Modern Engineering Thermodynamics

8.3 Systems Undergoing Irreversible Processes 257
Then,
1ðS P
Þ 2 = ð2:00 kg
Þ½2:8042 − 0:3687 kJ/ ðkg⋅KÞŠ−
= 0:449 kJ/K = 449 J/K
1650 kJ
100: + 273:15 K
Notice that the process described in this example also has a minimum isothermal system boundary temperature, like that
described in Example 8.3. In this case, 1 (S P ) 2 = 0, and the minimum boundary temperature is
ðT b Þ minimum =
1Q 2
= 339 K = 65:6°C
ms ð 2 − s 1 Þ
WHAT HAPPENS IF YOU TRY TO HEAT THE WATER IN EXAMPLE 8.5
WITH AN ISOTHERMAL BOUNDARY TEMPERATURE LESS THAN 65.2°C?
Any attempt to carry out the heating process in Example 8.5 with an isothermal boundary temperature lower than the minimum
of 65.6°C fails because it violates the second law by requiring a negative entropy production. But, what would happen
if you tried? You could add heat to the system with an isothermal boundary temperature of less than 65.6°C, but it
would not reach 95.0°C.
However, cooling the tank from 95.0°C back to 20.0°C with an isothermal boundary reverses all the signs in the entropy
production calculation so that 2 (S P ) 1 = 1 (S P ) 2 . This cooling process requires an isothermal boundary temperature less than
65.6°C to satisfy the second law.
Exercises
7. Determine the entropy production that occurs in Example 8.5 if the surface temperature of the container is maintained
at 80.0°C rather than 100.°C. Answer: 1 (S P ) 2 = 199 J/K.
8. Determine the entropy production that occurs in Example 8.5 if the two states are fixed as T 1 = 20.0°C and x 1 = 0.00%,
T 2 = 95.0°C and x 2 = 100.%. Answer: 1 (S P ) 2 = 9.83 kJ/K.
EXAMPLE 8.6 A CONTINUATION OF EXAMPLE 5.2, WITH
THE ADDED MATERIAL SHOW IN ITALIC TYPE
An incandescent lightbulb is a simple electrical device. Using the energy rate balance and the entropy 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.
c. The value of the entropy production rate for part a if the bulb has an isothermal surface temperature of 110.°C.
d. An expression for the entropy production rate as a function of time for part b.
Solution
First, draw a sketch of the system (Figure 8.6).
Insulated box
W· = 100. watts
W· = 100. watts
System boundary
(at a constant 110.°C)
(a)
Q· = ?
S·P = ?
(b)
S·P = ?
System boundary
(temperature not
constant)
FIGURE 8.6
Example 8.6.
(Continued )