10. single phase flow in porous media - piazza

PE 120 lecture notes

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10. Single phase flow in porous media Accurately simulating the flow of reservoir fluids is a challenging task. It would be far too expensive to realistically represent the porous material in a computer simulation. Therefore, researchers look for acceptable simplifications. In this course we stick to the very basic models for single phase and multi-phase flow. In later courses you will discuss more complicated versions. We start by deriving Darcy's law. This law is valid for steady flow in a homogeneous material and gives an average volumetric flow rate as function of the difference in hydraulic head of the flow. We look at Darcy's law for horizontal, vertical and radial flows, in multiple dimensions, and in parallel and serial beds. We also discuss extensions to Darcy's law for compressible flow, real gas flow, and turbulent flow. Finally we look at several models for porous medium flow derived from physical analogues. 10.1 Darcy's law

In 1856, French engineer Henry Darcy (Inspector General of Bridges and Roads), derived his famous Darcy's law experimentally and described it in appendix D of a report written on the public water works of the City of Dijon, France. Darcy needed to know how large a sand filter would be required to filter a given quantity of water per day. Unable to find this information, he proceeded to obtain it experimentally. In the experiments, Darcy's objective was to determine the relation between the volumetric flow rate of water through sand beds and the hydraulic head loss. A total of 35 experiments were conducted by him and his colleagues and reported on. The results of his tests are also available online and can be found simply by googling. His apparatus (shown in figure 10-1) consisted of a steel column with an inside diameter of 0.35 m and sealed on both ends by bolted plates. It's total height is reported as 2.5 m. At the bottom, an outlet reservoir was created by supporting a set of screens above the bottom, which in turn supported the sand. Water flow rate was determined by timing the accumulation of water in the box drawn in the bottom left of the picture. You can also see the supply line which tapped into the house supply line of a hospital (which induced considerable oscillations as users in the hospital turned faucets on or off). Mercury U-tube manometers (pressure meters) were connected to both the inlet and the outlet reservoir.

Darcy's tested several different types of sand, which of course changed the permeability k of the porous material. Darcy only tested water, not other fluids. Therefore, he did not study the effect of fluid density and viscosity.

Fig 10-1 Darcy and his famous apparatus J


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To understand the results of Darcy's experiments we must first study the concept of hydraulic head, which is a measure of energy of the fluid. Hydraulic head The hydraulic head is a measure of the energy of the fluid. We will see that the hydraulic head is equal to the sum of the potential energy, also known as elevation pressure, and the pressure. Now, energy of a moving fluid (not a stationary fluid) actually has three components, not two: potential energy; pressure; and kinetic energy. Kinetic energy, as you may remember, is proportional to the velocity squared. In fluids moving through porous media, the velocity is typically very small relative to the pressure or elevation pressure, and so this third term is readily neglected. Let's look at a very simple case first. We have a column of liquid, say water (see figure 10-2). The water is not flowing. In the column we identify two different positions: 1 and 2. Now imagine water at position 1. This volume is at height y1 above the indicated datum level (let's assume for now that that is ground level). Therefore its elevation energy is equal to 1gyρ . At position 2, the elevation energy is equal to 2gyρ . What about the pressure? If we compare positions 1 and 2, we know that the pressure at position 2 will be much higher than the pressure at position 1. The difference is the hydrostatic pressure formed by the column of water between positions 1 and 2. If the pressure at position 1 is p1, the pressure at position 2 is p1 + ( )21 yyg −ρ . Since pressure enables the fluid to do work, it is a form of energy, just as potential energy. We add pressure to the elevation energy to get a measure of the total energy of the fluid at a location, that is the hydraulic head. The hydraulic head, or head for short, is denoted by h (this is standard notation. Please don’t confuse with height). In this simple case we get

12211222

111

hgy)yy(gpgyph,gyph

=ρ+−ρ+=ρ+=

ρ+=

We see that in the above case, the hydraulic head in the two positions is identical. This is a direct result of the fluid not moving. If it is not moving, it cannot loose any energy to the surroundings through friction. Head loss Now suppose that we have a moving viscous liquid. Because of viscosity, heat is generated by friction. This means that the energy of the water will decrease as it moves along, in other words, the hydraulic head decreases. You could look at this in another way: in order to keep a viscous fluid moving at a constant rate, you must have a head difference to compensate for friction. This loss in hydraulic head or energy is called head loss.

