section 04 - capillary pressure

MSc Drilling & Well Engineering Formation Evaluation MSc Drilling & Well Engineering Formation Evaluation

CAPILLARY PRESSURE

The amount of hydrocarbon and water present at any level in a particular reservoir is strongly dependent on the interfacial forces acting both between the various fluids in the reservoir, and between these fluids and the rock matrix itself. Capillary pressure is caused by surface energy and wettability effects, and is defined as the difference in pressure across a curved interface between two fluids. Figures 1 and 2 illustrate this.

Fig.. 1 : Surface energy and Fig. 2 : Capillary pressure. wettability effect.

Surface Energy.

The fundamental property of a liquid surface is that it tends to contract to the smallest possible surface area. This tendency may be explained by the properties of molecules in a liquid. In all fluids the molecules are free to move. In liquids they are kept close together by attractive cohesive forces. A molecule in the interior of the liquid is completely surrounded by other molecules and is attracted equally in all directions. However, the molecular concentration in the vapour phase surrounding the liquid is relatively low. Consequently, a molecule at the liquid surface lacks outward attraction. In order to increase the surface area of the liquid, work must be done in moving the molecules from the interior. Therefore, it is evident that the surface possesses free surface energy.

Issued Oct 2003 Capillary Pressure Theo Grupping

MSc Drilling & Well Engineering Formation Evaluation

Figure 3 gives a schematic presentation of surface energy. In the liquid drop there are COHESIVE

attractive forces between the molecules. The molecules at the surface have anunbalanced pull inwards. The result is that the surface of the drop contracts to the smallest possible size. P1 is greater than P2, therefore the surface has free energy. The drop behaves as though it is contained in a skin.

Fig. 3 : Surface energy.

To increase the size of the drop by a unit surface area, work must be done to move the molecules outward from the interior of the drop.

Surface energy equals work per unit area increase. work done F x D F = = =

area increase L x D L

= Force per Unit Length = σ (surface tension). In Figure 4 it is shown that surface energy (work per unit areas increase) is numerically and dimensionally equal to the surface tension (force per unit length), acting at right angles to a line on the surface of the liquid. The unit of measurement may be expressed in dynes/cm, or Newton/m, and the symbol is sigma (σ).

. Fig. 4 : Surface tension.



As a consequence of the surface tension in a curved fluid interface, the pressure on the side containing the centre of curvature will be greater than that on the other side, by an amount which is inversely proportional to the radius of curvature of the interface. This is best explained as follows: Supply enough work to increase the radius of the sphere from r to (r + δr) The surface area increases from 4 π r2 to 4 π (r2 + 2r δr + δr2) The change in surface area is then 8 π r δr Since work done per unit area increase = surface tension, then: work done = (change in surface area) x (surface tension) = 8 π r δr x σ but: work done = (outward force) x (distance moved) = (difference in pressure) x (surface area) x (distance moved) = (P1 - P2) x 4 π r2 x δr At equilibrium: 8 π r δr x σ = (P1 - P2) x 4 π r2 x δr 2 σ or: (P1 - P2) = r The surface tension can be determined by the "ring method" in which the force F, required to break the two annular films is measured, as shown in Figure 5.

Fig. 5 : Surface tension measurement.

F σ = dynes/cm (10-3 Newton/m) 2 π D where: σ = surface tension F = force in dynes (Newton) D = diameter in cm (m)

Surface tensions between some common fluids and air at 20 oC are: Water 72.6 dynes/cm = 0.0726 Newton/meter Benzene 28.9 dynes/cm = 0.0289 Newton/meter Cyclohexane 25.3 dynes/cm = 0.0253 Newton/meter N-Hexane 18.4 dynes/cm = 0.0184 Newton/meter N-Octane 21.8 dynes/cm = 0.0218 Newton/meter The interfacial tension between water and oil at 20 oC is about 30 dynes/cm (0.030 N/m).



Interfacial tension between a liquid and its vapour phase decreases with increased temperature, until at the critical point the surface tension becomes zero, and differentiation between fluid and vapour phase ceases to exist. Wettability. The tendency of one liquid to displace another fluid from a solid surface is determined by the relative wettability of the fluids and the solid.

If the work of adhesion, of a fluid A, is greater than the work of adhesion of fluid B, then fluid A will displace fluid B from the surface until an equilibrium contact angle of θ is attained, as shown in Figure 6.

Fig. 6 : Contact angle wettability. The contact angle may be used to explain wettability. If the contact angle, measured through liquid A is acute, then the meniscus between the liquids will be concave towards liquid B, and liquid A will displace liquid B from a solid surface. Liquid A is a wetting phase with respect to the solid surface and liquid B. When considering the behavior of wetting fluids (like water) and non-wetting fluids (like mercury) in capillary tubes, the contact angle developed between the fluid and the wall of the tube determines the shape of the meniscus. Examples are shown in Figure 7.

