8-21
8-36 An insulated cylinder is initially filled with saturated liquid water at a specified pressure. The water is heated
electrically at constant pressure. The minimum work by which this process can be accomplished and the exergy destroyed
are to be determined.
Assumptions 1 The kinetic and potential energy changes are negligible. 2 The cylinder is well-insulated and thus heat
transfer is negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 The compression or expansion
process is quasi-equilibrium.
Analysis (a) From the steam tables (Tables A-4 through A-6),
u1  u f @120 kPa = 439.27 kJ / kg
3
P1  120 kPa  v 1  v f @120 kPa = 0.001047 m /kg

sat. liquid  h1  h f @120 kPa = 439.36 kJ/kg
s1  s f @120 kPa = 1.3609 kJ/kg  K
Saturated
Liquid
H2O
P = 120 kPa
The mass of the steam is
We
3
0.008 m
V
m

 7.639 kg
v 1 0.001047 m 3 / kg
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

E system
inout




Net energy transfer
by heat, work, and mass
Change in internal, kinetic,
potential, etc. energies
We,in  W b,out  U
We,in  m(h2  h1 )
since U + Wb = H during a constant pressure quasi-equilibrium process. Solving for h2,
We,in
1400 kJ
h2  h1 
 439.36 
 622.63 kJ/kg
m
7.639 kg
Thus,
h2  h f 622.63  439.36
x2 

 0.08168
h fg
2243.7
P2  120 kPa

 s 2  s f  x 2 s fg  1.3609  0.08168  5.93687  1.8459 kJ/kg  K
h2  622.63 kJ/kg 
u 2  u f  x 2 u fg  439.24  0.08168  2072.4  608.52 kJ/kg
v 2  v f  x 2v fg  0.001047  0.08168  (1.4285  0.001047)  0.1176 m 3 /kg
The reversible work input, which represents the minimum work input Wrev,in in this case can be determined from the exergy
balance by setting the exergy destruction equal to zero,
X  X out
in

Net exergy transfer
by heat, work,and mass
 X destroyed0 (reversible)  X system  Wrev,in  X 2  X1
 


Exergy
destruction
Change
in exergy
Substituting the closed system exergy relation, the reversible work input during this process is determined to be
Wrev,in   m(u1  u 2 )  T0 ( s1  s 2 )  P0 (v 1  v 2 )
 (7.639 kg){(439.27  608.52) kJ/kg  (298 K)(1.3609  1.8459) kJ/kg  K
+ (100 kPa)(0.001047  0.1176)m 3 / kg[1 kJ/1 kPa  m 3 ]}
 278 kJ
(b) The exergy destruction (or irreversibility) associated with this process can be determined from its definition Xdestroyed =
T0Sgen where the entropy generation is determined from an entropy balance on the cylinder, which is an insulated closed
system,
S in  S out  S gen  S system






Net entropy transfer
by heat and mass
Entropy
generation
Change
in entropy
S gen  S system  m( s 2  s1 )
Substituting,
X destroyed  T0 S gen  mT0 ( s 2  s1 )  (298 K)(7.639 kg)(1.8459  1.3609)kJ/kg  K = 1104 kJ
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