D (I-1C) Darcys Law

Darcy’s Law

1

Basics of Reservoir Reservoir Engine Engineering ering – Module I

I.1.C I.1.C – Darcy’ Darcy’ss Law Law

Copyright 2008, NExT, All rights reserved

Darcy’s Law OBJECTIVES

Upon completion of this section, you will be able to: •Understand and apply Darcy’s law of fluid flow through porous media to problems of relevance in reservoir engineering, •Write equations to calculate average permeability for the flow geometries that occur in practice for petroleum reservoirs.

Darcy’s Law

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Historical Basis for Permeability Darcy’s Experiments Darcy’s Paper h1-h2

Q

A

L

h4 h2

Henry Darcy, “The Public Fountains of the City of Dijon”, 1856


Darcy was a civil engineer and he was commissioned by his hometown (Dijon, Southern France) to design the local distribution of water to public fountains, hospitals, and major buildings. His “Law” appears in the appendix of his publication “The Public Fountains Fountains of the the City of Dijon - Determination Determination of the laws of flow of water through sand”. Darcy is regarded as the first experimental reservoir engineer.

Darcy’s Law

3

Historical Basis for Permeability Darcy’s Experiments

h1-h2

Q

A

,

L

h4 h2

Darcy was the first to point out (1856), while conducting experiments in his sand filters (built for the purpose of water purification for Dijon’s water distribution system), that for a porous material: “One can conclude that output volume is proportional to the head and inversely related to the thickness of a layer traversed”, or:

q∝

∆h L


Henry Darcy investigated the flow of water in a vertical, saturated, homogeneous sand filter or column depicted above. From his experiments, varying the length and diameter of the column, the porous material in it, and the water levels in the inlet and outlet reservoirs, he concluded that the rate of flow or volume of water passing per unit time, Q, through a sand column of length L and constant cross-sectional area, A, is: •proportional to the cross-sectional area, A of the column, •proportional to the difference in water level elevations, h1 and h2 , in the inflow and outflow reservoirs of the column, respectively, and •inversely proportional to the column length, L.

Darcy’s Law

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Darcy’s Findings

In other words, Darcy concluded that:

q=

K A (h1− h2 ) L


where q

= Flow rate, cm3/sec

A

= Cross-sectional area, cm2

h1 - h2

= Height difference, cm

L

= Length of column, cm

K

= Constant of proportionality

- 100% water saturated - K was the “constant” needed to make the equation valid

Darcy’s Law

5

Permeability and “K”

Darcy’s “K”, the constant of proportionality in the equation, was determined to be a combination of: •

k, permeability of the sand pack (or rock), and

•

µ, viscosity of the liquid

K =

k µ


Permeability is a property of the rock to transmit fluids. As such it is a derived property (like electric resistivity). Permeability can not be measured in the absence of flow. Permeability, k, is a measure of the fluid conductivity of a porous material (analogous to electrical resistivity of a material). Permeability usually correlates with other rock properties such as porosity and saturation. Absolute permeability - the permeability of a porous medium with only one fluid present (single-phase flow). When two or more fluids are present permeability of the rock to a flowing fluid is called effective permeability (k o, kg, kw). Relative permeability is the ratio of absolute and effective permeabilities kro=ko/k, krg=kg/k, krw=kw/k. Effective and relative permeability will be discussed in more detail in a later section of this course.

Darcy’s Law

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Darcy’s Findings

In piezometric head difference

Finding

Constant of proportionality

q∝

In pressure difference

∆h

q∝

L

∆P L

k A

K ⋅ A

µ

Darcy’s Law

q = −K ⋅ A

dh dS

v = − K

dh dS

q=−

k µ

A

dP dS

v=−

k dP µ

dS


Darcy’s empirical law has been extended to describe most fluid phenomena in a porous medium. For the flow of a homogeneous fluid in a porous medium, we write: v=−

k dP µ

dS

“Darcy’s Law for a fluid of constant density flowing in a porous system under negligible gravitational effects”

Darcy’s Law

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Dimensional Analysis of Darcy’s Law

v=−

k dp

µ

ds

v=−

k dP µ

Base units: L, length or distance m, mass t, time

dS

[v] = L / t [ µ ] =

m Lt

= dynamic viscosity (dynamic viscosity=absolute viscosity and kinematic viscosity=dynamic viscosity/density)

[ s ] = L [ p ] =

m Lt 2

 dp  =  ds 

m 2 2

Lt

, recall

p =

F A

=

ma L2

=

mL L2t 2

=

m Lt 2

Thus, what are the units of permeability?

 ds  L m [ k ] = vµ  =  dp  t Lt

L2t 2 m

= L

2

Permeability has dimensions of LENGTH 2 (AREA).


