0
FLUID FLOW THROUGH POROUS MEDIA
In 1856, Darcy investigated the flow of water through sand filters
for water purification purposes. The following assumptions are
implicit in Darcy’s experiment: Single-phase flow (only water),
Homogeneous porous medium (sand), Vertical flow, Non-
reactive fluid (water), and Single geometry.
Hassan S. Naji,
Professor,
www.hbsnumerics.com
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1. Darcy's Law Derived
In 1856, Darcy investigated the flow of water through sand filters for water
purification purposes. His experimental apparatus is shown below:
Where 𝑞 is the volume flow rate of water downward through the cylindrical sand
pack. The sand pack has a length 𝐿 and a cross-sectional area 𝐴. ℎ1 is the height
above a datum of water in a manometer located at the input face. ℎ2 is the height
above a datum of water in a manometer located at the output face. The following
assumptions are implicit in Darcy’s experiment:
San
d P
ack
A
L
Water
Water
Water
h1
h2
h1-h2
Datum
Input manometerOutput manometer
Water in
Water outat a rate q
Flo
w d
irecti
on
1
2
2
Single-phase flow (only water)
Homogeneous porous medium (sand)
Vertical flow.
Non-reactive fluid (water)
Single geometry.
From this experiment, Darcy concluded the following points: The volume flow rate is
directly proportional to the difference of water level in the two manometers; i.e.:
𝑞 ∝ ℎ1 − ℎ2 (1)
The volume flow rate is directly proportional to the cross-sectional area of the sand
pack; i.e.:
𝑞 ∝ 𝐴 (2)
The volume flow rate is inversely proportional to the length of the sand pack; i.e.:
𝑞 ∝1
𝐿 (3)
Thus we write:
𝑞 = 𝐶𝐴
𝐿(ℎ1 − ℎ2) (4)
Where 𝐶 is the proportionality constant, which depends on the rock and fluid
properties. For the fluid effect, 𝐶 is directly proportional to the fluid specific weight;
i.e.
𝐶 ∝ 𝛾 (5)
It is inversely proportional to the fluid viscosity; i.e.:
𝐶 ∝1
𝜇 (6)
Thus:
𝐶 ∝𝛾
𝜇 (7)
For the rock effect, 𝐶 is directly proportional to the square of grain size; i.e.:
𝐶 ∝ (𝑔𝑟𝑎𝑖𝑛 𝑠𝑖𝑧𝑒)2 = 𝑑2 (8)
3
It is inversely proportional to tortuosity; i.e.:
𝐶 ∝1
𝑡𝑜𝑟𝑡𝑢𝑜𝑠𝑖𝑡𝑦=1
𝜏 (9)
and inversely proportional to the specific surface; i.e.:
𝐶 ∝1
𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑠𝑢𝑟𝑓𝑎𝑐𝑒=1
𝑆𝑆 (10)
where 𝑆𝑆 is given by:
𝑆𝑆 =𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎
𝑏𝑢𝑙𝑘 𝑣𝑜𝑙𝑢𝑚𝑒 (11)
Combine the above un-measurable rock properties into one property, call it
permeability, and denote it by 𝑘, we get:
𝑞 = 𝑘𝛾
𝜇
𝐴
𝐿(ℎ1 − ℎ2) (12)
Since:
𝑞 = 𝜈𝐴 (13)
Thus we can write:
𝜈 =𝑞
𝐴= 𝑘
𝛾
𝜇
(ℎ1 − ℎ2)
𝐿 (14)
Now, let us consider the more realistic flow; i.e. the tilted flow for the same sand
pack:
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Note that the fluid flows from point 1 to point 2, which means that the pressure at
point 1 is higher than the pressure at point 2. Since:
ℎ1 = 𝐷1 −𝑝1𝛾 & ℎ2 = 𝐷2 −
𝑝2𝛾
(15)
Thus:
ℎ1 − ℎ2 = (𝐷1 −𝑝1𝛾) − (𝐷2 −
𝑝2𝛾) = (𝐷1 − 𝐷2) −
1
𝛾(𝑝1 − 𝑝2) (16)
which can be written in a difference form as:
ℎ1 − ℎ2 = ∆𝐷 −1
𝛾∆𝑝 (17)
Substituting (17) into (14) and rearranging yields:
𝜈 = 𝑘𝛾
𝜇(∆𝐷
𝐿−1
𝛾
∆𝑝
𝐿) = −
𝑘
𝜇(∆𝑝
𝐿− 𝛾
∆𝐷
𝐿) (18)
Flow Dire
ction
2
1
A
L
Datum
P2, D2
P1, D1
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The differential form of Darcy’s equation for single-phase flow is written as follows:
𝜈 = −𝑘
𝜇(𝜕𝑝
𝜕𝐿− 𝛾
𝜕𝐷
𝜕𝐿) (19)
For multi-phase flow, we write Darcy’s equation as follows:
𝜈 = −𝑘 (𝑘𝑟𝜇) (𝜕𝑝
𝜕𝐿− 𝛾
𝜕𝐷
𝜕𝐿) (20)
More conveniently, the compact differential form of Darcy’s equation is written as
follows:
�⃗� = −𝑘 (𝑘𝑟𝜇) (∇𝑝 − 𝛾∇𝐷) (21)
Where 𝑘 is the absolute permeability tensor which must be determined
