Physical Meaning

2. Physical Meaning

We introduce the fugacity f, which has the same units as pressure. Following Lewis (1875–1946), we define the fugacity f as

dg = v dP = RT d ln f. (103) Noting the similitude with Eq. (99), d((g – g o )/RT) = (Z – 1) dP/P = d ln(f/P) = d ln φ

(104) Upon comparing Eqs. (104) and (102) we realize that φ = f/P

(105) Integrating Eq. (104) at constant temperature

((g – g o )/RT) = ln φ + F(T).

As g →g o (i.e., P → 0), f → P (i.e., φ → 1). (The ideal gas equation of state can, therefore, be expressed in the form fv = RT where f= P for ideal gas). Hence F(T) = 0, and

((g – g o )/RT) = ln φ. The fugacity coefficient φ is a measure of the deviation of the Gibbs function from its ideal gas

value. One may express real gas equation of state Pv =ZRT as fv=Z’ (T R ,P R )RT where Z’ (T R ,P R )= φ(T R ,P R ) Z(T R ,P R ). In the presence of intermolecular attraction forces, typically, for a real gas h < h o and s < s o . The corresponding value of g (= h–Ts) is less than or greater than g o

(= h o –Ts o ), when, respectively, φ is smaller than or larger than unity.

Accounting for the Pitzer factor Z = Z (o) + ωZ (1) , Eq. (105) assumes the form

ln φ = ln φ (o) + w ln (1) φ (106) where

ln = φ

(0) P R

() 1 ∫ R

dP

, and ln = φ

dP

P → 0 (Z - 1)

a. Phase Equilibrium In general, during boiling, the pressure and temperature remain constant. Since dg = –s dT + v dP, for this process dg = 0. Consequently, as a fluid of unit mass undergoes change from the saturated liquid state to the saturated vapor state f ,g = g g . Using Eq. (105), we note

that the differences

f f g (g g (T,P) – g

o (T,P))/(RT) = ln φ , and (g (T,P) – g o (T,P))/(RT) = ln φ

are identical, since g f =g g . Therefore,

f = g , and f f =f φ g φ =f sat . (108) For a phase change from an

α (say, the solid phase) to a β phase (say, vapor) g α = g β (for a single component g is also called the chemical potential µ so that µ α = µ β ), φ α = φ β , and f α =f β .

This implies the existence of a single saturation curve along which the fugacities for both the saturated liquid and vapor states are the same at given T. The inflexion of the curve occurs at the critical point.

b. Subcooled Liquid Integrating Eq. (103) along an isotherm from the saturated liquid state (T,P sat ) at which f = f f to a compressed liquid state (T, P)

( P g(T,P) g(T,P ) RT = sat − )/ ln( f (T,P) f (T,P / sat

f )) = P sat (/ v RT dP )

The Poynting correction factor POY is related to the RHS of Eq. (109), namely,

sat

POY(T,P) = f (T,P) f (T,P / f ) = exp ( P sat (/ v RT dP ) )

In general, along an isotherm v ≈v f so that POY = exp(v f (P – P sat (T))/RT), i.e.,

(g(T,P) – g (T,P sat ))/RT = ln (f/f

sat

f )=v f (P – P )/RT.

A similar procedure can also be adopted for solids.

Treating the liquid as incompressible, i.e., u (T,P)

≈ u(T, P sat ), s(T,P) ≈ s(T,P ), and h(T,P) = u(T,P) + Pv(T,P)

sat

≈ u(T,P sat ) + Pv (T,P sat

f ).

c. Supercooled Vapor Sometimes a vapor can be cooled to a temperature below saturation temperature without causing condensation (Chapter 10). In the case of super-cooled vapor the thermody- namic properties can be related to the saturation properties, At low pressure u(T,P)

≈ u (T,P sat ), h(T,P)

≈ h (T,P sat ), s(T,P) ≈ s(T,P ) – R ln (P/P sat ), and g(T,P) ≈g o (T,P). Therefore, g(T,P) = h–Ts

sat

≈ h(T,P sat ) – T(s (T,P ) – R ln(P/P )) = g(T,P ) + RT ln(P/P ). where P sat is at T.

z. Example 26 Determine the fugacity of pure water for the following cases: Saturated vapor at 100ºC, Saturated liquid at 100ºC, Compressed liquid at 100ºC, and 200 bar. Superheated vapor at 100ºC, and 0.5 bar. Saturated vapor at 350ºC. Super-cooled vapor at 90 °C, 1 bar, assume ideal gas behavior

Solution The saturation pressure at 100ºC is P sat = 1 bar. Since P c = 220.9 bars, P R =P sat /P c « 1, at this state water vapor H 2 O(g) behaves as an ideal gas. Therefore, f=P sat = 100 kPa or 1 bar. Since P and T are constant, f is unchanged during the phase change. Therefore, for the

saturated liquid H 2 O( l) f = 100 kPa or 1 bar.

