- Preface
- Introduction to Transport Phenomena
- Thermodynamics of Multiphase Systems
- Generalized Governing Equations in Mutliphase Systems: Local Instance Formulations
- Generalized Governing Equations for Multiphase Systems : Averaging Formulations
- Solid-Liquid-Vapor Phenomena and Interfacial Heat and Mass Transfer
- Melting and Solidification
- Sublimation and Vapor Deposition
- Condensation
- Evaporation
- Boiling
- Two-Phase Flow and Heat Filter
- Appendix A : Constants, Units and Conversion Factors
- Appendix B: Transport Properties
- Appendix C: Vectors and Tensors
- Index

2
THERMODYNAMICS OF MULTIPHASE SYSTEMS
2.1 Introduction
The primary objective of this chapter is to define the concept of thermodynamic equilibrium, and to describe the conditions under which equilibrium exists in thermodynamic systems. The system considered may consist of either a single phase or multiple phases and may include one or more components. In its simplest form, a system is said to be in equilibrium when its measurable properties do not change over time. It must be clearly pointed out that steadystate open systems that exchange mass, heat, or work with the surroundings do not meet this criterion and therefore are not in equilibrium. There are no unbalanced driving potentials within a system in the thermodynamic equilibrium state, and the system in equilibrium experiences no change when it is isolated from its surroundings. The system is said to be at equilibrium if conditions for all types of thermodynamic equilibrium are satisfied. The equilibriums that can be encountered in a multiphase system include (1) thermal equilibrium (no change occurs in temperature), (2) mechanical equilibrium (no change occurs in pressure), (3) chemical equilibrium (no change occurs in chemical composition), and (4) phase equilibrium (no phase change occurs). For a thermodynamic system to be in equilibrium, a mathematical treatment of the combination of the first two laws of thermodynamics must ascertain that certain intensive properties are uniform throughout the system. These intensive properties include the temperature, pressure, and chemical potential. In other words, this uniformity implies no heat, mass, or mechanical work transfer between the system and surroundings. To describe a system that is in equilibrium, a certain number of independent intensive thermodynamic variables must be specified. Another related, but different, concept is stability, which characterizes the result of small perturbations to a system in thermodynamic equilibrium. The system in thermodynamic equilibrium is said to be thermodynamically stable if equilibrium can be maintained after a small perturbation.
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The Gibbs phase rule describes the relationship between the number of variables required to describe a system and the number of phases in the system. The relationships among different thermodynamic variables are represented by thermodynamic surfaces, phase diagrams, and equations of state. In practice, the equilibrium properties are used in solving engineering problems where mass and energy are continuously exchanged; in these cases local thermodynamic equilibrium is assumed. The thermodynamic laws and fundamental thermodynamic relations based on the first two laws of thermodynamics are briefly reviewed in Section 2.2. Also discussed in Section 2.2 is the Gibbs phase rule, which relates the number of phases and components present in a system to the number of intensive thermodynamic properties needed to specify the state of the system. Section 2.3 starts with a discussion of the general criteria for equilibrium in a closed system with a single-phase substance. The discussion of equilibrium criteria is then extended to a closed system with a multicomponent single-phase substance; this is followed by a discussion of the thermal, mechanical, and chemical potential stability for a single-phase system. Section 2.4 presents thermodynamic surfaces and phase diagrams for single and multicomponent substances, as well as the equations of state that can be used to obtain other thermodynamic variables based on the independent thermodynamic variables. Section 2.5 extends the discussion of the equilibrium criteria to a system consisting of two phases and a single component, and then develops the Clapeyron equation, which relates the temperature and pressure of a thermodynamic system in the saturation region. In addition, equilibrium criteria for a closed system with a multicomponent, multiphase substance are presented. Section 2.5 also addresses the question of how far from equilibrium a two-phase system can deviate before the system becomes unstable. This small region, termed the metastable region, is important to designers because it affects safety. Section 2.6 presents a discussion on the thermodynamic equilibrium conditions at an interface, followed by thermodynamic definitions of surface tension pressure, a discussion of the surface tension effects at microscale vapor bubble/liquid droplets, and a thermodynamic definition of disjoining pressure.
2.2 Fundamentals of Thermodynamics
2.2.1 Thermodynamic Laws
The first law of thermodynamics simply describes the conservation of energy. Max Planck put this more adeptly by stating that the first law of thermodynamics is “nothing more than the principle of the conservation of energy applied to the phenomena involving the production or absorption of heat.” The first law of thermodynamics states that during any system cycle, the production and absorption of heat must equal the work done by the system. Beyond mandating
108 Chapter 2 Thermodynamics of Multiphase Systems
this equality, however, the first law puts no restriction on the direction of the flow of heat or work. As is known from everyday experiences, there is only one direction in which real system processes may proceed. For example, the energy that feeds a sapling allows the sapling to grow into a tree; however, the energy cannot be removed from a mature tree in such a way that the tree is transformed back to a sapling. Therefore, the second law of thermodynamics, which is a statement of the irreversibility of real-life processes, was developed. There are two common statements of the second law of thermodynamics. The KelvinPlanck statement is: “it is impossible for any device that operates in a cycle to receive heat from a single reservoir and produce a net amount of work.” Another common statement is the Clausius statement: “It is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a lower-temperature body to a higher-temperature body.” It can be proven that the Kelvin-Planck statement and the Clausius statement are equivalent (Cengel and Boles, 2002). The second law of thermodynamics allows us to calculate the change of entropy for a system undergoing a process from one state to another state. The change in entropy of a thermodynamic system, during a reversible process in which an amount of heat δQ (J) is introduced at constant temperature T, is defined as dS = δ Q / T . However, it reveals nothing about the absolute value of entropy. The third law of thermodynamics differs from the first two in that it requires knowledge of the microscopic nature of the system and therefore requires a good understanding of statistical and quantum mechanics. Simply stated, however, the third law of thermodynamics is: “The entropy of a perfect crystalline substance is zero at zero absolute temperature.” It can be directly inferred from this statement that a crystalline structure has a maximum degree of order and, since it is at absolute zero temperature, the thermal energy is at its minimum value. It further follows that a nonperfect crystalline structure has a finite value of entropy at absolute zero temperature. In a practical engineering sense, the third law of thermodynamics provides an absolute base from which the entropy of a substance can be measured in various states and tabulated; this has been done for many substances. These absolute-zero temperature values of entropy are based on chemical reaction data at low temperatures, and on measurements of heat capacity at temperatures approaching absolute zero. With this information, it is then a relatively straightforward process to find the increase in entropy between absolute zero and any given state by using calorimetric data or statistical thermodynamic procedures.
2.2.2 Thermodynamic Relations
For a single-component closed system (fixed mass), the first law of thermodynamics gives us ˆ (2.1) dE = δ Q − δ W
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ˆ where E is the total energy of the closed system, δ Q is heat transferred to a system and δ W is the work done by the system to the surroundings. The contribution to the total energy is due to internal (E), kinetic, potential, electromagnetic, surface tension or other form of energies. If change of all other ˆ form of energies can be neglected, then E = E . Heat transfer to a system is positive (system receives heat), whereas heat transfer from the system is negative (system loses heat). In contrast, work done by a system is positive (system loses work), and the work done to the system is negative (system receives work). The mechanical work for a closed system is usually expressed as δ W = pdV , where p is the pressure and V is the volume of the system – both are thermodynamic properties of the system. Change of a thermodynamic property depends on initial and final states only and does not depend on the path by which the change occurred. Therefore, thermodynamic properties are path-independent and its infinitesimal change is represented by exact differential d (such as dE or dV). The heat transfer, Q, and work, W, on the other hand, are path-dependent functions; infinitesimal heat transfer and work are represented by δ Q and δ W , respectively, in order to distinguish them from the change of a path-independent function. In arriving at eq. (2.1), it is assumed that the only work done is by volume change, and that potential and kinetic energies are negligible. The second law of thermodynamics for the single-component closed system can be described by the Clausius inequality, i.e., δQ dS ≥ (2.2) T where the equal sign designates a reversible process, which is defined as an ideal process that after taking place can be reversed without leaving any change to either system or surroundings. The greater-than sign denotes an irreversible process. dS is the change of entropy of the closed system. Combining these general forms of the first two laws of thermodynamics results in an expression that is very useful for determining the conditions for equilibrium and stability of systems, namely, the fundamental relation of thermodynamics: dE ≤ TdS − δ W (2.3) where the inequality is used for irreversible processes and the equality for reversible processes. For a finite change in a system, the fundamental thermodynamic relationship becomes ΔE ≤ T ΔS − W (2.4)
where W = ³ pdV is the work done by the system to the surroundings. The first
V1 V2
and second laws of thermodynamics for open systems will be discussed in Chapter 3.
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2.2.3 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: Π+ f =N +2 (2.5) 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.
2.3 Equilibrium and Stability of Single-Phase Systems
The equilibrium criteria for a single-phase closed system under the following three sets of constraints are discussed in this section: (1) constant-volume isolated system with no heat or work transfer between the system and its surroundings, (2) constant-volume, constant-temperature system, and (3) constant-pressure, constant-temperature system.
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2.3.1 Equilibrium Criteria for Pure Substances
Constant-Volume Isolated System
For a constant-volume isolated system that exchanges neither heat nor work with its surroundings, Q=0 (2.6) ΔV = 0 (2.7) (2.8) W =0 It follows from the first law of thermodynamics that a system that has no heat or work interaction with the surroundings also has no change in internal energy. Thus ΔE = 0 (2.9) Applying eqs. (2.6) – (2.9) with the Clausius inequality, eq. (2.4), gives ΔS E ,V ≥ 0 (2.10) Equation (2.10) asserts that system entropy always increases for a spontaneous and irreversible finite process occurring in a system with constant internal energy, E, and constant volume V. These spontaneous processes continuously move the system toward an equilibrium state where the entropy will reach a maximum value. When the system reaches an equilibrium state, any infinitesimal change in the system will result in a zero change of entropy, i.e., dS E ,V = 0 (2.11)
Constant-Temperature and Volume System
Since the temperature and volume of the closed system are constants, we have ΔT = 0 (2.12) ΔV = 0 (2.13) Assuming that the only work present in this closed system is of type pV, the work exchange between the system and the surroundings must be zero, i.e., W = 0 . The fundamental thermodynamic relationship, eq. (2.4), simplifies as (2.14) ΔE − T ΔS ≤ 0 Recalling the well known Helmholtz free energy function, F, F = E − TS (2.15) and expanding eq. (2.15) to define a finite change in the system yields ΔF = ΔE − T ΔS − S ΔT (2.16) Substituting eqs. (2.16) and (2.12) into eq. (2.14), a second equilibrium criterion is obtained: (2.17) ΔFT ,V ≤ 0 Therefore, for a closed system at constant temperature and volume, the Helmholtz free energy must decrease with any spontaneous system change and
112 Chapter 2 Thermodynamics of Multiphase Systems
be minimal at equilibrium. At equilibrium conditions any infinitesimal change from constant-temperature, constant-volume equilibrium must result in zero change in the Helmholtz free energy. dFT ,V = 0 (2.18)
Constant Temperature and Pressure System
For a closed system with constant temperature and constant pressure, ΔT = 0 (2.19) Δp = 0 (2.20) Once again, the goal is to determine the equilibrium criteria for such a system, with the assumption that the only work is of the pV type. The work exchange between the system and its surroundings is W = pΔV . The fundamental thermodynamic relationship, eq. (2.4), can be written as ΔE − T ΔS + pΔV ≤ 0 (2.21) To determine a useful equilibrium criterion for such a system, another common thermodynamic property, the Gibbs free energy, is recalled: G = E − TS + pV (2.22) Expanding eq. (2.22) for a system undergoing a finite system change results in ΔG = ΔE − T ΔS − S ΔT + pΔV + V Δp (2.23) Substituting eqs. (2.23) and (2.19) – (2.20) into eq. (2.21) results in the well known criterion of equilibrium at constant temperature and pressure: ΔGT , p ≤ 0 (2.24) Thus, for a closed system at constant temperature and pressure, the Gibbs free energy of the system must decrease with any spontaneous finite system change and will be at its minimum value at equilibrium. Finally, if a system of constant temperature and pressure is at equilibrium, any infinitesimal system change will result in zero change in the system’s Gibbs free energy. dGT , p ≤ 0 (2.25)
Summary of the Equilibrium Criteria
Other sets of inequality constraints exist for system equilibrium and can be found in a manner similar to the one detailed above, but have limited applications. In summary, the equilibrium constraints for a system undergoing a finite change as determined above are as follows: ΔS E ,V ≥ 0 (2.26)
ΔFT ,V ≤ 0
(2.27) (2.28)
113
ΔGT , p ≤ 0
Transport Phenomena in Multiphase Systems
ΔES ,V ≤ 0 ΔS H , p ≥ 0 ΔH S , p ≤ 0
(2.29) (2.30) (2.31)
where H = E + pV is the enthalpy of the system. During an infinitesimal change in the system, this change can be assumed to be reversible ( dS = 0 ). Therefore, the criteria of equilibrium for a system undergoing an infinitesimal change from equilibrium conditions become dS E ,V = 0 (2.32) dFT ,V = 0 dGT , p = 0
dU S ,V = 0
(2.33) (2.34) (2.35) (2.36) (2.37)
dS H , p = 0 dH S , p = 0
Example 2.1: Show that the equilibrium criterion for a system with constant entropy and volume is ΔES ,V ≤ 0 . Solution: For a system with constant entropy and volume, we have ΔS = 0 and ΔV = 0 . The work exchange between the system and its surrounding is W = 0 . The fundamental relationship, eq. (2.4), becomes ΔE ≤ T ΔS (2.38) Since entropy is constant, i.e., ΔS = 0 , it follows that ΔES ,V ≤ 0 (2.39)
2.3.2 Maxwell Relations
The fundamental thermodynamic relation for a reversible process in a singlecomponent system, where the only work term considered is pdV, is obtained from eq. (2.3), i.e., dE = TdS − pdV (2.40) which can also be rewritten in terms of enthalpy ( H = E + pV ), Helmholtz free energy ( F = E − TS ), and Gibbs free energy ( G = H − TS ) as dH = TdS + Vdp (2.41) dF = − SdT − pdV (2.42) dG = − SdT + Vdp (2.43) which all have the form of dz = Mdx + Ndy (2.44)
114 Chapter 2 Thermodynamics of Multiphase Systems
where
§ ∂z · M =¨ ¸ © ∂x ¹ y
(2.45)
§ ∂z · (2.46) N =¨ ¸ © ∂y ¹ x and dz is an exact differential, as thermodynamic properties like E, H, F, and G are path-independent functions. Since eq. (2.44) is the total differential of function z, M and N are related by § ∂M · § ∂N · ∂2 z (2.47) =¨ = ¨ ¸ ¸ © ∂y ¹ x © ∂x ¹ y ∂x∂y
Applying eq. (2.47) to eqs. (2.40) – (2.43), the following relationships are obtained: § ∂T · § ∂p · (2.48) ¨ ¸ = −¨ ¸ ∂V ¹ S © © ∂S ¹V
§ ∂T · § ∂V · ¨ ¸ =¨ ¸ © ∂p ¹ S © ∂S ¹ p § ∂S · § ∂p · ¨ ¸ =¨ ¸ © ∂V ¹T © ∂T ¹V
(2.49) (2.50)
§ ∂S · § ∂V · (2.51) ¨ ¸ = −¨ ¸ © ∂T ¹ p © ∂p ¹T which are referred to as Maxwell relations. The goal of Maxwell relations is to find equivalent partial derivatives containing p, T, and V that can be physically measured and therefore provide a means of determining the change of entropy, which cannot be measured directly.