position 2

position 1

y1

y2

Fig 10-2 A simple example


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We can write this as lossfrictionhhh 21 =−=Δ . Because ( ) ( )212121 yygpphh −ρ+−=− , we get ( ) ( ) lossfrictionyygpp 2121 =−ρ+− . (We note that only the difference in elevation between positions 1 and 2 is important. That means that we can choose the datum level anywhere we want.) In porous media the head loss needed to compensate for friction is significant. Explain why in porous media the head loss (the friction effect) is significant The question is now: can we find a simple expression for this head loss? You may think that the head loss (total friction experienced by the moving fluid) would be very difficult to measure accurately. The porous medium is, after all, extremely complex. The viscous fluid moves through a complicated system of pore and pore throats of varying dimensions and the properties of the surface (its roughness and composition) also changes constantly. Fortunately for us, the overall effect of friction over representative volumes of the rock is relatively simple – which leads us back to Darcy’s experiments. Darcy's measurements Let's go back to Darcy's experiment. A sketch is given in figure 10-3. The length of the sand column is L, and the surface area A. Let's choose the datum level at position 2. Darcy measured the relation between head loss and volumetric flow rate q for several different sands. He kept q constant. He found that

,

Lhh

KAq

or,qKALhh

21

21

⎟⎠

⎞⎜⎝

⎛ −=

=−

where K is constant for each sand (but varies from sand to sand). We will see later that K is proportional to the permeability k. In other words, Darcy found that for porous media, the heat generated as a result of friction is proportional to the flow rate. This is often referred to as a linear viscosity rule: friction is a linear function of flux (or if you will velocity). In fluid dynamics courses, this is often discussed for creeping flow also. For faster moving flows, this linear relation no longer holds and we will discuss this later. We can interpret the above equation also slightly differently: the flux q is proportional to the gradient of h. Do you see where this observation comes from?


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It is important to understand that Darcy's law is really an energy equation, and that the flow rate in fact models the friction term. This will help you understand extensions to Darcy's equation that we will be discussing later. Darcy’s law, or relation, can also be interpreted as an average form of a momentum equation. We note that the pressures p1 and p2 can be determined from the level of the fluid in the manometers. The heads with respect to the given datum can then be found to be equal to

.gLh,gLh

22

11

ρ=

ρ=

10.2 Generalization of Darcy's experiment We now investigate the situation where the column of sand is not orientated vertically, but instead is buried and at an angle to the horizontal, as in figure 10-4. The length of the column again is L. Again we take the datum level at position 2. The height of the fluid in the top manometer is z1. The height of the fluid in the bottom manometer is z2-Δh. The head h1 at position 1 is given by ( )1211 zzggzh −ρ+ρ= , and at position 2 we have a head h2 equal to ( ).bzgh 22 −ρ= So, .gbhh 21 ρ=−

q

L L1

L2

q

Fig 10-3 Sketch Darcy's experiment

datum (position 2)

position1


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We can also write the distance b as

( )2121 zz

gppb −−

ρ

−=

Substitution in Darcy's law gives

.gLzz

LppKA

LhhKAq 212121

⎥⎦

⎤⎢⎣

⎡ρ⎟⎠

⎞⎜⎝

⎛ −−

−=

−=

Let's put a coordinate axis aligned with the direction of flow. We call this the s-axis. Then, we can write

Lzz

Lzz

dsdzand,

Lpp

Lpp

dsdp 21122112 −

−=−

=−

−=−

= .

So, Darcy's law can be written as

.dsdzg

dsdpKAq ⎥⎦

⎤⎢⎣

⎡ ρ−−=

Nutting (1930) found the constant K to be inversely proportional to the viscosity of the fluid. Therefore it was suggested to set K equal to

µ

=kK ,

where k is the proportionality constant. This k was then defined as the permeability of the porous material, which we discussed in chapter 9. Does it make sense that K is inversely proportional to the viscosity of the fluid?