At constant temperature,

d (ln f) = vdP/(RT).

3 For liquids, v –1 ≈ constant. For this problem, v = 0.001 m kg . Integrating from the saturated liquid state at 100ºC and 1 bar to the compressed liquid state at 100ºC and 2

bar,

ln (f(T,P)/f sat (T)) = v(P – P )/(RT), i.e., ln (f(100ºC, 200 bar))/(f sat (100ºC)) = ((0.001 ×(20,000–100))/(8.314×373/18.02))

sat

= 0.116. Therefore, (f(100ºC, 200 bar))/(f sat (100ºC)) = POY = exp (0.116) = 1.123, and

f(100ºC, 200 bar)) ≈ 1.122×(f sat (100 ºC) = 1.123 ×f sat (100ºC) = 1.123 bar. Superheated vapor behaves as an ideal gas at low pressures. Therefore, the fugacity

equals the pressure, i.e., f = 0.5 bar. At 350ºC, P sat = 165 bar, P R = 165 ÷220.9= 0.75, T R = 623 ÷641 = 0.98, and Z ≈ 0.3.

Therefore, under these conditions water vapor behaves as a real gas and f ≠ P. Using the fugacity coefficient charts, φ = 0.74, and f = 0.74×165 = 124 bar. Since the behavior a ideal gas, f = P sat at 90 °C = 0.7014 bars.

Remarks The example illustrates that when the pressure is increased by a factor of 200, in the case of liquids f changes by only 12 %. At a specified temperature the changes in the

values of f with respect to pressure (d lnf = v f dP) are small, since the liquid specific volume v f is small. However, for the gaseous state, v is much larger than v f (often- values of f with respect to pressure (d lnf = v f dP) are small, since the liquid specific volume v f is small. However, for the gaseous state, v is much larger than v f (often-

aa. Example 27 Employ the RK equation of state for the following problems. Determine the value of g for H 2 O(g) at its triple point of (T TP = 273 K, P TP = 0.0061 bar). Determine the corresponding value of g o .

Determine the value of g for H 2 O(g) at P = 250 bar, and T = 873 K if c p ,o = 28.85 +

0.01206 T+100,600/T –1 kJ kmole K . Solution

For water at its triple point, g(f, T TP ,P TP )=h f,TP –Ts f,TP = 0. Since the vapor behaves as an ideal gas at the triple point,

h o – h(T TP ,P TP ) ≈ 0, and s o – s(T TP ,P TP ) ≈0

Therefore, h o – Ts o ≈ h – Ts so that g(T TP ,P TP )=g o .

Consequently, φ =1, and f = P TP = 0.006 bar. We can get g(250, 873) by using the definition g = h–Ts and using correction charts

to determine h o – h and s o –s entropy, ideal gas enthalpy and entropy at 250 bar, 873 K or we can use

g (T,P) = g o (T,P) + RT ln φ Referring to Example 17,

h (873K) = 3706 kJ kg o –1 . Referring to Example 16 s (873, 250bar) = 6.566 kJ kg –1 o K –1 .

g –1 o (873,250) = h o (T) – T s o (T,P) = 3706–873 ×6.566 = –2,026.1 kJ kg . g(873,250) = g o (873,250) + RT ln φ From example 26,

φ = 0.854, f = 0.854 × 250 = 214 bars. g(873,250) = –2,026.1 + (8.314/18.02)

×873 ln 0.854 = –2089.7 kJ kg –1 .

bb. Example 28 Determine the properties u, h and f for liquid water at 120ºC and 250 kPa. Assume that data for u sat (120ºC), and v sat (120ºC) are available. Determine the properties u, h, s, g, and f for liquid water at120ºC and 100 kPa.