2.3.3 Closed Systems with Compositional Change
The internal energy in eq. (2.40) is a function of only two independent variables, E = E ( S ,V ), when dealing with a single phase, single-component system. When a compositional change is possible, i.e., for multicomponent systems, internal energy must also be a function of the number of moles of each of the N components: E = E ( S ,V , n1 , n2 ,...nN ) (2.52) Expanding eq. (2.52) in terms of each independent variable, while holding all other properties constant, produces the following: N § ∂E · § ∂E · § ∂E · (2.53) dE = ¨ dni ¸ ¸ dS + ¨ ¸ dV + ¦ ¨ © ∂S ¹V ,ni © ∂V ¹ S , ni i =1 © ∂ni ¹ S ,V , n
j ≠i
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where j i. The first two terms on the right side of eq. (2.53) refer to conditions of constant composition, as represented by eq. (2.40). Comparing eqs. (2.53) and (2.40), the coefficients of the first two terms in eq. (2.53) are § ∂E · (2.54) ¨ ¸ =T © ∂S ¹V ,ni
§ ∂E · (2.55) ¨ ¸ = −p © ∂V ¹ S ,ni The third term on the right-hand side of eq. (2.53) corresponds to the effects of the presence of multiple components. The chemical potential can be defined as § ∂E · μi = ¨ (2.56) ¸ © ∂ni ¹ S ,V , n
j ≠i
Therefore, the above expanded fundamental equation for a multicomponent system, as seen in eq. (2.53), can be rewritten as
dE = TdS − pdV + ¦ μi dni
i =1
N
(2.57)
which is known as the internal energy representation of the fundamental thermodynamic equation of multi-component systems. Other representations can be directly obtained from eq. (2.57) by using the definitions of enthalpy ( H = E + pV ) , Helmholtz free energy ( F = E − TS ) and Gibbs free energy (G = E − TS + pV ) , i.e.,
dH = Vdp + TdS + ¦ μi dni
dF = − SdT − pdV + ¦ μi dni dG = Vdp − SdT + ¦ μi dni
i =1 i =1 N
N
(2.58) (2.59) (2.60)
i =1 N
It is therefore readily determined from eqs. (2.58) – (2.60) that other expressions of chemical equilibrium exist; these are § ∂H · § ∂F · § ∂G · =¨ =¨ μi = ¨ (2.61) ¸ ¸ ¸ © ∂ni ¹ p , S ,n © ∂ni ¹T ,V ,n © ∂ni ¹T , p ,n
j ≠i j ≠i j ≠i
In addition, the following expressions for the fundamental thermodynamic properties are valid: § ∂E · § ∂H · T =¨ (2.62) ¸ =¨ ¸ © ∂S ¹V ,ni © ∂S ¹ p , ni
§ ∂E · § ∂F · −p =¨ ¸ =¨ ¸ © ∂V ¹ S ,ni © ∂V ¹T , ni
(2.63)
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§ ∂H · § ∂G · V =¨ ¸ =¨ ¸ © ∂p ¹ S ,ni © ∂p ¹T , ni which will be very useful in stability analysis in the next subsection.
(2.64)
2.3.4 Stability Criteria
Equation (2.11) demonstrates that for a constant-volume isolated system, the equilibrium condition requires that the entropy of the system must be stationary, i.e., dS = 0. Equation (2.11) is also valid if the entropy of the system is at either maximum or minimum. To ensure that the system is at stable equilibrium, i.e., equilibrium in the system can be maintained after a small perturbation, it is necessary for the system to satisfy eq. (2.10) as well. All other equilibrium criteria listed in eqs. (2.33) – (2.37) are similarly inadequate to ensure stable equilibrium, and the inequalities stated in eqs. (2.27)– (2.31) are necessary in addition. We will now focus on the use of the energy minimum principle (in E, F, G, and H representation) to address the stability of a simple system in equilibrium. For a simple system to be in equilibrium, as analyzed in detail above, the system must be stable. If the system is not stable, it will spontaneously change state to become stable. Stability for a single-phase system can be broken down into three distinct types: (1) thermal, (2) mechanical, and (3) chemical.
Thermal Stability
A simple system (homogeneous, single-phase) confined by an adiabatic, rigid, and impermeable boundary is shown in Fig. 2.1. Under these constraints, the energy and volume of the system remain constant during any process. In its initial state, the system is divided by an adiabatic partition into two halves of equal volume, with the left side at a slightly higher temperature than the right. This temperature difference is accounted for in the representation of the internal energy by ( E + ΔE ) for the left half and ( E − ΔE ) for the right half. At some arbitrary time, heat is allowed to be exchanged between the two halves and sufficient time elapses for the system to reach equilibrium. Since no energy has been added to the system, the total energy in the initial state matches the total energy in the final state, i.e., (2.65) Ei = ( E + ΔE ) + ( E − ΔE ) = 2 E = E f If the entropies of the left and right halves in the final state are SL and SR, respectively, they must be identical, because the two halves possess the same internal energy and volume at the final state, i.e., S L = S R = S ( E ,V ) (2.66)
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(a) Initial state, i.
Figure 2.1 Isolated system illustrating thermal stability.
(b) Final state, f.
Since the volume of the left half is not changed between the initial and final states, the entropy of the left half in the initial state can be found by Taylor series expansion as follows: 1 § ∂2 S · 2 § ∂S · (2.67) S L , i = S ( E + ΔE , V ) = S + ¨ ( ΔE ) + ¨ 2 ¸ ( ΔE ) + ... ¸ 2 © ∂E ¹V © ∂E ¹V Similarly, the entropy in the right half at the initial state can be represented by
1 § ∂2S · 2 § ∂S · S R ,i = S ( E − ΔE ,V ) = S − ¨ (2.68) ( ΔE ) + ¨ 2 ¸ ( ΔE ) − ... ¸ 2 © ∂E ¹V © ∂E ¹V According to eq. (2.10), the final system entropy must be equal to or greater than the initial total system entropy. In other words, the entropy generated by this process must not be negative. S gen = ( S L + S R ) − ( S L ,i + S R ,i ) ≥ 0 (2.69) Combining eqs. (2.66) – (2.69) results in the following expression: § ∂2S · 2 (2.70) S gen = − ¨ 2 ¸ ( ΔE ) ≥ 0 © ∂E ¹V Since ( E)2 will always be finite positive, approaching zero as the two halves steadily approach uniform total internal energy, the second order partial derivative in eq. (2.70) must be negative and finite. § ∂2S · (2.71) ¨ 2 ¸ <0 © ∂E ¹V Equation (2.71) can be rewritten as follows with the use of eq. (2.54): ∂ §1· 1 § ∂T · (2.72) ¨ ¸ =− 2¨ ¸ <0 ∂E © T ¹V T © ∂E ¹V i.e.,
118 Chapter 2 Thermodynamics of Multiphase Systems
1 § ∂T · (2.73) ¨ ¸ >0 T 2 © ∂E ¹V so thermal stability requires that internal energy of the system must be an increasing function of temperature. Considering the definition of the specific heat at constant volume, 1 § ∂E · cv = ¨ (2.74) ¸ m © ∂T ¹V eq. (2.73) can be written as 1 (2.75) >0 2 T mcv Since both T2 and m in eq. (2.75) are positive, eq. (2.75) implies that the specific heat of the system at constant volume must always be positive in order for the system to move from its initial state – as defined above – to its final state, rather than in the opposite direction. cv > 0 (2.76) In other words, positive cv ensures that the system cannot spontaneously segregate itself into two thermally-dissimilar regions.
Mechanical Stability
The mechanical stability criteria can be determined by considering a simple system that initially has an internal pressure discontinuity. Fig. 2.2 shows a simple system with constant total volume and temperature. The constant temperature is maintained by contact with a thermal reservoir at a constant temperature, T. Initially, the system is divided into two parts of slightly-different volume by an off-center partition held in place by a locking mechanism. The left side is at a slightly higher pressure and lower volume. When the locking mechanism is disengaged, the partition will gradually float to the midpoint of the system, at which point the pressures and volumes of the two halves are equalized. In a manner similar to the analysis of thermal stability, the total volume of the system in the initial and final states can be represented as Vi = (V − ΔV ) + (V + ΔV ) = V + V = V f (2.77) Since the temperature and total volume of this system remain constant, the minimum Helmholtz free energy principle, eq. (2.17), can be applied to this problem. In the final state, the volumes and temperatures of the two halves are the same; therefore, the Helmholtz free energies of the left and right halves in the final state must be the same, i.e., FL = FR = F (V , T ) (2.78) where F is the final Helmholtz free energy of either half of the system. As was the case with entropy in the preceding thermal stability analysis, the Helmholtz
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(a) Initial state, i.
(b) Final states, f.
Figure 2.2 Constant temperature and constant volume system illustrating mechanical stability.
free energies of the initial two parts can be related to the final values by noting the constant-temperature volume changes experienced by each of the two parts: 1 § ∂2 F · 2 § ∂F · (2.79) FL ,i = F (V − ΔV , T ) = F − ¨ ΔV ) + ¨ 2 ¸ ( ΔV ) − ... ¸( 2 © ∂V ¹T © ∂V ¹T 1 § ∂2 F · 2 § ∂F · FR ,i = F (V + ΔV , T ) = F + ¨ ( ΔV ) + ¨ 2 ¸ ( ΔV ) + ... (2.80) ¸ 2 © ∂V ¹T © ∂V ¹T From the equilibrium analysis above, it can be concluded that the Helmholtz free energy of the final system must be less than or equal to the initial total system Helmholtz free energy: (2.81) ( FL + FR ) − ( FL,i + FR ,i ) ≤ 0
Combining eqs. (2.78) – (2.81) results in the following expression: § ∂2 F · 2 (2.82) ¨ 2 ¸ ( ΔV ) ≥ 0 © ∂V ¹T Since ( V)2 will always be finite positive, approaching zero as the Helmholtz free energies of the two portions steadily approach uniformity, the second order partial derivative of eq. (2.82) must be positive and finite: § ∂2 F · (2.83) ¨ 2 ¸ >0 © ∂V ¹T Equation (2.83) can be rewritten in the following form by considering eq. (2.63): 1 § ∂p · −¨ >0 (2.84) ¸= © ∂V ¹T κ T V where κ T is isothermal compressibility:
κT = − ¨
1 § ∂V · ¸ V © ∂p ¹T
(2.85)
120 Chapter 2 Thermodynamics of Multiphase Systems
From eq. (2.84), the criterion for mechanical stability for the system shown in Fig. 2.2 is found to be κT > 0 (2.86) Therefore, a simple system at equilibrium is mechanically stable if the isothermal compressibility factor is positive, i.e., the volume of the system shrinks with increasing pressure.
Chemical Stability
For a simple system in equilibrium to be stable, the system must also have chemical stability. In other words, certain conditions will prevent the system from spontaneously separating into two or more subsystems of varying chemical composition. As an aid to analyzing the criteria for chemical stability, the system shown in Fig. 2.3 is presented. The simple system depicted is in contact with a constant temperature reservoir and its boundary is impermeable to all species present. Two frictionless pistons ensure that the pressure of the system is always a constant value. Internally, the system in its initial state consists of a semipermeable membrane partition that prevents the movements of only species i between the two portions of the system. Initially, the left portion of the system contains more moles of species i than the right portion. At some arbitrary time, the membrane is made permeable to permit the flow of species i between the two compartments, and the system reaches an equilibrium condition with respect to matter flow. Conservation of the number of moles of species i during the process shown in Fig. 2.3 can be written as ni ,i = ( ni + Δni ) + ( ni − Δni ) = 2ni = ni , f (2.87) Since this system maintains constant temperature and pressure, the energy minimum principle that governs this system is the Gibbs free energy principle shown in eq. (2.24). Therefore, denoting the Gibbs free energy of the left and
(a) Initial state, i.
(b) Final state, f.
