Modern Engineering Thermodynamics

7.16 Phase Change Entropy Production 239
1.00 × 10 −3 m
T = 600. K
5.00 × 10 −3 m
10.0 × 10 −3 m
i = 0.10 amp (S P ) W-electric = ?
FIGURE 7.23
Example 7.12.
The unknown is the entropy production rate of the chip. Since the chip has uniform properties, is isothermal, and has a
constant current, we can use Eq. (7.74) to determine its electrical work mode entropy production rate as:
ð_S P Þ W
elect
= I2 R e
T
where the electrical resistance R e = ρ e (L/A). The electrical resistivity can be found for silicon at 600. K from Figure 7.22 as
(ρ e ) silicon = 0.10 Ω · m. Then, from the dimensions given for the chip, we have A = 0.00100 × 0.00500 = 5.00 × 10 −6 m 2 ,
L = 0.0100 m, and
Then, Eq. (7.75) gives
R e = ð0:100 ohm.mÞð0:0100 mÞ=ð5:00 × 10 − 6 m 2 Þ = 200: Ω = 200: W=A 2
ð_S P Þ W
elect
= ð0:10 AÞ2 ð200 W/A 2 Þ
600: K
= 0:0033 W/K
Exercises
22. Determine the percent reduction in the entropy production rate in Example 7.12 if germanium (see Figure 7.22) is
used in place of silicon in the chip and all the other parameters in the example remain the same. Answer:
% reduction = 99%.
23. An incandescent lightbulb has a filament 0.13 mm in diameter and 0.076 m long. It is made of tungsten wire and
operates at 900. K while carrying 0.90 A. Use Figure 7.22 to determine the entropy production rate due to the electrical
work mode in the filament. Answer: ð_S P Þ W-elect
= 0:0052 W/K.
24. A solenoid is made of copper wire 0.0500 mm in diameter and 200. m long. The solenoid operates at 300. K and carries a
current of 0.500 A. Determine the entropy production rate inside the solenoid due to the electrical work mode. Answer:
_S P = 0:424 W/K.
W-elect
7.16 PHASE CHANGE ENTROPY PRODUCTION
In any process, the change in entropy is independent of the actual process path used, because entropy is a state
property (or a point function). Therefore, we may write ðΔSÞ rev
= ðΔSÞ act
or ðΔsÞ rev
= ðΔsÞ act
for any process
whatsoever. The entropy change produced by an actual process can occasionally be found by assuming that the
system has undergone a hypothetical reversible process that is easier to evaluate than the actual irreversible process.
As an example of this technique, consider the entropy change and associated entropy production that occur
in a phase change. For a reversible phase change carried out in a closed system, we have (S P ) rev = 0, and the
entropy balance for an isothermal system then gives
or
ðS 2 − S 1 Þ rev = ðS 2 − S 1 Þ act = ðQ/T b
Þ rev = ðQ/T b
ðS P Þ phase change
= Q rev − Q act
> 0
T b
Þ act + ðS P Þ phase change
Thus, for an exothermic (heat liberating) phase change (e.g., a condensation or solidification process), the heat
transfers are negative and it follows that jQ act j > jQ rev j and for an endothermic phase change (e.g.,