In reservoir engineering lab tests we DO NOT use m^2 or cm^2 or to express permeabillities. WHY? Typical permeabilities would be in the order of micro to nano units (10-6-10-9 cm2). Even worse if we used m^2. 1 cm^2 is too large of a permeability! Especially if you think of permeability as the mean square pore diameter of the rock. Therefore, “Darcy units” is the set of units customarily used in reservoir engineering based on the c.g.s unit system.

Darcy’s Law

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Darcy Units

Darcy units cm

Velocity

s

Pressure

at m

of fluid of

Viscosity

cp = 1 ⋅ 1 0

Distance

−3

N ⋅s m2

cm

.

Permeability darcy

 cm  s 

3

q µ k = A = dp ds Copyright 2008, NExT, All rights reserved

“A permeability of 1 darcy is defined as that permeability which allows the flow of 1 cm3/s of fluid of 1 cp viscosity through an area of 1 cm2 under a pressure gradient of 1 atm/cm”

   ( cp )

( cm ) 2

 atm     cm 

=

cm 2 ⋅ cp atm ⋅ s

= darcy

Muskat (1937) proposed to call this unit “Darcy” after the pioneer investigator. Since then, it has been widely accepted in the petroleum literature.

Darcy’s Law

9

Darcy’s Law

General form of Darcy’s law: Introducing the effect of gravity

v s

v s

=−

k d Φ µ

=−

ds

k d Φ µ

Φ = Hubbert’s Potential k  dp

d z 

ρ g

=−  −  1.0133 x 10 d s  ds µ  ds 6


In the equation above: vs

= Velocity along path, s, cm/sec

k

= Absolute permeability, darcies

dp/ds

= Pressure gradient along flow path, s, atm/cm

ρ

= Density of flowing fluid, g/cm3

g

= Acceleration due to gravity, 980 cm/sec 2

dz/ds

= Vertical depth gradient, cm/cm

-z vs

Darcy Coordinate System

s +y +z

+x

Darcy’s Law

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Darcy’s Law For Various Flow Geometries and Fluids

Various Reservoir Fluids: •

Liquids

•

Gas

Geometries: •

Horizontal and Vertical

•

Linear, Radial, Spherical

Geological Considerations: •

Flow parallel to depositional bedding planes

•

Flow perpendicular to bedding planes


Darcy’s Law was empirically derived in a laboratory using a fluid, flowing through a linear sandpack. This empirical formulation must be modified for the different conditions that exist in an actual formation. These modifications include various geometries, reservoir fluids, and flow directions. Darcy’s law sometimes requires other modifications that are beyond the scope of this course. For example, at high flow rates, Darcy’s equation has been modified by adding a quadratic or polynomial term to it. The quadratic form of Darcy’s law has been called the Forscheimer equation and it accounts for high velocity flow (also referred to as turbulent flow or non-Darcy flow). Other modifications have been developed for non-Newtonian fluids, requiring a numerical modeling method of solution rather than a simple analytical expression.

Darcy’s Law

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Horizontal, Linear Flow Incompressible Liquid

q

A

q

p1

p2 L


Basic Assumptions :

Horizontal system (No reservoir dip, Ignore gravity) Linear system with constant cross-sectional flow area Steady-state, laminar flow of an incompressible fluid Reservoir fluid is non-reactive with reservoir rock Porous medium is 100% saturated with single fluid Constant reservoir temperature

Darcy’s Law

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For Linear and Horizontal Flow (Liquid) =0

v s

=−

v s

=−

k  d p µ

 d s 

k d p µ

d s

−

d z 

ρ g

1.0133 x 10

=−

6



d s 

k d p d x

µ

Integrating:

q=

kA

( p − p ) 1

µL

2


vs

q A

k dp dx

k A dp dx

q

Separate variables and integrate: p2

L

q dx o

q L

p1

0 q

where

k A dp

kA

p2

kA p1 L

p1 p2

q

=

Flow rate, cm3/sec

k

=

Permeability, darcies

µ

=

Viscosity, centipoise

L

=

Length, cm

p1, p2

=

Pressure, atm

Darcy’s Law

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Modification of Darcy’s Law for Gas Properties

