Mixing Rules for Equations of State

1. Mixing Rules for Equations of State

The simplest model is the ideal solution model. The specific volume of species k within the mixture is assumed to be same as the specific volume of pure species k, i.e.,

ˆv k = v k .

a. General Rule The equation of state for a mixture can be obtained by implementing mixing rules (e.g., in order to the VW equation of state to a mixture, the constants a and can be determined by applying these rules). It is possible to formulate various mixing rules, i.e.,

β=Σ k X k β k , (127a) 1/2 β = (Σ 2

k X k β k ) , (127b) 1/3 β = (Σ 3

k X k β k ) , (127c)

(127d) β=Σ k X j X k β kj ,

k X k β k + (3/4) ( Σ k X k αβ k )( Σ k X k β k ), or

(127e) where β k represents appropriate constant in a state equation. In Kay’s rule to be discussed later,

β k could represent either T c,k or P c,k in Eq. (127a). Consider the Clausius–I Equation of State for a pure substance (Chapter 6). The same equation can be applied by replacing “b” with “b m ”, and

P= R T/( v – b m ),

where “b m ” can be defined using one of the mixing rules (Eqs. (127a) to (127e)) with b m = β, and b k = β k .

l. Example 14

A piston–cylinder assembly contains two sections. Section A contains 2 kmole of acetylene, while section B contains 1 kmole of CO 2 . The body (or collisional) volume

3 of CO –1

2 and acetylene are, respectively, 0.043 and 0.0326 m kmole . Determine the volumes of both gases in their pure states at 100 bar and 320 K using the state equa- tion

P= R T/( v – b ),

(A) where b denotes the body volume. What is the total volume V A +V B under those

conditions? If the partition between the two sections is removed and the gases al- lowed to mix, what is the partial molal volume of the two gases if they are maintained at 100 bar and 320 K? What is the total volume at this state?

Solution For a pure fluid,

v = R T/P + b .

Therefore, for CO 2 ,

CO 2 = 0.08314 × 320 ÷ 100 + 0.043 = 0.306 m kmole –1 , and for acetylene, v

CH 22 = 0.08314

× 320 ÷ 100 + 0.036 = 0.302 m –1 kmole .

The total volume

A +V B = 0.302 × 2 + 0.306 = 0.910 m . For the mixture

v = R T/P+ b m , i.e., V = N R T/P+N b m = (N 1 +N 2 +...) R T/P + (N 1 b 1 +N 2 b 2 +...), and

ˆv i = (dV/dN i ) T, P, N N N 1 , 2 ,..., ji ≠ ,..., N K = R T/P + b i .

The partial volumes of the gases in the mixture are

3 ˆv –1

CO 2 = (0.08314 × 320 ÷ 100) + 0.043 = 0.306 m kmole , and

ˆv

= 0.198 m CH 3 22 kmole –1 .

We find that the partial molal volumes are the same as the pure volumes, since the mixture is an ideal mixture in context of the given state equation (B). Since

V= ˆv CO 2 N CO 2 + ˆv CH 22 N CH 22 , then

3 V=2 id × 0.302+ 1 × 0.306 = 0.91 m =V .

b. Kay’s Rule Kay’s Rule is based on a pseudo–critical temperature T cm and a like pressure P cm for a gas mixture, while adopting the same equation of state as that used for the pure components of the mixture. The pseudo–critical temperature for a mixture is obtained by applying a linear mixing rule (cf. Eq. (127a)). The pseudo–critical properties

(128) The assumption is that the real gas state equations for pure components are also valid for mix-

T cm =X 1 T c1 +X 2 T c2 + ..., and P cm =X 1 T c1 +X 2 T c2 +….

tures, but with these pseudo–critical properties. In the context of the RK state equation

(129) The constants a m and b m are evaluated by using the pseudo–critical properties, i.e.,

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

(130) Other state equations can be similarly applied.

