Modern Engineering Thermodynamics

5.4 Power Plants 151
Step 7. Calculate the value(s) of the unknowns. Since we are assuming a constant bulb temperature in part a,
U = constant and _U = 0. Then, our calculations give
a. _Q = _W = − 100: W
(the minus sign tells us that heat is leaving the system). Thus, all the electrical work put into a lighting system ends
up as heat. Architects use the lighting within a building to supply part of the heating requirements of the building.
In the second part of this problem, the bulb is inside a small insulated box, so it cannot transport any heat energy
through its boundaries. Therefore, _Q = 0 here (the bulb undergoes an adiabatic process), and the reduced energy
rate balance yields
b. _U = − _W = 100: W
Step 8. A check of the algebra, calculations, and units shows that they are correct.
Note that the internal energy of the lightbulb must increase at a rate of 100. J/s. This means that its temperature
must continually increase. Treating the bulb as a simple incompressible substance, we can write (see Eq. (3.33))
_U = mc _T = 100: W
where m is the mass of the bulb, c is its specific heat, and _T is the time rate of change of its temperature. So long
as _U is constant and positive, the temperature of the bulb continually increases until the glass or the filament eventually
melts.
5.4 POWER PLANTS
An electrical power generating facility is a very complex set of open and closed systems. However, if the entire
facility is taken to be the system and the system boundaries are carefully chosen, then it can be modeled as a
closed system. We call such systems power plants, and a simple thermodynamic analysis can provide important
information about their operation, as the next example illustrates.
EXAMPLE 5.3
Read the problem statement. 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 this power plant (to three significant figures), 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 a 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.
Solution
Step 1. Identify and sketch the system. Sketch the system as the entire power plant (Figure 5.3). If we choose only the
turbine as the system, it would be an open system; and we do not wish to deal with open systems until Chapter 6.
Step 2. Identify the unknowns. The unknown here is ð _W T Þ net
:
Step 3. It is a closed system. This is closed system, and no specific information is given to identify the thermodynamic
states of the system. Presumably, they are not needed in the solution.
Step 4. Identify the process connecting the system states. We assume a steady state process with no changes in kinetic or
potential energy. Then, U, KE, and PE are all constants.
Step 5. Write down the basic equations. The only basic equation applicable here is the general closed system energy rate
balance, Eq. (4.21):
which reduces to
_Q − _W = _U 0
↘ 0
+ _ KE
↘0
+ _ PE
↘0
= 0
_W net = _Q net = ð _W T Þ net
Write any relevant auxiliary equations. No auxiliary equations are needed here.
(Continued )