experimentally. It is written as follows:
𝑘 = [
𝑘𝑥𝑥 𝑘𝑥𝑦 𝑘𝑥𝑧𝑘𝑦𝑥 𝑘𝑦𝑦 𝑘𝑦𝑧𝑘𝑧𝑥 𝑘𝑧𝑦 𝑘𝑧𝑧
] (22)
Substitute (22) into (21), yields:
{
𝜈𝑥𝜈𝑦𝜈𝑧} = −(
𝑘𝑟𝜇) [
𝑘𝑥𝑥 𝑘𝑥𝑦 𝑘𝑥𝑧𝑘𝑦𝑥 𝑘𝑦𝑦 𝑘𝑦𝑧𝑘𝑧𝑥 𝑘𝑧𝑦 𝑘𝑧𝑧
]
{
(𝜕𝑝
𝜕𝑥− 𝛾
𝜕𝐷
𝜕𝑥)
(𝜕𝑝
𝜕𝑦− 𝛾
𝜕𝐷
𝜕𝑦)
(𝜕𝑝
𝜕𝑧− 𝛾
𝜕𝐷
𝜕𝑧)}
(23)
Solve for velocity components yields:
𝜈𝑥 = −(𝑘𝑟𝜇) [𝑘𝑥𝑥 (
𝜕𝑝
𝜕𝑥− 𝛾
𝜕𝐷
𝜕𝑥) + 𝑘𝑥𝑦 (
𝜕𝑝
𝜕𝑦− 𝛾
𝜕𝐷
𝜕𝑦) + 𝑘𝑥𝑧 (
𝜕𝑝
𝜕𝑧− 𝛾
𝜕𝐷
𝜕𝑧)] (24)
𝜈𝑦 = −(𝑘𝑟𝜇) [𝑘𝑦𝑥 (
𝜕𝑝
𝜕𝑥− 𝛾
𝜕𝐷
𝜕𝑥) + 𝑘𝑦𝑦 (
𝜕𝑝
𝜕𝑦− 𝛾
𝜕𝐷
𝜕𝑦) + 𝑘𝑦𝑧 (
𝜕𝑝
𝜕𝑧− 𝛾
𝜕𝐷
𝜕𝑧)] (25)
𝜈𝑧 = −(𝑘𝑟𝜇) [𝑘𝑧𝑥 (
𝜕𝑝
𝜕𝑥− 𝛾
𝜕𝐷
𝜕𝑥) + 𝑘𝑧𝑦 (
𝜕𝑝
𝜕𝑦− 𝛾
𝜕𝐷
𝜕𝑦) + 𝑘𝑧𝑧 (
𝜕𝑝
𝜕𝑧− 𝛾
𝜕𝐷
𝜕𝑧)] (26)
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In most practical problems, it is necessary to assume that K is a diagonal tensor; i.e.:
𝑘 = [
𝑘𝑥𝑥 0 00 𝑘𝑦𝑦 0
0 0 𝑘𝑧𝑧
] (27)
Thus the velocity components of equations (24), (25) and (26) become:
𝜈𝑥 = −𝑘𝑥𝑥 (𝑘𝑟𝜇) (𝜕𝑝
𝜕𝑥− 𝛾
𝜕𝐷
𝜕𝑥) (28)
𝜈𝑦 = −𝑘𝑦𝑦 (𝑘𝑟𝜇) (𝜕𝑝
𝜕𝑦− 𝛾
𝜕𝐷
𝜕𝑦) (29)
𝜈𝑧 = −𝑘𝑧𝑧 (𝑘𝑟𝜇) (𝜕𝑝
𝜕𝑧− 𝛾
𝜕𝐷
𝜕𝑧) (30)
2. Darcy's Law Unit Analysis
Let's start with:
𝑞 =𝑘𝐴
𝜇
∆𝑝
∆𝐿 (31)
Solving for 𝑘 yields:
𝑘 =𝑞𝜇
𝐴
∆𝐿
∆𝑝 (32)
Where 𝑞 in cc/sec, 𝜇 in cp, ∆𝐿 in cm, 𝐴 in cm2, and ∆𝑝 in atm, then 𝑘 is in the units of
Darcy. Therefore, we simply write:
1𝐷𝑎𝑟𝑐𝑦 =(1𝑐𝑚3)(1𝑐𝑝)(1𝑐𝑚)
(1𝑠)(1𝑐𝑚2)(1𝑎𝑡𝑚)=(1𝑐𝑚2)(1𝑐𝑝)
(1𝑠)(1𝑎𝑡𝑚) (33)
It can be shown, however, that the Darcy is a unit of area as follows:
1𝑎𝑡𝑚 = 1,013,250𝑑𝑦𝑛𝑒
𝑐𝑚2 (34)
1𝑐𝑝 = 0.01𝑑𝑦𝑛𝑒 𝑠
𝑐𝑚2 (35)
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Plugging (34) and (35) into (33) yields:
1𝐷𝑎𝑟𝑐𝑦 =(1𝑐𝑚2) (0.01
𝑑𝑦𝑛𝑒 𝑠𝑐𝑚2 )
(1𝑠) (1,013,250𝑑𝑦𝑛𝑒𝑐𝑚2 )
=1
101,325,000𝑐𝑚2 (36)
Rearranging yields:
1𝑐𝑚2 = 101,325,000 𝐷 (37)
To get the conversion factor of Darcy's Law from laboratory to field units, we write:
1𝐷𝑎𝑟𝑐𝑦 =(1𝑐𝑚3)(1𝑐𝑝)(1𝑐𝑚)
(1𝑠)(1𝑐𝑚2)(1𝑎𝑡𝑚) (38)
1𝐷𝑎𝑟𝑐𝑦 = 1000 𝑚𝐷 (39)
1𝑐𝑚 = (1
2.54𝑥12) 𝑓𝑡 =
1
30.48𝑓𝑡 (40)
1𝑐𝑚2 = (1
2.54𝑥12)2
𝑓𝑡2 =1
929.0304𝑓𝑡2 (41)
1𝑐𝑚3 = (1
2.54𝑥12)3
𝑓𝑡3 =1
28316.846592𝑓𝑡3 (42)
1𝑠 =1
86,400𝑑𝑎𝑦 (43)
1𝑎𝑡𝑚 = 14.696 𝑝𝑠𝑖 (44)
Plugging (39) to (44) into (38) yields:
(1000)𝑚𝐷 =(
128316.846592
)𝑓𝑡3 (1
5.6146)𝑏𝑏𝑙𝑓𝑡3
(1𝑐𝑝) (1
30.48) 𝑓𝑡
(1
86,400) 𝑑𝑎𝑦 (1
929.0304) 𝑓𝑡2(14.696) 𝑝𝑠𝑖
(45)
This yield:
1𝑚𝐷 =
((86,400)(929.0304)
(1000)(28316.846592)(5.6146)(30.48)(14.696))(𝑏𝑏𝑙)(𝑐𝑝)(𝑓𝑡)
(𝑑𝑎𝑦)(𝑓𝑡2)(𝑝𝑠𝑖)
= 0.00112711(𝑏𝑏𝑙)(𝑐𝑝)(𝑓𝑡)
(𝑑𝑎𝑦)(𝑓𝑡2)(𝑝𝑠𝑖)
(46)
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3. Primary Reservoir Characteristics
The primary criteria for characterizing petroleum reservoirs are:
A. Types of fluids in the reservoir
B. Flow regimes
C. Reservoir geometry
D. Number of flowing fluids in the reservoir
A. Types of Fluids in the Reservoir
In terms of compressibility, fluids in the reservoir are classified as:
Incompressible Fluids; i.e. 𝜕𝑉
𝜕𝑝= 0 and
𝜕𝜌
𝜕𝑝= 0
Slightly Compressible Fluids; i.e. 𝜕𝑉
𝜕𝑝= 𝐶 and
𝜕𝜌
𝜕𝑝= 𝐶
Compressible Fluids; i.e. 𝜕𝑉
𝜕𝑝= 𝑓(𝑝) and
𝜕𝜌
𝜕𝑝= 𝑓(𝑝)
B. Flow Regimes in the Reservoir
There are basically three types of flow regimes:
Steady-State Flow; i.e. (𝜕𝑝
𝜕𝑡)𝑖= 0
Pseudo Steady-State Flow; i.e. (𝜕𝑝
𝜕𝑡)𝑖= constant
Unsteady-State Flow; i.e. (𝜕𝑝
𝜕𝑡) = 𝑓(𝑖, 𝑡)
C. Reservoir Geometry
For many engineering purposes, the actual flow geometry may be represented by one
of the following flow geometries:
Linear Flow
Radial Flow
Spherical and Hemispherical Flow
D. Number of Flowing Fluids in the Reservoir
There are three cases of flowing systems:
Single phase flow (oil, gas, or water)
Two phase flow (oil-gas, oil-water, or gas-water)
Three phase flow (oil, gas, and water)
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4. Isothermal Compressibility
In terms of fluid volume:
𝑐 = −1
𝑉
𝜕𝑉
𝜕𝑝 (47)
In terms of fluid density:
𝑐 =1
𝜌
𝜕𝜌
𝜕𝑝 (48)
5. Slightly Compressible Fluids
For slightly compressible fluids such as oils and waters, 𝜕𝑐
𝜕𝑝= 𝐶; i.e.
