Modern Engineering Thermodynamics

344 CHAPTER 10: Availability Analysis
EXAMPLE 10.12 (Continued )
Q boiler =1.00 × 10 6 kJ/s
Boiler
3.00 × 10 5 kJ/s
Electrical power
Pump
Turbine
Generator
Condenser
Ground state:
T 0 = 5.00°C
p 0 = 0.101 MPa
Q condenser = 7.00 × 10 5 kJ/s
T condenser = 40.0°C
FIGURE 10.17
Example 10.12.
The unknowns are the first law thermal efficiency of the power plant, the rate at which available energy enters the boiler, the
rate at which available energy enters the condenser, and the second law efficiency of the power plant.
a. The first law thermal efficiency of a heat engine is given by Eq. (7.5) as
η T =
Net work output
Total heat input =
so, the overall thermal efficiency of the power plant is
jW outj net
jQ input j total
=
η T = 3:00 × 105 kJ/s
1:00 × 10 6 = 0:300 = 30:0%
kJ/s
b. The rate at which available energy enters the boiler is given by
_A boiler
= 1 − T
0 _Q
T H =
input
H
1 −
5:00 + 273:15 K
700: + 273:15 K
c. The rate at which available energy enters the condenser is given by
_A condenser
input
= 1 − T
0
j _Q
T L j =
H
1 −
j _W out j net
j _Q input j total
ð1:00 × 10 6 kJ/sÞ = 7:14 × 10 5 kJ/s
5:00 + 273:15 K
ð7 × 10 6 kJ/sÞ = 0:78 × 10 5 kJ/s
40:0 + 273:15 K
d. The second law availability efficiency of the power plant is given by Eq. (10.30) as
ε HE
=
1 − T
0
T H
=
1 −
_W
_Q H − 1 − T 0
T L
5:00 + 273:15 K
700: + 273:15 K
=
j _Q L j
_A turbine
input
_W
− _A condenser
input
3:00 × 10 5 kJ/s
ð1:00 × 10 6 5:00 + 273:15 K
kJ/sÞ − 1 − ð7:00 × 10 5 kJ/sÞ
40:0 + 273:15 K
= 0:472 = 47:2%