Microsoft Word - Cengel and Boles TOC _2-03-05_.doc

Chapter 8 | 449

determine (a) the exergy of the steam at the initial and the final states, (b) the
exergy change of the steam, (c) the exergy destroyed, and (d) the second-law
efficiency for the process.

Solution Steam in a piston–cylinder device expands to a specified state. The
exergies of steam at the initial and final states, the exergy change, the exergy
destroyed, and the second-law efficiency for this process are to be determined.
Assumptions The kinetic and potential energies are negligible.
Analysis We take the steamcontained within the piston–cylinder device as
the system (Fig. 8–37). This is a closed systemsince no mass crosses the
system boundary during the process. We note that boundary work is done by
the system and heat is lost from the system during the process.
(a) First we determine the properties of the steam at the initial and final
states as well as the state of the surroundings:

State 1:

State 2:

Dead state:

The exergies of the system at the initial state X 1 and the final state X 2 are
determined from Eq. 8–15 to be

and

That is, steam initially has an exergy content of 35 kJ, which drops to 25.4
kJ at the end of the process. In other words, if the steam were allowed to
undergo a reversible process from the initial state to the state of the environ-
ment, it would produce 35 kJ of useful work.

(b) The exergy change for a process is simply the difference between the
exergy at the initial and final states of the process,

¢X
X 2 X 1 
25.435.0

9.6 kJ

25.4 kJ

¬ 1 100 kPa 231 0.959860.00103 2 m^3 >kg 461 kJ>kPa#m^32

¬ 1 298 K 231 7.28100.3672 2 kJ>kg#K 4

 1 0.05 kg 251 2577.1104.83 2 kJ>kg

X 2 
m 31 u 2 u 02 T 01 s 2 s 02 P 01 v 2 v 024

35.0 kJ

¬ 1 100 kPa 231 0.257990.00103 2 m^3 >kg 461 kJ>kPa#m^32

¬ 1 298 K 231 7.12460.3672 2 kJ>kg#K 4

 1 0.05 kg 251 2793.7104.83 2 kJ>kg

X 1 
m 31 u 1 u 02 T 01 s 1 s 02 P 01 v 1 v 024

P 0 
100 kPa

T 0 
25°C

f¬

u 0 uf @ 25°C
104.83 kJ>kg

v 0 vf @ 25°C
0.00103 m^3 >kg

s 0 sf @ 25°C
0.3672 kJ>kg#K

¬¬ 1 Table A–4 2

P 2 
200 kPa

T 2 
150°C

f¬

u 2 
2577.1 kJ>kg

v 2 
0.95986 m^3 >kg¬¬ 1 Table A–6 2

s 2 
7.2810 kJ>kg#K

P 1 
1 MPa

T 1 
300°C

f¬

u 1 
2793.7 kJ>kg

v 1 
0.25799 m^3 >kg

s 1 
7.1246 kJ>kg#K

¬¬ 1 Table A–6 2 2 kJ

P 1 = 1 MP

T 1 = 300°C

P 2 = 200 kPa

T 2 = 150°C

Steam

P 0 = 100 kPa

T 0 = 25°C

State 1 State 2

FIGURE 8–37

Schematic for Example 8–11.