Modern Engineering Thermodynamics

300 CHAPTER 9: Second Law Open System Applications
and the series expansion for the logarithm of k for 0 < k < 2is
ln k = ðk − 1Þ − ðk − 1Þ 2 /2 + ðk − 1Þ 3 /3 −
clearly kR/c p > ln k, which produces the following result when the tanks are filled to the same pressure (but not
with the same amount of mass):
1ðS P
Þ 2 ideal
gas
ðadiabaticÞ
< 1 ðS P
Þ 2 ideal
gas
ðisothermalÞ
In the dissipative hydraulic flows studied earlier in this chapter, we find that less entropy is produced in both incompressible
and ideal gas systems if the flows are carried out adiabatically as opposed to isothermally. The preceding
analysis shows that this is also true in the filling of a rigid vessel. Note, however, that these two filling processes do
not normally produce the same final system state. For example, adiabatic filling clearly produces a higher final temperature
than isothermal filling when starting from the same initial temperature. A complete entropy production
analysis has to include any additional processes required to reduce both systems to the same final state.
EXAMPLE 9.8
A 3.00 ft 3 rigid container is filled with oxygen entering at 70.0°F to a final pressure of 2000. psia. Assuming the container is
initially evacuated and that the oxygen behaves as an ideal gas with constant specific heats, determine the amount of
entropy produced when the container is filled
a. Adiabatically by insulating it.
b. Isothermally by submerging it in a water bath at 70.0°F while filling.
Solution
First, draw a sketch of the system (Figure 9.16).
The unknowns are the amount of entropy produced when the
container is filled adiabatically by insulating it and isothermally
by submerging it in a water bath at 70.0°F while filling.
The material is the oxygen gas in the tank.
O 2 at 70.0°F
O 2 at 70.0°F
From Table C.13a of Thermodynamic Tables to accompany Modern
Engineering Thermodynamics, we find for oxygen c p = 0.219 Btu/
(lbm·R), R = 48.29 ft·lbf/(lbm·R), k=1.39. The final temperature
after filling adiabatically is given by Eq. (6.36) as
ðT 2 Þ adiabatic
filling
= kT in = 1:39ð70:0 + 459:67 RÞ = 736 R = 277°F
and, for isothermal filling,
ðT 2 Þ isothermal
filling
= T in = 70:0 + 459:67 = 530: R
∀=3.00 ft 3
P final = 2000. psia
FIGURE 9.16
Example 9.8.
∀=3.00 ft 3
P final = 2000. psia
Insulation
Water at 70.0°F
(a) 1 (S P ) 2 =? (b) 1 (S P ) 2 = ?
The final mass of oxygen in the container can be found from the ideal gas equation of state as
so
and
ðm 2
ðm 2
Þ adiabatic
filling
Þ isothermal
filling
=
=
p 2 V 2
RðT 2 Þadiabatic
filling
p 2 V 2
RðT 2 Þisothermal
filling
m 2 = p 2V 2
RT 2
ð2000: lbf/in:2Þ 144 in: 2 /ft 2 3:00 ft 3
= = 24:3 lbm
½48:29 ft.lbf/ ðlbm.RÞŠð736 RÞ
ð2000: lbf/in:2Þ 144 in: 2 /ft 2 3:00 ft 3
= = 33:8 lbm
½48:29 ft.lbf/ ðlbm.RÞŠð530: RÞ