Irreversibility or Lost Work

2. Irreversibility or Lost Work

During a reversible process (cf. Figure 2b ) for the same change in state as for an irre- versible process (cf. Figure 2 ) the work W = W u =W u,opt . The irreversibility I or lost work LW During a reversible process (cf. Figure 2b ) for the same change in state as for an irre- versible process (cf. Figure 2 ) the work W = W u =W u,opt . The irreversibility I or lost work LW

LW = I = (W opt – W), or (30) LW = I = (W u,opt – W).

(31) Applying the First and Second Laws to the system shown within the control surface cs2 in

Figure 2a ,W=Q 0 – (U 2 –U 1 ) and the entropy generation σ=S 2 –S 1 –Q 0 /T 0 . Eliminating Q 0 from these two relations ,

W = (U 2 –U 1 )–T 0 (S 2 –S 1 )–T 0 σ=W opt –T 0 σ, (32) which has a form similar to Eq. (5) with T b =T 0 . In the context of Eqs. (30) and (31) LW = I = W opt –W=W u,opt –W=T 0 σ. (33) Since σ > 0, LW > 0 and W opt ≥ W.

a. Comments For the reversible process Q opt,,0 –W opt = ∆U while for irreversible process ∆U = Q 0

– W. Therefore, W opt – W = LW = Q opt,,0 –Q 0 . Since LW > 0, Q opt,,0 >Q 0 . Due to our convention, the heat rejection carries a negative sign. For example, if Q ,0 = -100 kJ, Q opt,,0 = -50 kJ then Q opt,,0 >Q 0 criterion is satisfied. The implication is that an irre- versible process rejects a larger amount of heat and further raises the internal energy of the ambient A due to the irreversibility it must overcome. Recall that in Chapter 3 we have discussed the entropy maximum principle at condi- tions corresponding to fixed U iso ,V iso and m iso , and the energy minimum principle at fixed S iso ,V iso , and m iso . Now, consider the hot matter M as it interacts with the ambi-

ent A. If M is cooled by A, the combined entropy of M and A,S iso =S M+A =S+S 0 , in- creases at fixed U iso (= U + U 0 ), V iso (= V + V 0 ), and m iso =m+m 0 , and eventually reaches equilibrium at state 2U by travelling along the path 1–A–B–2U that is shown in Figure 4 . The point 1 represents the initial state of the composite system in a U M+A –S M+A –U coordinate system. If matter M delivers work and transfer heat to its ambient then the irreversible process is represented by 1–2. If, at state 2, the system is insulated and its piston is restrained from moving (i.e., a constrained equilibrium state is established), S M+A will increase, U M+A will decrease (since work is delivered), while

V M+A and m M+A will remain unchanged. If the process is continued until the “dead state” is reached (0,I) , U will have decreased at fixed V and m along the path 1–2–O I . Now let us consider the optimum process 1-2opt for which no entropy is generated during a totally reversible process of the composite system. Consequently, S M+A is unchanged while U M+A decreases with V M+A and m M+A fixed as depicted by the path 1–2 opt in Figure 4 . If a constraint is placed between M and A once the composite sys- tem reaches the state 2 opt , this, once again, results in a constrained equilibrium condi-

tion. The entropy change for the process 1–2 opt dS M+A = dS M + dS 0 = 0. Even if the state of M is identical at points 2 and 2 opt so that U M,2 > U M ,2 opt ,U M+A,2 > U MA + ,2 opt ,

since U A,2 > U A ,2 opt due to more heat rejection in the irreversible process. With S, V and m for the composite system being fixed, the maximum optimum work is obtained

when the system M reaches a dead state along the path 1–2 opt –0 at which thermo- mechanical equilibrium is achieved within the ambient. At this point the energy U M+A the minimum and no more work can be delivered by M.

