ADVANCED THERMODYNAMICS ENGINEERING phần 7 ppt

Raoult’s Law for the Vapor Phase of a Real Gas If a liquid mixture exists at a high pressure and low temperature, its vapor phase must be treated as a real gas mixture, i.e., ˆfk1T,P,Xk,

In air XN

2/XO

2 = 3.76 Furthermore,(XO

2( ) l /XN2( ) l ) = (XO

2/XN

2)/(PNsat2(T)/POsat2(T)) (C)Since PNsat2(T) ≈ POsat2(T),

In case of liquid water,

For water, vl = 0.001053 m3 kmole–1, Psat = 1.5 bar, and for this case P = 2 bar, and T

= 384.56 K Therefore, the partial pressure of H2O in vapor phase,

pH O

2 = 1.500445 bar

a value close to saturation pressure at T= 384.6K since vf is small Further

A 20-liter rigid volume consists of 80% liquid and 20% vapor by mass at 111.4ºC and1.5 bar A pin is placed on piston to prevent its motion Gaseous nitrogen is isother-mally injected into the volume until the pressure reaches 2 bar What is the nitrogenmole fraction in the gas phase? Assume that N2 does not dissolve in the liquid

What happens if there is no pin during the injection of N2 Instead of adding N2,

dis-XH O

2 = 0.75022, and XN2 = 0.24798

The vapor mass

mv = pvVv/RT = pv(V – Vf)/RT, andthe liquid mass

mf = Vf/vf

Adding the two masses,

m = pv(V – Vf)/RT + Vf/vf.Therefore,

Vf = (m – pvV/RT)/(1/vf – pv/RT) ≈ vf(m – pvV/RT) (F)Since Pv after N2 injection is slightly higher than Pv before N2 injection, there should

be more vapor; thus the volume of liquid decreases According to Le Chatelier, thesystem counteracts the pressure increase by increasing the volume of the vapor phase

If we ignore the term (vf (P – Psat))/(RT), in Eq (A) this implies that pH2O = Psat and Xv

Similarly when we add salt in water( or an impurity), the Gibbs function of the liquid

H2O decreases which causes the vapor molecules to cross over from the vapor into theliquid phase

Figure 7: The variation of the mole fraction of heptane vapor with droplet temperaturefor a mixture containing 60 % n–heptane and 40 % hexadecane at 100 kPa

ergy increase must then cause the entropy of vapor to decrease which corresponds to

an increase in the partial pressure of vapor

Consider a component k of a liquid mixture that exists in equilibrium with a vaporphase that also contains a mixture of insoluble inert gases In this case,

a Immiscible Liquids and Miscible Gas Phase

This case is illustrated through the following example

Upon adding Eqs (C) and (D), we obtain the expression

Figure 8 shows the T- Xk(l)-Xk diagram

Figure 8 The lines BME and EJGA in that figure are called the dew lines for species

1 and 2, respectively The region above the curve BMEJGA is the superheated vapormixture region while that below the curve CELD is the compressed liquid region.Consider the following scenario A vapor mixture is contained in a pis-ton–cylinder–weight assembly, such that P = 1 bar, X2 = 0.6, and T = 100ºC (cf pointS) Species 2 exists in the form of superheated vapor, since p2 = 0.6 bar at T = 100ºC.The cylinder is now cooled The saturation temperatures Tsat

2 = 86.5ºC at p2 = 0.6 bar,and T1sat = 43.87ºC at p1 = 0.4 bar The assembly contains a vapor mixture only, aslong as T>86.5ºC As the vapor mixture is cooled, a liquid drop appears at T = 86.5ºCConsider binary vapor mixture of methanol (species 1) and water (species 2) that areassumed to be immiscible in the liquid phase Illustrate their behavior with respect to

(point G) (If the gas phase composition is changed to X2 = 0.2, in that case the firstliquid drop appears at 61ºC (point E)) If the mixture is cooled to 70ºC (cf point H),phase equilibrium – that is manifested in the form of Eq (C) – implies that vaporphase mole fraction must reduce to X2 = 0.3 (cf point J), i.e., more of species 2 mustcondense It also implies that X1 must increase to 0.7 from the initial mole fraction of0.4 Eqs (D) and (B) dictate that Tsat

1 = 52ºC, so that species 1 at 70ºC exists in theform of a superheated vapor Upon further cooling to 60ºC, phase equilibrium re-quires that X2 = 0.19 (cf point E), and Tsat

1 increases to 60ºC Any further coolingcauses both species 1 and 2 to condense, where the condensate phase is an immisciblebinary mixture Within the region EJGADE (i.e., for X2 > 0.19, 60ºC <T < 100ºC),liquid species 2 and vapor mixture must coexist In region BMEC (X2 < 0.19, 60ºC <

T <64.7ºC), liquid species 1 and vapor mixture must coexist At 60ºC, X2 = 0.19,and both the liquid and vapor mixture coexist

At point E there are two liquid phases and one vapor phase According to Gibbs phaserule, F = K + 2 – π = 2 + 2 – 3 = 1 Therefore, there is one independent variable in theset (P, T, X2) In case the pressure is fixed, then the temperature and X2 are fixed (i.e.,60ºC, and X2 = 0.19) for coexistence the three phases to coexist If the mixture iscooled from 100ºC (cf point K), species 1 condenses, increasing the mole fraction ofspecies 2 until X2 = 0.19

Now assume that the liquid mixture is heated at the condition X2(l) = 0.6 and P = 1 bar

in a piston–cylinder–weight assembly At low temperatures, the sum of the saturationpressures (cf Eq (E)) is insufficient to create the imposed 1 bar pressure Therefore,

at T < 60ºC (point Q), the fluid exists as a compressed liquid At ≈60ºC (cf point L),the sum of the saturation pressures is roughly 1 bar The temperature at this conditioncan be predicted using Eqs (A), (B), and (E) as 60ºC Consequently, the values of X1and X2 can be determined as 0.81 and 0.19 using Eqs (C) and (D) Thus first vaporbubble at a 1 bar pressure appears at 60ºC At this point there are three phases (two

Separate Liquid Species 1 and 2

D C

Figure 8: A T–Xl–X diagram for an immiscible liquid solution

immiscible liquid phases, since they are immiscible, and a vapor phase) As more heat

is isobarically added, the temperature cannot rise according to Eq (E), but the vaporbubble can grow If the heating process begins with 0.4 kmole of species 1 and 0.6kmole of species 2 and vaporization occurs until the vapor phase is at state E (i.e.,

T1sat = 60ºC), since the vapor phase mole fraction of species 1 is 0.81, the ratio of themoles of species 2 that are vaporized to those of species 1 is 0.19÷0.81 Therefore, forevery 0.4 kmole of species 1 that are vaporized, the moles of species 2 that are va-porized equal 0.4×0.19÷0.81 = 0.094 kmole Hence, the vapor mixture contains 0.4kmole of species 1, 0.094 kmole of species 2, and 0.6–0.094 = 0.506 kmole of species

2 remain in the liquid phase Now the species 1 from liquid have been completely porized Once T>60ºC, further vaporization of species 2 occurs, thereby increasingthe mole fraction of species 2 in the vapor state, and it is possible to determine thevalue of X2 along the curve EJGA As the temperature reaches 86.5ºC (cf point G),all of the initial 0.6 kmole of species 2 in the liquid phase vaporize so that X2= 0.6

va-b Miscible Liquids and Immiscible Solid Phase

Oftentimes two species 1 and 2 are miscible in the liquid phase, but are immiscible inthe solid phase and each species forms its own aggregate in the solid phase (i.e., upon cooling

of the liquid mixture, the two species form two separate solid phases) In this case, at phaseequilibrium,

ˆf1( l) = ˆf1(s), and ˆf2( l) = ˆf2(s) (17a)Under the ideal solution assumption and since X1(s) = 1 due to immiscibility

tempera-3 Partially Miscible Liquids

a Liquid and Gas Mixtures

Many liquids are miscible within a certain range of concentrations The solubility ofliquids with one another generally increases with the temperature The corresponding pres-sure–temperature relationships are a combination of the corresponding relationships for misci-ble and immiscible liquids Figure 9 illustrates the T-Xk(l) -Xk diagram for a partially miscibleliquid

In the context of Figure 9 assume that methanol and water are partially miscible LetX1(l) denote the methanol mole fraction and X2,l the water mole fraction in the liquid Suppose

that water (species 2) is soluble in methanol (species 1) up to 10% by mole fraction at 40ºC.

