APPLICATION TO BOILING AND CONDENSATION

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.

j. Example 10 Water is contained in a piston–cylinder–weight assembly that is immersed in a con- stant–temperature bath at 593 K. Assume that the fluid obeys the RK equation of state irrespective of its phase, i.e.,

P = RT/( ¯v – ¯b )– ¯a /(T 1/2 ¯v ( ¯v + ¯b ))

(A)

Obtain a plot for P vs. ¯v ; Using the relation dg T = v dP, obtain a plot for ¯g vs.

¯v . Solution

This problem was solved in Chapter 7. A brief summary is presented. Using the values ¯a = 142.64 bar m 3

0.5 K –1 kmole , ¯b

3 Figure 10: Variation of = 0.02110 m ∆G with mole fraction in a binary mix- k mole –2 , and T =

ture.

593 K in Eq. (A),

a plot of P vs. ¯v is readily obtained as shown in Figure 11 . The first term in Eq. (A) occurs due to the collisions of high velocity molecules, while the second term appears due to attractive forces that result in a pressure reduction. As the fluid is compressed from state B, the pressure increases along the path BECGKN. If the fluid at state N is compressed fur- ther, it instantaneously condenses into a liquid state L. Similarly, the fluid that is ini- tially at state L can be expanded to a low pressure along the path LFARM. If the fluid at M is expanded further, it vaporizes instantaneously. Since d g_ T = ¯v dP, it is possible to integrate Eq. (A) between the limits v and v ref ,

i.e., ∫d ¯g T = ¯g (T, ¯v ) – ¯g (T, ¯v ref )= ∫ ¯v dP, to obtain

¯g (T, ¯v ) – ¯g (T, ¯v )=P ¯v – P ref ref ¯v ref –( R T ln(( ¯v – ¯b )/( ¯v ref – ¯b )) –

ref )( ¯v ref – ¯b )/( ¯v – ¯b )). We will now arbitrarily set ¯v

¯a/( 1/2 ¯b T )) ln (( ( ¯v / ¯v

= 4.83 m 3 kmole ref –1 (so that P ref = 10 bar at 593 K) and ¯g ref = 0. This enables us to produce a plot of ¯g vs. ¯v (cf. Figure 11 ). Using the same values of v, one can obtain g vs. P as shown in Figure 12 .

Remarks The fluid is in a saturation state at states G and F at which the vapor and liquid coex-

ist. The Gibbs free energy for both phases is equal. (The saturation pressure according 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 lowest

value of ¯g at specified values of T and P (cf. Figure 12 ) indicates that P < P sat = 133 bar, i.e., the vapor has a lower free energy when it is compared to the liquid curve QRAF. 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 finite constraint). 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.

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.