟
⎟
2
2
4π kH ⎢⎣
⎝ φμ cRw ⎠
⎝ φμc4d ⎠ ⎥⎦
⎡
⎤
⎛ 2.246k ⎞
⎛ 2.246k ⎞
μQ ⎢
⎟ + ln⎜
⎟⎥.
= Pi −
2ln t + ln⎜⎜
⎜
2⎟
2⎟
4π kH ⎢⎣
⎝ φμcRw ⎠
⎝ φμc4d ⎠ ⎥⎦
(10)
This equation will also yield a straight line on a plot of Pw vs. lnt,
but with a slope that is twice that of the earlier slope:
dPw ΔPw −2μ Q −μ Q
=
=
=
.
d lnt Δ lnt 4π kH 2π kH
(11)
→ The occurrence of a doubling of the slope of Pw vs. lnt is an
indication of a nearby impermeable boundary! Moreover, the
distance from the production well to the fault can be estimated
from the transition time between the regime when eq. (7) is
applicable, to the regime when eq. (10) becomes applicable (see
module 2.1.6).
Physical Explanation: Because of the impermeable boundary, oil
is being produced from a “semi-infinite” reservoir, rather than an
infinite reservoir. Hence, we expect that only half as much oil will
be produced, for a given wellbore pressure. So, in order to
maintain a constant flowrate, we must double the pressure
drawdown!
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Section 4
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4.3. Two intersecting impermeable faults
The method of images that was used in section 4.2 to study
the effect of a single fault can also be used to study the effect of
two or more intersecting faults. For example, consider a well
located equidistant from two impermeable faults that intersect at a
right angle, as in Fig. 4.3.1:
Reservoir: k,φ,μ ,c,H
Well
d
Impermeable faults
d
Fig. 4.3.1. Well located near two intersecting faults.
To solve this problem, imagine that the reservoir is laterally infinite
in extent, and has no impermeable boundaries. We want the
locations of the actual impermeable boundaries to correspond to
“no-flow” boundaries in our fictitious unbounded reservoir. This
will be the case if we have vertical and a horizontal lines of
symmetry located at distances d from the actual well. This can be
accomplished by introducing the following three images wells:
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Actual Well
Image Well 1
d
d
d
Symmetry lines d
d
d
d
Image Well 2
d
Image Well 3
Fig. 4.3.2. Use of image wells to solve problem of a well near two
intersecting impermeable faults.
Image wells 1 and 3 are both at a distance of 2d from the
actual well, whereas the distance to image well 2 is 2√2d. Hence,
to the line-source solution for the actual well we must add three
more line-source solutions, as follows:
⎛ −φμcR2 ⎞
⎛
⎛
2⎞
2⎞
4π kH[Pw (t)−Pi ]
w ⎟ + 2Ei ⎜ − φμc4d ⎟ + Ei ⎜ −φμc8d ⎟ . (1)
= Ei ⎜⎜
⎟
⎝ 4kt ⎠
⎝ 4kt ⎠
μQ
⎝ 4kt ⎠
Recalling the definitions of dimensionless time and
dimensionless drawdown given in section 2.2, we can write eq.
(1) in dimensionless form as
ΔPDw
⎛ −(d / R )2 ⎞ 1 ⎛ −2(d /R ) 2 ⎞
1 ⎛ −1 ⎞
w ⎟
w ⎟
⎟⎟ − Ei ⎜⎜
= − Ei ⎜⎜
− Ei ⎜⎜
⎟
⎟.
2 ⎝ 4t Dw ⎠
2
t
t
⎝
⎠
⎝
⎠
Dw
Dw
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Section 4
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The wellbore pressure will exhibit an early stage in which the
effect of the image wells is not yet apparent, and the semi-log
slope of the wellbore pressure vs. time curve will be -Qμ/4πkH.
Eventually, at a sufficiently long time such that the drawdown in
the actual well due to production from all four wells can be
approximated by the logarithmic approximation, the semi-log
slope will be four times greater in magnitude, i.e., -Qμ/πkH.
Note that application of the method of images is not quite as
simple as saying “if we had one impermeable boundary we added
in one image well, so for two impermeable boundaries we add
two image wells”; this statement might seem logical, but it is not
correct. The trick is to place image wells so as to create a
fictitious unbounded reservoir that has the proper symmetry in its
flow field.
Several other examples of the use of the method of images
(two infinite parallel faults, two faults intersecting at 45o, etc.) can
be found in the book Well Testing in Heterogeneous Formations,
by T. D. Streltsova (Wiley, 1988).
4.4. Linear constant-pressure boundaries
Another situation that sometimes arises in reservoirs is that
of a linear constant-pressure boundary. This boundary may, for
example, be formed by a gas cap that borders a slightly dipping
oil reservoir. The effect that such a boundary has on a drawdown
test can also be studied by using the method of images.
Imagine an image well, located as in Fig. 4.2.1, but injecting
fluid at a rate Q. Again, we “ignore” the actual constant-pressure
boundary, and assume that the actual well and the image well are
located in an infinite reservoir. By superposition, the drawdown at
a generic location in the reservoir will be
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Section 4
Page 69
⎛ −φμcR2 ⎞
⎛ −φμcR 2 ⎞
μQ
μQ
1
2⎟.
⎜
⎟
⎜
P(R,t) = Pi +
Ei ⎜
Ei ⎜
−
⎟
4π kH ⎝ 4kt ⎠ 4π kH ⎝ 4kt ⎟⎠
(1)
Now consider a point on the mid-plane between the two
wells. At such a point, R1 = R2 , and so eq. (1) gives
⎡
⎤
⎛ −φμcR 2 ⎞
μQ ⎢ ⎜⎛ −φμcR12 ⎟⎞
1 ⎟⎥.
P(midplane,t) = Pi +
Ei ⎜
− Ei ⎜⎜
⎟
⎟
4π kH ⎢⎣ ⎝ 4kt ⎠
⎝ 4kt ⎠ ⎦⎥
= Pi
for all t,
(2)
proving that the midplane is indeed a constant-pressure
boundary.
Now, consider the pressure drawdown in the actual well,
which is found by setting R1 = Rw and R2 = 2d - Rw = 2d:
⎛ −φμcR2 ⎞
⎛ −φμc(2d)2 ⎞
μQ
μ
Q
w
⎟.
⎟−
P(R,t) = Pi +
Ei ⎜
Ei ⎜
⎟
4π kH ⎜⎝ 4kt ⎟⎠ 4π kH ⎜⎝
4kt
⎠
(3)
●
Again, aside from a very-early time regime defined by eq.
(4.2.4), which we ignore, we have, in analogy with eqs. (4.2.5,6),
a regime defined by tDw > 25 and
0.3φμcd 2
t<
,
k
or
t Dw < 0.3(d / Rw )2 ,
(4)
during which the effect of the image well has not yet been felt in
the actual well, and so the pressure in the well is given by
μQ ⎜⎛ 2.246kt ⎟⎞
Pw (t ) = Pi −
ln
.
4π kH ⎜⎝ φμcRw2 ⎟⎠
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Flow in Porous Media
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Section 4
Page 70
●
There is a late-time regime during which the logarithmic
approximation is valid (in the production well) for both the actual
line-source solution and the image line-source solution. In
analogy with eq. (4.2.9), this regime is defined by
25φμc(2d)2
t>
,
k
●
or
t Dw > 100(d /Rw ) 2 .
(6)
In this regime, the pressure in the well is given by
μQ ⎜⎛ 2.246kt ⎟⎞
μ Q ⎜⎛ 2.246kt ⎟⎞
Pw (t ) = Pi −
ln
ln
+
4π kH ⎜⎝ φμcRw2 ⎟⎠ 4π kH ⎜⎝ φμc(2d )2 ⎟⎠
⎛ 2.246k ⎞
⎛ 2.246k ⎞ ⎤
μ Q ⎢⎡
⎟⎟ − lnt − ln⎜⎜
⎟⎟ ⎥
= Pi −
lnt + ln ⎜⎜
2
2
4π kH ⎢⎣
⎝ φμ cRw ⎠
⎝ φμc(2d) ⎠ ⎥⎦
⎛ 2.246k ⎞
μ Q ⎢⎡ ⎛ 2.246k ⎞
⎟⎟ − ln(Rw2 ) − ln⎜⎜
⎟⎟ + ln (2d )2
= Pi −
ln⎜⎜
4π kH ⎢⎣ ⎝ φμc ⎠
⎝ φμc ⎠
⎤
⎥
⎦⎥
{ }
μQ
μQ ⎛ 2d ⎞
2
⎟.
= Pi −
ln (2d / Rw ) = Pi −
ln⎜
4π kH
2π kH ⎜⎝ Rw ⎟⎠
{
}
(7)
In this case, the pressure in the production well eventually
stabilises to a constant value. The steady-state value of the
drawdown is a function of the dimensionless distance to the
boundary (relative to the wellbore radius).
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Section 4
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Dimensionless drawdown, Δ Pw
Recall that the slope of Pw vs. lnt was -Qμ/4πkH for a well in
an infinite reservoir. An impermeable linear boundary eventually
causes an additional slope of -Qμ/4πkH, leading to a total latetime slope of -Qμ/2πkH. On the other hand, a constant-pressure
linear boundary causes and “additional” slope of +Qμ/4πkH,
leading to a late-time slope of 0:
0
t Dw ~ 1.781(d/Rw ) 2
5
10
15
20
10-1
infinite reservoir
impermeable fault
constant-pressure boundary
101
103
105
107
109
Dimensionless time, tDw
Fig. 4.4.1. Wellbore pressure for a well near a vertical fault.
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Flow in Porous Media
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Section 4
Page 72
____________________________________________________
Tutorial Sheet 4:
(1) As explained in Section 4.2, a doubling of the slope on a semilog plot of drawdown vs. time indicates the presence of an
impermeable linear fault. The drawdown data can also be used to
find the distance from the well to the fault, as follows. If we plot
the data and then fit two straight lines through the early-time and
late-time data, the time at which these lines intersect is called tD′ w .
Show that the distance to the fault can then be found from the
following equation:
d = 0.749 (t D′ w )1/2Rw .
(2) The curves in Fig. 4.4.1 were actually drawn for the case d/Rw
= 200. What would the curves look like for the case of a fault
located at a distance d = 400Rw?
(3) Consider a well located equidistant from two orthogonal
boundaries, as in Fig. 4.3.1, but imagine that the boundaries are
constant-pressure boundaries, rather than impermeable
boundaries. How would you utilise the method of images to find
the drawdown in this well?
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5.
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Section 5
Page 73
Inner (Wellbore) Boundary Conditions
In the line-source solution that we have been discussing thus
far, the effect of the borehole itself is actually ignored - the
borehole is idealised as an infinitely-thin “line”. In this section, we
will examine two physical phenomena associated with the
borehole that necessitate a more detailed analysis of the diffusion
equation. One of these phenomena is the existence of a zone of
altered permeability around the borehole; the second is the effect
of transient fluid storage within the borehole itself.
5.1. Wellbore skin concept - steady-state model
The solutions to the diffusion equation that were presented in
previous sections were all based on the assumption that the
permeability of the reservoir is uniform in space. However, it is
often the case that the rock immediately surrounding the wellbore
has a lower permeability than the remainder of the reservoir, for
various reasons:
(1)
Infiltration of drilling mud into the formation, which clogs up
the pores of the rock.
(2)
Swelling of clays in the formation due to contact with drilling
fluid.
(3)
Incomplete perforation of the casing, which causes the
flowpaths of the fluid to be constricted as they approach the
wellbore. This has an effect that is similar to that caused by
a low-permeability zone near the wellbore.
To account for these possibilities, we introduce the concept
of “wellbore skin”, which can be thought of as a thin annular
region surrounding the borehole in which the permeability ks is
lower than that of the undamaged reservoir rock, as illustrated
schematically in Fig. 5.1.1.
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Section 5
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P=Po
Reservoir, kR
Skin Zone, ks
Rw
Pw
Rs
Ro
Fig. 5.1.1. “Skin” surrounding a wellbore.
The permeability in this idealised composite reservoir is
discontinuous, with k(R) given by
Rw < R < Rs
⇒
k = ks ,
(1)
Rs < R < Ro
⇒
k = kR .
(2)
To quantify the effect of the wellbore skin, consider the
steady-state radial flow model presented in section 1.4. The
governing equation is the radial version of Darcy’s law:
Q=
2πkH
μ
R
dP
,
dR
(3)
where we use the convention that Q > 0 for production.
We can separate the variables of eq. (3), and then integrate
from R = Rw to R = Ro , bearing in mind that k depends on R:
1 dR 2πH
=
dP
μQ
k R
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Section 5
Page 75
1 dR Ro 1 dR Po 2πH
+ ∫
= ∫
dP
∫
μ
Q
k
k
R
R
Rw s
Rs R
Pw
Rs
1 Ro 2πH
1 Rs
ln
+
ln
=
(P − Pw )
ks Rw k R Rs
μQ o
1
kR
1
kR
⎡ R
k
R ⎤ 2πH
⎢ln o + R ln s ⎥ =
(P − Pw )
⎢⎣ Rs ks Rw ⎥⎦ μ Q o
⎡ R
Rs k R Rs
R s ⎤ 2π H
o
⎢ln
⎥=
+ ln
+
ln
− ln
(P − Pw )
⎢⎣ Rs
Rw ks Rw
Rw ⎥⎦ μQ o
1
kR
⎡ ⎛
⎤
⎞ ⎛
⎞
⎢ln⎜⎜ Ro Rs ⎟⎟ + ⎜⎜ kR − 1⎟⎟ ln Rs ⎥ = 2πH ( Po − Pw )
⎢⎣ ⎝ Rs Rw ⎠ ⎝ ks
⎠ Rw ⎥⎦ μQ
→
⎞ R
2πkR H( Po − Pw )
Ro ⎛ kR
⎜
⎟
= ln
+⎜
− 1⎟ ln s .
μQ
Rw ⎝ k s
⎠ Rw
(4)
Equation (4) is identical to eq. (1.4.3), except for the second term
on the RHS. This excess dimensionless pressure drop, which is
due to the presence of the skin effect, is denoted by s, i.e.,
2πkR H( Po − Pw )
R
= ln o + s .
μQ
Rw
(5)
The effect of the skin is therefore to add an additional component
to the pressure drawdown, over-and-above that which is due to
the hydraulic resistance of the reservoir itself.
The dimensional form of the pressure drawdown in the well
can be written as
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Pw = Po −
Section 5
Page 76
⎤
μ Q ⎡ Ro
⎢ln
+ s⎥ ,
2πk RH ⎢⎣ Rw
⎦⎥
⎛k
⎞ R
where s = ⎜⎜ R − 1⎟⎟ ln s .
⎝ ks
⎠ Rw
(6)
(7)
The additional pressure drop caused by the skin is
ΔPs =
μQs
.
2πk RH
(8)
Eq. (7) shows that if the permeability in the damaged zone is
reduced, or the thickness of the damaged zone is increased, the
skin effect will increase. The value of the skin factor is usually less
than 20. On the other hand, well stimulation (i.e., acidising) can
increase the near-wellbore permeability, and give rise to a
negative skin factor; a practical lower limit to s is about -5.
The following schematic illustration of the wellbore skin effect
is based on a figure from Pressure Transient Analysis by
Stanislav and Kabir (Prentice-Hall, 1990):
Wellbore
Damaged
Zone
w/o skin
w/ skin
P
ΔPs
Rw
Rs
Fig. 5.1.2. Effect of skin on pressure drawdown near a well.