Figure 2.3 Constant temperature and constant pressure system with composition change illustrating chemical stability. Transport Phenomena in Multiphase Systems 121
right halves at the equilibrium final state as GL and GR respectively, the following can be stated: GL = GR = G (ni , T , p ) (2.88) where G is the final Gibbs free energy of either half of the system. As demonstrated above with entropy in the thermal stability analysis, the Gibbs free energies of the initial two parts can be related to the final values by noting the change of the Gibbs free energy with the change in the mole number i of the leftand right-hand sides when temperature, pressure, and all other mole numbers are held constant: § ∂G · 1 § ∂ 2G · 2 GL ,i = G (ni + Δni , T , p ) = G + ¨ Δni ) + ¨ 2 ¸ ( ( Δni ) + ... (2.89) ¸ 2 © ∂ni ¹T , p ,n © ∂ni ¹T , p ,n
j j
§ ∂G · 1§∂ G· 2 GR ,i = G (ni + Δni , T , p) = G − ¨ ( Δni ) + ¨ 2 ¸ ( Δni ) − ... (2.90) ¸ 2 © ∂ni ¹T , p ,n © ∂ni ¹T , p ,n j j
2
In the equilibrium analysis followed above, it was established that the Gibbs free energy of the final system must be less than or equal to the initial total system Gibbs free energy, i.e., (2.91) ( GL + GR ) − ( GL ,i + GR ,i ) ≤ 0 Combining eqs. (2.87) – (2.91) results in the following expression: § ∂ 2G · 2 (2.92) ( Δni ) ≥ 0 ¨ 2¸ ∂ni ¹T , p , n ©
j
Since ( ni)2 will always be finite positive, approaching zero as the Gibbs free energy of the two portions steadily approaches uniformity, the second-order partial derivative in eq. (2.92) must be positive and finite: § ∂ 2G · >0 (2.93) ¨ 2¸ © ∂ni ¹T , p ,n
j
Substituting eq. (2.61) into eq. (2.93) results in the chemical stability criterion for the simple system presented in Fig. 2.3: § ∂μi · >0 (2.94) ¨ ¸ © ∂ni ¹T , p ,n
j
Therefore, a simple system in equilibrium, such as the final state of the system shown in Fig. 2.3, is chemically stable if the chemical potential of the ith species increases with an increase in mole number of the ith species.
122 Chapter 2 Thermodynamics of Multiphase Systems
2.3.5 System with Chemical Reactions
Chemical Reaction and Combustion
The change in the chemical composition in a system discussed in the preceding subsection can result from a chemical reaction. During a chemical reaction, some of the chemical bonds binding the atoms into molecules are broken, and new ones are formed. Since the chemical energies associated with the chemical bonds in reactants and products are generally different, the resulting change in chemical energy and its effect on the overall energy balance of the system must be accounted for. For a single-phase reacting system, the change of energy can be due to the change of sensible internal energy associated with temperature and pressure change. It may also reflect change of chemical energy associated with chemical reactions. The reacting system starts with the mixture of reactants, and the chemical reaction in the system produces new components that will coexist in the mixture. Therefore, a system undergoing chemical reaction can be considered as a mixture that contains both reactants and products. In practical applications, one particular kind of chemical reaction, namely combustion, is particularly important. Combustion is a chemical reaction during which a fuel is oxidized and a large amount of thermal energy is released. Most fuels (such as coal, gasoline, and diesel fuel) consist of hydrogen and carbon, and are called hydrocarbon fuels. The oxidant gas for most combustion processes is air, which can be treated as a mixture of 21% oxygen and 79% nitrogen (each kmol of oxygen in air is accompanied by 0.79/0.21=3.76 kmol of nitrogen). For a given reaction, the chemical equation establishes the relationship between the mole numbers of the reactants consumed and the mole numbers of the products generated. For example, combustion of 1 kmol of methane with air that contains 2 kmol of oxygen can be represented by the following chemical equation: CH 4 +2O 2 +7.52N 2 → CO 2 +2H 2 O+7.52N 2 (2.95) where nitrogen is present on both sides of the equation and is a non-reacting species that is carried to the products. At high combustion temperature, a small amount of nitrogen N2 is oxidized to nitrogen oxides (NO, NO2). However, this amount is negligibly small as far as the overall combustion reaction is concerned. Combustion is complete if all of the carbon in the fuel burns to carbon dioxide and all of the hydrogen burns to water. The combustion in eq. (2.95) is a complete combustion. On the other hand, the combustion is incomplete if the product contains any unburned fuel or C, H2, or CO. Common causes of incomplete combustion include an insufficient supply of oxygen and inadequate mixing between fuel and oxygen. The first law of thermodynamics is a general law that applies to any process, including combustion. The first law of thermodynamics for a closed system is expressed as
Transport Phenomena in Multiphase Systems
123
δ Q = dE + pdV
or, for a finite process,
Q = ( E2 − E1 ) + ³ pdV
V1 V2
(2.96)
(2.97)
V
where Q is the heat transfer between the system and its surroundings, ³V12 pdV is the work done by the system on its surroundings, and E1 and E2 are total internal energies, including sensible and latent internal energy and chemical energy, before and after chemical reaction. Equation (2.97) is a general expression that is applicable to any combustion process. Combustion occurring under two specific conditions is of the greatest interest for practical applications: constant-volume and constant-pressure. For combustion at constant volume, the heat transfer is Q = E2 − E1 (2.98) When combustion occurs at constant pressure, eq. (2.97) becomes Q = ( E2 − E1 ) + p (V2 − V1 ) = H 2 − H1 (2.99) Equation (2.99) is also valid for combustion in an open system provided that H1 and H2 represent enthalpy at the inlet and exit, respectively, of the system.
Chemical Equilibrium
Before the chemical equilibrium theory was established, it was believed that all chemical reactions would proceed until all reactants were completely converted into products. In fact, chemical reactions proceed only until they reach an equilibrium state, referred to as chemical equilibrium. At the equilibrium state, the chemical reaction then proceeds incrementally in both directions, so that there is no net change in composition. Under these specific conditions, and if the chemical equilibrium is stable, the equilibrium will not change with time. Chemical equilibrium is another cause of incomplete combustion, and cannot be prevented. At chemical equilibrium, the combustion of methane can be represented by CH 4 +2O2 +7.52N 2 R CO 2 +2H 2 O+7.52N 2 (2.100) where the two arrows in eq. (2.100) indicate that the reaction takes place in both directions at any equilibrium state. Therefore, chemical reaction does not cease at chemical equilibrium, but the reaction rate is the same in both directions. Consequently, there is no notable change of composition for a reacting system at chemical equilibrium. For a system containing Nr reactants and Np products, the generalized chemical equation can be expressed as ar1 Ar1 + ar 2 Ar 2 + " + arNr ArNr R a p1 Ap1 + a p 2 Ap 2 + " + a pN p ApN p (2.101) where Ari (i = 1, 2" N r ) are chemical symbols for the reactants and Api (i = 1, 2" N p ) are chemical symbols for the products, with their corresponding
124 Chapter 2 Thermodynamics of Multiphase Systems
stoichiometric coefficients ar ,i and a p ,i . In the chemical reaction shown in eq. (2.100), the total numbers of reactants and products are N r = 3 and N p = 3 , respectively. The chemical symbols of reactants and products are Ar1 = CH 4 , Ar 2 = O 2 , Ar 3 = N 2 , Ap1 = CO 2 , Ap 2 = H 2 O , and Ap 3 = N 2 . The coefficients preceding the chemical symbols in eq. (2.101) are referred to as stoichiometric coefficients. These coefficients describe the proportion of the mole numbers of reactants disappearing and mole numbers of products appearing during the reaction process. The stoichiometric coefficients for reactants CH4, O2 and N2 are respectively 1, 2, and 7.52. Equation (2.101) can also be written as a more compact form (Bejan, 1997): 0R where
Nr + N p
¦
i =1
ai Ai
(2.102)
−ari ° ai = ® °a p (i − Nr ) ¯
and
i = 1, 2," N r i = N r + 1, N r + 2," N r + N p
(2.103)
i = 1, 2," N r Ari ° Ai = ® (2.104) ° Ap ( i − Nr ) i = N r + 1, N r + 2," N r + N p ¯ During a chemical reaction process, the decreasing mole number of reactants and increasing mole number of products must be proportional to the corresponding stoichiometric coefficients. If a chemically-reacting system initially contains Nr reactants and the mole number of ith reactant is ni0 , when the chemical reaction reaches chemical equilibrium, the mole number of ith reactants 0 becomes nri = nri − ariζ (i = 1, 2," , N r ) and the mole number of ith product is n pi = a piζ (i = 1, 2," , N p ). In the above notation, ζ indicates the degree of
advancement of the chemical reaction (i.e., ζ = 0 means no reaction and a very large ζ represents a large mole number of products). The maximum value of ζ max is reached when at least one of the reactants is exhausted. When the system is at chemical equilibrium with a degree of advancement ζ , and the mole number of each of the reactants and products is represented by ni (i = 1, 2," N r , N r + 1," N r + N p ) , a slight advancement of the chemical reaction will bring the system to a new equilibrium state represented by ζ + d ζ , in which state the new mole numbers of each of the components become ni + dni (i = 1, 2," N r , N r + 1," N r + N p ). The change of mole number of each component is then dni = ai d ζ
i = 1, 2," N r , N r + 1," N r + N p
(2.105)
The change in internal energy for the chemically-reacting system can be obtained by substituting eq. (2.105) into eq. (2.57), i.e.,
Transport Phenomena in Multiphase Systems 125
§ Nr + N p · dE = TdS − pdV + ¨ ¦ μi ai ¸ d ζ ¨ i =1 ¸ © ¹ Introducing De Donder’s affinity function (Bejan, 1997),
Y =−
Nr + N p
(2.106)
¦
i =1
μi ai
(2.107)
eq. (2.106) becomes
dE = TdS − pdV − Yd ζ (2.108) The affinity function is a linear combination of the chemical potentials of reactants and products; therefore, the affinity function itself is a property of the chemically reacting system. Equation (2.108) suggests that the internal energy of a chemically reactive system is a function of entropy, volume, and degree of affinity, i.e., E = E ( S ,V , ζ ) (2.109) Expanding eq. (2.109) in terms of each independent variable while holding all other properties constant produces the following: § ∂E · § ∂E · § ∂E · (2.110) dE = ¨ ¸ dζ ¸ dS + ¨ ¸ dV + ¨ © ∂S ¹V ,ζ © ∂V ¹ S ,ζ © ∂ζ ¹ S ,V
Comparing the third terms of eqs. (2.108) and (2.110) yields § ∂E · Y = −¨ ¸ © ∂ζ ¹ S ,V
(2.111)
Other representations of the fundamental relation for chemically-reactive systems can be directly obtained from eq. (2.108) by using the definitions of enthalpy, Helmholtz free energy, and Gibbs free energy, i.e., dH = Vdp + TdS − Yd ζ (2.112) dF = − SdT − pdV − Yd ζ (2.113) dG = Vdp − SdT − Yd ζ (2.114) It can be readily determined from eqs. (2.112) – (2.114) that other expressions of chemical equilibrium are § ∂H · § ∂F · § ∂G · (2.115) Y = −¨ ¸ = −¨ ¸ = −¨ ¸ © ∂ζ ¹ S , p © ∂ζ ¹T ,V © ∂ζ ¹T , p For a typical process wherein pressure and temperature are constant, the equilibrium condition requires that [see eq. (2.34)] dGT , p = 0 (2.116) Comparison of eqs. (2.116) and (2.114) reveals that for a chemical reaction occurring at constant pressure and temperature, the degree of affinity at equilibrium is zero.
Y =−
Nr + N p
¦
i =1
μi ai = 0
(2.117)
126 Chapter 2 Thermodynamics of Multiphase Systems
Gibbs free energy, G
Equilibrium (dG=0) dG<0 dG>0
possible direction
impossible
0
ζ
ζmax
Figure 2.4 Chemical equilibrium at constant temperature and pressure.
In order for the chemical equilibrium of the reacting system with constant pressure and temperature to be stable, it is necessary for the Gibbs free energy to satisfy eq. (2.28) as well, i.e., the Gibbs free energy must be at its minimum as shown in Fig. 2.4. Mathematically, the condition for stability can be expressed as § ∂ 2G · (2.118) ¨ 2 ¸ >0 © ∂ζ ¹T , p Substituting eq. (2.115) into eq. (2.118), the condition for stability becomes § ∂Y · (2.119) ¨ ¸ <0 © ∂ζ ¹T , p
Example 2.2 At chemical equilibrium, the combustion of methane can be represented by eq. (2.100). What is the relationship among the chemical potential of the reactants and products? Solution: The total number of reactants is N r = 3 and the total number of products is N p = 3 . The stoichiometric coefficients for the reactants
are
aCH4 = −1 ,
aO2 = −2
and
aN2 = −7.52 ,
respectively.
The
stoichiometric coefficients for the products are aCO2 = 1 , aH2O = 2 and
aN2 = 7.52 , respectively. At equilibrium, the degree of affinity must be
zero, i.e.,
Transport Phenomena in Multiphase Systems
127
Y = −¦ μi ai
i =1
3+ 3
= −( μCH4 aCH4 + μO2 aO2 + μ N2 aN2 + μCO2 aCO2 + μH2O aH2 O + μ N 2 aN 2 ) =μCH4 + 2μO2 + 7.52 μ N2 − μCO2 − 2 μH2 O − 7.52 μ N2 = 0 Thus the chemical potentials of the reactants and products at equilibrium satisfy μCH4 + 2 μO2 + 7.52μ N2 = μCO2 + 2 μH2O + 7.52μ N2
2.4 Thermodynamic Surfaces and Equations of State
2.4.1 Thermodynamic Surfaces of a Single-Component Substance
Thermodynamic surfaces are three-dimensional diagrams that describe every equilibrium point of a pure substance, including the vapor, liquid, and solid phases. Two types of thermodynamic surfaces – pressure-volume-temperature (pv-T) surfaces and temperature-entropy-pressure (T-s-p) surfaces – are discussed here. Figure 2.5 shows a typical stable equilibrium p-v-T surface for a pure substance that contracts upon freezing. The primary point of interest on this surface is the critical point, which describes the equilibrium point of the substance where the saturated-liquid and saturated-vapor states are identical. This point exists at the inflection point of the two-phase hump that delineates the phase change from liquid to vapor. The phase change from solid to liquid (melting) is also described by the narrow surface between the solid and liquid phases. One last interesting feature of the p-v-T surface is the surface that shows
Figure 2.5 p-v-T surface of a pure substance that contracts upon freezing.
