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Phase behavior of pure fluids

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An accurate characterization of phase behavior is critical to the prediction of oil recovery. Often, sufficient pressure-volume-temperature (PVT) experimental data is not available, and mathematical models that are "tuned" to experimental data are needed. Equation of state (EOS) calculations are used for this purpose. EOS models are typically easy to implement in a numerical simulator.

Pressure-temperature diagrams

The phase behavior of a typical pure fluid can be represented on a pressure-temperature diagram, as illustrated in Fig. 1. From the Gibbs phase rule, the number of degrees of freedom are 3 - np, which means that one, two, or three phases can be present at equilibrium. For simplicity, these phases are shown as a solid, liquid, and a vapor, although numerous additional solid and liquid phases are possible, as long as no more than three of those phases coexist at equilibrium at the same temperature and pressure. Water, for example, has nine different solid phases, each of which has a different crystalline structure. At the pressure and temperature of the Earth’s surface, however, we experience only one solid, liquid, and vapor phase of water.

According to the Gibbs phase rule, there are no degrees of freedom when three phases are in equilibrium. This necessarily implies that three phases must be in equilibrium only at one temperature and pressure; this is the triple point indicated in Fig. 1. That is, we are not free to choose the temperature and pressure at which three phases can coexist. The temperature and pressure of the triple point are determined.

For two phases, however, the degrees of freedom are one, and we can set either the temperature or the pressure, but not both. Once the temperature is specified, the pressure is determined. Thus, two equilibrium phases are represented by curves on a pressure-temperature diagram.

• The sublimation curve gives the locus of points where solid and vapor are in equilibrium.
• The melting or fusion curve gives the locus of points where solid and liquid are in equilibrium
• The vaporization or saturation curve gives the locus of points where the liquid and vapor are in equilibrium.

The pressure along the vaporization curve is called the vapor or saturation pressure. For example, the pure fluid at temperature T1 in Fig. 1 has a vapor pressure of .

A single solid, liquid, or vapor phase can exist over a range of temperatures and pressures (degrees of freedom are two). This is indicated in Fig. 1 by the three single-phase regions of:

• Solid
• Liquid
• Vapor

When the temperature and pressure along the vaporization curve is increased to the critical pressure, pc, and temperature, Tc, the interface between the liquid and vapor phases becomes indistinct. This occurs at the critical point shown in Fig. 1. Beyond the critical point, fluids become "supercritical" and no phase interface is visible. For example, the vapor at point "A" in Fig. 1 would become more liquid-like as the pressure and temperature are varied along the semicircle to point "B," a liquid. The fluid would never exhibit two phases during the process, as identified by an interface. Because there is no actual boundary that defines the supercritical region, fluids should be described in terms of liquid-like or vapor-like.

One important characteristic of critical fluids is that thermodynamic properties approach each other; that is, the densities/viscosities of the vapor and liquid phases become the same at the critical point. We often inject supercritical fluids, such as carbon dioxide, into reservoirs so that, as the supercritical fluid and the reservoir fluid mix, phase interfaces disappear, facilitating production of hydrocarbon components previously not recovered. Thus, an accurate characterization of critical behavior is very important.

PVT experiment

A pressure-temperature diagram contains no information about molar volume—just phase boundaries. Consider an experiment in a closed system that is held at constant temperature (ignoring the solid phase). The experiment is designed to measure pressure as the volume is changed and is generally referred to as a PVT experiment. Fig. 2 illustrates several phase behavior states that occur in a PVT experiment. The article on phase diagrams outlines similar experiments for multicomponent mixtures. The steps in the PVT experiment for a pure fluid are listed next.