Fig 10-4 Generalization of Darcy's experiment

q

q

b z2 z1

L

position 1

position 2


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How good is Darcy's law? As stated before, Darcy's law is only valid for steady state flows in homogeneous materials (k constant). In actual situations k is of course not constant, and the flow is often unsteady. The relationship between head loss and flow rate that Darcy found is only linear for small flow rates (and small head losses). At higher flow rates, non-Darcy effects become important. This is illustrated in figure 10-5.

Darcy's law holds

non-Darcy effects

q

head loss

Fig 10-5 Relation between q and head loss


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Flux velocity and interstitial velocity So far, we have looked at the volumetric flow rate q only. We can define a velocity u as

.Aqu =

This velocity is called the flux velocity. It is an average velocity measured over the total surface area the fluid flows through (the cross sectional area) covering both solid material and void space. The so called interstitial velocity v is also frequently used. This velocity is defined by

φ

=φ

=u

Aqv .

It gives the average velocity of the liquid in the pores and is therefore a good measure of the actual flow velocity. 10.3 Horizontal flow In horizontal flow, the elevation energy is everywhere the same. Therefore the head loss is a function of the pressure gradient only. Also, the s-coordinate will now be the same as the x-coordinate. We can write

.dxdpkAq

µ−=

The flux velocity is given by

.dxdpku

µ−=

Units of permeability Because the last equation is so simple, we can use it to easily find the units of the permeability k. The units

of u are length/time, or L/T. The unit for pressure gradient is 22TL

m , where m is the unit of mass. The unit

for viscosity is LTm . This gives the unit of L2 for k. So, k is measured in feet squared or meter squared.

Check that the units for pressure and viscosity are correct. The values for k are generally of the order of 10-14 m sq. It is usually expressed in terms of "darcies". 1 darcy is equal to 10-12 m sq. The symbol for a darcy is d. The permeability is 1 darcy when u=1 cm/sec, the viscosity is equal to 1 centipoise and the pressure gradient is equal to 1 atm/cm. Clearly, rocks that have a permeability of 1d or higher or highly permeable rocks. Unconsolidated rocks fall into this category, for example.


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10.4 Vertical flow We now look at a few examples of vertical flow. Vertical free flow Suppose that we have a core of porous material "suspended" in air as in figure 10-6a. A liquid is poured in to the core at location 1 and leaves the core at location 2. We assume it flows at a steady rate. In this case, s=z. Both location 1 and 2 are exposed to the air and therefore the pressure in these locations is the atmospheric pressure. This means that dp/ds=dp/dz=0 in this particular flow problem. Also, we know that dz/ds=1. Therefore

.gkAq ρµ

=

Vertical flow with a driving head This situation is depicted in figure 10-6b. Again position 2 is exposed to the air, but now a layer of liquid is situated on top of the porous material. The thickness of the layer is b. The pressure in position 1 is now equal to .gbpp atm1 ρ+= We could interpret this as the layer of liquid providing an overburden pressure. So,

.Lgb

dzdp

dsdp

or,gbpgbppp atmatm21

ρ−==

ρ=−ρ+=−

With dz/ds=1, we get

.1LbgkAq ⎟

⎠

⎞⎜⎝

⎛ +ρµ

=

q

q

L

1

2

L

q

q

L

1

2

b

q

L

1

2

b q

q

Fig 10-6 Various types of vertical flows. a: free flow, b: driving head, c: upwards flow

(a) (b) (c)


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Vertical upward flow This situation is depicted in figure 10-6c. Position 1 is exposed to the air. Therefore, .pp atm1 = The pressure in position 2 is equal to ( ).bLgpp atm2 +ρ+= Now, dz/ds=-1 and therefore