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Section 5
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Another interpretation of the skin effect is based on the
observation that the effect of skin is the same as the effect of
having a smaller wellbore radius, i.e., it reduces the flow, for a
given pressure drawdown. Hence, we can write eq. (6) as
⎤
μ Q ⎡ Ro
−s
⎢ln
Pw = Po −
− ln(e ) ⎥
2πk RH ⎢⎣ Rw
⎦⎥
⎛ R ⎞
μQ
o ⎟
= Po −
ln⎜
2πkR H ⎜⎝ Rwe −s ⎟⎠
⎛ R ⎞
μQ
⎜
= Po −
ln o ⎟ ,
2πkR H ⎜⎝ Rweff ⎟⎠
eff
where Rw = Rwe
−s
.
(9)
(10)
eff
Rw can be thought of as the hypothetical wellbore radius (in an
undamaged formation) that would lead to the same drawdown as
is actually observed in the well that is surrounded by the damaged
formation.
[However, this interpretation
general it is not correct to
radius with the effective
interpretation only applies
reservoir.]
Department of Earth Science and Engineering
is potentially misleading, because in
merely replace the actual wellbore
radius given in eq. (10). This
to drawdown tests in an infinite
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Section 5
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5.2. Effect of skin on drawdown tests
The preceding derivation was carried out under the steadystate assumption. However, the thickness of the skin region is
usually small, and so transient effects due to the skin region will
die out rapidly.
The time required for “transient effects” to die out within the
skin region can be estimated using eq. (1.6.12), with the distance
taken to be Rs:
(φμc ) s Rs2
ts =
.
4ks
(1)
If we assume reasonable values of the parameters, such as
φ ≈ 10 −1,
μ ≈ 10−3 Pas ,
c ≈ 10 −9 /Pa ,
Rs ≈ 100 m ,
and
ks ≈ 10 −15 m2 , we find that t s ≈ 25 s .
So, after a short period of time, the skin region will be in a
quasi-steady regime. If the flowrate into the wellbore is constant,
the pressure drop through the skin region will be constant.
Therefore, it is usually assumed that, even during a transient well
test, the effect of the skin is to contribute an additional pressure
drop in the wellbore, as given by eq. (5.1.8).
For example, the drawdown in an infinite reservoir producing
at a constant rate, with a skin region around the wellbore, is found
by subtracting the additional skin-related pressure drop given by
eq. (5.1.8) from the usual line-source solution:
⎤
μQ ⎢⎡ ⎜⎛ φμcRw2 ⎟⎞
+ 2s⎥ ,
Pw = Pi −
−Ei ⎜
⎟
⎢
4πkH ⎣
⎝ 4kt ⎠
⎦⎥
(2)
where we now revert to using k instead of kR to denote the
reservoir permeability, since the permeability kS is absorbed into
the skin factor, s.
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Section 5
Page 79
If t is large enough that the logarithmic approximation can be
used for the Ei function, the drawdown will be given by
⎡
⎤
μQ ⎢ ⎜⎛ kt ⎟⎞
Pw = Pi −
ln⎜
+ 0.80907 + 2s ⎥ ,
⎟
2
⎥
4πkH ⎢⎣ ⎝ φμcRw ⎠
⎦
⎛
⎞
μQ ⎜ 2.246e 2s kt ⎟
.
Pw = Pi −
ln
4πkH ⎜⎝ φμcRw2 ⎟⎠
or,
(3)
(4)
It is clear from eq. (3) that the skin will have no effect on the
semi-log slope of pressure vs. lnt, but it will shift the entire
drawdown curve downward by a constant amount.
The skin factor can be found by proper interpretation of a
pressure buildup test. Incorporating the skin effect into the
equation for the buildup pressure, eq. (3.2.5), gives
⎡ ⎛
⎞
⎞⎤
⎛
2s
2.246e kΔt ⎟ ⎥
μQ ⎢ ⎜ 2.246e 2s k(t + Δt) ⎟
⎜
− ln⎜
Pw = Pi −
ln
2
2
⎟
⎟⎥
4πkH ⎢⎣ ⎜⎝
φμcRw
⎠
⎝ φμcR w ⎠ ⎦
⇒
μQ ⎛ t + Δt ⎞
⎟.
Pw = Pi −
ln ⎜
4πkH ⎜⎝ Δt ⎟⎠
(5)
The two skin terms cancel out, and do not appear in the equation
for the wellbore pressure.
Now consider the difference between the wellbore pressure
−
immediately before shut-in, which we will call Pw , and the
+
wellbore pressure a short time after shut-in, Pw . Using eq. (4) for
−
+
Pw , and eq. (5) for Pw , we find
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+
Pw
−
− Pw
Flow in Porous Media
Dr. R. W. Zimmerman
Section 5
Page 80
⎛
⎞
μQ ⎛ t + Δt ⎞
μQ ⎜ 2.246e 2s kt ⎟
⎟+
=−
ln⎜
ln
4πkH ⎜⎝ Δt ⎟⎠ 4πkH ⎜⎝ φμcRw2 ⎟⎠
⎡
⎤
⎛ 2.246k ⎞
− μQ ⎢ ⎛ t + Δt ⎞
⎥
⎟ − 2s .
⎟⎟ − ln⎜
=
ln⎜⎜
2
⎜
⎟
⎢
⎥
4πkH ⎣ ⎝ tΔt ⎠
⎝ φμcRw ⎠
⎦
(6)
For times shortly after shut-in, Δt is small, and (t + Δt ) / t ≈ 1, so
eq. (6) reduces to
+
Pw
−
− Pw
⎡
⎤
⎛ 2.246k ⎞
μQ ⎢
⎜
⎟ + 2s ⎥ .
=
ln Δt + ln⎜
2
⎟
⎢
⎥
4πkH ⎣
⎝ φμcRw ⎠
⎦
(7)
Eq. (7) gives us an equation that can be solved for s.
In practice, however, we can’t use the pressure at a time
immediately after shut-in in eq. (7), for several reasons:
(a) Wellbore storage effects, (called “afterproduction” in the case
of a buildup test), will cause the actual pressure drawdown
measured at the bottom of the borehole to deviate from that
predicted by the equations used above, for a certain period of
time (see section 5.3 below).
(b) Eq. (5) made use of the logarithmic approximation to the
exponential integral function; this approximation is only valid after
a sufficiently long elapsed time after shut-in.
+
It is traditional practice to use eq. (7) with Pw evaluated after
one-hour of shut-in time; this value was originally chosen
because, when using “oilfield” units, lnΔt = ln1 = 0 when Δt = 1
hour. However, in many cases one hour is not enough time for
afterproduction to cease and radial flow to be established.
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Section 5
Page 81
Therefore, we must extrapolate the straight-line portion of the
+
Horner plot back to Δt = 1 hour, and use that pressure for Pw .
Use of buildup data to estimate the skin factor will be illustrated in
more detail in the Well Test Analysis module.
5.3. Wellbore storage phenomena
The pressure drawdown is usually measured at the bottom
of the borehole, whereas the flowrate is usually measured at the
wellhead. The flowrate Q in our equations, however, refers to the
sandface flowrate from the reservoir into the borehole. In a quasisteady situation, the measured wellhead flowrate will be identical
to the influx from the reservoir.
Immediately after a change in flowrate, such as after the
start of a drawdown test, or after shut-in, these two flowrates will
be different. The difference is due to the fact that the fluid in the
borehole is compressing (or expanding) due to the changing
pressure to which it is subjected.
For example, imagine that we begin to withdraw fluid at time
t = 0 at rate Qwh that is measured at the wellhead. Initially, the
fluid that flows at the wellhead is taken mainly from the wellbore
itself; only gradually does this fluid begin to be supplied by the
reservoir. Eventually, the fluid in the wellbore reaches a quasisteady state, and all of the flow measured at the wellhead does in
fact emanate from the reservoir. After this time, Qwh = Qsf .
Conversely, if a well that is producing at a constant rate is
shut in, fluid will continue to enter the borehole from the reservoir,
even though it is not allowed to flow out at the wellhead. This
additional fluid will be restricted to the wellbore; as the mass of
the fluid in the wellbore increases, the wellbottom pressure will
gradually increase. These two situations are illustrated in Fig.
5.3.1, modified from p. 43 of Stanislav and Kabir:
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Dr. R. W. Zimmerman
Drawdown
Buildup
Flow
Rate
Flow
Rate
Qwh
Section 5
Page 82
Qsf
Qwh
Qsf
Time
Time
Fig. 5.3.1. Effect of wellbore storage on pressure tests.
We can begin to analyse the effect of wellbore storage
effect by taking a mass balance on the fluid in the wellbore, along
the lines of that given for the reservoir in section 1.5. Assuming
that the wellbore is completely filled with a single-phase liquid,
then:
Mass flux in - Mass flux out = rate of change of mass storage
ρf Qsf − ρf Qwh =
i.e.,
d( ρf Vw )
,
dt
(1)
where Vw is the volume of the wellbore.
If we assume uniform pressure and uniform density of the
fluid within the wellbore, we have
ρf (Qsf − Qwh ) = Vw
⇒
dρ f
dρ dPw
dPw
= Vw f
= Vw ρf cf
,
dt
dPw dt
dt
Qsf − Qwh = Vwcf
dPw
dP
≡ Cs w ,
dt
dt
(2)
where Cs is the wellbore storage coefficient., and Vw is the total
volume of fluid within the wellbore, from the wellbottom to the
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Section 5
Page 83
surface. [If the wellbore is not fully filled with liquid, the
appropriate expression for Cs can be found in Advances in Well
Test Analysis by Earlougher (SPE, 1977)].
At early times after the start of a constant-rate drawdown
test, Qwh = Q (the nominal flowrate), but Qsf ≅ 0 , so eq. (2) takes
the form
dPw
.
(3)
dt
Eq. (3) can be integrated, using the initial condition that Pw = Pi
when t = 0:
−Q = Cs
t
Pw (t )
Q
−∫
dt = ∫ dPw
C
P
0 s
⇒
Pw = Pi −
i
Qt
.
Cs
(4)
Eq. (4) can be written in dimensionless form as follows:
(1)
Rewrite eq. (4) in terms of the dimensionless pressure:
PDw =
(2)
2πkH(Pi − Pw ) 2πkHQt
=
.
μQ
μQCs
(5)
Express the RHS of eq. (5) in terms of tD:
PDw
2πkHQ φμct Rw2 t Dw t Dw
=
≡
,
μQCs
k
CD
(6)
where the dimensionless wellbore storage coefficient CD is
defined by
CD =
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Cs
2πHφct Rw2
,
(7)
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Section 5
Page 84
with ct being the total reservoir compressibility. Aside from the
factor of 2, CD is the ratio of the storativity of the borehole to the
storativity of the rock that had occupied the borehole before it was
drilled.
On a log-log plot, eq. (6) takes the form
lnPDw = ln t Dw − lnCD ,
⇒
d ln PDw
= 1.
d lnt Dw
(8)
→ A slope of unity occurring at early times on a log-log plot
(dimensionless or not) of wellbore pressure vs. time is therefore
an indicator of a pressure response that is dominated by wellborestorage effects.
5.4. Effect of wellbore storage on pressure tests
The preceding analysis focused on fluid produced from
wellbore storage, and neglected flow from the reservoir into the
wellbore. In reality, fluid is supplied to the wellhead by wellbore
storage at early times, but these effects gradually die out as the
flow regime in the wellbore reaches a steady-state. An analysis
that includes both contributions to the measured well-head
flowrate Qwh, i.e., from the wellbore storage and from the
reservoir, can be carried out by solving eq. (2.1.1), with the inner
boundary condition, eq. (2.1.3), replaced by (see eq. (5.3.2)):
⎛ 2πkH dP ⎞
⎛ dP ⎞
⎜⎜
⎟⎟ = Qwh + Cs ⎜⎜ ⎟⎟ .
R
μ
dR
⎝
⎠ Rw
⎝ dt ⎠ Rw
(10)
This problem has been solved by Wattenbarger and Ramey
(SPEJ, 1970) using finite differences, and by Agarwal et al.
(SPEJ, 1970) using Laplace transforms. The analysis shows that
there are three regimes of behaviour for a well in an infinite
reservoir, with skin and wellbore storage:
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Section 5
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• An early-time regime that is dominated by wellbore storage,
during which the drawdown is given by eq. (6). The duration of
this regime is defined by
t Dw < CD (0.04 + 0.02s) .
(11)
• A transition regime during which both wellbore storage and
reservoir inflow contribute to the wellhead flowrate. The duration
of this regime is defined by
CD (0.04 + 0.02s) < t Dw < CD (60 + 3.5s) .
(12)
• A third regime during which wellbore storage effects have died
off, and the drawdown is given by eq. (5.2.4). The duration of
this regime is defined by
t Dw > CD (60 + 3.5s) .
(13)
Criterion (13) can be used to estimate when the drawdown
curve can be aproximated by the logarithmic equation (5.2.4).
During a build-up test, however, the time required in order to
reach a “straight line” on the Horner plot is best estimated by the
following equation, found by Chen and Brigham (JPT, 1978):
tDw > 50 CD exp(0.14s).
(14)
10
10
102
1
ΔP Dw
103
0.1
104
CD=105
(s = 0)
0.01
102
103
104
105
106
107
tDw
Fig. 5.4.1. Effect of wellbore storage on drawdown.
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Section 6
Page 86
Outer Boundary Conditions
6.1. Well at the centre of a circular reservoir with constant
pressure on its outer boundary and constant wellbore
pressure
Thus far, we have assumed that our wells are located in an
infinite (or at least semi-infinite) reservoir. In reality, the reservoir
is always of finite size, and the pressure disturbance will
eventually reach some outer boundary that effectively encloses
the reservoir. This outer boundary may be an “aquifer”, i.e., a vast
expanse of water-filled rock that surrounds the hydrocarbon-filled
“reservoir”. In this case, the appropriate outer boundary condition
may be one of constant pressure.
On the other hand, if a number of wells are producing from
the same reservoir, each will drain fluid from only a finite region,
and so each well will effectively behave as if it were surrounded
by a no-flow boundary, as illustrated in Fig. 6.1.1. (The effective
drainage area of each well will depend on the production rate of
that well, and of all nearby wells.)
Outer
boundary
of reservoir
Wells
No-flow
boundaries
Fig. 6.1.1. Reservoir with several production wells.
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Section 6
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Problems involving wells producing from a finite, bounded
drainage area can be solved using either of two mathematical
techniques: Laplace transforms, or the method of eigenfunction
expansions. Laplace transforms will be discussed in section 7. In
this section we use the method of eigenfunction expansions to
solve the problem of a well located at the centre of a circular
reservoir, with its outer boundary maintained at constant
pressure, and with a constant borehole pressure. This is by no
means the most important of such problems, but it will allow us to
develop the method of eigenfunction expansions in a (relatively)
simple context.
The first point to note is that the Boltzmann transformation is
of no use in bounded-reservoir problems. The reason that it
“worked” in the infinite-reservoir case is that the initial condition, at
t = 0, and the outer boundary condition, at R → ∞ , both
correspond to the same limit η → ∞ , and both also correspond to
the initial pressure, Pi. This “collapsing” of two conditions into one
equivalent condition allowed us to replace a 2nd-order PDE
(which requires 2 BCs + 1 IC = 3 conditions) with a 2nd-order
ODE (which requires a total of 2 subsidiary conditions).
This does not occur in the bounded-reservoir problem,
because the outer BC is imposed at R = Re , and R = Re does not
correspond to a fixed value of the Boltzmann variable, η. (In
general, a Boltzmann-type transformation will not work in a
problem that contains a natural length scale, such as a wellbore
radius or an outer radius; the problem of an infinitely-small linesource in an infinite reservoir had no natural length scale
associated with it.)
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Section 6
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So, now let us imagine that we have a well at the centre of a
circular reservoir, which is initially at pressure Pi . At time t = 0,
the pressure in the wellbore is immediately lowered to some value
Pw , and it is maintained at that value (see Fig. 6.1.2). (In contrast
to previous problems, in which the flowrate was controlled, and
we calculated the drawdown, in this problem the drawdown in the
well is specified, and we must calculate the flowrate).