As the temperature is increased, the increased “ve” can overcome the attractive forces betweenmethanol molecules and hence its solubility increases as the temperature approaches 60ºC.Line FLC represents the boundary between miscibility and immiscibility When solubility re-mains constant, the line is vertically oriented (cf line VF) Water is insoluble from X2(l) = 0.1

to X2(l) = 0.8 say at temperatures less than 40ºC, but at 60ºC it is immiscible for values of X2(l)

< 0.7 Region I is a miscible liquid mixture but rich in species 2, while region II is a miscibleliquid mixture but rich in species 1 The boundary DQG represents the variation of miscibilitywith temperature in the region richer in X2,l The region above line CED is similar to the im-miscible case we have just discussed

Consider the vapor mixture at a 90% water vapor concentration (point K) As we coolthe vapor from state K to M, first a liquid drop appears containing both species that has a com-position corresponding to point R, while the vapor has a composition corresponding to point M

as discussed for miscible liquids As the temperature is decreased to point N, the last liquidwill have composition at N (at the bubble line) while the vapor is at state T If temperature isfurther decreased, a liquid mixture fixed at a composition N forms

If we start at point S, then we obtain the first drop at point T with drop compositioncorresponding to N, which is in the miscible region As we cool further to point U, the liquidcomposition is at D (miscible limit at 60ºC) while the vapor is at E However there is still wa-ter and methanol vapor left in the mixture Condensation will occur at a constant vapor com-position with the liquid-I composition at D (rich in species 2) and liquid-II composition C (rich

in species 1) If the temperature drops below 60ºC, there are two separate liquid phases I(composition rich in species 2 along DQG) and II (composition lean in species 2 along FLC).However, the fraction of species 2 in liquid–II will increase since the solubility of species 2increases (DQG) while the fraction of species 2 in liquid–II will decrease (FLC)

b Liquid and Solid Mixtures

When a solid (a solute, such as salt) is dissolved in a liquid (a solvent, e.g., water), thedissolved solid can be considered as a liquid in the liquid solution It is pertinent to know themaximum amount of solute that can be dissolved in a solvent We will denote the salt in solidphase as s(s) and that in the liquid as s(l) At the equilibrium state of a saturated liquid solution

with a solid salt,

ˆfs(l) = Xsl fs(l)(T,P) = fs(s)(T,P), where (21)

fs(s)(T,P) = fs(s)(T,Psub) POYs(s), and (22)

The Psub denotes the saturation pressure for the sublimation of a salt at a specified temperature.Since fs(s)(T,P) = fs(g)(T,Psub),

Employing Eqs (21) and (26)

Xs(l) φs( l)(T,P) P = Psub(T) POYs(s), and (27)

Xs(l) = (Psub(T) POYs(s))/(P φs( l)) (28)

At low pressures, an increase in the pressure causes the solubility to decrease while at higherpressures the value of φs(l) may decrease and, consequently, the solubility may also increase

D DISSOLVED GASES IN LIQUIDS

Gases dissolve in liquid solutions through a process called absorption (This shouldnot be confused with adsorption, which is a process during which molecules are attached to athe surface of a solid material due to strong intermolecular forces.) The solubility of a compo-nent in a mixture is expressed as a ratio of the maximum amount of solute that can be present

in a specified amount of solvent In case of gases, the solubility is typically expressed in units

of ppm We will treat dissolved gaseous species within a liquid as though they behave likeliquids

→

60 C

40 C

Vapor Liq

S T

U

N

M K

regions (From Smith and Van Ness, Introduction to Chemical Engineering

Thermody-namics, 4th Edition, McGraw Hill Book Company, 1987, p 455 With permission.)

1 Single Component Gas

As carbon dioxide is dissolved in water, at some concentration a vapor bubble taining pure CO2 will start to form At that saturated condition

imply-The relation shown in Eq (31) presumes Raoult’s law or ideal solution behavior Atlow carbon dioxide mole fractions, since a relatively large number of water molecules sur-round the molecules of carbon dioxide, the liquid water molecules dominate the intermolecularattraction forces If the attractive forces between water molecules significantly differ fromthose between the CO2 molecules, the ideal solution model breaks down The ideal solutionmodel is also not applicable at higher pressures, since the dioxide no longer behaves as anideal gas

2 ( l) POYCO

2 ( l) PCO2 = XCO

Deaera-Another example pertains to diving in deep water The human body contains air ties (e.g., the sinuses and lungs) As a diver proceeds to greater depths, the surrounding pres-sure increases In order to prevent the air cavities from collapsing at greater depths, the diversmust adjust the air pressure they breathe in They do so by manipulating their diving equip-ment to equalize the cavity pressures with the surrounding water pressure Consequently, thepressurized air gets dissolved in the blood (Eq (31)) Upon rapid depressurization, in theprocess of reaching phase equilibrium, the dissolved air is released into the blood stream in theform of bubbles that can be very harmful to human health Raoult’s Law may be applied toestimate the concentration of air in blood Similarly when a person develops high blood pres-sure, the amount of soluble O2 and CO2 may increase

cavi-If we assume blood to have the same properties as water, we can determine the bility of oxygen at a 310 K temperature and 1 atm pressure as follows The vapor pressure data

solu-of oxygen can be extrapolated from a known or reference condition to 310 K using sius–Clayperon equation (which is valid if (hfg/Zfg) is constant), namely,

Clau-(Pk

sat

/Pref) = exp ((hfg,k/(RkZfg,k))(1/Tref – 1/T)) (34)The saturation pressure at 310 K can be determined using the relation ln (Psat) = 9.102 – 821/T(K) bar, i.e., Psat(310 K) = 635 bar In air, at 1 atm pO

2 = 0.21 bar, and the resulting solubility

of O2 in water is 300 ppm

Another example pertains to hydrocarbon liquid fuels (e.g., fuel injected engines)that are injected into a combustion chamber at high pressures (≈ 30 bar) The gaseous carbondioxide concentration in these chambers is of the order of 10% At 25ºC, the solubility of thedioxide in the fuels is ≈0.1×3 MPa÷61MPa = 0.005 This solubility increases as the pressure isincreased

3 Approximate Solution–Henry’s Law

Rewrite Eq (33) as,

Where “k is the solute dissolved in a liquid solvent The symbol Hk denotes Henry’s constantfor the k–th gaseous species dissolved in the liquid solution The units used for Hk are typicallythose of pressure Since vf has a relatively small value, POY ≈ 1 Therefore,