( ) .LgbkAg

LbLgkAq ρ

µ=⎥⎦

⎤⎢⎣

⎡ ρ++ρ

−µ

−=


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Example (from Amyx, ex 2-8) In a city water-filtration plant, it was desired to filter 5,000 gallons of water per hour through a sand filter bed to remove all the suspended matter and solids from the water. A vertical cross-sectional view of the filtration unit is shown in figure 10-7. What level of water must be kept in the pit above the sand filter bed in order to filter these 5,000 gallons of water per hour? Assume that the solids removed from the water do not alter the permeability of the bed. The data for this problem are Thickness of sand filter bed 4ft Permeability of sand filter bed 1,200 md Cross-sectional area 1,800 sq ft Viscosity of water 1.0 centipoise Barometric pressure 14.7 psia Ans.: 6.88 ft 10.5 Radial flow Radial flows are very important in petroleum engineering because they are analogous to the flow into a well bore. We will find the Darcy’s equation governing the radial flow into a cylindrical well of thickness b (figure 10-8). Now, s = -r and dz/ds = 0. This gives

b filter bed 4 ft

Fig 10-7 Vertical flow through filter bed with constant head b

redge

rwell

Fig 10-8 Radial flow to central well bore. Topview and side view

b


90

.drdpkAq

µ=

The surface area A is given by .rb2A π= . Substitution in Darcy’s equation gives

.drdpk

rb2q

µ=

π

Separating variables and integration gives

.dpkdrr1

b2q

∫∫ µ=

π

We integrate from the edge of the circular domain to the well bore and get

)r/rln(

ppb2kqwelledge

welledge −

µ

π= .

10.6 Extensions to Darcy’s law: heterogeneous materials In the previous sections we assumed that the porous material was homogeneous. In reality this is of course not the case. In this course we will not consider the fully heterogeneous case, but here we make a first step by considering materials that consists of distinct layers of material, each with a constant permeability. (a) (b) Parallel beds First, we will look at a porous material with several parallel beds, as illustrated in picture 10-9a.

k1 k2 k3 q

Δp1 Δp2 Δp3

L1 L2 L3

Fig 10-9 a: Parallel beds; b: series of beds

b1

b2

b3

k1

k2

k3

q1

q2

q3

b

L

w


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We assume that the beds are separated by an infinitely thin impermeable barrier. The pressure at the left boundaries of the layers is pl, and at the right boundary is pr. For each of the layers, we can set up an equation of Darcy in the form

( ) ( ).ppLwbkpp

LAkq rl

iirl

iii −

µ=−

µ=

The total flow rate q is equal to the sum of the three qi’s. We can write

( ),ppLwbkq rl −µ

=

where

,b

bkk

ii

iii

∑∑

=

is an average permeability for the layered system. Check this Series of beds Show that for the series of beds in figure 10-9b we have the following equation

( ) .

kL

Lkwith,pp

Lwbkq

i i

i

ii

rl

∑

∑=−

µ=

Show that for the situation depicted in figure 10-10a the average permeability is the same as for the parallel beds discussed above. Show that for the situation in figure 10-10b the average permeability is given by

.

k)r/rln()r/rln(

k

i i

1ii

welledge

∑ −=


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(a) (b) Fig 10-10 a: parallel beds in cylindrical domain; b: series of beds in cylindrical domain


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10.7 Multi-dimensional flow If we have flow in the x-, y- and z-direction, then we can set up a Darcy equation for each coordinate direction. The elevation term will of course only enter into the equation for the vertical coordinate, in this case z. In a homogeneous material, we simply

.gzpku

,ypku

,xpku

z

y

x

⎟⎠

⎞⎜⎝

⎛ ρ−∂

∂

µ−=

⎟⎟⎠

⎞⎜⎜⎝

⎛

∂

∂

µ−=

⎟⎠

⎞⎜⎝

⎛∂

∂

µ−=

Here, ux, uy and uz are the components of the velocity vector. These three equations can also be written in vector notation as

( ),.gpku ρ−∇µ

−=

where the superscript arrow is used to denote a vector. This is of course quite a simple case. If there is anisotropy, that is if the material is not homogeneous and the permeability changes as a function of space, then the equation complicates. What would the equation look like if the permeabilities in the x, y and z direction are, respectively, kx, ky and kz, and these permeabilities are constant? What would the equation look like if the permeability changes in all directions? How to deal with such cases of anisotropy will be discussed in more advanced courses. 10.8 Compressible flow In the derivation of Darcy's law it was assumed that the density of the fluid is constant. Darcy conducted his experiments with water, for which this incompressibility assumption is certainly true. Can we extend Darcy's law to take into account compressibility? We'll look at a simple horizontal flow (figure 10-11).


