Modern Engineering Thermodynamics

5.8 Closed System Unsteady State Processes 157
then
2Q 3 = mðu 3 − u 2 Þ + mp 3 ðv 3 − v 2 Þ
= ð0:100 lbmÞð31:7 − 115:47 Btu/lbmÞ
+ ð0:100 lbmÞð30:0 lbf/in: 2 Þð144 in: 2 /ft 2 Þð0:3105 − 1:9662 ft 3 /lbmÞ
1 Btu
778:17 ft⋅lbf
= − 9:31 Btu
c. Since state 3 is saturated (a mixture of liquid and vapor), T 3 must be equal to the saturation temperature at 30.0 psia,
which, from Table C.7b, is T 3 = 15:38°F.
WHY AREN’T THE VALUES OF u AND h IN THE TABLES THE SAME
AS THOSE YOU GET FROM THERMO COMPUTER PROGRAMS?
To develop a table or thermodynamic program of values of properties like internal energy and enthalpy, a zero reference
point for these properties has to be chosen. This is called a reference state, and whoever develops the table or computer program
is free to chose his or her own reference state. So the values of u and h for a particular material may not be the same
from table to table or the same as those given by computer programs. However, it turns out that this is not important
because the first law uses only the differences, u 2 − u 1 or h 2 − h 1 ,andwhenyousubtracttwovalues, the reference state
used (whatever it was) cancels out. This is like calculating T 2 − T 1 . You can use either ºF or R (or ºC or K) and get the same
answer.
Using EES, the answers to Example 5.6 are (a) 1 W 2 = − 1:02 Btu, (b) 2 Q 3 = − 9:29 Btu, and (c) T 3 = 15:37°F. While these
are not exactly the same as those given in Example 5.6, they differ by less than 3%, which is quite acceptable for this
calculation.
Exercises
7. Find the work transport of energy in part a of Example 5.6 if the working fluid is air (an ideal gas) instead of
Refrigerant-134a. Answer: ð1W 2 Þ a = − 1:03 Btu.
8. Find the final temperature in part c of Example 5.6 if the working fluid is an ideal gas. Answer: T 3 = 96 R.
9. Find the heat transport of energy in part b of Example 5.6 if we set v 3 = v g ð30:0 psiaÞ instead of v 1 /2. Answer:
ð 2 Q 3 Þ b
= −2:20 Btu.
5.8 CLOSED SYSTEM UNSTEADY STATE PROCESSES
One of the most difficult thermodynamic processes to analyze is an unsteady state process. This is largely due to
the fact that there are many more unknowns in these processes. In addition, they usually involve integrating the
rate form of the basic equations, so some knowledge of the solution techniques for ordinary differential equations
is essential before a complete thermodynamic analysis can be carried out. The following example illustrates
this type of problem.
EXAMPLE 5.7
A microwave antenna for a space station consists of a 0.100 m diameter rigid, hollow, steel sphere of negligible wall thickness.
During its fabrication the sphere undergoes a heat-treating operation in which it is initially filled with helium at 0.140 MPa
and 200.°C, then it is plunged into cold water at 15.0°C for exactly 5.00 seconds. The convective heat transfer coefficient of the
sphere in the water is 3.50 W/(m 2 · K). Neglecting any changes in kinetic or potential energy and assuming the helium behaves
as an ideal gas, determine
a. The final temperature of the helium.
b. The change in total internal energy of the helium.
Solution
First, draw a sketch of the system (Figure 5.7).
(Continued )