3      CAPILLARY PRESSURE

# 3         CAPILLARY PRESSURE

Reservoir rock typically contains the immiscible phases: oil, water, and gas. The forces that hold these fluids in equilibrium with each other and with the rock are expressions of capillary forces. During waterflooding, these forces may act together with frictional forces to resist the flow of oil. It is therefore advantageous to gain an understanding of the nature of these capillary forces.

Definition : Capillary pressure is the pressure difference existing across the interface separating two immiscible fluids.

If the wettability of the system is known, then the capillary pressure will always be positive if it is defined as the difference between the pressures in the non-wetting and wetting phases. That is:

Thus for an oil-water system (water wet)

For a gas-oil system (oil-wet)

## What Causes Capillary Pressure?

Capillary pressure is as a result of the interfacial tension existing at the interface separating two immiscible fluids. The interfacial tension itself is caused by the imbalance in the molecular forces of attraction experienced by the molecules at the surface as shown below.

For molecules in the interior:

Net forces = 0 since there are enough molecules                                                                        around to balance out.

For molecules on the surface:

Net result of forces is a pull toward the interior                                                              causing a tangential tension on the surface.

The net effect of the interfacial tension is to try to minimize the interfacial area in a manner analogous to the tension in a stretched membrane. To balance these forces and to keep the interface in equilibrium, the pressure inside the interface needs to be higher than that on the outside.

Forces reducing interface are due to:                 a) Interfacial tension

b) External pressure

The effect of interfacial tension is to compress the non-wetting phase relative to the wetting phase. The force created by the internal pressure is balancing it.

## 3.1        EXPRESSIONS FOR CAPILLARY PRESSURE UNDER STATIC    CONDITIONS

### 3.1.1       Pc In terms of radius of capillary tube

Since the interface is in equilibrium, force can be balanced on any segment. The interfacial forces are eliminated by taking as a free body, that part of the interface not in direct contact with the solid. A force balance would give:

(Internal pressure - External pressure) * Cross-sectional area = Interfacial tension * Circumference

Thus,

Therefore,

And since by definition, , we have:

For an air-water system, the air is the non-wetting phase and

This equation is referred to as Laplace's Equation in some texts.

### 3.1.2       Pc In terms of height of fluid column.

because there is no capillary pressure across a horizontal interface.

but

therefore,

Since  then

Oil-water system

because no capillary exists across any interface that is horizontal.

Since , then,

Therefore,

That is,

 Cgs Units: Field Units:

The two expressions for capillary pressure in a tube, one in terms of height of a fluid column and the other in terms of the radius of the capillary tube can be combined to give an expression for the height of a fluid column in terms of the radius of the tube as follows:

Therefore for an air-water system,

Similarly, for an oil-water system,

These two equations show an inverse relationship between fluid height and capillary radius. The smaller the radius is, the higher the height of the fluid column will be.

Example 3.1

a) Derive the expression for the pressure at the bottom of a capillary tube containing oil and water and exposed to the atmosphere as shown below.

b) If , ,  and , and the radius of the tube is 1 cm. What is the value of the pressure at the bottom of the tube ?

.

Solution

Consider points (1) and (2) at the air/oil and water/oil interfaces respectively.

(1)

(2)

From  . we obtain

Since,

Solving for Pcow gives;

(3)

b) Substituting values,

### 3.1.3       Pc In terms of radii of curvature of interface

The dependence of  on the curvature of the interface is analyzed with reference to the figure below. This figure represents a small segment of a curved interface containing the point p. The point is at the center of the segment, which is approximately square in shape. The edges of the segment are each of length . The angles  and  are those subtended by each arc of half length  on orthogonal planes normal to the segment at p, with radii of curvature R1 and R2 respectively.

NOTE:  and  are the radii of curvature of the interface itself and have nothing to do with the radius  of any tube.

By balancing forces

therefore,

This is the general expression for capillary pressure that is applicable to all systems regardless of shape. For example, it can be applied to the case of two parallel plates standing in water.

### 3.1.4       Application to Parallel Plates

Consider two parallel glass plates separated by a gap to a fluid standing in water. The expression for capillary pressure and the height to which the fluid will rise between the plates can be obtained from the general expression for capillary pressure in terms of the radii of the interface as follows:

In general,

For the case of parallel plates, , and

Therefore,

That is, , where  is the contact angle.

Please note that even though this equation looks similar to that of a tube, the denominators are not the same. In this case, b is the separation between the plates, and not the radius of any tube.

The height to which the fluid will rise can be obtained by equating the two expressions:

Therefore,

Example 3.2

Consider three capillary tubes having respectively:

a) a circular cross-section

b) a square cross-section

c) a rectangular cross-section with one side having twice the dimension of the other.

