Modern Engineering Thermodynamics

6.10 Open System Unsteady State Processes 193
Air at 20.0°C
and 1.40 MPa Air T Final = ?
FIGURE 6.16
Example 6.8.
The unknown is the final temperature in the tank immediately after filling. The material is air.
The final temperature in the scuba tank immediately after it has been filled is given by Eq. (6.36) as
T final
= kT in
filling
In this problem, T in = 20.0°C and Table C.13b gives k = 1.40 for air. Then, Eq. (6.36) gives
T final
= kT in = 1:40ð20:0 + 273:15Þ = 410 K = 137°C
filling
Therefore, bottled gas tanks can get quite hot during their filling process and should be cooled to minimize any rupture
potential.
Exercises
20. If the scuba tank in Example 6.8 were filled with air at 70.0°F and 2000. psia, determine the final temperature,
neglecting any air initially in the tank and assuming the tank is insulated during the filling process.
Answer: T final = 742 R = 282°F.
21. How do you account for the fact that the final temperature given by Eq. (6.36) is independent of the ideal gas
pressure? Answer: Since the internal energy of an ideal gas is independent of pressure (see Chapter 3) and the
property that defines the final state after the filling is complete is u 2 , the final pressure cannot affect the final
temperature when an ideal gas is used in the filling operation.
22. If the scuba tank in Example 6.8 is filled very quickly, it appears to be insulated whether it is actually insulated or not
(why?) and the air inside gets quite hot. However, the tank is a heavy walled steel vessel with a large capacity to absorb
the heat produced in filling the tank. If the tank itself has a mass of 10.0 kg, a specific heat of 0.503 kJ/kg · K, and
initially is at 20.0°C, and it contains 0.500 kg of air with (c v ) air = 0.718 kJ/kg · K that is initially at 137°C, determine the
final equilibrium temperature of the tank-air combined system. Answer: T combined = 27.8°C.
The tank-emptying process is illustrated in Figure 6.17. Again we
neglect flow stream specific kinetic and potential energies and
require that the tank remain stationaryandhavenoworktransport
of energy.
Valve
When the valve is opened, the initially pressurized rigid tank
empties into the environment. The generalized energy rate balance
for this process reduces to
_Q − 0 + 0 − _m out ðh out + 0 + 0Þ = ðdU/dt + 0 + 0
Þ tank
or
FIGURE 6.17
Emptying a rigid vessel.
Rigid tank (vessel)
System boundary
_Q − _m out h out = ½dðmuÞ/dtŠ tank
Again, multiplying through by dt and integrating gives
1Q 2 −
Z 2
1
h out dm out = ðm 2 u 2 − m 1 u 1 Þ tank
(6.37)
where state 1 is the filled state and state 2 is the empty state (the
reverse of the filling process). Unlike the filling process described
earlier, the emptying process has no constant flow stream specific