128 Chapter 2 Thermodynamics of Multiphase Systems
Figure 2.6 T-s-p surface of a pure substance.
sublimation, solid-to-vapor phase change. Sublimation occurs if the pressure of the substance at a constant temperature is low enough. The solid-vapor surface exists below the narrow solid-liquid region and the liquid-vapor two-phase hump. All of these phenomena are more easily observed on the phase diagram that will be described in Section 2.4.2. It should be noted that the thermodynamic surface in Fig. 2.5 describes a pure substance that contracts upon freezing. For a substance that expands upon freezing, such as water-ice, the solid-liquid two-phase region would have a negative slope in the pressure-temperature plane, instead of the positive slope shown here. Very similar to the p-v-T surface is the T-s-p surface shown in Figure 2.6. As in the p-v-T surface, the critical point lies at the top of the liquid-vapor two-phase hump. However, the two-phase surfaces show how entropy of the substance changes during phase change, instead of how the volume changes, as in the p-v-T surface. This diagram also demonstrates that the entropy of the substance increases from the densest phase (solid) to the least dense phase (vapor).
2.4.2 p-T, p-v and T-s Phase Diagrams for a Pure Substance
Phase diagrams are created by taking a slice from a thermodynamic surface; the slice is perpendicular to one of its principal axes. Phase diagrams are usually more convenient to use than thermodynamic surfaces because they show the behavior of a substance more clearly in a two-dimensional diagram, i.e., one thermodynamic property from the thermodynamic surface is assumed constant. The three most common phase diagrams for pure substances are: (1) the pressuretemperature phase diagram from the p-v-T surface, (2) the pressure-volume phase
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129
(a) p-T diagram.
Figure 2.7 p-T and p-v phase diagrams.
(b) p-v diagram.
p=pcr Vapor
Temperature Liquid
Critical Point
T=Tcr Vapor
Solid -Liquid
Solid
Liquid-Vapor Triple line Solid-Vapor Entropy
Figure 2.8 T-s phase diagram.
diagram from the p-v-T surface, and (3) the temperature-entropy phase diagram from the T-s-p surface. Figure 2.7 shows the p-T and p-v phase diagrams for a pure substance that contracts upon freezing. The p-T diagram is relatively straightforward and shows the three phases separated by saturation curves. Along the saturation curves, the two adjacent phases are at equilibrium. The p-T diagram also shows the triple point where all three phases coexist at equilibrium. The critical point denotes the highest pressure and temperature at which the distinction between the liquid and vapor phase can be made. The p-v-T diagram for a pure substance that expands upon freezing is similar to that shown in Fig. 2.7(a), except that the solid-liquid saturation curve has a negative slope instead of a positive slope.
130 Chapter 2 Thermodynamics of Multiphase Systems
The more interesting phase diagram shown here is the p-v diagram. The liquid-vapor two-phase dome and solid-vapor two-phase regions are clearly shown in this diagram. Also shown are the critical point and the triple line, the latter of which coincides with the triple point in the p-T diagram. However, here, the range of volumes over which all three phases are present is shown. These are the two most common phase diagrams for a pure substance. Additional phase diagrams can be obtained by combination of two-principle axes in the thermodynamic surfaces discussed above. One example is a T-s phase diagram, shown in Fig. 2.8, which is used mostly in investigations of power or refrigeration cycles. Similar to the p-v diagram, the liquid-vapor two-phase region, solid-vapor two-phase region, critical point, and triple line are clearly shown in Fig. 2.8.
2.4.3 Equations of State for Pure Substances
Proceeding from the state postulate for simple, compressible, pure substances, any intensive property is solely a function of two other independent, intensive properties. In general, a functional relationship among any three properties could be called an equation of state. However, in common usage, the expression equation of state usually refers only to equilibrium relationships involving the pressure p, the temperature T, and the specific volume v having the functional form of f ( p , v, T ) = 0 (2.120) An equation of state serves two useful purposes. Its most obvious use is for predicting p-v-T behavior over a desired range of values. The equilibrium states of a simple compressible substance can be represented by a surface on p-v-T Cartesian coordinates. The other major use of p-v-T data is in the evaluation of thermodynamic property data that are not directly measurable; these include internal energy, enthalpy, entropy, Gibbs function, and Helmholtz function. The theoretical relationships for these properties contain first and second derivatives involving p, v, and T, so it is important that the mathematical format of an equation of state lend itself to the required differentiation and subsequent integration. If the vapor or gas can be approximated as an ideal gas, the equation of state is simply pV = nRuT or pv = Rg T (2.121) where Ru = 8.3143 kJ/kmol-K is the universal gas constant, n is mole number, and Rg = Ru / M (M is molecular mass in kg/kmol) is the particular gas constant. The equation of state for an ideal gas is only applicable to a situation where the pressure of the gas is very low. At higher pressures, the behavior of a gas or vapor deviates substantially from that of the ideal gas. In addition, the ideal gas law is also invalid in the two-phase region represented by the “hump” on a p-v diagram, where liquid and vapor phases coexist. As is shown in the p-v diagram,
Transport Phenomena in Multiphase Systems
131
the properties experience a discontinuous change when liquid-vapor phase change takes place. It was known that if the critical isotherm – which passed through the critical point on the p-v diagram – was followed, the change from vapor to liquid would be continuous. It is noted that continuous transition isotherms below the critical isotherm must exist in order to bridge the gap between ideal gas and incompressible liquid. This idea was realized in the theoretical van der Waals equation constructed by Johannes Diderik van der Waals in 1873. The theoretical equation of state developed by van der Waals is a cubic polynomial: Rg T a p= − (2.122) v − b v2 where the constant b represents the minimum volume occupied by the pure substance in the limit as p ∞, i.e., the volume occupied by the molecules of the substance. The a/v2 term is the additional pressure term that accounts for mutual attraction between the molecules and is proportional to the density squared. The van der Waals equation bridges the gap between ideal gas behavior and incompressible liquid behavior, which can be shown in two limiting cases. The first is the limit as the specific volume of the substance approaches infinity, i.e., the very dilute gas limit, where the van der Waals equation reduces to the ideal gas equation of state, i.e., pv = Rg T v→∞ (2.123) The second is the limit as the pressure of the system approaches infinity, and the volume of the system approaches the minimum volume that the molecules of the substance can occupy, i.e., v=b p→∞ (2.124) The van der Waals equation is a cubic polynomial that provides three roots for v at any given pressure. This can be seen in Fig. 2.9, which shows the shape of the van der Waals isotherms below the critical temperature of the substance. While an isotherm passing the two-phase hump on a p-v diagram makes a straight line, the van der Waals isotherm has minimum and maximum points below and above the straight line, respectively. It will be shown in the next section that this wavy line is inherently unstable, and therefore the van der Waals equation of state is not a good approximation in the two-phase region. However, it is a very helpful equation because on either side of the two-phase “hump,” it very accurately represents the ideal gas and incompressible liquid behavior of a given substance. It is necessary to know the constants a and b for different substances in order to use the van der Waals equation to evaluate the p-v-T relation. Because the critical isotherm passes through a point of inflection at the critical point, and the slope is zero at this point, the van der Waals equation can be differentiated with respect to v at constant temperature:
132 Chapter 2 Thermodynamics of Multiphase Systems
Figure 2.9 van der Waals polynomial showing three roots for v at a given p.
Rg T 2a § ∂p · + 3 =0 ¨ ¸ =− 2 ∂v ¹T © (v − b) v
(2.125)
2 Rg T § ∂2 p · 6a − =0 (2.126) ¨ 2¸ = ∂v ¹T ( v − b )3 v 4 © These two derivatives, along with the van der Waals equation at the critical conditions, Rg Tc a pc = −2 (2.127) vc − b vc can be solved to find the two constants and the critical specific volume, which are not as amenable to measurement as are the critical temperature and pressure. These values are as follows:
Table 2.1 Critical point properties for selected fluids Substance Air Ammonia Carbon dioxide Carbon monoxide Nitrogen Oxygen Water Propane R-12 R-134a Formula -NH3 CO2 CO N2 O2 H2O C3H8 CCl2F2 CF3CH2F Molecular Mass 28.97 17.031 44.01 28.01 28.013 31.999 18.015 44.094 120.914 102.03 Critical Temperature (K) 133.2 405.5 304.1 132.9 126.2 154.6 647.3 369.8 385.0 374.2 Critical Pressure (MPa) 3.77 11.35 7.38 3.5 3.39 5.04 22.12 4.25 4.14 4.06
Transport Phenomena in Multiphase Systems
133
vc = 3b a=
22 gc
(2.128)
27 R T (2.129) 64 pc Rg Tc b= (2.130) 8 pc The critical properties of selected substances are tabulated in Table 2.1.
Example 2.3 A 1-m3 vessel is filled with 80 kg of propane at a temperature of 120 °C. The gas constant for propane is 0.188 kJ/kg-K. Find the pressure of the propane by using the van der Waals equation. What will the pressure of the propane be if the ideal gas law is used? Solution: The critical pressure and temperature for propane can be found from Table 2.1; they are pc = 4.25 MPa and Tc = 369.8 K. The constants a and b can be determined using eqs. (2.129) and (2.130): 22 27 Rg Tc 27 (0.188 × 103 ) 2 369.82 a= = = 479.78 Pa-m 6 /kg 2 64 pc 64 4.25 × 106 Rg Tc 0.188 × 103 × 369.8 b= = = 2.04 × 10−3 m3 /kg 6 8 pc 8 × 4.25 × 10 The specific volume of the propane is 1 V v= = = 0.0125 m3 /kg m 80 The pressure of the propane can be found by using the van der Waals equation, eq. (2.122), i.e. Rg T a 0.188 × 103 × (120 + 273.15) 479.78 −= − = 4.00 MPa p= v − b v2 0.0125 − 0.00204 0.01252 If the ideal gas law is used, the pressure of propane will be Rg T 0.188 × 103 × (120 + 273.15) = = 5.91 MPa p= v 0.0125 It can be seen that the ideal gas law significantly overpredicts the pressure in this case.
The van der Waals equation is one of the most compact equations of state since it has only two empirical constants. Many other equations of state have been developed in an effort to improve on the accuracy of the van der Waals equation. The Redlich-Kwong equation (1949) is generally considered to be the best among the two-constant equations of state. The Redlich-Kwong equation is Rg T a p= − (2.131) v − b v(v + b)T 1/ 2
134 Chapter 2 Thermodynamics of Multiphase Systems
where the two constants a and b are obtained by applying the critical point conditions: (∂p / ∂v)T = 0 and (∂ 2 p / ∂v 2 )T = 0. a = 0.4275
2 Rg Tc2.5
(2.133) pc Also widely used is the Beattie-Bridgeman equation of state Rg T § c· A (2.134) p = 2 ¨ 1 − 3 ¸ (v + B ) − 2 v © vT ¹ v where § a· (2.135) A = A0 ¨ 1 − ¸ © v¹ § b· (2.136) B = B0 ¨ 1 − ¸ © v¹ where the values of the five constants (A0, a, B0, b, and c) for selected substances are listed in Table 2.2 [note that a and b in eqs. (2.135) and (2.136) are different from a and b in eq. (2.131)], and v is specific molar volume (m3/kmol). The Beattie-Bridgeman equation of state is valid for densities up to 0.8 ρc , where ρc is density at the critical point.
Table 2.2 Constants for Beattie-Bridgeman equation of state (Bejan, 1997) Gas Air Argon Carbon dioxide Helium Hydrogen Nitrogen Oxygen Rg (J/kg-K) 286.95 208.14 188.93 2078.18 4115.47 296.69 259.79 A0 (N-m4/kg2) 157.12 81.99 262.07 136.79 4904.92 173.54 147.56 a×103 (m3/kg) 0.66674 0.58290 1.62129 1.49581 -2.5053 0.93394 0.80097 B0×103 (m3/kg) 1.5919 0.98451 2.3811 3.5004 10.376 1.8011 1.4452 b×103 (m3/kg) -0.03801 0.0 1.6444 0 -1.582 -0.24660 -0.13148 c×10-3 (m3-K3/kg) 1.498 1.499 14.997 9.955 0.24942 1.498 1.498
b = 0.08664
pc Rg Tc
(2.132)
An improved equation of state that is valid for densities up to 2.5 c is the Benedict-Webb-Rubin equation of state: Rg T § C ·1 1 + ¨ B0 Rg T − A0 − 0 ¸ 2 + ( bRg T − a ) 3 (v + B ) p= 2 v T ¹v v © (2.137) γ aα c§ γ · − v2 + 6 + 3 2 ¨1 + 2 ¸ e v vT © v ¹ which has eight constants (A0, B0, C0, a, b, c, α, γ) that are listed in Table 2.3.
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135
Table 2.3 Constants for Benedict-Webb-Rubin equation of state (Bejan, 1997) A0×10-2 Formula (N-m4 /kg2) CH4 7.31195 C2H4 C2H6 C3H6 C3H8 C4H10 C4H8 C4H10 C5H12 C5H12 C6H14 4.3055 4.66269 3.50217 3.58575 3.07308 2.88571 3.02865 2.49391 2.37376 1.97242 1.77041 B0×103 (m3/kg) C0×10-7 (N-m4K2 /kg2) 2.65735 0.889635 1.69071 2.01509 2.51642 2.65194 2.55256 2.98871 2.98168 3.40357 4.13424 4.53487 4.79543 c×10-5 a α×109 γ×105 b×105 7 (N-m7K2 (N-m 6 2 9 3 (m /kg ) (m /kg ) (m6/kg2) /kg2) /kg3) 1.21466 1.31523 0.62577 30.1853 2.33469 1.19119 1.28892 1.05482 1.12224 1.00195 0.97316 0.97334 1.01546 1.10159 1.12913 1.04602 1.09451 1.23191 1.05806 1.15892 1.25806 1.10774 1.18582 1.28545 1.28545 1.47181 1.51575 0.97139 1.22361 1.39829 1.52759 1.47891 1.58056 1.6361 1.87887 2.22807 2.40013 2.49275 8.08173 8.9722 6.13014 7.09776 1.17469 1.30701 1.03453 1.13317
Gas Methane Ethylene Ethane Propylene Propane i-Butane i-Butylene n-Butane i-Pentane n-Pentane n-Hexane
1.98649 2.08914 2.02308 2.20855 2.36826 2.06958 2.14127 2.22006 2.17426 2.06498 1.98756
5.48279 1.00799 5.16963 0.941616 5.62184 1.00799 4.53682 0.890805 4.83038 0.913893 4.40244 4.33982 0.899353 0.897754
n-Heptane C7H16
A more general form of the equation state can be expressed in the following series form: Rg T a(T ) b(T ) c(T ) d (T ) (2.138) p= + 2 + 3 + 4 + 5 +" v v v v v where its accuracy depends on the number of terms used. The coefficients a(T ), b(T ), c(T ), d (T )" are functions of temperature only and can be measured experimentally or derived from statistical mechanics. It should be noted that as v → ∞ or p → 0 , eq. (2.138) is reduced to the equation of state for ideal gas. The equations of state for most substances are too complex to be expressed by simple equations like those presented above. The alternative is to present the thermodynamic properties in the form of tables. While some thermodynamic properties can be measured directly, the others are not directly measurable and must be calculated using their relationships with the measurable properties. The results of these measurements and calculations are usually presented in the form of thermodynamic properties tables. Appendix B presents thermophysical properties of gases, solids, and phase change materials (PCMs), as well as liquid and vapor properties at saturation. Also presented in Appendix B are temperature-property relationships of saturated liquid and vapor for various substances.