Hence Eq (35) is written as,

pres-E DEVIATIONS FROM RAOULT’S LAW

Consider two species k and j that form a binary mixture The attraction force betweensimilar molecules of species k is denoted as Fkk and between dissimilar molecules as Fkj Thefollowing scenarios ensue: (1) Fkj = Fkk so that the ideal solution model and Raoult’s Law ap-ply, e.g., toluene–benzene mixtures and mixtures of adjacent homologous series; (2) Fkj > Fkkimplying a nonideal solution in which contraction occurs upon mixing, e.g., acetone–watermixtures and other examples of hydrogen bonding; and (3) Fkj < Fkk, which corresponds to anonideal solution in which the volume expands upon mixing, e.g., ethanol–hexane and otherpolar–non polar liquids In case of the second scenario, since the intermolecular attractionforces are stronger between k–j pairs than between k–k molecular pairs, the vapor pressure ofspecies k can be lower than that predicted using Raoult’s Law, which is referred to as a nega-tive deviation from the Law In case (3) the attraction forces are lower, and a larger amount ofvapor may be produced as compared with the Raoult’s Law prediction, i.e., both the secondand third scenarios suggest that we must involve activity coefficients, γk(l) It will now beshown that

where γk(l) = ˆf k(l)(T,P)/ˆfk(l)id(T,P) = ˆf k(l)(T,P)/(Xk(l) fk(l)(T,P))

1 Evaluation of the Activity Coefficient

We have previously employed the ideal solution model to predict the vapor pressure

of a component k in an ideal solution If the measured component vapor pressure differs fromthat prediction, then it is apparent that the ideal solution model is not valid We can determinethe activity coefficient (that represents the degree of non-ideality from the measured vaporpressure data) as follows

γk = ˆfk/ˆfk,id = ˆfk/Xk fk(T,P). (44)

At phase equilibrium,

ˆfk(g)= γk(g) Xk fk(g)(T,P) = ˆfk( l) and (45)

γk(g) Xk fk(g)(T,P)= ˆfk( l) = γk (l) Xk,lfk(l) (T,P) (46)Since the vapor is assume to be an ideal gas mixture γk(g) =1, and fk(g)(T,P) = P Therefore,

fk,l(T,P) = fk(l)(T,Psat) POY ≈ fk(l)(T,Psat) = Psat, i.e.,

pk = γk(l) Xk,l fk(l)(T,P) = γk(l) Xk,l Psat = γk(l) pk,Raoult (48)Thus, γk(l) is a measure of the deviation from Raoult’s law With respect to the measured vaporpressure at a specified value of Xk, namely,

γk(l) = pk/(Xk,lpsat), or γk,l = pk/pk,Raoult (49)Figure 10: Henry’s constant H(T) for the solubility of gases in water (From S S Zum-

dahl, Chemistry, DC Heath and Company, Lexington, Mass, 1986 With permission.)

Recall that the γk’s for any phase are related to (g /(RT)) (see Eqs (114) to (117), Chapter 8).If,

X1(l) X2(l) RT/gE = B´ + C´ (X1(l) - X2(l)), (50)then we obtain the Van Laar Equations:

At the azeotropic condition, at which Xk(l) = Xk, Eq (47 ) assumes the form

T and P Relations for dew point, bubble point, vapor/liquid composition, solubilities, etc., areobtained The methodology is extended to nonideal solutions

Eqs (B) represent two equations µα =µβ and µβ = µγ and, since there are two unknowns P and

T, the solutions are uniquely fixed so that F = 0

2 General Phase Rule for Multicomponent Fluids

From the phase equilibrium condition for any species k

µk( )1(X1 1( ),K,XK−1 1( ), , )P T =µk( )2 (X1 2( ),K,XK−1 2( ), , )P T =…=µK( )π (X1( )π ,K,XK−1( )π , , )P T , (D)which represents a set of (π–1) equations Overall, there are (π–1)K equations available thatrelate the composition variables (X , X11 , k−1 1( )), ( X1 2( ), , XK-1 2 ( )), X, ( 1 ( ) π, , XK-1 ( ) π) and thetwo intensive variables P and T The total number of variables then equals (π(K–1)+2) Sincethese variables are related by (π–1)K equations, the number of degrees of freedom is

which is known as Gibbs phase rule The phase rule is usually applied to a system of K ponents and π phases at specified values of P and T

If the system is reacting according to the chemical reaction

H2SO4(l) + Ca(s) → H2 (g) + CaSO4(s),the chemical equilibrium condition requires that

ˆgH SO

2 4 + ˆgCa(s) = ˆgH

2 + ˆgCaSO s

4 ( ).This restriction reduces the number of freedom by 1, and

3 Raoult’s Law for the Vapor Phase of a Real Gas

If a liquid mixture exists at a high pressure and low temperature, its vapor phase must

be treated as a real gas mixture, i.e.,

ˆfk1(T,P,Xk,l) = ˆfkg(T,P,Xk)

Applying an ideal solution and ideal mixture model,

Consider a binary mixture of species 1 and 2 in the liquid phase, which on coolingforms two separate pure solids, one for each component These solids are in equilib-

Consider a nonreacting system containing H2SO4(l), Ca(s), and H2(g) in two phases,

A single liquid phase is desired for a mixture of water and alcohol Determine the How many independent intensive variables are required to fix the state of superheated

de-Xk,lˆfk1(T,P) = Xkˆfkg(T,P).

A species may exist as a gas at a specified temperature and pressure when alone (e.g., water at110ºC, 100 kPa), but as a liquid in a liquid mixture with another higher boiling temperaturecomponent In that case,

ln (fk(T,P)/fk(T,Psat)) =

P P

sta-The various equilibrium states of mechanical systems are illustrated in Figure 1.States B, C, D, and F represent mechanical equilibrium positions (or states) while A is a non-equilibrium position (and, hence, not a state) The equilibrium states B, C, and D can be classi-fied according to their stability behavior by conducting perturbation tests as follows An equi-librium state is disturbed from its initial state by a small amplitude perturbation that changesthe potential energy If a ball returns to its original state, that state is stable The mechanicalstability behavior can be characterized by several states Figure 1 illustrates a stable state (e.g.,state D corresponding to a minimum potential energy), an unstable state (e.g., state C that has alocally maximum potential energy), a metastable state (e.g., state B that can be perturbed topotential energy levels, but which have finite constraints that require a relatively large distur-bance to overcome), and a neutrally stable state (e.g., state F that has an invariant potentialenergy) A disturbance at state C that is a point of maximum potential energy will cause a ballplaced there to move to either of positions B or D that are more stable Therefore, stability canalso be defined with respect to a disturbance from an equilibrium state State C is unstablesince any disturbance causes the ball to move towards either State B or State D which are thenext equilibrium states in the immediate vicinity of State C Thus State C is impossible toachieve in a practical system The state C is equivalent to a nickel standing on its edge in theabsence of a disturbance A slight disturbance, however, can cause the nickel to fall flat on thesurface.

