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

cold stream, provided that the cold stream is not at a lower temperature than

the surroundings. Then the second-law efficiency of the heat exchanger

becomes

(8–55)

where S

.

gen
m

.

hot(s 2 s 1 ) + m

.

cold(s 4 s 3 ). Perhaps you are wondering what

happens if the heat exchanger is not adiabatic; that is, it is losing some heat

to its surroundings at T 0. If the temperature of the boundary (the outer sur-

face of the heat exchanger) Tbis equal T 0 , the definition above still holds

(except the entropy generation term needs to be modified if the second defi-

nition is used). However, if TbT 0 , then the exergy of the lost heat at the

boundary should be included in the recovered exergy. Although no attempt is

made in practice to utilize this exergy and it is allowed to be destroyed, the

heat exchanger should not be held responsible for this destruction, which

occurs outside its boundaries. If we are interested in the exergy destroyed

during the process, not just within the boundaries of the device, then it

makes sense to consider an extended systemthat includes the immediate sur-

roundings of the device such that the boundaries of the new enlarged system

are at T 0. The second-law efficiency of the extended system reflects the

effects of the irreversibilities that occur within and just outside the device.

An interesting situation arises when the temperature of the cold stream

remains below the temperature of the surroundings at all times. In that case

the exergy of the cold stream actually decreases instead of increasing. In

such cases it is better to define the second-law efficiency as the ratio of the

sum of the exergies of the outgoing streams to the sum of the exergies of the

incoming streams.

For an adiabatic mixing chamberwhere a hot stream 1 is mixed with a cold

stream 2, forming a mixture 3, the exergy supplied is the sum of the exergies

of the hot and cold streams, and the exergy recovered is the exergy of the

mixture. Then the second-law efficiency of the mixing chamber becomes

(8–56)

where m

.

3 
m

.

1 + m

.

2 and S

.

gen
m

.

3 s 3 m

.

2 s 2 m

.

1 s 1.

hII,mix

m

#

3 c 3

m

#

1 c 1 m

#

2 c 2

¬or¬hII,mix
1 

T 0 S

#

gen

m

#

1 c 1 m

#

2 c 2

hII,HX

m#cold 1 c 4 c 32

m

#

hot^1 c 1 c 22

¬or¬hII,HX
1 

T 0 S

#

gen

m

#

hot^1 c 1 c 22

460 | Thermodynamics

EXAMPLE 8–15 Second-Law Analysis of a Steam Turbine

Steam enters a turbine steadily at 3 MPa and 450°C at a rate of 8 kg/s and

exits at 0.2 MPa and 150°C, (Fig. 8–45). The steam is losing heat to the sur-

rounding air at 100 kPa and 25°C at a rate of 300 kW, and the kinetic and

potential energy changes are negligible. Determine (a) the actual power output,

(b) the maximum possible power output, (c) the second-law efficiency, (d) the

exergy destroyed, and (e) the exergy of the steam at the inlet conditions.

Solution A steam turbine operating steadily between specified inlet and exit

states is considered. The actual and maximum power outputs, the second-law

efficiency, the exergy destroyed, and the inlet exergy are to be determined.

STEAM

TURBINE

W

T 0 = 25°C

P 0 = 100 kPa

3 MPa

450 °C 300 kW

0.2 MPa

150 °C

FIGURE 8–45

Schematic for Example 8–15.