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In the incompressible case (standard Darcy) we have

,dxdpk

Aqu

µ−==

where q is constant and the flow is steady. In the steady compressible case, this equation is still valid, but now q may be dependent on x. To see this we must think a little about what it means to have steady flow. It means that the mass flux must be constant. If it were not constant, the mass contained in the porous medium would change over time, but that would mean that the flow is no longer steady. The mass flux at any position x is equal to the density times

the velocity, .Aq

ρ Thus, in steady flow, Aq

ρ is constant, or qρ is constant. The density may not be

constant, which means that the flow rate q can not be constant anymore. Therefore q will be a function of x (and so will dp/dx). Multiplying the above equation by the density gives us the equation for the mass rate of flow:

.dxdpkAq

µ

ρ−=ρ

Liquids We know that for slightly compressible liquids we can write (see liquid properties notes) ( ).e 0ppc

0−ρ=ρ

Here p0 and 0ρ are appropriate reference values for the flow. This gives

( ) .dxdpec

dxd

0ppc0

−ρ=ρ

So, we can write

.dxdpc

dxd

ρ=ρ

Substitution into the mass rate of flow equation gives

L

q A

p1 p2

Fig 10-11 Horizontal flow


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.dxd

ckAq ρ

µ−=ρ

If we then integrate this equation from x=0 to x=L we get

( ).ckAqL 12 ρ−ρµ

−=ρ

If p-p0 is small we can write ( )( ),ppc1 00 −+ρ≈ρ so

ρ q = −kµcALρ0c p2 − p1( ).

This is very similar to what we had before for incompressible flows. Just keep in mind that now both the density and the flow rate can be functions of x, and that this equation only holds for slightly compressible fluids under a small pressure drop. Gases For gases, we can use the equation of state to find a relation between the density and the pressure. We know that

.znRTp=

ρ

If T is constant (isothermal flow), we see that the ratio of pressure over density is constant, or

.pp

00=

ρ

ρ

Let’s substitute this in the mass flux equation

.dxdpkAq

µ

ρ−=ρ

After some working we get

0

22

21

0 ppp

L2kAq

−

µ= .

Show this. We can play around with this a little more. Let’s create an average pressure defined by


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2ppp 21 +≡ .

Show that for the flux at this mean pressure, we have .LppkAq 12 −

µ−=

Again, this looks like the incompressible case for a volumetric flow rate defined at the algebraic mean pressure. Please note that the lecture slides for lectures 22/23 contain information about extensions to Darcy’s law, as well as porous media analogues, that are not in these lecture notes. So use these lecture notes, as well as the lecture slides to study for your final examination.


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11. Multi-phase flow in porous media In this chapter, we look at the flows through porous media with two or more phases present. Besides the rock-fluid interaction, we must now also consider the fluid-fluid interaction. A few conventions: A gas-liquid interface is referred to in these notes as a liquid interface. A liquid-liquid interface is referred to as a liquid-liquid interface J. In the gas-liquid interface we talk about surface tension. In the liquid-liquid interface we talk about interfacial tension. 11.1 Surface tension When a fluid element is located at the boundary of a fluid, where it comes in contact with a dissimilar fluid or a solid, its properties can be different from those of an interior element. Consider a water droplet surrounded by air, for example. In its interior, each water molecule is pulled equally in all directions by surrounding molecules. At the water surface, however, the water molecules experience larger pulls from surrounding water molecules than surrounding air molecules, and are therefore pulled inward. This creates some internal pressure (see below) and also forces liquid surfaces to contract to the minimal area possible (to minimize energy).

Surface molecules are subject to an attractive force from nearby surface molecules so that the surface is in a state of tension: it costs energy to increase the surface area. This tensile force per unit of length along the surface, called the surface tension for gas-liquid interfaces and interfacial tension for liquid-liquid interfaces, is a property of the fluid and its adjacent fluid or solid. The unit of surface tension is force per unit length. The surface tension makes the interface resemble a membrane: it takes a force to penetrate it. The water strider makes use of this to carry itself on the water. In figure 11-2a, you can see the dimples in the water surface.