Same assumptions as liquid flow Additional assumptions for gas properties •

Several forms of flow equation are available

•

Simplest form is shown here: – Valid at low reservoir pressures


Fluid properties are independent of pressure, i.e., µz = constant (determine µz at mean pressure). This assumption is valid only at low pressures, e.g., in the laboratory. The more complicated modifications to the Darcy’s law equation for gas flow are beyond the scope of this course. One of the more common modifications is to transform pressure into gas pseudo-pressure, incorporating the changes in the gas properties (also called the Real Gas Potential).

Darcy’s Law

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Modified Darcy’s Law for Gas q=

p sc q sc z T p T sc

=−

k A dp µ d x

Separating variables and integrating: L

p2

p sc qsc z T p dp q dx = − k Tsc O p

∫

∫

1

qsc

=−

k A Tsc  p 22  L T z p sc 

− p   2  2

1


where q

=

Flow rate at test conditions, cm 3/sec

psc

=

Pressure at standard conditions, atm

qsc

=

Flow rate at standard conditions, cm 3/sec

z

=

Gas deviation factor

T

=

Temperature at test conditions, °K

Tsc

=

Temperature at standard conditions, °K

k

=


µ

=

Viscosity, cp

L

=

Length, cm

p1, p2 =

Pressures, atm

psc vsc =

n R Tsc

psc qsc (time) = n r Tsc p q (time) = z n R T q = zT/p = qsc psc/Tsc

Darcy’s Law

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Horizontal, Radial Flow Liquid

pw

r e

pe

r w h


Modifications to Darcy’s Law for radial system: - A = 2 πrh - ds = - dr - Flow is inward during production or outward during injection

Darcy’s Law

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For Radial and Horizontal Flow =0 k  d p

v s

=−

−  µ  d s 1.0133 x 10

v s

=+

k d p

q

=

d r A

µ

d z 

ρ g

=

6



d s 

q 2 π

r h

Integrating:

q=

2 π

h k ( pe

µ ln

− pw )

(r e / r w )


Details of derivation:

k dp dv q

Separate variables:

2 Integrate:

q 2

r e

=

q 2 r h

dr k = dp h r

dr k h r w r

pe

dp

pw

q

=

Flow rate, cm3/sec

h

=

Height, cm

k

=


µ

=

Viscosity, cp

pe

=

Pressure at external boundary (re ), atm

pw

=

Pressure at wellbore, atm

r e

=

Radius of external boundary, cm

r w

=

Wellbore radius, cm

Darcy’s Law

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Exercise 1: Horizontal Flow Parallel to Depositional Bedding Planes

p1

p2 Layer A

qt

Layer B h

qt

Layer C w L


Many reservoirs have depositional layers that are parallel to each other. In some cases, they are near-horizontal. In other cases, the depositional dip and gravity must be included in the flow equation. Since many reservoirs are composed of multiple layers of rock with various rock properties, petroleum engineers must be able to calculate flow through these systems and to calculate average properties for these systems. Flow through the three rock layers shown in this schematic is similar to an electrical current flowing through three parallel resistors. What is the overall permeability of this system?

Darcy’s Law

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Exercise 1: Solution qt = q A + qB + qc

ht = h A + hB + hc

n

Flow area, A = wht

k is arithmetic average permeability of layered system n

∑ k h i

k =

i

i =1

ht


k

If is average permeability of system, and Darcy’s law is applied to total system kA k w ht

qt

Since

qt

= q A + qB + q C

k wh t p1 L

p2

L

p1

k A wh A p1 L

p2

k C whC p1 L

p2

p2

L

k B whB L

k ht k A h A

k B hB

k C hC

k A hA

k B hB ht

k C hC

k

where n = number of layers n

k

i

K i hi

1

ht

p1

p2

p1

p2

Darcy’s Law

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Exercise 2: Horizontal Flow Perpendicular to Depositional Bedding Planes

p2

p1 kA qt

pA LA

kB

kC

pB

pC

LB

h

qt

LC

w L


Many reservoirs have depositional layers that are parallel to each other. In most cases, they are not vertical as shown in the figure. However, flow in the vertical direction with near-horizontal depositional layers is relatively common. Gravity must usually be included, when considering vertical flow. Since many reservoirs are composed of multiple layers of rock with various rock properties, petroleum engineers must be able to calculate flow through these systems and to calculate average properties for these systems. Flow through the three rock layers shown in this schematic is similar to an electrical current flowing through three resistors in a series. What is the overall permeability of this system?