Reservoir
P=Pw
P=Pi
Rw
Re
R
Fig. 6.1.2. Circular reservoir with constant-pressure inner and
outer boundaries.
This problem can be stated mathematically as:
Governing PDE:
1 d ⎡ dP ⎤ φμc dP
⎢R
⎥=
,
R dR ⎢⎣ dR ⎥⎦
k dt
(1)
BC at wellbore:
P( R = Rw, t ) = Pw ,
(2)
BC at outer boundary:
P( R = Re, t ) = Pi ,
(3)
Initial condition:
P( R, t = 0) = Pi .
(4)
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Section 6
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The first step in solving this problem is to simplify its
appearance by introducing dimensionless variables. Define
Dimensionless radius:
RD = R / Rw ,
(5)
Dimensionless time:
tD =
kt
,
φμ cRw2
(6)
Dimensionless drawdown:
PD =
Pi − P
.
Pi − Pw
(7)
(Note that the production rate Q is unknown a priori, so we can’t
define PD as we did before, in terms of Q.)
In terms of these variables, eqs. (1-4) take the form
1 d ⎡
dP ⎤ dP
⎢RD D ⎥ = D ,
RD dRD ⎢⎣ dRD ⎥⎦ dt D
Governing PDE:
(8)
BC at wellbore:
PD (RD = Rw / Rw = 1, t D ) = 1,
BC at outer boundary:
PD (RD = Re / Rw ≡ RDe , t D ) = 0 ,
(10)
Initial condition:
PD (RD, t D = 0) = 0 .
(11)
(9)
We next invoke the principle of superposition to break up the
pressure into a steady-state part, PDs (RD ) , and a transient part,
pD (RD ,t D ) :
PD (RD,t D ) = PDs (RD ) + pD (RD ,t D ) .
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Section 6
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By definition, the steady-state pressure function must satisfy
eq. (8), as well as both BCs, (9) and (10). The time derivative is
zero for the steady-state part, so the actual governing differential
equation for PDs (RD ) is eq. (8) with the RHS set to zero.
Therefore, PDs (RD ) is governed by
dPDs ⎟⎞
1 d ⎜⎛
= 0,
RD
⎜
⎟
dRD ⎠
RD dRD ⎝
Governing ODE:
(13)
BC at wellbore:
PDs ( RD = 1) = 1 ,
(14)
BC at outer boundary:
PDs (RD = RDe ) = 0 .
(15)
♦ Integrate eq. (13) once to get (i.e., an indefinite integral):
dPDs
RD
= B = constant .
dRD
(16)
♦ Separate the variables in eq. (16) and integrate again:
PDs (RD )
∫
dPDs
=
RD
dR
∫ B R D + A,
D
⇒ PDs (RD ) = BlnRD + A .
(17)
♦ Comparison of eq. (17) with eqs. (14) and (15) shows that A =
1 and B = −1/ lnRDe , so eq. (17) can be written as
⇒ PDs ( RD ) = 1 −
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ln RD
− ln(RD / RDe )
=
.
ln RDe
ln RDe
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Section 6
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Note 1: The steady-state solution satisfies the diffusion
equation (8), and the boundary conditions (9,10), but it doesn’t
satisfy the initial condition (11); this is why we also need the
transient component of the solution.
Note 2: Eq. (18) is just the Thiem equation, eq. (1.4.3), in a
dimensionless form.
If we substitute (12) into eqs. (8-11), and make use of eq.
(18), we see that the transient pressure function must satisfy
Governing PDE:
1 d ⎡
dp ⎤ dp
⎢RD D ⎥ = D ,
RD dRD ⎢⎣ dRD ⎥⎦ dt D
(19)
BC at wellbore:
pD (RD = 1, t D ) = 0 ,
(20)
BC at outer boundary:
pD (RD = RDe , t D ) = 0 ,
(21)
Initial condition:
pD (RD , t = 0) =
ln(RD / RDe )
.
ln RDe
(22)
Thus far, we have transformed a diffusion equation with a
nonzero BC, eq. (9), and zero IC, eq. (11), into a diffusion
equation with zero BC and a nonzero IC, eq. (22). We may not
appear to be making progress, but we are, because the
eigenfunction method requires “zero” boundary conditions, but
does not require “zero” initial conditions.
To solve eqs. (19-22), we again make use of the
superposition principle, and first search for (as many functions as
we can find!) that each satisfy eq. (19) and the boundary
conditions (20,21); later, we will superpose these functions to
satisfy eq. (22).
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Section 6
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Assume that these functions can be written in the form
pD (RD , t D ) = F(RD )G (t D ) .
(b)
Insert eq. (23) into eq. (19) to find:
⎡ 1 d ⎛
dF ⎞⎤
dG
⎜⎜ RD
⎟⎟⎥ = F ( RD )
G (t D ) ⎢
.
dRD ⎠⎦
dt D
⎣ RD dRD ⎝
(c)
(23)
(24)
Divide through by F( RD )G (t D ) to find:
1
d ⎛
dF ⎞
1 dG
⎜⎜ RD
⎟⎟ =
.
RD F ( RD ) dRD ⎝
dRD ⎠ G (t D ) dt D
(25)
(d) The LHS of eq. (25) doesn’t depend on tD, and the RHS
doesn’t depend on RD; since they are equal to each other, they
cannot depend on either variable! Hence, each side of eq. (25)
must equal a constant, which we call − λ2 :
1
d
RD F ( RD ) dRD
(e)
⎛
dF ⎞
1 dG
⎜⎜ RD
⎟⎟ =
= −λ2 .
dRD ⎠ G (t D ) dt D
⎝
(26)
The space-dependent part of eq. (26) can be written as
F ′′(RD ) +
1
F ′(RD ) + λ2F( RD ) = 0 .
RD
(27)
We now make a change of variables x = λRD , after which eq. (27)
can be written as
F ′′(x ) +
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F ′(x) + F(x) = 0 .
x
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Section 6
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Eq. (28) is known as a Bessel equation of order zero. It is a
2nd-order ordinary differential equation, so it must have two
independent solutions. One solution is readily found by assuming
a power-series solution:
F( x) =
∞
∑ an x n .
(29)
n =0
♦ If we insert eq. (29) into eq. (28), we find
∞
∑ n(n − 1)a x
n−2
n
n =0
∞
+ ∑ nan x
n−2
n =0
∞
+ ∑ an x n = 0 .
n =0
(30)
♦ In order to have x appear to the nth power in each term, we let
n → n + 2 in the first two series, which, after combining the first
two series and taking the “unmatched” terms outside of the series,
leads to
∞
0a0 x + a1 x + ∑ [(n + 2) 2 an+2 + an ]x n = 0 .
−2
−1
n =0
(31)
♦ The coefficient of each power of x in eq. (31) must be zero.
The x −2 term already has a factor of zero, so it will vanish
regardless of the choice of a0 ; we may as well pick
a0 = 1.
(32)
♦ In order for the x −1 term to vanish, we must pick
a1 = 0 .
(33)
♦ For all higher-order terms to vanish, eq. (31) shows that the
coefficients must satisfy the following recursion relationship:
an +2 =
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.
(n + 2) 2
(34)
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Section 6
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♦ This recursion relationship allows us to generate all subsequent
coefficients, starting with the first two coefficients, given by eqs.
(32) and (33):
a3 = a5 = a7 = ... = 0 ,
a2 =
(35)
−1
1
−1
, a4 =
, a6 =
....
4
64
2304
(36)
The solution to eq. (28) that is given by eqs. (29), (35) and (36) is
known as the Bessel function of the first kind (of order zero), and
is denoted by Jo(x):
x2 x4
x6
J 0 ( x) = 1 − + −
+ ... ,
4 64 2304
or
x2
x4
x6
J 0 ( x) = 1 − 2 2 + 4
−
+ ... ,
2 (1!) 2 (2!) 2 26 (3!) 2
(37)
where n! ≡ 1× 2 × 3 × ... × n is the factorial function.
The function J0(x) is similar to a damped cosine function
(see Fig. 6.1.3). It “starts” at J0 (0) = 1, oscillates (but not quite
periodically), and then slowly decays to zero according to (see
Theory of Bessel Functions, G. N. Watson, 1922, or any book on
advanced applied mathematics)
J 0 ( x) ≈
2
π⎞
⎛
cos⎜ x − ⎟asx → ∞ .
πx
4⎠
⎝
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Jo(x) or Yo(x)
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Section 6
Page 95
Bessel Functions of Order Zero
0.5
0.0
-0.5
-1.0
Jo(x)
Yo(x)
0
2
4
6
8
10
x
Fig. 6.1.3. Bessel functions of order zero.
The derivation of the second independent solution to eq. (29)
entails a somewhat lengthier procedure, because it is not an
analytic function (i.e., it is not a pure power series). This solution,
called the Bessel function of the second kind (of order zero), is
defined by
Y0 (x) =
∞
ln(x / 2) + γ ]J0 (x ) − ∑
x 2n ,
[
π
π (n!) 2 22n
2
2
(−1)n hn
(39)
n=1
where γ ≈ 0.577 is Euler’s constant, and
1
1
+ ... + .
(40)
2
n
This function becomes infinitely negative as x → 0 , due to the
ln(x) term, then oscillates in a manner similar to J0 (x ) , and
eventually decays to zero according to (Fig. 6.1.3)
hn = 1+
Y0 (x) ≈
⎛
2
π⎞
sin⎜ x − ⎟⎟
πx ⎜⎝
4⎠
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as
x → ∞.
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Section 6
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The general solution to eq. (28) can be written as a linear
combination of these two kinds of Bessel functions:
F( x) = AJ0( x) + BY0 (x) .
(42)
Recalling that x = λRD , we have
F(RD ) = AJ0 ( λRD ) + BY0 ( λRD ) .
(43)
Eq. (43) will be a solution to eq. (28) for any value of λ.
However, it will satisfy the boundary conditions (20) and (21) only
for certain special values of λ, which we will now find. Insertion of
eq. (43) into BCs (20) and (21) yields
AJ0 ( λ ) + BY0 ( λ ) = 0 ,
(44)
AJ0 ( λRDe ) + BY0 (λRDe ) = 0 .
(45)
These two equations can be written in matrix form as
⎡ J 0( λ )
Y0 ( λ ) ⎤ ⎡ A⎤ ⎡0⎤
⎢
⎥⎢ ⎥ = ⎢ ⎥ .
⎢⎣J0 ( λRDe ) Y0 (λRDe )⎥⎦ ⎢⎣B⎥⎦ ⎢⎣0⎦⎥
(46)
In order for eq. (46) to have non-zero solutions for A and B,
the determinant of the matrix must be zero:
J0 ( λ )Y0 ( λRDe ) − Y0 ( λ )J0 ( λRDe ) = 0 .
(47)
The values of λ that satisfy eq. (47) are the eigenvalues of
this problem. They will depend on the dimensionless size of the
reservoir, RDe = RD / Rw . There will always be an infinite number
of eigenvalues, and they can be arranged in order as
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Section 6
Page 97
.
Each of these eigenvalues
eigenfunction (see eq. (43)):
is
associated
(48)
with
its
Fn (RD ) = AnJ0 ( λnRD ) + BnY0 ( λnRD ) .
own
(49)
From eq. (44), the constants An and Bn are related by
Bn =
− An J 0 (λn )
Y0 (λn ) ,
(50)
and so eq. (49) can be written as
Fn (RD ) = AnJ0 ( λnRD ) − An
=
J 0( λn )
Y0 ( λnRD )
Y0 ( λn )
[
]
[
]
An
Y0 ( λn )J0 (λnRD ) − J 0( λn )Y0 ( λnRD )
Y0 ( λn )
= Cn Y0 ( λn )J 0( λnRD ) − J0 ( λn )Y0 ( λnRD ) ,
(51)
where the Cn are arbitrary constants, and the functions inside the
brackets are the eigenfunctions of this problem.
We now return to the time-dependent part of the solution,
which, according to eq. (26), must satisfy
dGn
= − λ2nGn (t D ) .
dt D
(52)
The solution to eq. (52) is
Gn (t D ) = e −λnt D .
2
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Section 6
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Recalling eq. (23), the most general solution that satisfies the
diffusion equation (19) and the BCs (20,21) is given by
pD (RD ,t D ) =
∞
∑ Cn [Y0 ( λn )J 0( λnRD ) − J0 ( λn )Y0 ( λnRD ) ]e −λntD . (54)
2
n=1
All that remains now is to satisfy the initial condition, eq. (22).
Evaluating eq. (54) at t D = 0 , and invoking eq. (22), yields
∞
∑ Cn[Y0( λn )J0 ( λnRD ) − J0 (λn )Y0( λnRD )]=
n= 1
ln(RD / RDe )
. (55)
lnRDe
The constants Cn must be chosen so as to satisfy eq. (55)
for all values of the dimensionless radius. This can be done using
the orthogonality properties of eigenfunctions (see Muskat, The
Flow of Homogeneous Fluids through Porous Media, 1937,
chapter 10 for details). The result is
Cn =
πJ0 (λn )J0 ( λnRDe )
.
J02 ( λn ) − J 02 ( λnRDe )
(56)
We now combine the transient and steady-state pressures to
find, from eqs. (12), (24), (54) and (56), the complete solution to
this problem:
− ln( RD / RDe ) ∞ π J 0 (λn ) J 0 (λn RDe )
−λ2nt D
PD ( RD , t D ) =
U
(
λ
R
)
e
+∑ 2
n
n D
2
ln RDe
n =1 J 0 (λn ) − J 0 (λn RDe )
where
(57)
Un (λnRD ) = Y0 ( λn )J0 ( λnRD ) − J0 ( λn )Y0 ( λnRD ) .
This completes the solution to this problem. The dimensional
values can be found by recalling eqs. (5-7).
The flowrate into the well can be found by differentiating the
pressure, and applying Darcy’s law at the wellbore. A detailed
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Section 6
Page 99
analysis (not given here) of the flowrate shows that at early times
it is proportional to t -1/2, and it gradually approaches a steadystate value given by eq. (1.4.4).
6.2. Well at the centre of a circular reservoir with constant
pressure on its outer boundary and constant flowrate into
the wellbore
Consider a well at the centre of a circular reservoir,
producing fluid at a constant rate, with the pressure at the outer
boundary maintained at the initial pressure at all times (see Fig.
6.2.1):
Reservoir
Q
Rw
R
P=Pi
Re
Fig. 6.2.1. Circular reservoir with constant flowrate, constant
pressure at outer boundary.
The mathematical formulation of this problem is equivalent to
that discussed in section 6.1, except that the inner BC, eq. (6.1.2)
is replaced by
⎛ 2πkH dP ⎞
⎜
⎟⎟
BC at wellbore:
R
= Q,
(1)
⎜
dR ⎠ R =R
⎝ μ
w
where we take Q > 0 if the flow is into the borehole.
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Section 6
Page 100
This problem can also be solved using the method of
eigenfunction expansions, as in the previous section. The solution
is (Muskat, Flow of Homogeneous Fluids through Porous Media,
1937, p. 643):
⎛ R ⎞ ∞ π J 2 ( λnRDe )Un ( λnRD )
2
0
ΔPD (RD,t D ) = − ln⎜⎜ D ⎟⎟ − ∑
e −λntD ,
⎝ RDe ⎠ n =1λn [J02 ( λnRDe ) − J12 ( λn )]
(2)
where the eigenfunctions Un are given by
Un (λnRD ) = Y1( λn )J0 ( λnRD ) − J1( λn )Y0 ( λnRD ) ,
(3)
and the eigenvalues λn are defined implicitly by
Un (λnRDe ) = Y1( λn )J 0( λnRDe ) − J1( λn )Y0 ( λnRDe ) = 0 . (4)
The functions J1 and Y1 are Bessel functions of order one, and
are defined by
J1(x) ≡ −
dJ0 (x )
dY (x)
, Y1(x ) ≡ − 0
.
dx
dx
(5)
The dimensionless variables in eq. (2) are defined by
Dimensionless radius:
RD = R /Rw ,
Dimensionless time:
tD =
Dimensionless pressure:
ΔPD =
kt
φμ cRw2
(6)
,
(7)
2πkH(Pi − P)
.