The surface energy is equal to the surface tension times the surface area. In other words, the tension is defined as the energy it takes to increase the surface area by 1 unit. Note that the symbol used for surface tension is sometimes Y, and sometimes σ. We predominantly use the latter.

Why does it take energy to increase the surface area?

To minimize energy, the surface will try to take the shape with the smallest surface area (this explains why drops of water tend to be spherical and why your egg yolk on your fried egg in the morning is not flat.)

Fig 11-1 Net force on molecules near surface is inward, leading to surface tension


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The internal pressure created because of the inward pull on the surface molecules leads to a pressure jump across the interface. There are two ways to derive the size of this pressure difference. Method 1. The energy required to change the surface area is equal to the work done by the pressure difference. Remember that work = force x distance. Here force = pressure difference x surface area. This leads to the right hand side of the first equation below. The left hand side just states the difference in surface energy caused by the increase in surface area.

.drr2prdr4

or,drr2pdrdr)area(d

2

2

πΔ=πΥ

πΔ=Υ

Method 2. The excess pressure inside the drop, acting upon the cross-sectional area 2rπ (where r is the drop radius) is counterbalanced by the surface tension Υ acting on the circumference r2π of this cross-sectional area See figure 11-3b. Both methods lead to the same expressions:

r2p

or,r2rp 2

Υ=Δ

πΥ=πΔ

Fig 11-2 a: water strider; b: egg yolk

Fig 11-3 a: Method 1 - energy balance; b: Method 2 – force balance


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We see from this equation also that when an interface is plane (radius of curvature r is infinite), the pressure difference is zero; i.e., the pressure is equal on both sides of the plane interface despite the interfacial tension.

The interfacial tension of a liquid may change when impurities in the liquid congregate on its surface, thereby altering its surface properties. Detergents are surfactants that change the surface properties of oil, grease and dirty particles in contact with water, making it easier to clean them away in a washing process. Note that surface tension/interfacial tension depends on the temperature. For example, the surface tension of water at higher temperatures is less (why do you think that is?), which explains why it is more effective to wash clothes at a higher temperature.

Surface or interfacial tension can be important for fluid flows involving small droplets or bubbles where the pressure excess is comparable to pressure changes in the flow field and should be accounted for in reservoir simulations.

Surface tension is typically measured in dynes/cm, the force in dynes required to break a film of length 1 cm. Equivalently, it can be stated as surface energy in ergs per square centimeter. Water at 20°C has a surface tension of 72.8 dynes/cm compared to 22.3 for ethyl alcohol and 465 for mercury.

11.2 Wetting When an interface between two fluids meets a solid surface, the interface forms an angle with respect to the solid surface called the contact angle, see figure 11-4. This contact angle depends upon the nature of the two fluids and the solid. Note that the contact angle is measured through the denser phase. The values of the angle and the interfacial tension determine the effects of capillarity, such as the vertical rise height of a fluid in a capillary tube, which we will discuss below.

Rocks can be wetting or non-wetting with respect to certain fluids. For example, a water-wet rock prefers water over oil. An oil-wet rock prefers oil over water. It is also possible to have mixed-wet rocks, because the wetting depends on rock properties such as chemical constitution and grain sizes which may vary throughout the rock. In figure 11-5 are a few illustrations. Here, DNAPL stands for dense non-aqueous phase liquids and is a common term for water pollutants (the pictures were taken from a ground water pollution website). Which fluid will preferentially wet the solid is determined by the adhesion tension, which is the difference between the solid-water tension (or energy) and the solid-oil tension (or energy), see also further below. Let's look at a solid/oil/water interface for a water-wet rock (a rock that has a greater affinity with water than oil) illustrated in figure 11-6. The contact angle is determined by the relative sizes of the surface energy (surface tension times surface area). Here, we will start using the symbol σ for tension force, which is common in petroleum engineering. If the interface energy between the solid and the oil is very large and the interface energy between the solid and water is small, the water tends to wet the solid, and the contact angle should be smaller than 90.