Darcy’s Law

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Exercise 2: Solution p1 - p2 = ∆p A + ∆pB + ∆pB

qt = q A = qB = qc L = L A + LB + LC

k

is the harmonic average system permeability

qt =

k wh

( p − p ) 1

µ L

2

L n Li

k =

∑ K i =1

i


Total pressure drop p 1 - p2 Since p1 - p2 = ∆p A + ∆pB + ∆pc

p1

p2

qt L k wh

qA LA k A wh

Since qt = q A = qB = qC Then

L k

LA kA

LB kB

or

L

k

n i

1

Li ki

LC kC

qB LB kB wh

qC LC k c wh

Darcy’s Law

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Exercise 3: Horizontal Radial Flow Liquid Layered System Flow Parallel to Bedding Planes

pe

pw

r e qA

hA ht

hB

r w

qB qC

hC Wellbore Copyright 2008, NExT, All rights reserved

This radial system is similar to the earlier layered system where the flow was linear. The derivation on the following pages shows that the arithmetic average is the appropriate method for this system. This system is common for many oil or gas well completions. What is the overall permeability of this system?

Darcy’s Law

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Exercise 3: Solution

qt = q A + qB + qc ht = h A + hB + hc

k = average permeability of layered system qt =

2 π µ ln

k ht

(r e / r w )

( pe − pw )


where qt

=

Total flow rate, cm3/sec

k

=

Average permeability, darcies

ht

=

Total height, cm

µ

=

Viscosity, cp

pe

=

Pressure at re , atm

pw

=

Pressure at rw , atm

r e

=

Radius of reservoir, cm

r w

=

Wellbore Radius, cm

Darcy’s Law

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Exercise 3: Solution qt =

+

2 π µ ln

(r e / r w )

2 π µ ln

Simplifying

k ht

( pe − pw ) =

k B h B

(r e / r w )

( pe − pw ) +

k ht = k A h A k =

ht

∑ k h i


µ ln

k A h A

(r e / r w )

2 π µ ln

i =1

ht

i

( pe − pw )

k C hC

(r e / r w )

+ k B h B + k C hC k A h A + k B h B + k C hC n

k =

2 π

( pe − pw )

Darcy’s Law

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Exercise 4: Composite Radial Flow System

q

q

pe pC

pB pApw

q

q r w r C

r B

r A

ht


This composite system represents various rings around the wellbore with different rock properties. At first glance, this system would not appear to be very common. However, we use this system routinely by assuming a certain level of formation damage or stimulation around the wellbore. Formation damage or skin is discussed in the section on well testing. What is the overall permeability of this system?

Darcy’s Law

25

Exercise 4: Solution

qt = q A = qB = qC

n

ht = h A = hB = hC

pe - pw = ∆p A + ∆pB + ∆pC

k = average permeability of composite system qt =

2 π

k ht

( pe − pw ) µ ln (r e / r w )

k =

ln n

∑

(r e / r w ) ln (r i / r i- ) 1

k i

i =1


Details of derivation: qt ln r e / r w pe p w 2 k ht qB

ln r B / r A 2 k B ht

qA

ln r / r w 2 k Ah t

qC

ln r e / r B 2 k C ht

Divide through by the similar terms:

ln r e / r w k ln r e / r w k k

ln r A / r w kA ln r i / r i 1 ki 1

n i

ln r e / r w n ln r / r i i 1 ki i 1

ln r B / r A kB

ln r C / r B kC

Darcy’s Law

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References

1. Amyx, J.W., Bass, D.M., and Whiting, R.L.: Petroleum Reservoir Engineering , McGrow-Hill Book Company New York, 1960. 2. Tiab, D. and Donaldson, E.C.: Petrophysics, Gulf Publishing Company, Houston, TX. 1996. 3. Dandekar, A. “Petroleum Reservoir Rock and Fluid Properties”, Taylor and Francis, 2006.


D (I-1C) Darcys Law

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