exam 1 could look roughly like this:
exam 1 could look roughly like this:
“exam 1 practice set 2”
3-51
4-29
solutions
3-51 A rigid tank that is filled with saturated liquid-vapor mixture is heated. The
temperature at which the liquid in the tank is completely vaporized is to be determined,
and the T-v diagram is to be drawn.
Analysis This is a constant volume process (v = V /m = constant),
and the specific volume is determined to be
H2O
v
V
m

2.5 m 3
 0.1667 m 3 /kg
15 kg
75C
When the liquid is completely vaporized
the tank will contain saturated vapor only.
Thus,
T
2
v 2  v g  0.1667 m /kg
3
1
The temperature at this point is the temperature that
corresponds to this vg value,
T  Tsat@v 0.1667 m3 /kg  187.0C (Table A-4)
g
v
4-29 Saturated water vapor is isothermally condensed to a saturated liquid in a pistoncylinder device. The heat transfer and the work done are to be determined.
Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy
changes are zero. 2 There are no work interactions involved other than the boundary
work. 3 The thermal energy stored in the cylinder itself is negligible. 4 The compression
or expansion process is quasi-equilibrium.
Analysis We take the contents of the cylinder as the system. This is a closed system
since no mass enters or leaves. The energy balance for this stationary closed system can
be expressed as
E E
inout

Net energy transfer
by heat, work, and mass

E system



Change in internal,kinetic,
potential,etc. energies
Wb,in  Qout  U  m(u 2  u1 )
(since KE = PE = 0)
Qout  Wb,in  m(u 2  u1 )
The properties at the initial and final states are (Table A-4)
Water
200C
sat.
vapor
Heat
T1  20 0C  v 1  v g  0.12721 m 3 / kg

x1  1
 u1  u g  2594 .2 kJ/kg
P1  P2  1554 .9 kPa
T2  20 0C  v 2  v f  0.001157 m 3 / kg

x2  0
 u 2  u f  850 .46 kJ/kg
T
2
The work done during this process is
wb,out 

2
1
 1 kJ
PdV  P(v 2  v 1 )  (1554.9 kPa)(0.001 157  0.12721) m 3 /kg 
 1 kPa  m 3

That is,
wb,in  196.0 kJ/kg
Substituting the energy balance equation, we get
qout  wb,in  (u 2  u1 )  wb,in  u fg  196.0  1743.7  1940kJ/kg
1
v

  196 .0 kJ/kg