1. Place a known amount of a single-component fluid in a constant temperature PVT cell at vapor state one, illustrated in Fig. 2. The molar volume of the fluid can be calculated from . Measure the pressure.
2. Inject a volume of mercury into the cell. Calculate VV again, and monitor the corresponding pressure change. Because vapor is highly compressible and the mercury is highly incompressible, the pressure change with decreasing molar volume of the vapor phase will be relatively small. The fluid remains a vapor.
3. Continue to inject mercury until the vapor becomes saturated at state two. At this pressure (the vapor pressure, , in Fig. 1), a small amount of liquid forms (for multicomponent mixtures, this is the dew point). Calculate the saturated VV, and measure the pressure.
4. Continue to inject mercury to state three, where the molar fractions of vapor and liquid are about equal. The pressure is constant at the vapor pressure , corresponding to temperature T1. The pressure is constant because the temperature is constant (i.e., the degrees of freedom from the Gibbs phase rule are one for two phases in equilibrium; thus, the pressure is fixed if the temperature is constant). Because the pressure does not change with decreasing molar volume, the compressibility of the closed system is infinite.
5. Continue to inject mercury to state four until only a small drop of vapor remains (for multicomponent mixtures, this is the bubble point). Calculate the saturated liquid molar volume, VL.
6. Continue mercury injection to state five and calculate VL, while monitoring the pressure. During this step, the fluid remains a liquid, and the pressure rises very quickly for small changes in the molar volume. The pressure rises quickly with decreasing molar volume because liquids are nearly incompressible compared to vapors.

The PVT process just described generates an isotherm on a pressure-volume diagram, as illustrated in Fig. 3. States one through five correspond to those in the isothermal PVT experiment. A vapor pressure dome or two-phase envelope outlines the two-phase region. Along the left boundary of the dome, the liquid is completely saturated, for any reduction in pressure would cause a bubble of vapor to form. Along the right boundary of the dome, the vapor is saturated. To the left of the dome the liquid is subcooled, whereas to the right of the dome, the vapor is superheated. Two metastable states have been observed within the two-phase region:

• Superheated liquid
• Subcooled vapor

These states are unlikely to exist in a reservoir where nucleation is always present (see Firoozabadi on nucleation).

From the Gibbs phase rule, the pressure within the two-phase region along the T1 isotherm remains at its corresponding vapor pressure as molar volume changes from states two to four (i.e., VV to VL). VV in Fig. 3 is the molar volume of the saturated vapor, whereas VL is the saturated liquid molar volume. VC is the molar volume of the fluid at its critical point.

The critical point, in which the molar volumes of the liquid and vapor become equal, is at the apex of the dome in Fig. 3. The critical point must be at the apex because the isotherms in the two-phase region are horizontal, as required by the Gibbs phase rule. Thus, the slope and the inflection point of the isotherm at the critical temperature must be zero. For isotherms where T > TC, the molar volume (inverse of molar density) changes continuously from vapor-like values to liquid-like values as the pressure increases.

The isothermal compressibility of the fluid is given by the inverse of the slopes of the isotherms. For isotherms where T < TC, the slope is small in the superheated vapor region, indicating that the fluid compressibility is large. The slope is large in the subcooled liquid region, indicating that the fluid is nearly incompressible. At the boundary of the two-phase dome, the isothermal compressibility is discontinuous and is equal to infinity (zero slope). Compressibility is infinite in the two-phase region because, as the system volume is decreased, some of the vapor molecules, which occupy more space, are condensed into the denser liquid phase in which they occupy less space.

For a two-phase mixture at constant temperature, such as that shown by state three in Fig. 3, the molar phase volumes, VV and VL, do not change as the volume of the closed system changes. This occurs because the pressure is constant in the two-phase region for a pure fluid. Thus, as the volume is changed, the molar densities of the vapor and liquid phases do not change, but only the molar fractions (or amounts) of the phases change. The overall density of the two-phase mixture will change as the closed system is compressed or expanded.

For example, consider state three in the two-phase region at a temperature of T1. The vapor molar density is (at state two), whereas the liquid molar density is (at state four). The mole fraction of vapor at state two is V. Extensive parameters, such as the total molar volume of the two-phase mixture, can be calculated by ....................(1)

For total molar enthalpy, ....................(2)

where:

• HL is the molar enthalpy of the saturated liquid
• HV is the molar enthalpy of the saturated vapor

Eqs. 1 and 2 can be solved for V as ....................(3)

Eq. 3 is known as a lever rule. The fluid quality is the molar volume fraction V as a percentage. An illustration of quality lines is given in Fig. 4. A quality of 100% is a saturated vapor. State three in Fig. 4 has a quality of about 60%.