μQ
(8)
The pressure in the wellbore, ΔPDw(tD), is found by setting
RD = 1 in eq. (2), and then making use of the following Besselfunction identity (see Muskat, p. 631):
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Y1(x )J0 (x) − J1(x )Y0( x) =
Section 6
Page 101
−2
;
πx
(9)
the result is (Matthews and Russell, p. 12):
2 J 02 (λn RDe )e − λntD
ΔPDw (t ) = ln RDe + ∑ 2 2
.
2
n =1 λn [ J 0 (λn RDe ) − J 1 (λn )]
2
∞
(10)
A detailed analysis of this solution shows that at early times
the pressure agrees with that given by the Theis solution,
because the pressure pulse will not yet have reached the outer
boundary. Eventually, the wellbore reaches a steady-state
pressure (see Fig. 6.3.2) that is equivalent to the pressure found
in the Thiem problem.
6.3. Well at the centre of a circular reservoir with a no-flow
outer boundary and constant flowrate into the wellbore
This problem (see Fig. 6.3.1) is similar to the problem of
section 6.2, except that the outer BC is now replaced by
⎛ 2πkH dP ⎞
⎜⎜
⎟⎟
R
= 0.
dR ⎠ R =R
⎝ μ
e
BC at outer boundary:
Reservoir
Q
(1)
No-flow
Rw
R
Re
Fig. 6.3.1. Circular reservoir with constant flowrate into borehole,
no flow across the outer boundary.
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Section 6
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The solution to this problem was found by Muskat in 1937
using the eigenfunction method; it was re-derived by van
Everdingen and Hurst in 1949 (Trans. AIME, 1949), using Laplace
transforms (see section 7). The solution is
ΔPD ( RD , t D ) =
2
2
RDe
⎤
⎡ RD2
2
ln RD ⎥
+ t D − RDe
⎢
−1 ⎣ 4
⎦
4
4
2
⎡ 3R De
− 4 R De
− 1⎤
ln R De − 2 R De
−⎢
⎥
2
2
−
4
(
R
1
)
De
⎣
⎦
π J12 (λn RDe )U n (λn RD ) −λ t
e
+∑
,
2
2
n =1 λn [ J 1 (λn RDe ) − J 1 (λn )]
∞
2
n D
(2)
where the eigenfunctions Un are given by
U n (λn RD ) = J1 (λn )Y0 (λn RD ) − Y1 (λn ) J 0 (λn RD ) ,
(3)
the eigenvalues λn are defined implicitly by
J1 (λn )Y1 (λn RDe ) − Y1 (λn ) J1 (λn RDe ) = 0 ,
(4)
and the dimensionless variables are defined as in section 6.2.
Because of the importance of this problem, we will examine the
solution in detail.
The pressure in the wellbore is found by setting RD = 1 in eq.
(2), and again using eq. (6.2.9) to simplify:
ΔPDw (t D ) =
1
2
RDe
⎡1
⎤ ⎡ 3R 4 − 4R4 lnRDe − 2R 2 − 1⎤
De
De
⎥
⎢ + 2t D ⎥ − ⎢ De
2
2
⎢
⎥⎦ ⎣
4(RDe − 1)
− 1 ⎢⎣ 2
⎦⎥
2 J 12 (λn RDe )e − λntD
+∑ 2 2
.
2
n =1 λn [ J 1 (λn RDe ) − J 1 (λn )]
∞
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Section 6
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In most cases of practical interest, however, RDe >> 1 , and eq. (5)
can effectively be written as
2J12 ( λnRDe )e − λnt D
2t D
3 ∞
.
ΔPDw (t D ) = 2 + lnRDe − + ∑ 2 2
2
4
RDe
n=1λn[J1 ( λnRDe ) − J1 ( λn )]
2
(6)
There are three important time regimes in this problem:
(i) A regime in which t is sufficiently large that the logarithmic
approximation to the line-source solution is valid, but for which
the leading edge of the pressure pulse has not yet reached
the outer boundary of the reservoir. From eq. (2.4.7) and Fig.
2.5.1, we find that this regime is defined by:
2
25 < t Dw < 0.1RDe .
•
(7)
It is not easy to see that eq. (5) reduces to eq. (2.4.10) during
this time regime, but it does.
(ii) A second regime that is “late” enough that the presence of the
closed outer boundary is felt at the well, but still early enough
that the exponential terms in eq. (6) have not yet died out.
The start of this regime is given by the upper bound in eq. (7);
2 (see Tutorial Sheet 5). This regime
it ends when t Dw ≈ 0.3RDe
is therefore defined by
2 <t
2
0.1RDe
Dw < 0.3RDe .
(8)
In this regime we must use the entire series in eq. (6) to
calculate the wellbore pressure, and the ΔPw(tDw) curve has no
simple description.
(iii) A third regime that is late enough such that the exponential
terms in eq. (6) have died out. This regime is defined by
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Section 6
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2 .
t Dw > 0.3RDe
(9)
In this regime the wellbore pressure is given by
ΔPDw =
2t Dw
3
+
ln
R
−
.
De
2
4
RDe
(10)
NOTE: These regimes have been given various names, such as
early-transient, intermediate, late-transient, quasi-steady-state,
semi-steady-state, steady-state, etc. There is no consistent usage
of these terms, and most are mathematically incorrect. We will
refer to the first regime as the “infinite reservoir” regime, the
second as the “transitional regime”, and to the last as the “finite
reservoir” regime.
•
An important feature of the third time regime is that the
pressure in the well declines linearly with time. The rate of
pressure decline can be used to find the drainage area of the well,
as follows:
(1) First, rewrite eq. (10) in terms of the actual variables, rather
than the dimensionless variables:
⎛ Re ⎞ 3 ⎫⎪
Qμ ⎧⎪ 2kt
⎨
Pw (t ) = Pi −
+ ln ⎜⎜ ⎟⎟ − ⎬ .
2πkH ⎪⎩φμcRe2
⎝ Rw ⎠ 4 ⎪⎭
(2)
(11)
Now take the derivative of Pw with respect to t:
Qμ
2k
dPw
−Q
=−
⋅
=
.
dt
2π kH φμcRe2 π Re2Hφc
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The rate of increase of the late-time drawdown can therefore be
used to find the radius of the drainage area, or, equivalently, the
drainage area A, i.e.,
⎡
⎤1/ 2
−Q
⎥ ,
Re = ⎢
⎢⎣ π Hφ c(dPw /dt ) ⎥⎦
(13)
−Q
.
φ cH(dPw /dt )
(14)
A = πRe2 =
Eq. (12) can also be “derived” in the following, much simpler way,
by just doing a mass balance on the oil in the reservoir:
(a) Imagine that the entire reservoir is at a uniform pressure, P
(i.e., a zero-dimensional model, so to speak).
(b)
Let M = ρVφ be the total amount of oil in the reservoir.
(c)
Following the line of reasoning of section 1.6,
dM = ρVφ c dP .
t dt
dt
(15)
(d) - dM/dt is the mass flowrate out of the reservoir, and Q is the
volumetric flowrate, so dM/dt = -ρQ, in which case
-Q =Vφct dP .
dt
(16)
(e) The volume of the reservoir is V = πRe H , so eq. (16) is
equivalent to eq. (12)!
2
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Section 6
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The dimensionless wellbore pressure for a well located at
the centre of a bounded circular reservoir is plotted from eq. (6) in
Fig. 6.3.2, for various values of RDe = Re / Rw . Also shown is the
wellbore pressure for the case of constant flowrate into the
borehole and constant pressure at the outer boundary, from eq.
(6.2.10). Note that
a) In accordance with eq. (7), the semi-log straight line begins at
about t Dw = 25 .
Dimensionless Pressure, ΔPDw
b) In accordance with eq. (8), the “finite reservoir” regime begins
2 . For example, when R
when t Dw = 0.3RDe
De = 1000 , the curve
deviates from the semi-log straight line at about
2
5
t Dw = 0.3RDe = 3 × 10 .
0
2
Constant-pressure
outer boundary
4
RDe =
102
6
No-flow outer
boundary
8
103
RDe = 102 103
104
10
1
10
102 103 104 105 106 107
Dimensionless Time, t Dw
Fig. 6.3.2. Well at the centre of a circular reservoir.
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Section 6
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6.4. Non-Circular Drainage Regions
If the drainage region is not circular, and/or the well is not
located at the centre, then the problem treated in section 6.3 is
more difficult to solve. If the drainage region is bounded by a
polygon (as would occur if the reservoir was bounded by a set of
linear faults), then the problem can be solved by superposing
solutions from appropriately located image wells, as in section
4.3. For example, an infinite array of image wells arranged on a
square lattice will create a square reservoir that is bounded by
four no-flow boundaries, etc. Some examples of this type of
analysis can be found in Well Testing in Heterogeneous
Formations, by T. D. Streltsova, Wiley, 1988.
Without going into the details of this analysis, the following
can be said about a well producing from within a non-circular
drainage region:
• There is an “infinite reservoir” regime, during which the
pressure pulse has not yet reached any portion of the outer
boundary. The duration of this regime can be calculated using eq.
(1.6.11), if the geometry of the drainage area and the location of
the well is known. During this regime,
ΔPDw =
1
[ln(t Dw ) + 0.80907] .
2
(1)
At sufficiently long times, the mass-balance argument
presented in the previous section must hold, and so the rate at
which the wellbore drawdown changes must be given by eq.
2
(6.3.12), with Re replaced by A / π . Hence, an equation of the
same form as eq. (6.3.10) must hold for the drawdown, except
that the constant term may be different.
The duration of this regime is roughly given by eq. (6.3.9),
2
again with Re replaced by A / π , although the start of this regime
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is delayed if the well is located off-centre of the drainage region
by an appreciable amount. During this regime, the dimensionless
wellbore drawdown is traditionally written in the following form:
PDw
1 ⎜⎛ 4A ⎟⎞
= 2π t DA + ln⎜ 2 ⎟ ,
2 ⎝ γRwCA ⎠
(2)
where tDA is a dimensionless time based on the drainage area A:
t DA =
kt
,
φμcA
(3)
γ = 1.781, and CA is a dimensionless constant known as the
“Dietz shape factor”.
Comparison of eq. (2) and eq. (6.3.10) reveals that, for a well
at the centre of a circular reservoir,
CA =
4π
γ e −3 / 2
= 31.62 .
(4)
The shape factor tends to decrease as the drainage area
becomes more non-circular, or as the location of the well
becomes more off-centre. Shape factors for numerous geometries
can be found on p. 111 of Matthews & Russell; a few cases are
shown in Fig. 6.4.1.
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CA
Section 6
Page 109
CA
1
31.6
22.6
2
2.07
1
31.6
2
27.6
1
2.36
4
12.9
1
0.232
4
Fig. 6.4.1. Dietz shape factors for various geometries.
______________________________________________
Tutorial Sheet 5:
(1) Show that the transition regime for the closed circular
2 . Hints:
reservoir problem ends when t Dw = 0.3RDe
(a) e − x ≈ 0 when x > 4, so the terms in the series will be
negligible when λ2nt Dw > 4 for all n.
(b) When RDe is large, the smallest eigenvalue, λ1, i.e., the
smallest value of λ that satisfies eq. (6.3.4), is very small. This
fact should help to estimate the value of λ1 as a function of RDe .
Also, make use of eqs. (6.2.37,39,48), and Fig. 6.1.3.
(2) Starting with eq. (6.3.2), calculate the average pressure in
2 ), as
the reservoir during the finite reservoir regime ( t Dw > 0.3RDe
a function of time. Is your result consistent with the mass-balance
that is embodied in eq. (6.3.15)?
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Laplace Transform Methods
The diffusion equation for fluid flow in a porous medium, in
the linearised form that is often used, is a linear PDE. As such,
many of the classical methods of applied mathematics can be
used to solve it. In section 2.1 we used the Boltzmann
transformation to solve the problem of a single, fully-penetrating
well in an infinite, homogeneous reservoir, although we saw in
section 6.1 that this method is only occasionally applicable. In
section 6 we used the method of eigenfunction expansions to
solve problems involving a well in a circular reservoir. Although
this method is in general the most widely-used method in applied
mathematics, it has not been used very often in petroleum
engineering. Traditionally, reservoir engineers have solved the
diffusion equation using the method of Laplace transforms. In this
section we will describe the Laplace transform approach, and give
an example of its use in solving reservoir problems.
7.1. Laplace transforms
The method of Laplace transforms allows us to transform
PDEs, which are generally difficult to solve, into ODEs, which can
be solved in essentially all cases. Specifically, the function P(R,t),
which satisfies the PDE, is transformed into a different function
~
P(R, s) , which satisfies an ODE. This ODE is then solved to yield
~
P(R, s) . The only difficult part of the method of Laplace transforms
is the process of “inverting” the function from the “Laplace
domain” back into the “time domain”, to get P(R,t). This last step
can be achieved either by contour integration in the complex
plane, or by numerical integration along the real axis.
In this section we present the definition of the Laplace
transform, along with a few of its more important properties. A
fuller treatment can be found in essentially any textbook on
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Section 7
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advanced applied mathematics. Three books that are devoted
exclusively to Laplace (and related) transforms are
(1) Operational Mathematics, R. V. Churchill, McGraw-Hill, New
York, 1958.
(2) Operational Methods in Applied Mathematics, H. S. Carslaw
and J. C. Jaeger, Oxford University Press, Oxford, 1949.
(3) Integral Transforms in Mathematical Physics, C. J. Tranter,
Chapman & Hall, London, 1971.
Definition: If we have a function f (t) , its Laplace transform is
defined by
∞
~
L {f (t )} ≡ f (s ) ≡ ∫ f (t )e − st dt .
(1)
0
~
Both notations, L{ f (t) } and f (s ) , are useful, depending on the
context. In the general theory of Laplace transforms, s must be
regarded as a complex variable, but for our purposes this is
usually not necessary.
If we have a function of two variables, such as P(R,t), we
can defined its Laplace transform in the same manner:
∞
~
P (R, s ) ≡ ∫ P (R, t )e − st dt .
(2)
0
Note that the Laplace transform is taken with respect to the time
variable, not the spatial variable. To simplify the notation, we will
usually suppress the R variable when discussing the general
theory.
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The Laplace transform operator L is a linear operator,
because it follows directly from definition (1) that
∞
L{cf(t)} = ∫
cf (t)e −st dt
0
∞
= c ∫ f(t)e −st dt = cL{f (t )} ,
(3)
0
∞
L{f1(t) + f 2(t)} = ∫ [f1(t ) + f2 (t )]e −st dt
0
=
∞
∫ f1(t)e −st dt
0
(4)
∞
+ ∫ f2 (t)e − st dt = L{f1(t)} + L{f2 (t)}.
0
• The most important property of the Laplace transform is that
differentiation in the time domain corresponds to multiplication by
s in the Laplace domain. To prove this, consider the Laplace
transform of the time derivative of f (t) :
∞
L{ f ′(t)} ≡ ∫ f ′(t)e −st dt .
(5)
0
Now recall the formula for integration by parts:
∞
∞
∫ f ′(t)g(t)dt = f (t )g(t)]0 − ∫ f (t) g ′(t)dt .