If the ball at B is disturbed (e.g., by a potential energy disturbance) by a finiteamount, then ball may move to a more stable state D which has the lowest energy of all statesillustrated in Figure 1 The rate at which the ball returns to its original state depends upon thefriction between the ball and surface If the potential energy constraint mgZ is removed, theball will eventually roll to position D State D is the most stable equilibrium state, state C isunstable and state B is

a metastable

equilib-rium state As long as

constraints exist in the

mechanical system of

Figure 1 and the

distur-bances are minor, a

metastable system is

also a stable system

During a

me-chanical disturbance

test the potential energy

is changed from its

initial level in order to

determine if its value

increases or decreases

In that context, state C

is unstable since d(PE)

< 0, but state D is

ble since d(PE) > 0 with disturbance A

distur-bance to State D (cf Figure 1) creates a

non-equilibrium situation (i.e., towards higher

poten-tial energy locations as d(PE) > 0), which induces

the ball to roll back This process brings the ball

back to state D and decreases the potential energy

by converting it into kinetic energy Any

pertur-bation of a stable equilibrium state causes a

proc-ess that tends to attenuate the disturbance This is

also known as the Le Chatelier principle and is a

consequence of the Second Law

Consider a thermodynamic system with

water as working fluid (cf Figure 2) The path

AFGH is a 40 bar isobar and all states along

AFGH are stable analogous to the state D in

Figure 1 However, it is possible to reach superheated liquid states along FM (except the pointM) which are analogous to state B in Figure 1 Similarly, if vapor at state H is cooled Simi-larly if vapor at state G is cooled, it can be cooled to a vapor state to along GN in the form ofsubcooled vapor Again, states along path GN (except at point N) are analogous to state B.Likewise, the states along the path MN are analogous to state C in Figure 1 In a manner simi-lar to mechanical stability tests, we can test the stability of fluids along the paths HGN or AFM

by perturbing the system (e.g., by disturbing the volume from a value V to V + dV) and mining if the system returns to its original state The rate of return to the stable equilibriumstate for any fluid depends upon transport rate processes, e.g., heat and mass transfer, which isbeyond the scope of classical thermodynamics

deter-If one strikes a match in air, the match simply burns and extinguishes This impliesthat the constituents of air do not react at a significant rate On the other hand, if a match isignited in air in the presence of a significant amount of gasoline vapors (i.e., if the reactivemixture is in metastable equilibrium), small temperature disturbances can ignite the mixture Ingeneral, if a disturbance decreases the entropy of an isolated system at specified values of U, Vand m, then the system must initially have been at a stable equilibrium state (SES) Note thatwhat is known as “equilibrium” in the context of classical thermodynamics yields only an av-erage state (e.g., an average system temperature or pressure) while any real system is inces-santly dynamic at its microscopic level Thus, a system should be stable with inherent micro-scopic and small natural disturbances

B STABILITY CRITERIA

1 Isolated System

An example of an isolated system is one constrained by rigid adiabatic and impermeablewalls, i.e., with specified values of U, V, and m An isolated system is at equilibrium when itsentropy reaches a maximum value so that δS = 0 At this state, if the system is perturbed suchthat the values of U, V, and m of the system are unchanged (e.g., by increasing the temperature

or internal energy by an infinitesimal amount), the perturbations are dampened since the tem is stable The perturbations at fixed values of U, V, and m actually decrease the entropy,i.e., δS < 0 at stable equilibrium (e.g., consider adiabatic chemical reactions, Chapter 11)

sys-a Single Component

In Chapters 6 and 7 we employed the real gas state equation and evaluated variousthermodynamic properties Let us consider the compression of water in the context of the RKequation of state is, say, at 593 K and 1 bar The volume of water decreases (or its intermo-lecular spacing decreases) and the pressure first increases, then decreases with a further de-crease in the volume, and again increases Since the intermolecular spacing continuously de-creases, stability analysis can help determine the state at which the fluid becomes a liquid

T

v

Figure 2: Thermodynamic states of ter

wa-First, let us consider an isolated system with specified values of U, V and m and press S = S(U, V, m) From Chapter 7,

which is the entropy fundamental equation

duT = T(R/(v–b)) – (RT/(v–b) – a/v2) = (a/v2) dv

Integrating at constant temperature,

In case a = 0,

Eliminating f(T) between Eqs (C) and (D),

Assuming constant (ideal gas) specific heats and u = uref,0 at T = Tref,

Eq (E) assumes the form

Similarly, we can integrate the expression

ds = cvdT/T + ∂P/∂T dv

at constant temperature to obtain the relation

Using the ideal gas relation for s0(T,v),

s0(T,v) – sref,0(TRef,vref) = cv0 ln (T/TRef) + R ln (v/vref),

Consider the Van der Waals equation of state Obtain an expression for s = s(u,v) forwater assuming the specific heat to be constant and equal to the ideal gas value cv0 =

s(T,v) = cv0 ln (T/TRef) + R ln (v/vref) + sref,0(T,vref)– R ln (v/(v–b)) (J)Further, using Eqs (H) and (J) to eliminate the temperature,

s =cv0 ln((u–uref,0)/(cv0Tref)+(a/(vcv0Tref) + 1)+R ln((v–b)/vref +sref,0(TRef,vref) (K)Setting ,

sref,0(Tref ,vref) =cv0 ln(cv0 Tref) + R ln vref, uref,0 = cv0 Tref

we obtain the expression

For ideal gases, a = b =0, and Eq (K) leads to the relation

s0 = cv0 ln (u/(cv0 TRef)) + R ln (v/vRef) (M)

This expression leads to a plot of s vs both u and v for a real or ideal gas.

Using the values ¯a = 5.3 bar m6 kmole–2, ¯b = 0.0305 m3 kmole–1, u = 7000 kJkmole–1, Tref = 1 K, vref = 1 m3 kmole–1, cv0 = 28 kJ kmole–1 K–1, uref,0 = cv0 Tref = 28

kJ kmole–1 in Eq (K), a plot of s vs v at u = 7500 kJ kmole–1 is presented in Figure 3

Remarks

The relation s =s (u,v) is the entropy fundamental equation It is somewhat more cult to manipulate the RK equation and obtain an explicit expression for s = s(u,v)using Eq (L)

146 146.5 147 147.5 148 148.5 149 149.5 150 150.5 151

A N D

E

Liquid

Gas

y

Figure 3: Variation in the entropy vs volume for water

modeled as a VW fluid at u = 7500 kJ kmole–1

Consider a rigid and insulated 0.4 m3 volume tank filled with 4 kmole (24×1026 cules) of water, and an internal energy of 7500 kJ kmole–1 Divide the tank into twoequal 0.2 m3 parts Assume that water follows the VW state equation What is the en-tropy of each section? Assume that section 1 is slightly compressed to 0.14 m3 whilethe section 2 is expanded to 0.26 m3, but maintaining the same total volume and total

mole-entropy of each section is initially

SD = (S1 = N1 ¯s1) = (S2 = N2 ¯s2) = 2×149.83 =299.7 kJ (A)The total entropy of the tank is S = 2SD = 299.7 + 299.7 = 599.4 kJ After the pertur-bation,

v 1 = 0.14÷2 = 0.07 m3 kmole–1, u1 = 7500 kJ kmole–1, and

v 2 = 0 26÷2 = 0.13 m3 kmole–1, u2 = 7500 kJ kmole–1

The corresponding states are represented by points M and N in Figure 3, i.e.,

dis-2 Mathematical Criterion for Stability

a Perturbation of Volume

i Geometrical Criterion

Consider the state B illustrated in Figure 3 which undergoes a small disturbance ∆V at

a specified value of U Due to the disturbance, Section 1 of the system in Example 2 reachesstate L while Section 2 reaches state M (Figure 3) With respect to stability