Fig 11-4 Contact angle of gas at the gas/liquid/solid interface


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But we also have to deal with the interface energy of the oil/water interface. We will have equilibrium at a certain contact angle if the net energy change is zero. That is: energy required to increase oil-solid surface area –

energy gained by decreasing water-solid surface area = energy change because of change in oil-water surface area

The equation that results from this energy balance is called Young's equation. It leads to the following equation .)cos( swsowo σ−σ=θσ Here, the subscript sw indicates the surface tension for the solid-water interface and similar for the other subscripts. Note that here we use σ, not Y, to indicate tension and we’ll do this from now on.

The contact angle can be found from

wo

swso)cos(σ

σ−σ=θ .

water

oil

solid

Fig 12-6 solid/oil/water interface for a water-wet rock

Fig 11-5 Non-wetting, wetting and neutral DNAPL


101

The adhesion tension (or energy) AT is defined as the difference between the solid-water tension (energy) and the solid-oil tension (energy), or

swsoTA σ−σ= .


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11.3 Capillary pressure Rise of fluids in capillaries First we will conduct a wee class experiment. The class experiment shows that it is possible for a liquid to rise in a capillary tube. This ability is due to adhesion forces between the solid of the tube and the fluid and also depends on the interfacial forces between the two fluids present in the tube (in this case the liquid and air). In this section we will try to shed some light on two questions: 1. How high can a fluid rise in a capillary (what determines the height)? 2. What effects does the capillary rise have on the pressure distribution in the tube? The first question is very important in petroleum engineering because it controls imbibition, which can cause displacement of one phase by another (for example, oil by (injected) water, or (ground) water by oil). The second question is a natural question to ask if we think back to section 11.1 on surface tension. We know that if there is a curved interface between two fluids, there will be surface tension at the interface and this surface tension will be balanced by an increased pressure in the concave section (the non-wetting side). Figure 11-7 shows the experimental set-up. A narrow capillary tube is placed in a glass with solution. Liquid rises in the tube. The interface in this case is shown. We make a few observations: • The density of air is small. Therefore the difference in height between positions A' and B' will not lead

to any measurable change in air pressure. • The pressure at A is equal to the pressure at B + the hydrostatic pressure as a result of the column of

liquid of height h. • The pressures in A’ below and above the interface are equal. • When the liquid is in equilibrium, the pressures at A and A' must be the same. Do you agree with these observations? Mathematically we get

h

r

A A' A'

B B'

Fig 11-7 Schematic of capillary rise in tube


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.ghpp

,pp,ppp

liquidBA

'AA

atm'B'A

ρ=−

=

==

This gives ghppp liquidB'Bc ρ=−= , where pc is defined as the capillary pressure: the difference in pressure across an interface in a capillary tube as a result of capillary rise. As in section 11.1 we can again write down an energy or force balance on the liquid/air interface:

.

grcos2h

or,rA2cosr2rp

liquid

la

Tla2

c

ρ

θσ=

π=θσπ=π

Here, laσ is the surface tension at the liquid/air interface. Note that again

.cos

r2p

or,Ar2p

lac

Tc

θσ=

=

It is clear from these expressions, that the capillary pressure is lower in wider tubes and higher in narrower tubes as shown in figure 11-8. Please note that the above expressions were derived for cylindrical tubes. "The pressure is always greater in the non-wetting phase". True or false? If the liquid rises in oil, not in air, we cannot ignore the difference in pressure between A' and B'. We now get

.ghpp

,pp,ghpp

liquidBA

'AA

oil'B'A

ρ=−

=

ρ+=

So, gh)(ppp oilliquidB'Bc ρ−ρ=−= .

lower pc

higher pc

Fig 11-8 Capillary pressure depends on the size of the pores


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11.4 Two phases in the reservoir

Figure 11-9a, c and d show three possible scenarios in a water-wet reservoir rock with air and water present. In figure 11-9a the water saturation of the rock is low. The water crawls into the small pores and forms a set of pendular rings. Figure 11-9 b shows one of these rings. At this low saturation level, the rings are not connected and therefore water flow is not possible (the water cannot transmit pressure differences).

