# 6-140 allowed to escape slowly such that temperature remains constant. The... 6-167

```6-140
6-167 A tank initially contains saturated mixture of R-134a. A valve is opened and R-134a vapor only is
allowed to escape slowly such that temperature remains constant. The heat transfer necessary with the
surroundings to maintain the temperature and pressure of the R-134a constant is to be determined.
Assumptions 1 This is an unsteady process since the conditions within the device are changing during the
process, but it can be analyzed as a uniform-flow process since the state of fluid at the exit remains
constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved.
Analysis We take the tank as the system, which is a control volume since mass crosses the boundary.
Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and
internal energy u, respectively, the mass and energy balances for this uniform-flow system can be
expressed as
Mass balance:
min m out
'msystem
me
m 2 m1
me
m1 m 2
Energy balance:
E E
in
out
Net energy transfer
by heat, work, and mass
Qin m e he
Qin
'E system
Change in internal, kinetic,
potential, etc. energies
R-134a
0.4 kg, 1 L
25qC
m 2 u 2 m1u1
m 2 u 2 m1u1 m e he
Combining the two balances:
Qin
m 2 u 2 m1u1 (m1 m 2 )he
The specific volume at the initial state is
v1
V
m1
0.001 m 3
0.4 kg
0.0025 m 3 /kg
The initial state properties of R-134a in the tank are
T1
v1
26qC
&frac12;&deg; x1
&frac34;
3
0.0025 m /kg &deg;&iquest;
u1
v1 v f
0.0025 0.0008313
0.05726
(Table A-11)
v fg
0.029976 0.0008313
u f x1u fg 87.26 (0.05726)(156.87) 96.24 kJ/kg
The enthalpy of saturated vapor refrigerant leaving the bottle is
he
h g @ 26qC
264.68 kJ/kg
The specific volume at the final state is
v2
V
m2
0.001 m 3
0.1 kg
0.01 m 3 /kg
The internal energy at the final state is
T2
v2
26qC
&frac12;&deg; x 2
&frac34;
3
0.01 m /kg &deg;&iquest;
u2
v 2 v f
v fg
u f x1u fg
0.01 0.0008313
0.3146
(Table A-11)
0.029976 0.0008313
87.26 (0.3146)(156.87) 136.61 kJ/kg
Substituting into the energy balance equation,
Qin m 2 u 2 m1u1 ( m1 m 2 )he
(0.1 kg )(136.61 kJ/kg ) (0.4 kg )(96.24 kJ/kg ) (0.4 0.1 kg )(264.68 kJ/kg )
54.6 kJ
PROPRIETARY MATERIAL. &copy; 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
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