2.4.4 Phase Diagrams for Multicomponent Systems
We have looked at phase diagrams for pure substances, in which the p-T phase diagram distinctly shows the equilibrium lines between the various phases. For a multicomponent system, the unique relationship between saturation pressure and
136 Chapter 2 Thermodynamics of Multiphase Systems
temperature no longer exists, because phase change occurs over a range of temperature and pressure. The concentration of each component in the system will affect the range of temperature and pressure over which phase change occurs. For a binary system containing two components, one component – referred to as the solute – is dissolved into another component – referred to as the solvent. The saturation temperature, or melting point, is a function of both the pressure and the mass fraction of the solute. For practical use, the phase diagram is often presented as a temperature-concentration diagram under isobaric (constant pressure) conditions. Phase diagrams for both solid-liquid phase change and liquid-vapor phase change in a binary system are discussed below. Solid-liquid phase diagrams of binary alloys are extremely useful for material scientists and mechanical engineers. Phase diagrams for multicomponent substances differ considerably from single-component phase diagrams. A mixture of two metals is called a binary alloy and constitutes a twocomponent system, since each metallic element in an alloy is considered a separate component. The phase diagrams of binary alloy systems are usually presented with the temperature of the system as the ordinate and the chemical composition of the system as the abscissa. Binary alloys can be classified in two groups: (1) isomorphous alloys, and (2) eutectic alloys. The two components in an isomorphous alloy are completely soluble in each other in both liquid and solid states; therefore, the solid alloy can be characterized by a single type of crystal structure for different compositions of the components. In eutectic alloys, on the other hand, the solubility between the two components is limited. Consequently, the solid alloy can have different types of crystal structure depending on the chemical composition of the alloys. Figure 2.10 shows a phase diagram of an isomorphous alloy (copper-nickel) system, which is a slice of a thermodynamic surface taken at atmospheric pressure. The upper line in the diagram is the liquidus, above which lies a stable two-component liquid phase. The lower line in the diagram is the solidus, below which lies a stable solid phase. Between the liquidus and solidus exists a twophase region in which the alloy contains both liquid and solid phases of both components. However, the average composition of the solid is not the same as that of the liquid. This can be illustrated by focusing on a point where the composition of the alloy is 53 wt% Ni and 47 wt% Cu, and the temperature is 1300 °C. The alloy at this point contains both liquid and solid phases at 1300 °C but neither phase can have an average composition of 53 wt% Ni and 47 wt% Cu. The composition of the liquid phase corresponds to the composition at the point of intersection between the horizontal tie line and the liquidus, which is 45 wt% Ni and 55 wt% Cu. Similarly, the composition of the solid phase can be found by using the compositions at the point of intersection between the horizontal tie line and the solidus, which is 58 wt% Ni and 42 wt% Cu. It is necessary to point out that the diagram shown in Fig. 2.10 is obtained by slow cooling or heating of the alloy at equilibrium conditions. For rapidly-cooled or rapidly-heated alloys, the alloy system may experience nonequilibrium; in such cases the diagram of Fig. 2.10 is not applicable.
Transport Phenomena in Multiphase Systems 137
Figure 2.10 Phase diagram of copper-nickel isomorphous alloy at atmospheric pressure (Reproduced from Smith, W., 1995, Principles of Materials Science and Engineering, 3rd ed. with permission from McGraw-Hill Professional Book Group).
Figure 2.11 Phase diagram for eutectic binary solution (solid-liquid), aqueous ammonium chloride (NH4Cl-H2O) at constant pressure.
In a eutectic binary alloy, the two components have limited solid solubility in each other. Although some experimental results regarding solid-liquid phase change of binary metallic alloys appear in the literature, many researchers conduct experiments using transparent phase change materials (PCMs) such as
138 Chapter 2 Thermodynamics of Multiphase Systems
NH4Cl-H2O solution, because their solidification is quite similar to that of alloys and, moreover, it is easy to observe (Beckerman and Viskanta, 1988; Braga and Viskanta, 1990). The equilibrium phase diagram for aqueous ammonium chloride is shown in Fig. 2.11. At the eutectic point, the temperature and composition (NH4Cl-H2O mass fraction) are Te= –15.4 °C and e=19.7% respectively. The eutectic point is also the point of intersection of the two liquidus lines, above which the binary solution is in the liquid phase. When the temperature is below the eutectic temperature, or the temperature corresponding to the second solidus line, the solid phase is present. There are two mushy zones where solid and liquid coexist. Mushy zone 1 is for a subeutectic concentration of NH4Cl-H2O in water ( ω < ωe ) and is bounded by the solidus 1, the liquidus 1 and the eutectic line. The solid phase in the mushy zone 1 is pure ice. Mushy zone 2 is for the supereutectic concentration ( ω > ωe ) and is bounded by liquidus 2, solidus 2, and the eutectic line. The solid phase contained in the mushy zone 2 is solid NH4ClH2O. The phase diagram can be used, in conjunction with knowledge of the mixture concentration and temperature, to relate the phase concentrations to the mass fraction of the phase on the basis of local thermodynamic equilibrium.
Example 2.4: The liquidus 1 in the phase diagram, Fig. 2.11, can be approximated by a ω (T ) = 1.678 × 10−3 − 1.602 × 10−2 T − 2.857 × 10−4 T 2 −4.491 × 10−6 T 3 , where the unit of temperature is °C. The local temperature and concentration of NH4Cl at a point in the mushy zone formed by solidification are Tm= −10 °C and ω m=10%, respectively. What is the local solid fraction, f , i.e., mass fraction of the solid in the mushy zone? Solution: Since the local concentration of NH4Cl-H2O is less than the eutectic concentration, the mushy zone is in the subeutectic region, i.e., mushy zone 1, which contains pure ice and NH4Cl -H2O solution. The mass fraction of NH4Cl in the mushy zone 1 can be obtained by the concentration at the liquidus 1 corresponding to temperature Tm, i.e., 3 ω (Tm ) = 1.678 × 10−3 − 1.602 × 10−2 Tm − 2.857 × 10−4 Tm2 − 4.491 × 10−6 Tm
= 13.8% Since the solid phase in the mushy zone contains pure ice, the mass balance of NH4Cl requires that ωm = ω (Tm )(1 − f ) i.e., ω (Tm ) − ωm 13.8% − 10% f= = = 0.275 ω (Tm ) 13.8%
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av
Solute mass fraction ( 2)
Figure 2.12 Phase diagram for liquid-vapor phase change of binary system at constant pressure.
Our attention so far has been focused on the solid-liquid phase change of a binary solution. Liquid-vapor phase change is also very important, given its application in processes such as distillation or separation. Figure 2.12 shows a typical phase diagram of a miscible binary system (i.e., the two components in the binary system can dissolve into each other) undergoing liquid-vapor phase change. The system is in the vapor phase above the dew point line. If the temperature of the binary vapor falls below the dew point line, condensation of one component will take place. There is also a bubble point line, below which the system is in the liquid phase. If the temperature of the binary liquid is increased to a temperature above the bubble point line, vaporization of one component will take place and bubbles will be formed. The region between the dew point line and bubble point line is a two-phase mixture that includes both liquid and vapor phases. The characteristics of liquid-vapor phase change in a binary system can be demonstrated by analyzing evaporation and condensation of a binary system with an initial solute mass fraction of ω2 a . If the initial temperature of the binary system is below the bubble point, heating the binary liquid beyond the bubble point temperature (point aA ) will result in evaporation that produces vapor at point b. It should be noted that the temperature at b is the same as the bubble point temperature corresponding to its initial mass fraction. The mass fraction at b, however, is lower than that at point a. Since the vapor produced during the evaporation process contains less solute, the solute mass fraction in the remaining liquid will be increased. The temperature of the remaining liquid must be increased in order to continue the evaporation process. A similar analysis can be applied to condensation of the binary vapor with an initial mass fraction of ω2 a . Cooling of the binary vapor below the dew point
140 Chapter 2 Thermodynamics of Multiphase Systems
(point av ) will result in condensation that produces liquid at point c, the mass fraction of which is greater than that at point a. However, its temperature equals the dew point temperature corresponding to initial mass fraction ω 2a. The remaining vapor can condense only at a lower temperature because its solute mass fraction is lower.
2.5 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.
2.5.1 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., dS E ,V = 0 (2.139) For a closed system containing two phases, eq. (2.139) can be written as §2 · (2.140) ¨ ¦ dSk ¸ = 0 © k =1 ¹ E ,V where the subscript k identifies the individual phases. This generic subscript is chosen so that eq. (2.140) can represent any type of phase change combination, including liquid-vapor, solid-liquid, or solid-vapor. The fundamental relation, eq. (2.57), in terms of internal energy, where N i = 1 (single component) for both phases under consideration, is as follows: dEk = Tk dSk − pk dVk + μk dnk (k = 1, 2) (2.141) 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 = ¦ dEk = 0 dV = ¦ dVk = 0
k =1 k =1 2 2
(2.142) (2.143)
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 = ¦ dnk = 0
k =1
2
(2.144)
Combining eqs. (2.139) – (2.144), the following expression is obtained:
Transport Phenomena in Multiphase Systems 141
§1 1· §p p · §μ μ · dS = 0 = ¨ − ¸ dE1 + ¨ 1 − 2 ¸ dV1 − ¨ 1 − 2 ¸ dn1 (2.145) © T1 T2 ¹ © T1 T2 ¹ © T1 T2 ¹ It follows directly from eq. (2.145) that for equilibrium to exist between two phases of a single component, T1 = T2 (2.146) p1 = p2 (2.147) μ1 = μ2 (2.148) 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. (2.146) – (2.148) are derived by applying the fundamental relation, eq. (2.57), to a two-phase system with constant internal energy and volume, they are valid phase equilibrium conditions for any twophase systems.
2.5.2 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. (2.146) – (2.148). The two statements represented by eqs. (2.146) and (2.147) are the conditions for thermal and mechanical equilibrium. Equation (2.148) is automatically satisfied when eqs. (2.146) and (2.147) are satisfied, because the intensive chemical potential property, , 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 twodimensional phase diagram of a pure substance in Fig. 2.7(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 twophase 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: μ1 + d μ1 = μ2 + d μ2 (2.149) Substituting eq. (2.148) into eq. (2.149), one obtains d μ1 = d μ2 (2.150)
142 Chapter 2 Thermodynamics of Multiphase Systems
The fundamental relations, eq. (2.60), in terms of Gibbs free energy for both phases under consideration, are dG1 = V1dp − S1dT + μ1dn1 (2.151) dG2 = V2 dp − S 2 dT + μ2 dn2 (2.152) These equations can be considered in light of the following relationships between the extensive and intensive properties: V = nv , S = ns , G = ng (2.153) 3 where v , s , and g are specific molar volume (m /mol), specific molar entropy (J/mol-K), and specific molar Gibbs free energy (J/mol), respectively. Equations (2.151) and (2.152) can be rewritten as μ − g1 dg1 = v1dp − s1dT + 1 dn1 (2.154) n1 μ − g2 (2.155) dg 2 = v2 dp − s2 dT + 2 dn2 n2 For a single component system, eqs. (2.61) and (2.153) indicate that § ∂G · μ =¨ (2.156) ¸ =g © ∂n ¹T , p Thus, eqs. (2.154) and (2.155) can be rewritten as d μ1 = v1dp − s1dT (2.157) d μ2 = v2 dp − s2 dT (2.158) Substituting eqs. (2.157) and (2.158) into eq. (2.150) yields v1dp − s1dT = v2 dp − s2 dT (2.159) which can be rearranged as dp s1 − s2 (2.160) = dT v1 − v2 Equation (2.160) can be rewritten in term of specific entropy and specific volume per unit mass, i.e., dp s1 − s2 (2.161) = dT v1 − v2 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 g1 = g 2 or g1 = g 2 (2.162) i.e., h1 − Ts1 = h2 − Ts2 (2.163) Equation (2.163) can be rearranged to yield h −h s1 − s2 = 1 2 (2.164) T Substituting eq. (2.164) into eq. (2.160), one obtains
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143
h −h h dp = 1 2 = 21 dT ( v1 − v2 ) T v21T
(2.165)
where h21 is the latent heat of phase change and v21 is the change of specific volume during phase change. Equation (2.165), 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, twophase system at equilibrium. All of the properties in eq. (2.165) 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 superscripts 1 and 2. For a liquid-vapor system, the Clapeyron equation can be written as hAv dp (2.166) = dT (vv − vA )Tsat
Example 2.5: Water boils on the top of a mountain at 95 °C. Estimate the height of the mountain. Solution: At sea level, where the pressure is equal to 1 atm, the boiling point of the water is 100 °C. The latent heat of vaporization and the change of specific volume for water at 1 atm are hAv = 2257.03 kJ/kg
and vAv = 1.67185 m /kg . The Clapeyron equation (2.166) for a liquidvapor system can be approximated as h Δp = Av ΔT vAvTsat The height of the mountain, H, is related to the pressure difference by Δp = ρ air gH where the density of the air can be approximated as the density of the air at sea level at 25 °C, i.e., ρ air = 1.169 kg/m3 . Combining the above two equations and substituting the given values yields hAv ΔT 2257.03 × 103 × (100 − 95) H= = = 1573 m vAvTsat ρ air g 1.67185 × 373.15 × 1.169 × 9.8
3
For liquid-vapor equilibrium at low pressure, the specific volume of the liquid, vA , 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 = Rg T / p . In this case, the Clapeyron equation (2.166) reduces to
144 Chapter 2 Thermodynamics of Multiphase Systems
dp hAv p = dT Rg T 2
(2.167)
which is referred to as the Clausius-Clapeyron equation. 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. (2.167), i.e., h §1 1 · p (2.168) ln = − Av ¨ − ¸ p0 Rg © T T0 ¹ Rearranging eq. (2.168) yields the saturation pressure at temperature T: ª h § 1 1 ·º (2.169) p = p0 exp « − Av ¨ − ¸ » « Rg © T T0 ¹ » ¬ ¼ Equation (2.169) 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 pv = xv p (2.170) and the molar fraction of the saturated vapor in the mixture, x , is defined as §ρ ρ· ρ xv = v ¨ v + g ¸ (2.171) Mv ¨ Mv M g ¸ © ¹ where v and g 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 ª h § 1 1 ·º (2.172) xv = xv 0 exp « − Av ¨ − ¸ » « Rg © T T0 ¹ » ¬ ¼ where xv0 is the molar fraction of vapor at the reference temperature T0. Equation (2.172) can be rearranged using the ideal gas law: Rg T xv = ρv (2.173) p After substituting eq. (2.173) into eq. (2.171) and assuming the total pressure is constant, the concentration of vapor in the mixture becomes ª h § 1 1 ·º T (2.174) ρ v = ρ v 0 0 exp « − Av ¨ − ¸ » T « Rg © T T0 ¹ » ¬ ¼ where v0 is the density of vapor at the reference temperature T0.
