# Phase and chemical equilibrium

## Gibbs Phase Rule

The Gibbs phase rule identifies the degree of freedom of a multiphase system that is in thermodynamic equilibrium. It relates the number of intensive independent thermodynamic properties for each phase and the number of phases for a system. For a system that does not experience a chemical reaction, the Gibbs phase rule reads as follows:

$\Pi + f = N + 2\qquad \qquad(1)$

where Π is the number of phases present and N is the number of components present. The degrees of freedom, f, designates the number of intensive independent properties that must be specified to fix the state of a system for each phase. Application of the Gibbs phase rule can be illustrated by considering the pure substance water, where N=1. When one phase is present, Π = 1, so that for the case of a subcooled solid, for example, it can be determined that f = 2. This means that two intensive properties must be specified to fix the exact state of the system, i.e., the system can exist in equilibrium for any arbitrary combination of temperature and pressure. A system that must have two intensive properties specified is a system with two degrees of freedom. Another example is a pure substance that has two phases in equilibrium, such as saturated liquid and vapor. The number of phases in this case is Π = 2, and the number of the components is N = 1. Application of the Gibbs phase rule leads to f = 1, which means that only one intensive property must be specified to determine the state of the system in each phase. If the pressure is given, the temperature is directly known and therefore the state of the system in each phase is determined. However, the quality of the two-phase system, which is the fraction of vapor in the saturated two-phase mixture, is not known and is needed to find the relative amount of one phase with respect to the other. This type of substance has one degree of freedom. Finally, consider the triple point of a pure substance, where N = 1 and Π = 3, which leads to f = 0. This is a system with zero degrees of freedom because all intensive properties are fixed and therefore the state of the system is known.

# Equilibrium and Stability of Multiphase Systems

Thermodynamic equilibrium criteria for single-phase systems were discussed in Section 2.3. This section focuses on thermodynamic equilibrium and stability criteria for multiphase systems.

## Two-Phase Single-Component Systems

The criteria for equilibrium of two phases of a pure substance can be developed from any of the criteria for equilibrium equalities given in Section 2.3, along with a corresponding fundamental relationship. An infinitesimal departure from equilibrium will result in zero entropy change when the internal energy and volume are held constant, i.e.,

$d{S_{E,V}} = 0\qquad \qquad(2)$

For a closed system containing two phases, eq. (2) can be written as

${\left( {\sum\limits_{k = 1}^2 {d{S_k}} } \right)_{E,V}} = 0\qquad \qquad(3)$

where the subscript k identifies the individual phases. This generic subscript is chosen so that eq. (3) can represent any type of phase change combination, including liquid-vapor, solid-liquid, or solid-vapor. The fundamental relation, eq. $dE = TdS - pdV + \sum\limits_{i = 1}^N {{\mu _i}d{n_i}}$ from Closed Systems with Compositional Change, in terms of internal energy, where Ni = 1 (single component) for both phases under consideration, is as follows:

$d{E_k} = {T_k}d{S_k} - {p_k}d{V_k} + {\mu _k}d{n_k}\begin{array}{*{20}{c}} {} & {(k = 1,2)} \\ \end{array}\qquad \qquad(4)$

Since E and V are held constant in this analysis, the change in internal energy of the two phases must sum to zero, i.e.,

$dE = \sum\limits_{k = 1}^2 {d{E_k}} = 0\qquad \qquad(5)$

$dV = \sum\limits_{k = 1}^2 {d{V_k}} = 0\qquad \qquad(6)$

Also, because the system is closed by definition, the change in the number of moles resulting from the two-phase changes must sum to zero:

$dn = \sum\limits_{k = 1}^2 {d{n_k}} = 0\qquad \qquad(7)$

Combining eqs. (2) – (7), the following expression is obtained:

$dS = 0 = \left( {\frac{1}{{{T_1}}} - \frac{1}{{{T_2}}}} \right)d{E_1} + \left( {\frac{{{p_1}}}{{{T_1}}} - \frac{{{p_2}}}{{{T_2}}}} \right)d{V_1} - \left( {\frac{{{\mu _1}}}{{{T_1}}} - \frac{{{\mu _2}}}{{{T_2}}}} \right)d{n_1}\qquad \qquad(8)$

It follows directly from eq. (8) that for equilibrium to exist between two phases of a single component,

${T_1} = {T_2}\qquad \qquad(9)$

${p_1} = {p_2}\qquad \qquad(10)$

${\mu _1} = {\mu _2}\qquad \qquad(11)$

In other words, the pressure, temperature and chemical potential of the two phases must be identical in order for equilibrium to exist. Although the equilibrium conditions specified in eqs. (9) – (11) are derived by applying the fundamental relation, eq. $dE = TdS - pdV + \sum\limits_{i = 1}^N {{\mu _i}d{n_i}}$ from Closed Systems with Compositional Change, to a two-phase system with constant internal energy and volume, they are valid phase equilibrium conditions for any two-phase systems.

## Clapeyron Equation

If the temperature of a two-phase system in equilibrium is slightly changed, the pressure of the system will be affected; this relationship is described by the Clapeyron equation. This simple relation between pressure and temperature for two phases in equilibrium is derived in this subsection, and common forms are presented in this section. As shown in detail in the preceding subsection, the equilibrium conditions for two phases of a pure substance are represented by eqs. (9) – (11). The two statements represented by eqs. (9) and (10) are the conditions for thermal and mechanical equilibrium. Equation (11) is automatically satisfied when eqs. (9) and (10) are satisfied, because the intensive chemical potential property, μ, at equilibrium can be expressed as a function of two other intensive properties, T and p. Furthermore, the temperature, T, and pressure, p, are not independent of each other in a system that contains two phases in equilibrium. The relationship between T and p that describes all possible phase equilibrium states can be represented by any one of the three curves in the two-dimensional phase diagram of a pure substance in Fig. 7(a). An explicit expression for the slope of these equilibrium lines can be found in terms of easily-measurable variables, including temperature and pressure. Suppose a two-phase equilibrium system at temperature T and pressure p experiences an infinitesimal change of temperature to T + dT, so that the corresponding pressure changes to p + dp. Since the two-phase system is at equilibrium at the new temperature and pressure, the new chemical potentials of the two phases must also be equal:

${\mu _1} + d{\mu _1} = {\mu _2} + d{\mu _2}\qquad \qquad(12)$

Substituting eq. (11) into eq. (12), one obtains

$d{\mu _1} = d{\mu _2}\qquad \qquad(13)$

The fundamental relations, eq. $dG = Vdp - SdT + \sum\limits_{i = 1}^N {{\mu _i}d{n_i}}$ from Closed Systems with Compositional Change, in terms of Gibbs free energy for both phases under consideration, are

$d{G_1} = {V_1}dp - {S_1}dT + {\mu _1}d{n_1}\qquad \qquad(14)$

$d{G_2} = {V_2}dp - {S_2}dT + {\mu _2}d{n_2}\qquad \qquad(15)$

These equations can be considered in light of the following relationships between the extensive and intensive properties:

$V = n\bar v,{\rm{ }}S = n\bar s,{\rm{ }}G = n\bar g\qquad \qquad(16)$

where $\bar v,{\rm{ }}\bar s,{\rm{ and }}\bar g$ are specific molar volume (m3/mol), specific molar entropy (J/mol-K), and specific molar Gibbs free energy (J/mol), respectively. Equations (14) and (15) can be rewritten as

$d{\bar g_1} = {\bar v_1}dp - {\bar s_1}dT + \frac{{{\mu _1} - {{\bar g}_1}}}{{{n_1}}}d{n_1}\qquad \qquad(17)$

$d{\bar g_2} = {\bar v_2}dp - {\bar s_2}dT + \frac{{{\mu _2} - {{\bar g}_2}}}{{{n_2}}}d{n_2}\qquad \qquad(18)$

For a single component system, eqs. ${\mu _i} = {\left( {\frac{{\partial H}}{{\partial {n_i}}}} \right)_{p,S,{n_{j \ne i}}}} = {\left( {\frac{{\partial F}}{{\partial {n_i}}}} \right)_{T,V,{n_{j \ne i}}}} = {\left( {\frac{{\partial G}}{{\partial {n_i}}}} \right)_{T,p,{n_{j \ne i}}}}$ from Closed Systems with Compositional Change and (16) indicate that

$\mu = {\left( {\frac{{\partial G}}{{\partial n}}} \right)_{T,p}} = \bar g\qquad \qquad(19)$

Thus, eqs. (17) and (18) can be rewritten as

$d{\mu _1} = {\bar v_1}dp - {\bar s_1}dT\qquad \qquad(20)$

$d{\mu _2} = {\bar v_2}dp - {\bar s_2}dT\qquad \qquad(21)$

Substituting eqs. (20) and (21) into eq. (13) yields

${\bar v_1}dp - {\bar s_1}dT = {\bar v_2}dp - {\bar s_2}dT\qquad \qquad(22)$

which can be rearranged as

$\frac{{dp}}{{dT}} = \frac{{{{\bar s}_1} - {{\bar s}_2}}}{{{{\bar v}_1} - {{\bar v}_2}}}\qquad \qquad(23)$

Equation (23) can be rewritten in term of specific entropy and specific volume per unit mass, i.e.,

$\frac{{dp}}{{dT}} = \frac{{{s_1} - {s_2}}}{{{v_1} - {v_2}}}\qquad \qquad(24)$

The specific entropy may be advantageously replaced with the more usable term of specific enthalpy. The equilibrium condition in terms of the specific Gibbs free energy is

${\bar g_1} = {\bar g_2}{\rm{ or }}{g_1} = {g_2} \qquad \qquad (25)$

i.e.,

${h_1} - T{s_1} = {h_2} - T{s_2}\qquad \qquad(26)$

Equation (26) can be rearranged to yield

${s_1} - {s_2} = \frac{{{h_1} - {h_2}}}{T}\qquad \qquad(27)$

Substituting eq. (27) into eq. (23), one obtains

$\frac{{dp}}{{dT}} = \frac{{{h_1} - {h_2}}}{{\left( {{v_1} - {v_2}} \right)T}} = \frac{{{h_{21}}}}{{{v_{21}}T}}\qquad \qquad(28)$

where h21 is the change in specific enthalpy of phase change and v21 is the change of specific volume during phase change. Equation (28), which is referred to as the Clapeyron equation, describes a general relationship among the pressure, temperature, volume change, and enthalpy change for a single-component, two-phase system at equilibrium. All of the properties in eq. (28) are experimentally measurable; the equation itself has been repeatedly tested and found to be valid (Kyle, 1999). The Clapeyron equation applies to any two phases in equilibrium, such as solid/liquid, solid/vapor and liquid/vapor, which are signified by the general subscripts 1 and 2. For a liquid-vapor system, the Clapeyron equation can be written as

$\frac{{dp}}{{dT}} = \frac{{{h_{\ell v}}}}{{({v_v} - {v_\ell }){T_{sat}}}}\qquad \qquad(29)$

$\frac{{\Delta p}}{{\Delta T}} = \frac{{{h_{\ell v}}}}{{{v_{\ell v}}{T_{sat}}}}$

The height of the mountain, H, is related to the pressure difference by

Δp = ρairgH

where the density of the air can be approximated as the density of the air at sea level at 25 °C, i.e., ρair = 1.169kg / m3. Combining the above two equations and substituting the given values yields

$H = \frac{{{h_{\ell v}}\Delta T}}{{{v_{\ell v}}{T_{sat}}{\rho _{air}}g}} = \frac{{2257.03 \times {{10}^3} \times (100 - 95)}}{{1.67185 \times 373.15 \times 1.169 \times 9.8}} = 1573{\rm{ m}}$
$ds = \frac{{\partial s}}{{\partial v}}dv + \frac{{\partial s}}{{\partial T}}dT$

During a phase change, the temperature is constant (dT= 0), so

$ds = \frac{{\partial s}}{{\partial v}}dv$

Using the following Maxwell relation [eq. ${\left( {\frac{{\partial S}}{{\partial V}}} \right)_T} = {\left( {\frac{{\partial p}}{{\partial T}}} \right)_V}$ from Maxwell Relations]:

${\left( {\frac{{\partial s}}{{\partial v}}} \right)_T} = {\left( {\frac{{\partial p}}{{\partial T}}} \right)_V}$

gives:

$ds = \frac{{\partial p}}{{\partial T}}dv$

Since temperature and pressure are constant during a phase change, the derivative of pressure with respect to temperature is not a function of the specific volume.