∞
0
(6)
0
If we apply formula (6), with g(t) = e −st , we have
L{ f ′(t)} = f (t)e
−st
∞
] + s ∫ f (t )e
∞
0
−st
dt
0
= f (∞)e −s⋅∞ − f(0)e − s⋅0 + sL{f (t)} ,
so:
L{ f ′(t)} = sL{f (t)} − f(0) .
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Hence, the Laplace transform of f ′(t) is equal to the Laplace
transform of f (t) multiplied by s, minus the initial value of f (t) .
Eq. (7) illustrates two important properties of Laplace transforms:
(i) Differentiation in the time domain corresponds
(essentially) to multiplication by s in the Laplace domain;
(ii) The initial conditions get incorporated directly into the
governing equation - unlike in “time domain” methods, where
initial conditions must be considered separately after we have
found the general solution.
[For our purposes, since we often work with “drawdowns”,
which are defined so as to be zero when t = 0, the f(0) term will
usually drop out].
Differentiation with respect to time corresponds to
multiplication by s in Laplace space, so we might expect that
integration with respect to time corresponds to division by s in
Laplace space. To prove this, let F(t) be the integral of f(t), i.e.,
t
F(t) ≡ ∫ f (τ )dτ .
(8)
0
By definition, then,
F ′(t) = f (t),
0
F(0) ≡ ∫ f (τ )dτ = 0 .
(9)
0
Now, using property (7),
⎧⎪t
⎪⎫
L{F ′(t)} = sL{F(t)} − F(0) = sL{F(t)} = sL ⎨ f (τ )dτ ⎬ ,
⎪
⎪⎩
⎭
0
∫
so:
⎧⎪ t
⎫⎪
1
L⎨ f(τ )dτ ⎬ = L{F ′(t)} =
⎪⎩
⎪⎭ s
0
∫
1
1
L{f(t)} = f˜ (s) ,
s
s
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which proves that, in order to find the Laplace transform of the
integral of f (t) , we merely have to take the Laplace transform of
f (t) , and divide it by s.
Another useful property of Laplace transforms is the “timeshift” property. Consider a function f (t) , and the time-shifted
function f(t - t0), defined by the following graph:
f(t)
0
f(t-t0)
t
t0
Fig. 7.1.1. A function, f(t), and its time-shifted variant, f(t-t0).
Note that f(t - t0) = 0 if t < t0, which is consistent with the fact that,
when using Laplace transforms, we always consider all functions
f (t) to be zero when t < 0. (This is also consistent with the
physical fact that drawdowns will be zero before production
begins.).
The Laplace transform of f(t - t0) can be found directly from
the Laplace transform of f(t), as follows. First,
{
}
∞
L f (t − t 0 ) = ∫ f (t − t 0 )e −st dt .
(11)
0
Now, let t - t0 = τ, in which case dt = dτ. But the limits of
integration change, according to
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τ = - t 0 when t = 0,
Section 7
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τ = ∞ when t = ∞ .
(12)
Hence,
{
}
∞
L f (t − t 0 ) =
∫ f (τ )e −s(τ +t )dτ .
0
(13)
−t 0
But f(τ) = 0 when τ < 0, so
{
}
∞
L f (t − t 0 ) = ∫ f (τ )e
− s(τ +t0 )
0
=e
−st 0
∞
dτ = ∫ f (τ )e −sτ e − st0 dτ
0
∞
∫ f (τ )e −sτ dτ
= e −st 0 f÷(s) .
(14)
0
So, “delaying” a function f (t) by an amount of time t0 corresponds
to multiplying its Laplace transform by e −st 0 .
If we take a function f (t) , and “damp it out” by multiplying it
by e , this is equivalent to replacing s with s+a in the Laplace
transform of f (t) . The proof is as follows:
-at
L
{
∞
}= ∫ f (t)e
e −at f(t)
−at e −st dt
0
∞
(15)
= ∫ f (t)e −(s+a)t dt = f÷(s + a).
0
Finally, consider a function f(t), and the “stretched” function
f(at). The Laplace transform of f(at) is related to the Laplace
transform of f(t) as follows (see Tutorial Sheet 6):
1~
L{f (at )} = f (s/a) .
a
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If we think of a as a frequency, as in the function sin(at), eq. (16)
shows that “speeding up” a function by a factor a is essentially
equivalent to “slowing down” its Laplace transform by the factor
1/a.
The usefulness of the various rules presented above is that
they minimise the number of times we actually have to calculate a
Laplace transform via the basic definition (1). By knowing the
Laplace transforms of a few functions, we can use these rules to
calculate the Laplace transforms of other functions. For example,
first note that if f (t) = 1, then
{ } {}
∞
L f (t) = L 1 = ∫ 1e −st dt =
0
−1 −st
e
s
]
∞
0
=
1
.
s
(17)
[Note: f (t) = 1 is a very well-behaved mathematical function, but
~
its Laplace transform is not, since f (s ) → ∞ when s = 0. Points
~
where f (s ) becomes infinite, or is otherwise non-analytic (i.e.,
can’t be represented by a Taylor series), are known as
~
singularities. The singularities of f (s ) are actually the key to
“inverting” f÷(s) in order to find the actual function, f (t) . However,
we will not make use of these ideas, because they require
knowledge of complex variable theory].
Now suppose that we want to calculate L{ f (t) } for the case
of f(t) = t. We could use eq. (1), but it is easier to see from eq.
(10) that, since t is the integral of 1, then L{t} can be found by
dividing L{1} by s, i.e.,
⎧t
⎫ 1
1
L{}
t = L⎨ ∫ 1dτ ⎬ = L{}
1= 2.
s
s
⎩0
⎭
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A particular Laplace transform that is very useful in solving
diffusion problems is L{t -1/2}:
L{t
−1/ 2
∞
} = ∫ t −1/ 2e −st dt .
(19)
0
To evaluate this integral, we let st = u, in which case dt = du/s,
and the integral becomes
L{t
−1/ 2
∞
} = ∫ (u / s)
−1/ 2
e
−u
0
du
1 ∞ −1/ 2 −u
=
∫ u e du .
s
s0
(20)
Now put u = m2, in which case du = 2mdm = 2√udm, and so
L{t
−1/ 2
}=
2
s
∞
∫ e −m dm =
2
0
π
2
s 2
=
π
s
,
(21)
which can also be expressed as
L−1{π 1/ 2s −1/ 2 } = t −1/ 2 .
(22)
By applying the rule that division by s in the Laplace domain
corresponds to integration with respect to t in the time domain, we
can show that
L−1{π 1/ 2s −3 / 2 }
t
= ∫ τ −1/ 2dτ = 2t 1/ 2
0
⇒
L{t 1/ 2 } =
π
2s 3 / 2
(23)
.
This process can be repeated to give, for n = 1,2,...:
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L{t n− (1/ 2) } =
Section 7
Page 118
1⋅ 3 ⋅ 5 ⋅... ⋅ (2n − 1) π
.
2n s n+(1/ 2)
(24)
The most difficult aspect of using Laplace transforms is
~
“inverting” the Laplace transform f (s ) to find the function f (t) .
The difficulty lies in the fact that, even though we, as petroleum
engineers, are ultimately interested only in real-valued functions,
the standard inversion procedure involves integration in the
complex plane. The inversion formula, which allows us to recover
~
f (t) from f (s ) , is the following:
y
Singularity
α + i∞ ~
1
f (s ) =
∫ f (s )est ds
2πi α − i∞
6
6
6
6
6
x
x=α
(25)
where i = −1 is the imaginary unit number, and the
integration takes place along any vertical line in the complex
plane (see above) that lies to the right of all the singularities
~
of f (s ) . The proof of eq. (25), which not elementary, can be
found in the books by Carslaw and Jaeger, or Churchill.
One additional fact about Laplace transforms that is
needed in order to use them to solve partial differential
equations is the following rule:
~
• If P(R,t) is a function of R and t, and P (R, s ) is its Laplace
transform, as defined in eq. (2), then the LT of the partial
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Section 7
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derivative of P with respect to R is equal to the partial
derivative, with respect to R, of the LT of P, i.e.,
⎧dP ⎫ d
dP÷(R,s)
⎨
⎬
=
L
L P(R,t) =
.
⎪⎩dR ⎪⎭ dR
dR
[{
}]
(26)
The proof of this fact follows directly from Leibnitz’s theorem
for differentiating an integral with respect to a parameter that
appears in the integrand:
⎧dP(R,t ) ⎫ ∞ dP(R,t)
L⎨
e −st dt
⎬= ∫
⎪⎩ dR ⎪⎭ 0 dR
∞
d
dP÷(R,s)
−st
=
∫ P(R,t)e dt = dR .
dR 0
(27)
In the second term we first differentiate with respect to R,
and then integrate with respect to t, whereas in the third term
we first integrate with respect to t, and then differentiate with
respect to R. Leibnitz’ theorem says that the order in which
we carry out these two operations is immaterial.
7.2. Flow to a hydraulically-fractured well
In general, exact inversion (as opposed to numerical
inversion, which will be briefly discussed in section 7.4) of a
Laplace transform requires knowledge of complex variable
theory. One of the few important problems that we can solve
with Laplace transforms without knowledge of complex
integration is the one-dimensional diffusion equation in a
semi-infinite region.
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The one-dimensional diffusion problem is relevant to
flow to a “hydraulically-fractured” well, for example. In low
permeability reservoirs, we often introduce fractures into the
rock by pumping fluid into the borehole at high pressure (see
the geomechanics module). These “hydraulic fractures” then
provide very conductive paths for the oil to reach the
borehole. Oil first flows to the fracture, and then through the
fracture to the well.
Imagine that the thickness of the reservoir is H, and that
the hydraulic fracture extends out from the borehole a
distance L in each direction, as shown in Fig. 7.2.1:
Top View
L
Flow
Wellbore
Fracture
z
Fig. 7.2.1. Schematic diagram of one-dimensional flow to a
hydraulically-fractured well.
Initially, fluid flows straight to the nearest part of the
fracture, after which it travels through the hydraulic fracture
to the wellbore. If the length L is large, we can model this
process as uniform, one-dimensional, horizontal flow.
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The governing equation for this problem is the 1-D
diffusion equation in Cartesian co-ordinates, eq. (1.6.8):
PDE:
d 2P
dP
=D
,
dt
dz 2
(1)
where z is the coordinate normal to the fracture plane, and D
is the hydraulic diffusivity k / φμct .
If the total flowrate into the well is Q, and this flow is
distributed uniformly over an area of 4LH (two fractures with
two faces each), then the initial and boundary conditions are
P( z, t = 0) = Pi ,
(2)
far-field BC:
P( z → ∞, t ) = Pi ,
(3)
fracture BC:
dP
μQ
( z = 0, t ) =
.
dz
4kLH
(4)
IC:
(1)
To solve this problem, we first define the LT of P( z, t ) :
∞
~
P ( z,s ) ≡ ∫ P ( z,t )e − st dt .
(5)
0
(2) Next, we take the Laplace transform of both sides of eq.
(1). Using eq. (7.1.7), and eq. (2), the LHS of eq. (1) is
transformed into:
⎧dP ⎫
L ⎨ ⎬ = sL{ P( z, t )} − P( z, t = 0) = sP÷( z, s ) − Pi .
⎪⎩ dt ⎪⎭
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Applying rule (7.1.26) twice, the Laplace transform of
the RHS of eq. (1) is
⎧⎪ d 2P ⎫⎪
d2
d 2P÷( z, s )
L ⎨D 2 ⎬ = D 2 L P( z, t ) = D
.
2
⎪⎩ dz ⎪⎭
dz
dz
[{
}]
(7)
So, the transformed representation of eq. (1) is
ODE:
D
d 2P÷( z, s )
dz
2
− sP÷(z, s) = −Pi .
(8)
~
Although P ( z,s ) is a function of two variables, z and s, there
~
are no derivatives of P ( z,s ) with respect to s in eq. (8). So s
really appears in eq. (8) as a parameter, not a variable.
Consequently, eq. (8) is an ODE rather than a PDE.
The initial condition is already incorporated into eq. (8).
However, we must now take the Laplace transforms of the
two BCs, eqs. (3) and (4):
•
far-field BC:
•
P
L{ P( z = ∞, t )} = P÷(z = ∞, s) = L{ Pi } = i ,
s
fracture BC:
(9)
⎧dP
⎫ dP÷
⎧ μQ ⎫
μQ
⎬=
L ⎨ (z = 0, t )⎬ =
(z = 0, s ) = L ⎨
. (10)
⎪⎩ dz
⎪⎭ dz
⎪⎩ 4kLH ⎪⎭ 4kLHs
(3) Next, we solve the problem in the Laplace domain. The
general solution to eq. (8) is
P÷( z, s ) = Ae z
s/ D
+ Be − z
s /D
+
where A and B are arbitrary constants.
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,
s
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The far-field boundary condition, eq. (9), shows that A
must be zero. The fracture BC, eq. (10), implies that
dP÷
μQ
( z = 0, s ) = − B s / D =
,
dz
4kLHs
⇒
B=
− μQ
4kLHs s / D
(12)
.
So, the solution to this problem in the Laplace domain is
P
μQ D
P÷(z,s) = i −
e −z
s 4kLHs 3 / 2
s/ D
.
(13)
(4) Finally, we must invert P÷(z,s) to find P(z,t). To keep
the analysis simple and brief, we will only invert for the
pressure in the fracture, P(z=0,t).
Note: This illustrates another advantage of Laplace
transform methods: we can usually find the wellbore
pressure without having to find the pressure at every point in
the reservoir; with other methods, you must find P(z,t) first as
a complete function, and then set z = 0.
So,
P
μQ D
P÷(z = 0,s) = i −
.
s 4kLHs 3 / 2
(14)
We now “invert” to find P(z=0,t), using eqs. (7.1.17,23):
⎧⎪
μQ D ⎫⎪
−1⎨ Pi
÷
⎬
P(z = 0,t) = L { P(z = 0,s)} = L
−
⎪⎩ s 4kLHs 3 / 2 ⎪⎭
−1
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⎧ ⎫
⎪⎩ s ⎪⎭
1
Pi L−1⎨ ⎬ −
= Pi −
μQ D
4kLH
Section 7
Page 124
⎧ 1 ⎫
⎬
⎪⎩s 3 / 2 ⎪⎭
L−1⎨
μQ
Dt
2kLH
π
.
(15)
But, if the permeability of the fracture is much greater than
that of the reservoir (which is the idea behind hydraulic
fractures!), then there will be very little pressure drop in the
fracture itself, and the fracture pressure will be equal to the
wellbore pressure.
So, eq. (15) shows that the (early-time) drawdown in a
fractured well should increase in proportion to t1/2.
7.3. Convolution principle in Laplace space
The convolution integral that appeared in eq. (3.4.6) is
a particular case of a more general mathematical operation
that can be defined by
t
f ∗ g ≡ ∫ f (τ )g(t − τ )dτ ,
(1)
0
and which is known as the convolution of f and g.
If we let t - τ = x in eq. (1), then τ = t - x and dτ = - dx,
and the limits of integration are transformed into x = t and x =
0. Hence,
0
t
f ∗ g = −∫ f(t − x )g(x)dx = ∫ f (t − x)g(x )dx ≡ g ∗ f ,
t
0
which shows that f ∗ g = g ∗ f .
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Now, let’s look at the Laplace transform of f ∗ g :
t =∞ ⎡ τ =t
⎤
⎢
⎥
L{f ∗ g} = ∫ ∫ f(τ )g(t − τ )dτ e −st dt .
⎢
⎥
⎦
t =0 ⎣τ =0
τ
(3)
τ
t=τ
t=τ
t
t
Fig. 7.3.1. Two paths of integration for eq. (3).