If all three tubes have the same cross-sectional area and same wettability, which tube will have the highest capillary rise?

Solution

The formula to use is    ……………..….(a)

For the circular cross-section,

Since the radius of the tube =  = , then,

………………………(b)

For the square cross-section, the length of each side

………………………(c)

For the rectangular cross-section, since ,

Substituting for R1 and R2 in the general equation (equation (a)) gives:

……………………..(d)

Since all three have the same cross-sectional area ,

Substituting for  in equations (c) gives:

…………for the square

Similarly, substituting for  in equation (d) gives:

…………………for the rectangle

The appropriate equations are now:

 tube square rectangle

Since  and

Thus, the rectangle will have the highest capillary rise, followed by the square and the circle last.

## 3.2        APPLICATIONS OF CAPILLARY PRESSURE EXPRESSIONS IN             POROUS MEDIA

### 3.2.2       Porous media modelled as a bundle of capillaries

One of the earliest and simplest depiction of porous media was as a bundle of capillary tubes of arbitrarily varying diameters. By applying the applicable one of the equations:

or

The different water heights in such a system is illustrated in the figure below where if the number of tubes were numerous, a smooth curve will result as shown in the lower figure. That figure is for a three-phase gas-oil-water system. The figures also show the difference between the water-oil contact  (WOC) and the free-water table. The WOC is the depth at which  begins (moving downward) while the free-water table is the depth at which .

### 3.2.3       Porous media modelled as a packaging of uniform spheres

An even more realistic model is the depiction of porous media as a packaging of uniform spheres. Applying the two expressions for capillary pressure in terms of the radii of the interface and in terms of the height of fluid column, we have for this system:

From which,

In field units,

Unfortunately,  and are impossible to measure in porous media and so are usually determined empirically from other measurements in the porous media. For this reason, it is more convenient to explicitly measure capillary pressure and use the equation below to calculate the height of the fluid column.

Example 3.3

Using the drainage capillary pressure curve of the Venango Core (shown below). How many feet above the free water table is the water/oil contact? (1 ft = 30.48 cm)

Solution

From the figure, the capillary pressure at the water-oil contact can be read as 4 cm.

Since

Then,

=

## 3.3        Laboratory methods of measuring capillary pressure

Three generally accepted methods of measuring capillary pressure in the laboratory are:

a) The Porous Diaphragm (or restored state) Method

b) The Centrifugal Method

c) The Mercury Injection Method

All three tests are conducted on core plugs cut from reservoir whole core samples. Drilling fluids, coring fluids, coring procedure, core handling and transportation, storage and experimental processes can alter the natural state of the core. Therefore, special precautions are necessary to avoid altering the natural state of the core. If the natural state of saturation of the core had been altered, then it must be restored to its natural state before conducting any capillary pressure tests.

Fresh Core :

Samples from core taken with either water or oil-base muds that are preserved (with invaded fluids) and subsequently tested without cleaning and drying are referred to as fresh cores.

Native State Cores :

Samples from core recovered with lease crude or special oil base fluids known to have minimal influence on core wettability, and that are tested as fresh samples, are referred to as Native State. These cores are in their native state (i.e. without invaded fluids). Such cores coming from above the transition zone should have the same quantity and distribution of water as in the reservoir. These samples are preferred for water displacement tests.

Restored Cores :

Core samples cleaned and dried prior to testing are referred to as restored cores. An advantage is that air permeability and porosity are available to assist in sample selection. A disadvantage is that core wettability and spatial distribution of pore water may not match that in the reservoir.

The following precautions can be helpful in obtaining representative cores if the drilling conditions permit.

1. Use oil-base drilling mud to minimize clay swelling

2. Use non-oxidized lease crude as a coring fluid.

3. Suitable storage procedures include submersion under degassed water, and preservation with saran foil, and wax.

Refined oil versus crude oil

Refined oils are suitable for most tests, and are preferred when tests are at ambient conditions.

Crude oils to be used in ambient tests should be sampled from non-water producing wells upstream from chemical or heater treaters.

Crude oils often precipitate paraffin or asphaltenes at ambient conditions, resulting in invalid test data.

Reservoir condition test utilizing live crude oil at reservoir pressures and temperatures often overcome difficulties experienced with crude at ambient conditions.

Reservoir fluid samples for special core tests may be recovered using bottom-hole sampling techniques, or recombined from separator gas and oil samples.

### 3.3.1       Centrifugal Method

1.         Rotate at a fixed constant speed. The centrifugal force displaces some liquid, which        can be read at the window using a strobe light. Thus, the saturation can be obtained.