−𝑐∫ 𝜕𝑝𝑝
𝑝𝑟𝑒𝑓
= ∫𝜕𝑉
𝑉
𝑉
𝑉𝑟𝑒𝑓
(49)
𝑐(𝑝𝑟𝑒𝑓 − 𝑝) = 𝑙𝑛𝑉
𝑉𝑟𝑒𝑓 (50)
𝑉 = 𝑉𝑟𝑒𝑓𝑒𝑐(𝑝𝑟𝑒𝑓−𝑝) (51)
In terms of fluid density:
𝑐 ∫ 𝜕𝑝𝑝
𝑝𝑟𝑒𝑓
= ∫𝜕𝜌
𝜌
𝜌
𝜌𝑟𝑒𝑓
(52)
−𝑐(𝑝𝑟𝑒𝑓 − 𝑝) = 𝑙𝑛𝜌
𝜌𝑟𝑒𝑓 (53)
𝜌 = 𝜌𝑟𝑒𝑓𝑒−𝑐(𝑝𝑟𝑒𝑓−𝑝) (54)
Since we write:
𝑒𝑥 = 1 + 𝑥 +𝑥2
2!+𝑥3
3!+ ⋯+
𝑥𝑛
𝑛! (55)
Thus, in terms of fluid volume, we may write:
𝑉 ≈ 𝑉𝑟𝑒𝑓[1 + 𝑐(𝑝𝑟𝑒𝑓 − 𝑝)] (56)
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Similarly, in terms of fluid density, we may write:
𝜌 ≈ 𝜌𝑟𝑒𝑓[1 − 𝑐(𝑝𝑟𝑒𝑓 − 𝑝)] (57)
Similarly, in terms of fluid flow rate, we may write:
𝑞 ≈ 𝑞𝑟𝑒𝑓[1 − 𝑐(𝑝𝑟𝑒𝑓 − 𝑝)] (58)
6. Highly Compressible Fluids
For highly compressible fluids such as gases, the Universal Gas Law is written as:
𝑝𝑉 = 𝑍𝑛𝑅𝑇 (59)
𝑉 =𝑍𝑛𝑅𝑇
𝑝 (60)
𝜕𝑉
𝜕𝑝=𝑝 (𝑛𝑅𝑇
𝜕𝑍𝜕𝑝) − 𝑍𝑛𝑅𝑇
𝑝2
(61)
𝜕𝑉
𝜕𝑝=𝑛𝑅𝑇
𝑝2(𝑝𝜕𝑍
𝜕𝑝− 𝑍) (62)
𝜕𝑉
𝜕𝑝= 𝑛𝑅𝑇 (
1
𝑝
𝜕𝑍
𝜕𝑝−𝑍
𝑝2) (63)
Plugging (60) and (63) into the definition of compressibility (47) yields:
𝑐 = −𝑝𝑛𝑅𝑇
𝑍𝑛𝑅𝑇(1
𝑝
𝜕𝑍
𝜕𝑝−𝑍
𝑝2) (64)
Rearranging yields:
𝑐 = −𝑝
𝑍(1
𝑝
𝜕𝑍
𝜕𝑝−𝑍
𝑝2) (65)
𝑐 = −(1
𝑍
𝜕𝑍
𝜕𝑝−1
𝑝) (66)
𝑐 =1
𝑝−1
𝑍
𝜕𝑍
𝜕𝑝 (67)
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7. Fluid Flow Equations
The basic Darcy's Law is written as follows:
𝑞 =𝑘𝐴
𝜇𝐵
𝜕𝑝
𝜕𝑙 (68)
𝜐 = 𝑞
𝐴=𝑘
𝜇𝐵
𝜕𝑝
𝜕𝑙 (69)
The actual velocity model – Poiseuille's Law, is written as:
𝜈 =𝑟2
8𝜇𝐵
𝜕𝑝
𝜕𝑙 (70)
𝑘
𝜙=𝑟2
8 (71)
𝑘 =𝜙 𝑟2
8 (72)
8. Linear Horizontal Flow of Incompressible Fluid – Lab
For linear horizontal flow, Darcy's Law for laboratory units is written as follows:
𝑞 =𝑘𝐴
𝜇
𝜕𝑝
𝜕𝑙 (73)
∫ 𝑞𝜕𝑙𝑙2
𝑙1
=𝑘𝐴
𝜇∫ 𝜕𝑝𝑝2
𝑝1
(74)
𝑞(𝑙2 − 𝑙1) =𝑘𝐴
𝜇(𝑝2 − 𝑝1) (75)
𝑞 =𝑘𝐴
𝜇
∆𝑝
∆𝑙 (76)
In terms of fluid potential, if point 𝑖 is below the datum level, we write:
Φ𝑖 = 𝑝𝑖 − 𝛾∆𝐷𝑖 (77)
Φ𝑖 = 𝑝𝑖 −𝜌
144∆𝐷𝑖 (78)
12
Φ𝑖 = 𝑝𝑖 − 0.433 𝑆𝐺 ∆𝐷𝑖 (79)
If point 𝑖 is above the datum level, we write:
Φ𝑖 = 𝑝𝑖 + 𝛾∆𝐷𝑖 (80)
Φ𝑖 = 𝑝𝑖 +𝜌
144∆𝐷𝑖 (81)
Φ𝑖 = 𝑝𝑖 + 0.433 𝑆𝐺 ∆𝐷𝑖 (82)
Therefor Darcy's equation is written as:
𝑞 = 𝑘
𝜇
𝜕Φ
𝜕𝑙=
𝑘
𝜇𝐵
𝛥Φ
∆𝑙 (83)
9. Linear Horizontal Flow of Incompressible Fluids - Field
Start with the basic Darcy's Law:
𝑞 =𝑘𝐴
𝜇𝐵
𝜕𝑝
𝜕𝑙 (84)
For linear horizontal flow, Darcy's Law, for field units, is written as follows:
∫ 𝜕𝑙𝑙2
𝑙1
= 0.001127𝑘𝐴
𝑞𝜇𝐵∫ 𝜕𝑝𝑝2
𝑝1
(85)
(𝑙2 − 𝑙1) = 0.001127𝑘𝐴
𝑞𝜇𝐵(𝑝2 − 𝑝1) (86)
𝑞 = 0.001127𝑘𝐴
𝜇𝐵
∆𝑝
∆𝑙= 0.001127
𝑘𝐴
𝜇𝐵
∆Φ
∆𝑙 (87)
Since:
𝑉 ≈ 𝑉𝑟𝑒𝑓[1 + 𝑐(𝑝𝑟𝑒𝑓 − 𝑝)] (88)
This equation can be written in terms of flow rate as follows:
𝑞 ≈ 𝑞𝑟𝑒𝑓[1 + 𝑐(𝑝𝑟𝑒𝑓 − 𝑝)] (89)
13
Therefore:
𝑞𝑟𝑒𝑓[1 + 𝑐(𝑝𝑟𝑒𝑓 − 𝑝)] = 0.001127𝑘𝐴
𝜇𝐵
∆𝑝
∆𝑙 (90)
∫ 𝜕𝑙𝐿
0
= 0.001127𝑘𝐴
𝑞𝑟𝑒𝑓𝜇𝐵∫
𝜕𝑝
[1 + 𝑐(𝑝𝑟𝑒𝑓 − 𝑝)]
𝑝2
𝑝1
(91)
Integrating yields:
𝑞𝑟𝑒𝑓 = [0.001127𝑘𝐴
𝜇𝑐𝐿𝐵] 𝑙𝑛 [
[1 + 𝑐(𝑝𝑟𝑒𝑓 − 𝑝2)]
[1 + 𝑐(𝑝𝑟𝑒𝑓 − 𝑝1)]] (92)
Selecting the upstream pressure 𝑝1 as the reference pressure, 𝑝𝑟𝑒𝑓 gives the flow rate
at point 1 as:
𝑞1 = [0.001127𝑘𝐴
𝜇𝑐𝐿𝐵] 𝑙𝑛[1 + 𝑐(𝑝1 − 𝑝2)] (93)
