Closed System

1. Closed System

We now discuss thermo–mechanical–chemical equilibrium. In the following we pre- sent a methodology of achieving TMC equilibrium followed by brief derivation. Consider a cylinder containing an ideal gas mixture consisting of 40% N 2 and 60% O 2 by volume which has expanded from initial state to the final restricted dead state ( Figure 16 ), i.e., in thermo- mechanical equilibrium with the ambient. Once in the TM state the cylinder can be divided into two chambers A and B by a partition and constrained by two pistons placed on either side of the partition that can move independently, as shown in Figure 16 . The partial pressures of oxygen and nitrogen in the cylinder are p O2,0 = 0.6 and p N2,0 = 0.4 bars, respectively, while the corresponding ambient pressures are p N2, ∞ = 0.79 and p O2, ∞ = 0.21 bars. Next, a semi-

permeable membrane that is only permeable to O 2 replaces the partition. One piston, say A, is then moved so as to decrease the pressure in chamber A without altering its temperature. Therefore, the partial pressure of O 2 in chamber A decreases below its corresponding value in chamber B and, consequently, oxygen molecules migrate across the membrane from B to A (Chapter 1 and Chapter 3). By maneuvering the piston to achieve very low pressure, virtually

all of the oxygen can be transferred from chamber B into A. Next, the O 2 –permeable mem- brane can be replaced with one permeable to nitrogen and all N 2 molecules can be transferred from A into chamber B. By so manipulating the two pistons and the semipermeable mem- branes, the two components can be separated so that chamber A consists of only O 2 and cham- ber B consists of only N 2 Once the separation process is completed semipermeable membrane is replaced with

a partition impermeable to either of the species, the pistons A and B can be moved so that p N2,B

Semipermeable Membrane

p O2 = 0.21

p O2 = 0.6

p N2 = 0.4

p N2 = 0.79

Piston A

Piston B

a) Initial State

d) TM State with P,T,V

b) TM State

c) TM State with

0 P ,T 0 ,V 0 Semipermeable

Semipermeable

membrane

membrane P k =P k , ∞ ∞ ∞ ∞

Figure 16: Illustration of a device that may be used to achieve thermo–mechanical–chemical equilibrium of a system with its environment. a) Gas mixture at initial state; b): gas mixture at dead state; c) partition with semi-permeable membrane with two separate pistons; d) separation of components and adjustments to partial pressures of ambient.

=p N2, ∞ and p O2,A =p O2, ∞ ( Figure 16d ), i.e., same as the partial pressures of the two species in the environment. Note that T 0 = T ∞ in this section of the chapter. Isothermal work called chemical work W ch,02 is obtained from chamber A, since p O2, ∞ <p O2,0 (the initial partial pressure of oxygen), but compression work must be performed on chamber B for which p N2, ∞ >p N2,0 .

The pistons A and B can now be removed and replaced with rigid semipermeable membrane pistons that are, respectively, permeable to O 2 and N 2 . There can not be transfer of species across this membrane since partial pressures are the same. In this manner chemical equilibrium will be achieved in the combined isolated system consisting of the chambers A and B and the ambient.

Thus, the work can be obtained through a two–step process consisting of (1) the work obtained during expansion from the initial state to that in the thermo–mechanical equilibrium W u,0 , and (2) the chemical work obtained in proceeding from the thermo–mechanical state to

the thermo–mechanical–chemical equilibrium state W ch .

Therefore, for a closed system in TMC equilibrium,

W u,opt, ∞ ∞ ∞ ∞ =W u,opt,0 +W ch , where

(85) W ch = (U 0 –U ∞ ) –T 0 (S 0 –S ∞ )+P 0 (V 0 –V ∞ ) = (H 0 –H ∞ )–T 0 (S 0 –S ∞ ) == G 0 –G ∞ .

(86) Where properties U ∞ , S ∞, V ∞ are determined for O 2 and N 2 contained within the sections A and

B at T 0 =T ∞, p N2, ∞ and p O2, ∞ . The work w u,opt,0 can be obtained using Eq. (28). On a unit mass basis,

w ch = (h 0 –h ∞ )–T 0 (s 0 –s ∞ )=g 0 –g ∞ .

