8-101
8-111 Two rigid tanks that contain water at different states are connected by a valve. The valve is opened and steam flows
from tank A to tank B until the pressure in tank A drops to a specified value. Tank B loses heat to the surroundings. The
final temperature in each tank and the work potential wasted during this process are to be determined.
Assumptions 1 Tank A is insulated and thus heat transfer is negligible. 2 The water that remains in tank A undergoes a
reversible adiabatic process. 3 The thermal energy stored in the tanks themselves is negligible. 4 The system is stationary
and thus kinetic and potential energy changes are negligible. 5 There are no work interactions.
Analysis (a) The steam in tank A undergoes a reversible, adiabatic process, and thus s2 = s1. From the steam tables (Tables
A-4 through A-6),
Tank A :
v 1, A  v f  x1v fg  0.001084  0.80.46242  0.001084  0.37015 m 3 /kg
P1  400 kPa 
 u1, A  u f  x1u fg  604.22  0.81948.9   2163.3 kJ/kg
x1  0.8
 s  s  x s  1.7765  0.85.1191  5.8717 kJ/kg  K
f
1, A
1 fg
T2, A  Tsat @300kPa  133.52 C
P2  300 kPa  x  s 2, A  s f  5.8717  1.6717  0.7895
 2, A
s fg
5.3200
s 2  s1

sat. mixture  v  v  x v  0.001073  0.78950.60582  0.001073  0.47850 m 3 /kg
2, A
f
2, A
fg
u 2, A  u f  x 2, A u fg  561.11  0.78951982.1 kJ/kg   2125.9 kJ/kg
Tank B :
v  1.1989 m 3 /kg
P1  200 kPa  1, B
 u1, B  2731.4 kJ/kg
T1  250C 
s1, B  7.7100 kJ/kg  K
900 kJ
A
V = 0.2 m3
The initial and the final masses in tank A are
m1, A
steam
P = 400 kPa
x = 0.8
V
0.2 m 3
 A 
 0.5403 kg
v 1, A 0.37015 m 3 /kg

B
m = 3 kg
steam
T = 250C
P = 200 kPa
and
m 2, A 
VA
0.2m 3

 0.4180 kg
v 2, A 0.479m 3 /kg
Thus, 0.540 - 0.418 = 0.122 kg of mass flows into tank B. Then,
m2, B  m1, B  0122
.
 3  0122
.
 3.122 kg
The final specific volume of steam in tank B is determined from
v 2, B 
VB
m 2, B

m1v1 B
m 2, B

3 kg 1.1989 m 3 /kg   1.152 m 3 /kg
3.122 m 3
We take the entire contents of both tanks as the system, which is a closed system. 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
 Qout  U  (U ) A  (U ) B
(since W  KE = PE = 0)
 Qout  m 2 u 2  m1u1  A  m 2 u 2  m1u1  B
Substituting,


 900  0.4182125.9  0.54032163.3  3.122u 2, B  32731.4
u 2, B  2425.9 kJ/kg
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.
8-102
Thus,
v 2, B  1.152 m 3 /kg 
u 2, B
T2, B  110.1C

s
 2425.9 kJ/kg  2, B  6.9772 kJ/kg  K
(b) The total entropy generation during this process is determined by applying the entropy balance on an extended system
that includes both tanks and their immediate surroundings so that the boundary temperature of the extended system is the
temperature of the surroundings at all times. It gives
S  S out
in

Net entropy transfer
by heat and mass

 S gen  S system




Entropy
generation
Change
in entropy
Qout
 S gen  S A  S B
Tb,surr
Rearranging and substituting, the total entropy generated during this process is determined to be
S gen  S A  S B 
Qout
Q
 m 2 s 2  m1 s1  A  m 2 s 2  m1 s1  B  out
Tb,surr
Tb,surr
 0.4185.8717   0.54035.8717   3.1226.9772   37.7100  
900 kJ
273 K
 1.234 kJ/K
The work potential wasted is equivalent to the exergy destroyed during a process, which can be determined from an exergy
balance or directly from its definition X destroyed  T0 S gen ,
X destroyed  T0 S gen  (273 K)(1.234 kJ/K )  337 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.