${s_2} - {s_1} = \frac{{dp}}{{dT}}\left( {{v_2} - {v_1}} \right)$

Or

$\frac{{dp}}{{dT}} = \frac{{{s_2} - {s_1}}}{{{v_2} - {v_1}}} = \frac{{\Delta s}}{{\Delta v}}$

Substitution of eq. (27) gives:

$\frac{{dp}}{{dT}} = \frac{{{h_2} - {h_1}}}{{\left( {{v_2} - {v_1}} \right)T}} = \frac{{{h_{21}}}}{{{v_{21}}T}}$

which is identical to eq. (28).

For liquid-vapor equilibrium at low pressure, the specific volume of the liquid, ${v_\ell }$, is negligible in comparison with the specific volume of the vapor, vv. It is further assumed that the vapor behaves like an ideal gas at low pressure, and therefore, the specific volume of vapor can be obtained using the ideal gas law vv = RgT / p. In this case, the Clapeyron equation (29) reduces to

$\frac{{dp}}{{dT}} = \frac{{{h_{\ell v}}p}}{{{R_g}{T^2}}}\qquad \qquad(30)$

which is referred to as the Clausius-Clapeyron equation. Since the conditions that eq. (30) was obtained was that ${v_v} \gg {v_\ell }$ and vapor phase behavior like idea gas, it should not be used when these conditions are not satisfied. If the saturation temperature corresponding to any reference pressure, p0, is T0, the relationship between the saturation temperature and pressure at the vicinity of a point (p0,T0) can be obtained by integrating eq. (30), i.e.,

$\ln \frac{p}{{{p_0}}} = - \frac{{{h_{\ell v}}}}{{{R_g}}}\left( {\frac{1}{T} - \frac{1}{{{T_0}}}} \right)\qquad \qquad(31)$

Rearranging eq. (31) yields the saturation pressure at temperature T:

$p = {p_0}\exp \left[ { - \frac{{{h_{\ell v}}}}{{{R_g}}}\left( {\frac{1}{T} - \frac{1}{{{T_0}}}} \right)} \right]\qquad \qquad(32)$

Equation (32) is also applicable to a mixture of vapor and gas, provided that the pressure is the partial pressure of the vapor in the mixture; this is related to the total pressure, p, by

${p_v} = {x_v}p\qquad \qquad(33)$

and the molar fraction of the saturated vapor in the mixture, x, is defined as

${x_v} = \frac{{{\rho _v}}}{{{M_v}}}{\left( {\frac{{{\rho _v}}}{{{M_v}}} + \frac{{{\rho _g}}}{{{M_g}}}} \right)^{ - 1}}\qquad \qquad(34)$

where ρv and ρg are concentrations of vapor and gas in the mixture, respectively. If the total pressure of the vapor-gas mixture remains constant (as is the case with fog in air) when the temperature of the mixture is changed from T0 to T, the vapor molar fraction consistent with the new saturated vapor state becomes

${x_v} = {x_{v0}}\exp \left[ { - \frac{{{h_{\ell v}}}}{{{R_g}}}\left( {\frac{1}{T} - \frac{1}{{{T_0}}}} \right)} \right]\qquad \qquad(35)$

where xv0 is the molar fraction of vapor at the reference temperature T0. Equation (35) can be rearranged using the ideal gas law:

${x_v} = \frac{{{R_g}T}}{p}{\rho _v}\qquad \qquad(36)$

After substituting eq. (36) into eq. (34) and assuming the total pressure is constant, the concentration of vapor in the mixture becomes

${\rho _v} = {\rho _{v0}}\frac{{{T_0}}}{T}\exp \left[ { - \frac{{{h_{\ell v}}}}{{{R_g}}}\left( {\frac{1}{T} - \frac{1}{{{T_0}}}} \right)} \right]\qquad \qquad(37)$

where ρv0 is the density of vapor at the reference temperature T0.

## Multiphase Multicomponent Systems

The requirements for equilibrium can be expressed in terms of different thermodynamic variables in a number of ways. One of the more common thermodynamic situations is a system with constant temperature and pressure. The equilibrium criterion for a closed system with compositional changes in terms of the Gibbs free energy is expressed by eq. $dG = Vdp - SdT + \sum\limits_{i = 1}^N {{\mu _i}d{n_i}}$ from Closed Systems with Compositional Change. For a simplified case of a two-component, two-phase system, eq. $dG = Vdp - SdT + \sum\limits_{i = 1}^N {{\mu _i}d{n_i}}$ can be written as follows:

$d{G_k} = - {S_k}dT + {V_k}dp + {\mu _{k,A}}d{n_{k,A}} + {\mu _{k,B}}d{n_{k,B}}\begin{array}{*{20}{c}} {} & {(k = 1,2} \\ \end{array})\qquad \qquad(38)$

where the subscript k denotes the kth phase, and the subscripts A and B denote the components A and B. As stated above, the temperature and pressure of the system are assumed to be constant, which simplifies eq. (38). The system is defined further by allowing a very small amount of components A and B to be transferred from phase 1 to phase 2.

$d{n_{2,A}} = - d{n_{1,A}}\qquad \qquad(39)$

$d{n_{2,B}} = - d{n_{1,B}}\qquad \qquad(40)$

Since the system is assumed to be in equilibrium, dGT,p = 0, and therefore

$dG = \sum\limits_{k = 1}^2 {d{G_k}} = 0\qquad \qquad(41)$

Equations (38) and (41) can be combined to create the following expression:

$\begin{array}{l} dG = \sum\limits_{k = 1}^2 {{\mu _{k,A}}d{n_{k,A}}} + \sum\limits_{k = 1}^2 {{\mu _{k,B}}d{n_{k,B}}} \\ = d{n_{1,A}}\left( {{\mu _{1,A}} - {\mu _{2,A}}} \right) + d{n_{1,B}}\left( {{\mu _{1,B}} - {\mu _{2,B}}} \right) = 0 \\ \end{array}\qquad \qquad(42)$

which takes into account that T and p are constant. Since dn1,A and dn1,B are independent and are not necessarily equal to zero, it follows that at equilibrium,

${\mu _{1,A}} = {\mu _{2,A}}\qquad \qquad(43)$

${\mu _{1,B}} = {\mu _{2,B}}\qquad \qquad(44)$

Thus, the equilibrium requires that the chemical potential of each component be the same in all phases. As stated above, there are many ways of arriving at the most general form of equilibrium criteria. This principle may be easily extended to a system that includes multiple phases and components, and where all components may be transferred from one phase to another. Therefore, eq. (43) and (44) can be extended to state that the chemical potential of each component must be identical in all phases for systems to be in equilibrium at constant temperature and pressure, i.e.,

$\begin{array}{l} {\mu _{1,A}} = {\mu _{2,A}} = {\mu _{3,A}} \\ {\mu _{1,B}} = {\mu _{2,B}} = {\mu _{3,B}} \\ {\mu _{1,C}} = {\mu _{2,C}} = {\mu _{3,C}} \\ \end{array} \qquad \qquad (45)$

which can be repeated up to N components. This means that the chemical potential for a particular component must be equal in all phases at equilibrium. The number of independent intensive thermodynamic variables and the number of phases for a system are related by the Gibbs phase rule, eq. (1). Having already discussed the thermodynamic equilibrium of a multiphase system, we can offer proof of the Gibbs phase rule. Consider a system that has N components and Π phases in equilibrium at a given temperature and pressure, and assume that each component can exist in each phase. The system in each phase could be completely specified if the concentration of each component in each phase, the temperature, and the pressure were specified, i.e., the number of degrees of freedom is

$f = N\Pi + 2\qquad \qquad(46)$

However, we know that at equilibrium the chemical potential of each component is the same in all the phases, so we reduce the degrees of freedom by $N\left( {\Pi - 1} \right)$. Finally, we also recognize that since the sum of the mole fractions equals unity in each of the Π phases, we may also reduce the degrees of freedom by Π additional intensive properties. Subtracting these two corrections from the original degrees of freedom in eq. (46) gives

$f = N\Pi + 2 - \Pi - N(\Pi - 1) = N + 2 - \Pi \qquad \qquad(47)$

which can be rearranged to give the Gibbs phase rule, eq. (1).

## Metastable Equilibrium and Nucleation

Equilibriums can be classified as (a) stable equilibrium, (b) metastable equilibrium, and (c) unstable equilibrium; these equilibriums can be illustrated using analogous examples of the mechanical equilibrium of a ball shown in Fig. 1. The ball in Fig. 1(a) is in stable equilibrium because it can always return to equilibrium after displacement. The ball in Fig. 1(b) is in metastable

Figure 1 Schematic of stability.

equilibrium because it can return to equilibrium after small displacement. If the displacement is large, the ball will move to a new equilibrium position. The ball in Fig. 1(c) is in unstable equilibrium because equilibrium cannot be maintained after any displacement. The nature of the metastable equilibrium is defined as stable equilibrium restricted to small systematic and environmental changes. If the changes of the systematic or environmental variables exceed the restricted range, the metastable system becomes unstable. When imbalances in the intensive variables are large enough, a spontaneous change must occur in the system to bring the system to a new equilibrium state. However, many situations arise in which the changes proceed slowly enough that departures from stable equilibrium are small. Consequently, the unstable intermediate states may closely approximate a stable equilibrium path and time is no longer an important factor. All of the thermodynamic surfaces presented in Section 2.7 satisfy these conditions. In thermodynamics, metastable regions play an important role in determining equilibrium states. Figure 2 shows a pv diagram for a pure substance – an isothermal slice through a surface on the pvT diagram. Liquid-vapor phase change occurs along an isotherm (1→2→4→5) that consists of three states: liquid, two-phase mixture, and vapor. Under stable conditions, the liquid phase at point 1 may expand along the isotherm 1→2. At point 2, the fluid reaches the saturated liquid state, and continued expansion under stable conditions results in vaporization, represented by the path 2→4. On the other hand, the superheated vapor phase at point 5 may be compressed along the same isotherm 5→4. At point 4, the fluid becomes saturated vapor and further compression results in condensation, represented by path 4→2. These single-phase paths (1→2 and 5→4), as well as the phase change path 2<->4, are completely reversible under stable conditions. However, the volume of the liquid can be increased along line 2→2' instead of going through process 2→3. Therefore, it is possible in the absence of vapor bubble nucleation to superheat the liquid above the saturation temperature. The volume of the vapor can also be decreased along line 4→4', which means that, in the absence of liquid droplet nucleation, the vapor can be subcooled below its saturation temperature. The superheated liquid and subcooled vapor are both in metastable equilibrium because the criterion for mechanical stability represented by eq. $- {\left( {\frac{{\partial p}}{{\partial V}}} \right)_T} = \frac{1}{{{\kappa _T}V}} > 0$ from Stability Criteria is satisfied. However, the states along the path 2'→3→4' are completely unstable because when moving along this path, ${{(\partial p / \partial v )}_T} > 0$, which violates eq. $- {\left( {\frac{{\partial p}}{{\partial V}}} \right)_T} = \frac{1}{{{\kappa _T}V}} > 0$ from Stability Criteria. Therefore, the path 2'<->4' is not accessible for boiling and condensation. The loci of the limiting points 2' and 4', where ${(\partial p/\partial v)_T} = 0$, are called liquid and vapor spinodals, respectively. Since the states along the path 2'→3→4' are not in equilibrium condition, the equations of state presented in Section 2.4 are not valid to describe them. However, this path is very similar to the isotherm obtained by using the van der Waals equation. If it is assumed that the van der Waals equation, eq. $p = \frac{{{R_g}T}}{{v - b}} - \frac{a}{{{v^2}}}$ from Properties of pure substances, is

Figure 2 pv diagram for a pure substance illustrating metastable equilibrium.

valid to describe this path, one can estimate the parameters on the spinodal by using

${\left( {\frac{{\partial p}}{{\partial v}}} \right)_T} = - \frac{{{R_g}T}}{{{{(v - b)}^2}}} + \frac{{2a}}{{{v^3}}} = 0\qquad \qquad(48)$

The thermodynamic parameters on the spinodal can be determined by using eqs. $p = \frac{{{R_g}T}}{{v - b}} - \frac{a}{{{v^2}}}$ from Properties of pure substances and (48).