The region of integration covers the first octant of the t τ plane, as can be seen from Fig. 7.3.1. The thick line on the
left runs from τ = 0 to τ = t, for fixed t, which corresponds to
the inner integral in eq. (3). If we sweep this line across from
left to right, then t will cover the range from t = 0 to t = ∞ ,
which corresponds to the outer integral.
But this octant can also be covered by first letting t run
from t = τ out to t = ∞ , for fixed τ (see the thick line on right),
and then letting τ run from τ = 0 to τ = ∞ . Hence,
τ = ∞ ⎡t = ∞
⎤
−st ⎥
⎢
L{f ∗ g} = ∫
g(t − τ )e dt f (τ )dτ .
⎢ ∫
⎥
⎦
τ = 0 ⎣ t =τ
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If we now let t - τ = x in eq. (4), then t = τ + x, dt = dx, and
the limits of integration of the inner integral are transformed
into x = 0 and x = ∞ :
τ = ∞ ⎡ x= ∞
⎤
−s(
τ
+
x)
⎢
⎥
L{f ∗ g} = ∫
dx f(τ )dτ
∫ g( x)e
⎢
⎥⎦
τ = 0 ⎣ x= 0
⎤
⎡ x= ∞
−
s
τ
−sx
⎢
= ∫ f (τ )e dτ ∫ g( x)e dx ⎥ = f÷(s) g÷(s).
⎥
⎢
⎦
⎣ x= 0
τ=0
τ=∞
(5)
We have therefore proven that convolution of two
functions in the time domain corresponds to multiplication of
the their respective Laplace transforms. This is extremely
useful, because
(a) It implies that, if we can break up a difficult Laplace
transform into the product of two simpler transforms whose
inverses are known, then we can find the complete inverse
function by convolution.
(b) More importantly, it implies that, for any given reservoir
geometry, if we can solve the diffusion equation for the case
of a constant production rate, then the solution for an
arbitrary production schedule can be found by convolution.
Actually, we already knew this from section 3.4. However,
the concept is more general, since, for example, we can also
use it to find the solution for the case of varying wellbore
pressure from the solution for the case of constant wellbore
pressure.
As an example of the use of convolution in the Laplace
domain, consider the hydraulic fracture problem of section
7.2, but with an arbitrary time-dependent flowrate into the
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fracture, Q(t). The only change in the formulation of the
problem will be in the fracture boundary condition, eq.
(7.2.4), which now becomes
dP
μQ(t )
(z = 0,t ) =
.
dz
4kLH
(6)
If we follow the solution procedure used in section 7.2,
the only change occurs when we take the Laplace transform
of this boundary condition, in which case μQ / 4kLHs is
replaced with the more general expression μQ÷(s) / 4kLH .
The solution in Laplace space then becomes
P μQ÷(s) D − z
P÷(z,s) = i −
e
s 4kLHs 1/ 2
s /D
.
(7)
We again restrict our attention to the fracture, where z = 0:
Pi μQ÷(s) D
÷
P(z = 0,s) = −
.
s 4kLHs1/ 2
(8)
We now take the inverse Laplace transform, and in doing so
make use of linearity and convolution:
−1
P(z = 0,t) = L
=
⎧⎪
÷(s) D ⎫⎪
Q
μ
⎬
P÷(z = 0,s) = L
−
1/
2
⎪⎩ s 4kLHs ⎪⎭
{
⎧ ⎫
⎪⎩ s ⎪⎭
1
Pi L−1⎨ ⎬ −
= Pi −
}
−1⎨Pi
μ D
⎧⎪ ÷
⎫⎪
s
⎭
Q(s)
⎬
L−1⎨
⎪⎩ 1/ 2 ⎪
4kLH
[L
4kLH
μ D
−1{Q
÷(s)}
]∗ [L
].
−1{s −1/ 2 }
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But, L−1{Q÷(s)} = Q(t ) , by definition, and L−1{s −1/ 2 } = (πt) −1/ 2
by eq. (7.1.22), so
P(z = 0,t) = Pi − μ D [Q(t)]∗[(πt)−1/2]
4kLH
= Pi −
μ D/π
4kLH
t
Q(τ )
0
t −τ
∫
dτ
1
μ t Q(τ )
dτ
= Pi −
∫
4LH πkφc 0 t − τ
1
μ t Q(τ )
dτ
= Pi −
∫
A πkφc 0 t − τ
μ t q(τ )
dτ .
= Pi −
∫
πkφc 0 t − τ
(10)
Eq. (10) gives us the pressure in the fracture as a function of
the time-varying flowrate per unit area, q(t) = Q(t)/A.
Although it may be difficult to evaluate this integral
analytically for some particular q(t), it will always be
straightforward to evaluate it numerically. Convolution leads
to an integral over time, which is a real variable, and such
integrals are generally easier to evaluate than the complex
integral specified in eq. (7.1.24).
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7.4. Numerical inversion of Laplace transforms
The complex integral that allows us to invert the
Laplace transform of the pressure to obtain the pressure as
a function of time is often very difficult to evaluate in closed
form. Consequently, numerous algorithms have been
devised to carry out this inversion numerically. Their use in
petroleum engineering problems is reviewed in the book
Fundamental and Applied Pressure Analysis by Daltaban
and Wall (Imperial College Press, 1998).
The most widely-used numerical algorithm for inverting
a Laplace transform is the Stehfest algorithm. Although the
derivation of this algorithm is very lengthy, and beyond the
scope of this course, we can gain a rough understanding of it
as follows.
In general, any integral can be approximated by a
summation whose terms consist of the integrand evaluated
at various discrete points, which each functional evaluation
multiplied by an appropriate weighting factor.
For
example,
consider
a
simple
numerical
approximation of an integral based on the trapezium rule.
Imagine a function f(x), which we want to integrate from x = a
to x = b. Recall that an integral can be interpreted as the
area under the graph of the function.
The area under the graph of f(x) can be approximated
by the area of the trapezium shown in Fig. 7.4.1, i.e.,
⎡ f(a) + f (b) ⎤
⎥(b − a) .
∫ f (x)dx ≈ ⎢⎢
⎥⎦
2
⎣
a
b
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This formula approximates the integral by evaluating the
integrand at x = a and x = b, multiplying the two function
values by the weighting factor (b - a)/2, and then summing.
f(x)
f(b)
f(a)
a
b
x
Fig. 7.4.1. Approximation of an integral.
More accurate numerical approximations of an integral
can be made by
(a)
evaluating the integrand at a larger number of points,
(b) carefully choosing the points at which the integrand is
evaluated (i.e., they need not be equally-spaced),
(c)
carefully choosing the weighting factors.
With these ideas in mind, we can think of the Stehfest
algorithm as an optimised method for approximating the
complex inversion integral by a weighted finite sum of
function evaluations.
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●
Actually, the situation is a bit more subtle than this,
because the Stehfest method also manages to avoid
~
evaluating f (s ) at complex values of s, as would seem to be
required by the path of integration shown in eq. (7.1.25);
~
instead, the method requires only that the function f (s ) be
evaluated at real values of s.
●
If we could perform the complex inversion integration
analytically, we would arrive at a single mathematical
equation that is valid for all values of t. On the other hand, if
we use a numerical procedure, we must perform a new
calculation for each value of t at which we want to know the
pressure.
If we have the Laplace transform of, say, the wellbore
~
pressure Pw (s ) , the actual wellbore pressure at time t is
given, according to Stehfest, by
Pw (t ) =
ln 2 2N ~ ⎛
nln 2 ⎞
∑Vn Pw ⎜ s =
⎟,
t n =1
t
⎠
⎝
(2)
where the weighting factors Vn are defined by
Vn =
(−1) n
min(n,N )
∑
k= (n+1) / 2
k N (2k)!
.
(N − k)!k!(k − 1)!(n − k)!(2k − n)!
(3)
In principle, the accuracy of the approximation should
increase as the number of terms in the series, which is 2N,
increases. In practice, if too many terms are taken,
accumulated round-off error begins to overshadow the
additional accuracy. It has generally been found that an
optimum value of 2N is about 18.
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______________________________________________
Tutorial Sheet 6:
(1)
What is the Laplace transform of the function f(t) = e-at?
(2) Starting with the basic definition of the Laplace
transform, eq. (7.1.1), verify eq. (7.1.16).
(3) Following the steps that were taken in section 7.2, use
Laplace transforms to try to solve the one-dimensional
diffusion problem with a constant-pressure boundary
condition:
1 dP d 2P
=
,
2
D dt
dz
(i)
P(z,t = 0) = Pi ,
(ii)
far-field BC:
P(z → ∞,t) = Pi ,
(iii)
fracture BC:
P(z = 0,t) = Pf .
(iv)
PDE:
IC:
(a) Find the solution for the pressure in the Laplace
~
domain, i.e., find P(z,s) .
(b) Find, in the Laplace domain, an expression for the
~
flowrate into the fracture; call it Qf (s) .
~
(c) Invert Qf (s) to find the flowrate into the fracture as a
function of time, Qf (t) .
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Section 8
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Naturally-Fractured Reservoirs
Many reservoirs contain a network of interconnected
naturally occurring fractures that provide most of the reservoirscale permeability. In fact, it has been estimated that about 40%
of the world’s known reserves occur in fractured reservoirs. Flow
through such reservoirs is more complicated than through
unfractured reservoirs, and cannot be analysed with the equations
that we have developed and solved in the previous sections.
The basic mathematical model used for fractured reservoirs
was first developed by Barenblatt, Zheltov and Kochina in 1960,
and by Warren and Root in 1963 (SPEJ, 1963). Several
refinements have been made since then, and the resulting
equations have been solved for many different reservoir
geometries, with and without wellbore storage, skin effects, etc. In
this section we will briefly describe this model, and present and
discuss the solution for flow to a well in an unbounded dualporosity reservoir.
8.1. Dual-porosity model of Barenblatt et al.
In this model, the permeability of the reservoir is assumed to
be provided solely by the fracture network. The regions of rock
between the fractures, called “matrix blocks”, store most of the
fluid, and feed this fluid into the fractures.
The first step in constructing a mathematical model for dualporosity media is to start with a pressure diffusion equation for the
fracture network. In radial co-ordinates, this equation takes the
form
dPf kf 1 d ⎛ dPf ⎞
⎜R
⎟ + qmf ,
(φc)f
=
dt
μ R dR ⎜⎝ dR ⎟⎠
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where the subscript f denotes the properties of the fracture
network, and the term qmf represents the volumetric flux of fluid
from the matrix blocks into the fractures, per unit time and per unit
volume of the reservoir. The “fracture-matrix interaction term” qmf
therefore has dimensions of [m3 /m 3s] , or [1/s].
Note: in a dual-porosity model, every point “R” is supposed to
represent an REV (see section 1.3) that is large enough to
encompass several fractures and matrix blocks.
The fracture-matrix flow term could be found by solving a
flow equation in the matrix blocks, as has been done by Kazemi
(SPEJ, 1969) and others. However, most dual-porosity
formulations follow Barenblatt et al. and assume that the flow from
the matrix blocks to the fractures, at point R in the reservoir, is
proportional to the difference between the pressure in the
fractures and the average pressure in the matrix blocks, Pm ; this
is called “pseudo-steady-state fracture/matrix flow”. The flow
should also be proportional to the permeability of the matrix, and
inversely proportional to the fluid viscosity, so qmf is assumed to
have the form
qmf =
αkm
(Pm − Pf ) ,
μ
(2)
where α is a “shape factor” that accounts for the size and shape
of the matrix block.
Dimensional analysis shows that α must have dimensions of
[1/L2 ] , or [1/ m2 ] in SI units. The precise value of α for a matrix
block of a given shape is found by calculating the smallest
eigenvalue of the diffusion equation inside the matrix block. The
shape factors for a few simple, idealised geometries are as
follows (Zimmerman et al., Water Resources Research, 1993):
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• spherical block of radius a:
α=
π2
a
,
2
(3)
• long cylindrical block of radius a:
α=
5.78
a2
,
(4)
,
(5)
• cubical block of side L:
α=
3π 2
2
L
• rectangular block of sides {Lx ,Ly ,Lz } :
⎡
⎤
1⎥
α =π 2 + 2 + 2 .
⎢L
⎥
⎣ x Ly Lz ⎦
2⎢
1
1
(6)
If the fracture system in the reservoir consists of three
mutually orthogonal sets of parallel fractures, with spacings
{Lx ,Ly ,Lz } , then the matrix blocks would be rectangular slabs;
hence this last model, although simplified, is not unrealistic.
Eqs. (1,2) give two equations for three unknowns,
{Pf ,Pm ,qmf } . Three equations are needed in order to find a
solution. The third equation is found by performing a massbalance on the matrix blocks. In analogy with eq. (6.3.16), we find
qmf = −(φc)m
dPm
.
dt
(7)
Eqs. (1,2,7) give a complete set of equations that can be solved
for the unknown pressures and flowrates.
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8.2. Dual-porosity equations in dimensionless form
Before presenting the solution to the problem of constant
flowrate to a well in a dual-porosity reservoir, it is convenient to
first write the governing equations in dimensionless form.
We define the following dimensionless variables:
z
Dimensionless time:
tD =
kf t
;
(φ f c f + φ mcm ) μRw2
(1)
this is the “standard” definition, but based on the reservoir-scale
permeability, which is kf, and the total storativity, which is the sum
of fracture storativity plus the matrix storativity.
z
Dimensionless pressure in the fractures:
2πkf H(Pi − Pf )
;
μQ
PDf =
(2)
this is again the standard definition, as in section 2.2.
z
Dimensionless pressure in the matrix blocks:
PDm =
2πk f H(Pi − Pm )
;
μQ
(3)
this definition is based on kf, rather than km, so as to be consistent
with the definition of PDf.
z
Dimensionless radius:
RD =
R
;
Rw
(4)
which is the standard definition we have used previously for
single-porosity reservoirs.
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We now use these definitions, along with the chain rule, to
transform eqs. (8.1.1,2,7) into dimensionless form. We start with
the left-hand side of eq. (8.1.1):
(φc)f
dPf
dPf dPDf dt D
= (φ c) f
⋅
⋅
dt
dPDf dt D dt
⎡ − μQ ⎤ dPDf
⎥
= (φc) f ⎢
⎢⎣ 2πkf H ⎥⎦ dt D
⎡
⎤
kf
⎥
⎢
2
⎢⎣ (φ f cf + φ mcm )μ Rw ⎥⎦
⎡
⎤ dP
φ
c
Q
f
f
⎥ Df .
= −⎢
⎢⎣(φ f c f + φ mcm )2πHRw2 ⎥⎦ dt D
(5)
Applying the same procedure to the spatial derivative term in eq.
(8.1.1) gives
−Q
kf 1 d ⎛ dPf ⎞
1 d ⎛
dPDf ⎞
⎜R
⎟=
⎜ RD
⎟.
μ R dR ⎜⎝ dR ⎟⎠ 2πHRw2 RD dR D ⎜⎝
dRD ⎟⎠
(6)
Lastly, we transform the qmf term in eq. (8.1.1), using eq. 8.1.7).
In analogy with eq. (5), this term becomes
qmf
⎤ dP
dPm ⎢⎡
φ mcmQ
Dm
⎥
= −(φc)m
=
.
dt
⎣⎢ (φ f cf + φ mcm )2πHRw2 ⎥⎦ dt D
(7)
Inserting eqs. (5,6,7) into eq. (8.1.1) yields
1 d ⎛
dPDf ⎞
φf cf
dPDf
φmcm
dPDm
⎜⎜ RD
⎟⎟ =
. (8)
+
RD dRD ⎝
dRD ⎠ (φf cf +φmcm ) dt D (φ f cf +φ mcm ) dt D
Eq. (8) is the dimensionless form of eq. (8.1.1).