It is seen that the chemical work per unit mass is also the same as the difference in stream availability between the restricted dead state, i.e., TM to TMC equilibrium state (see next sec- It is seen that the chemical work per unit mass is also the same as the difference in stream availability between the restricted dead state, i.e., TM to TMC equilibrium state (see next sec-

gTp k (, 0 k ) = hT k () 0 − TsT 0 0 (() − R ln( p k /) 1 = gTP k (, 0 0 ) + RT 0 ln( p k /) 1 .

Therefore,

(87) and writing in terms of g

w ch =∑ k X k φ k , 0 ( Tp 0 , k , 0 ) −∑ X k φ k , ∞ ( Tp 0 , k , ∞ ) .

(88) w ch = φ 0 − φ ∞ , where

φ 0 =ΣXg k , 0 k , 0 (, TP 0 0 ), X k =X k,0 =X k,e , and

(91) k. Example 11

φ ∞ =ΣXg k , 0 k , ∞ ( TP ∞ , ∞ ) .

Determine the maximum work that can be performed if a gas mixture consisting of 40% O 2 and 60% N 2 is expanded from 2000 K and 60 bars to the dead state at which it is at thermo–mechanical–chemical equilibrium.

Solution The maximum work

w max, ∞ = w max, 0 + w ch , where

(A)

w max,0 = φ – φ 0 .

Now, u – T

0 s +P 0 v , where v = 0.08314 × 2000 ÷ 60 = 2.77 m kmole . The specific entropy, s N = s 0 2 N 2 – R ln (p N2 /1) = 251.6 – 8.314 (ln(0.4 ×60)/1) = 225.2 kJ kmole –1 K –1 , and s O 2 = s 0 O 2 – R ln (P O2 /1) = 268.7 – 8.314 (ln(0.6 ×60)/1) = 238.9 kJ

kmole –1 K –1 . The mixture initial specific entropy and internal energy are s

) = 233.4 9 kJ kmole –1 K = (0.4 s –1 N 2 s O 2 , and

u = (0.4 × 48,181+ 0.6 × 51,253) = 50,024 kJ kmole –1 . Therefore,

φ = 50,024 – 298 × 233.4 + 100 × 2.77 = –19252.2 kJ kmole –1 ,

u –1 × 6190 + 0.6 × 6203 = 6197.8 kJ kmole , s 0 = 4(191.52–8.314 ×ln (0.4/1))+0.6(205.0–8.314

×ln (0.6/1))=205.2 kJ kmole –1 K –1 ,

0 = 0.08314 × 298/1 = 24.78 m kmole , φ –1

3 v –1

0 = 6197.8 – 298 × 205.2 + 100 × 24.78 = –52473.8 kJ kmole . Therefore, w

= –19252.2 – (–52473.8) = 33,221.6 kJ kmole max,0 –1 . Since

w ch = φ 0 – φ ∞ and φ ∞ = ΣX k φ k, ∞

φ O 2 , ∞ (T 0 ,P O2 , ∞ )= u O 2 (T 0 ,p O2 , ∞ )–T 0 s O 2 (T 0 ,p O2 , ∞ )+P 0 v O 2 (T 0 ,p O2 , ∞ )

= 6203 – 298 × (205.0 – 8.314 × ln (0.21/1)) + 100 × (0.08314 × 298/0.21)

= –46981 kJ kmole –1 .

Likewise,

φ N 2 , ∞ (T 0 ,p N2 , ∞ )

= 6190 – 298 × (191.5 – 8.314 × ln (0.79/1)) + 100 × (0.8314 × 298/0.79)

× 193.5 + 100 × 31.4 = –48,333 kJ kmole –1 . Therefore,

× (–46981) + 0.4 × (–48333) = –47521.8 kJ kmole –1 , and

w = –52473.8 – (–47521.8) = –4952 kJ kmole –1 ch .

w u, max, ∞ = 33,221.6+ (-4952) = 28,269.6 kJ/mole Remarks

The chemical work per kmole of mixture is negative, since a larger work input is nec- essary to compress 0.4 kmoles of N 2 from 0.4 bars to 0.79 bars as compared to the work output obtained from the expansion of 0.6 kmoles of O 2 from 0.6 bar to 0.21 bar. It will be shown later that the expression for w ch for an open system is similar to that for a closed system. Several cases involving water, air–vapor mixtures, and product gas mix- tures will be dealt with later in the context of open systems.