$v = \frac{V}{m} = \frac{1}{{80}} = 0.0125{\rm{ }}{{\rm{m}}^{\rm{3}}}{\rm{/kg}}$

When the vessel is cooled isochorically, the propane gas becomes supercooled and enters a metastable state. Continued cooling below the temperature corresponding to the temperature at the vapor spinodal will make the system unstable and result in condensation. This temperature can be found using eq. (48), i.e.,

$T = \frac{{2a{{(v - b)}^2}}}{{{R_g}{v^3}}}$

The constants a and b are a = 479.78Pa − m6 / kg2 and b$= 2.04 \times {10^{ - 3}}$ m3 / kg. Therefore, the temperature is $T = \frac{{2a{{(v - b)}^2}}}{{{R_g}{v^3}}} = \frac{{2 \times 479.78 \times {{(0.0125 - 0.00204)}^2}}}{{0.188 \times {{10}^3} \times {{0.0125}^3}}} = 285.92{\rm{ K}} = {12.77^{\rm{o}}}{\rm{C}}$

The corresponding pressure of the propane can be found using the van der Waals equation, eq. $p = \frac{{{R_g}T}}{{v - b}} - \frac{a}{{{v^2}}}$ from Properties of pure substances, i.e.,

$p = \frac{{{R_g}T}}{{v - b}} - \frac{a}{{{v^2}}} = \frac{{0.188 \times {{10}^3} \times 285.92}}{{0.0125 - 0.00204}} - \frac{{479.78}}{{{{0.0125}^2}}} = 2.07{\rm{ MPa}}$

For liquid heated at constant pressure above its corresponding saturation temperature, the liquid spinodal (point 2') represents a maximum upper limit of superheat based on thermodynamic consideration; it is referred to as the thermodynamic limit of superheat. Similarly, the spinodal limit for supercooled vapor (point 4') is the maximum thermodynamic limit for supercooling of vapor. While the spinodal limits provide maximum limits on the superheat or supercooling, nucleation of new phases occurs in temperature ranges defined by the saturation temperature and the spinodal limits. Nucleation of vapor that occurs completely in liquid, or nucleation of liquid that occurs completely in vapor, is referred to as homogeneous nucleation. On the other hand, if nucleation occurs at an interface between the metastable phase (liquid or vapor) and solid, it is called heterogeneous nucleation. The conditions for nucleation of the liquid phase in vapor (condensation) and nucleation of the vapor phase in liquid (boiling) are discussed in Chapters 7 and 8, respectively.

# Thermodynamics at the Interfaces

## Equilibrium at the Interface

Two bulk fluids of large extent, separated by an interfacial region, constitute a system in equilibrium. This very general description can be used to consider, for example, the cases of two immiscible liquids in contact with each other; a single substance in two phases; or a mixture of gases in contact with a solid, with a chemical reaction occurring on the surface of the solid. When analyzing such systems, the interface is a unique region that requires special attention. When mass transport occurs between the two bulk substances, the interface problem becomes significantly more complicated. Any mass exchange between the two bulk substances also requires consideration of momentum and energy exchange. A single substance undergoing a phase change is the simplest case of mass transport across an interfacial surface. To develop an understanding of the unique and significant effects of interfacial surfaces on the interaction of two bulk systems, a thermodynamic analysis will be performed. Surfaces 1 and 2 constitute a demarcation of the region that possesses all of the properties of the bulk fluids. The dividing surface I is at an arbitrary location within the region between surfaces 1 and 2 (see Fig. 3). For a single-component system, the fundamental thermodynamic relation represented by eq. $dE = TdS - pdV + \sum\limits_{i = 1}^N {{\mu _i}d{n_i}}$ from Closed Systems with Compositional Change can be simplified as

$dE = TdS - pdV + \mu dn\qquad \qquad(49)$

Suppose the configuration of the volume and surfaces 1, 2, and Iare fixed. In this case, the internal energy is only a function of S and n, and

$dE = TdS + \mu dn\qquad \qquad(50)$

for each 1, 2, and I within the interfacial region. Now consider a process wherein some mass and energy exchange occurs between the bulk fluid 2 and the interfacial surface I, with bulk fluid 1 remaining unchanged. The total energy of the entire (larger) system, comprised of bulk fluids 1 and 2 and the interface, remains constant. Thus, an energy balance for the system requires of the new equilibrium that

${E_{total}} = {E_I} + {E_1} + {E_2} = {\rm{constant}}\qquad \qquad(51)$

The energy in the region between 2 and I, with the energy at 1 held constant, are then related by

$d{E_{total}} = d{E_I} + d{E_2} = 0\qquad \qquad(52)$

i.e.,

$0 = {T_I}d{S_I} + {\mu _I}d{n_I} + {T_2}d{S_2} + {\mu _2}d{n_2}\qquad \qquad(53)$

Figure 3 Interfacial region between two fluids.

When mass and energy are exchanged between bulk region 2 and surface I, with fluid 1 remaining unchanged, the total mole number for bulk region 2 and surface I satisfies

${n_{total}} = {n_I} + {n_2} = {\rm{constant}}\qquad \qquad(54)$

i.e.,

$d{n_{total}} = 0 = d{n_I} + d{n_2}\qquad \qquad(55)$

$d{n_I} = - d{n_2}\qquad \qquad(56)$

The entropy balance for the system requires that

${S_{total}} = {S_I} + {S_1} + {S_2} = {\rm{constant}}\qquad \qquad(57)$

$d{S_{total}} = 0 = d{S_I} + d{S_2}\qquad \qquad(58)$

$d{S_I} = - d{S_2}\qquad \qquad(59)$

Substituting these into the energy-accounting eq. (53) gives the conditions for equilibrium of the process described above as:

$\left( {{T_2} - {T_I}} \right)d{S_2} + \left( {{\mu _2} - {\mu _I}} \right)d{n_2} = 0\qquad \qquad(60)$

Since eq. (60) must be valid for any dS2 and dn2, the following conditions must be satisfied:

${T_2} = {T_I}\qquad \qquad(61)$

${\mu _2} = {\mu _I}\qquad \qquad(62)$

The same procedure can be used to show that

${T_1} = {T_I}\qquad \qquad(63)$

${\mu _1} = {\mu _I}\qquad \qquad(64)$

Thus, at equilibrium and constant volume, the temperatures and chemical potentials for each of the three regions must be equal, i.e.,

${T_1} = {T_I} = {T_2}\qquad \qquad(65)$

${\mu _1} = {\mu _I} = {\mu _2}\qquad \qquad(66)$

## Surface Tension: Thermodynamic Definitions

The liquid-vapor (gas) interface is often treated as a sharp discontinuity in macroscale thermodynamics and heat transfer. However, the change of properties between different phases actually occurs over a very thin but finite region, as shown in Fig. 3. Since the density of the liquid is higher than that of the vapor, the molecules in the liquid are closer to each other and the intermolecular forces are attractive in nature. By formulating the problem of the phase interface in terms of the surface excess quantities (Carey, 1992), classical thermodynamics can then be used to determine the relationship between surface tension and other macroscopic variables. Equilibrium conditions establish the equality of T and μ for the three regions, i.e., the bulk fluids and the interface. The total mole number of the two-phase system shown in Fig. 15 can be written as

${n_{total}} = {n_I} + {n_1} + {n_2}\qquad \qquad(67)$

The mole number of the interface is then

${n_I} = {n_{total}} - {n_1} - {n_2}\qquad \qquad(68)$

Likewise, for internal energy,

${E_I} = {E_{total}} - {E_1} - {E_2}\qquad \qquad(69)$

The surface I possesses all the information required to analyze this region. The exact location of this surface – referred to as the dividing surface – does not need to be known at this time. An imaginary surface defined to possess these properties is sufficient.

As stated by eq. (50), the internal energy of the surface is

${E_I} = {E_I}\left( {{S_I},{n_I}} \right)\qquad \qquad(70)$

which is valid when the shape and location of the interface are fixed and flat. If the interface is deformable, both the shape and the area of the interface will affect the internal energy of the interface. The internal energy of the deformable interface becomes

${E_I} = {E_I}\left( {{S_I},{n_I},A,K} \right)\qquad \qquad(71)$

where A and K are the surface area and the curvature of the interface, respectively.

The change in E is then

$dE = {\left( {\frac{{\partial E}}{{\partial S}}} \right)_{n,A,K}}dS + {\left( {\frac{{\partial E}}{{\partial n}}} \right)_{S,A,K}}dn + {\left( {\frac{{\partial E}}{{\partial A}}} \right)_{S,n,K}}dA + {\left( {\frac{{\partial E}}{{\partial K}}} \right)_{S,n,A}}dK\qquad \qquad(72)$

where the subscript I is dropped for ease of notation. Considering eq. (50), eq. (72) can be rewritten in the following form:

$dE = TdS + \mu dn + {\left( {\frac{{\partial E}}{{\partial A}}} \right)_{S,n,K}}dA + {\left( {\frac{{\partial E}}{{\partial K}}} \right)_{S,n,A}}dK\qquad \qquad(73)$

If the variation of the curvature effect is negligible, eq. (73) simplifies to

$dE = TdS + \mu dn + \sigma dA\qquad \qquad(74)$

where σ is the surface tension, defined as

$\sigma = {\left( {\frac{{\partial E}}{{\partial A}}} \right)_{S,n}}\qquad \qquad(75)$

Another representation of the surface tension uses the Helmholtz free energy for the surface region, which is defined as

${F_I} = {E_I} - T{S_I}\qquad \qquad(76)$

Differentiating the above equation gives

$d{F_I} = d{E_I} - Td{S_I} - {S_I}dT\qquad \qquad(77)$

Substituting eq. (74) into eq. (77), one obtains

$dF = - SdT + \mu dn + \sigma dA\qquad \qquad(78)$

If the differential of the Helmholtz function is as given above, then the function can be considered as

$F = F(T,n,A)\qquad \qquad(79)$

which implies that

$\sigma = {\left( {\frac{{\partial F}}{{\partial A}}} \right)_{T,n}}\qquad \qquad(80)$

As the force acting on the interface, surface tension tends to resist an increase in the interfacial area. The work done on the system to increase the area of the interface is given by

${\mathop{\rm work}} = \sigma dA\qquad \qquad(81)$

which supplies a second definition of surface tension: the work per unit area required to produce a new surface. When the interface area is increased by dA, the perimeter of the interface is increased by dP. Therefore, the work done on the system, eq. (81), can also be expressed as a product of the force per unit length of the perimeter and increase of the perimeter, dP. It follows that there are two equivalent interpretations of surface tension, σ: (1) energy per unit area of the surface, and (2) force per unit perimeter of the surface. Equation (81) relates σ to the work required to increase the area of a surface. Thermodynamics establishes that work is a path-dependent function. It is convenient to shift our emphasis from work done on the system to work done by the system when considering change of area. If work done by the system when its area is changed is defined as δWσ, eq. (81) becomes

$\delta {W_\sigma } = - \sigma dA\qquad \qquad(82)$

According to eq. (82), a decrease in area (dA negative) corresponds to work done by the system, whereas an increase in area requires work to be done on the system (dA positive and δWσ negative). The quantity δWσ can also be related to other thermodynamic variables. According to the first law, the change in the internal energy E of the system equals

$dE = \delta Q - \delta W\qquad \qquad(83)$

in which δW is the work done by the system and δQ is the thermal energy absorbed by the system. The quantity δW is conveniently divided into a pressure-volume term and a non-pressure-volume term:

$\delta W = \delta {W_{pV}} + \delta {W_{non - pV}} = pdV + \delta {W_{non - pV}}\qquad \qquad(84)$

The non-pressure-volume types of work include electrical and chemical, as well as other non-pV types of mechanical work. The work defined by eq. (82) may also be classified as non-pressure-volume work. The second law of thermodynamics indicates that for reversible processes,

$\delta {Q_{rev}} = TdS\qquad \qquad(85)$

Substituting eqs. (84) and (85) into eq. (83), with the stipulation of reversibility as required by eq. (85), one obtains

$dE = TdS - pdV - \delta {W_{non - pV}}\qquad \qquad(86)$

Recalling the definition of Gibbs free energy G,

$G = H - TS = E + pV – TS \qquad \qquad(87)$

Differentiating eq. (87) yields

$dG = dE + pdV + Vdp - TdS - SdT\qquad \qquad(88)$

Substituting eq. (86) into eq. (88) gives

$dG = Vdp - SdT - \delta {W_{non - pV}}\qquad \qquad(89)$

which shows that for a constant temperature, constant pressure, reversible process,

$dG = - \delta {W_{non - pV}}\qquad \qquad(90)$

that is, dG equals the maximum non-pressure-volume work derivable from such a process, since maximum work is associated with reversible processes. We have already seen through eq. (82) that changes in surface area entail non-pressure-volume work. Therefore we can identify δWσ from eq. (82) with δWnonpV in eq. (90) and write

$dG = \sigma dA\qquad \qquad(91)$

Considering the stipulations made in going from eq. (88) to eq. (90), we obtain

$\sigma = {\left( {\frac{{\partial G}}{{\partial A}}} \right)_{T,p}}\qquad \qquad(92)$

which identifies the surface tension as the increment in Gibbs free energy per unit increment in area. The path-dependent variable δWσ is replaced by a state variable as a result of this analysis. The three definitions of surface tension given by eqs. (75), (80) and (92) are equivalent to each other, and different definitions can be applied to different systems. It is worthwhile to discuss surface tension from the molecular perspective in order to understand the mechanism of surface tension for different substances. Surface tension can be considered as the summation of two parts: one part is due to dispersion force, and the other part is due to specific forces, like metallic or hydrogen bonding (Fowkes, 1965). Surface tension force in a nonpolar liquid is due only to the dispersion force; therefore, the surface tension for a nonpolar fluid is very low. For a hydrogen-bonded liquid, surface tension is slightly higher because the surface tension is due to both dispersion forces and hydrogen bonding. The surface tension for a liquid metal is highest because the surface tension is due to a combination of dispersion forces and metallic bonding, and metallic bonding is much stronger than hydrogen bonding. The surface tensions for different liquids are quantitatively demonstrated in Table 9.