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We now write eq. (8.1.2) in terms of the dimensionless
variables:
qmf =
αkm
−αkmQ
(Pm − Pf ) =
(PDm − PDf ) .
μ
2πkf H
(9)
Equating this result to eq. (7) gives
φmcm
dPDm αkmRw2
=
(PDf − PDm ) .
(φf cf + φmcm ) dt D
kf
(10)
Eqs. (8,10) are two coupled differential equations for the two
dimensionless pressures, PDf and PDm .
These two equations contain two dimensionless parameters.
The first is the ratio of fracture storativity to total (fracture + matrix)
storativity, known as ω:
ω=
φ f cf
.
(φf cf + φmcm )
(11)
The second dimensionless parameter, λ, is essentially a ratio of
matrix permeability to fracture permeability:
λ=
αkmRw2
kf
.
(12)
In practice, we usually have ω < 0.1, and λ < 0.001.
In terms of these dimensionless parameters ω and λ, and the
dimensionless variables defined in eqs. (1-4), the two governing
equations for a dual-porosity reservoir are
1 d ⎛
dPDf ⎞
dP
dP
⎜⎜ RD
⎟⎟ = ω Df + (1− ω ) Dm ,
RD dRD ⎝
dRD ⎠
dt D
dt D
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dPDm
= λ(PDf − PDm ) .
dt D
(14)
If the reservoir is actually a “single-porosity” reservoir, then
the storativity ratio ω would be 1, and these two equations, (13)
and (14), would reduce to the standard pressure diffusion
equation, (6.1.8).
In the usual form of the dual-porosity model, fluid does not
flow from one matrix block to another. Only the fractures provide
the macroscopic, reservoir-scale permeability. Hence, the
pressure that we measure in the wellbore represents the pressure
in those fractures that are nearest the wellbore. Therefore, the
drawdown in the wellbore is found from
ΔPDw (t D ) ≡ PDf (RD =1,t D ) .
(15)
A more general model of naturally fractured reservoirs, in
which fluid can also flow from one matrix block to another, is
known as the “dual-permeability model”; it is used by hydrologists,
but is not yet widely used in petroleum engineering.
8.3. Solution for a well in a dual-porosity reservoir
Warren and Root (SPEJ, 1963) solved the problem of a well
producing at constant flowrate Q in an unbounded dual-porosity
reservoir, using Laplace transforms. They found that, if the
dimensionless time satisfies the condition
t D > 100 ω,
(1)
(which usually covers the times that we are interested in), the
dimensionless drawdown in the well is described by
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⎡ − λt D ⎤
⎡ −λt D ⎤⎫⎪
1 ⎧⎪
⎨
⎢
⎥
⎢
⎥⎬ .
=
lnt D + 0.8091+ Ei
− Ei
⎪
⎢
⎥
⎢
2⎩
⎣ω (1− ω ) ⎦
⎣ (1− ω ) ⎥⎦⎪⎭
(2)
As usual, it is instructive to analyse the behaviour of the
solution in the various time regimes. When t D is > 100ω, but is not
“too large”, the variables inside the Ei functions in eq. (2) will still
be small enough to use eq. (2.4.3), which says that, if x < 0.01,
−Ei(−x) ≈ −0.5772 − ln x
(3)
Use of eq. (3) in eq. (2) yields
ΔPDw
⎡ λt D ⎤
⎡ λt D ⎤ ⎫⎪
1 ⎧⎪
⎥ − ln ⎢
⎥⎬
= ⎨lnt D + 0.8091+ ln⎢
⎢⎣ ω (1− ω ) ⎥⎦
⎢⎣ (1− ω ) ⎥⎦ ⎪⎭
2 ⎪⎩
⎡ λt D ⎤
⎡ λt D ⎤ ⎫⎪
1 ⎧⎪
⎥ − ln ω − ln⎢
⎥⎬
= ⎨lnt D + 0.8091+ ln⎢
⎢⎣ (1− ω ) ⎥⎦
⎢⎣ (1− ω ) ⎥⎦ ⎪⎭
2 ⎪⎩
=
{
}
1
ln(t D / ω ) + 0.8091 .
2
(4)
If we use eqs. (8.2.1,2,11) to convert the drawdown to
dimensional form, we find
⎫⎪
2πkf HΔPw 1 ⎧⎪ ⎢⎡
kf t
[(φc)f + (φ c) m ] ⎥⎤
⎬
= ⎨ln
⋅
+
0.8091
μQ
⎥⎦
⎪⎭
2 ⎪⎩ ⎢⎣ [(φ c) f + (φc) m ] μRw2
( φc)f
i.e.,
⎫⎪
⎤
μ Q ⎧⎪ ⎢⎡ kf t
⎥ + 0.8091⎬ ,
⎨ln
ΔPw =
2
4πk f H ⎪⎩ ⎢⎣ (φμc)f Rw ⎥⎦
⎪⎭
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which is precisely the drawdown that would occur in a singleporosity system consisting only of the fractures (i.e., without
matrix blocks)! Note also that in the early-time regime, the
drawdown will be a straight line on a semi-log plot.
The physical explanation of this behaviour is as follows. At
early times, fluid flows to the well only through the fractures; fluid
has not yet had time to flow out of the matrix blocks, because of
their relatively low permeability. Hence, in this regime the matrix
storativity is irrelevant.
Now let’s consider “very large” times. Recall from Table 2.1.1
that when x is large, Ei( −x ) → 0 . So, in this regime the two Ei
terms in eq. (2) drop out, and the drawdown is
ΔPDw =
{
}
1
ln t D + 0.8091 ,
2
(6)
which in dimensional form is
⎫⎪
⎞
kf t
μ Q ⎧⎪ ⎜⎛
⎟ + 0.8091⎬ .
⎨ln
ΔPw =
4πk f H ⎪⎩ ⎜⎝ [(φc)f + (φ c) m ]μ Rw2 ⎟⎠
⎪⎭
(7)
This is the drawdown that would occur in a single porosity
reservoir whose permeability is that of the fracture system, but
whose storativity is that of the fractures plus the matrix blocks.
The physical explanation of this behaviour is (essentially) as
follows: At very large times, the matrix blocks have had sufficient
time for their pressures to “equilibrate” with the pressure in the
fractures. So at large times, the reservoir behaves like a
homogeneous, single-porosity reservoir whose storativity is
composed of the fracture storativity plus the matrix storativity. The
overall permeability of the reservoir is always the permeability of
the fracture system, because the matrix blocks are assumed not
to be interconnected to each other.
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Both the early-time solution given by eq. (5) and the late-time
solution given by eq. (7) will give straight lines when plotted on a
semi-log plot, with exactly the same slope,
dΔPw
μQ
=
.
d lnt
4πk f H
(8)
The vertical offset of these two lines, denoted by δΔPDw , can be
found by comparing eqs. (4) and (6):
1
2
1
2
δΔPDw = − ln ω = ln(1/ ω ) .
(9)
In dimensional form, the vertical offset is
μQ
ln(1/ ω ) .
4πk f H
δPw =
(10)
The full curve for the drawdown in a well in an infinite dualporosity reservoir is shown below, in schematic form, adapted
from Gringarten (SPEJ, 1984):
ΔPDw
m
se
g
i-lo
r
st
ht
ai g
li n
e
fo r
c
fra
r
tu
te
ys
s
e
m
i-lo
m
se
tD1
ln(tD)
δΔPDw
g
a ig
str
li
ht
ne
a
tot
r
fo
te
ys
ls
m
tD2
Fig. 8.3.1. Drawdown in a well in a dual-porosity reservoir.
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Fig. 8.3.1 shows that there is a transition regime between the
two semi-log straight lines. In this regime, the pressure drawdown
changes very gradually. This regime is dominated by the effect of
fluid beginning to flow out of the matrix blocks and into the
fractures. During this regime, the flow from the matrix blocks into
the fractures nearly compensates for the flow from the fractures
into the borehole, and the wellbore pressure is (nearly) constant.
Bourdet and Gringarten (SPE paper 9293, 1984) showed
that if you draw a horizontal line through the drawdown curve at
the midpoint between the two semi-log straight lines, this line will
intersect the first semi-log straight line at a dimensionless time t D1
that is given by
tD1 = ω ,
(11)
γλ
where γ = 1.78 is Euler’s constant, and ω and λ are the two dualporosity parameters defined in eqs. (8.2.11,12).
This horizontal line intersects the late-time semi-log straight
line at a dimensionless time t D2 given by
t D2 =
1
γλ
.
(12)
Eqs. (11,12) show that the horizontal offset of the two asymptotic
semi-log straight lines is equal to ln(1/ω).
In dimensional form, these intercept times are
( φμc )f
,
α γ km
(13)
μ[(φc )f + ( φc )m ]
.
α γ km
(14)
t1 =
t2 =
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The first intercept time, t D1, is essentially a measure of the
time required for the volume of fluid that has been depleted from
the matrix blocks near the wellbore to become of the same order
of magnitude as the volume depleted from the fractures. The
second intercept time, t D2 , is a measure of the time required for
the pressure in the matrix blocks nearest the well to come into
equilibrium with the pressure in the surrounding fractures.
Eqs. (11) and (12) allow one to calculate the dual-porosity
parameters ω and λ from well-test data, as will be demonstrated
in the module on well test analysis.
______________________________________________
Tutorial Sheet 7:
(1) Without looking at the paper by Warren and Root, sketch the
forms that you think the drawdown curve in Fig. 8.3.1 would have
if
(a) The storativity ratio ω increased (or decreased) by a factor of
10.
(b) The permeability ratio λ increased (or decreased) by a factor
of 10.
(2) By examining eq. (8.3.2), and making use of either eq.
(2.1.22) or Table 2.1.1, derive an expression for the
dimensionless time that must elapse in order for the
approximation (8.3.6) to be accurate to within about 1%. Your
answer should be in terms of the parameter λ.
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Flow of Gases in Porous Media
9.1. Diffusion equation for gas flow
The main difference between analysing flow in gas
reservoirs as opposed to liquid reservoirs is that, for gases, the
governing partial differential equation becomes unavoidably
nonlinear. The main reason for this nonlinearity is that, whereas
the compressibility of a liquid can be assumed to be constant, and
relatively small (in the sense that was quantified in section 1.6),
the compressibility of a gas varies greatly with pressure.
Furthermore, the viscosity of a gas also usually varies with
pressure. The result is that the governing equation for gas flow
cannot usually be well approximated by a linear diffusion equation
with constant coefficients, as was the case for liquids.
To derive the governing equation for gas flow, we return to
the analysis given in section 1.7, and note that, up to eq. (1.7.5),
no assumptions were made about the magnitude of the
compressibility. Hence, we can begin our analysis of gas flow with
eq. (1.7.5):
d( ρqR)
d( ρφ )
=R
.
(1)
dR
dt
We now use Darcy’s law, in the form given by eq. (1.4.1), for
the flux term q on the left-hand side, i.e.,
−
q=−
k dP
,
μ dR
(2)
in which case eq. (1) takes the form
1 d ⎛ ρk dP ⎞ d( ρφ )
⎜ R
⎟=
.
R dR ⎜⎝ μ dR ⎟⎠
dt
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We next note that the process of expressing the RHS of eq.
(3) in terms of the pressure, which was carried out section 1.6, did
not require any assumptions about the magnitude of the
compressibility terms, etc. Hence, we can use eq. (1.6.1) to
rewrite eq. (3) as
dP
1 d ⎛ ρk dP ⎞
⎜⎜ R
⎟⎟ = ρφ (Cf + Cφ )
.
dt
R dR ⎝ μ dR ⎠
(4)
Eq. (4) is completely general, in that it contains no
assumption that the compressibility of the pore fluid is small. It
also allows for the possibility that the density, permeability, and
viscosity vary with pressure. As such, it is too general to be
useful, except as the basis of a numerical solution. However,
there are many cases of practical interest in which it can be either
linearised, or “almost” linearised, in which case the standard
solutions presented in the previous sections can be used.
9.2. Ideal gas, constant reservoir properties
One situation in which the nonlinear diffusion equation for
gas flow can be effectively linearised is if the gas can be assumed
to obey the ideal gas law:
P = ρRT ,
(1)
where R is the gas constant, and T is the absolute temperature. If
we use eq. (1) to express the density in terms of the pressure, eq.
(9.1.4) takes the form
dP ⎞ φ (Cf + Cφ ) dP
1 d ⎛ kP
⎜⎜
⎟⎟ =
R
P
.
dt
R dR ⎝ μRT dR ⎠
RT
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The simplest model of gas flow can be derived from eq. (2)
by making the following assumptions.
(a)
Assume that the temperature is constant.
(b) Assume that the viscosity of the gas is independent of
pressure. This is actually rigorously true for an ideal gas, for
which the viscosity depends only on temperature. This
assumption allows us to take μ outside of the derivative on the
left-hand side of eq. (2).
(c) Assume that the compressibility of the gas is much larger than
that of the formation, as is usually the case. In fact, the
compressibility of an ideal gas is (1/P), since
1 ⎛ dρ ⎞
RT d ⎛ P ⎞
RT 1
1
⎜
⎟
⎜⎜
⎟⎟ =
= .
Cf = ⎜ ⎟ =
ρ ⎝ dP ⎠ T
P dP ⎝ RT ⎠ T
P RT P
(3)
As the reservoir pressure is usually less than 10,000 psi, the
compressibility is at least 10 −4 /psi. The formation compressibility,
on the other hand, is usually less than 10 −5 /psi (see Matthews
and Russell, p. 159; Zimmerman, Compressibility of Sandstones,
Elsevier, 1991). Hence, we can usually neglect Cφ relative to Cf .
(d) Assume that the permeability of the formation is independent
of the pore pressure. This is reasonably accurate for many
reservoirs, but is sometimes not a very good assumption in
fractured reservoirs or in tight gas sands, for example. This
assumption allows us to take k outside of the derivative on the
left-hand side of eq. (2).
Under these four assumptions, eq. (2) can be written as
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1 d ⎛
dP ⎞ φμ dP
⎜⎜ RP
⎟=
P
.
R dR ⎝
dR ⎟⎠ kP dt
(4)
We now note that
dP 1 d(P 2 )
P
=
,
dR 2 dR
dP 1 d(P 2 )
and P
=
,
dt 2 dt
(5)
which allows us to write eq. (4) as
⎛
⎞
1 d ⎜ d(P 2 ) ⎟ φμ d(P 2 )
.
=
R
R dR ⎜⎝
dR ⎟⎠ kP dt
(6)
Eq. (6) is the standard diffusion equation in radial coordinates,
except that
(1)
The dependent variable is P 2 rather than P;
(2) The compressibility
pressure.
term,
Ct ≈ Cf = 1/ P ,
varies
with
Because of point (2), eq. (6) is still nonlinear, and is not quite
equivalent to the diffusion equation used for liquid flow. However,
we can linearise eq. (6) by evaluating the term P in the
denominator of the RHS at either the initial pressure, Pi, or at the
mean wellbore pressure during the test, i.e., at
Pm =
Pi + Pw (end of test)
.
2
(7)
If this is done, then eq. (6) becomes a linear diffusion equation,
with P2 rather than P as the dependent variable.
P
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Real gas, variable reservoir properties
The ideal gas law is a good approximation to gas behaviour
only at low pressures. It is also generally more accurate for
monatomic or diatomic gases than for larger molecules such as
gaseous hydrocarbons.
Hence, for gas reservoirs, which are usually at relatively high
pressures, it is more accurate to replace the ideal gas law, eq.
(9.2.1), with the “real gas” equation of state,
P = ρzRT, or
ρ=
P
,
zRT
(1)
where z is the (dimensionless) gas deviation factor.
