Modern Engineering Thermodynamics

260 CHAPTER 8: Second Law Closed System Applications
EXAMPLE 8.7 (Continued )
The surface area of our system can be divided into three major parts: the boiler’s surface area, the condenser’s surfacearea,
and all the remaining surface areas. Therefore, the surface area Σ system of the system is composed of the boiler, the condenser,
and everything else, or
∑ system
= ∑ boiler
+∑ condenser
+∑ everything else
Now, both the boiling and the condensing processes are isothermal phase changes; and since no heat transfer occurs at any
other point in the system, we can write
Z
d dQ
= d dt Σ T b dt
Z
Σ boiler
dQ
T b
+
Z
Σ condenser
dQ
T b
+
Z
!
dQ
Σ T b remainder
where
and
Z
d dQ
=
dt Σ T b boiler
Z
d dQ
=
dt Σ T b condenser
_Q boiler
T boiler
_Q condenser
T condenser
Z
d
dQ
= 0 ðno heat transfer across the remaining surface areaÞ
dt Σ T b remainder
Then, we have
_S P = _S −
and, for steady state operation ð _S = 0Þ, this reduces to
_Q boiler
T boiler
−
_Q condenser
T condenser
!
_Q
_S p = − boiler
_Q
+ condenser
T boiler
T condenser
950: × 10
= −
5 − 600: × 105 kJ/h
+
500: + 273:15 10:0 + 273:15 K
Note that the actual thermal efficiency of this power plant is given by Eq. (7.9) as
= 89:0 × 10 3 kJ/ ðh⋅KÞ
ðη T Þ act
= 1 − j _Q out j
= 1 −
600: × 105
= 0:368 = 36:8%
5 _Q in 950: × 10
whereas its theoretical reversible (Carnot) efficiency is given by Eq. (7.16) as
ðη T Þ rev
= 1 − T condenser
= 1 −
10:0 + 273:15
= 0:634 = 63:4%
T boiler 500: + 273:15
Therefore, the actual efficiency is less than the theoretical maximum (reversible) efficiency, as it should be.
Exercises
11. Determine the condenser temperature that would cause the entropy production rate of the power plant in Example 8.7
to be 500. kJ/h·K. Answer: T condenser = 486 K = 213°C.
12. Determine the condenser temperature that would cause the entropy production rate of the power plant in Example 8.7
to equal zero. Why can’t this occur? Answer: T condenser = 488 K = 215°C. This cannot occur because it would require the
plant to be reversible throughout (i.e., it could have no internal friction, heat transfer, chemical reactions, or anything
else that would naturally be irreversible).
13. If the heat transfer rate into the boiler is doubled in Example 8.7 and all the other parameters remain constant,
determine the new entropy production rate of the power plant. Is this possible? Answer: _S P = −33,900 kJ=h⋅K: This is
not possible because it violates the second law of thermodynamics _S P must be > 0 .