Table 1 Surface tensions for different liquids at liquid-vapor interface

 Types of liquid Liquid Temperature (°C) Surface tension (mN/m) Nonpolar liquid Helium -271 0.26 Nitrogen -153 0.20 Hydrogen-bonded liquid (polar) Ammonia -40 35.4 Water 20 72.9 Metallic liquid Mercury 20 484 Silver 1100 878

## Effects of Interfacial Tension Gradients

Since surface tension depends on temperature, a permanent nonuniformity of temperature or concentration (for a multicomponent system) at a liquid-vapor interface causes a surface tension gradient. The interfacial area with small surface tension expands at the expense of an area with greater surface tension, which in turn establishes a steady flow pattern in the liquid; this flow caused by the surface tension gradient along the liquid-vapor (gas) interface is referred to as the Marangoni effect. The surface tension of a multicomponent liquid that is in equilibrium with the vapor is a function of temperature and composition of the mixture, i.e.,

$\sigma = \sigma (T,{x_1},{x_2}, \cdots ,{x_{N - 1}})\qquad \qquad(93)$

where xi is the molar fraction of the ith component in the liquid phase and N is the total number of components in the liquid phase. The change of surface tension can be caused by either change of temperature or composition, i.e.,

$d\sigma = {\left( {\frac{{\partial \sigma }}{{\partial T}}} \right)_{{x_i}}}dT + \sum\limits_{i = 1}^{N - 1} {{{\left( {\frac{{\partial \sigma }}{{\partial {x_i}}}} \right)}_{T,{x_{j \ne i}}}}d{x_i}} \qquad \qquad(94)$

Recall from thermodynamics that as the critical temperature for a given fluid is approached, the properties of the liquid and vapor phases of the fluid become identical, i.e., σ vanishes. It therefore follows that the surface tension decreases with temperature, i.e., [${(\partial \sigma /\partial T)_{{x_i}}} < 0$]. For a pure substance, surface tension is a function of temperature only. The curve-fit equations for surface tension are almost linear for most fluids and can have the form

$\sigma = {C_0} - {C_1}T\qquad \qquad(95)$

where C0 and C1 are empirical constants that vary between substances. For water, ${C_0} = 75.83 \times {10^{ - 3}}{\rm{N/m}}$ and ${C_1} = 0.1477 \times {10^{ - 3}}{\rm{N/m - }}{}^{\rm{o}}{\rm{C}}$. The unit of the temperature T in eq. (95) is °C. Since surface tension varies with temperature, the surface tension will not be uniform if the temperature, along the liquid-vapor interface, is nonuniform. Consequently, the liquid in the lower surface tension region near the interface will be pulled toward the region with higher surface tension. Furthermore, because the surface tension usually decreases with increasing temperature, the flow driven by surface tension moves away from the interface with high temperature and towards the interface with low temperature. As noted before, this motion of a liquid caused by a surface tension gradient at the interface is referred to as the Marangoni effect. The most well-known example of surface-tension-driven flow is Bernard cellular flow, which occurs in a thin horizontal liquid layer heated from below. Figure 4 illustrates steady cellular flow driven by the Marangoni effect. Once a steady Marangoni flow is established, the liquid velocity is upward at point A and downward at point B. Since the liquid at surface point A comes directly from the hot surface, the temperature at point A is higher than that at point B. Consequently, the surface tension at point A is smaller

Figure 4 Marangoni effect: cellular flow driven by surface tension gradient.

than that at point B, and the fluid at point A is pulled toward point B. Although a temperature gradient exists in the vertical direction, the actual driving force is the surface tension gradient in the horizontal direction. The liquid surface loses heat to the gas phase at temperature, TG, through convection with a heat transfer coefficient of hδ. An important boundary condition at the interface explains the liquid flow field caused by surface tension variation along the interface:

${\left. {{\tau _{yx}}} \right|_{y = \delta }} = {\mu _\ell }{\left( {\frac{{\partial u}}{{\partial y}}} \right)_{y = \delta }} = \left( {\frac{{\partial \sigma }}{{\partial T}}} \right){\left( {\frac{{\partial T}}{{\partial x}}} \right)_{y = \delta }}\qquad \qquad(96)$

The Marangoni number may be regarded as the proportion of surface tension forces to viscous forces. The relationship between temperature, viscosity and the surface tension gradient is evident in the Marangoni number, which is:

$Ma = \frac{{T_{\delta}}-{T_w}}{\delta}{(d\sigma / dt){{\delta}^2}}{ {{\alpha}_{\ell}}{{\mu}_{\ell}}} \qquad \qquad (97)$

where δ is the film thickness, Tδ is the temperature at the liquid-vapor interface, and Tw is the surface temperature. Further discussion of the Marangoni number and its applications is provided later in this book.

## Microscale Vapor Bubbles and Liquid Droplets

Surface tension effects on system equilibrium between phases are very important. However, analysis of these systems usually requires individual attention; it is difficult to arrive at a simple form of expression that covers all cases. In this subsection, the effect of interfacial surface tension between a liquid and its vapor is considered.

Figure 5 Vapor bubble suspending in a liquid phase in a rigid vessel.

For the development of this problem, we will consider phase change in an isolated rigid system which is occupied by a mixture of liquid and vapor (see Fig. 5). The vapor at temperature Tv and pressure pv is contained in a microscale spherical vapor bubble of radius Rb; liquid at a constant temperature ${T_\ell }$ and pressure ${p_\ell }$ surrounds the vapor bubble. At phase equilibrium, the temperatures of the two phases are the same, i.e., ${T_\ell } = {T_v} = T.$ According to eq. (11), the chemical potentials of the liquid and vapor phases are also the same at equilibrium, i.e., ${\mu _\ell } = {\mu _v} = \mu .$ The condition for equilibrium is chosen to be that the total Helmholtz free energy function will be at its minimum. By applying the fundamental thermodynamic relation of Helmholtz free energy function, eq. $dF = - SdT - pdV + \sum\limits_{i = 1}^N {{\mu _i}d{n_i}}$ from Closed Systems with Compositional Change, in the liquid and vapor phases, one obtains

$d{F_\ell } = - {S_\ell }dT - {p_\ell }d{V_\ell } + \mu d{n_\ell }\qquad \qquad(98)$

$d{F_v} = - {S_v}dT - {p_v}d{V_v} + \mu d{n_v}\qquad \qquad(99)$

The fundamental relation for Helmholtz free energy at the interface, as indicated by eq. (78), is

$d{F_I} = - {S_I}dT + \sigma dA + {\mu _I}d{n_I}\qquad \qquad(100)$

For a reversible phase change process under constant volume and temperature, eq. (91) must be satisfied, i.e.,

$dF = d{F_\ell } + d{F_v} + d{F_I} = 0\qquad \qquad(101)$

Substituting eqs. (98) – (100) into eq. (101), the equilibrium condition becomes

$- ({S_\ell } + {S_v})dT - ({p_\ell }d{V_\ell } + {p_v}d{V_v}) + \sigma dA + \mu (d{n_\ell } + d{n_v} + d{n_I}) = 0\qquad \qquad(102)$

The total volume does not change because the system is rigid, i.e.,

$dV = d{V_\ell } + d{V_v} = 0\qquad \qquad(103)$

Conservation of mass requires that

$dn = d{n_\ell } + d{n_v} + d{n_I} = 0\qquad \qquad(104)$

Since the phase change occurs at constant temperature, we have

$dT = 0\qquad \qquad(105)$

Substituting eqs. (103) – (105) into eq. (102) yields a relationship between the pressures in the liquid and vapor phases at equilibrium:

${p_v} - {p_\ell } = \sigma \frac{{dA}}{{d{V_v}}}\qquad \qquad(106)$

Since the surface area and volume of the vapor bubbles are, respectively, $A = 4\pi R_b^2$ and ${V_v} = 4\pi R_b^3/3,$ eq. (106) becomes

${p_v} - {p_\ell } = \frac{{2\sigma }}{{{R_b}}}\qquad \qquad(107)$

which is the Laplace-Young equation. Although eq. (107) was obtained by analyzing a vapor bubble suspended in a liquid within a rigid system, it can be demonstrated that it is also valid for any other system (Faghri and Zhang, 2006). At phase equilibrium, the chemical potentials of both phases must be equal. As indicated by eq. (19), the chemical potential for the two-phase system is the specific Gibbs energy. Therefore, the specific Gibbs free energy functions must be equal for the liquid and vapor phases, i.e.,

${g_\ell }\left( {{p_\ell },T} \right) = {g_v}\left( {{p_v},T} \right)\qquad \qquad(108)$

By differentiating eq. (108) and considering the fact that the temperature is constant in the phase change process, one can obtain

${\left( {\frac{{\partial {g_\ell }}}{{\partial {p_\ell }}}} \right)_T}d{p_\ell } = {\left( {\frac{{\partial {g_v}}}{{\partial {p_v}}}} \right)_T}d{p_v}\qquad \qquad(109)$

Considering eq. $V = {\left( {\frac{{\partial H}}{{\partial p}}} \right)_{S,{n_i}}} = {\left( {\frac{{\partial G}}{{\partial p}}} \right)_{T,{n_i}}}\qquad \qquad(13)$ from Closed Systems with Compositional Change, eq. (109) can be rewritten as

${v_\ell }d{p_\ell } = {v_v}d{p_v}\qquad \qquad(110)$

The relationship between changes in the liquid and vapor pressures, due to an infinitesimally-small change of the vapor bubble radius, can be obtained by differentiating the Laplace-Young equation:

$d{p_v} - d{p_\ell } = - \frac{{2\sigma }}{{R_b^2}}d{R_b}\qquad \qquad(111)$

Substituting eq. (111) into eq. (110), the pressure in the vapor phase can be eliminated:

${v_\ell }\left( {d{p_v} + \frac{{2\sigma }}{{R_b^2}}d{R_b}} \right) = {v_v}d{p_v}\qquad \qquad(112)$

If the vapor behaves like an ideal gas (vv = RgT / pv), eq. (112) can be rewritten as

${R_g}T\frac{{d{p_v}}}{{{p_v}}} - {v_\ell }d{p_v} = \frac{{2\sigma }}{{R_b^2}}{v_\ell }d{R_b}\qquad \qquad (113)$

where Rg is the gas constant of the vapor. If the radius of the vapor bubble goes to infinity ($1/{R_b} \to 0$), the vapor pressure equals the saturation pressure corresponding to the temperature, psat(T). Integrating eq. (113) from an equilibrium state for a flat surface, one obtains

${R_g}T\ln \left[ {\frac{{{p_v}}}{{{p_{sat}}(T)}}} \right] - {v_\ell }[{p_v} - {p_{sat}}(T)] = - \frac{{2\sigma }}{{{R_b}}}{v_\ell } \qquad \qquad (114)$

i.e.,

${p_v} = {p_{sat}}(T)\exp \left\{ {\frac{{{v_\ell }[{p_v} - {p_{sat}}(T) - 2\sigma /{R_b}]}}{{{R_g}T}}} \right\}\qquad \qquad(115)$

which indicates that the bubble is in equilibrium only if the pressure of the vapor phase exceeds the saturation pressure psat(T). In other words, the vapor phase must be superheated. If the pressure inside the vapor bubble is below the pressure required by eq. (115), the vapor bubble will shrink by condensation. On the other hand, the vapor bubble will grow by evaporation if the pressure inside the bubble is higher than that required by eq. (115). Equation (115) can be inverted to obtain the equilibrium bubble radius:

