Modern Engineering Thermodynamics

8.3 Systems Undergoing Irreversible Processes 259
Exercises
9. If the surface temperature of the bulb in Example 8.6 were 40.0°C (which is typical of a fluorescent light), determine the
entropy production rate of the bulb. Answer: _S p = 0:319 W=K:
10. Take _U = 100: watts, T 0 = T b = 110:°C, and for a glass bulb take m = 0.050 kg and c = 0.80 kJ/kg · K. Then, using the
last expression in Example 8.6 for the decay in entropy production rate for an insulated lightbulb, determine the entropy
production rate at t = 0, 300., 600., and 1000. s. Answers: _S p = 0:26, 0:088, 0:053, and 0:035 W=K:
EXAMPLE 8.7 A CONTINUATION OF EXAMPLE 5.3, WITH
THE ADDITION SHOWN IN ITALIC TYPE
A basic vapor cycle power plant consists of the following four parts:
a. The boiler, where high-pressure vapor is produced.
b. The turbine, where energy is removed from the high-pressure vapor as shaft work.
c. The condenser, where the low-pressure vapor leaving the turbine is condensed into a liquid.
d. The boiler feed pump, which pumps the condensed liquid back into the high-pressure boiler for reheating.
In such a power plant, the boiler receives 950. × 10 5 kJ/h from the burning fuel, and the condenser rejects 600. × 10 5 kJ/h
to the environment. The boiler feed pump requires 23.0 kW input, which it receives directly from the turbine. Assuming
that the turbine, pump, and connecting pipes are all insulated, determine the net power of the turbine and the rate of entropy
production of the plant if the boiler temperature is 500.°C and the condenser temperature is 10.0°C.
Solution
First, draw a sketch of the system (Figure 8.7).
Q B = 950. × 10 5 kJ/h
T B = 500.°C
S P = ?
Boiler
Boiler
feed
pump
W p =−23.0 kW
Turbine
(W T )net = ?
Condenser
Q c =−600. × 10 5 kJ/h
T c = 10.0°C
System
boundary
FIGURE 8.7
Example 8.7.
The unknowns here are _W T
net and _S P for the entire power plant. Therefore, the system is the entire power plant. Since we
have few specific details on the internal operation of the plant, the thermodynamic properties within the power plant are
apparently not needed in the solution.
The net turbine work output is determined in Example 5.3 to be
_W T
net
= 9720 kW
The answer to the second part of this problem can be obtained by the indirect method from Eq. (8.2) as
_S P = _S − d Z
dQ
dt T b
Σ
act
(Continued )