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

The final expression is called flow(or stream) exergy,and is denoted by c

(Fig. 8–23).

Flow exergy: (8–22)

Then the exergy changeof a fluid stream as it undergoes a process from

state 1 to state 2 becomes

(8–23)

For fluid streams with negligible kinetic and potential energies, the kinetic

and potential energy terms drop out.

Note that the exergy changeof a closed system or a fluid stream represents

the maximumamount of useful work that can be done (or the minimum

amount of useful work that needs to be supplied if it is negative) as the sys-

tem changes from state 1 to state 2 in a specified environment, and repre-

sents the reversible work Wrev. It is independent of the type of process

executed, the kind of system used, and the nature of energy interactions with

the surroundings. Also note that the exergy of a closed system cannot be neg-

ative, but the exergy of a flow stream can at pressures below the environment

pressure P 0.

¢c
c 2 c 1 
 1 h 2 h 12 T 01 s 2 s 12 

V^22 V^21

2

g 1 z 2 z 12

c
 1 hh 02 T 01 ss 02 

V^2

2

gz

 1 hh 02 T 01 ss 02 

V^2

2

gz

 1 uPv 2  1 u 0 P 0 v 02 T 01 ss 02 

V^2

2

gz

438 | Thermodynamics

COMPRESSED

AIR

1 MPa

300 K

FIGURE 8–24

Schematic for Example 8–7.

Energy:Energy:

Exergy:Exergy:

(a) A fixed mass (nonflowing)) A fixed mass (nonflowing)

e = u u + V + gzgz

2

2

FixedFixed

massmass

V^2

f = (= (uu – u 0 ) + + P 0 (v – v 0 ) – T 0 (ss – s 0 ) + 2 + gzgz

Energy:Energy:

Exergy:Exergy:

(b) A fluid ) A fluid streamstream (flowing) (flowing)

u = h h + V + gzgz

2

2

V^2

2

c = (= (h h – h 0 ) + + T 0 (s s – s 0 ) + + gzgz

FluidFluid

streamstream

FIGURE 8–23

The energyand exergycontents of
(a) a fixed mass and (b) a fluid stream.

EXAMPLE 8–7 Work Potential of Compressed Air in a Tank

A 200-m^3 rigid tank contains compressed air at 1 MPa and 300 K. Deter-

mine how much work can be obtained from this air if the environment condi-

tions are 100 kPa and 300 K.

Solution Compressed air stored in a large tank is considered. The work

potential of this air is to be determined.

Assumptions 1 Air is an ideal gas. 2 The kinetic and potential energies are

negligible.

Analysis We take the air in the rigid tank as the system (Fig. 8–24). This is

a closed systemsince no mass crosses the system boundary during the

process. Here the question is the work potential of a fixed mass, which is

the nonflow exergy by definition.

Taking the state of the air in the tank to be state 1 and noting that T 1 


T 0 
300 K, the mass of air in the tank is

m 1 

P 1 V

RT 1

1 1000 kPa 21 200 m^32

1 0.287 kPa#m^3 >kg#K 21 300 K 2

2323 kg