Fig. 7 : Contact angles for wetting and non-wetting fluids.



Capillary Pressure. The pressure differential across a curved interface is: P1 - P2 = 2 σ / r

Fig. 8 : Capillary tube

For all practical purposes encountered in rocks, the liquid surface will be a portion of a hemisphere. In Figure 8 the radius of the capillary tube is given by: 2 σ cos θ rt = r cos θ , thus: P1 - P2 = rt

This relationship defines the capillary pressure in terms ofthe radius of the tube. It is this pressure difference whichcauses the rise of a wetting liquid in a capillary tube.

When liquid wets the surface of a glass capillary tube, the surface tension around the periphery of the contact pulls the liquid interface up into the tube, until equilibrium is reached with the downward force due to he height of the liquid column. The equation for this pressure equilibrium is given by:

surface tension x circumference = column hydrostatic pressure σ cos θ (2 π r) = π r2 h (ρL - ρV) g where: σ = surface tension, dynes/cm (N/m) r = radius of the tube, cm (m) h = capillary rise, cm (m) ρL = density of liquid, gram/cm3 (kg/m3)

ρV = density of vapour, gram/cm3 (kg/m3) g = acceleration due to gravity, 981 cm/sec2 (9.81 m/sec2) θ = contact angle between liquid and solid cos θ = -1 when θ = 180 o , for complete wetting.

A pressure applied above the liquid in a capillary, sufficient to lower the meniscus to the same level as in the reservoir, is equal to the capillary pressure. The capillary pressure in this equilibrium condition, PC = P1 - P2 , is given in Figure 9.

2 σ cos θ PC = P1 - P2 = rcap

Fig. 9 : Measurement of capillary pressure by depression of rise.



When the two fluids are air and water, the following relationships are apparent as shown in Figure 10.

P1 = P3 - h . g . ρρair

P2 = P3 - h . g . ρwater So: PC = P1 - P2 = h . g . (ρwater - ρair) Or, the height to which the water rises is: h = PC / g . (ρwater - ρair) Since: PC = 2 σ cos θ / r h = 2 σ cos θ / r . g . (ρwater - ρair)

Fig. 10 : Capillary tube with air/water system. Because surface tension, contact angle and fluid densities are constants for given fluids, it is evident that the height (h) to which the water will rise is inversely proportional to the radius (r) of the tube.

If air is replaced by oil, then the height to which the water will rise is: h = 2 σ cos θ / r . g . (ρwater - ρoil) This situation is shown in Figure 11. Free Water Level (PC = 0) Fig. 11 : Capillary tube with oil/water system.

For a very wide tube (r tends to infinity) both PC and h will be equal to zero. The water level will then be the same inside and outside the tube. This is called the "Free Water Level" (FWL). The Free Water Level is thus defined as the level at which PC = 0.



Since interfacial tensions and contact angles of various fluid/solid systems are known from laboratory measurements, it is possible to approximate the capillary pressure value. The data commonly used are given in Table 1.

Table 1

System θ cos θ σ 10-3 Nm-1

at 20 oC (dyne/cm) w/g/s 0 1 72 w/o/s 0 1 35 a/Hg/s 0 1 368 *

* using weathered mercury

The air/mercury/solid system (a/Hg/s) is included because mercury is often injected into air filled pores in core plugs in the laboratory, as a method of measuring capillary pressure. This technique will be described later. Capillary Pressure in Reservoir Rock. Reservoir rock contains numerous pores of different sizes. These pores are connected, directly or indirectly, to other pores. The whole pore system in hydrocarbon accumulations is normally in a water-wet condition. For discussion purposes it is assumed that all pores are represented by straight capillary tubes of only three different sizes: small, medium and large. Since the capillary pressure, PC, and the height to which the fluid rises, h, are inversely proportional to the pore radius, r, the fluid distribution for a water/gas/solid system will be as shown in Figure 12. The capillary pressure for one particular size of capillary tube with radius r can be read from the height versus pressure plot. It is simply the difference in water gradient and gas gradient pressure.



Fig. 12 : Capillary pressure - height versus pressure plot. Displacement Pressure. Before a non-wetting phase can penetrate a capillary tube, or a porous medium saturated with a wetting phase, a minimum threshold pressure must be exerted. This is shown in Figure 13.