${R_b} = \frac{{2\sigma }}{{{R_g}T\ln [{p_{sat}}(T)/{p_v}]/{v_\ell } + {p_v} - {p_{sat}}(T)}}\qquad \qquad(116)$

Only vapor bubbles with radii equal to that given by eq. (116) will be in equilibrium with the surrounding superheated liquid at ${T_\ell }$ and ${p_\ell }$. For most cases, ${p_v} - {p_{sat}}(T) \ll 2\sigma /{R_b}$, eqs. (115) and (116) can be simplified as

${p_v} = {p_{sat}}(T)\exp \left\{ { - \frac{{2\sigma {v_\ell }}}{{{R_b}{R_g}T}}} \right\} = {p_{sat}}(T)\exp \left\{ { - \frac{{2\sigma /{R_b}}}{{{p_v}}}\frac{{{v_\ell }}}{{{v_v}}}} \right\}\qquad \qquad(117)$

${R_b} = \frac{{2\sigma {v_\ell }}}{{{R_g}T\ln [{p_{sat}}(T)/{p_v}]}}\qquad \qquad(118)$

Since $2\sigma /{R_b} \ll {p_v}$ and ${v_\ell } \ll {v_v}$, eq. (117) can be simplified as

${p_v} \simeq {p_{sat}}(T)\qquad \qquad(119)$

The pressure in the liquid phase can be readily obtained by using the Laplace-Young equation, eq. (107), i.e.,

${p_\ell } = {p_{sat}}(T) - \frac{{2\sigma }}{{{R_b}}}\qquad \qquad(120)$

which means that the liquid pressure must be less than the saturation pressure corresponding to the system temperature T, i.e., the liquid phase must be superheated in order to maintain phase equilibrium. The superheated liquid is in metastable equilibrium, as represented by the region between 2 and 2' in Fig. 2.

${p_v} = {p_{sat}}(T) = 1.1102 \times {10^5}{\rm{Pa}}$

The surface tension of water at 102 °C is $\sigma = 58.52 \times {10^{ - 3}}{\rm{N/m}}.$ The pressure in the liquid pool can be obtained by using eq. (120):

${p_\ell } = {p_{sat}}(T) - \frac{{2\sigma }}{{{R_b}}} = 1.1102 \times {10^5} - \frac{{2 \times 58.52 \times {{10}^{ - 3}}}}{{0.5 \times {{10}^{ - 3}}}} = 1.1079 \times {10^5}{\rm{Pa}}$

The ratio of the vapor pressure over the liquid pressure is

$\frac{{{p_v}}}{{{p_\ell }}} = \frac{{1.1102 \times {{10}^5}}}{{1.1079 \times {{10}^5}}} = 1.002$

For a liquid droplet suspended in the vapor phase (Fig. 6), the pressure inside the liquid droplet is related to the vapor pressure by the Laplace-Young equation

${p_\ell } - {p_v} = \frac{{2\sigma }}{{{R_d}}}\qquad \qquad(121)$

which can be differentiated to yield

$d{p_\ell } - d{p_v} = - \frac{{2\sigma }}{{R_d^2}}d{R_d}\qquad \qquad(122)$

At phase equilibrium, the specific Gibbs free energy functions must be equal for the liquid and vapor phases [see eq. (108)]. By following a procedure similar to that for the case of the vapor bubble, it can be shown that eq. (110) is also valid for the case of a liquid droplet in the vapor phase. Substituting eq. (122) into eq. (110), the pressure in the vapor phase can be eliminated:

${v_\ell }\left( {d{p_v} - \frac{{2\sigma }}{{R_d^2}}d{R_d}} \right) = {v_v}d{p_v}\qquad \qquad(123)$

Applying ideal gas laws to the vapor phase, eq. (123) can be rewritten as

${R_g}T\frac{{d{p_v}}}{{{p_v}}} - {v_\ell }d{p_v} = - \frac{{2\sigma }}{{R_d^2}}{v_\ell }d{R_d}\qquad \qquad(124)$

If the radius of the liquid droplet equals infinity ($1/{R_d} \to 0$), the vapor pressure equals the saturation pressure corresponding to the temperature, psat(T). Integrating eq. (124) from an equilibrium state for a planar surface, the pressure in the vapor phase is obtained:

Figure 6 Liquid droplet suspending in vapor phase.
${p_v} = {p_{sat}}(T)\exp \left\{ {\frac{{{v_\ell }[{p_v} - {p_{sat}}(T) + 2\sigma /{R_d}]}}{{{R_g}T}}} \right\}\qquad \qquad(125)$

If the vapor pressure is below the pressure required by eq. (125), the liquid droplet will shrink by evaporation. The liquid droplet will grow by condensation if the vapor pressure is higher than that required by eq. (125). For most cases, ${p_v} - {p_{sat}}(T) \ll 2\sigma /{R_d}$, eq. (125) can be simplified as

${p_v} = {p_{sat}}(T)\exp \left( {\frac{{2\sigma {v_\ell }}}{{{R_d}{R_g}T}}} \right)\qquad \qquad(126)$

The equilibrium droplet radius can be obtained by reverting eq. (126), i.e.,

${R_d} = \frac{{2\sigma {v_\ell }}}{{{R_g}T\ln [{p_v}/{p_{sat}}(T)]}}\qquad \qquad(127)$

which is similar to eq. (118) except the pressure ratio in the denominator is different. It can be seen that equilibrium of a liquid droplet requires that the vapor phase be supersaturated. The degree of supersaturation, as measured by pv / psat(T), is dependent on the size of the vapor bubble or liquid droplet as indicated by eqs. (117) and (126).

$\begin{array}{l} \frac{{{p_v}}}{{{p_{sat}}}} = \exp \left( {\frac{{2\sigma {v_\ell }}}{{{R_d}{R_g}T}}} \right) \\ {\rm{ }} = \exp \left( {\frac{{2 \times 58.91 \times {{10}^{ - 3}} \times 1.043 \times {{10}^{ - 3}}}}{{{R_d} \times 461.9 \times 373}}} \right) = \exp \left( {\frac{{7.13 \times {{10}^{ - 10}}}}{{{R_d}}}} \right) \\ \end{array}$

The number of molecules in the droplet is estimated by

$N = 1000{N_A}\frac{m}{M} = \frac{{1000{N_A}}}{M}{\rho _\ell }\frac{4}{3}\pi R_d^3\qquad \qquad(128)$

where ${N_A} = 6.022 \times {10^{23}}$ is Avogadro’s number, which is the number of molecules per mole; m is the mass of the liquid droplet; and M = 18.0kg / kmol is the molecular mass of water. Thus

$N = \frac{{1000 \times 6.022 \times {{10}^{23}}}}{{18.0}} \times 958.77 \times \frac{4}{3}\pi R_d^3 = 1.34 \times {10^{29}}R_d^3$

The dependence of the degree of supersaturation on the size of the droplet and the number of molecules per droplet can be tabulated in Table 2. The results in Table 5 can be interpreted as follows. If the degree of supersaturation is very negligible (e.g., pv / psat = 1.000007), 1.34×1017 water molecules must come together spontaneously for the liquid phase pressure to be nucleated. As the degree of the supersaturation increases, the number of molecules needed to come together to nucleate the liquid phase will be significantly reduced. In reality, foreign nuclei often act as the seed for the liquid droplet.

Table 2 Dependence of supersaturation pressure ratio of water at 1 atm on the size of the droplet

 Droplet radii(m) Number of moleculesper droplet Supersaturation pressure ratio (pv/psat) 1.0×10-9 1.34×102 2.040612 1.0×10-8 1.34×105 1.07393 1.0×10-7 1.34×108 1.007158 1.0×10-6 1.34×1011 1.000714 1.0×10-5 1.34×1014 1.000071 1.0×10-4 1.34×1017 1.000007

## Capillary Pressure: The Young-Laplace Equation

Since the distance between the molecules in the vapor phase is much greater than that in the liquid phase, the intermolecular force between the molecules in the vapor phase is very weak. The intermolecular attractive force in the liquid phase holds the molecules in the liquid close to each other. For the molecules within the liquid phase, the intermolecular forces from all directions are balanced. Although the forces acting on the molecules at the liquid-vapor interface are balanced along the tangential direction, the attractive force from the molecules in the liquid phase, Fi (in normal direction), tends to pull the molecules at the liquid-vapor interface toward the liquid phase because the attractive force from the vapor phase, F0, is much weaker. The net inward force FiFo causes movement of the liquid molecules until the maximum number of molecules is in the interior, which leads to an interface of minimum area (Fig. 7).

Figure 7 Origin of surface tension at liquid-vapor interface.
Figure 8 Arbitrarily-curved surface with two radii of curvature RI and RII.

It is generally necessary to specify two radii of curvature to describe an arbitrarily-curved surface, RI and RII, as shown in Fig. 8. The surface section is taken to be small enough that RI and RII are approximately constant. If the surface is now displaced outward by a small distance, the change in area is

$\Delta A = \left( {x + dx} \right)\left( {y + dy} \right) - xy\qquad \qquad(129)$

If $dxdy \approx 0$, then

$\Delta A = y\,dx + x\,dy\qquad \qquad(130)$

The work required to displace the surface is obtained from eq. (82), i.e.,

$\delta W = \sigma (x\,dy + y\,dx)\qquad \qquad(131)$

Displacement acting on the area xy over the distance dz also creates a pressure difference Δp across the surface – capillary pressure (pcap). The work attributed to generating this pressure difference is

$\delta W = \Delta p\,xy\,dz = {p_{cap}}xy\,dz\qquad \qquad(132)$

From the geometry of Fig. 8, it follows that

$\frac{{x + dx}}{{{R_I} + dz}} = \frac{x}{{{R_I}}}\qquad \qquad(133)$

Or

$dx = \frac{{x\,dz}}{{{R_I}}}\qquad \qquad(134)$

Similarly,

$dy = \frac{{y\,dz}}{{{R_{II}}}}\qquad \qquad(135)$

For the surface to be in equilibrium across this differential change, the two expressions for the work must be equal:

$\sigma (x\,dy + y\,dx) = \Delta p\,xy\,dz\qquad \qquad(136)$

i.e.,

$\sigma \left( {\frac{{xy\,dz}}{{{R_I}}} + \frac{{xy\,dz}}{{{R_{II}}}}} \right) = \Delta p\,xy\,dz\qquad \qquad(137)$

The pressure difference between two phases becomes

${p_{cap}} = \Delta p = \sigma \left( {\frac{1}{{{R_I}}} + \frac{1}{{{R_{II}}}}} \right) = \sigma ({K_1} + {K_2})\qquad \qquad(138)$

where K1 and K2 are curvatures of the surface. This expression is called the Young-Laplace equation, and it is the fundamental equation for capillary pressure. It can be seen that when the two curvature radii are equal, in which case the curved surface is spherical, eq. (138) can be reduced to eq. (107).

## Wetting Phenomena and Contact Angles

In addition to the surface tension at a liquid-vapor ($\ell v$) interface discussed above, surface tensions can also exist at a solid-liquid interface ($s\ell$) and solid-vapor interface (sv); this can be demonstrated using a liquid-vapor-solid system as in Fig. 9. The contact line is the locus of points where the three phases intersect. The contact angle, θ, is the angle through the liquid between the tangent to the liquid-vapor interface and the tangent to the solid surface. The contact angle is defined for the equilibrium condition. In 1805, Young published the basic equation for the contact angle on a smooth, insoluble, and homogeneous solid:

$\cos \theta = \frac{{{\sigma _{sv}} - {\sigma _{s\ell }}}}{{{\sigma _{\ell v}}}}\qquad \qquad(139)$

which follows from a balance of the horizontal force components (Faghri, 1995), as shown in Fig. 9. When there is relative motion of a liquid drop over a solid surface, a different contact angle can be expected. When the relative motion stops, an angle different from the apparent (equilibrium) contact angle is seen; it depends upon the direction of the previous motion, i.e., whether it was a receding

Figure 9 Drop of liquid on a planar surface.
Figure 10 Schematic of apparent contact angle θ: (a) stationary liquid, (b) liquid flows upward, (c) liquid flows downward.

or advancing surface, as shown in Fig. 10. The minimum wetting contact angles for different solid-liquid combinations obtained by Stepanov et al. (1977) are reproduced in Table 3. All contact angle approaches are based on the following assumptions (Kwok and Neumann, 1999; Yang et al., 2003): (1) validity of Young’s equation (139), (2) pure liquid, (3) constant values of ${\sigma _{sv}}{\rm{, }}{\sigma _{s\ell }}{\rm{, and }}{\sigma _{\ell v}}$, (4) the value of liquid surface tension should be higher than the anticipated solid surface tension, and (5) a value of σsv independent of the liquid used. For a rough surface, the contact angle θrough is related to the contact angle on a smooth surface θ by

$\cos {\theta _{rough}} = \gamma \cos \theta \qquad \qquad(140)$

where γ is the ratio of the rough surface area to the smooth surface area. Since γ is always greater than 1, cosθrough is greater than cosθ. Therefore, the contact angle θrough is less than the contact angle on a smooth surface, θ. Equilibrium contact angles can vary depending on the motion history of the contact line, particularly for rough surfaces. Equilibrium contact angles may be used in calculations for various heat transfer devices.