2 a 2.5

m = (0.4275 R T cm /P cm , and b m = 0.08664 R T c /P c .

m. Example 15

3 A 2 m rigid tank contains a mixture of 40% carbon dioxide and 60% acetylene by volume at 320 K and 100 bar. Determine N CO 2 , and P CO 2 using Kay’s rule and ap-

plying both the RK equation and the compressibility charts. Solution

The pseudo–critical temperature and pressure are, respectively, T cm =X CH 22 T c, CH 22 +X CO 2 T c, CO 2 = 0.60 × 308.3 + 0.4 × 304 = 307 K, and

(A) P c =X CH 22 P c, CH 22 +X CO 2 P c, CO 2 = 0.6 × 61.4 + 0.4 × 73.9 = 67 bar.

(B) Therefore, in the context of the RK state equation

P= 1/2 R T/( v – b

m )– a m /(T v ( v + b )),

(C)

b m = 0.08664 R T cm /P = 0.08664

3 kmole –1

cm

×0.08314×307÷67 = 0.033 m , and (D) ×0.08314×307÷67 = 0.033 m , and (D)

2 2.5 6 0.5 –2 m

a =0.4275 R 2 T 2.5 /P =0.4275

cm

with the highest value, v = 0.08655 m 3 kmole –1 , i.e., N = V/ v =2 ÷ 0.08655 = 23.2 kmole.

Therefore, N CO 2 =X CO 2 × 23.2 = 0.4 × 23.2 = 9.28 kmole. Now, consider the state equation

PV = NZ R T, (F) where Z is obtained from the compressibility charts using the reduced properties P R =

P/P cm = 100 ÷ 67 = 1.49, and T R = T/T cm = 320 ÷ 308 = 1.04. At these conditions, from the charts, Z = 0.32. Using this result in Eq. (F), 100.0 × 2 = N × 0.32 × 0.08314 × 320, i.e., N = 23.6 kmole, and

N CO 2 = 0.4 × 23.6 = 9.4 kmole. n. Example 16 3

A2m rigid tank contains a mixture of 40% carbon dioxide and 60% acetylene by volume at 320 K and 100 bar. Use Kay’s rule and the RK equation to determine the

internal energy and enthalpy of the mixture. Assume that c p,o, CO 2 and c p,o, CH 22 remain constant and select values at 300 K. In order to determine u and h of real gas mixture,

assume that each component exists in the ideal gas state and that the enthalpy is

zero–valued at 0 K. Furthermore, c –1 v,o, CH 22 = 36.12 kJ kmole K (or see Tables A- 6C )

Solution Treating the mixture as an ideal gas, its internal energy

(A) where u ko = c vk,o T. Therefore,

u .o (T) = X CO 2 u o, CO 2 (T) + X CH 22 u o, CH 22 (T),

u –1

= c T = 36.12

o, CH 22 v,o, CH 22 × 320 = 11559 kJ kmole .

× 44.01× 320= 28.91 × 320 = 9253 kJ kmole –1 . u –1

u o, CO 2 = c v,o, CO 2 T = 0.657

o = 0.6 ×11559 + 0.4× 9253 = 10,637 kJ kmole of mixture. The mixture behaves like a pure component with a pseudo-critical temperature and

pressure. We will assume that the relations derived for pure components are valid for mixtures, i.e., ( u o – u )/ R T cm = 7.4013/T 1/2 R ln (1 + 0.08664/v R ´). Recall that

c ´ = 0.08314 × 307 ÷ 67 = 0.381 m kmole . Therefore, v R ´ = 0.08655 ÷ 0.381 = 0.227, and T R = 320 ÷ 307 = 1.042, and ( 1/2 u

3 v –1 = 0.08655 m kmole , and v

o – u )/ R T cm = 7.4013 ÷ (1.042) ln(1 + 0.08664 ÷ 0.227) = 2.344, or ( –1 u

o – u ) = 2.344 R T cm = 2.344 × 8.314 × 307 = 5983 kJ kmole , i.e., u –1 = 10637 – 5983 = 4654 kJ kmole .

Similarly,

h = u +P v = 4654 + 67

× 100 × 0.08655 = 5234 kJ kmole –1 .

c. Empirical Mixing Rules Various other mixing rules are available. For instance, we can use a square rule (cf.