Selecting the downstream pressure 𝑝2 as the reference pressure, 𝑝𝑟𝑒𝑓 gives the flow
rate at point 2 as:
𝑞2 = [0.001127𝑘𝐴
𝜇𝑐𝐿𝐵] 𝑙𝑛 [
1
1 + 𝑐(𝑝2 − 𝑝1)] (94)
10. Linear Tilted Flow of Incompressible Fluids
Start with the basic Darcy's Law:
𝑞 =𝑘𝐴
𝜇𝐵
𝜕𝑝
𝜕𝑙 (95)
For tilted flow, Darcy's Law is written as follows:
𝑞 = −𝑘𝐴
𝜇𝐵(𝜕𝑝
𝜕𝑙± 𝛾
𝜕𝐷
𝜕𝑙) (96)
𝑞 = −0.001127𝑘𝐴
𝜇𝐵(∆𝑝
∆𝑙± 𝛾
∆𝐷
∆𝑙) (97)
𝑞 = −0.001127𝑘𝐴
𝜇𝐵(∆𝑝
∆𝑙± 0.433 𝑆𝐺 sin 𝛼) (98)
14
11. Radial Flow of Incompressible Fluids
Start with the basic Darcy's Law:
𝑞 =𝑘𝐴
𝜇𝐵
𝜕𝑝
𝜕𝑙 (99)
Since:
𝐴 = 2𝜋𝑟ℎ (100)
Thus, we write:
𝑞 = (0.001127)(2𝜋)𝑘ℎ
𝜇𝐵
𝑟
𝜕𝑟𝜕𝑝 (101)
𝜕𝑟
𝑟= 0.00708
𝑘ℎ
𝑞𝜇𝐵𝜕𝑝 (102)
∫𝜕𝑟
𝑟
𝑟𝑒
𝑟𝑤
= 0.00708∫𝑘ℎ
𝑞𝜇𝐵𝜕𝑝
𝑝𝑒
𝑝𝑤
(103)
ln𝑟𝑒𝑟𝑤= 0.00708
𝑘ℎ
𝑞𝜇𝐵(𝑝𝑒 − 𝑝𝑤) (104)
𝑞 = 0.00708𝑘ℎ
𝜇𝐵
(𝑝𝑒 − 𝑝𝑤)
ln𝑟𝑒𝑟𝑤
(105)
Including the skin effect, the equation is written as follows:
𝑞 = 0.00708𝑘ℎ
𝜇𝐵
(𝑝𝑒 − 𝑝𝑤)
[ln𝑟𝑒𝑟𝑤− 0.75 + 𝑠]
(106)
This equation may be arranged as follows:
𝑝 = 𝑝𝑤 + 141.22𝑞𝜇𝐵
𝑘ℎ[ln
𝑟𝑒𝑟𝑤− 0.75 + 𝑠] (107)
15
12. Radial Flow of Slightly Compressible Fluids
Start with the basic Darcy's Law:
𝑞 =𝑘𝐴
𝜇𝐵
𝜕𝑝
𝜕𝑙 (108)
Since:
𝐴 = 2𝜋𝑟ℎ (109)
Thus, we write:
𝑞 = (0.001127)(2𝜋)𝑘ℎ
𝜇𝐵
𝑟
𝜕𝑟𝜕𝑝 (110)
𝜕𝑟
𝑟= 0.00708
𝑘ℎ
𝑞𝜇𝐵𝜕𝑝 (111)
Since:
𝑉 ≈ 𝑉𝑟𝑒𝑓[1 + 𝑐(𝑝𝑟𝑒𝑓 − 𝑝)] (112)
This equation can be written in terms of flow rate as follows:
𝑞 ≈ 𝑞𝑟𝑒𝑓[1 + 𝑐(𝑝𝑟𝑒𝑓 − 𝑝)] (113)
Therefore:
𝑞𝑟𝑒𝑓[1 + 𝑐(𝑝𝑟𝑒𝑓 − 𝑝)] = 0.00708𝑘ℎ
𝜇𝐵
𝑟
𝜕𝑟𝜕𝑝 (114)
∫𝜕𝑟
𝑟
𝑟
𝑟𝑤
= 0.00708𝑘ℎ
𝑞𝑟𝑒𝑓𝜇𝐵∫
𝜕𝑝
[1 + 𝑐(𝑝𝑟𝑒𝑓 − 𝑝)]
𝑝
𝑝𝑤
(115)
Integrating yields:
𝑞𝑟𝑒𝑓 = [0.00708𝑘ℎ
𝜇𝑐𝐵𝑙𝑛 (𝑟𝑟𝑤)] 𝑙𝑛 [
[1 + 𝑐(𝑝𝑟𝑒𝑓 − 𝑝)]
[1 + 𝑐(𝑝𝑟𝑒𝑓 − 𝑝𝑤)]] (116)
Selecting the bottom-hole flowing pressure 𝑝𝑤 as the reference pressure, 𝑝𝑟𝑒𝑓 gives
the flow rate at the well bore as:
16
𝑞𝑜 = [0.00708𝑘ℎ
𝜇𝑜𝑐𝑜𝐵𝑜] [𝑙𝑛[1 + 𝑐𝑜(𝑝 − 𝑝𝑤)]
𝑙𝑛 (𝑟𝑟𝑤)
] (117)
Selecting the downstream pressure 𝑝2 as the reference pressure, 𝑝𝑟𝑒𝑓 gives the flow
rate at point 2 as:
𝑞2 = [0.001127𝑘𝐴
𝜇𝑐𝐿𝐵] 𝑙𝑛 [
1
1 + 𝑐(𝑝2 − 𝑝1)] (118)
13. Linear Horizontal Flow of Compressible Fluids
Start with the basic Darcy's Law:
𝑞 =𝑘𝐴
𝜇𝐵
𝜕𝑝
𝜕𝑙 (119)
Since:
𝑝𝑉 = 𝑧𝑛𝑅𝑇 (120)
𝑉 =𝑧𝑛𝑅𝑇
𝑝 (121)
𝐵 =𝑉𝑔,𝑟
𝑉𝑔,𝑠=𝑍𝑟𝑛𝑅𝑇𝑟𝑝𝑟
𝑝𝑠𝑍𝑠𝑛𝑅𝑇𝑠
=𝑍𝑟𝑇𝑟𝑝𝑟
𝑝𝑠𝑍𝑠𝑇𝑠
=14.696
(1)(519.67)
𝑍𝑇
𝑝= 0.02828
𝑍𝑇
𝑝 (122)
Thus, we write:
𝑞 =(0.001127)(5.6146)
0.02828
𝑘𝐴
𝜇𝑍𝑇𝑝𝜕𝑝
𝜕𝑙 (123)
𝑞𝜕𝑙 = 0.223750148𝑘𝐴
𝜇𝑍𝑇𝑝𝜕𝑝 (124)
∫ 𝜕𝑙𝑙2
𝑙1
= 0.223750148𝑘𝐴
𝑞�̅��̅�𝑇∫ 𝑝𝜕𝑝𝑝2
𝑝1
(125)
𝑞∆𝑙 = 0.111875𝑘𝐴
�̅��̅�𝑇(𝑝2
2 − 𝑝12) (126)
17
𝑞 = (111.875)10−3𝑘𝐴
�̅��̅�𝑇
(𝑝22 − 𝑝1
2)
∆𝑙 𝑆𝐶𝐹 (127)
𝑞 = (111.875)10−6𝑘𝐴
�̅��̅�𝑇
(𝑝22 − 𝑝1
2)
∆𝑙 𝑀𝑆𝐶𝐹 (128)
𝑞 = (111.875)10−9𝑘𝐴
�̅��̅�𝑇
(𝑝22 − 𝑝1
2)
∆𝑙 𝑀𝑀𝑆𝐶𝐹 (129)
14. Radial Flow of Compressible Fluids
Start with the basic Darcy's Law:
𝑞 =𝑘𝐴
𝜇𝐵
𝜕𝑝
𝜕𝑙 (130)
Since:
𝐴 = 2𝜋𝑟ℎ (131)
𝐵 =𝑉𝑔,𝑟
𝑉𝑔,𝑠=𝑍𝑟𝑛𝑅𝑇𝑟𝑝𝑟
𝑝𝑠𝑍𝑠𝑛𝑅𝑇𝑠
=𝑍𝑟𝑇𝑟𝑝𝑟
𝑝𝑠𝑍𝑠𝑇𝑠
=14.696
(1)(519.67)
𝑍𝑇
𝑝= 0.02828
𝑍𝑇
𝑝 (132)
Thus, we write:
𝑞 =(0.001127)(5.6146)(2𝜋)