Phase behavior models of pure fluids

One of the first EOS was Boyle’s and Charles’s law. These laws were combined into the ideal gas equation we use today, pV = RT. The ideal gas equation is generally satisfactory for vapors at pressures below a few atmospheres. Numerous other types of EOS were developed through the years, but the most widely used EOS type in the petroleum industry is the cubic EOS.

Van der Waals developed the first cubic EOS. Unlike the ideal gas equation, which is limited to low pressure vapors, the van der Waals EOS attempted to provide good phase behavior estimates for both liquids and vapors by using only one equation. He also developed the principle of corresponding states, which is frequently used in the petroleum industry today. Numerous cubic EOS models are available today that give better accuracy than the van der Waals’ EOS. The two most widely used cubic EOS are:

• The Peng-Robinson EOS
• Modified versions of the Redlich-Kwong EOS

Prediction of the phase behavior of real reservoir fluids is difficult because of the complex interaction of molecules. Intermolecular forces of attraction and repulsion determine thermodynamic properties for any mixture of molecules. The attraction forces allow fluids to form liquid and solid phases, whereas repulsions are responsible for resistance to compression.

The accuracy of any EOS depends on its ability to model the attractions and repulsions between molecules over a wide range of temperatures and pressures. EOS models are empirical in that they do not attempt to model the detailed physics but only the cumulative effect in terms of a small number of empirical parameters. Generally, EOS models are more accurate when attractions are small, which explains why water, a polar substance, is more difficult to model.

Ideal gas equation

The simplest and most fundamental EOS is the ideal gas equation, in which the pressure, volume, and temperature of a fluid are related by ....................(4)

As stated previously, the behavior of a gas may be approximated by Eq. 4 if the pressure is relatively low (i.e., less than a few atmospheres). A gas is ideal if molecular interactions are negligible, something that could only occur at zero pressure. Thus, molecular interactions are negligible at zero pressure; therefore, thermodynamic properties, such as the molar internal energy of an ideal gas, are only a function of temperature.

Real fluid equation

Most fluids are not ideal. Real fluids, whether vapor or liquid, can be defined by the compressibility factor ....................(5)

Intensive thermodynamic properties, such as the molar internal energy, are a function of both temperature and pressure for a real fluid. A common mistake is to believe that Eq. 5 is only applicable to gases. Eq. 5 will be used to represent the behavior of any phase, whether liquid or vapor.

Coefficient of isothermal and isobaric compressibility

For a single-phase fluid, we often employ very simple equations-of-state that describe the relationships between pressure, temperature, and molar volume. As stated previously, molar volume is a state function—it is not determined by the process or path taken to that state. Thus, for every temperature and pressure, there corresponds only one value of molar volume, and hence, we can write an equation that describes differential changes in molar volume as ....................(6)

Division of Eq. 6 by the molar volume gives ....................(7)

where

The minus sign is introduced, per convention, to make the compressibilities positive.

Eq. 7 describes the fractional change in the molar volume for small changes in temperature and pressure. Because reservoirs are usually thought to be isothermal, the isothermal compressibility is most often used in reservoir engineering. It is often written in terms of density or formation volume factor (FVF) as

Several special cases of Eq. 7 are useful to consider. First, when the volume under consideration is closed (dV = 0), Eq. 7 gives . Under the assumption of constant compressibilities, this result can be integrated to , which gives the pressure change in a closed volume resulting from a temperature change.

Second, when the process is isothermal, Eq. 7 reduces to . This result can be integrated under the assumption of constant compressibility to obtain ....................(8)

where subscript o is a reference state. Eq. 8 is often referred to as the EOS for a constant compressibility fluid. It can be rewritten in a more familiar form as or . This result is often used in well testing.

Third, when the compressibility of the fluid is both constant and small, and the pressure change is not too large, the exponential term in Eq. 8 can be simplified so that V = Vo(1 - cΔp) or ρ = ρo(1 + cΔp) . This result is often referred to as the EOS for a fluid with a slight, but constant, compressibility.

Last, substitution of the definition of compressibility factor (Eq. 5) into Eq. 7 gives ....................(9)

Eq. 9 is exact, whether the fluid is a vapor or liquid. For an ideal gas, the compressibility factor is constant and equal to 1.0, and Eq. 9 reduces to . Thus, the isothermal compressibility for an ideal gas is inversely proportional to pressure.