δS = (SL(U,V–∆V,N) – SB(U,V,N)) + (SM(U,V+∆V,N) – SB(U,V,N)) < 0

Since

δS = SL(U,V–∆V,N) + SM(U,V–∆V,N) – 2SB(U,V,N) < 0, i.e.,

(SL(U,V–∆V,N) + SM(U,V–∆V,N))/2 < (SB(U,V,N) (3)The entropy after a disturbance decreases in order that the initial state of the system is stable

In the context of Figure 3, the ordinate of the midpoint C of the chord LCM that connects thepoints L and M is represented by the LHS of Eq (3) while the RHS represents the ordinate ofthe point B Therefore, the chord LCM must lie below the curve LBM for the system to bestable The curve LKBHM satisfying the criteria given by Eq (3), is a concave curve with re-spect to the chord LCM On the other hand, the midpoint of the chord MAN lies above theconvex curve MDN, thereby violating this stability criterion We find that for a system to bestable the fundamental relation for S = S (U, V, N) one must satisfy the concave condition,which is established by Eq (3)

ii Differential Criterion

The discussion so far pertains to disturbances, which are of large magnitude ∆V sider a disturbance is in the neighborhood of state B (cf Figure 3) that extends from state K at(U, V–∆V, N) to state H at (U, V + ∆V, N) In that case

Con-(δS = (SK(U,V–∆V,N)–SB(U,V,N)) + (SH(U,V+∆V,N)–SB(U,V,N))) < 0, (4)where δS denotes the entropy change due to the disturbance The entropy should decrease fol-lowing disturbance at the stable points A, B and E illustrated in Figure 4 Expanding SK and

must be satisfied in the neighborhood of an equilibrium state for it to be stable If case the ond derivatives are zero, the third derivatives in Taylor series are included to obtain the stabil-ity condition d3S < 0, and so on

At constant internal energy

Assume that water follows the ideal gas state equation and show that for a unit mass,

(∂s/∂v)u = P/T, and

since for ideal gases, P/T = R/v,

sv = (∂s/∂v)u = R/v = f(v) alone, and

svv =(∂2s/∂v2) u = –R/v2 < 0

Remark

This illustrates that ideal gases are stable at all states

Figure 5 contains plots of ¯s vs ¯v at various values of ¯u At very high values of

¯u (e.g., at temperatures larger than the critical temperature), the fluid is stable for any ume, i.e., there is no increase in entropy in the presence of a disturbance at any state For asolid there are certain regimes in which stability criteria are not satisfied and disturbances re-sult in phase transitions e.g., in case of water from ice–phase I, to ice–phase II

vol-Since, in our illustrations, U, V and N (or m) were specified during a disturbance, the

δWopt = – dU + T0δS,with disturbance at a stable state yields

which is identical to the “concavity condition” described previously

Consider the plot of entropy vs internal energy at specified values of N (or m)

illus-trated by the curve ABCDEFGH in Figure 6 for a fluid following a real gas equation of state

At some equilibrium states d2S >0 (i.e., these states are unstable) and at others, d2S < 0 (stablestates) In the context of Figure 6 consider the fluid at state B at which the curve ABC is con-cave with respect to U Here, the stability criterion is satisfied On the other hand, curve DEF

is convex, the criterion is violated

A plot of (∂S/∂U)V = 1/T vs U is presented in Figure 6b and that of T vs U is

illus-trated in Figure 6c It is apparent that for certain ranges of U (UD to UF), T decreases with creasing U Mathematically, the stability condition ∂2S/∂U2 < 0 is equivalent to the condition

(∂/∂U)(∂S/∂U) < 0, i.e., ∂/∂U(1/T) < 0 or –1/T2(∂T/∂U) < 0 (9)

In the context of Eq (9), since (∂U/∂T) = mcv, then mcv > 0, and at a specified value of either

8500

8000 10000

Figure 5: Variation of s vs v in case of water according to the VW state equation.

for a state to be stable In Figure 6c, cv has positive values along the branch ABCD but

nega-tive values along DEF, which violates the thermal stability criterion provided by Eq (10) The

negative cv values imply that temperature decreases when energy is added and increase whenheat is removed Consider water at 25ºC and 1 bar in an insulated cup and assume that it hasnegative specific heat Divide the water into two equal sections A and B We now discuss themeaning of thermal stability If the water is thermally disturbed with a small amount of heattransfer δQ from A to B, then A will continue to become warmer than B (due to larger storage

of energy as intermolecular potential energy and less as vibrational energy) with the bance The entropy generation is still positive for the adiabatic cup since δσ = δQ (1/TA–1//TB) > 0 as TA > TB It still satisfies Second Law with entropy generation indicating that theinitial state is thermally unstable

distur-The specific heat cv > cvo for fluids following the RK equation, where cvo > 0 This

implies that a curve representing variations in S vs U must be concave (i.e., it must have a

decreasing radius of curvature) with respect

to U at a specified value of V In other

words, thermal stability is always satisfied

for satisfying most of the real gas state

equations Read Example 17 for instances

when this may be violated

d Example 4

(suu = ∂2

s/∂u2) < 0 at all volumes

Solution

Consider the relation T ds – P dv =

du At fixed volume

∂s/∂u = 1/T

Differentiating this expression,

∂2s/∂u2 = –1/T2(∂T/∂u)v.Since

(∂u/∂T)v = cv,(∂2s/∂u2 = –1/(cvT2)) < 0

Remarks

For a real gas,

(∂cv/∂v)T = T(∂P/∂T2)v,i.e., for a VW gas

(∂cv/∂v)T = 0,which implies that cv = f(T) = cv0(T) >0

A VW gas is thermally stable at all points

c Perturbation with Energy and Volume

The discussion so far has pertained to disturbance in the internal energy at specifiedvolume or in the volume at specified values of U If the volume and energy are both perturbed,the following conditions must be satisfied, i.e.,

δS = S(U+dU, V+dV, N) + S(U–dU, V–dV, N) – 2 S(U, V, N) < 0, i.e.,

S(U+dU, V+dV, N) + S(U–dU, V–dV, N) < 2 S(U, V, N)

Expanding this expression in a Taylor series, and retaining terms up to the second derivative,

G

H

H F

F G

Figure 6: Qualitative illustration of the tion of a) entropy, b) ∂S/∂U=1/T and c) tem-perature with respect to the internal energy.Consider a VW gas and show that

varia-δS = (∂S/∂U dU + ∂S/∂V dV + (∂/∂U + ∂/∂V) S) +

(–∂S/∂U dU – ∂S/∂VdV +(∂/∂U + ∂/∂V)2

S) < 0Since,

dS = (∂S/∂U dU + ∂S/∂V (–dU)) + (∂S/∂V dV + ∂S/∂V (–dV)) = 0, then

(d2S = SUU2 dU2 + 2 SUU SUV dU dV + SUU SVV dV2) > 0, i.e., (12)(SUU dU + SUV dV)2 + (SUU SVV – SUV2) dV2 > 0 (13)Now, (SUU dU + SUV dV)2 > 0, and since dV2 > 0, it is apparent from the perspective of stabilitythat

which is the stability condition in the presence of volumetric and energetic fluctuations within

a system The stability criterion at a given state can be summarized as

D1,U = SUU = ∂2S/∂U2 < 0, and D1,V = SVV = ∂2S/∂V2 < 0, (15)