Eq. (127b)) to determine a , and a linear rule (cf. Eq. (127a)) for b .

a m 1/2 =( 2 Σ k X k a k )

b m = Σ k X k b k (132) Therefore, the RK equation assumes the form

3 Z 2 –Z +( a

m ∗ – b m ∗2 – b m ∗ )Z– a m ∗ b m ∗ = 0, where

a m ∗ =( Σ ∗1/2

(134a, b)

(134c,d) P R,k =P/P c,k , and T R,k =T/T c,k .

k =0.4275P R,k /T

R,k

, b k =0.08664P R,k /T R,k ,

Once a m ∗ and a m ∗ are defined, one can derive pseudo–critical temperature T cm and pressure P cm at which Eqs. (134a) and (134b) are satisfied, i.e.,

a m ∗ = 0.4275 R 2 T 2.5 cm /P cm , and b m ∗ = 0.08664 R T cm /P cm .

Using Eqs (134c) and (134d) in Eqs. (134a) and (134b), we obtain the relations

T = (T 5/2 /P 1/2 ) cm 2 ΣX k ck ck / ΣX k (T ck /P ck ), and P cm =T cm / ΣX k (T ck /P ck ).

Another mixing rule for the RK equation involves using Eq. (127e) for a and Eq. (127a) for b , i.e., .

a m = Σ i Σ j X i X j a ij , b m = Σ i X i b i , where

2 a 2.5 ij = 0.42748 R T c,ij /P c,ij ,P c,ij =Z c,ij R T c,ij / v c,ij , (136)

Z 3 c,ij = (Z c,i +Z c,i )/2, and v c,ij = (( v c,i + v c,j )/2) . (137)

d. Peng Robinson Equation of State The Peng Robinson state equation can be written in the form

P= R T/( v – b m )– a m /( v 2 +2 b m v – b m 2 ), where

a 0.5

ij = (1– δ ij )( a i a j ) ,i ≠j, a ii = a i ,

and the mixing rule of Eqs (127a) and (127e) apply. i.e., a m = Σ k X j X k a kj and b m = Σ k X k b k .

e. Martin Hou Equation of State The Martin Hou state equation has the form

(139) (–T/ T c F )

P= av R T/( v – b )– ∑

i = 2 F i /( v – b ) +F 6 (T)/e , where

i (T) = A i +B i T+C i e .

Z = 1 + BP/ R T, where (141) B=

(142) ω ij =( ω i + ω j )/2, and T c,ij =(T c,i T c,i ) 1/2 (1 – k ij ).

i Σ j i j ij ,B ij =( R T c,ij /P c,ij )(B + ω ij B ),

(143) The parameter k ij ≈ 0 for most cases and is zero for a pure component. o. Example 17

The compressibility factor Z can be determined using the following relation Z = 1 + (BP/( R T)), where

B ij = (RT c,ij /P c,ij )B o , (C)

(D) T c,ij = (T c,i T c,j ) (1/2) (1 – k ij ),

B 1.6 = 0.083 – 0.422/T

(E) P cij =Z c,ij R T cij /v c,ij ,

(F) Z c,ij = (Z c,i +Z c,j )/2, and

(G) v

= (v c,i +v c,j )/2. (H) In a binary mixture, show that the partial molal volume of component 1 is

ˆv 1 =( ∂V/∂N 1 ) T,P, N 2 = R T/P + b´, where

(I)

(J) Determine the partial molal volume of component 1 in a binary mixture.

b´ = d(NB)/dN 1 =B 11 (1 – X 1 )X 2 (2B 12 –B 11 –B 22 ).

Solution The total volume

V = NZ R T/P = (1 + BP/ R T) N R T/P = N R T/P + NB. The partial molal volume ˆv 1 = ∂V/∂N 1 = R T/P + b´, where

b´ = ∂(NB)/∂N 1 . Using Eq. (B),

b´ = ( ∂/∂N 1 ) (N(X 1 X 1 B 11 +X 1 X 2 B 12 +X 2 X 1 B 21 +X 2 X 2 B 22 )). We assume that B 21 =B 12 so that ∂(NB)/∂N 1 =B 11 (1 – X 1 )X 2 (2B 12 –B 11 –B 22 ), i.e., ˆv 1 = (dV/dN 1 ) T,P, N 2 = R T/P + b´.

The specific volume of the pure component is

v 1 = R T/P + B 11 , i.e., ˆv 1 – v 1 = (1 – X 1 )X 2 (2B 12 –B 11 –B 22 ).