Closed System (Non–Flow Systems)

6. Closed System (Non–Flow Systems)

In this section we will further illustrate the use of the availability balance equation Eq. (47), particularly the boundary volume changes resulting in deformation work ( Figure 11 ).

a. Multiple Reservoirs For closed systems, ˙ m i = m ˙ e = 0, and the work W ˙ cv = W ˙ shaft + W ˙ u + P dV o cyl / dt . For a closed system containing multiple thermal energy reservoirs, the balance equation assumes the

form

0 cv )/ dt = ∑ j = 1 Q ˙ Rj , ( 1 − T 0 / T Rj , ) − W ˙ cv − I .

(61) If this relation is applied to an automobile piston–cylinder assembly with negligible shaft work

dE ( − TS

cv

( δW shaft =0) and with the inlet and exhaust valves closed, the useful optimum work delivered to the wheels over a period of time dt is δW u. The work

δ W u =− ( dE + T dS 0 ) − P dV 0 +∑ δ Q R ( 1 − T 0 / T Rj , ) w − δ I , i.e.,

W u =− ( ∆ E + TS 0 ∆ ) − PV 0 ∆ +∑ Q R ( 1 − T 0 / T Rj , ) − I (63)

where ∆E=E 2 -E 1 , ∆S=S 2 -S 1 , ∆V= V 2 -V 1 . Dividing the above relation by the mass m,

w u =− ( ∆ e + Ts 0 ∆ ) − Pv 0 ∆ +∑ q R ( 1 − T 0 / T Rj , ) − i .

b. Interaction with the Ambient Only With values for q R = 0, i = 0, and e = u, Eq. (64) simplifies as

w u,opt = φ 1 – φ. (27) When φ 2 = φ 0 ,

w u,opt,0 = φ 1 ′ = φ 1 – φ 0 .

The term φ´ is called closed system exergy or closed system relative availability. Consider the cooling of coffee in a room, which is a spontaneous process (i.e., those that occur without out-

side intervention). The availability is completely destroyed during such a process that brings the system and its ambient to a dead state. Thus, w u = 0 and i = w u,opt,0 = φ 1 – φ 0 .

c. Mixtures If a mixture is involved, Eq. (63) is generalized as,

W u =− ( Σ ( Ne k , 2 ˆ k , 2 − Ne k , 1 ˆ) k , 1 + T 0 ( Σ Ns k , 2 ˆ k , 2 − Ns k , 1 ˆ) k , 1 )

− P 0 Σ ( Nv k , 2 ˆ k , 2 − Nv k , 1 ˆ) k , 1 +∑ Q R ( 1 − T 0 / T Rj , ) − I

where, typically, ˆ e ≈≈ e u and s , ,ˆ k = s o k − R ln p k / P ref for a mixture of ideal gases and P ref =1 bar

f. Example 6 This example illustrates the interaction of a closed system with its ambient. A closed tank contains 100 kg of hot liquid water at a temperature T 1 = 600 K. A heat engine transfers heat from the water to its environment that exists at a uniform temperature T 0 = 300 K. Consequently, the water temperature changes from T 1 to T 0 over a finite time period. What is the maximum possible (optimum) work output from the engine? The specific heat of the water c = 4.184 kJ kg –1 K –1 .

Solution Consider the combined closed system to consist of both the hot water and the heat en- gine. Since there are no thermal energy reservoirs within the system and, for optimum work, I = 0,

dE ( cv − TS 0 cv )/ dt = W ˙ cv opt , , or (A)

R,1 T

R,1 Q

Figure 11. Application of the availability balance for a piston-cylinder assembly.

W cv,opt = (E

– (E

cv –T 0 S cv ) 1 cv –T 0 S cv ) 2 , where

(B) (E cv ) 1 =U 1 =mcT 1 , (E cv ) 2 =U 2 =mcT 2 , and (S cv ) 1 – (S cv ) 2 = m c ln(T 1 /T 0 ). (C)

Substituting Eq. (C) into Eq. (B), we obtain

W cv,opt = m c (T 1 –T 0 )–T 0 m c ln(T 1 /T 0 )=

100 × 4.184 × (600 – 300 – 300 × ln (600/300)) = 38520 kJ. Remarks

If only the heat engine is considered to be part of the system, it interacts with both the hot water and the ambient. In this case the hot water is a variable–temperature thermal energy reservoir. Since the heat engine and, therefore, the system, is a cyclical device, there is no energy accumulation within it. Therefore, for an infinitesimal time period

(D) where the hot water temperature T R,w decreases as it loses heat. Applying the First and

δQ R,w (1–T 0 /T R,w )= δW cv,opt ,

Second laws to the variable–temperature thermal energy reservoir, δQ R,w = –dU R,w =– m w c w dT R,w and δQ R,w /T R,w = dS R,w . Using these relations in the context of Eq. (D) we obtain the same answer as before.