In determinant form, the stability criterion is

From Example 1 we note that svv < 0, and from Example 2 that suu < 0

Since ∂s/∂u = 1/T, then svu = ∂2

suu suu – svu svu > 0,which satisfies the stability criterion

f Example 6

Apply the RK equation of state

Differentiating this expression with respect to v at constant u, we obtain the relation

svv=(∂2s/∂v2)u =– R/(v–b)2+a(2v+b)/(T3/2 v2(v+b)2) +(3/2)a(∂T/∂v)u/(T5/2 v(v+b)) (E)Recall that

B E C

G K N

D U

H

J M

F A

Figure 7: Plot of pressure and Gibbs energy with respect to volume for water using the RKequation of state

to water at 593 K (a) Plot the pressure with respect to volume; (b) determine sVV, sUUand sUV at v = 0.1 and 0.4 m3 kmole–1; and (c) check whether the stability criteria are

Similarly differentiating Eq (C) with respect to u, we obtain the relation

suu = ∂ 2s/∂u2 = – (∂T/∂u)v/T2 = – 1/(cvT2), (G)and differentiating Eq (C) with respect to v, we have

Thus, all of the differentials involved with the stability criterion represented by Eq.(16) can be evaluated

For v = 0.4 m3 kmole–1; and T = 593 K, P = 95.4 bar, cv = 30.09 kJ kmole–1 K, a =

142.64 bar m3 kJ kmole–2, b = 0.02110 m3 kmole–1, (∂T/∂v)u = –173.04 kmole K m–3,

∂2s/∂v2 = –31.92 kJ kmole m– 6 K– 1, ∂2s/∂u2 = –9.4×10–8 kmole kJ–1 K–1, and

where the terms within the parentheses () are all intensive properties Rewriting this sion, we obtain

expres-d2S U,V,N = – d(T dS – P dV + Σkµk dNk) < 0 (19)The term within the parentheses in Eq (19) is dU and it is apparent that U = U(S,V,N), i.e.,

(–d(dA)= –d2AT,V,N) < 0, or d2AT,V,N > 0 (22b)which is the criterion for A having a minimum value at specified values of T, V and N Like-wise

Σ kd(µk) dNk = Σk(Σj(∂µk/∂Nj) dNj) dNk (24)

ii Criterion for Binary Mixture

Consider a binary mixture at specified values of U, V, and N (= N1 + N2) that goes disturbances in U, V, N (i.e., of the third order) We can expand Eq (16) as a third orderdeterminant to obtain the stability criterion, i.e.,

e System With Specified Values of S, V, and m

In this case the criterion for a system to be stable is that the internal energy U mustreach a minimum value, i.e., d2U > 0 For a perturbation in volume at a specified entropy, since

molecular velocity, which cannot be negative, Chapter 1.) On a plot of U vs V the slope

∂U/∂V represents (–P) provided the entropy is held fixed (isentropic expansion or sion)

compres-Recall from Chapter 7 that

((∂P/∂V)S = k(∂P/∂V)T) < 0, and cp = cv +T v βP /βT, (29)where βT = –(1/v)(∂v/∂P)T and βP = (1/v)(∂v/∂T)P Using the two relations in Eqs (29) we ob-tain the expression

cp > cv.

The criterion given by Eq (29) is called the mechanical stability condition (i.e., lated to a mechanical or volumetric disturbance) in the context of which (∂P/∂V)T < 0 Due tothis condition a fluid does not generally break down into two phases upon mechanical pertur-bation (one a dense liquid and the other a dilute vapor phase) Even a minute disturbance willcause unstable behavior in case (∂P/∂V)T < 0 Therefore, if the volume decreases during anisothermal process, the fluid pressure must increase in order for the fluid to be stable

An RK gas behaves according to the equation

(∂P/∂v)T = –RT/(v–b)2

+ a(2v + b)/((T1/2v2(v+b)2) (B)For stability,

f Perturbation in Entropy at Specified Volume

For a system to be stable, the internal energy must be at a local minimum when theentropy is disturbed but the volume is held fixed Recall that

(∂U/∂S)V = T(S,V),

so that for a local minimum,

D1 =USS = (∂2U/∂S2) > 0, where (∂2U/∂S2) = (∂T/∂S)V = T/Cv, (30)which implies that Cv or cv > 0, since the temperature is a positive quantity The condition cv >

0 implies that the internal energy and entropy must be a monotonic function of temperaturewith positive slope at a specified volume The temperature will increase if the entropy is in-creased for a specified volume of a fluid

g Perturbation in Entropy and Volume

Again, we will employ the relation dU = TdS – PdV so that

d(dU) = d2U = d(T(S,V)) dS – d(P(S,V)) dV

Since T = US and P = UV,

dT =USS dS + USV dV, and – dP =UVV dV + UVS dS, so that

(d2U = USS dS2 + 2USV dS dV + UVV dV2) > 0

Since USS > 0, multiplication by USS yields

(USS dS + USV dV)2 + (UVV USS – USV2)dV2 > 0 (31)Hence, the condition for stability is

If (UVVUSS – USV2) < 0, Eq (31) still holds if |(USS dS + USV dV)2| > (UVVUSS – USV2)dV2, i.e.,depending upon strength of the disturbances dS and dV, Eq (31) may or may not be satisfiedwhen Eq (32) is violated

Writing Eq (32) in determinant form

In case USS > 0, then dividing Eq (32) by USS, we obtain the inequality (UVV – USV2/USS) > 0

or UVV > USV2/USS Since USV2 > 0, if USS > 0, this implies that UVV > 0 Therefore, if D1 > 0and D2 > 0, we satisfy the condition that UVV > 0 At the limit of intrinsic stability, D1 = D2 =0

i Binary and Multicomponent Mixtures

For a multicomponent mixture,

1 1 1

0

At the limits of intrinsic stability, D1 = D2 = D3 = 0 for a binary mixture Extending the result

to a k–component mixture, it can be shown that the determinant of (k+1)th order must be tive, i.e.,

−

−

− 1

h System With Specified Values of S, P, and m

Consider the relation

dH = T dS + V dP

Since intensive properties are not additive, we cannot obtain enthalpy variations by perturbing

P to P±∆P at a specified entropy and mass, but keeping the overall pressure constant If theentropy is specified

(∂H/∂P)S = V, and (∂2

H/∂P2)S = (∂V/∂P)S

We have previously shown that (∂P/∂V)S < 0, i.e., (∂2H/∂P2)S < 0 In the case of entropy turbations

which implies that Cp or cp > 0 Thus the enthalpy and entropy increase monotonically with

increasing temperatures at a specified pressure A plot of H vs S at specified P shows

convex-ity with respect to S for stable states The derivation for a multicomponent system is left as anexercise for the reader

i System With Specified Values of T, V, and m

In this case A is to be minimized with respect to the disturbance at specified values of

T, V, and m Figure 8 illustrates plots of A with respect to V and T For a single component,

AVV = – (∂P/∂V)T > 0, or (∂P/∂V)T < 0

The resultant curve of A vs V will be convex with respect to V for a system to be stable.