0.02828
𝑟
𝜕𝑟
𝑘ℎ
𝜇𝑍𝑇𝑝𝜕𝑝 (133)
𝑞 = 1.406𝑟
𝜕𝑟
𝑘ℎ
𝜇𝑍𝑇𝑝𝜕𝑝 (134)
𝑞∫𝜕𝑟
𝑟
𝑟𝑒
𝑟𝑤
= 1.406∫𝑘ℎ
𝜇𝑍𝑇𝑝𝜕𝑝
𝑝𝑒
𝑝𝑤
(135)
𝑞∫𝜕𝑟
𝑟
𝑟𝑒
𝑟𝑤
= 1.406𝑘ℎ
�̅��̅�𝑇∫ 𝑝𝜕𝑝𝑝𝑒
𝑝𝑤
(136)
𝑞 ln𝑟𝑒𝑟𝑤= 0.703
𝑘ℎ
�̅��̅�𝑇(𝑝𝑒
2 − 𝑝𝑤2 ) (137)
𝑞 = 0.703𝑘ℎ
�̅��̅�𝑇
(𝑝𝑒2 − 𝑝𝑤
2 )
ln𝑟𝑒𝑟𝑤
𝑆𝐶𝐹 (138)
18
𝑞 = (703)10−6𝑘ℎ
�̅��̅�𝑇
(𝑝𝑒2 − 𝑝𝑤
2 )
ln𝑟𝑒𝑟𝑤
𝑀𝑆𝐶𝐹 (139)
𝑞 = (703)10−9𝑘ℎ
�̅��̅�𝑇
(𝑝𝑒2 − 𝑝𝑤
2 )
ln𝑟𝑒𝑟𝑤
𝑀𝑀𝑆𝐶𝐹 (140)
Including the skin effect, the equation is written as follows:
𝑞 = (703)10−9𝑘ℎ
�̅��̅�𝑇
(𝑝𝑒2 − 𝑝𝑤
2 )
[ln𝑟𝑒𝑟𝑤− 0.75 + 𝑠]
𝑀𝑀𝑆𝐶𝐹 (141)
However, the above formulation may be rigorously formulated as follows:
𝑞∫𝜕𝑟
𝑟
𝑟𝑒
𝑟𝑤
= 1.406∫𝑘ℎ
𝜇𝑍𝑇𝑝𝜕𝑝
𝑝𝑒
𝑝𝑤
(142)
𝑞 = 0.703𝑘ℎ
𝑇 ln𝑟𝑟𝑤
(2∫𝑝
𝜇𝑍𝜕𝑝
𝑝
𝑝𝑤
) (143)
The term:
2∫𝑝
𝜇𝑍𝜕𝑝
𝑝
𝑝𝑤
= 2∫𝑝
𝜇𝑍𝜕𝑝
𝑝
0
− 2∫𝑝
𝜇𝑍𝜕𝑝
𝑝𝑤
0
(144)
Defining the real gas pseudo-potential as:
𝑚(𝑝) = 𝜓 = 2∫𝑝
𝜇𝑍𝜕𝑝
𝑝
0
(145)
Similarly:
𝑚𝑤(𝑝) = 𝜓𝑤 = 2∫𝑝
𝜇𝑍𝜕𝑝
𝑝𝑤
0
(146)
Thus:
2∫𝑝
𝜇𝑍𝜕𝑝
𝑝
𝑝𝑤
= 𝜓 − 𝜓𝑤 (147)
𝑞 = 0.703𝑘ℎ
𝑇 ln𝑟𝑟𝑤
(𝜓 − 𝜓𝑤) (148)
19
15. Continuity Equation
The continuity equation is expressed as follows:
[𝑀𝑎𝑠𝑠]𝑖𝑛 − [𝑀𝑎𝑠𝑠]𝑜𝑢𝑡 = [𝑀𝑎𝑠𝑠]𝑎𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑒𝑑 (149)
Since:
𝑀𝑎𝑠𝑠 = 𝑉𝜌 = ∆𝑡𝑞𝜌 = ∆𝑡𝐴𝜈𝜌 (150)
Therefore:
[𝑀𝑎𝑠𝑠]𝑖𝑛 = ∆𝑡[𝐴𝜈𝜌]𝑟+d𝑟 (151)
[𝑀𝑎𝑠𝑠]𝑜𝑢𝑡 = ∆𝑡[𝐴𝜈𝜌]𝑟 (152)
[𝑀𝑎𝑠𝑠]𝑎𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑒𝑑 = 𝑑𝑉[(𝜙𝜌)𝑡+∆𝑡 − (𝜙𝜌)𝑡] (153)
Since:
𝐴𝑟 = 2𝜋𝑟ℎ (154)
𝐴𝑟+d𝑟 = 2𝜋(𝑟 + 𝑑𝑟)ℎ (155)
𝑉 = 𝜋𝑟2ℎ (156)
𝑑𝑉
𝑑𝑟= 2𝜋𝑟ℎ (157)
𝑑𝑉 = (2𝜋𝑟ℎ)𝑑𝑟 (158)
Therefore:
[𝑀𝑎𝑠𝑠]𝑖𝑛 = 2𝜋(𝑟 + 𝑑𝑟)∆𝑡ℎ[𝜈𝜌]𝑟+d𝑟 (159)
[𝑀𝑎𝑠𝑠]𝑜𝑢𝑡 = 2𝜋𝑟∆𝑡ℎ[𝜈𝜌]𝑟 (160)
[𝑀𝑎𝑠𝑠]𝑎𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑒𝑑 = (2𝜋𝑟ℎ)𝑑𝑟[(𝜙𝜌)𝑡+∆𝑡 − (𝜙𝜌)𝑡] (161)
Plug (159), (160), and (161) into (149) yields:
2𝜋(𝑟 + 𝑑𝑟)∆𝑡ℎ[𝜈𝜌]𝑟+d𝑟 − 2𝜋𝑟∆𝑡ℎ[𝜈𝜌]𝑟 = (2𝜋𝑟ℎ)𝑑𝑟[(𝜙𝜌)𝑡+∆𝑡 − (𝜙𝜌)𝑡] (162)
(𝑟 + 𝑑𝑟)∆𝑡[𝜈𝜌]𝑟+d𝑟 − 𝑟∆𝑡[𝜈𝜌]𝑟 = 𝑟𝑑𝑟[(𝜙𝜌)𝑡+∆𝑡 − (𝜙𝜌)𝑡] (163)
1
𝑟𝑑𝑟[(𝑟 + 𝑑𝑟)(𝜈𝜌)𝑟+d𝑟 − 𝑟(𝜈𝜌)𝑟] =
1
∆𝑡[(𝜙𝜌)𝑡+∆𝑡 − (𝜙𝜌)𝑡] (164)
20
Which is written as:
1
𝑟
𝜕
𝜕𝑟[𝑟(𝜈𝜌)] =
𝜕
𝜕𝑡(𝜙𝜌) (165)
Since:
𝑣 = (5.6146)(0.001127)𝑘
𝜇𝐵
𝜕𝑝
𝜕𝑟= 0.006328
𝑘
𝜇𝐵
𝜕𝑝
𝜕𝑟 (166)
Plug (166) into (165) yields:
1
𝑟
𝜕
𝜕𝑟[𝑟 (0.006328
𝑘
𝜇𝐵
𝜕𝑝
𝜕𝑟𝜌)] =
𝜕
𝜕𝑡(𝜙𝜌) (167)
(0.006328𝑘
𝜇𝐵)1
𝑟
𝜕
𝜕𝑟(𝜌𝑟
𝜕𝑝
𝜕𝑟) =
𝜕
𝜕𝑡(𝜙𝜌) (168)
1
𝑟
𝜕
𝜕𝑟(𝜌𝑟
𝜕𝑝
𝜕𝑟) = (
1
0.006328
𝜇𝐵
𝑘)𝜕
𝜕𝑡(𝜙𝜌) (169)
Since:
𝜕
𝜕𝑡(𝜙𝜌) = 𝜙
𝜕𝜌
𝜕𝑡+ 𝜌
𝜕𝜙
𝜕𝑡= 𝜙
𝜕𝜌
𝜕𝑝
𝜕𝑝
𝜕𝑡+ 𝜌
𝜕𝜙
𝜕𝑝
𝜕𝑝
𝜕𝑡 (170)
Which is written as:
𝜕
𝜕𝑡(𝜙𝜌) = [𝜙
𝜕𝜌
𝜕𝑝+ 𝜌
𝜕𝜙
𝜕𝑝]𝜕𝑝
𝜕𝑡 (171)
Since the fluid compressibility is expressed as:
𝑐𝑓 =1
𝜌
𝜕𝜌
𝜕𝑝 ⇒
𝜕𝜌
𝜕𝑝= 𝜌𝑐𝑓 (172)
Similarly the rock compressibility is expressed as:
𝑐𝑟 =1
𝜙
𝜕𝜙
𝜕𝑝 ⇒
𝜕𝜙
𝜕𝑝= 𝜙𝑐𝑟 (173)
Plug (172) and (173) into (171) yields:
𝜕
𝜕𝑡(𝜙𝜌) = [𝜙𝜌𝑐𝑓 + 𝜙𝜌𝑐𝑟]
𝜕𝑝
𝜕𝑡= 𝜙𝜌𝑐𝑡
𝜕𝑝
𝜕𝑡 (174)
21
Plug (174) into (169) yields:
1
𝑟
𝜕
𝜕𝑟(𝜌𝑟
𝜕𝑝
𝜕𝑟) =
𝜙𝜇𝐵𝑐𝑡0.006328𝑘
𝜌𝜕𝑝
𝜕𝑡 (175)
This equation represents the general partial differential equation used to describe the
radial fluid flow in porous media.