2.         The speed of rotation is converted to capillary pressure using appropriate equations.

3.         Repeat for several speeds and plot capillary pressure with saturation.

### 3.3.2       Mercury Injection Method

1.         Place core sample in a chamber and evacuate it.

2.         Force mercury in under pressure. The amount of mercury injected divided by the           pore volume is the non-wetting phase saturation. The capillary pressure is the    injection pressure.

3.         Continue for several pressures and plot the pressure against the mercury saturation.

Advantages:      1. Fast (minutes)

2. No threshold pressure limitation

Disadvantage:   1. Can only be used for shaped cores.

### 3.3.3       Porous Diaphragm Method

1.         Saturate both the core sample and the diaphragm with the fluid to be displaced.

2.         Place the core in the apparatus as shown

3.         Apply a level of pressure, wait for the core to reach static equilibrium.

The capillary pressure = height of liquid column + applied pressure

 Equilibrium

Production

 Time

4.         Increase the pressure and repeat step (3)

5.         Plot capillary pressure versus saturation

Disadvantages:              1. Have to work within threshold pressure of the diaphragm

2. Takes too long to reach the equilibrium, therefore a complete                                                 curve takes from 10 - 40 days

Mercury injection technique was developed to reduce this time.

## 3.4        Other methods

Dynamic method:

1.         Simultaneous steady flow of two fluids is established in the core

2.         Using special welted discs, the pressure of the two fluids in the core is measured.

The difference = Capillary pressure

3.         Change the rate of one fluid and the saturation changes

4.         Plot  versus saturation.

### 3.4.1       Field Method:

A long column of porous medium put in contact with a wetting fluid at its base and suspended in the earths gravitational field. It is left to reach equilibrium. Samples are taken at different heights and the capillary pressure calculated using

Disadvantage:   May take very long to reach equilibrium

### 3.4.3       Explanations for capillary hysteresis

1.         The advancing and receding contact angles are different. If the contact angle during imbibition is the advancing contact angle, it differs from the contact angle drainage (receding). This may explain the phenomenon of hysteresis.

2. "Ink bottle effect

For porous media modelled as a bundle of tubes with varying diameters, a given capillary pressure exhibits a higher fluid saturation on the drainage curve than on the imbibition curve.

### 3.4.4       The effect of pore size distribution on capillary pressure curve

The more uniform the pore sizes, the flatter the transition zone of the capillary pressure curve.

### 3.4.5       Conversion of Laboratory Capillary Data to Reservoir Capillary Data

Water (brine) - oil capillary pressure data are difficult to measure in the laboratory. Generally, air - brine or air - mercury data are measured instead and it becomes necessary to convert these data to equivalent oil - water data representative of reservoir fluids. If we denote  or  as , and  as  the conversion equation can be derived as follows:

From , we obtain

From , we obtain

Assuming that the same porous medium applies in both laboratory and field, we equate the  to obtain,

ignoring the contact angles,

An identical equation would be obtained by starting from the two equations:

Assuming the radii of curvature in the laboratory is the same as that in the reservoir, the RHS's can be equated and :

### 3.4.6       Calculating Average water saturation

If a reservoir average capillary pressure curve (or even a laboratory curve) is available, it can be converted to a height versus water saturation curve and used to calculate the average water saturation for any desired interval. One simply needs to put a new scale for the height on the y-axis of the Pc graph. The average water saturation between any two height intervals can be evaluated as the area enclosed between them divided by the distance between the height intervals. An example illustrates the procedure.

Example 3.4

For a pay zone whose top and bottom are 45 ft and 25 ft from the free water table, use the laboratory Pc graph below to calculate the average water saturation for this pay interval.

Solution to Example 3.4

First, convert the Pc lab to Pc res:

Next, convert Pc res to height above the free water table and plot on the right axis by putting

a new scale on the RHS for “h”.  Its scale is 5.0 ´ the scale for Pc lab.  This is shown below.

Mark off the top (45 ft) and bottom (25 ft) on the h-axis.

Since the area in this case can be approximated by a single trapezoid, the shaded area

Therefore,

= 0.28

The area under more complex shaded areas can be calculated after sub-division into a number of trapezoids and applying the trapezoidal rule.

## 3.5        Averaging Capillary Pressure Curves

Consider a reservoir cross-section from which four core samples are taken at different depth as shown below. Each core will generate its own complete capillary pressure curve in the laboratory which can be converted to a reservoir capillary pressure curve. Thus four different laboratory capillary pressure curves are obtained as shown below. The question then arises:

How do we get a single  curve to represent the reservoir?