Table 3 Minimum wetting contact angle in arc degree (the upper and lower values are for an advancing and receding liquid front, respectively; Stepanov et al. 1977)

 ' Acetone Water Ethanol R-113 Aluminum 73/34 Beryllium 25/11 63/7 0/0 Brass 82/35 18/8 Copper 84/33 15/7 Nickel 16/7 79/34 16/7 Silver 63/38 14/7 Steel 14/6 72/40 19/8 16/5 Titanium 73/40 18/8

Depending on the contact angle, liquids can be classified as nonwetting (${90^ \circ } < \theta < {180^ \circ }$), partially wetting (${0^ \circ } < \theta < {90^ \circ }$), or completely wetting ($\theta = {0^ \circ }$). When a small amount of liquid is brought into contact with an initially dry solid surface, the liquid behaves in one of two ways: (1) if the liquid does not wet the solid it may break up into small droplets, or (2) if the liquid wets the solid it may spread over the solid surface and form a thin liquid film. Wettability can be attributed to a strong intermolecular attractive force near the interface between the solid and liquid. The thermodynamic definition of surface tension, eq. (75), establishes that there is a significant decrease in the surface free energy per unit area in a wetting liquid. Spreading of a liquid on a solid surface can be described by the spreading coefficient Sp, defined as

$S{p_{\ell s}} = {\sigma _{sv}} - {\sigma _{\ell v}} - {\sigma _{s\ell }}\qquad \qquad(141)$

which is a measure of the ability of a liquid to spread over a solid surface. The spreading coefficient defined by eq. (141) is difficult to evaluate, because data for σsv and ${\sigma _{s\ell }}$ are not available for most substances. Substituting eq. (139) into eq. (141), one obtains

$S{p_{\ell s}} = - {\sigma _{\ell v}}(1 - \cos \theta )\qquad \qquad(142)$

For a partially wetting liquid (${0^ \circ } \le \theta \le {90^ \circ }$), $\cos \theta \le 1$ and the spreading coefficient $S{p_{\ell s}}$ is always negative, which means that the liquid will partially wet the solid surface and an equilibrium contact angle can be established. The discussion thus far has treated the three phases – solid, liquid, and vapor – as though their boundaries were sharply-delineated lines or surfaces. This idealization, which serves as a useful analytical device at the macroscopic level, does not hold at the microscopic level. At that level, the interfaces between phases appear as regions over which properties vary continuously, rather than as lines or surfaces with discontinuous property changes. Intermolecular forces of both repulsion and attraction influence how material in the various phases is distributed throughout these interfacial regions. Adsorption – which is one of the consequences of this intermolecular action – occurs when a liquid or solid phase adjacent to a second phase (solid, liquid, or gas) retains molecules, atoms, or ions of the second phase at the shared interface. Adsorption affects the wetting process because it alters the interfacial tension of the solid-liquid interface. Introducing the surface pressure of the adsorbed material on the solid surface, ps:

${p_s} = {\sigma _{sv}} - {\sigma _{sv,a}}\qquad \qquad(143)$

where σsv,a is the interfacial tension with the absorbed substance present. Young’s equation, eq. (139), can be rewritten as

${\sigma _{\ell v}}\cos \theta = ({\sigma _{sv}} - {p_s}) - {\sigma _{s\ell }}\qquad \qquad(144)$

which indicates that adsorption changes equilibrium contact angles and surface tension. Surface tension, equilibrium contact angles, and capillary pressure determine the behavior of liquids in small-diameter tubes, slots, and porous media. The best example is the capillary rise of a wetting liquid in a small tube (see Fig. 11), in which case the capillary force is balanced by gravitational force. The pressure difference across the liquid-vapor interface in such a tube is given by eq. (138). For tubes with a very small radius, the two radii of curvature RI and RII are the same, i.e.,

${R_I} = {R_{II}} = \frac{r}{{\cos \theta }}\qquad \qquad(145)$

Substituting eq. (145) into eq. (138), one obtains the pressure difference between liquid and vapor at point C

${({p_v} - {p_\ell })_C} = {p_{cap,C}} = \frac{{2\sigma }}{r}\cos \theta \qquad \qquad(146)$

The pressure at the flat surface A [see Fig. 11 (a)] is related to the vapor pressure by

${p_A} = {p_v} + {\rho _v}gH\qquad \qquad(147)$

(a) wetting liquid (b) nonwetting liquid
Figure 11 Capillary phenomenon in an open tube (Faghri, 1995; Reproduced by permission of Routledge/Taylor & Francis Group, LLC).

where H is the height of the meniscus above the flat surface A. The pressure at point B inside the tube must be equal to that at point A because they are on the same horizontal surface, i.e.,

${p_B} = {p_v} + {\rho _\ell }gH - {p_{cap,C}} = {p_A}\qquad \qquad(148)$

Combining eqs. (146) – (148), one obtains

$\frac{{2\sigma \cos \theta }}{r} = ({\rho _\ell } - {\rho _v})gH\qquad \qquad(149)$

Capillary rise phenomena can be observed when the liquid wets the tube wall ($\theta < {90^ \circ }$). If the liquid cannot wet the tube wall ($\theta > {90^ \circ }$), the capillary rise H obtained by eq. (149) is negative, which indicates that there is a capillary depression, as shown in Fig. 5(b). In general, solid materials have only two types of behavior when they interact with water. They are either hydrophobic or hydrophilic. Hydrophobic materials have little or no tendency to absorb water, while hydrophilic materials have an affinity for water and readily absorb it. The criteria for hydrophobic or hydrophilic properties of a material is based on its contact angle

${\theta _{Hydrophilic}} < \frac{\pi }{2},{\rm{ }}{\theta _{Hydrophobic}} \ge \frac{\pi }{2}$

Hydrophobic materials can be observed as beading of water on a surface, such as a freshly waxed car surface. Hydrophilic materials allow water to wet its surface forming a film or coating. Hydrophilic materials are usually charged or have polar side groups to their structure that attract water. There are many cases of hydrophobic surfaces in nature, including some plant leaves, butterfly wings, duck feathers and some insects’ exoskeletons. There are many synthetic hydrophobic materials available including waxes, alkanes, oils, Teflon, and Gortex. There are numerous applications for using these materials such as protection of stone, wood, and cement from the effects of rain, waterproofing fabrics and the removal of water from glass surfaces, such as a windshield, to increase transparency. Hydrophobic materials are also used for cleaning up oil spills, removal of oil from water and for chemical separation processes to remove nonpolar from polar compounds. A hydrophilic material’s ability to absorb and transport water gives it numerous applications in cleaners, housings, cables, tubes and hoses, waterproofing, catheters, surgical garments, etc. A hydrophilic coating on a tube or hose eliminates the need for other lubricants, which is useful to prevent cross-contamination. Hydrophilic coatings on plugs and o-rings increase their ability to stop leaks; this is the basis for sealants. Another use for hydrophobic and hydrophilic materials is the storage and distribution of water and methanol in miniature direct methanol fuel cells (DMFCs). For the distribution of methanol, a material is chosen that wets methanol but is hydrophobic to water. This type of a material is a preferential wicking material. This allows neat methanol to be stored and distributed in a DMFC without water diffusing into the methanol storage layers. The water storage layer at the anode of the fuel cell is a hydrophilic material. It is not preferential to either water or methanol and provides a layer in which they can mix.

## Disjoining Pressure: Thermodynamic and Hydrodynamic Definitions

Disjoining pressure is a phenomenon that occurs in thin liquid films (Israelachvilli, 1992). When ultra-thin liquid films are in contact with a solid surface, there is attraction between the liquid molecules and the solid molecules. The pressure in the liquid must balance the ambient pressure and the attractive forces between the liquid and solid. When a film is very thin, the liquid-solid attractive forces act to pull the liquid away, and the balancing pressure that counteracts this force is the disjoining pressure. Disjoining pressure theory has been used in ultra-thin films on solid surfaces to model the molecular force interactions between the liquid-solid interfaces. This has been used most extensively in modeling thin-film transport in micro heat pipes, axially grooved evaporators and micro heat pump loops (Khrustalev and Faghri, 1995; Faghri, 1995). The idea of disjoining pressure is well known as an explanation of the effect of wall-fluid force interaction in thin films. Carey and Wemhoff (2005) analyzed the effects of disjoining pressure in ultra-thin layers and films, as discussed below. An example of a situation that involves disjoining pressure in micropassages is shown in Fig. 12. The deeper Region A carries most of the liquid while Region B carries a thin film flow with

Figure 12 Analysis of disjoining pressure in a cross-section of a micropassage containing thin liquid films for a stratified configuration.

a thickness δ. Most of the interface (Region A) is flat and separates the deeper liquid flow from the vapor at equilibrium

${p_{v,}}_\delta = {p_\ell }\qquad \qquad(150)$

where ${p_\ell }$ is the liquid pressure at the interface in the absence of attractive wall forces and pv is the vapor pressure at the interface. The pressure in the thin film at Region B is changed by force interactions between the liquid and solid. The general disjoining pressure is derived using the potential energy from the forces that the wall exerts on the fluid. To find the pressure variation in the film, the interface between solid and fluid as shown in Fig. 12 is considered. The interactions between fluid and metallic solid molecules are modeled using the Lennard-Jones interaction potential

${\phi _{fs}}\left( r \right) = - \frac{{{C_{\phi ,fs}}}}{{{r^6}}}\left[ {1 - \frac{{D_m^6}}{{{r^6}}}} \right]\qquad \qquad(151)$

where φfs is the solid-fluid intermolecular potential and r is the distance between molecules. The long-range attraction between a pair of two fluid molecules is assumed to have a similar form.

${\phi _{ff}}\left( r \right) = - \frac{{{C_{\phi ,ff}}}}{{{r^6}}}\left[ {1 - \frac{{D_f^6}}{{{r^6}}}} \right]\qquad \qquad(152)$

where φff is the fluid-fluid intermolecular potential. The constants Dm and Df are the closest approach distance of fluid to solid molecules and two fluid molecules, respectively. To find the total effect of all the solid molecules on a

Figure 13 Schematic used for derivation of disjoining pressure (Carey and Wemhoff, 2005)

free fluid molecule, the product of density and molecular potential is integrated. The mean-field potential energy, Φfmf, felt by the free fluid molecule due to interactions with all the solid molecules, as seen in Fig. 13, is

${{\Phi}_{fmf}}= \int_{{z_s}=z}^{\infty} \int_{x=0}^{\infty} {{\mathfrak{N}}_s}{{\phi}_{fs}}(2 \pi x)dxd{z_s} \qquad \qquad (153)$

where ${{\mathfrak{N}}_s}$ is the solid wall molecular number density. Substituting eq. (151) and x2 + z2 for r2 reduces eq. (153) to

${{\Phi}_{fmf}} = - \frac{\pi {{\mathfrak{N}}_s}{C_{\phi , fs}}}{6{z^3}} + \frac{\pi {{\mathfrak{N}}_s}{C_{\phi , fs}}{{D_m}^6} }{45{z^9}} \qquad \qquad (154)$

The above relation is reorganized in terms of a modified Hamaker constant AH

${A_H} = {{\pi}^2}{{\mathfrak{N}}_f}{{\mathfrak{N}}_s}{C_{\phi , fs}} \qquad \qquad (155)$

where ${{\mathfrak{N}}_f}$ is the fluid molecular number density. Considering eq. (155), eq. (154) becomes

${{\Phi}_{fmf}}=\frac{{A_H}}{6 \pi {{\mathfrak{N}}_f}{{D_m}^3}}[ \frac{2}{15} {{(\frac{{D_m}}{z})}^9} - {{(\frac{{D_m}}{z})}^3}] \qquad \qquad (156)$

which is equivalent to a body force similar to the hydrostatic variation of pressure caused by gravity. A force balance is used to obtain the pressure gradient