16. Radial Flow of Slightly Compressible Fluids
Expanding the left hand side of (175):
1
𝑟
𝜕
𝜕𝑟(𝜌𝑟
𝜕𝑝
𝜕𝑟) =
1
𝑟[(𝜌𝑟)
𝜕2𝑝
𝜕𝑟2+ (
𝜕𝑝
𝜕𝑟)𝜌] = 𝜌 [
𝜕2𝑝
𝜕𝑟2+1
𝑟
𝜕𝑝
𝜕𝑟] (176)
Plugging (176) into (175), diving all through by ρ and rearranging yields:
𝜕2𝑝
𝜕𝑟2+1
𝑟
𝜕𝑝
𝜕𝑟=
𝜙𝜇𝐵𝑐𝑡0.006328𝑘
𝜕𝑝
𝜕𝑡 (177)
Expressing time in hours, we may write:
𝜕2𝑝
𝜕𝑟2+1
𝑟
𝜕𝑝
𝜕𝑟=
𝜙𝜇𝐵𝑐𝑡0.006328
24 𝑘
𝜕𝑝
𝜕𝑡=
𝜙𝜇𝐵𝑐𝑡0.0002637𝑘
𝜕𝑝
𝜕𝑡 (178)
In terms of diffusivity, we may write the Diffusivity Equation as follows:
𝜕2𝑝
𝜕𝑟2+1
𝑟
𝜕𝑝
𝜕𝑟=1
𝜂
𝜕𝑝
𝜕𝑡 (179)
The term 𝜂 is called the diffusivity constant and is given by:
𝜂 =0.0002637𝑘
𝜙𝜇𝐵𝑐𝑡 (180)
Where 𝑘 is the permeability in md, 𝑟 is the radial position in feet, 𝑝 is the pressure in
psia, 𝑐𝑡 is the total compressibility in psi-1, 𝑡 is the time in hours, 𝜙 is the porosity in
fraction, and 𝜇 is the viscosity in cp. For the steady-state flow, the right hand side of
(179) is set to zero and the equation is called the Laplace Equation for steady-state
flow.
22
17. Principles of Well Testing
Since we have:
𝑞 = 0.00708𝑘ℎ
𝜇𝐵
(𝑝𝑒 − 𝑝𝑤)
ln𝑟𝑒𝑟𝑤
(105)
Solving for 𝑝𝑒 − 𝑝𝑤 yields:
𝑝𝑒 − 𝑝𝑤 = 141.22𝑞𝜇𝐵
𝑘ℎln𝑟𝑒𝑟𝑤
(181)
Rewriting (181) to dimensionless form yields:
𝑝𝑒 − 𝑝𝑤 = 141.22𝑞𝜇𝐵
𝑘ℎ𝑝𝐷 (182)
The advantage of using the dimensionless form (182) over the classic form (181) is
that for steady state flow (SSF), pseudo-steady state flow (PSSF), or transient flow
(TF), only 𝑝𝐷 is different. In addition to its use for more complex situations.
Generally we have:
𝑝𝑒 − 𝑝𝑤 = 141.22𝑞𝜇𝐵
𝑘ℎ𝑝𝐷(𝑟𝐷 , 𝑡𝐷 , 𝐶𝐷 , … . ) (183)
𝑝𝐷 is the dimensionless pressure solution to (181) for the appropriate boundary
conditions. In transient flow, 𝑝𝐷 is a function of dimensionless time which is given as
follows:
𝑡𝐷 =0.0002637𝑘
𝜙𝜇𝐵𝑐𝑡𝑟𝑤2𝑡 (184)
𝑡𝐷𝐴 =0.0002637𝑘
𝜙𝜇𝐵𝑐𝑡𝐴𝑡 (185)
𝑟𝐷 =𝑟
𝑟𝑤 (186)
The exponential integral solution to the flow equation (183) is:
𝑝𝐷(𝑟𝐷, 𝑡𝐷) = −1
2𝐸𝑖 (−0.25
𝑟𝐷2
𝑡𝐷) (187)
23
Since:
𝐸𝑖(−𝑥) = −∫𝑒−𝑡
𝑡
∞
𝑥
𝑑𝑡 = ln 𝑥 + 𝛾 +∑𝑥𝑛
𝑛𝑛!
∞
𝑛=1
(188)
The exponential integral in (188) may be approximated as follows:
𝐸𝑖(−𝑥) ≈ ln 𝑥 + 0.5772 (189)
Therefore:
𝑝𝐷(𝑟𝐷, 𝑡𝐷) = −1
2𝐸𝑖 (−0.25
𝑟𝐷2
𝑡𝐷) = −
1
2[ln (0.25
𝑟𝐷2
𝑡𝐷) + 0.5772]
= −1
2[ln (
𝑟𝐷2
𝑡𝐷) + ln(0.25) + 0.5772]
= −1
2[ln (
𝑟𝐷2
𝑡𝐷) − 0.80909] =
1
2[ln (
𝑡𝐷
𝑟𝐷2) + 0.80909]
(190)
Therefore (182) is written as:
𝑝𝑒 − 𝑝𝑤 = 70.61𝑞𝜇𝐵
𝑘ℎ[ln (
𝑡𝐷
𝑟𝐷2) + 0.80909] (191)
18. Volumetric-Average Reservoir Pressure
Equation (181) may be expressed in terms of 𝑝 as follows:
𝑝 = 𝑝𝑤 + 141.22𝑞𝜇𝐵
𝑘ℎln𝑟
𝑟𝑤 (181)
Craft and Hawkins showed that, for the steady-state flow (SSF), the volumetric-
average reservoir pressure is located at 61% of 𝑟𝑒; i.e.:
𝑝𝑅 = 𝑝𝑤 + 141.22𝑞𝜇𝐵
𝑘ℎln (0.61
𝑟𝑒𝑟𝑤) (192)
Since:
24
ln (0.61𝑟𝑒𝑟𝑤) = ln (
𝑟𝑒𝑟𝑤) + ln(0.61) = ln (
𝑟𝑒𝑟𝑤) − 0.494 (193)
Therefore the volumetric-average reservoir pressure and flow rate for the steady-state
flow (SSF) are given by:
𝑝𝑅 = 𝑝𝑤 + 141.22𝑞𝜇𝐵
𝑘ℎ[ln (
𝑟𝑒𝑟𝑤) − 0.494] (194)
𝑞𝑜 =𝑘ℎ
141.22𝜇𝑜𝐵𝑜
(𝑝𝑅 − 𝑝𝑤)
[ln (𝑟𝑒𝑟𝑤) − 0.494]
(195)
They also showed that, for the pseudo-steady state flow (PSSF), the volumetric-
average reservoir pressure is located at 47.2% of 𝑟𝑒; i.e.:
𝑝𝑅 = 𝑝𝑤 + 141.22𝑞𝜇𝐵
𝑘ℎln (0.472
𝑟𝑒𝑟𝑤) (196)
Since:
ln (0.472𝑟𝑒𝑟𝑤) = ln (
𝑟𝑒𝑟𝑤) + ln(0.472) = ln (
𝑟𝑒𝑟𝑤) − 0.751 (197)
Therefore the volumetric-average reservoir pressure and flow rate for the pseudo-
steady state flow (PSSF) are given by:
𝑝𝑅 = 𝑝𝑤 + 141.22𝑞𝜇𝐵
𝑘ℎ[ln (
𝑟𝑒𝑟𝑤) − 0.751] (198)
𝑞𝑜 =𝑘ℎ
141.22𝜇𝑜𝐵𝑜
(𝑝𝑅 − 𝑝𝑤)
[ln (𝑟𝑒𝑟𝑤) − 0.751]
(199)
Either equation (195) or (199) may be written as:
𝑞𝑜 =𝑘ℎ
141.22𝜇𝑜𝐵𝑜
(𝑝𝑅 − 𝑝𝑤)
[ln (𝑟𝑒𝑟𝑤) − 0.751]
= 𝐽(𝑝𝑅 − 𝑝𝑤) (200)
Where 𝐽 is the productivity index and is written as:
𝐽 =𝑞𝑜
(𝑝𝑅 − 𝑝𝑤)=
𝑘ℎ
141.22𝜇𝑜𝐵𝑜 [ln (𝑟𝑒𝑟𝑤) − 0.751]
(201)
25
Where 𝜇𝑜𝐵𝑜 are evaluated at 0.5(𝑝𝑅 + 𝑝𝑤). Equation (200) is called the Inflow
Performance Rate (IPR) equation.
19. Multiple-Phase Two-Dimensional Model
Equation (105) is written as:
𝑞 =𝑘ℎ
141.22𝜇𝐵
(𝑝𝑒 − 𝑝𝑤)
ln𝑟𝑒𝑟𝑤
(105)
Peaceman has presented the following conventional well equation:
𝑞 =𝑘ℎ
141.22𝜇𝐵
(𝑝𝑜 − 𝑝𝑤)
ln𝑟𝑒𝑞,𝑜𝑟𝑤
(202)
Where 𝑟𝑒𝑞,𝑜 is the equivalent well block radius at which the steady state flowing
pressure equals 𝑝𝑜. On the other hand, Su has presented the following well equation
for multiple-phase 2D models:
𝑞 =1
141.22∑{𝐹𝑖 (
𝑘𝑟𝑘ℎ
𝜇𝐵)𝑖
(𝑝𝑖 − 𝑝𝑤)
ln𝑟𝑖𝑟𝑤
}
4
𝑖=1
(203)
Where 𝑖 is the running index for the four neighboring cells; i.e.:
26
Example #1: A core (L = 1.5 inches, d = 0.75 inches) was 100% saturated with water
and placed in a liquid permeameter. The water has a viscosity of 1.0 cp. The linear
flow rate was measured at 5cc/90sec with a pressure drop of 0.5 atm. Calculate the
absolute permeability.