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2.5.3 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. (2.61). For a simplified case of a two-component, two-phase system, eq. (2.61) can be written as follows: dGk = − Sk dT + Vk dp + μk , A dnk , A + μk , B dnk , B (k = 1, 2) (2.175) th where the subscript k denotes the k 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. (2.175). 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. dn2, A = −dn1, A (2.176)
dn2, B = − dn1, B
(2.177)
Since the system is assumed to be in equilibrium, dGT , p = 0 , and therefore
dG = ¦ dGk = 0
k =1
2
(2.178)
Equations (2.175) and (2.178) can be combined to create the following expression:
dG = ¦ μk , A dnk , A + ¦ μk , B dnk , B = dn1, A ( μ1, A − μ2, A ) + dn1, B ( μ1, B − μ2, B ) = 0
k =1 k =1 2 2
(2.179)
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, μ1, A = μ 2, A (2.180) (2.181) 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. (2.180) and (2.181) 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., μ1, A = μ2, A = μ3, A
μ1, B = μ2, B
μ1, B = μ2, B = μ3, B μ1,C = μ2,C = μ3,C
(2.182)
146 Chapter 2 Thermodynamics of Multiphase Systems
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. (2.5). 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 = NΠ + 2 (2.183) 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 ( Π − 1) . 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. (2.183) gives f = N Π + 2 − Π − N (Π − 1) = N + 2 − Π (2.184) which can be rearranged to give the Gibbs phase rule, eq. (2.5).
2.5.4 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 analogy examples in the mechanical equilibrium of a ball shown in Fig. 2.13. The ball in Fig. 2.13(a) is in stable equilibrium because it can always return to equilibrium after displacement. The ball in Fig. 2.13(b) is in metastable
(a) stable equilibrium. (b) metastable equilibrium.
Figure 2.13 Schematic of stability.
(c) unstable equilibrium.
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147
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. 2.13(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 2.4 satisfy these conditions. In thermodynamics, metastable regions play an important role in determining equilibrium states. Figure 2.14 shows a p-v diagram for a pure substance – an isothermal slice through a surface on the p-v-T 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.
Figure 2.14 p-v diagram for a pure substance illustrating metastable equilibrium.
148 Chapter 2 Thermodynamics of Multiphase Systems
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. (2.84) is satisfied. However, the states along the path 2′→3→4′ are completely unstable because when moving along this path, ( ∂p ∂v )T > 0 , which violates eq. (2.84). Therefore, the path 2′↔4′ is not accessible for boiling and condensation. The loci of the limiting points 2′ and 4′, where (∂p / ∂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. (2.122), is valid to describe this path, one can estimate the parameters on the spinodal by using Rg T 2a § ∂p · + 3 =0 (2.185) ¨ ¸ =− 2 (v − b ) v © ∂v ¹T The thermodynamic parameters on the spinodal can be determined by using eqs. (2.122) and (2.185).
Example 2.6 A 1-m3 rigid vessel is filled with 80 kg of propane. The vessel is cooled in order to condense the propane. Determine the temperature at which condensation will occur. What is the corresponding pressure? Solution: Since the vessel is rigid, the specific volume of the propane will remain constant in the cooling process. The specific volume of the propane is V 1 v= = = 0.0125 m3 /kg m 80 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. (2.185), i.e.,
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T=
2a (v − b ) 2 Rg v3
The constants a and b are a = 479.78 Pa-m6 /kg 2 and b = 2.04 × 10−3 m3 (see Example 2.3). Therefore, the temperature is 2a(v − b) 2 2 × 479.78 × (0.0125 − 0.00204) 2 = = 285.92 K = 12.77o C T= 3 3 3 Rg v 0.188 × 10 × 0.0125
The corresponding pressure of the propane can be found using the van der Waals equation, eq. (2.122), i.e., Rg T a 0.188 × 103 × 285.92 479.78 −= − = 2.07 MPa p= v − b v2 0.0125 − 0.00204 0.01252 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 range 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) will be discussed in detail in Chapters 8 and 10, respectively.
2.6 Thermodynamics at the Interfaces
2.6.1 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.
150 Chapter 2 Thermodynamics of Multiphase Systems
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 here. 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. 2.15). For a singlecomponent system, the fundamental thermodynamic relation represented by eq. (2.57) can be simplified as dE = TdS − pdV + μ dn (2.186) Suppose the configuration of the volume and surfaces 1, 2, and I are fixed. In this case, the internal energy is only a function of S and n, and dE = TdS + μ dn (2.187) 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 Etotal = EI + E1 + E2 = constant (2.188) The energy in the region between 2 and I, with the energy at 1 held constant, are then related by dEtotal = dEI + dE2 = 0 (2.189) i.e., 0 = TI dS I + μ I dnI + T2 dS2 + μ2 dn2 (2.190)
Figure 2.15 Interfacial region between two fluids. Transport Phenomena in Multiphase Systems 151
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 ntotal = nI + n2 = constant (2.191) i.e., dntotal = 0 = dnI + dn2 (2.192) dnI = −dn2 (2.193) The entropy balance for the system requires that Stotal = S I + S1 + S2 = constant (2.194) dStotal = 0 = dS I + dS 2 (2.195) dS I = − dS 2 (2.196) Substituting these into the energy-accounting eq. (2.190) gives the conditions for equilibrium of the process described above as: (2.197) (T2 − TI ) dS 2 + ( μ 2 − μ I ) dn2 = 0 Since eq. (2.197) must be valid for any dS2 and dn2, the following conditions must be satisfied: T2 = TI (2.198) μ2 = μI (2.199) The same procedure can be used to show that T1 = TI (2.200) μ1 = μ I (2.201) Thus, at equilibrium, the temperatures and chemical potentials for each of the three regions must be equal, i.e., T1 = TI = T2 (2.202) μ1 = μ I = μ 2 (2.203)
2.6.2 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. 2.15. 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 twophase system shown in Fig. 2.12 can be written as
152 Chapter 2 Thermodynamics of Multiphase Systems
ntotal = nI + n1 + n2 (2.204) The mole number of the interface is then nI = ntotal − n1 − n2 (2.205) Likewise, for internal energy, EI = Etotal − E1 − E2 (2.206) 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. (2.187), the internal energy of the surface is EI = EI ( S I , nI ) (2.207) 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 EI = EI ( S I , nI , A, K ) (2.208) where A and K are the surface area and the curvature of the interface, respectively. The change in E is then § ∂E · § ∂E · § ∂E · § ∂E · (2.209) dE = ¨ dS + ¨ dn + ¨ dA + ¨ dK ¸ ¸ ¸ ¸ ∂S ¹ n , A, K ∂n ¹ S , A, K ∂A ¹ S ,n , K © © © © ∂K ¹ S ,n , A where the subscript I is dropped for ease of notation. Considering eq. (2.187), eq. (2.209) can be rewritten in the following form: § ∂E · § ∂E · (2.210) dE = TdS + μ dn + ¨ dA + ¨ dK ¸ ¸ © ∂A ¹ S , n , K © ∂K ¹ S ,n , A If the variation of the curvature effect is negligible, eq. (2.210) simplifies to dE = TdS + μ dn + σ dA (2.211) where σ is the surface tension, defined as § ∂E · σ =¨ ¸ (2.212) © ∂A ¹ S ,n Another representation of the surface tension uses the Helmholtz free energy for the surface region, which is defined as FI = EI − TS I (2.213) Differentiating the above equation gives dFI = dEI − TdS I − S I dT (2.214) Substituting eq. (2.211) into eq. (2.214), one obtains dF = − SdT + μ dn + σ dA (2.215) If the differential of the Helmholtz function is as given above, then the function can be considered as F = F (T , n, A) (2.216)
Transport Phenomena in Multiphase Systems 153
which implies that
σ =¨ ¸ © ∂A ¹T ,n
§ ∂F ·
(2.217)
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 (2.218) work = σ dA 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. (2.218), 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 (2.218) 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 δ Wσ , eq. (2.218) becomes δ Wσ = −σ dA (2.219) According to eq. (2.219), 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 δ Wσ negative). The quantity δ Wσ 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 = δ Q − δ W (2.220) in which W is the work done by the system and Q is the thermal energy absorbed by the system. The quantity W is conveniently divided into a pressure-volume term and a non-pressure-volume term: δ W = δ W pV + δ Wnon − pV = pdV + δ Wnon − pV (2.221) 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. (2.219) may also be classified as non-pressure-volume work. The second law of thermodynamics indicates that for reversible processes, δ Qrev = TdS (2.222) Substituting eqs. (2.221) and (2.222) into eq. (2.220), with the stipulation of reversibility as required by eq. (2.222), one obtains dE = TdS − pdV − δ Wnon − pV (2.223) Recalling the definition of the Gibbs free energy G, G = H − TS = E + pV − TS Differentiating eq. (2.224) yields dG = dE + pdV + Vdp − TdS − SdT (2.224) (2.225)
154 Chapter 2 Thermodynamics of Multiphase Systems
Substituting eq. (2.223) into eq. (2.225) gives dG = Vdp − SdT − δ Wnon − pV
(2.226)
which shows that for a constant temperature, constant pressure, reversible process, dG = −δ Wnon − pV (2.227) 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. (2.219) that changes in surface area entail non-pressurevolume work. Therefore we can identify δ Wσ from eq. (2.219) with δ Wnon − pV in
dG = σ dA (2.228) Considering the stipulations made in going from eq. (2.225) to eq. (2.227), we obtain § ∂G · σ =¨ (2.229) ¸ © ∂A ¹T , p which identifies the surface tension as the increment in Gibbs free energy per unit increment in area. The path-dependent variable δ Wσ is replaced by a state variable as a result of this analysis. The three definitions of surface tension given by eqs. (2.212), (2.217) and (2.229) 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 the hydrogen bonding. The surface tensions for different liquids are quantitively demonstrated in Table 2.4.
Table 2.4 Surface tensions for different liquids at liquid-vapor interface Types of liquid Nonpolar liquid Hydrogen-bonded liquid (polar) Metallic liquid Liquid Helium Nitrogen Ammonia Water Mercury Silver Temperature (°C) -271 -153 -40 20 20 1100 Surface tension (mN/m) 0.26 0.20 35.4 72.9 484 878
eq. (2.227) and write
Transport Phenomena in Multiphase Systems
155
2.6.3 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. 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. 2.16). The vapor at temperature Tv and pressure pv is contained in a microscale spherical vapor bubble of radius Rb; liquid at a constant temperature TA and pressure pA surrounds the vapor bubble. At phase equilibrium, the temperatures of the two phases are the same, i.e., TA = Tv = T . According to eq. (2.148), the chemical potentials of the liquid and vapor phases are also the same at equilibrium, i.e., μA = μv = μ . 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. (2.59), in the liquid and vapor phases, one obtains dFA = − SA dT − pA dVA + μ dnA (2.230) dFv = − Sv dT − pv dVv + μ dnv (2.231) The fundamental relation for Helmholtz free energy at the interface, as indicated by eq. (2.215), is dFI = − S I dT + σ dA + μ I dnI (2.232) For a reversible phase change process under constant volume and temperature, eq. (2.33) must be satisfied, i.e., dF = dFA + dFv + dFI = 0 (2.233)
pA , TA
Vapor bubble
pv, Tv Rb
Liquid
Tv = TA = TI
Figure 2.16 Vapor bubble suspending in a liquid phase in a rigid vessel.