$\frac{dp}{dz}{n_z} = - \frac{{{\mathfrak{N}}_f}M}{{N_A}}{f_{fs}} \qquad \qquad (157)$

where nz is the unit vector in the z-direction and ffs is the force per unit mass on the fluid system. The force exerted on a single molecule by the entire wall is given by

${F_{fs}}= - \frac{d {{\Phi}_{fmf}}}{dz} = \frac{A_H}{2 \pi {{\mathfrak{N}}_f}{{D_m}^4}}[ \frac{2}{5}{{(\frac{{D_m}}{z})}^{10}} - {{(\frac{{D_m}}{z})}^4}] {n_z} \qquad \qquad (158)$

The corresponding force per unit mass is

${f_{fs}} = \frac{{N_A}{F_{fs}}}{M} = \frac{{N_A}{A_H}}{2 \pi M {{\mathfrak{N}}_f}{{D_M}^4}} [ \frac{2}{5} {(\frac{D_m}{z})^{10}} – {(\frac{D_m}{z})^4}]{n_z} \qquad \qquad (159)$

where NA is Avogadro’s number. Substituting eq. (159) into eq. (157) and considering that the force only acts in the z-direction,

$\frac{{dp}}{{dz}} = - \frac{{{A_H}}}{{2\pi D_m^4}}\left[ {\frac{2}{5}{{\left( {\frac{{{D_m}}}{z}} \right)}^{10}} - {{\left( {\frac{{{D_m}}}{z}} \right)}^4}} \right]\qquad \qquad(160)$

Integrating both sides of eq. (160) from z to ∞, and considering the pressure at ∞ is ${p_\ell }$, the pressure profile close to the wall is

$p\left( z \right) = {p_\ell } + \frac{{{A_H}}}{{6\pi D_m^3}}\left[ {\frac{2}{{15}}{{\left( {\frac{{{D_m}}}{z}} \right)}^9} - {{\left( {\frac{{{D_m}}}{z}} \right)}^3}} \right]\qquad \qquad(161)$

The z − 9 term in eq. (161) may be neglected because Dm is on the order of a molecular diameter. Equation (161) simplifies to

$p\left( z \right) = {p_\ell } - \frac{{{A_H}}}{{6\pi {z^3}}}\qquad \qquad(162)$

The pressure in the thin liquid film at Region B in Fig. 30 is expected to vary with distance from the lower wall. At the interface $\left( {z = \delta } \right)$, the pressure in the liquid${p_{\ell ,\delta }}$ must be

${p_{\ell ,\delta }} = p\left( \delta \right) = {p_\ell } - \frac{{{A_H}}}{{6\pi {\delta ^3}}}\qquad \qquad(163)$

Combining eqs. (150) and (163) and solving for pv${p_{\ell ,\delta }}$ yields

${p_{v,\delta }} - {p_{\ell ,\delta }} = \frac{{{A_H}}}{{6\pi {\delta ^3}}}\qquad \qquad(164)$

The disjoining pressure pd is the amount that pv differs from ${p_{\ell ,\delta }}$

${p_d} = - \frac{{{A_H}}}{{6\pi {\delta ^3}}}\qquad \qquad(165)$

The pressure difference across the interface in Fig. 12 is equal to the disjoining pressure, which quickly increases in magnitude as the film thins. The disjoining pressure has been found to alter thermodynamic equilibrium conditions at the liquid-vapor interface of thin films. The change in vapor pressure versus temperature relation must be considered when modeling thin film evaporation and condensation in micropassages of micro heat pipes and micro capillary-pumped loops. Disjoining pressure can also be developed using the classical thermodynamic analysis by integrating the Gibbs-Duhem equation at a constant temperature from saturation conditions to an arbitrary state in the liquid and vapor phases.

$d\bar \mu = - \bar sdT + \bar vdp\qquad \qquad(166)$

where $\bar s$ is the molar entropy and $\bar v$ is the molar specific volume. The liquid is assumed incompressible and the vapor is an ideal gas. This yields the following for liquid and vapor molar chemical potentials, ${\bar \mu _\ell }$ and ${\bar \mu _v}$ respectively, without wall attraction effects:

${\bar \mu _v} = {\bar \mu _{v,sat}} + {R_u}T\ln \left( {{p_{v,\delta }}/{p_{sat}}} \right)\qquad \qquad(167)$

${\bar \mu _\ell } = {\bar \mu _{\ell ,sat}} + {\bar v_\ell }\ln \left( {{p_\ell } - {p_{sat}}} \right)\qquad \qquad(168)$</center>

where ${\bar \mu _{\ell ,sat}}$ and ${\bar \mu _{v,sat}}$ are the molar chemical potential of bulk liquid and vapor at saturation, respectively. The chemical potential is equal to the specific Gibbs function for a pure liquid:

${\bar \mu _\ell } = {\bar g_\ell }\qquad \qquad(169)$

When considering wall attraction effects in the liquid film, the potential energy associated with the interaction of fluid and surface molecules is added to the Gibb’s function. The liquid chemical potential becomes

${\bar \mu _\ell } = {\bar \mu _{\ell ,sat}} + {\bar v_\ell }\ln \left( {{p_\ell } - {p_{sat}}} \right) + {N_A}{\Phi _{fmf}}\qquad \qquad(170)$

where Φfmf is the potential energy per fluid molecule due to interactions with the wall, and ${\bar v_\ell }$ is the liquid molar specific volume. For equilibrium conditions at the interface, the liquid pressure must be equal to the vapor pressure

${p_\ell } = {p_{v,\delta }}\qquad \qquad(171)$

Setting ${p_{\ell}}$ equal to pv and equating the right sides of eqs. (167) and (170)

${R_u}T ln ({p_{v, \delta}} / {p_{sat}}) = {{\overline{v}}_{\ell}}({p_{v, \delta}} - {p_{sat}} ) + {N_A} {{\Phi}_{fmf}}\qquad \qquad(172)$

where psat is the normal saturation pressure corresponding to vapor bulk of the system. Rearranging eq. (172), one obtains

${p_{v,\delta }}/{p_{sat}} = \exp \left[ {\frac{{{p_{sat}}{{\bar v}_\ell }}}{{{R_u}T}}\left( {\frac{{{p_{v,\delta }}}}{{{p_{sat}}}} - 1} \right) + \frac{{{N_A}{\Phi _{fmf}}}}{{{R_u}T}}} \right]\qquad \qquad(173)$

Neglecting the first term in the brackets at the right-hand side of eq. (173) gives

${p_{v,\delta }}/{p_{sat}} = \exp \left( {\frac{{{N_A}{\Phi _{fmf}}}}{{{k_b}T}}} \right)\qquad \qquad(174)$

Substituting eq. (156) into (174), neglecting z − 9 and setting z equal to δ at the interface gives,

${p_{v,\delta }}/{p_{sat}} = \exp \left( { - \frac{{{A_H}}}{{6\pi {\rho _f}{\delta ^3}{k_b}T}}} \right)\qquad \qquad(175)$

${\bar \mu _v} = {\bar \mu _{v,sat}} + {R_u}T\ln \left( {{p_{v,\delta }}/{p_{sat}}} \right)\qquad \qquad(176)$

${\bar \mu _\ell } = {\bar \mu _{\ell ,sat}} + {\bar v_\ell }\ln \left( {{p_{\ell ,\delta }} - {p_{sat}}} \right)\qquad \qquad(177)$

The vapor and liquid pressures are related by

${p_{v,\delta }} - {p_{\ell ,\delta }} = {p_{cap}} - {p_d}\qquad \qquad(178)$

where pcap is the capillary pressure. If the liquid film is flat, the capillary pressure can be neglected and eq. (178) is simplified as

${p_{v,\delta }} - {p_{\ell ,\delta }} = - {p_d}\qquad \qquad(179)$

Substituting eq. (179) into eq. (177) and equating the right sides of eqs. (176) and (177) yields

${R_u}T ln ({p_{v, \delta}} / {p_{sat}}) = {{\overline{v}}_{\ell}}({p_{v, \delta}} + {p_d} - {p_{sat}})\qquad \qquad(180)$

which can be rearranged to obtain

${p_{v,\delta }}/{p_{sat}} = \exp \left[ {\frac{{{p_{sat}}{{\bar v}_\ell }}}{{{R_u}T}}\left( {\frac{{{p_{v,\delta }}}}{{{p_{sat}}}} - 1} \right) + \frac{{{p_d}{{\bar v}_\ell }}}{{{R_u}T}}} \right]\qquad \qquad(181)$

Neglecting the first term in the brackets on the right-hand side of eq. (181) gives

${p_{v,\delta }}/{p_{sat}} = \exp \left( {\frac{{{p_d}{{\bar v}_\ell }}}{{{R_u}T}}} \right)\qquad \qquad(182)$

Comparing eqs. (175) and (182), we have

${p_d} = - \frac{{A_H}}{6 \pi {{\delta}^3}} \frac{{N_A}}{{\mathfrak{N}_f} {\overline{v}_{\ell}}} \qquad \qquad (183)$

where ${N_A}/ {{\mathfrak{N}}_f} {v_{\ell}} = 1$, and eq. (183) will become identical to eq. (165).

## Superheat-Thermodynamic and Kinetic Limit Definitions

From the classical thermodynamic point of view, phase transformation occurs at the equilibrium normal saturation condition as a quasi-equilibrium process. However, real phase transformation usually occurs as a non-equilibrium process. For example, in vaporization processes a superheated liquid may exist in part of the system. Similarly, in the condensation process, part of the vapor usually has been supercooled below its equilibrium normal saturation temperature. In Fig. 2, it was shown that metastable conditions correspond to situations where vapor is supercooled below its normal saturation temperature or liquid is superheated above its normal equilibrium temperature. As shown in Section 2.7.4, mechanical stability requires that

${\left. {\frac{{\partial p}}{{\partial \nu }}} \right|_T} \le 0\qquad \qquad(184)$

Based on the above, the liquid or vapor in a metastable domain is not mechanically unstable, even though it is not in thermodynamic equilibrium. In Fig. 2 between 2' and 4', ${(\partial p/\partial v)_T} > 0$ and this domain corresponds to a region between the liquid and vapor spinodal where it does violate mechanical stability. The superheat limit is the maximum temperature that a liquid can be heated before it homogeneously nucleates. This superheat limit can be determined thermodynamically or kinetically using kinetic and nucleation theory. The degree of superheat (TtTsat) ranges from less than one to a few hundred degrees, and depends on factors such as the type and amount of liquid, surface conditions and the type and rate of heating. A superheated liquid is one that does not follow its normal or saturated equilibrium phase boundary. Normal refers to a special case of equilibrium across a flat-plate boundary, $R \to \infty$, where R is the radius of curvature of the phase boundary. Initial bubbles in superheated liquids are in mechanical equilibrium.

${p_v} - {p_\ell } = \frac{{2\sigma }}{R} > 0\qquad \qquad(185)$

Vapor pressure, pv, is not the same as equilibrium vapor pressure, psat at temperature T. This is due to equilibrium across flat and curved boundaries, Section 2.10.3.

${p_v} = {p_{sat}}\left( T \right)\exp \left\{ {\frac{{{v_\ell }}}{{{R_g}T}}\left[ {{p_v} - {p_{sat}}\left( T \right) - \frac{{2\sigma }}{R}} \right]} \right\}\qquad \qquad(186)$

The external liquid pressure, ${p_\ell }$, may be either compressive or extensive. However, for this problem it will be a compressive pressure on the bubble wall. For a vapor bubble in which ${p_v} > {p_\ell }$ and $R < \infty$, $T\left( {{p_\ell }} \right) > {T_{sat}}\left( {{p_\ell }} \right)$.