Solution #1: We use Darcy's Law for lab units, (76), as follows:
𝑞 =𝑘𝐴
𝜇
∆𝑝
∆𝑙 (76)
Solving for 𝑘 yields:
𝑘 =𝑞𝜇
𝐴
∆𝑙
∆𝑝
Since: 𝐿 = 1.5 𝑖𝑛𝑐ℎ𝑒𝑠 = 1.5(2.54) = 3.81 𝑐𝑚.
Since: d = 0.75 inches; i.e.
𝐴 = 𝜋 (𝐷
2)2
= 𝜋 (0.75(2.54)
2)2
=𝜋
4[0.75(2.54)]2 = 2.85 𝑐𝑚2
Plug all values and solve for 𝑘 yields:
𝑘 =(5)(1)(3.81)
(90)(2.85)(0.5)= 0.148538 𝐷 = 148.538 𝑚𝐷
Example #2: The core was cleaned and 100% saturated with oil having a viscosity of
4 cp. Calculate the flow rate in cc/sec with a pressure drop of 0.5 atm.
Solution #2:
𝑞 =(0.148538)(2.85)(0.5)
(4)(3.81)= 0.01389 𝑐𝑐/𝑠𝑒𝑐
Example #3: Redo #2 using a diameter of 1.5 inches.
Solution #3:
𝐴 = 𝜋 (𝐷
2)2
= 𝜋 (1.5(2.54)
2)2
=𝜋
4[1.5(2.54)]2 = 11.4 𝑐𝑚2
𝑞 =(0.148538)(11.4)(0.5)
(4)(3.81)= 0.05556 𝑐𝑐/𝑠𝑒𝑐
27
Example #4: Redo #2 using a pressure drop of 0.25 atm.
Solution #4:
𝑞 =(0.148538)(2.85)(0.25)
(4)(3.81)= 0.00694 𝑐𝑐/𝑠𝑒𝑐
Example #5: Redo #2 using a length of 3 inches.
Solution #5:
𝑞 =(0.148538)(2.85)(0.5)
(4)(7.62)= 0.00694 𝑐𝑐/𝑠𝑒𝑐
Example #6: A horizontal core (L = 2 feet, D = 6 inches) is shown in the following
sketch. Calculate the flow rate 𝑞 for the following cases:
𝜇 = 1.0 𝑐𝑝, SG = 1.0
𝜇 = 10.0 𝑐𝑝, SG = 0.8654
𝜇 = 1000.0 𝑐𝑝, SG = 15.0
Solution #6: We use Darcy's Law for fields units, (87), as follows (Note that SG or
fluid specific weight is not used in horizontal flow):
𝑞 = 0.001127𝑘𝐴
𝜇𝐵
∆𝑝
∆𝑙= 0.001127
(320)𝜋 (6
(2)(12))2
(1)(1)
(50 − 48)
(2)= 0.07081841 𝑏𝑏𝑙/𝑑
(87)
𝑞 = 0.001127𝑘𝐴
𝜇𝐵
∆𝑝
∆𝑙= 0.001127
(320)𝜋 (6
(2)(12))2
(10)(1)
(50 − 48)
(2)= 0.007081841 𝑏𝑏𝑙/𝑑
𝑞 = 0.001127𝑘𝐴
𝜇𝐵
∆𝑝
∆𝑙= 0.001127
(320)𝜋 (6
(2)(12))2
(1000)(1)
(50 − 48)
(2)= 0.00007081841 𝑏𝑏𝑙/𝑑
28
Example #7: Redo #6, if the core is tilted so that end "B" is one foot above end "A".
Work in terms of both pressure and head. State direction of flow; i.e. A to B or B to
A.
Solution #7: We use Darcy's Law for tilted flow in field units, (98), as follows (Note
that SG or fluid specific weight is used in tilted flow):
𝑞 = −0.00112711𝑘𝐴
𝜇𝐵(∆𝑝
∆𝑙± 0.433 𝑆𝐺 sin 𝛼) (98)
𝑞 = −0.001127(320)𝜋 (
6(2)(12)
)2
(1)(1)((48 − 50)
(2)+ 0.433(1)(0.5))
= −0.07081841
(1)(1)(−1 + 0.2167)
= −0.07081841(−0.7833) = 0.0555 𝑏𝑏𝑙/𝑑
𝑞 = −0.07081841
(10)(−1 + 0.433(0.8654)(0.5))
= −0.007081841(−1 + 0.18750) = 0.005754 𝑏𝑏𝑙/𝑑
𝑞 = −0.07081841
(1000)(−1 + 0.433(15.0)(0.5)) = −0.00007081841(2.25)
= −0.000159 𝑏𝑏𝑙/𝑑
Example #8: Redo #6, if end "B" is rotated 90 and is above end "A". State direction
of flow. Work in terms of pressure or head.
Solution #8: Use Darcy's Law as above:
𝑞 = −0.07081841
(1)(−1 + 0.433(1)(1)) = −0.07081841(−1 + 0.433)
= 0.040154 𝑏𝑏𝑙/𝑑
𝑞 = −0.07081841
(10)(−1 + 0.433(0.8654)(1)) = −0.007081841(−0.6252818)
= 0.004428 𝑏𝑏𝑙/𝑑
𝑞 = −0.07081841
(1000)(−1 + 0.433(15)(1)) = −0.00007081841(5.495)
= −0.000389 𝑏𝑏𝑙/𝑑
29
Example #9: A radial flow system has the following properties:
Drainage area 80 acres
Wellbore diameter 5 inches
Average formation thickness 25 𝑓𝑡 Pressure at reservoir boundary 2500 psig
Pressure at wellbore 1000 psig
Oil volume flow rate 300 STBOPD
Oil viscosity 1.35 𝑐𝑝
Oil formation volume factor 1.35 RB/STB
Average porosity 21.5%
A. Calculate the average reservoir permeability.
B. What will the flow rate increase to if the pressure at the wellbore is reduced to 500
psig?
C. What diameter must the wellbore be increased to if the wellbore pressure is
maintained at 1000 psig but a flow rate equal to the flow rate in b above is desired?
D. What must the viscosity be reduced to if the wellbore pressure is maintained at 1000
psig but a flow rate equal to the flow rate in b above is desired?
E. What must the external pressure be increased to if the wellbore pressure is
maintained at 1000 psig but a flow rate equal to the flow rate in b above is desired?
F. What must the permeability be increased to if the wellbore pressure is maintained at
1000 psig but a flow rate equal to the flow rate in b above is desired?