Fig. 13 : Comparison of displacement from a capil To displace water, the wetting phase, with oil, the ntube, a pressure slightly higher than the capillary pres splacement pressure. Fig. 14 If the radius of a large pore, a medium pore and a sm0.25 x 10-5 m respectively, then an oil pressure of sland 28 x 103 Pa (which is equivalent to 1, 2 and 4 prespective pore.

Issued Oct 2003 Capillary Pressure

2 σ PC = (cos θ = 1) r 2 σ Pd =

req

lary tube and granular packs.

on-wetting phase, from a single capillary sure must be applied to the oil.

Likewise, to displace water from a granular pack, a pressure slightly greater than the threshold pressure must be applied to the oil. The displacement pressure is controlled by the size of the pore openings, req, as defined by the relationship in Figure 13. If the sample in Figure 14 is fully water-bearing, and if oil is forced into the pores to displace the water, then a pressure just exceeding the capillary

all pore is 1.0 x 10-5 m, 0.5 x 10-5 m and ightly more than 7 x 103 Pa, 14 x 103 Pa si) has to be applied to force oil into the

Theo Grupping


The contribution of the volume of each capillary tube to the total porosity volume in this case is 76 %, 19 % and 5 % respectively.

If it is assumed that the number of pores (straight capillary tubes) in Figure 14 is equal for each size, then one may plot the pore volume of each tube against the required injection pressure PC. The pressure required for the initial displacement of water by oil is the initial displacement pressure, or the 100 % water level. This is the height above the Free Water Level (FWL) at which the rock is still 100 % water saturated as the result of capillary forces, as shown in Figure 15.

Fig. 15 : Displacement vs. pressure.

If the sample has a continuous spectrum of tubes of different radii with a similar distribution, the displacement history can be represented by a curve which is called the "capillary pressure curve". During deposition, reservoir rocks are completely saturated and water-wet, as hydrocarbons migrate and accumulate in the reservoir rock, a portion of this connate water is displaced. Both silica (SiO2) and calcite (CaCO3) have a strong tendency to remain water-wet. Therefore, the connate water may only be displaced by hydrocarbons migrating into the reservoir to the extent of attaining an equilibrium, when the pressure arising from fluid density differences is equal to the capillary pressure between the fluid phases at a particular level. Taking the level where the water saturation is 100 %, the pressure difference between the phases at some elevation "h" above this level will be: Po - Pw = g . h . (ρw - ρo)



This pressure difference, which is equal to the capillary pressure, increases with height above the FWL and a gradient in the capillary pressure implies a gradient in the water saturation above the hydrocarbon, though the capillary pressure curve becomes almost vertical at great heights above the FWL, indicating an apparently irreducible water saturation (Swi). Laboratory Capillary Pressure Measurements. 1. Capillary pressures can be measured using a pressure vessel filled with oil in which the

sample, saturated with water, is immersed, as shown in Figure 16. Oil is admitted stepwise with small pressure increments, and the displaced water is measured at each step in a graduated cylinder. A fritted glass disk of very low permeability, and strongly water-wet, allows only water to pass, provided that the oil pressure does not exceed about 7 bar (the entry pressure to the glass disk).

The capillary pressure curve of the sample is obtained by plotting the pressure applied against the volume of water displaced, as illustrated in Figure 17. The volume of water is represented as a percentage of the previously carefully measured interconnected pore volume.

Fig. 16 : Capillary pressure vessel. Fig. 17 : Capillary pressure curve.

The advantage of this method is that reservoir fluid conditions may be simulated; the disadvantage is that the method is normally very time consuming. A very long period of time is needed for the oil to displace water from every pore of a given radius throughout the entire core volume, before proceeding to the next pressure increment. Another disadvantage is the limited pressure range.

2 Capillary pressures can also be derived using the centrifuge technique. A sample, initially

saturated with wetting fluid (e.g. water or kerosene), is placed in a centrifuge, which is then rotated at successfully higher speeds. The amount of wetting fluid, displaced from the sample as a result of a measured increase in the rotational speed, is carefully measured at each step. The end of the sample furthest from the centre of rotation will remain completely saturated. The average saturation of the sample will be a function of the pressure differential along it, and thus of the rotational speed. Only when the sample is completely homogeneous, can the capillary pressure be calculated from the centrifugal speed and the



average saturation of the sample. The advantage of the method is that it is fast and that water can be used as the wetting fluid, but a limitation is that only gas can be used as the displacing medium.

3. The mercury injection method was developed by Purcell. The dry sample is placed in a

chamber, which is then evacuated and filled with mercury. The mercury will not enter the sample at atmospheric pressure. Pressure is then applied to the mercury and will build up until the initial displacement pressure for the sample is exceeded. Thereafter the mercury will enter the pores with the greatest diameter. As the pressure continue to increase, the mercury will occupy essentially the same pore space as the hydrocarbons in-situ.

