Multicomponent Mixtures

2. Multicomponent Mixtures

The spinodal analyses can also be applied to mixtures. n. Example 14

Consider a mixture that contains 60% water (species 1) and 40% methyl alcohol (spe- cies 2). Determine the spinodal curves for the mixture as a function of pressure and temperature. Assume that the mixture follows the RK equation of state.

Solution The stability of two components require that

D 1 =A VV ≥ 0, and (A)

A VV A VN 1

D 2 = ≥ 0 (B)

A NV 1 A NN 11

Selecting the mixing rule for the RK equation of state

a m =( ΣY k 1/2 ¯a 2 k ) , so that

(C)

(D) The pseudo critical temperature and pressure can be expressed as

m =a m N , ¯b m = ΣY k ¯b k , and B m =b m N.

T c ´ = (0.08664/(0.4275 R)(a m /b m )) 2/3 , and P c ´ = 0.08664 RT c ´/b m , i.e.,

a m = (0.6 × 142.6 0.5 + 0.4 × 220 1/2 ) 2 = 171.6 bar k 1/2 m 6 kmole -2 ,

b = (0.6 × 0.0211 + 0.4 × 0.0462) = 0.03115 m 3 kmole m -1 , and

T c ´ = 564.9 K and P c ´ = 130.6 bar. Using the condition A VV = 0 we determine that ∂P/∂V = 0. In Figure 20 the vapor spi- nodal points are represented by the curve VGC and the liquid spinodal points by the

curve CFL. However, this condition alone does not satisfy the stability criteria for a mixture. The following additional spinodal condition must be satisfied, i.e.,

(E) Since

A V N 1 = ∂/∂N 1 (A V )=– ∂P/∂N 1 .

(F) where B m = Nb ,A

P = NRT/(V–B 1/2

m )–A m /(T V(V+B m )),

m =N a m , ∂B m / ∂N 1 =b 1 , ∂A m / ∂N 1 = 2 (a

1 a) (cf. Chapter 8), ∂P/∂N 1 = RT/(V–B m ) + NRT ( ∂B m / ∂N 1 )/(V–B m ) 2 –( ∂A m / ∂N )/(T 1/2 1 V(V+B m ))

(G) Substituting Eq. (G) in Eq. (E), and multiplying the resultant expression by N, we

m ( ∂B +A m / ∂N )/(T 1/2 1 V(V+B m ) 2 )

obtain the relation NA 2

V N 1 = –RT/(v – b m ) – RTb 1 /(v – b m )

Z P, bars

Spin-Liquid Sat

Figure 20: Spinodal curves for a binary mixture.

(H) An additional spinodal condition for a mixture is obtained by using the following

equality, i.e.,

A V N 1 = 0, since N > 0. (I) Substituting Eq. (I) in Eq. (H), we obtain a relation for the temperature, i.e.,

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

m ) – ab 1 )) . (J) Thereafter, using Eq. (F) the pressure at which A V N 1 = 0 is obtained (cf. Figure 20 -

curve MNC´ for the liquid and C´HB for vapor). The pressure along which A VV = 0 is represented by the curves LFC (for the liquid) and CGV (for the vapor). Figure 20 also illustrates the bubble point J along the saturation curve YJC at 50 bar for T sat = 475 K.

We now discuss the criteria. When A VV = 0, the mixture can be superheated to 520 K (point F) without a bubble forming, and a vapor mixture can be subcooled to 430 K (point G) without condensation. The presumption is that any minor disturbance within the system occurs due to volumetric changes alone (i.e., there is a uniform composi- tion within the disturbed space). However, if the composition is also locally nonuni-

form due to a disturbance (i.e., due to fluctuations in N 1 ) then the spinodal condition corresponds to A V N 1 = 0. Accordingly, the curve BHC´RNM is the spinodal vapor

curve.

Within HG, 2 2 2 ∂ 2 A/ ∂N 1 > 0 but ∂ A/ ∂V < 0. If a disturbance occurs due to changes in both V and N 1 , then the conditions 2 A/ 2 2 ∂ 2 ∂V > 0 (i.e., ∂P/∂V < 0) and ∂ A/ ∂N 1 > 0 re- quire that the system lie within GF at 50 bar.