## Thermodynamic limit of superheat

There is a limit to the extent of isobaric heating that the liquid can undergo. At this limit the liquid is unstable and any perturbation will cause a phase transition. The limit of stability, also called the thermodynamic limit of superheat, occurs when the entropy of an isolated system is at its maximum point in a stable equilibrium state with respect to small variations of its natural variables. The Helmholtz function, F, assumes a minimum stable equilibrium state for an open system with respect to variations. For variations from a stable state

$\Delta F > 0 \qquad \qquad (187)$

The thermodynamic limit of superheat is the limit of mechanical stability since it is already thermally stable, [see eq. (142)]

${\left. {\frac{{\partial p}}{{\partial v}}} \right|_T} \le 0\qquad \qquad(188)$

The calculated value of superheat depends upon the equation of state used for the calculation of stability. For example, consider van der Waals equation of state for a pure substance [see eq. (185)]

$\left( {p + \frac{a}{{{v^2}}}} \right)\left( {v - b} \right) = {R_g}T\qquad \qquad(189)$

where a and b are constants. This equation inaccurately represents the saturation state of most substances. The spinodal curve for the van der Waals equation of state is

$p = \frac{a}{{{v^2}}} - \frac{{2ab}}{{{v^3}}}\qquad \qquad(190)$

If a pressure is given, then v may be eliminated from equations (189) and (190) to give the thermodynamic limit of superheat, $T \to {T_t}$. This generally requires an iterative solution except when $p \to 0$, in which case Tt = 27Tc / 32. A simple correlation for Tt is given by Lienhard (1976) to eliminate the need for iterations:

${T_t} \simeq {T_c}\left[ {\frac{{27}}{{32}} + \frac{5}{{32}}{{\left( {\frac{{{T_{sat}}}}{{{T_c}}}} \right)}^{5.16}}} \right]\qquad \qquad(191)$

Table 4 lists the thermodynamic limit of superheat for some substances at 0.10 MPa (Avedisian 1986). These values are significantly higher than their respective boiling temperatures, which prove that they can undergo substantial superheating. This can be proved experimentally; however, the best experiments can only be expected to yield maximum temperatures of

${T_t}\left( {{p_\ell }} \right) > {T_m}\left( {{p_\ell }} \right)\qquad \qquad(192)$

The van der Waals limit is not valid for calculations because calculated superheat limits would fall into unstable regions, violating the second law. Using the Peng-Robinson (1976) equation of state a different thermodynamic limit of superheat is obtained.

$p = \frac{{{R_g}T}}{{\left( {v - b} \right)}} - \frac{a}{{{v^2} + 2bv + {b^2}}}\qquad \qquad(193)$

where a and b are constants. The spinodal curve of eq. (193) is

$\frac{{{R_g}T}}{{\left( {v - {b^2}} \right)}} - \frac{{2a\left( {v + b} \right)}}{{{{\left( {{v^2} + 2bv - {b^2}} \right)}^2}}} = 0\qquad \qquad(194)$

The superheat temperature derived by the above equation, Tt2, in Table 4 is appropriate since it is higher than measured data. Use of another equation of state would yield another thermodynamic limit of superheat. This shows the challenges involved in trying to solve the thermodynamic limit of superheat. The thermodynamic superheat limit is the upper limit of stability of superheated liquid.

Table 4 Thermodynamic Limit of Superheat of Some Pure Liquids at Atmosphere Pressure* (Avedisian 1986)

 Substance Tsat Tt1 Tt2 Tm Tc J(Tt2) n-pentane 309 405 431 426 470 $8 \times {10^{24}}$ n-heptane 372 468 499 494 540 $8 \times {10^{26}}$ n-octane 399 494 525 514 569 $2 \times {10^{26}}$ Methanol 338 442 477 466 513 1029 Ethanol 352 447 482 472 516 1030 water 373 552 596 575 647 $9 \times {10^{28}}$
• Tsat-Normal boiling point (K) at 0.101 MPa.

Tt1-Calculated thermodynamic limit of superheat (K) at 0.101 MPa using van der Waals equation of state. Tt2-Calculated thermodynamic limit of superheat (K) at 0.101 MPa using Peng-Robinson equation of state. Tm-Highest measured liquid phase temperature (K) at 0.101 MPa

J-Nucleation rate (nuclei/cm3-s) at Tt2 and 0.101 MPa.

## Kinetic limit of superheat

Superheated liquids are not quiescent at the microscopic level. Random molecular motion creates local variations in density. The fluctuation in density creates “holes” or “nuclei” within which the molecules may resemble gas in terms of spacing and potential energy. These nuclei grow and decay until a certain size nucleus is created that is in unstable equilibrium with its surroundings. These bubbles are the initial condition for bubble growth within a liquid and the critical size nuclei is known.

Homogeneous nucleation theory can predict the rate of formation of a critical size nucleus at a given pressure, temperature and composition. The nucleation rate is the mean rate at which nuclei are formed and grow to macroscopic size. Kinetic theory mechanically formulates the critical nucleus. The theory states that the steady-state nucleation rate is proportional to the exponential of the formation energy

$J = \Gamma {k_{f(N*)}} {\mathfrak{N}_{\ell}} exp [- \frac{\Delta \Phi *}{{k_b}{T_{\ell}}}] \qquad \qquad (195)$

where, kf(N * ) is the molecular condensation rate in a critical size nucleus (with N* molecules), ${{\mathfrak{N}}_{\ell}}$ is the total number density of molecules. In terms of ${T_\ell }$

${T_{\ell}} = {T_k} = \frac{\Delta \Phi *}{{k_b}}{{[ln(\frac{\Gamma {k_{f(N*)}} {\mathfrak{N}_{\ell}}}{J})]}^{-1}} \qquad \qquad(196)$

Tk is the kinetic limit of superheat and ΔΦ * is the minimum energy necessary for formation of a critical size nucleus and is given as

$\Delta \Phi * = \frac{{16\pi {\sigma ^3}}}{{3{{\left( {{p_{sat\left( {{T_\ell }} \right)}} - {p_\ell }} \right)}^2}}}\qquad \qquad(197)$

The factor Γ takes into account the detailed mechanism by which critical nuclei are formed within the molecular network of the liquid. To determine Γ, the following issues must be solved.

1. The energy of a nucleus is a function of the number of molecules in it.

2. The exponential dependence of J on ΔΦ* must be determined.

3. The mechanism by which critical size nuclei form, must be described.

The theory of homogeneous nucleation, including the determination of the superheat and appropriate experiments, is investigated extensively (Blander and Katz, 1975; Skripov, 1974; Avedisian, 1986; Debenedetti, 1996; and Iida et al., 1997). In nucleation theory, the net rate of embryo growth from the size N(N molecules) to size N + 1, per unit volume per unit time is (Skripov, 1974)

$J = \frac{{{p_v}/{{\left( {2\pi m{k_b}{T_\ell }} \right)}^{1/2}}}}{{\int_0^\infty {{{\left( {A{N_s}} \right)}^{ - 1}}dN} }}\qquad \qquad(198)$

where m is mass of a single molecule, kb is the Boltzman constant, A is the interfacial area of the embryo and Ns is the number density of embryos containing N molecules. The numerator in Eq. (198) is the vaporization rate for the surface of the embryo. Debenedetti (1996) simplified the above relation in terms of easily measured properties

$J = {N_\ell }{\left( {\frac{{2\sigma }}{{\pi mB}}} \right)^{1/2}}\exp \left[ { - \frac{{16\pi }}{{3{k_b}T}}\frac{{{\sigma ^3}}}{{{\delta ^2}{{\left( {{p_{sat\left( {{T_\ell }} \right)}} - {p_\ell }} \right)}^2}}}} \right]\qquad \qquad(199)$

Table 5 Limit of Superheat and Nucleation Rate of Water at Atmospheric Pressure* (Avedisian 1986)

 T pv psat R × 107 J Waiting time/cm3 (~1/J) 500 25.8 25.2 25.2 <10-99 > 1091 years 550 59.1 61.0 6.76 <10-99 < 1091 years 560 68.3 71.0 5.2 $1.7 \times {10^{ - 76}}$ $1.2 \times {10^{68}}$ years 570 78.5 82.0 3.9 $8.5 \times {10^{ - 20}}$ $3.7 \times {10^{11}}$ years 575 83.9 88.0 3.4 $5.7 \times {10^{ - 3}}$ $1.8 \times {10^2}$ sec 580 89.6 94.4 2.9 $4.3 \times {10^9}$ $2.3 \times {10^{ - 10}}$ sec 590 101.6 108.9 2.1 $4.3 \times {10^{23}}$ $2.3 \times {10^{ - 24}}$ sec
• T-temperature (K), pv-pressure in vapor nucleus (atm), psat-equilibrium vapor pressure (atm), R-radius of critical nucleus (cm), J-nucleation rate (nuclei/cm3-sec).
$\delta = 1 - \frac{{{p_v}{v_\ell }}}{{{k_b}{T_\ell }}}\qquad \qquad(200)$

$B = \frac{2}{3} + \frac{1}{3}\frac{{{p_\ell }}}{{{p_{sat\left( {{T_\ell }} \right)}}}}\qquad \qquad(201)$

The effect of temperature is significant in rate of nucleation considering the exponential form of eq. (199). Avedisian (1986) showed the variation of J with change in temperature for superheated water at 1 atm in Table 5. The waiting time (the reciprocal of J assuming volume of liquid 1 cm3) is also shown in the same table. The data for water at 1 atm pressure indicates that the bubble nucleation is rare at temperatures less than 570K. At higher temperatures, J increases while waiting time decreases with increasing temperature. This is a range in which homogeneous nucleation does not occur and above that temperature it occurs spontaneously. In practice to use eq. (199) a suitable threshold is chosen. It is found that a threshold of J = 1030 nuclei/cm3-sec is appropriate for most working fluids and operating conditions. The detailed discussion of superheat limit can be found in Avedisian (1986).

## References

Avedisian, C.T., 1986, “Bubble Growth in Superheated Liquid Droplets,” Encyclopedia of Fluid Mechanics, Chapter 8, Gulf Publishing Company, Houston, TX.

Blander, M and Katz, J.P., 1975, “Bubble Nucleation in Liquids,” AIChE Journal, Vol. 21, No. 5, pp. 833-848.

Carey, V.P., 1992, Liquid-Vapor Phase-Change Phenomena: An Introduction to the Thermophysics of Vaporization and Condensation Processes in Heat Transfer Equipment, Hemisphere Publishing Corp., Washington, D. C.

Carey, V.P. and Wemhoff, A.P., 2005, “Disjoining Pressure Effects in Ultra-Thin Liquid Films in Micropassages-Comparison of Thermodynamic Theory with Predictions of Molecular Dynamics Simulations,” IMECE2005-80234, Proceedings of 2005 ASME International Mechanical Engineering Congress and Exposition, Orlando, FL.

Debenedetti, P.G., 1996, Metastable Liquids –Concepts and Principles, Princeton University Press, Princeton, NJ.

Faghri, A., 1995, Heat Pipe Science and Technology, Taylor and Francis, Washington D.C.

Faghri, A., and Zhang, Y., 2006, Transport Phenomena in Multiphase Systems, Elsevier, Burlington, MA.

Fowkes, F.M., 1965, “Attractive Forces at Interfaces,” Chemistry and Physics of Interfaces, American Chemical Society, Washington, DC.

Iida, Y., Okuyama, K., and Nishizawa, T., 1997, “Heat Transfer During Boiling Initiated by Fluctuation Nucleation on a Platinum Film Rapidly Heated to the Limit of Liquid Superheat,” Journal of the Japan Society of Mechanical Engineers, B 63 (613) pp. 3048-3054.

Israelachvilli, J., 1992, Intermolecular & Surface Forces, 2nd ed., Academic Press, London.

Khrustalev, D. and Faghri, A., 1995, “Heat Transfer During Evaporation on Capillary Grooved Structures of Heat Pipes,” ASME Journal Heat Transfer, Vol. 117, pp. 740-747.

Kwok, D.Y., and Neumann, A. W., 1999, “Contact Angle Measurement and Contact Angle Interpretation,” Advances in Colloid and Interfaces Science, Vol. 81, pp. 167-249.

Kyle, B.G., 1999, Chemical and Process Thermodynamics, 3rd ed., Prentice-Hall Inc., Englewood Cliffs, New Jersey.

Lienhard, J.H., 1976, “Correlation for the Limiting Liquid Superheat,” Chemical Engineering Science, Vol. 31, pp. 847-849.

Peng, D.Y. and Robinson, D.B., 1976, “A New Two Constant Equation of State,” Industrial Engineering Fundamentals, Vol. 15, pp. 59-64.

Skripov, V.P., 1974, Metastable Liquids, New York, John Wiley and Sons.

Stepanov, V.G., Volyak, L.D., and Tarlakov, Y.V., 1977, “Wetting Contact Angles for Some Systems,” Journal of Engineering Physics and Thermophysics, Vol. 32, No. 6, pp. 1000-1003.

Yang, J., Han, J., Isaacson, K., and Kwok, D. Y., 2003, “Effects of Surface Defects, Polycrystallinity, and Nanostructure of Self-Assembled Monolayers for Octadecanethiol Adsorbed on to Au on Wetting and its Surface Energetic Interpretation,” Langmuir, Vol. 19, pp. 9231-9238.