Infinitesimal Form

1. Infinitesimal Form

a. Uniform Temperature within a System The entropy generated is the difference between the entropy change and the transit entropy during a process. For a closed system, the differential relation of Eq. (25) may be re- written to explicitly include the entropy generation s, i.e.,

Figure 15: a. Irreversible process; b. Irreversible state changes.

dS = δQ/T + δσ. (26) This relation is also known as Gibbs’ equation. The entropy generation δσ > 0 for internally

irreversible processes and is zero for internally reversible processes. Although the boundary temperature may be uniform, thereby indicating thermal reversibility, other irreversibilities, such as those due to chemical reactions, can contribute to σ, as will be discussed in Chapter 11.

i. Example 9 Assume a large primary system to consist of a vessel containing warm water at a sys-

tem temperature of 350.001 K (T 1 ). An infinitesimal amount of its heat (1050 J) is transferred to a secondary system consisting of room air at a temperature of 300 K (T 2 ). Consequently, the water temperature drops to 349.999 K. What is the entropy generation: If the system is cooled in air (as illustrated through process (a) (as shown in Figure

16a )? If the heat removed from the primary system is used to run a Carnot engine (process

(b)), and that rejected by the engine is transferred to the secondary system (cf. Figure 16b )?

Solution Assuming an internally reversible cooling and heating process for the water and air, using Eq. (26), δσ = 0 for both systems. Therefore,

dS = δQ/T (A) Since the temperature is approximately constant ( ≈350 K), upon integrating Eq. (A)

(B) which is represented by path AB in Figure 16c . Similarly, ∆S 2 =1050

∆S 1 = –1050

÷350 = –3 J K –1 .

÷300 =3.5 J K –1 . (C) The entropy gain for the air is represented by the path C–D–E in Figure 16c . If a

boundary is placed around systems 1 and 2, as illustrated by the dashed line in Figure 16a , there is no heat or work transfer across the composite isolated system. However, an irreversible process occurs within the composite system, and

(D) ∆S 1 + ∆S 2 –0= σ.

dS – (0/T) = δσ, i.e., dS 1 + dS 2 = δσ, or

(E) Employing Eqs. (B) and (C),

σ = –3.0 + 3.5 – 0 = 0.5 J K –1 . The path D–E, shown in Figure 16c illustrates the net entropy gain for the isolated

system that occurs since entropy is generated due to irreversible heat transfer between the two subsystems 1 and 2.

The second scenario is illustrated in Figure 16d . In this case subsystem 2 un- dergoes the same entropy change as does subsystem 1 so that

∆S 1 = – 3.0 J K –1 , and ∆S 2 =Q 2 /T 2 .

(F)

Therefore, Q L /Q H =Q 2 /Q 1 =T 2 /T 1 = 300÷350 = 0.857.

Since Q 2 = 1050 × 0.857 = 900 J, and ∆S –1

2 = 900÷300 = + 3.0 J K . (G) For the composite system ∆S 1 + ∆S 2 –0= σ. Employing Eqs. (F) and (G),

In this particular case | ∆S 1 |= | ∆S 2 |.

Remarks For the first case 1050 J of thermal energy is transferred from the warm water to the

ambient and ∆U 1 = – ∆U 2 =

–1050 J. It is impossible to transfer the 1050 J back from

the ambient to restore the water 350 K

350 K

1 1 S 1 = -3.0 J/K

to its original state. Further,

S 1 = -3.0 J/K

1050 J

1050 J

∆U = ∆U W=150 J

1 + ∆U 2 =0

For the second case work is S 2 = +3.0 J/K

900 J

produced and used to lift a

T =300 K

weight (e.g., lift an elevator) T 2 2 =300 K

S 2 = +3.0 J/K 2 2

with the consequence that a smaller amount of heat is re-

S 1 +S 2 > 0 S 1 +S 2 = 0

jected to the ambient while ac-

(a)

(b)

complishing the same change in state of the water. In this

case, ∆U 1 = – 1050 J, and ∆U 2

= + 900 J so that ∆U (=∆U 1 +

Figure 16: a). Illustration of a) direct cooling; b). cooling using a Carnot engine.

(c)

(d)