Real Gas Equations

2. Real Gas Equations

The previous section required a knowledge of “c”, c p0 ,h fg and relations for “s”. In- stead, a real gas state equation can be used to determine a relation for P sat (T) and by applying the equality g Ρ =g v . In order to do so, the critical conditions of the fluid must be known.

a. Graphical Solution It is possible to characterize the variation in pressure with respect to specific volume at T<T c , e.g., the curve BECGNDUHJMAFL contained in Figure 16 . At constant temperature,

dg = v dP. (113) where “v’ is given by any real gas state equation presented in Chapter 6. As the fluid is com-

pressed from a large volume at low pressure, the pressure first increases (e.g., the curve BECGN along which dP >0). Therefore, due to the work input, the Gibbs energy, which is a measure of availability, also increases (cf.. Eq. (113)). With a further decrease in volume past point N along the curve NHM, dP < 0, i.e., dg < 0 and hence “g’ decreases. The pressure de- pressed from a large volume at low pressure, the pressure first increases (e.g., the curve BECGN along which dP >0). Therefore, due to the work input, the Gibbs energy, which is a measure of availability, also increases (cf.. Eq. (113)). With a further decrease in volume past point N along the curve NHM, dP < 0, i.e., dg < 0 and hence “g’ decreases. The pressure de-

dg = d(Pv) – Pdv. (114) Using the RK state equation, P = RT/(v–b) –a/(T 1/2 v(v+b)) and integrating between

the resultant expression between the limits v ref and v, we obtain the relation g–g ref = (Pv – P ref v ref ) – RT ln((v – b)/(v ref – b)) +

(116) We arbitrarily set g ref (v ref ) = 0. The resultant plot is presented in Figure 16 . The vapor–like

1/2 ) ln((v/(v+b))((v (a/bT

ref +b)/v ref )).

curve BECKN and the liquid–like curve MFL are apparent. The unstable branch NDUHJM (along which the pressure decreases with decreasing specific volume) will be discussed in

Chapter 10. At the points F and G, g l =g g , which, according to the phase equilibrium condition, must correspond to the saturation pressure. We note from Eq. (113) that the slopes ∂g/∂P = v f along the liquid–like curve and

F U D K G 1500

v, m 3 /kmol

H A Stable,liquid /kmole

Stable,Vapor B 125 133 140

Figure 15. Variation in the Gibbs energy with respect to pressure for water vapor modeled as an RK gas at 593 K.

∂g/∂P = v g along the vapor–like curve. (Further discussion of this behavior is contained in Chapter 10.) A geometrical interpretation of the equality of g l =g g follows. Since, dg = d(Pv) –

Pdv,

dg = P ( v − v ) − v α Pdv = 0 .

sat

Since, at equilibrium, g α (T, P sat )=g β (T, P sat ),

sat

v α P(T,v) dv = 0 .

If v α =v ,v β g =v l , the LHS of this relation represents the area FCEDGMF in Figure 17 , while the RHS represents the area FCEDGCMAF. This implies that the area FAMF = area MCGM.

The equality of the two areas is called Maxwells equal area rule Apply Eq. (115) for the two phases α and β. At the equilibrium condition,

(g α (T,v α )–g ref (T,v ref )) = ((RTv α /(v α – b)–(a/T 1/2 )/(v α +b)) – (RTv ref /(v ref – b) – (a/T 1/2 )/(v

ref +b))) – RT ln ((v α –b)/( vref –b)) + (a/(T 1/2 b)) (ln((v α (v

ref +b))/(v ref (v α +b)))), and,

(118a)

(g β (T,v β )–g ref (T,v ref )) = ((RTv β /(v β – b) – (a/T 1/2 )/(v β + b)) – (RTv ref /(v ref – b) – (a/T 1/2 )/(v ref + b))) – RT ln((v α – b)/(v ref – b))

ref +b))/(v ref (v β +b))). (118b) Subtracting the first of Eqs. (118a) from Eq.(118b), and using the condition g α =g β , after sim-

1/2 b)) (ln ((v + (a/(T β (v

plification we obtain the expression

(RTv β /(v β – b) – (a/T 1/2 )/(v β + b)) – (RTv α /(v α – b) – (a/T 1/2 )/(v α + b)) – RT ln ((v β – b)/(v α – b)) – (a/(T 1/2 b)) ln((v β (v α + b))/(v α (v β + b))) = F = 0. (119)

Eq. (119) is satisfied at P = P sat .

ee. Example 31 Determine the saturation pressure for water at 593 K. Solution We will use an iteration method. First, we assume that P sat = 150 bar, a = 142.64 bar m 6 kmole –2 K –1

, b = 0.0211 m 3 kmole –1 . Using the RK state equation, v α = 0.0418 m 3 kmole –1 ,v β = 0.174 m 3 kmole –1 . Therefore, Eq. (119) implies that F = 3.0123.

Now, we will assume that P sat = 133 bars, so that v α

3 = 0.0427 m –1 kmole ,v β = 0.237

3 m –1 kmole , and F = 0.238 which is almost zero. The saturation pressure can be identified in this manner.

Upon normalizing Eq.(119) we obtain the relation

T=Const

Figure 16: Maxwell equal area rule.

(T 1/2 R v R ´ β /(v R ´ β – 0.08664) – 0.4275/(T R (v R ´ β + 0.08664))) – (T 1/2

R v R ´ α /(v R ´ α –0.08664) – 0.4275/(T R (v R ´ α +0.08664))) =T R ln((v R ´ β – 0.08664)/(v α R ´ – 0.08664)) – (4.9342/(T 1/2 R )) ln((v R ´ β (v R ´ α + 0.08664))/(v R ´ α (v R ´ β + 0.08664))).

This equation can be applied to determine the behavior of P sat R with respect to T R for RK flu- ids.

b. Approximate Solution The RK relation suggests that (cf. Chapter 6)

R /dT R ) v ′ R = 1/(v R ´ – 0.08664) + 0.2138/(T R v R ´(v R ´ + 0.08664)). (120) Along the critical isometric curve v´ R,C = 0.3333, so that

(dP

(121) At the critical point ,T R = 1, and (dP R /dT R ) c = 5.582.

(dP /dT R )

= ′v RC , = 4.054 + 1.528/T R .

The behavior of many substances can be described by the relation ln P R = A – B/T R .

Differentiating this relation, d ln(P R )/d(1/T R ) = – B. At the critical point P R =T R = 1, so that A= B= 5.582. Therefore,

ln P R = 5.582(1 – 1/T R ). (122)

1.00E 00

1.00E-01

1.00E-02

Approximate,RK

1.00E-03

Exact,RK

1.00E-04

1.00E-05