156 Chapter 2 Thermodynamics of Multiphase Systems
Substituting eqs. (2.230) – (2.232) into eq. (2.233), the equilibrium condition becomes −( SA + Sv ) dT − ( pA dVA + pv dVv ) + σ dA + μ (dnA + dnv + dnI ) = 0 (2.234) The total volume does not change because the system is rigid, i.e., dV = dVA + dVv = 0 (2.235) Conservation of mass requires that dn = dnA + dnv + dnI = 0 (2.236) Since the phase change occurs at constant temperature, we have dT = 0 (2.237) Substituting eqs. (2.235) – (2.237) into eq. (2.234) yields a relationship between the pressures in the liquid and vapor phases at equilibrium: dA (2.238) pv − pA = σ dVv Since the surface area and volume of the vapor bubbles are, respectively, 3 A = 4π Rb2 and Vv = 4π Rb / 3, eq. (2.238) becomes 2σ pv − pA = (2.239) Rb which is the Laplace-Young equation. Although eq. (2.239) 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 (see Chapter 5). At phase equilibrium, the chemical potentials of both phases must be equal. As indicated by eq. (2.156), 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 A ( pA , T ) = g v ( pv , T ) (2.240) By differentiating eq. (2.240) and considering the fact that the temperature is constant in the phase change process, one can obtain § ∂g v · § ∂g A · (2.241) ¸ dpv ¨ ¸ dpA = ¨ © ∂pA ¹T © ∂pv ¹T Considering eq. (2.64), eq. (2.241) can be rewritten as vA dpA = vv dpv (2.242) 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: 2σ dpv − dpA = − 2 dRb (2.243) Rb Substituting eq. (2.243) into eq. (2.242), the pressure in the vapor phase can be eliminated: § · 2σ vA ¨ dpv + 2 dRb ¸ = vv dpv (2.244) Rb © ¹
Transport Phenomena in Multiphase Systems 157
If the vapor behaves like an ideal gas ( vv = Rg T / pv ), eq. (2.244) can be rewritten as
dpv 2σ − vA dpv = 2 vA dRb (2.245) pv Rb where Rg is the gas constant of the vapor. If the radius of the vapor bubble goes to infinity ( 1/ Rb → 0 ), the vapor pressure equals the saturation pressure corresponding to the temperature, psat (T ) . Integrating eq. (2.245) from an equilibrium state for a flat surface, one obtains ª pv º 2σ (2.246) Rg T ln « vA » − vA [ pv − psat (T )] = − Rb ¬ psat (T ) ¼ Rg T
i.e.,
v [ p − psat (T ) − 2σ / Rb ] ½ ° ° (2.247) pv = psat (T )exp ® A v ¾ Rg T ° ° ¯ ¿ which indicates that the bubble is in equilibrium only if the pressure of the vapor phase exceeds the saturation pressure psat (T ). In another words, the vapor phase must be superheated. If the pressure inside the vapor bubble is below the pressure required by eq. (2.247), 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. (2.247). Equation (2.247) can be inverted to obtain the equilibrium bubble radius: 2σ Rb = (2.248) Rg T ln[ psat (T ) / pv ]/ vA + pv − psat (T )
Only vapor bubbles with radii equal to that given by eq. (2.248) will be in equilibrium with the surrounding superheated liquid at TA and pA . For most cases, pv − psat (T ) 2σ / Rb , eqs. (2.247) and (2.248) can be simplified as ½ 2σ / Rb vA ½ ° 2σ vA ° (2.249) pv = psat (T ) exp ® − ¾ = psat (T ) exp ®− ¾ pv vv ¿ ° Rb Rg T ¿ ° ¯ ¯ 2σ vA (2.250) Rb = Rg T ln[ psat (T ) / pv ] Since 2σ / Rb pv and vA vv , eq. (2.249) can be simplified as pv psat (T ) (2.251) The pressure in the liquid phase can be readily obtained by using the LaplaceYoung equation, eq. (2.239), i.e., 2σ pA = psat (T ) − (2.252) Rb 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
158 Chapter 2 Thermodynamics of Multiphase Systems
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. 2.14. Example 2.7 Nucleate boiling is characterized as generation, growth, and departure of vapor bubbles. A 0.5-mm-diameter vapor bubble is observed in superheated liquid water at a temperature of 102 °C. Find the pressures in the vapor bubble and in the liquid pool. Solution: According to eq. (2.251), the pressure in the vapor bubble is equal to the saturation pressure corresponding to the vapor temperature. Therefore,
pv = psat (T ) = 1.1102 × 105 Pa The surface tension of water at 102 °C is σ = 58.52 × 10−3 N/m. The
pressure in the liquid pool can be obtained by using eq. (2.252): 2σ 2 × 58.52 × 10−3 pA = psat (T ) − = 1.1102 × 105 − = 1.1079 × 105 Pa Rb 0.5 × 10−3 The ratio of the vapor pressure over the liquid pressure is pv 1.1102 × 105 = = 1.002 pA 1.1079 × 105 For a liquid droplet suspended in the vapor phase (Fig. 2.17), the pressure inside the liquid droplet is related to the vapor pressure by the Laplace-Young equation 2σ pA − pv = (2.253) Rd
Liquid droplet pA , TA
Rd
Vapor pv , Tv
Tv = TA = TI
Figure 2.17 Liquid droplet suspending in vapor phase.
Transport Phenomena in Multiphase Systems
159
which can be differentiated to yield 2σ dRd (2.254) 2 Rd At phase equilibrium, the specific Gibbs free energy functions must be equal for the liquid and vapor phases [see eq. (2.240)]. By following a procedure similar to that for the case of the vapor bubble, it can be shown that eq. (2.242) is also valid for the case of liquid droplet in vapor phase. Substituting eq. (2.254) into eq. (2.242), the pressure in the vapor phase can be eliminated: § · 2σ vA ¨ dpv − 2 dRd ¸ = vv dpv (2.255) Rd © ¹
dpA − dpv = −
Applying ideal gas laws to the vapor phase, eq. (2.255) can be rewritten as dp 2σ Rg T v − vA dpv = − 2 vA dRd (2.256) pv Rd If the radius of the liquid droplet equals infinity (1/ Rd → 0 ), the vapor pressure equals the saturation pressure corresponding to the temperature, psat (T ) . Integrating eq. (2.256)from an equilibrium state for a planar surface, the pressure in the vapor phase is obtained: v [ p − psat (T ) + 2σ / Rd ] ½ ° ° (2.257) pv = psat (T ) exp ® A v ¾ Rg T ° ° ¯ ¿ If the vapor pressure is below the pressure required by eq. (2.257), 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. (2.257). For most cases, pv − psat (T ) 2σ / Rd , eq. (2.257) can be simplified as
§ 2σ vA · (2.258) pv = psat (T ) exp ¨ ¨ Rd Rg T ¸ ¸ © ¹ The equilibrium droplet radius can be obtained by reverting eq. (2.258), i.e., 2σ vA (2.259) Rd = Rg T ln[ pv / psat (T )]
which is similar to eq. (2.250) 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 from the size of the vapor bubble or liquid droplet as indicated by eqs. (2.249) and (2.258).
Example 2.8 A liquid water droplet with radius, Rd, is suspended in steam at 1 atm. Quantitatively demonstrate the dependence of the degree of
160 Chapter 2 Thermodynamics of Multiphase Systems
supersaturation on the radius of the droplet and the number of molecules in each droplet.
Solution: At 1 atm, the properties of the water can be found from Table B. 48 as vA = 1/ 958.77 = 1.043 × 10−3 m3 /kg , σ = 58.91 × 10−3 N/m . The
gas constant of the water vapor is Rg = Ru / M = 8.3143 × 103 /18.0
461.9J/kg-K . The degree of supersaturation is therefore expressed as:
§ 2σ vA pv = exp ¨ ¨R R T psat ©dg · ¸ ¸ ¹
§ 2 × 58.91 × 10−3 × 1.043 × 10−3 · § 7.13 × 10−10 · = exp ¨ = exp ¨ ¸ ¸ Rd × 461.9 × 373 Rd © ¹ © ¹ The number of molecules in the droplet is estimated by m 1000 N A 4 3 N = 1000 N A = ρ A π Rd (2.260) M M 3 where N A = 6.022 × 1023 is the Avogadro’s number, which is the number of molecules per mole; m is the mass of the liquid droplet; and M = 18.0 kg/kmol is the molecular mass of the water. Thus
1000 × 6.022 × 1023 43 3 × 958.77 × π Rd = 1.34 × 1029 Rd 18.0 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 2.5.
N=
Table 2.5 Dependence of supersaturation pressure ratio of water at 1 atm on the size of the droplet Droplet radii (m) 1.0×10-9 1.0×10-8 1.0×10-7 1.0×10-6 1.0×10-5 1.0×10-4 Number of molecules per droplet 1.34×102 1.34×105 1.34×108 1.34×1011 1.34×1014 1.34×1017 Supersaturation pressure ratio (pv/psat) 2.040612 1.07393 1.007158 1.000714 1.000071 1.000007
The results in Table 2.5 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
Transport Phenomena in Multiphase Systems
161
phase will be significantly reduced. In reality, foreign nuclei often act as the seed for the liquid droplet.
2.6.4 Disjoining Definitions
Pressure:
Thermodynamic
and
Hydrodynamic
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. 2.18. The deeper Region A carries most of the liquid while Region B carries a thin film flow with
Region B
Region A vapor
Region B
pv ,δ pA
liquid solid
pv ,δ pA ,δ
Figure 2.18 Analysis of disjoining pressure in a cross-section of a micropassage containing thin liquid films for a stratified configuration.
162 Chapter 2 Thermodynamics of Multiphase Systems
a thickness . Most of the interface (Region A) is flat and separates the deeper liquid flow from the vapor at equilibrium pv ,δ = pA (2.261) where pA 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. 2.19 is considered. The interactions between fluid and metallic solid molecules are modeled using the Lennard-Jones interaction potential 6 C , ª Dm º (2.262) φ fs ( r ) = − φ6 fs «1 − 6 » r¬ r¼ where φ fs 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. C , ª D6 º (2.263) φ ff ( r ) = − φ 6ff «1 − 6f » r« r» ¬ ¼ where φ ff 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 free fluid molecule, the product of density and molecular potential is integrated. The mean-field potential energy, Φ fmf , felt by the free fluid molecule due to interactions with all the solid molecules, as seen in Fig. 2.19, is
Figure 2.19 Schematic used for derivation of disjoining pressure (Carey and Wemhoff, 2005).
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163
Φ fmf =
∞
zs = z x = 0
³ ³ N φ ( 2π x ) dxdz
s fs
∞
s
(2.264)
where N s is the solid wall molecular number density. Substituting eq. (2.262) and x2 + z2 for r2 reduces eq. (2.264) to 6 πN s Cφ , fs πN s Cφ , fs Dm Φ fmf = − + (2.265) 6z3 45 z 9 The above relation is reorganized in terms of a modified Hamaker constant AH AH = π 2N f N s Cφ , fs (2.266) where N f is the fluid molecular number density. Considering eq. (2.266), eq.
AH 3 6πN f Dm
(2.265) becomes
ª 2 § Dm ·9 § Dm ·3 º (2.267) «¨ ¸ −¨ ¸» «15 © z ¹ © z ¹ » ¬ ¼ 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 NfM dp nz = − f fs (2.268) dz NA where n z is the unit vector in the z-direction and f fs is the force per unit mass on Φ fmf =
the fluid system. The force exerted on a single molecule by the entire wall is given by ª 2 § Dm ·10 § Dm · 4 º d Φ fmf AH (2.269) F fs = − = −¨ ¸ » nz 4« dz 2πN f Dm « 5 ¨ z ¸ ¹ © z¹» ¬© ¼ The corresponding force per unit mass is ª 2 § Dm ·10 § Dm · 4 º N AF fs N A AH (2.270) = −¨ f fs = ¸ » nz 4« M 2π M N f Dm « 5 ¨ z ¸ © ¹ © z¹» ¬ ¼ where NA is Avogadro’s number. Substituting eq. (2.270) into eq. (2.268) and considering that the force only acts in the z-direction, 10 4 A ª2§ D · dp §D · º (2.271) = − H 4 « ¨ m ¸ −¨ m ¸ » 2π Dm « 5 © z ¹ dz © z¹» ¬ ¼ Integrating both sides of eq. (2.271) from z to , and considering the pressure at is pA , the pressure profile close to the wall is
A p ( z ) = pA + H 3 6π Dm
ª 2 § Dm ·9 § Dm ·3 º «¨ ¸ −¨ ¸» «15 © z ¹ © z ¹ » ¬ ¼
(2.272)
The z −9 term in eq. (2.272) may be neglected because Dm is on the order of a molecular diameter. Equation (2.272) simplifies to
164 Chapter 2 Thermodynamics of Multiphase Systems
AH (2.273) 6π z 3 The pressure in the thin liquid film at Region B in Fig. 2.18 is expected to vary with distance from the lower wall. At the interface ( z = δ ) , the pressure in p ( z ) = pA −
the liquid pA ,δ must be
AH (2.274) 6πδ 3 Combining eqs. (2.261) and (2.274) and solving for pv, – pA ,δ yields pA ,δ = p (δ ) = pA − AH (2.275) 6πδ 3 The disjoining pressure pd is the amount that pv, differs from pA ,δ pv ,δ − pA ,δ =
AH (2.276) 6πδ 3 The pressure difference across the interface in Fig. 2.18 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 μ = − sdT + vdp (2.277) where s is the molar entropy and 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, μA and μv respectively, without wall attraction effects: μv = μv , sat + RuT ln ( pv ,δ / psat ) (2.278) pd = −
where μA , sat at saturation, respectively. The chemical potential is equal to the specific Gibbs function for a pure liquid: μA = g A (2.280) 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 μA = μA , sat + vA ln ( pA − psat ) + N AΦ fmf (2.281)
μA = μA , sat + vA ln ( pA − psat ) (2.279) and μv , sat are the molar chemical potential of bulk liquid and vapor
Transport Phenomena in Multiphase Systems
165
where fmf is the potential energy per fluid molecule due to interactions with the wall, and vA is the liquid molar specific volume. For equilibrium conditions at the interface, the liquid pressure must be equal to the vapor pressure pA = pv ,δ (2.282) Setting pA equal to pv, and equating the right sides of eqs. (2.278) and (2.281) RuT ln ( pv ,δ psat ) = vA ( pv ,δ − psat ) + N A Φ fmf (2.283) where psat is the normal saturation pressure corresponding to vapor bulk of the system. Rearranging eq. (2.283), one obtains ªp v § p · N A Φ fmf º (2.284) pv ,δ / psat = exp « sat A ¨ v ,δ − 1¸ + » RuT » « RuT © psat ¹ ¬ ¼ Neglecting the first term in the brackets at the right-hand of eq. (2.284) gives § N AΦ fmf · (2.285) pv ,δ / psat = exp ¨ ¸ © k bT ¹
Substituting eq. (2.267) into (2.285), neglecting z-9 and setting z equal to at the interface, § · AH pv ,δ / psat = exp ¨ − (2.286) ¨ 6πρ f δ 3 kbT ¸ ¸ © ¹
Example 2.9 Show that the disjoining pressure obtained by thermodynamic analysis is consistent with eq. (2.276). Solution: The liquid and vapor molar chemical potentials, μv and μA , with attraction effects are μv = μv , sat + RuT ln ( pv ,δ / psat ) (2.287)
μA = μA , sat + vA ln ( pA ,δ − psat )
(2.288) (2.289)
The vapor and liquid pressures are related by pv ,δ − pA ,δ = pcap − pd
where pcap is the capillary pressure. If the liquid film is flat, the capillary pressure can be neglected and eq. (2.289) is simplified as pv ,δ − pA ,δ = − pd (2.290) Substituting eq. (2.290) into eq. (2.288) and equating the right sides of eqs. (2.287) and (2.288) yields RuT ln ( pv ,δ psat ) = vA ( pv ,δ + pd − psat ) (2.291) which can be rearranged to obtain ªp v § p · pvº pv ,δ / psat = exp « sat A ¨ v ,δ − 1¸ + d A » « RuT © psat ¹ RuT » ¬ ¼
(2.292)
166 Chapter 2 Thermodynamics of Multiphase Systems
Neglecting the first term in the brackets at the right-hand of eq. (2.292) gives §pv · pv ,δ / psat = exp ¨ d A ¸ (2.293) © RuT ¹ Comparing eqs. (2.286) and (2.293), we have A NA pd = − H 3 6πδ N f vA where (2.276). (2.294)
N A / N f vA = 1 , and eq. (2.294) will become identical to eq.