Figure 8b and Figure 8d illustrate the stable and unstable branches At the critical point ∂P/∂V

= 0, ∂2P/∂V2 = 0 and, hence, AVV = 0 Here, ∂3P/∂V3 must have a negative value in order forstability conditions to be satisfied

At T = Tc, and v = vc, ∂P/∂v = 0, ∂P/∂v = 0 The third derivative w.r.t v,

(∂3

P/∂v3

= – 6RTc/(vc–b)4 + 24 a/vc5 = – 6RTc/((2/3)vc)4 + 24(9/8)(RTcvc)/vc5 = – (243/8) RTc/vc4 + 27 RTc/vc4 = – (27/8) RTc/vc4) < 0 (39)This stability condition is satisfied at the critical point

Remark

The specific heat cv is finite at the critical point, but (cp–cv) = –T(∂P/∂T)2/(∂P/∂v)→∞

at that state

i Perturbations With Respect to Temperature

Since the temperature is an intensive property, if two sections of a specified volumeare disturbed, δT ≠ δT1 + δT2 In this case,

AT = (∂A/∂T)V = –S

Recall that, from the energy minimum principle, ((∂S/∂T)V = cv/T) > 0 Consequently,

(ATT = (∂2

A/∂T2)V = – (∂S/∂T)V = - cv/T) < 0

ii Binary and Multicomponent Mixtures

For a binary mixture,

∂∂∂∂2

A/ ∂∂∂∂T 2

Figure 8: Qualitative variation of a) A with respect to T, b) -(∂2A/∂V2)T= (∂P/∂V)T with spect to V, c) A with respect to T, and d) (∂2A/∂T2) with respect to T

1 1 1

0

These relations are readily extended for multicomponent fluids

j System With Specified Values of T, P, and m

In this case, for a system to be stable, the Gibbs energy must be minimized in thepresence of disturbances in P and T Consider the relation,

i Perturbations With Respect to Pressure

Considering perturbations with respect to the temperature,

GP = (∂G/∂P)T = V, and (GPP = (∂2G/∂P2)T = (∂V/∂P)T) < 0 (43a)

At the critical point,

(∂P/∂v)T = 0, and (∂2P/∂v2)T = 0, i.e.,

both (∂v/∂P)T and (∂2g/∂P2)P tend to infinity

ii Perturbation With Respect to temperature

Similarly,

GT = (∂G/∂T)P = –S, GTT = (∂2G/∂T2)P = – (∂S/∂T)P = – CP/T (43b)Since Cp or cP > 0, (∂2G/∂T2)P < 0

iii Perturbations With Respect to P and T

Since G is to be minimized at stable conditions, one can show that

Disturbances in the mixture state can place the local properties along either curve (a)

or (b) If there is local disturbance at state C on curve (a), a negative increment in Z1 causes alarger Gibbs energy drop dG– (= GA–GC) to state A as compared to a positive increment tostate D, i.e., dG+ (= GD–GC) Therefore, in this hypothetical case, the net disturbance (dG– +

dG+) > 0, causing the mixture to return to its original state C Curve (b) also satisfies the dition ∆G < 0 However, we have introduced a hypothetical kink in the curve at states M and

con-N If there is a local disturbance in N1 (or Z1) at state E, then |dG +| < |dG–| so that (dG–+dG+) <

0, which implies that at fixed temperature and pressure the disturbance is undamped This canresult in the formation of two phases or two components, depending upon the local state

In the context of this discussion, it is apparent that

C APPLICATION TO BOILING AND CONDENSATION

We will illustrate an application of stability criteria to boiling and condensation (i.e,the formation of two phases) through the following example

Figure 9: Variation of a) G with T, b) ∂G/∂T, c) G with P, d) ∂2G/∂T2

Water is contained in a piston–cylinder–weight assembly that is immersed in a stant–temperature bath at 593 K Assume that the fluid obeys the RK equation of state

at M is expanded further, it vaporizes instantaneously.

Since dg _T = ¯v dP, it is possible to integrate Eq (A) between the limits v and vref,i.e.,

coex-to the RK equation is 133 bar at 593 K.)

Path QBECGN is a stable vapor branch, since ∂P/∂ ¯v<0 (or ∂ ¯v/∂P <0) The lowestvalue of ¯g at specified values of T and P (cf Figure 12) indicates that P < Psat = 133bar, i.e., the vapor has a lower free energy when it is compared to the liquid curveQRAF

Path FL is a stable liquid branch since ∂P/∂ ¯v<0 (or ∂ ¯v/∂P <0)

Path GKN represents metastable vapor (i.e., an equilibrium condition with a finiteconstraint)

The state N represents an intrinsic stability limit for the vapor at which ∂P/∂v = 0.The state M is an intrinsic stability limit for the liquid at which ∂P/∂v = 0

Figure 10: Variation of ∆G with mole fraction in a binary ture

mix-Obtain a plot for

P v s

the relation dgT =

v dP, obtain a

plot for

Path MRAF corresponds to metastable liquid with intermediate values of ¯g.

Path NDHJM is an unstable branch since ∂P/∂ ¯v > 0 and the highest values of ¯g are

to be found here

1 Physical Processes and Stability

Consider the isothermal compression of water at 593 K (cf Figure 11) As the fluid iscompressed from state B towards states E, C, G, etc., the volume decreases with the increase inpressure, since the intermolecular spacing decreases The intermolecular attraction forcesslowly increase as the states E, C, G, etc., are approached At larger volumes (i.e lower pres-sures) the first term in RK equation dominates, i.e., b « v, P ≈ RT/v so that P ∝ 1/v at a speci-fied temperature The Gibbs energy value is lower at larger volumes, and gradually increases

as the pressure is raised (BECG in Figure 11), indicating that the fluid accumulates a largerpotential to perform work The rate of pressure increase with decreasing volume is lower atsmaller volumes due to the larger intermolecular attraction forces and the second term in the

RK equation a/(T1/2v(v+b)) becomes significant Beyond a maximum pressure at state N, theintermolecular attraction forces are so large that this second term dominates and tends to lowerthe pressure Consequently, the pressure starts to decrease with compression at smaller vol-umes due to the very small intermolecular spacing (The pressure can sometimes be negativeindicating that the fluid is under tension In case of water the tension can be as high as –40 barwithout evaporation occurring.) Consequently the “g” decreases along NDUHJM Upon com-pression beyond states M, R, A, etc., the body volume effect (which reduces the space avail-able for the movement of molecules) becomes dominant and results in a higher number density

so that the first term in the RK equation again dominates Thereupon, the pressure again creases rapidly and P ≈ RT/(v–b), i.e., P ∝ 1/(v–b) and the Gibbs energy again increases(MRAFL)

in-All fluid states along the path BECGKNDHJMRFL are equilibrium states The bility criterion ∂P/∂v < 0 suggests that the path NDUHJM is an unstable branch along whichinherent disturbances exist and hence a uniform intermolecular spacing or stable state cannot

sta-be maintained

a Physical Explanation

At T = 593 K, gF= gG at P = 133 bars Hence Psat = 133 bar; the fluid in the context

of Example 10 typically changes its state from vapor to liquid (from state G to F along lineGHF (cf Figure

the adverse

poten-tials at the

satura-tion condisatura-tion that

the fluid has to

v, m 3 /kmole

0 500 1000 1500 2000 2500

B E C G K N

D U H J

M A F L

R

y

Figure 11: Plot of P (and ¯g) vs volume using the RK equation.

state G (Figure 13a) We note from Figure 11 or Figure 12 that when P < 50 bar, a single state

is possible (i.e., superheated vapor) If 50 < P < 155 bar, there are three possible states for thesame Gibbs energy value, and when P > 155 bar there is a single liquid solution (i.e., com-pressed liquid)

Consider a constant T, P, and m system at an arbitrary state J A reduction in the fluidvolume from a value vJ reduces the internal pressure exerted by the fluid further compressing

the fluid Assume that equilibrium is achieved at a liquid volume vA at which PA = PJ (Figure

12) Likewise a corresponding vapor state C exists at which PC = PJ with a volume vC Theimplication is that at the pressure PJ there are three plausible solutions for the Gibbs energy.The questions are as follows Which are stable states? Which are nonstable?

