Modern Engineering Thermodynamics

256 CHAPTER 8: Second Law Closed System Applications
8.3 SYSTEMS UNDERGOING IRREVERSIBLE PROCESSES
We begin our treatment of irreversible processes by extending Examples 5.1 through 5.7 of Chapter 5 using the
entropy balance or entropy rate balance equations in their analysis to obtain additional results from the problem.
To do this effectively we have to add some information to these problem statements, in the way of additional
unknowns or additional data. When additional wording has been added to these problem statements it
appears in italic type so that you can clearly see what changes have been made. Also, the analysis in these examples
contains less commentary, as they are designed to be straightforward applications.
EXAMPLE 8.5 A CONTINUATION OF EXAMPLE 5.1, WITH
THE ADDED MATERIAL SHOWN IN ITALIC TYPE
A sealed, rigid container whose volume is 1.00 m 3 contains 2.00 kg of liquid water plus water vapor at 20.0°C. The
container is heated until the temperature inside is 95.0°C. Determine
a. The quality in the container when the water is 20.0°C.
b. The quality in the container when the water is at 95.0°C.
c. The heat transport of energy required to raise the temperature of the contents from 20.0 to 95.0°C.
d. The entropy production that occurs if the boundary of the container is maintained isothermal at 100.°C during the heat transfer
process by condensing steam at atmospheric pressure on the outside of the tank.
Solution
First, draw a sketch of the system (Figure 8.5).
Sealed
rigid
container
= 1.00 m 3
m = 2.00 kg
A
System boundary
maintained at a constant
temperature of 100.°C
Vapor
Liquid
1 (S P ) 2 = ?
FIGURE 8.5
Example 8.5.
The unknowns are (a) x 1 , (b) x 2 , (c) 1 Q 2 and (d) 1 (S p ) 2 . The system is the water and is closed. The thermodynamic states are
The answers to a, b, and c can be found in Example 5.1 as
a). x 1 = 0:863%:
b). x 2 = 25:2%:
c). 1Q 2 = 1650 kJ:
Isochoric
State 1 ƒƒ!
State 2
v2 = v1
T 1 = 20:0 °C T 2 = 95:0 °C
v 1 = 0:500 m 3 =kg
v 2 = v 1 = 0:500 m 3 =kg
The answer to part d is obtained from the entropy rate balance or the indirect method since we lack the detailed information
about the interior of the system required to use the direct method. Equation (8.1) for an isothermal boundary becomes
where
and
1ðS P Þ 2 = ms ð 2 − s 1 Þ = 1 Q 2
T b
s 1 = s f 1 + x 1 s fg1
= 0:2965 + ð0:00863Þð8:3715Þ = 0:3687 kJ/ ðkg⋅KÞ
s 2 = s f 2 + x 2 s fg2
= 1:2503 + ð0:252Þð6:1664Þ = 2:8042 kJ/ ðkg⋅KÞ