Modern Engineering Thermodynamics

156 CHAPTER 5: First Law Closed System Applications
EXAMPLE 5.6 (Continued )
Cylinder
1 W 2 = ?
Vapor dome
1
Adiabatic
expansion
Piston
P
System
0.100 lbm
boundary
3
2
Isobaric
compression
Refrigerant-134a
v
2 Q 3 = ?
FIGURE 5.6
Example 5.6.
and
2Q 3 − 2 W 3 = mðu 3 − u 2 Þ + KE 3 − KE 2 + PE 3 − PE 2
Since we are not given any potential or kinetic energy information, we assume that no changes occur in these variables. As
auxiliary equations we have 1 Q 2 = 0 (because the process from state 1 to state 2 is adiabatic); consequently, the resulting
energy balance equation for the solution to part a) is
1W 2 = − mðu 2 − u 1 Þ
Also, since the process from state 2 to state 3 is isobaric, the work for this process is given by 2 W 3 = m
and the energy balance for the process from 2 to 3 gives the equation for the solution to part b as
2Q 3 = mðu 3 − u 2 Þ + mp 3 ðv 3 − v 2 Þ
Z 3
2
pdv = mp 3 ðv 3 − v 2 Þ
The solution to part c must be determined from the values of the independent properties p 3 and v 3 and the use of the
R-134a tables. From Table C.7e of Thermodynamic Tables to accompany Modern Engineering Thermodynamics, wefindthatat
p 1 = 100: psia and T 1 = 180:°F, v 1 = 0:6210 ft 3 /lbm, u 1 = 125:99 Btu/lbm. Similarly, at p 2 = 30:0 psia and T 2 = 120°F, we
find from Table C.7e that v 2 = 1:966 ft 3 /lbm and u 2 = 115:47 Btu/lbm.
The final answers are
a.
1W 2 = − mðu 2 − u 1 Þ
= − ð0:100 lbmÞð115:47 − 125:99 Btu/lbmÞ = 1:05 Btu
b. 2Q 3 = mðu 3 − u 2 Þ + mp 3 ðv 3 − v 2 Þ, where we already have numerical values for m, u 2 , p 3 , and v 2 , but we also need values
for v 3 and u 3 . From the problem statement for the process from 2 to 3, we find that
v 3 = v 1
2 = 0:6210 = 0:3105 ft 3 /lbm
2
From Table C.7b, we find that, since state 3 is at 30.0 psia, v f ð30:0 psiaÞ < v 3 < v g ð30:0 psiaÞ, state 3 is a mixture of
liquid plus vapor. It therefore has a quality given by
x 3 = v 3 − v f 3
v fg3
Table C.7b at 30.0 psia gives v f 3 = 0:01209 ft 3 /lbm, v g3 = 1:5408 ft 3 /lbm, u f 3 = 16:24 Btu/lbm, and u g3 = 95:40 Btu/lbm.
We can now calculate the quality at state 3 as
Finally, we get the other state 3 properties as
x 3 = v 3 − v f 3
= 0:3105 ft3 /lbm − 0:01209 ft 3 /lbm
v g3 − v f 3 1:5408 ft 3 /lbm − 0:01209 ft 3 = 0:195 = 19:5%
/lbm
u 3 = u f 3 + x 3 ðu g3 − u f 3 Þ
= 16:24 Btu/lbm + 0:1952ð95:40 − 16:24 Btu/lbmÞ = 31:7 Btu/lbm