The answer is to use the Leverett J-function

### 3.5.1       Leverett J-function

The Leverett J-function is defined as:

where,              = permeability

= interfacial tension

= contact angle

= porosity

The J-function has the effect of normalizing all curves to approach a single curve and is based on the assumption that the porous medium can be modelled as a bundle of non-connecting capillaries (Slider pp 279-280). Obviously the more capillary bundle assumption deviates from reality, the less effective the J-function correlation becomes. This correlation is not unique, but seems to work better when the rocks are classed as to rock types, eg; limestone, dolomite, etc.

Given several capillary pressure curves, with corresponding values of permeability  and porosity , the procedure for obtaining J-function curve is as follows:

a)         Pick several values of  from 0 to 1 and read the corresponding values of. There will be as many  values as there are curves.

b)         For each  value, calculate J and plot versus . Repeat for all  values.

c)         Put your best correlation curve through the data.

This J-Curve is now a master curve that can be used to represent that reservoir and in the absence of other data can be used for other reservoirs of similar rock type. The graphs below, taken from (Amyx et.al.) shows the J-function curve for the Edwards formation showing classification as to rock types.

Fig. J-function correlation of capillary pressure data in the Edwards formation, Jourdanton Field. J-curve for (a) all cores; (b) limestone cores; (c) dolomite cores; (d) microgranular limestone cores; (e) coarse-grained limestone cores. (Source Amyx et.al.)

### 3.5.2       How to use the Leverett J-function to calculate Average Water Saturation

Values for average initial or connate water saturation are required in many petroleum engineering calculations. Examples are: (a) average water saturation in a section of reservoir in order to fix effective fluid permeabilities,  and (b) average water saturation in the whole reservoir in order to fix the initial hydrocarbon volume in place,

Under capillary equilibrium conditions, the water saturation of a particular piece, or sample of rock not depends on several factors. It has been shown with certain limitations, that a properly determined Leverett J-function versus water saturation curve can be used to obtain an average water saturation from a number of capillary pressure curves. It is assumed that a Leverett J-function curve is available and applies to the reservoir. The objective here is to show how to use the J-function to obtain the best possible estimate of average saturation. Recall that the J-function is defined as:

By expressing the  term in terms of height and fluid densities the equivalent equation is:

It is important to note while applying this equation that its units are not important. Mixed units can be used without appropriate conversion factors. It is only important to be sure to use the same units that went into determining the values of J making up the original plot. In other words, find out what units were used to calculate the J-function curve and stay consistent with those units whether they are mixed or not.

Note also that J=constant*h. Therefore, the shape of a J-function versus  curve would be similar to that of a height versus  curve. The difference is a displacement by a factor equal to the constant. Thus, a  curve can be converted to a height curve simply by adding a new y-axis having its abscissa equal to the constant*.

#### 3.5.2.1     Case 1: Permeability, Porosity, and Elevation are known for each sample

This figure illustrates four reservoir samples having different values of permeability and porosity and located at different heights above a  datum. Assuming fluid properties are the same in all pieces, the J-function equation can be simplified to:

#### 3.5.2.2     The correct method

The correct method of obtaining the average saturation,  for the four pieces is to calculate J for each piece, determine the corresponding water saturation,  of each piece by using the J-curve and then taking the arithmetic average of the saturations with the equation:

Note that this procedure correctly takes into account the vertical position of the pieces and their corresponding permeability and porosity.

Less correct methods

These methods first calculate average values of  and , substitute them into the J equation to get an average J value, and then read the average water saturation  from the J-function versus  graph. The only advantage of these methods is that the amount of calculations is reduced. The resulting  will always have error in it. How much error depends on the specific condition being calculated. The figure below illustrates the concept behind using average values in order to obtain an average J value.

There are two ways:

Method (a):      Calculate  for each sample and obtain the arithmetic average for all four. Also, obtain the arithmetic average . It is assumed that the average  is located at the average height . The average J-function equation then becomes:

where,

This is the easiest of the averaging methods to do.

Method (b): The geometric average permeability and porosity are used to get the average J-function:

= geometric mean permeability =

= geometric mean porosity =

Zero values of  and  are not permitted when evaluating the geometric averages. Because porosity values usually show very limited range, the geometric average porosity, , can be replaced by the easier to calculate arithmetic average, , with little loss of accuracy. Therefore, the form used by most engineers is

where,                          is the arithmetic average

#### 3.5.2.3       Errors due to using average values of  and

Standing(4) discusses the amount of error in  introduced by using average values of  and  and states that the error depends on several factors.

One factor is the distribution of  in a vertical sense. If the  are distributed randomly, no error will be involved. On the other hand, if high permeabilities predominate in one portion of the section and low permeabilities in another, some error will be introduced.