2.6.5 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. 2.14, 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 2.3.4, mechanical stability requires that ∂p ≤0 (2.295) ∂ν T Based on the above, the liquid or vapor in metastable domain is not mechanically unstable, even though it is not in thermodynamic equilibrium. In Fig. 2.14 between 2′ and 4′ , (∂p / ∂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 (Section 2.6.5.1) or kinetically using kinetic and nucleation theory (Section 2.6.5.2). The degree of superheat (Tt − Tsat ) 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 → ∞ , where R is the radius of curvature of the phase boundary. Initial bubbles in superheated liquids are in mechanical equilibrium. 2σ pv − pA = >0 (2.296) R
Transport Phenomena in Multiphase Systems 167
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 2.6.3. 2σ º ½ °v ª ° (2.297) pv = psat (T ) exp ® A « pv − psat (T ) − »¾ R ¼¿ ° Rg T ¬ ° ¯ The external liquid pressure, pA , 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 pv > pA and R < ∞ , T ( pA ) > Tsat ( pA ) .
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 (2.298) ΔF > 0 The thermodynamic limit of superheat is the limit of mechanical stability since it is already thermally stable, [see eq. (2.85)] ∂p ≤0 (2.299) ∂v T 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. (2.122)] a· § (2.300) ¨ p + 2 ¸ ( v − b ) = Rg T v¹ © 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 a 2ab p= 2 − 3 (2.301) v v If a pressure is given, then v may be eliminated from equations (2.300) and (2.301) to give the thermodynamic limit of superheat, T → Tt . This generally requires an iterative solution except when p → 0 , in which case Tt = 27Tc / 32 . A simple correlation for Tt is given by Lienhard (1976) to eliminate iteration needed, as
168 Chapter 2 Thermodynamics of Multiphase Systems
ª 27 5 § T ·5.16 º Tt Tc « + ¨ sat ¸ » (2.302) « 32 32 © Tc ¹ » ¬ ¼ Table 2.6 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 Tt ( pA ) > Tm ( pA ) (2.303) 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 PengRobinson (1976) equation of state a different thermodynamic limit of superheat is obtained. Rg T a (2.304) p= −2 ( v − b ) v + 2bv + b 2 where a and b are constants. The spinodal curve of eq. (2.304) is Rg T 2a ( v + b ) (2.305) − =0 2 2 2 v−b v + 2bv − b 2
(
)(
)
The superheat temperature derived by the above equation, Tt2, in Table 2.6 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 2.6 Thermodynamic Limit of Superheat of Some Pure Liquids at Atmosphere Pressure* (Avedisian 1986) Substance n-pentane n-heptane n-octane Methanol Ethanol water Tsat 309 372 399 338 352 373 Tt1 405 468 494 442 447 552 Tt2 431 499 525 477 482 596 Tm 426 494 514 466 472 575 Tc 470 540 569 513 516 647 J(Tt2) 8 × 1024 8 × 1026 2 × 1026 1029 1030 9 × 1028
*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.
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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 ª ΔΦ * º J = Γk f ( N *)N A exp « − (2.306) » ¬ kbTA ¼ where, k f ( N *) is the molecular condensation rate in a critical size nucleus (with N* molecules), N A is the total number density of molecules. In terms of TA
·º (2.307) ¸» ¸» ¹¼ 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 16πσ 3 ΔΦ* = (2.308) 2 3 psat (TA ) − pA ΔΦ * ª § Γk f ( N *)N A TA ≡ Tk = « ln ¨ kb « ¨ J ¬©
−1
(
)
Factor takes into account the detailed mechanism by which critical nuclei are formed within the molecular network of the liquid. To determine , the following issues must be solved. 1. The energy of a nucleus is a function of the number of molecules in it. * must be determined. 2. The exponential dependence of J on 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)
170 Chapter 2 Thermodynamics of Multiphase Systems
J=
pv / ( 2π mkbTA )
1/ 2
³ ( AN )
0 s
∞
−1
(2.309)
dN
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. (2.309) is the vaporization rate for the surface of the embryo (see Section 5.5.2). Debenedetti (1996) simplified the above relation in terms of easily measured properties ª º 1/ 2 σ3 § 2σ · « 16π » exp « − J = NA ¨ (2.310) 2» 3kbT δ 2 p π mB ¸ © ¹ « » sat (TA ) − pA ¬ ¼
(
)
pv vA kbTA 2 1 pA B= + 3 3 psat (TA )
δ =1−
(2.311) (2.312)
The effect of temperature is significant in rate of nucleation considering the exponential form of eq. (2.310). Avedisian (1986) showed the variation of J with change in temperature for superheated water at 1 atm in Table 2.7. 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. (2.310) 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).
Table 2.7 Limit of Superheat and Nucleation Rate of Water at Atmospheric Pressure* (Avedisian 1986) T 500 550 560 570 575 580 590 pv 25.8 59.1 68.3 78.5 83.9 89.6 101.6 psat 25.2 61.0 71.0 82.0 88.0 94.4 108.9 R × 1 07 25.2 6.76 5.2 3.9 3.4 2.9 2.1 J <10-99 <10-99 1.7 × 10−76 8.5 × 10−20 5.7 × 10−3 4.3 × 109 Waiting time/cm3 (~1/J) > 1091 years < 1091 years 1.2 × 1068 years 3.7 × 1011 years 1.8 × 102 sec 2.3 × 10−10 sec
4.3 × 1023 2.3 × 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).
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References
Avedisian, C.T., 1986, “Bubble Growth in Superheated Liquid Droplets,” Encyclopedia of Fluid Mechanics, Chapter 8, Gulf Publishing Company, Houston, TX. Beckerman, C., and Viskanta, R., 1988, “An Experimental Study of Solidification of Binary Mixture with Double Diffusion in the Liquid,” Proceedings of the 1988 National Heat Transfer Conference, Vol. 3, pp. 67-79. Bejan, A., 1997, Advanced Engineering Thermodynamics, 2nd ed., John Wiley & Sons, New York, NY. Blander, M and Katz, J.P., 1975, “Bubble Nucleation in Liquids,” AIChE Journal, Vol. 21, No. 5, pp. 833-848. Braga, S.L., and Viskanta, R., 1990, “Solidification of a Binary Solution on a Cold Isothermal Surface,” International Journal of Heat and Mass Transfer, Vol. 33, pp. 745-754. 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. Cengel, Y.A., and Boles, M.A., 2002, Thermodynamics – An Engineering Approach, 4th ed., McGraw-Hill, New York, NY. 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. 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.
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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. 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. Redlich, O., and Kwong, J.N.S., 1949, “On the Thermodynamics of Solutions,” Chemical Review, Vol. 44, pp. 233-244. Skripov, V.P., 1974, Metastable Liquids, New York, John Wiley and Sons. Smith, W., 1995, Principles of Materials Science and Engineering, 3rd ed., McGraw-Hill, New York, New York.
Problems
2.1. Show that the equilibrium criterion for a system with constant entropy and pressure is ΔH S , p ≤ 0 . 2.2. A simple system is confined by an adiabatic, rigid, and impermeable boundary. In its initial state, the system is divided by a diathermal partition (a partition that allows heat to penetrate) into two equal halves. The internal energies of the left and right halves satisfy 3 EL = nL RuTL 2 3 ER = nR RuTR 2
where the mole numbers of the left and right sides are nL=2 kmol and nR=3 kmol, respectively. The initial temperatures of the two halves are respectively TL=250K and TR=350K. After thermal equilibrium has been established, find (a) the values of EL and ER, and (b) the equilibrium temperature. 2.3. Two cylinders are fitted with two pistons coupled together so that the linear displacement of the two pistons must be the same. The ratio of the cross-sectional area of the two cylinders is AL/AR=0.5. A bar with high thermal conductivity connects the two cylinders so that the final temperatures of the two cylinders are the same. Determine the ratio of the pressures in the two cylinders, pL / pR , at equilibrium.
Transport Phenomena in Multiphase Systems
173
Figure P2.1
2.4. A closed system is initially divided into two halves. The pressures of the two halves are identically equal to p. The initial entropies of the left and right sides are both equal to S, and the total entropy of the system is 2S. After an infinitesimal change, the entropy of the left and right halves becomes S L = S + ΔS and S R = S − ΔS . The pressures of the two halves remain unchanged throughout the process. Using the enthalpy minimum principle, show that the system is stable if the specific heat at constant pressure is positive, c p > 0 . 2.5. Show that the specific heat at constant pressure is always greater than that at constant volume for any substance, i.e., c p > cv . 2.6. The following two equations are valid for an ideal gas: pV = nRuT E = cnRuT where Ru is the universal gas constant and c is a constant which depends on the molecular structure of the ideal gas. What is the entropy of the ideal gas? 2.7. A 1-m3 vessel is filled with propane at room temperature of 20 °C and pressure of 100 kPa. Find the mass of the propane by using (a) the ideal gas law, and (b) the van der Waals equation. 2.8. Reduced pressure, temperature, and specific volume can be defined as p T v pr = Tr = vr = pc Tc vc where pc, Tc, and vc are the pressure, temperature and specific volume at the critical point. Show that the van der Waals equation in terms of reduced pressure, temperature, and specific volume is 8Tr 3 pr = −2 3vr − 1 vr
174 Chapter 2 Thermodynamics of Multiphase Systems
2.9. For a fluid that does not satisfy the ideal gas law, a compressibility factor Z is introduced: Z = pv / RT ; this can also be written in terms of reduced pressure, temperature and specific volume, i.e., Z = Z c pr vr / Tr , where Zc is the compressibility factor at the critical point. Determine the compressibility of the fluid that satisfies the van der Waals equation of state. 2.10. For a gas that satisfies the van der Waals equation, show that its internal energy can be expressed as § 1 1· e = e0 + cv (T − T0 ) + a ¨ − ¸ © v0 v ¹ if the specific heat at constant volume, cv, is constant. The subscript 0 in the equation denotes a reference state. 2.11. Show that the two constants a and b in the Redlich-Kwong equation can be obtained by applying the critical point conditions: (∂p / ∂v)T = 0 and
(∂ 2 p / ∂v 2 )T = 0.
2.12. An NH4Cl-H2O solution is used as a phase change material (PCM) in a cold storage system designed to be operated at –10 °C. In order to maintain the mobility of the mushy PCM, the maximum allowable solid fraction is 0.5. What is the appropriate concentration of NH4Cl? 2.13. A system with constant temperature and pressure contains two phases of the same substance. Show that the two phases in the system are in equilibrium if eqs. (2.146) – (2.148) are satisfied. 2.14. The temperature of ice in an ice rink is –5 °C. An ice skater is standing on one foot and the contact area between the skate and ice is 280×2 mm2. The weight of the ice skater is 60 kg. What is the melting point of the ice underneath the skate? 2.15. The saturation temperature of water at 1 atm is 100 °C. Use the ClausiusClapeyron equation to find the saturation pressure if the temperature is increased to 110 °C and compare your result with that obtained by using a steam table. 2.16. The chemical potential of a single-component system is its Gibbs free energy, g [see eq. (2.156)]. According to eq. (2.148), the equilibrium chemical potential of the liquid and vapor at the liquid-vapor interface must be equal: g v = g A . Start from this relation and show that the latent heat of vaporization is hAv = T ( sv − sA ) where sv and sA are the entropies of saturated vapor and liquid, respectively. 2.17. A mixture of saturated liquid and vapor are in a piston-cylinder system as shown in Fig. P2.2. The piston is frictionless so the piston-cylinder system
Transport Phenomena in Multiphase Systems
175
is a system with constant pressure and temperature. If the masses of liquid and vapor are respectively mA and mv , the Gibbs free energy of the mixture is G = mA g A + mv g v . Show that the condition for the liquid and vapor phases at phase equilibrium is g A = g v . 2.18. A rigid tank filled with saturated nitrogen vapor at T = 100K is cooled to condense the vapor. The system is in a metastable state before condensation starts. Determine the temperature and the corresponding pressure at which condensation will occur.
Vapor mv
Liquid mA
Figure P2.2
Figure P2.3
2.19. A mixture of liquid and vapor water fills a chamber with the wall temperature maintained at a constant level, Tw (see Fig. P2.3). A valve near the bottom of the chamber is opened and 1 kg of water is drained from the chamber. Find an expression, in terms of the saturation properties of liquid and vapor, for the increase in the volume of the vapor phase. 2.20. Surface tension has been represented in terms of internal energy [eq. (2.212)], Helmholtz free energy [eq. (2.217)], and Gibbs free energy [eq. (2.229)]. Find the representation of the surface tension using enthalpy. 2.21. The degree of supersaturation was expressed in terms of the ratio of the vapor pressure and saturation pressure in Example 2.8. Express the degree of supersaturation in Example 2.8 in terms of temperature using the Clapeyron-Clausius equation.
176 Chapter 2 Thermodynamics of Multiphase Systems