Using this equation of state, eq. (9.1.3) takes the form
1 d ⎛
d⎛ P ⎞
k
dP ⎞
⎟.
⎜R
⎟ = φ ⎜⎜
P
R dR ⎜⎝ μzRT dR ⎟⎠
dt ⎝ zRT ⎟⎠
(2)
Under isothermal conditions, this becomes
1 d ⎛ k dP ⎞
d ⎛ P⎞
⎜R
⎟ = φ ⎜⎜ ⎟⎟ .
P
R dR ⎜⎝ μz dR ⎟⎠
dt ⎝ z ⎠
(3)
Eq. (3) is highly nonlinear, but we can “partially linearise” it
by defining a new variable, the “real gas pseudopressure”, as
follows (see G. L. Chierici, Principles of Petroleum Reservoir
Engineering, Vol. 1, Springer-Verlag, 1994, chapter 7):
P
k(P)P
dP .
μ
(P)z(P)
0
m(P) = 2 ∫
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The real gas pseudopressure m(P) is just a generalisation of
the P 2 parameter that was used for ideal gases; this can be seen
by noting that if we assume permeability and viscosity to be
independent of pressure, and recall that z = 1 for an ideal gas,
then
k 2
2k P
m(P) →
∫ PdP = P .
μ
0
μ
(5)
Hence, for an ideal gas in a stress-insensitive reservoir, the
pseudopressure m is, aside from a multiplicative constant, equal
to P 2 .
Returning to the real gas case, we differentiate eq. (4) with
respect to P, to find
dm( P )
2k (P )P
=
,
dP
μ( P )z( P )
(6)
which implies that
dm dm dP
2k( P )P dP
=
=
.
dR dP dR μ ( P )z( P ) dR
(7)
Substituting eq. (7) into the left-hand side of eq. (3) leads to
1 d ⎛ dm ⎞
d ⎛ P⎞
⎜R
⎟ = φ ⎜⎜ ⎟⎟ .
dt ⎝ z ⎠
2R dR ⎜⎝ dR ⎟⎠
(8)
The left-hand side of eq. (8) is essentially in the standard
form for the diffusion equation. On the right-hand side, we recall
that P / z = ρRT , so that under isothermal conditions,
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d ⎛ P ⎞ d ( ρRT )
dρ
dρ dP
⎜⎜ ⎟⎟ =
= RT
= RT
.
dt
dt
dP dt
dt ⎝ z ⎠
(9)
But from the definition of compressibility,
Cg =
1 ⎛ dρ ⎞
⎜ ⎟ ,
ρ ⎜⎝ dP ⎟⎠ T
so
⎛ dρ ⎞
⎜⎜ ⎟⎟ = ρCg ,
⎝ dP ⎠ T
(10)
in which case eq. (9) can be written as
d ⎛ P⎞
dP
⎜⎜ ⎟⎟ = R Tρ( P )Cg (P )
.
dt ⎝ z ⎠
dt
(11)
Next we recall from eq. (1) that ρ = P / zRT , and rewrite eq. (11)
as
d ⎛ P ⎞ Cg ( P) P dP
⎜ ⎟=
.
dt ⎜⎝ z ⎟⎠
z (P ) dt
(12)
Combining eq. (12) with eq. (8) yields
1 d ⎛ dm ⎞ φ Cg ( P )P dP
⎜R
⎟=
.
2R dR ⎜⎝ dR ⎟⎠
z (P ) dt
(13)
But by analogy with eq. (7), we have
dm
2k( P )P dP
=
,
μ ( P) z( P) dt
dt
(14)
so that
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Section 9
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dP μ( P )z( P ) dm
=
.
dt
2k ( P) P dt
(15)
Combining eq. (15) with eq. (13) yields
1 d ⎛ dm ⎞ φμ (P )Cg ( P) dm
⎜R
⎟=
.
dt
R dR ⎜⎝ dR ⎟⎠
k (P )
(16)
Eq. (16) is in the form of a standard diffusion equation in radial
coordinates, except that the diffusivity term on the RHS is
pressure-dependent.
As a final step in linearising the equation for gas flow to a
well, we make the approximation that the diffusivity terms on the
right can be replaced by some representative constant value,
such as the value at the mean wellbore pressure encountered
during a well test, i.e.,
1 d ⎛ dm ⎞ φμ (Pm )Cg (Pm ) dm
⎜R
⎟=
,
dt
R dR ⎜⎝ dR ⎟⎠
k( Pm )
(17)
where Pm is defined by eq. (9.2.7).
Eq. (17) is the (approximate) governing equation for the flow
of real gases to a well. If the pressure-dependent terms on the
RHS are evaluated at Pm , we can then use all of the standard
solutions that have been developed for liquid flow, provided that
we use m instead of P as the dependent variable.
The transformation between P and m is made by evaluating
the integral that is given in eq. (4). If the reservoir is stresssensitive, we would need to know k(P) to do this. Usually, we do
not have this information. If we ignore the pressure-dependence
of permeability, then the relationship between P and m becomes
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2003-2004
Flow in Porous Media
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Section 9
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P
P
dP .
0 μ ( P ) z( P )
m ( P) = 2k ∫
(18)
Since z(P) and μ(P) can be measured in the laboratory on
samples of the reservoir gas, this integral can be evaluated a
priori. The result is usually a relationship between m and P in
tabular form, which is used to transform the measured pressures
into values of m that are used in conjunction with our solutions to
eq. (17).
9.4. Non-Darcy flow effects
All of the previous analyses have been based on Darcy’s
law, which states that the flowrate is proportional to the pressure
gradient. Darcy’s law is an empirical law that is known to hold only
at low flowrates, which can be defined, roughly, as flows for which
the Reynolds number is less than one. The Reynolds number is a
dimensionless measure of the relative strengths of inertial forces
relative to viscous forces. Using the definition of Reynolds
number, this condition can be written as
Re =
ρvd
< 1,
μ
(1)
where ρ is the density of the fluid, μ is the viscosity, d is a mean
pore diameter, and v is the mean (microscopic) velocity.
It would be useful to be able to express this criterion in terms
of macroscopically measurable quantities. If we use any of the
common correlations between pore diameter and permeability,
such as the following simplified form of the Kozeny-Carman
equation,
k=
φd 2
96
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≈
φd 2
100
,
(2)
Imperial College London
M.Sc. in Petroleum Engineering
2003-2004
Flow in Porous Media
Dr. R. W. Zimmerman
Section 9
Page 154
we can express the pore diameter as
d = 10 k / φ .
(3)
The criterion for Darcy’s law to be valid can therefore be written
as
v<
μ φ
.
10 ρ k
(4)
Next, we note that the macroscopic flowrate q must be equal
to the microscopic flowrate v, multiplied by the porosity, i.e.,
q = vφ , so that
v=
q
φ
.
(5)
Hence, the criterion (4) for the validity of Darcy’s law can be
written as
μφ 3 / 2
q<
.
10 ρ k
(6)
The total flowrate towards the well is related to the flux by
Q = Aq = 2πRHq ,
(7)
so condition (6) can be written as
Q<
2πRH μφ 3 / 2
10 ρ k
≈
RHμφ 3 / 2
ρ k
,
(8)
which is expressed solely in terms of our usual reservoir
parameters.
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2003-2004
Flow in Porous Media
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Section 9
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Criterion (8) becomes more stringent near the wellbore,
because as a fixed volumetric flowrate gets channelled through a
smaller area, the velocity must increase. If we use “typical”
reservoir values in eq. (8), we would find that this criterion is
usually violated for gas wells in the vicinity of the wellbore, but not
for liquids, unless the flow is through fractures rather than pores.
If the fluid is flowing through fractures, then the actual area
available for flow is smaller, and the velocity of the fluid must be
greater. In this case, it is more likely that the flow will be
turbulent.
If criterion (8) is violated, Darcy’s law must be replaced with
a nonlinear law such as Forchheimer’s equation:
dP μ q
=
+ βρq 2 ,
dR
k
(9)
in which the coefficient β accounts for non-Darcy (inertial) effects.
[In eq. (9), we assume that q is positive if it is towards the well].
Dimensional analysis of eq. (9) shows that the factor β has
dimensions of L-1. As k has dimensions of L2 (recall eq. 2), it is
roughly the case that β ≈ 1/ k . If this is true, then the magnitude
of the nonlinear term in eq. (9) relative to that of the linear term is
βρq 2
βρqk ρq k
ratio =
=
≈
.
( μ q / k)
μ
μ
(10)
The term on the right of eq. (10) is essentially a Reynolds number
based on a length scale of k . Hence, if the Reynolds number is
much less than one, the nonlinear terms in eq. (9) are negligible,
and we recover Darcy’s law.
The Forchheimer equation can be justified heuristically by
noting that ρ q 2 is related to the kinetic energy per unit volume of
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Flow in Porous Media
Dr. R. W. Zimmerman
Section 9
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fluid, and the kinetic energy is not negligible at higher flowrates
(recall eq. 1.1.2). Alternatively, one can interpret the Forchheimer
equation as an empirical law.
Eq. (9) implies that the flowrate under a given pressure
gradient will actually be less than would be predicted on the basis
of Darcy’s law. Another way of interpreting eq. (9) is that, for a
given flowrate, the pressure drop will be larger than that which
would be predicted by Darcy’s law. Hence, non-Darcy effects can
be said to contribute an “additional” pressure drop.
As non-Darcy flow is confined to a region near the wellbore,
its effect during a well test is therefore somewhat similar to a
“skin” effect. In fact, the main consequence of non-Darcy flow
near the wellbore is that the skin factor gets “replaced” by an
apparent skin factor, s ′ , which is given by
s ′ = s + DQ,
(11)
where D is the “non-Darcy skin coefficient”. This coefficient is
related to the coefficient β that appears in Forchheimer’s equation
(see Chierici, p. 241, for the details).
9.5. Klinkenberg effect
Liquids, as well as gases at typical reservoir pressures,
behave like continua, in the sense that we can ignore the motions
of individual molecules, and instead work with mean velocities
that are averaged over a (very) large number of molecules. This
leads to continuum-type theories such as Darcy’s law for flow
through porous media, or the Navier-Stokes equations for fluid
flow on the pore scale.
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2003-2004
Flow in Porous Media
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Section 9
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A fundamental aspect of the Navier-Stokes equations is that
the fluid velocity is assumed to be zero at the boundary with a
solid - such as at the pore walls. This obviously must be true for
the normal component of the velocity, but it is in fact also true for
the tangential component. This boundary condition is the main
reason why the N-S equations average out (“upscale”) to yield
Darcy’s law.
However, this micro-scale boundary condition will not hold
for a gas at very low densities, which is to say at very low
pressures. The reasons are complicated, but they boil down to the
fact that in order for the gas to behave like a continuum, a given
gas molecule must collide much more frequently with other gas
molecules than with the pore walls.
To quantify whether or not this will be the case, we must
consider the concept of the “mean free path”, which is essentially
the mean distance travelled by a molecule between subsequent
collisions with other molecules. According to the kinetic theory of
gases (see Molecular Theory of Gases and Liquids, Hirschfelder,
Curtiss, and Bird, Wiley & Sons, 1954), the mean free path λ is
essentially given by
λ=
1
=
2nπσ 2
kBT
2πσ 2P
,
(1)
where n is the molecular density in molecules/volume, kB is the
Boltzmann constant (i.e., the gas constant per molecule), and σ is
an effective molecular diameter.
If the pore size is smaller than the mean free path, collisions
with the pore walls will be much more frequent than collisions with
other molecules, and the gas will essentially flow through the
pores as individual molecules, rather than as a fluid continuum.
This type of flow is known as “Knudsen flow”, or “slip flow”.
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Flow in Porous Media
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Section 9
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Klinkenberg (1941) assumed that gas flow through a porous
medium could be modelled as Knudsen flow through a capillary
tube, and showed that the “apparent” permeability measured
during gas flow will be related to the “true” absolute permeability k
by
kgas
⎡ 8cλ ⎤
⎥,
= k ⎢1 +
⎢⎣
d ⎦⎥
(2)
where λ is the mean free path, d is the pore diameter, and c ≈ 1 is
a dimensionless coefficient.
If we combine eqs. (1) and (2), we find
kgas
⎡
8c k BT 1 ⎤
⎥.
= k ⎢1+
⎢⎣
2π dσ 2 P ⎥⎦
(3)
If we now use eq. (9.4.3) to relate the pore diameter to the
permeability, we find
kgas
⎡ 4c φ
⎢
= k 1+
⎢⎣ 5 2π
kBT ⎤⎥
.
2
⎥
kσ P⎦
(4)
The second term in the brackets in eq. (4) is the relative
discrepancy due to the Klinkenberg effect. Noting that absolute
temperature, molecular diameter, and √φ will not vary by very
much, for all cases of practical interest, we see that the
parameters that control the strength of the Klinkenberg effect are
pressure and permeability. Eq. (4) shows that the Klinkenberg
effect is enhanced if either (i) the pressure is low, or (ii) the
permeability is low.
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Section 9
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For a given gas flowing through a given rock at a fixed
temperature, we can evaluate all the terms inside the bracket
except P, and call the result P*. Eq. (4) can then be written as
kgas
⎡ P *⎤
⎥,
= k ⎢1+
⎢⎣
P ⎦⎥
(5)
where P* (which is usually written as b) is a characteristic
pressure that has the following significance: the Klinkenberg
effect will have a noticeable effect on the measured permeability if
(roughly) P < 10P*. Alternatively, we can say that the Klinkenberg
effect will be negligible if P > 10P*.
To quantify the magnitude of the Klinkenberg effect, consider
a rock having a porosity of φ = 0.10, with a gas flowing through it
at 300°K. Boltzmann’s constant is 1.38×10-23 J/°K, and typical
molecular diameters are on the order of 4Å, so we find from eq.
(4) that
for k = 10 mD:
P* ≈ 15 kPa ≈ 2 psi ,
for k = 1000 mD:
P* ≈ 1.5 kPa ≈ 0.2 psi .
We see that reservoir pressures will be much larger than P*
for typical reservoir rocks, and so the Klinkenberg effect is usually
negligible for flow in the reservoir. However, laboratory
permeability tests are often conducted using gas at low pressures
(because these measurements are quicker, safer and cheaper
than measurements made at simulated reservoir pressures), and
so the lab-measured permeabilities must be corrected to eliminate
the Klinkenberg effect. (Note that it is the actual permeability k
that we want to know, not the apparent gas permeability kgas,
which is an experimental artefact!).
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Imperial College London
M.Sc. in Petroleum Engineering
2003-2004
Flow in Porous Media
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Section 9
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This is done by measuring kgas over a range of pressures,
and then plotting the results as a function of 1/P. According to eq.
(5), the data should fall on a straight line with a positive slope.
Often, we do not make measurements at sufficiently high
pressures to be able to avoid the Klinkenberg effect. But, if we fit
a straight line through the k vs. 1/P data, and extrapolate this line
back to 1/P = 0, we find the absolute permeability, k. This
procedure is illustrated in Fig. 9.5.1 on the next page, taken from
Chierici, p. 78. Note that 1/P = 0 corresponds to very large values
of the pressure, at which the gas does behave like a continuum.
Klinkenberg Effect
Permeability, kGAS (mD)
300
250
Kgas data
Fitted line
200
150
100
50
kabs = kgas (line, at 1/P = 0) = 30 mD
0
0
200
400
600
800
1000
1/Pressure (atm -1)
Fig. 9.5.1. Apparent “gas permeability” at low pressures,
extrapolated to 1/P = 0 to find true permeability.
Note that the parameter P* contains k in it as part of its
definition, and we don’t know k before we measure it! However,
the extrapolation procedure does not require knowledge of k. The
data can be plotted against 1/P in any units, dimensionless or not;
extrapolation to 1/P = 0 will give the “actual” permeability.
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