A second factor is the shape of the  curve. Where log J is linear to , no error will result from geometric average . Where J is linear with , some error will result.

A third factor is the range of permeability values. Little error is introduced when the range is small and more error is introduced when the range is large. The best way to minimize errors of averaging is not to average. Use the correct method.

#### 3.5.2.4     Case 2: Permeability and porosity are unknown as functions of elevation. Distance from  (distance from free-water table) is known

The petroleum engineer often needs to develop a value for average water saturation but does not have detailed information on permeability and porosity as a function of elevation. (Many wells are not core analyzed). However, he may know from results of pressure buildup tests that the average permeability in the region of the wellbore is, say, 100 md. Also, he may know from well logs that an average porosity is, say 18%. With these average permeability and porosity values plus information on the distance to the appropriate  datum and information on fluid properties, he can make a reasonable calculation of the average water saturation.

To illustrate the method of getting , consider the sketch above. At the wellbore location, the bottom and top of the formation are  and  from the  datum. For given values of , calculate . Shade the area enclosed by  and  on the  curve and calculate the average initial water saturation .

The simplest way of determining  is by graphical integration. Thus, determine the area under the curve, divide this area by the value  and the result will give . That is:

Example 3.5

Example calculation of the use of Capillary Pressure Data to Obtain Average Water Saturation Using J-Function

It is desired to calculate the initial oil in place for an oil reservoir having a gas cap as illustrated below. There is no prior J-function curve available and no well logs to give permeability, porosity, and saturation data with depth. All we have are old cores from storage.

The bulk volume of the oil zone is 1,000 acre-ft. The thickness of the oil zone is 20 ft. Four core samples were taken from the oil zone in the middle of 5 ft. intervals. From laboratory measurements of porosity and permeability, the data are:

 Interval depth Permeability Porosity 4,000 - 4,005 11.2 0.147 4,005 - 4,010 34.0 0.174 4,010 - 4,015 157.0 0.208 4,015 - 4,020 569.0 0.275

The free-water table is at a depth of 4030 ft. In addition to porosity and permeability, the capillary pressure for each sample was measured using air displacing water in a centrifuge. These laboratory derived capillary pressure curves are shown below. The water/oil interfacial tension for this reservoir is estimated to be 28 dynes/cm, the reservoir (water/oil) wetting angle is 0.0. The air/water interfacial tension is 70 dynes/cm with a wetting angle of 0.0 also, . Calculate the average water saturation and the initial oil in place.

Solution

a)      Convert the  data to data and calculate the J-function curve using:

This has been calculated and plotted below.

b) Calculate the value of  at each "" of each core and read the corresponding water saturation from the  curve.

 27.5 0.738 0.37 22.5 0.967 0.35 17.5 1.478 0.29 12.5 1.748 0.27

c) Obtain the arithmetic average water saturation.

The average water saturation

Using the Less correct method

Therefore,  by reading on the J-function curve at

Only  and  available

Suppose there are no cores, there are no well logs, an average  is available from well tests, and  can be estimated from correlations.

In this case, use the heights of the top and bottom of the pay zone from the free-water table to obtain  and .

Similarly,

Plot these and find the average water saturation graphically.

Exercises

1. Give a possible reason why  for a soap solution is about 40 dynes/cm and not about 70 dynes/cm as would be the case for fresh water.

2. Show that the expression  derived for a tube is a special form of the general expression:

3. Given that and , would you expect a water/air interface or an oil/water interface to have a smaller contact angle assuming the same capillary pressure applies to both, in the same capillary tube.

4. Calculate the entry pressure for natural gas into a pore throat having the following sizes and shapes.

a) A cylindrical shape pore throat of 0.0001 inch diameter

b)      An elliptical shape pore throat of d1 = 0.0001 inches and d2 = 0.001 inches

c) An infinite horizontal fracture of fracture width = 0.0001 inches.

Use s = 35.2 dynes/cm, and q = 0.0

References

1. Clark, Norman J.  "Elements of Petroleum  Reservoirs" Henry L. Doherty Series, Society of Petroleum Engineers of AIME, Dallas, 1960.

2. Slider H.C., “Worldwide Practical Petroleum Reservoir Engineering Methods”, Penwell Books, 1983

3. Wilhite, G.P. : “Waterflooding”, SPE Textbook Series, Vol. 3, 1986.

4. Standing, M.B.: Lecture notes, Stanford University, 1977

5. Amyx, J. W., Bass, Jnr. D. M., Whiting, R. L. : Petroleum Reservoir Engineering, McGraw-Hill, 1960