Modern Engineering Thermodynamics

8.3 Systems Undergoing Irreversible Processes 263
c. The final temperature at the end of the isobaric compression.
d. The total entropy production for both processes if the heat transport of energy and the boundary temperature are related by the
formula dQ = KdT b , where K = 5.00 Btu/R and T b is in R.
Solution
First, draw a sketch of the system (Figure 8.10).
1W 2 = ?
Vapor dome
1
Cylinder
0.100 lbm
Piston
System
boundary
Refrigerant–134a
P
3
Isobaric
compression
Adiabatic
expansion
2
2Q 3 = ?
(S P ) total = 1 (S P ) 2 + 2 (S P ) 3 = ?
ν
FIGURE 8.10
Example 8.10.
The unknowns here are (a) 1 W 2 , (b) 2 Q 3 , (c) T 3 , and (d) 1 (S P ) 3 . The system is just the R-134a, and the state variables are as
follows:
State 1
Adiabatic
ƒƒƒ!
pansion
State 2
Isobaric
ƒƒƒƒƒ!
compression
State 3
p 1 = 100: psia p 2 = 30:0 psia p 3 = p 2 = 30:0 psia
T 1 = 180:°F T 2 = 120:°F v 3 = v 1 /2
v 1 = 0:6210 ft 3 /lbm v 2 = 1:9662 ft 3 /lbm v 3 = 0:3105 ft 3 /lbm
h 1 = 137:49 Btu/lbm h 2 = 126:39 Btu/lbm x 3 = 0:1952
s 1 = 0:2595 Btu/ðlbm⋅RÞ s 2 = 0:2635 Btu/ðlbm⋅RÞ s 3 = 0:07241 Btu/ðlbm⋅RÞ
The solutions to the first three parts of this problem are given in Example 5.6 as
a. 1W 2 = 1.05 Btu.
b. 2Q 3 = –9.31 Btu.
c. T 3 = 15.4°F.
The solution to part d can be determined by the indirect method (entropy balance) as follows. From Eq. (8.1), we have
Z
dQ
1ðS P Þ 2 = ms ð 2 − s 1 Þ−
From Table C.7e, we find that
and
Σ
T b
⎵
act
s 1 = 0:2595 Btu/ ðlbm⋅RÞ
0 ðadiabatic processÞ
s 2 = 0:2635 Btu/ ðlbm⋅RÞ
Since the process from 1 to 2 is adiabatic, dQ = 0: and
1ðS P
Þ 2
= ð0:100 lbm
Þ½0:2635 − 0:2595 Btu/ ðlbm⋅RÞŠ− 0 = 4:00 × 10 −4 Btu/R
(Continued )