Since dGT =VdP, dGembryo = (V(∂P/∂V)) dV An increased volume (e.g., duringevaporation) implies that ∂P/∂V < 0 so that ∂G/∂V < 0 Suppose a disturbance at J causes theembryo phase to expand to a volume slightly higher than vJ from state J, the volume increasetends to increase the embryo phase pressure Since the embryo phase pressure is higher thanthe mother phase pressure, which is held fixed, the embryo expands to larger and larger vol-umes, eventually to the vapor state B The first bubble during boiling is formed through thisprocess The embryo phase bubble is associated with a lower Gibb’s free energy (Figure 12)

as compared to the mother phase that is still at state J In Chapter 3 we discussed that a Gibbsenergy gradient produces a species flow from a system at a higher Gibbs energy to that at alower Gibbs energy In that context, the molecules from the mother phase migrate to the vaporphase during vaporization as long as the pressure and temperature are maintained constant Ifthe disturbance at J results in reduction of volume, the embryo pressure decreases; since themother phase is at higher pressure, embryo is compressed further until it forms a liquid droplet

at A In the case of flow processes, a sudden condensation or vaporization produces a severepressure disturbance or a sudden acceleration of the flow, leading to local turbulence (e.g., inboiler tubes and in clouds)

0 500

M

D U

J H

In the context of the above discussion, we now consider the H2O at 140 bar and 593 Kfor which the specific volume vD is at an unstable state An embryo may form, but a decrease

in the embryo volume causes the pressure to decrease below 140 bar, which results in pression of the embryo by the mother phase that is still at 140 bar This accelerates the forma-tion of a drop at state L Since the Gibbs energy of the compressed liquid at state L is lowerthan that of the mother phase at state K, the fluid molecules will tend to migrate to the liquidstate State H is also unstable and only a microscopic disturbance is required to drive the state

com-to either of states F or G At saturation, the two minima are equal, i.e., gF = gG = gsat Thisprocess is called a first order phase transition during which both the liquid and vapor states areprobable, which is a consequence of boiling at a specified temperature and pressure In thiscase, the liquid and vapor molecules can exchange phase if a disturbance is strong enough toovercome the potential (gH – gG), which results in a wet mixture Note that thermodynamics donot specify the time scales (called relaxation time scale) required to effect the change frommeta-stable or unstable state to stable state Constitute equations for the transport processes arerequired to determine those time scales

These examples pertain to phase equilibrium An analogous situation exists duringchemical equilibrium where, at a specified temperature and pressure, the Gibbs energy ofproducts reaches a minimum value

2 Constant Temperature and Volume

Consider a fluid of mass m within a rigid

tank of volume V that is immersed in a bath at a

temperature T for which the real gas state

equa-tion yields the P-v diagram presented in Figure

11 The states F and G represent the saturated

states at a specified temperature At state K

(Figure 14) the fluid is in the form of slightly

sub–cooled vapor We will divide the system into

two parts 1 and 2, and disturb the volume such

that portion 1 expands slightly to state K″, while

portion 2 shrinks to K′, but the total volume is

fixed Then, PK' > PK" and portion 2 will expand

back so that the system eventually reverts to its

original state K

Consider the fluid at State D (cf Figure

14) If the fluid is disturbed so that the mother

phase volume (say, portion 1) slightly expands to

vD" while the embryo volume (portion 2)

con-tracts to vD' with the total volume held fixed,

since vD" ≈ vD and vD' « vD, PD"≈ PD, while PD >

PD' The mother phase is at a higher pressure than

the embryo phase, which undergoes increasing

compression The pressure in the embryo first

decreases along path HJM (cf Figure 11) but

starts increasing again along path MRL (the

liq-uid path) until state L is reached at which the

pressure in the embryo equals PD A disturbance

can occur at several locations inside the system

creating a large number of drops that can combine

to form a continuous liquid phase However,

since liquid drops generally occupy a smaller

volume than the gas phase in a fixed volume

sys-(c )

(b )

(a ) C

Figure 13: Variation in the value of g at

a specified value of T, but for differentpressures (a) P = Psat; (b) P>Psat; (c) P <

Psat

tem, the entire vapor body may not condense, and a wet mixture may form.

At equilibrium, there cannot be any potential gradients between the liquid and vaporphases so that gliquid = gvapor At a specified temperature, this condition occurs at a particularvalue of the saturation pressure Psat, which allows us to determine the volume or quality x, i.e.,

The volume at a stable state v > vG For the condition vN < v < vG, a metastable vaporstate exists and a mixture of vapor and liquid is formed, but at a higher quality If vF < v < vM,then a metastable liquid state exists and a vapor-liquid mixture of lower quality is formed Thecondition vM < v < vN is unstable and a mixture of vapor and liquid of medium quality isformed

When phase transformation from a metastable state occurs at a specified temperatureand volume, the Helmholtz energy is minimized Figure 15 presents a plot of a¯ and P vs v¯ State D on the P-¯v diagram (P = 140 bar, vD = 0.1 m3 kmole-1, T = 593) corresponds to thepoint Q on the ¯a- ¯v curve A disturbance increases ¯a within the mother phase if the volume

decreases and vice versa However, the value of the Helmholtz energy increase is smaller than

its decrease with the result that ¯a

decreases (at fixed T and V)

Conse-quently, the fluid eventually reaches

a vapor state and ¯a is minimized

At a specified temperature and

vol-ume the pressure (i.e., P = 133 bar, v

= 0.1 m3 kmole-1, T = 593 K) and

quality adjust such that the

Helm-holtz energy is at a minimum value

(e.g., with vapor at State G where a¯

= 20.6 MJ and liquid at State F with

a higher value of ¯a = 23.2 MJ)

k Example 11

ues of ¯s, ¯u , and

¯s, ¯u , and ¯a at this state?Assume that cΡ = 5.96 kJ kg-1 K-1

Solution

¯v = 2.662/27 = 0.0986 m3 kmole-1.Applying the RK equation at 593 K and 140 bar (State D in Figure 11),

Psat = 133 bar

Since P > Psat at 593 K, and P < PN = 155 bars the state is metastable

We will use the method described in Chapter 7 to determine the fluid properties Forinstance,

¯h(320ºC, 0.0986 m3 kmole-1) = 44633 kJ kmole-1 or 2477 kJ kg-1,

¯u (320C, 0.0986 m3 kmole-1) = h – P¯v

= 43252 kJ kmole-1 or 2400 kJ kg-1, and

¯s (320C, 0.0986 m3 kmole-1) = 91.34 kJ kmole-1 or 5.07 kJ kg-1.Thereafter,

D

”

K K

2.662 kmole of water are

contained in a rigid tank of

volume 27 m3 at 320ºC and

140 bar What are the

val-¯a atthis state? If the tempera-

ture and volume are

main-tained constant, what is the

most stable equilibrium state, and what are the values of