In figure 11-9 c the water saturation is much higher and the pendular rings expand and touch to form a continuous water phase allowing water flow. In this figure, the air may also still be a continuous phase (note that we are only looking at a cross section of the rock). In figure 11-9 d the water saturation is higher still. Now, the air is broken into non-connected bubbles that are located in the larger pores. If we want to be fancy we can say that the rock is in a state of insular air saturation. We note that the air can only move if the surrounding water applies a pressure difference sufficient to squeeze it through the constrictions between the grains. Of course it is possible to have parts of the rock as in a, parts as in c and parts as in d. Then we can talk about parts of the water being immobile and parts mobile. Note that over time, the volume of air at insular saturation may decrease due to air solubility in water. Similarly, the volume of water in pendular rings may decrease with

time due to evaporation.

Mr Plateau (appr. 1865) derived an equation for the capillary pressure for a pendular ring as shown in figure 11-9 b. This pendular ring is also shown in figure 11-10. The interface formed is one of the "Plateau's problems", a famous collection of topology and geometry problems that many mathematicians ponder over worldwide (by the way, Plateau first got interested in such

r2 r1

Fig 11-9 Various saturation scenarios in a water-wet porous rock

Fig 11-10 Capillary pressure for a pendular ring. Cross-sectional view.


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interface problems when he tried to find expressions for surfaces of connecting soapbubbles). Plateau's equation says that

⎟⎟⎠

⎞⎜⎜⎝

⎛+σ=

21c r

1r1p ,

where σ is the interfacial tension. This can be derived by setting up the force balance over an infinitesimally small surface area. This is however not that trivial an exercise and we will leave it for now. Here we merely use the above equation to provide an approximate measure of capillary pressure in the pores of a reservoir rock. It is practically impossible to measure the radii of curvature. In general therefore people take about the mean radius Rm determined by

21m r1

r1

R1

+= .

This "mean radius" is determined by some other measurements on a porous medium. Note that the capillary pressure increases if any of the radii of curvature decreases. When the radii are the same, we get

rrrfor,r2p 21c ==σ

= .

which is the same as for capillary rise with zero contact angle. 11.5 Drainage, imbibition and hysteresis

Suppose that we are draining the pore shown in figure 11-11. We start with the interface with lowest curvature. This corresponds to a higher saturation of the pore. We now apply a pressure to the interface. As a result the interface moves down into areas with smaller ratio until the higher pressure is balanced by the interfacial tension. Remember that to sustain a higher pressure difference across the surface the curvature must go up. As the interface goes down, the saturation goes down as well. So, the lower the saturation, the smaller the radii of curvature in the porous rock, and the higher the capillary pressure. So here, at low saturation, the wetting-phase material (eg. water) will exist in the smaller crevices and openings of the system, leaving the large open channels to the nonwetting phase.

We can picture this relation between saturation and capillary pressure in a graph as shown in the drainage curve of figure 11-12. We note that when draining there is always a thin layer of water/liquid that remains on the rock of tens of molecules thickness. This has effect on subsequent wetting phases, for example, the contact angle will be different in re-wetting.

lower pc

higher pc

Fig 11-11 Draining a typical pore


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In figure 11-12 another saturation-capillary pressure curve is given for re-wetting (imbibition), after drainage. As you can see, this curve is slightly different and, in general has lower saturation for the same capillary pressure.

Can you explain the difference?

The fact that there is a difference in the drainage and wetting relations is called hysteresis.

The figure shows that there is a threshold saturation St and a maximum saturation Sm in the wetting process.

Figure 11-13 shows imbibition of water in a sandstone core (just because I like this picture!). The core was initially filled with air. Of course the level and speed of imbibition will depend on several factors such as the type of liquid and rock and the other fluid present.

Fig 11-13 Imbibition of water in an air-filled core over time. Lighter colors indicate higher water saturation.

Fig 11-12 Drainage and wetting saturation/capillary pressure relations


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An oil bubble is trapped in a conical pore as shown in figure 11-14.

For this problem, .m20r,30r,cm/dynes24,15,30 21wo µ=µ==σ=α=θ !! Calculate the pressure in the oil bubble and at position A if the pressure at B is 1 atmosphere.

Fig 11-14 Oil bubble trapped in conical pore

10. single phase flow in porous media - piazza

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