Single Component

1. Single Component

For the branches ECGKN and MRAFL in Figure 15 , ∂P/∂v < 0 and (d 2 A) T,V,m > 0. Along the branch NDHJM, 2 ∂P/∂v > 0 and (d A)

< 0. The points M and N at which ( ∂P/∂v) T = 0, are the limits of intrinsic stability and are called “spinodal points”. At the spi- nodal points M and N, ∂P/∂v = 0 and (d 2 A) T,V,m = 0. At a specified temperature, the spinodal points yield the maximum pressure at which the fluid exists as vapor (cf. state N) and the

T,V,m

minimum pressure at which it exists as liquid (state M). The spinodal points can be predicted by applying the state equations as illustrated through the following example.

l. Example 12 Use the relation

P = RT/(v–b) – a/(T 1/2 v(v+b)) (A) To obtain an expression for P vs. v and T vs. v for water along the spinodal points plot

P vs. v at 567, 593 and 615 K for water. Obtain an expression for the reduced values P R and T R vs. v R ' along the spinodal points, and plot P R and T R vs. v R ' and P R vs. T R along these points.

Solution At the spinodal points ∂P/∂v = 0. Differentiating Eq. (A),

( ∂P/∂v = – RT/(v–b) 2 + (a/(T 1/2 v(v+b))) (1/v + 1/(v+b))) = 0. Thereafter, solving for the temperature,

(B) Using Eqs. (A) and (B) we obtain the relation

T = ((a(v–b) 2 /(R v(v+b))) (1/v + 1/(v+b))) 2/3.

P = (R(v–b)) 1/3 ((a/(v(v+b))) (1/v + 1/(v+b))) 2/3

– (a/(v(v+b))) 2/3 (R/((v–b) 2 (1/v + 1/(v+b)))) 1/3

(C) With the values a = 142.59 bar m 6 K 1/2 kmole -2 and ¯b = 0.0211 m 3 kmole -1 in the

context of Eq. (B), we obtain a plot of T vs. ¯v , and using Eq. (C) we can obtain a plot of P vs. ¯v . Using Eq. (C), one can also obtain the spinodal pressure vs. volume for water at the temperatures 567, 593 and 615 K (cf.). We can normalize Equations (B) and (C) so that

R '(v R '+0.08664))) (1/v R '+1/(v R '+0.08664))) , and (D) P 1/2

T R =((0.4275(v R '–0.08664) 2 /(v

R =T R /(v R ' – 0.08664) – 0.4275/(T R v R ' (v R ' + 0.08664)) (E) Figure 16 presents a plot of P R vs. v R ' and T R vs. v R ' and Figure 17 contains a plot of

P R vs. T R along the spinodal points. The curves CGV and CFL denote the vapor and liquid spinodal curves. Figure 18 presents plots of Z vs P R with T R as a parameter for nonpolar fluids for the spinodal and saturation conditions at any given state.

Remarks Along the spinodal points on the curves CGV and CFL, c > 0. Therefore, ( 2 U/ v 2 ∂ ∂S )=

(T/mc v ) ≠ 0, but the condition (∂ 2 U/ ∂ 2 V) S,m =( ∂P/∂V) T = 0 is satisfied. As the critical temperature T c is approached, the vapor and liquid spinodal volumes

merge into a single value v c so that d 2 A = 0 without exhibiting a maxima or a minima.

JT = 0 0.6 Inversion

P 0.5

0.3 0.2 V Spin-Liquid 0.1

Figure 17: The spinodal, inversion, and saturation values of P R vs .T R for an RK fluid.

Figure 18. Z curves: spinodal, saturation and other isotherms. (From W. G. Don- gand, J. H. Lienhard, Can. J. Chem.Eng., 64, pp. 158-161, 1986. With permis- sion.)