Consider a constant volume system, e.g., a car battery if T 0 =T 1 =T 2 = T then

W u,opt = (U 1 –U 2 ) – T (S 1 –S 2 ), i.e.,

W u,opt = (U 1 –TS 1 ) – (U 2 –TS 2 ), or

(35) where A = U – T S denotes the Helmholtz function or free energy. The magnitude of

W u,opt = (A 1 –A 2 ),

A represents the capability of a closed system to deliver work.

a. Example 1 Air is expanded to perform work in a piston–cylinder assembly. The air is initially at P 1 = 35 bar and T 1 = 2000 K, and the expansion ratio r v (= v 2 /v 1 ) is 7. The ambient

temperature T 0 = 298 K and pressure P 0 = 1 bar.

Determine the useful work that is delivered for an isentropic process. If process is nonadiabatic and P 2 = 2.5 bar, what are the absolute closed system avail- abilities at the initial and final states? Determine the optimum and useful optimum work. If a dynamometer measures the useful work to be 0.8 kJ for non-adiabatic process and

the initial volume V 1 = 0.000205 m 3 , determine the heat loss, irreversibility (or lost work), and the entropy generated during the process.

Solution

1 =1679 kJ kg ,s 1 = 3.799 kJ kg K –1 ,v r1 = 2.776, and P r1 = 2068. For the isentropic processes,

0 From the tables for air ( –1 Table A-7 ) at 2000 K, u

v 2s /v 1 =v r2s /v r1 = 7. ∴v r2s = 19.43, and from the tables P r2 = 161, T 2 = 1090 K, and u 2s = 835 kJ kg –1 . Hence,

P 2 /P 1 =P r2 /P r1 = 161/2068 = 0.0779, and P 2 = 2.725 bar,

and the isentropic work w –1

s =u 1 –u 2s = (1679-835) = 844 kJ kg . The specific volumes

3 v –1

1 = RT 1 /P 1 = 0.164 m kg , and v 2 = 1.148 m kg (using the expansion ratio). Hence, the useful work w –1

u,s = 844 – 100 × (1.148 – 0.164) = 745.6 kJ kg . Since the cylinder mass is constant, applying the ideal gas law, T 2 =T 1 P 2 v 2 /(P 1 v 1 ),

i.e., T 2 = 2.5 × 7 × 2000 ÷ 35 = 1000 K. The initial entropy s 1 =s 0 1 – R ln (P

1 /P 0 ) = 3.799 – 0.287 ln (35/1) = (3.799 – 1.020) = 2.779 kJ kg K , and the final entropy

2 =s 2 2 /P 0 ) = 2.97 – 0.287 ln (2.5/1) = 2.71 kJ kg K . The absolute availabilities φ 1 =u 1 –T 0 s

1 +P 0 v 1 = 1679 – 298 × 2.779 + 100 × 0.164 = 867.23 kJ kg , and =u –T s +P v = 758.9 – 298

2 2 0 2 0 2 × 2.71+ 100 × 1.148 = 66.1 kJ kg . Therefore, w u,opt = φ 1 –

2 = 867.2 – 66.1 = 801.1 kJ kg . Now,

× (1.148 – 0.164) = 899.5 kJ kg –1 . Hence, the optimal heat transfer

w opt =w u,opt +P 0 (v 2 –v 1 ) = 801.1 + 100

q opt = (u 2 –u )+w = 758.9 – 1679 + 899.5 = –20.6 kJ kg –1 1 opt .

Applying the relation V /V =v /v = 7, since V = 0.000205 m 3 2 1 2 1 1 ,

V 2 =7 × 0.000205= 0.00144 m 3

× 10 2 × 0.000205/(0.287×2000) = 0.00125 kg. Since the total useful work W u = 0.8 kJ, on a mass basis w u = 0.8/0.00125 = 640 kJ

The mass m = P 1 V 1 /RT 1 = 35

kg –1 . Therefore, the work w = 640 + 100

× (1.148 – 0.164) = 738.4 kJ kg –1 .

ty Method )

U+U o Actual Process

(2,opt) (O, I)