Cubic equations of state

The van der Waals EOS was the first EOS capable of representing both the liquid and vapor. The van der Waals EOS, however, is not used because of its lack of accuracy near critical points. The Peng-Robinson EOS and a modified version of the Redlich-Kwong EOS are generally used (see Table 1 for comparison). Nevertheless, the simplicity of the van der Waals EOS makes it useful in demonstrating several key concepts.

As for other cubic EOS, the van der Waals EOS describes the pressure as a function of molar volume and temperature, which is written as ....................(10)

where:

• b is the repulsion parameter
• a is the attraction parameter

The first term on the right side of Eq. 10 attempts to represent the pressure deviation from an ideal gas that results from molecules occupying and competing for space. The effective volume available for movement of the molecules (on a molar basis) is Vb—not V, as it would be for an ideal gas. Thus, b represents the smallest possible volume that one mole of molecules can occupy (no space would exist between the molecules).

As V approaches b, the denominator in the first term on the right side becomes small so that pressure increases very rapidly. Because b is based on the effective molecule size, the value for b will change with the nature of the pure fluid and will determine the lower boundary for the region of interest on a pressure-volume diagram (i.e., the physical region of interest is only where V > b).

The second term on the right side of Eq. 10 accounts for the attractive forces between molecules. The attractive forces will be proportional to the square of the number of molecules present and, thus, on a macroscopic scale. The proportionality constant, a, depends on the nature and strength of the forces between the molecules and, therefore, the fluid type. As V becomes large, the contribution of the attractive forces becomes small, and the second term on the right side of Eq. 10 becomes negligible. Under these conditions, the van der Waals EOS approaches the ideal gas equation (Eq. 4). This is also true for the other EOS in Table 1.

The van der Waals EOS can be rewritten in cubic form as ....................(11)

or in terms of the compressibility factor,

Determination of a and b from van der Waals EOS

We previously demonstrated that an isotherm on a pressure-volume diagram must have zero slope at the critical point. There is also an inflection point there. Thus, any cubic EOS must be constrained by ....................(12)

Eq. 12 consists of two equations that can be solved simultaneously to determine the unknowns, a and b. This procedure can be somewhat tedious, especially for more advanced cubic EOS.

A simpler method is to recognize that Eq. 12 implies that only one volume root of the cubic EOS exists at the critical point. Mathematically, this can be expressed as (V - VC)3 = 0, which upon expansion is . Comparison of this expansion to Eq. 13 term by term at the critical point gives , , and . These three relationships are easily solved to obtain a and b in terms of only the critical pressure and temperature, as well as to determine the relationship between critical parameters. That is, ; ; and . A similar procedure can be used for any cubic EOS.

The result shows that the critical compressibility factor is constant and is independent of the fluid. This is true for the other cubic EOS models in Table 1. For the van der Waals EOS, we obtained . In reality, the critical compressibility factor is not constant for different fluids and is generally smaller than 0.3. For example, water has a critical compressibility of about 0.23, and that of carbon dioxide is 0.27. The poor match of the van der Waals EOS at the critical point explains why it is not used today. Both the Redlich-Kwong EOS and the Peng-Robinson EOS have critical compressibilities closer to measured values (see Table 1).

A constant critical compressibility factor means that only two of the three critical properties can be satisfied at the critical point. For example, if critical pressure and temperature are specified, the critical volume will not be correctly predicted. For the van der Waals EOS, . Thus, when the critical pressure and temperature are specified, the critical volume or critical density of the fluid is likely in error. In general, liquid densities predicted by cubic EOS exhibit greater error than do vapor densities. Firoozabadi gives a good description of how density predictions can be improved using volume translation parameters.

We determined a and b only at the critical point. Their values could be different away from the critical point and could be functions of temperature. Therefore, in general, , and , where α and β are functions of temperature; these functions must approach 1.0 at the critical point. For most cubic EOS, β is taken to be 1.0, and α is adjusted to give the correct vapor pressure (see Table 1).