Solution P sat (120ºC) = 199 kPa. Since P = 250 kPa, the liquid is in a compressed state. For this

state u(120ºC, 199 kPa) = u sat (120ºC) = 503.5 kJ kg –1 . Recall that du = c v dT + (T(dP/dT) v – P) dv

(A) Assume that v = v(T) so that its value (v = 0.001063 m 3 kg –1 ) along the 120ºC iso-

therm does not change. Therefore, since dv = dT = 0, du = 0, and u sat (120ºC, 250 kPa)

≈ u (120ºC, 199 kPa) = 503.5 kJ kg –1 .

Furthermore,

h = u + Pv = 503.5 + 250 –1 × 0.001063 = 503.5 kJ kg . Using Eq. (112)

ln (f(T,P)/f sat (T)) = v

f (P – P )/(RT)

sat

= 0.001063 ×(250 – 199)/((8.314/18.02)×393) = 0.000179, POY = f(T,P)/f(T, P sat ) = 1.000179, i.e., f(T,P)

≈f sat(g) (T).

Since the pressure is relatively low, f sat(g) =P sat = 199 kPa, i.e., Thus f sat(l) =f sat(g) = 199 kPa. Water boils at 100ºC when P ≈ 100 kPa. Therefore, we can expect the water vapor at

120ºC to be superheated. However, under some circumstances (to be discussed in Chapter 10) water can exist as a superheated liquid at 120ºC and 100 kPa, instead of as a superheated vapor. Since u is a function temperature alone for liquids,

u(120ºC,100 kPa) –1 ≈u (120ºC) = 503.5 kJ kg . Therefore,

sat

h(120ºC,100 kPa) = u sat (120ºC) + Pv = 503.5 + 100 ×0.001063 = 503.51 kJ kg –1 ,

s(120ºC,100 kPa) = s –1 (120ºC) = 1.5276 kJ kg K ,

sat

g(120ºC,100 kPa) = g sat (120ºC) + v f (P – P sat )

= (503.71 – 393 –1 ×1.5276) + 0.00106×(199–100) = –96.53 kJ kg , and

f (P – P )/RT = 0.001063(100–199) ÷(0.4614×393) = –0.000578, i.e., f/f sat = 0.9994 or f = f sat × 0.9994 = 199 × 0.9994 = 198.4 kPa. Remark

ln (f/f sat )=v

sat

Note that the fugacity of the superheated liquid is lower than that of the corresponding saturated liquid at the same temperature.

cc. Example 29 Methane is reversibly compressed at 230 K in a steady state steady flow (sssf) device from 150 bar to 1000 bar. Using the fugacity charts, determine work done in kJ kmole –1 .

Solution The sssf energy balance for a reversible process has the form

δw = – vdP. (A) For an isothermal process,

dg T = v dP. (B) From Eqs. (A) and (B) we determine that – δw = dg T , i.e.,

(C)

(D) Since dg = RT d ln f, integrating this relation at constant temperature,

–w = g 2 –g 1 .

–w = g 2 –g 1 = RT ln f 2 /f 1 = RT ln ( φ 2 P 2 /( φ 1 P 1 )).

(E) From the fugacity charts, P R,2 =P 2 /P c = 250 ÷46.4= 5.4, P R,1 =P 1 /P c = 150 ÷46.4 = 3.2,

and T R,1 =T R,2 =T 2 /T c = 230 ÷190.7 = 1.2, i.e., log 1o φ 2 = –0.358, log 1o φ 1 = –0.275.

Therefore, φ 2 = 0.439, and φ 1 = 0.531 so that

– w = R T ln(0.439 ×250÷(0.531×150)) = 8.314 × 230 × ln 1.378 = 613 kJ kmole –1 . Remarks

If the gas behaves like an ideal gas, w 12 =– ∫vdP = – R T ln(P 2 /P 1 ) = –8.314 × 230 × ln(250

÷150) = –977 kJ kmole –1 . The magnitude for the work done in case of an ideal gas is much larger since the ideal gas involves higher pressure for the same volume

and hence requires more boundary work during compression.

If gas is compressed isothermally in a closed system w 12 = ∫ P dv = (P 2 v 2 -P 1 v 1 - ∫vdP) where second part on the right can be determined using fugacity charts