Solution #9: Use Darcy's Law for radial flow of incompressible fluids as follows:
𝑞 = 0.00708𝑘ℎ
𝜇𝐵
(𝑝𝑒 − 𝑝𝑤)
ln𝑟𝑒𝑟𝑤
(105)
Solution #9A: Solve for 𝑘 yields:
𝑘 = 141.22𝑞𝜇𝐵
ℎ
ln𝑟𝑒𝑟𝑤
(𝑝𝑒 − 𝑝𝑤)
= 141.22(300)(1.35)(1.35)
(25)
𝑙𝑛
(
√80(43560)
𝜋512 )
(2500 − 1000)
= 3088.4814𝑙𝑛 (
1053.2080.4167 )
1500= (2.0589876)(7.834985)
= 16.13 𝑚𝐷
30
Solution #9B:
𝑞 = 0.00708(16.13)(25)
(1.35)(1.35)
(2500 − 500)
𝑙𝑛 (1053.2080.4167 )
=3133.07
7.834985= 400 𝐵/𝐷
Solution #9C:
400 = 0.00708(16.13)(25)
(1.35)(1.35)
(2500 − 1000)
𝑙𝑛 (1053.208
𝑟𝑤)
Rearranging and solving for ln yields:
𝑙𝑛 (1053.208
𝑟𝑤) = 0.00708
(16.13)(25)
(1.35)(1.35)
(1500)
400= 5.87451
1053.208
𝑟𝑤= 355.85
𝑟𝑤 = 2.96 𝑓𝑡 → 𝑑 = 5.92 𝑓𝑡 = 71 𝑖𝑛𝑐ℎ𝑒𝑠
Solution #9D:
400 = 0.00708(16.13)(25)
(𝜇)(1.35)
(2500 − 1000)
𝑙𝑛 (1053.2080.4167 )
Rearranging and solving for 𝜇 yields:
𝜇 = 0.00708(16.13)(25)
(400)(1.35)
(1500)
𝑙𝑛 (1053.2080.4167 )
=7.93
𝑙𝑛 (1053.2080.4167 )
= 1.01 𝑐𝑝
Solution #9E:
400 = 0.00708(16.13)(25)
(1.35)(1.35)
(𝑝𝑒 − 1000)
𝑙𝑛 (1053.2080.4167 )
Rearranging and solving for 𝑝𝑒 yields:
31
𝑝𝑒 = 1000 +(400)(1.35)(1.35)𝑙𝑛 (
1053.2080.4167 )
(0.00708)(16.13)(25)= 3000 𝑝𝑠𝑖𝑔
Solution #9F:
400 = 0.00708(𝑘)(25)
(1.35)(1.35)
(2500 − 1000)
𝑙𝑛 (1053.2080.4167 )
Rearranging and solving for 𝑘 yields:
𝑘 =(400)(1.35)(1.35)
(0.00708)(25)(1500)𝑙𝑛 (
1053.208
0.4167) = 21.5 𝑚𝐷
Example #10: An oil well, after cleanup following a light acid treatment, produced at a
stabilized rate of 1325 STB/D for 22 hours. Downhole pressure gauges run into the well
indicate that the flowing pressure had stabilized at about 1850 𝑝𝑠𝑖𝑎. The flow period
was followed by a buildup test. Interpretation of the buildup data gave a permeability-
thickness product (𝑘ℎ) of 3640 md-ft. The bit diameter was 8.5 inches. Bottom hole
samples were analyzed by a PVT lab, which reported bubble-point pressure of 1610
𝑝𝑠𝑖𝑎 and undersaturated oil FVF ranging from 1.2352 BBL/STB at 2600 𝑝𝑠𝑖𝑎 to 1.2450
𝑏𝑏𝑙 /STB at the bubble-point pressure. Undersaturated oil viscosity of 1.62 𝑐𝑝 was
measured with a rolling-ball viscometer.
Calculate the theoretical pressure distribution at r = 1.354 𝑓𝑡, 4 𝑓𝑡, 5 𝑓𝑡, 49 𝑓𝑡, 50
𝑓𝑡, 499 𝑓𝑡, and 500 𝑓𝑡. Plot pressure vs. radius r.
Plot pressure vs. 𝑙𝑛𝑟
𝑟𝑤.
Solution #10: Use MS Excel to solve (107) as follows:
𝑝 = 𝑝𝑤 + 141.22𝑞𝜇𝐵
𝑘ℎ[ln
𝑟𝑒𝑟𝑤− 0.75 + 𝑠] (107)
q = 1325 STB/D pwf = 1850 psia kh = 3640 md-ft rw = 8.5 inches 0.354167 ft
Pb = 1610 psia Bo = 1.2352 at 2600 Bo = 1.245 at 1610 Bo avg= 1.2401
u = 1.62 cp
32
r, ft p, psia ln(r/rw)
0.354167 1850 0
1 1957.195 1.037988
2 2028.778 1.731135
3 2070.651 2.1366
4 2100.36 2.424282
5 2123.405 2.647426
6 2142.234 2.829747
7 2158.153 2.983898
8 2171.943 3.117429
9 2184.107 3.235212
10 2194.987 3.340573
11 2204.83 3.435883
12 2213.816 3.522894
13 2222.082 3.602937
14 2229.736 3.677045
15 2236.861 3.746038
16 2243.526 3.810576
17 2249.787 3.871201
18 2255.689 3.928359
19 2261.273 3.982427
20 2266.57 4.03372
21 2271.609 4.08251
22 2276.413 4.12903
23 2281.004 4.173482
24 2285.399 4.216041
25 2289.615 4.256863
26 2293.665 4.296084
27 2297.563 4.333825
28 2301.318 4.370192
29 2304.942 4.405283
30 2308.443 4.439185
31 2311.83 4.471975
32 2315.108 4.503724
33 2318.286 4.534495
34 2321.369 4.564348
35 2324.363 4.593336
36 2327.272 4.621507
37 2330.102 4.648906
38 2332.856 4.675574
39 2335.538 4.701549
40 2338.153 4.726867
41 2340.703 4.75156
33
1850
1950
2050
2150
2250
2350
2450
2550
2650
0 100 200 300 400 500 600
Pre
ssu
re, p
sia
Radius, ft
Pressure versus External Radius
1850
1950
2050
2150
2250
2350
2450
2550
2650
0 1 2 3 4 5 6 7 8
Pre
ssu
re, p
sia
Ln(r/rw)
Pressure versus Ln(r/rw)
34
Example #11: An oil well has the following data:
Drainage area 40 acres
Wellbore diameter 12 inches
Average formation thickness 47 𝑓𝑡 Pressure at reservoir boundary 3265 psia
Oil volume flow rate 135 STBOPD
Oil viscosity 13.2 𝑐𝑝
Oil permeability 90 mD
Oil formation volume factor 1.02 RB/STB
Average porosity 17%
Total compressibility 2.0E-5 psia-1
Skin effect 0
Plot pressure vs. time at the wellbore.
Solution #11: Use MS Excel to solve (191) as follows:
𝑝𝑒 − 𝑝𝑤 = 70.61𝑞𝜇𝐵
𝑘ℎ[ln (
𝑡𝐷
𝑟𝐷2) + 0.80909] (191)
Where:
𝑡𝐷 =0.0002637𝑘
𝜙𝜇𝐵𝑐𝑡𝑟𝑤2𝑡 (184)
𝑟𝐷 =𝑟
𝑟𝑤 (186)
A = 40 acre re = 744.73025 ft q = 135 STB/D pi = 3265 psia k = 90 mD h = 47 ft rw = 6 inches 0.5 ft
Bo = 1.02 at 2600 u = 13.2 cp phi = 0.17
Ct = 2.00E-05 psia-1 tD = 2115.24064(t)
rD = 1.00000
35
time, h tD pwf
1 2115.24064 3008.13
2 4230.48128 2987.099
3 6345.72193 2974.797
4 8460.96257 2966.068
5 10576.2032 2959.298
6 12691.4439 2953.766
7 14806.6845 2949.089
8 16921.9251 2945.038
9 19037.1658 2941.464
10 21152.4064 2938.267
11 23267.6471 2935.375
12 25382.8877 2932.735
13 27498.1283 2930.307
14 29613.369 2928.058
15 31728.6096 2925.965
16 33843.8503 2924.007
17 35959.0909 2922.167
18 38074.3316 2920.433
19 40189.5722 2918.792
20 42304.8128 2917.236
2910
2920
2930
2940
2950
2960
2970
2980
2990
3000
3010
3020
0 5 10 15 20 25
Pw
f, p
sia
Time, hours
Pwf vs Time
36
Example #12: An oil well has the following data:
Drainage area 80 acres
Wellbore diameter 8.5 inches
Average formation thickness 35 𝑓𝑡 Average reservoir pressure 2608 psia
Well bore pressure 1850 psia
Oil volume flow rate 1325 STBOPD
Oil viscosity 1.24 𝑐𝑝
Oil permeability 90 mD
Oil formation volume factor 1.6 RB/STB
Permeability 104
Calculate the IPR of the well.
The IPR of the well is calculated using (201) as follows:
𝐽 =𝑞𝑜
(𝑝𝑅 − 𝑝𝑤)=
1325
(2608 − 1850)= 1.75
𝑆𝑇𝐵
𝐷/𝑝𝑠𝑖
Also we may use:
𝐽 =𝑘ℎ
141.22𝜇𝑜𝐵𝑜 [ln (𝑟𝑒𝑟𝑤) − 0.751]
=3640
141.22(1.24)(1.6) [ln (10530.354
) − 0.75]= 1.793
𝑆𝑇𝐵
𝐷/𝑝𝑠𝑖