Since ∂P/∂v = –(∂T/∂v) P ( ∂P/∂T) v , if ∂P/∂v = 0 (i.e., β T →∞), then ( ∂T/∂v) P = 0. Therefore, the coefficient of thermal expansion β P →∞ at the spinodal point (since, typically, ∂P/∂T ≠ 0). Applying the results from Chapter 7, we can plot the inversion pressure with respect

to the temperature (curve IQ in Figure 17 ). The regimes where µ JT > 0, µ JT < 0, and µ JT = 0 are indicated in the figure. The spinodal curves are useful to predict the degree of liquid superheat and vapor

subcooling of vapors at a specified pressure. Consider a fluid which has saturation temperature of T sat

1 = T at P 1 (cf. Figure 19 , 593 K at P =133 bar). Figure 19 plots three curves of pressure with respect to volume for the temperatures T 1 , T 2 , and T 3 . The temperatures T 2 and T 3 are selected such that the vapor spinodal pressure at N corresponding to T 2 coincides with the liquid spinodal pressure at M corresponding to T 3 . The points F and G represent the saturated liquid and vapor states at the condition (P

1 , T 1 = T ). The liquid at a pressure P 1 can be superheated to the temperature T 3 (i.e., from state F to M) without boiling and the vapor can be likewise subcooled to T 2 (i.e., from state G to N) without condensation. This phenomenon is illustrated through the following example.

sat

m. Example 13 Consider the RK equation of state

P = RT/(v–b) – a/(T 1/2 v(v+b)). (A) Determine the maximum temperature to which water can be superheated at 133 bar

without boiling and the temperature to which water vapor can be subcooled at the same pressure without condensation occurring. Assume that T sat = 593 K at P = 133 bar for water when it is modeled by the RK equation of state. What is the fluid state at 615.001ºC and 133 bar?

Solution

From Example 12, the temperatures at which ∂P/∂v = 0 are represented by the relation T = ((a (v–b) 2 /(R v(v+b))) (1/v + 1/(v+b))) 2/3 .

(B) There are two spinodal vapor and liquid volumes at particular temperature, e.g., the

states M and N at 133 bar in Figure 19 . Using (B) and the RK equation to eliminate the temperature, we obtain the expression

P = (R (v–b)) 1/3 ((a/(v(v+b)))(1/v + 1/(v+b))) 2/3

(C) Eq. (C) is useful for obtaining the vapor spinodal curves AN and liquid spinodal ON

– (a/(v(v+b))) 2/3 (R/((v–b) 2 (1/v + 1/(v+b)))) 1/3 .

(cf. Figure 19 ). Using the values R = 0.08314 bar m 3 kmole -1 K -1 , a = 142.59 bar m 6 K 1/2 kmole -2 , b

3 = 0.0211 m -1 kmole , P = 133 bar in Eq. (C), v M = 0.0597 m kmole (cf. Figure 19 , liquid–like spinodal point) and v = 0.158 m 3 kmole N -1 (vapor-like spinodal point). and

3 -1

3 v -1 N = 0.158 m kmole (vapor-like spinodal point). Thereafter, applying these results in Eq. (B), for the liquid T Sp,l = 615 K and for the vapor T Sp,v = 567 K. Therefore, water can be superheated at 133 bar by 615–593 = 22 K and the vapor subcooled by 567–593 = –26 K.

One can use Charts in Fig. 17 to predict the degree of superheating and sub- cooling at any given pressure. With P = 133 bars, P R = 0.602, then T R along spinodal liquid is 0.95 while along spinodal vapor is 0.88 while T R along saturated line is

0.92. The corresponding T’s are 614.5, 569, and 593 K. Sometimes the RK equation may be crude to determine the degree of superheating. Recall from Chapter 6 that a =c 3 R 2 T 2.5 c /P c , and b =c 4 R T c /P c where from theory c 3 = 0.4275, c 4 = 0.08664. If

A Vapor Spinodal Curv

P, bars

T 2 = 567

50 D Liquid Spinodal Curv

Liquid

v, m 3 /kmole

Figure 19: Spinodal curves at 567, 593, and 615 K for water mod- eled according to the RK equation of state. If the pressure is main- tained at 133 bar, T sat = 593 K according to the relation with vapor at state G and liquid at state F. The vapor can be subcooled to 567 K (cf. the spinodal point N) and the liquid superheated to 615 K (cf. spinodal point M) at the same value of P sat = 133 bar.

c 3 and c 4 are selected to be different to fit the experimental data for P,v,and T, it is equivalent to modifying T c and P c ’s (Chapter 6) . The charts in Figure 17 can still be used with modified critical properties (Chapter 6). Subcooling also occurs during ice formation and in sublimation processes. Pure distilled water can be cooled to –8ºC without forming ice.