The capillary pressure curve obtained with the mercury injection method should be the same as that using hydrocarbons under in-situ conditions, providing correction is made for the difference in contact angles and interfacial tensions between the two systems. A pressure conversion must, therefore, be made in order to make a direct comparison. The pressure conversion is done as follows: PC(a/Hg/s) = 2 σ cos θ/r = 2 x 368 x 1/r PC(o/w/s) = 2 σ cos θ/r = 2 x 35 x 1/r Therefore: PC(a/Hg/s) 368 = = 10.5 PC(o/w/s) 35 Similarly: PC(a/Hg/s) 368 = = 5.1 PC(g/w/s) 72 (conditions at 20 oC. At higher temperatures and pressures encountered in a reservoir, values for σoil/water = 20 dynes/cm and σgas/water = 40 dynes/cm are not uncommon). Application of mercury capillary pressure data in reservoir evaluations requires only this straight conversion. Air-mercury capillary pressure curves are the most commonly measured, and unless specifically stated otherwise, it may be assumed that any capillary pressure curve given is an air-mercury measurement. The values of 10.5 and 5.1 are often simplified to 10 and 5 for the "average" reservoir oil and gas respectively. This is justified on the grounds of the inaccuracy of the values of theta (θ) and sigma (σ) actually used, because of assumptions made concerning in-situ reservoir conditions.



In many cases, however, actual measurements on the formation fluids in the reservoir show that if this simplification had been employed instead of the measured values, a very considerable error would have resulted. Practical Application of Capillary Pressure Data. It has been shown that the capillary pressure is related to the height above the FWL by the equation: PC = ∆ρ . g . h This can be written as follows: PC(a/Hg/s) = 10 . PC(o/w/s) = 10 . ∆ρ . g . h and: PC(a/Hg/s) = 5 . PC(g/w/s) = 5 . ∆ρ . g . h In practice various systems are in use; the c.g.s. system (cm/gram/second), the imperial system, and the system using S.I. units. Because it may be necessary to convert from one system to another, consistent sets of units and conversion factors are given below: Parameter Symbol C.G.S. Unit Imperial Unit S.I. Unit Density difference ∆ρ g . cm-3 psi / ft. kg . m-3

Gravity acceleration g 981 cm . sec-2 1 * 9.81 m . sec-2

Height above FWL h cm ft. m Capillary pressure PC dynes . cm-2 psi Pa * Since pressures are quoted as weight per unit area above, as opposed to force per unit

area, the gravitational constant is already included.



The most commonly used conversion factors are: 1 g . cm-3 = 62.4 / 144 psi / ft. = 0.433 psi / ft. = 103 kg . m-3 69,000 dynes . cm-2 = 1 psi

1.01 . 106 dynes . cm-2 = 14.7 psi = 1 atmosphere

1 kg. cm-2 = 14.22 psi = 0.981 bar 1 bar = 14.5 psi

Fluid Distribution in Heterogeneous Reservoirs. Rock properties can vary widely across a continuous reservoir. A common example is in deltaic-type deposits, where sands may be laid down in fan-like bodies. Towards the edges of the fan the grain size will decrease as the velocity of the water carrying the sand was reduced, and fine clay particles will be present in increasing proportions as the sands themselves become finer grained. The clay will occupy space between the sand grains, reducing the porosity. Because the sand grains are very small, the pores between them will be very narrow, and thus have a very small radius. This causes the capillary pressure to increase, and also the water saturation in the case that the reservoir is hydrocarbon-bearing. Such a situation is shown in Figure 18, demonstrating the use of capillary pressure measurements on core material to predict water saturation at any given height in the reservoir. Ideally, the same water saturation should be obtained from wireline logs. Where this is the case, well logs can be used with confidence to derive saturations in the wells in the same reservoir which have not been cored. If there is a discrepancy between the two methods, the quality of core measurements and the well logs and rock properties used in the calculations have to be critically examined.



Fig. 18 : Fluid distribution in a heterogeneous reservoir.

If as shown in Figure 19, uniform sand layers, each of different grain size, are laid down in sequence, each layer will have a different capillary pressure curve. The hydrocarbon saturation, and the height of the 100 % water saturation level with regard to the FWL, will be different for each layer. This explains why it is possible to drill through 100 % water saturated layers in an oil or gas reservoir structurally above the hydrocarbon/water contact for the field, as sketched in Figure 19 (Well C).


Fig. 19 : Schematic reservoir.

section 04 - capillary pressure

Documents

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engineering

straight capillary

capillary

initial displacement

mercury injection

free water

contact angle