Modern Engineering Thermodynamics

150 CHAPTER 5: First Law Closed System Applications
5.3 ELECTRICAL DEVICES
In a vast number of closed and open systems the primary work mode is electrical. We call these systems electrical
devices and recognize that one of the appropriate auxiliary equations to be used in their analysis is the electrical
work or power mode equation introduced earlier. The following example illustrates a typical closed system electrical
device problem.
EXAMPLE 5.2
Read the problem statement. An incandescent lightbulb is a simple electrical device. Using the energy rate balance on a
lightbulb determine
a. The heat transfer rate of an illuminated 100. W incandescent lightbulb in a room.
b. The rate of change of its internal energy if this bulb were put into a small sealed insulated box.
Solution
Step 1. Identify and sketch the system. We imagine and sketch the system in each case as the entire lightbulb, the glass
bulb plus its contents, rather than just the contents as in Example 5.1 (Figure 5.2).
W = 100. watts
Insulated box
W = 100. watts
System
boundary
Q = ?
System
boundary
(a)
(b)
FIGURE 5.2
Example 5.2.
Step 2. Identify the unknowns. The unknowns are (a) _Q and (b) _U :
Step 3. It is a closed system. This is a closed system for which we are not given specific thermodynamic properties in the
problem statement. Presumably, they are not needed in the solution.
Step 4. Identify the process connecting the system states. The following processes occur:
■ In part a, the bulb does not change its thermodynamic state, so its properties must remain constant. In
particular, the process path (after the bulb has warmed to its operating temperature) is U = constant.
■ In part b, the bulb is insulated, so it undergoes an adiabatic (i.e., _Q = 0) process.
Step 5. Write down the basic equations. The only basic equation thus far available for a closed system rate process is
Eq. (4.21), the general closed system energy rate balance (ERB) equation:
_Q − _W = d dt ðmuÞ + d
mV 2
+ d
mgZ
= _U + KE _ + PE _
dt 2g c dt g c
Since both parts a and b imply that the lightbulb is to be stationary during analysis, we assume KE _ ¼ PE _ ¼ 0: This
reduces the governing ERB equation for this problem to
_Q − _W = _U = d dt ðmuÞ
Write any relevant auxiliary equations. The only relevant auxiliary equation needed here is the recognition that the
lightbulb has an electrical work input of 100. W, so that
_W = − 100: W
Step 6. Algebraically solve for the unknown(s). Algebraically solving for the unknowns, we have, for part a)
_Q = _U + _W
and, for part b,
_U = _Q − _W