8-75
8-84 An insulated cylinder initially contains saturated liquid-vapor mixture of water. The cylinder is connected to a supply
line, and the steam is allowed to enter the cylinder until all the liquid is vaporized. The amount of steam that entered the
cylinder and the exergy destroyed are 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 inlet remains constant. 2 The expansion process is
quasi-equilibrium. 3 Kinetic and potential energies are negligible. 4 The device is insulated and thus heat transfer is
negligible.
Properties The properties of steam are (Tables A-4 through A-6)
 h1  h f  x1 h fg  561.43  0.8667  2163.5  2436.5 kJ/kg

x1  13 / 15  0.8667  s1  s f  x1 s fg  1.6716  0.8667  5.3200  6.2824 kJ/kg  K
P1  300 kPa
P2  300 kPa  h2  h g @ 300 kPa  2724.9 kJ/kg

sat.vapor
 s 2  s g @ 300 kPa  6.9917 kJ/kg  K
Pi  2 MPa  hi  3248.4 kJ/kg

Ti  400C  s i  7.1292 kJ/kg  K
Analysis (a) We take the cylinder as the system, which is a control volume.
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 unsteady-flow system can be expressed as
H2O
300 kPa
P = const.
2 MPa
400C
m in  m out  m system  m i  m 2  m1
Mass balance:
Energy balance:
E E
inout


Net energy transfer
by heat, work, and mass
E system



Change in internal, kinetic,
potential, etc. energies
mi hi  W b,out  m 2 u 2  m1u1 (since Q  ke  pe  0)
Combining the two relations gives 0  W b,out  m 2  m1 hi  m 2 u 2  m1u1
0  m 2  m1 hi  m 2 h2  m1 h1
or,
since the boundary work and U combine into H for constant pressure expansion and compression processes. Solving for
m2 and substituting,
m2 
Thus,
hi  h1
(3248.4  2436.5)kJ/kg
m1 
(15 kg) = 23.27 kg
hi  h2
(3248.4  2724.9)kJ/kg
mi  m 2  m1  23.27  15  8.27 kg
(b) The exergy destroyed during a process can be determined from an exergy balance or directly from its
definition X destroyed  T0 S gen where the entropy generation Sgen is determined from an entropy balance on the insulated
cylinder,
S  S out
in

Net entropy transfer
by heat and mass
 S gen  S system




Entropy
generation
Change
in entropy
mi s i  S gen  S system = m 2 s 2  m1 s1
S gen  m 2 s 2  m1 s1  mi s i
Substituting, the exergy destruction is determined to be
X destroyed  T0 S gen  T0 [m 2 s 2  m1 s1  mi s i ]
 (298 K)(23.27  6.9917  15  6.2824  8.27  7.1292)
 2832 kJ
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
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line, and the steam is allowed to enter the cylinder... 8-84