Construction of pressure-volume diagram from cubic EOS

Once a and b are defined, a cubic EOS can be used to generate the pressure-volume diagram, as illustrated in Fig. 3. From the Gibbs phase rule, there is only one degree of freedom within the two-phase region for pure fluids. We assume that temperature has been specified but not the pressure. Our goal, therefore, is to solve for the vapor pressure, given temperature and the critical properties of the fluid.

The procedure used to estimate a and b ensures that the cubic EOS gives the experimentally measured shapes of the isotherms for temperatures greater than or equal to the critical point. Fig. 5 illustrates the isotherms generated from a cubic EOS. As shown by the loop in the curve, cubic EOS can have three roots for isotherms below the critical temperature. The loop is not physical because the pressure must be constant in the two-phase region.

The nonphysical condition must be removed to achieve the correct physical response in the two-phase region. That is, the loop within the two-phase region must be discarded and replaced by the correct vapor pressure. The procedure is relatively simple. Within the two-phase region, there are three roots along an isotherm. At the vapor pressure, the root with the largest molar volume is taken to be the molar volume of the saturated vapor VV, whereas the smallest root is VL. The middle root is discarded because choosing that root would lead to unstable phases (see equilibrium, stability, and reversible thermodynamic systems). The middle root is clearly nonphysical in that it is located on the isotherm where (i.e., pressure increases as density decreases).

Principle of corresponding states

Correlations for reservoir fluids, such as the generalized compressibility factor charts for natural gases, use reduced temperature, pressure, and volume, where , , and . A cubic EOS shows why these parameters give good correlations. For example, substitution of the reduced parameters into the van der Waals EOS (Eq. 10), along with the definitions of a and b, gives ....................(13)

Eq. 13 is dimensionless and is often called the reduced form of the van der Waals EOS. The reduced form leads directly to the principle of corresponding states. The two-parameter principle of corresponding states says that all fluids, when compared at the same reduced temperature and pressure, have approximately the same compressibility factor, and all deviate from ideal-gas behavior by about the same degree. The reduced compressibility factor is given by . Because ZC is constant for a cubic EOS, the compressibility factor is constant for the same reduced temperature and pressure (reduced volume is related to reduced pressure and temperature through Eq. 13).

The principle of corresponding states is a powerful idea even though it is only qualitatively correct. Experiments show that ZC is not constant for different fluids. Nevertheless, it demonstrates that, to obtain reasonable estimates of fluid properties, only the reduced pressure and temperature must be known. This is why most fluid correlations use reduced temperature and pressure.

In reality, fluids can deviate from the principle of corresponding states. Pitzer noted that a plot of vs. for simple fluids (molecules that are roughly spherical in shape such as the Noble gases) collapse onto a straight line (see Fig. 6). The parameter, , is the reduced pressure at the vapor pressure. Other more complex and nonspherical molecules such as hydrocarbon-chained molecules, however, do not plot on that same line. Thus, Pitzer defined an additional correlation variable called the acentric factor, where

The acentric factor measures the deviation of complex fluids from the simple fluids at a reduced temperature of 0.70 (see Fig. 6). Hydrocarbons with longer chains generally have greater acentric factors. For example, methane has an acentric factor of 0.008, while n -butane has an acentric factor of 0.193.

Because the acentric factor is simple to measure, it is often used to improve phase-behavior calculations from cubic EOS. The three-parameter principle of corresponding states is that a fluid will have about the same compressibility factor as another fluid, if the reduced pressure, reduced temperature, and acentric factors are similar.

Calculation of vapor pressure ....................(14) ....................(15)

Although the shape of an isotherm from a cubic EOS can be made qualitatively correct, the problem remains that the vapor pressure is unknown for a given temperature. The vapor pressure is determined using the equilibrium criterion of Eq. 14. For example, substitution of the Soave Redlich-Kwong EOS into Eq. 15 and subsequent integration gives the fugacity as a function of pressure, molar volume, and temperature. That is,

The fugacity of the vapor is computed using the molar volume of the vapor phase, VV, whereas the liquid fugacity is determined using VL. Thus, the fugacity for the vapor phase from the Soave Redlich-Kwong EOS is ....................(16)

where ; and for the liquid phase, ....................(17)

The problem of calculating the vapor pressure reduces to finding the pressure that gives the required phase molar volumes so that the fugacities of the phases are equal. Fig. 7 illustrates the procedure. The procedure works well as long as the initial guess for the pressure is in the range of the cubic EOS where three roots exist (i.e., the pressure is within the loop of the cubic EOS). If the pressure is above the critical pressure, only one root exists for the molar volume. This is also true if the pressure is below the minimum value of the loop (the minimum pressure of the loop could be negative).

