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Modern Engineering Thermodynamics

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8.3 Systems Undergoing Irreversible Processes 257<br />

Then,<br />

1ðS P<br />

Þ 2 = ð2:00 kg<br />

Þ½2:8042 − 0:3687 kJ/ ðkg⋅KÞŠ−<br />

= 0:449 kJ/K = 449 J/K<br />

1650 kJ<br />

100: + 273:15 K<br />

Notice that the process described in this example also has a minimum isothermal system boundary temperature, like that<br />

described in Example 8.3. In this case, 1 (S P ) 2 = 0, and the minimum boundary temperature is<br />

ðT b Þ minimum =<br />

1Q 2<br />

= 339 K = 65:6°C<br />

ms ð 2 − s 1 Þ<br />

WHAT HAPPENS IF YOU TRY TO HEAT THE WATER IN EXAMPLE 8.5<br />

WITH AN ISOTHERMAL BOUNDARY TEMPERATURE LESS THAN 65.2°C?<br />

Any attempt to carry out the heating process in Example 8.5 with an isothermal boundary temperature lower than the minimum<br />

of 65.6°C fails because it violates the second law by requiring a negative entropy production. But, what would happen<br />

if you tried? You could add heat to the system with an isothermal boundary temperature of less than 65.6°C, but it<br />

would not reach 95.0°C.<br />

However, cooling the tank from 95.0°C back to 20.0°C with an isothermal boundary reverses all the signs in the entropy<br />

production calculation so that 2 (S P ) 1 = 1 (S P ) 2 . This cooling process requires an isothermal boundary temperature less than<br />

65.6°C to satisfy the second law.<br />

Exercises<br />

7. Determine the entropy production that occurs in Example 8.5 if the surface temperature of the container is maintained<br />

at 80.0°C rather than 100.°C. Answer: 1 (S P ) 2 = 199 J/K.<br />

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%,<br />

T 2 = 95.0°C and x 2 = 100.%. Answer: 1 (S P ) 2 = 9.83 kJ/K.<br />

EXAMPLE 8.6 A CONTINUATION OF EXAMPLE 5.2, WITH<br />

THE ADDED MATERIAL SHOW IN ITALIC TYPE<br />

An incandescent lightbulb is a simple electrical device. Using the energy rate balance and the entropy rate balance on a lightbulb,<br />

determine<br />

a. The heat transfer rate of an illuminated 100. W incandescent lightbulb in a room.<br />

b. The rate of change of its internal energy if this bulb were put into a small, sealed, insulated box.<br />

c. The value of the entropy production rate for part a if the bulb has an isothermal surface temperature of 110.°C.<br />

d. An expression for the entropy production rate as a function of time for part b.<br />

Solution<br />

First, draw a sketch of the system (Figure 8.6).<br />

Insulated box<br />

W· = 100. watts<br />

W· = 100. watts<br />

System boundary<br />

(at a constant 110.°C)<br />

(a)<br />

Q· = ?<br />

S·P = ?<br />

(b)<br />

S·P = ?<br />

System boundary<br />

(temperature not<br />

constant)<br />

FIGURE 8.6<br />

Example 8.6.<br />

(Continued )

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