For a pure fluid, the vapor pressure can also be determined graphically with Maxwell’s equal area construction. Fig. 5 shows that the vapor pressure is the pressure required so that the areas bounded by the vapor pressure line and the loop of the cubic EOS must be equal. The equal area construction results from the equality of Gibbs energy (or fugacities). This method is less accurate but serves as a useful check to the calculated vapor pressure. Firoozabadi outlines Maxwell’s construction method in detail for pure fluids and mixtures.

Example calculation of two-phase envelope

This page demonstrates a calculation of vapor pressure and the two-phase envelope for a pure fluid using the procedure outlined in Fig. 7. Propane is selected as the pure fluid at a temperature of 40°C (313°K). We also select the Soave-Redlich-Kwong EOS to model the phase behavior. The properties for propane are a critical temperature of 370°K; a critical pressure of 42.5 bars; and an acentric factor of 0.152. The gas constant in consistent units is 83.1 cm3-bar/mol/K.

With these values, the parameter "a" in Table 1 is calculated to be 9.51E6 cm6-bar/mol2, and parameter "b" is 62.7 cm3/mol. The value for α from Table 1 is found to be 1.05 (the reduced temperature at 40°C is 0.913). Fig. 8 shows the isotherm at 40°C calculated with the SRKEOS.

Based on the calculated isotherm, we select an initial value of 10 bars for the vapor pressure. Any value within the S -loop of the isotherm would be satisfactory as an initial guess for the vapor pressure. A vapor pressure of 13.8 bars is calculated with the iteration procedure of Fig. 7. The calculated vapor pressure is the pressure at which the fugacities of the vapor and liquid phases are equal (illustrated in Figs. 5 and 8 by the Maxwell equal-area rule). The fugacities are 11.3 bars at the vapor pressure, which are calculated with Eqs. 16 and 17. The equilibrium liquid molar volume is 105 cm3/mol, and the vapor molar volume is 1462 cm3/mol.

Fig. 8 also illustrates the phase behavior with the SRKEOS at a higher temperature of 70°C. The calculated vapor pressure at this temperature is 26.2 bars. The equilibrium liquid molar volume is 128 cm3/mol, and the vapor molar volume is 691 cm3/mol. Thus, as the temperature is increased, the size of the two-phase region shrinks. Fig. 8 shows the two-phase envelope that connects the liquid and vapor molar volumes. At the critical temperature, the two-phase region disappears.

The values for vapor pressure and molar volumes are calculated parameters only. Using the critical temperature and pressure, the critical volume from the SRKEOS is approximately 241 cm3/mol (ZC = 1/3 always for the SRKEOS). The actual critical volume from experimental data is 200 cm3/mol, which is about 20 percent smaller than the calculated value. Firoozabadi outlines a more complex method to improve the calculated match of cubic EOS to experimental data.

Nomenclature

 c = isothermal compressibility of a fluid, 1/pressure, 1/Pa cp = isobaric compressibility of a fluid, 1/pressure, 1/Pa f = fugacity of a pure fluid, pressure, Pa H = molar enthalpy of fluid, energy/mole, J/mole L = liquid mole fraction, moles liquid/total moles, dimensionless p = pressure, force/area, Pa R = gas constant, pressure-volume/temperature/mole, Pa-m3/(Kelvin-mole) T = temperature, Kelvin V = vapor mole fraction, moles vapor/total moles, dimensionless or molar volume of fluid, volume/mole, m3/mole Z = compressibility factor of a fluid, dimensionless α = temperature dependence function in cubic EOS in Table 1, dimensionless β = temperature dependence function in cubic EOS, typically set to 1.0, dimensionless ρ = molar density of fluid, moles/volume, mole/m3 ω = acentric factor, dimensionless