4-37
4-46 Two tanks initially separated by a partition contain steam at different states. Now the partition is removed and they are
allowed to mix until equilibrium is established. The temperature and quality of the steam at the final state and the amount of
heat lost from the tanks are to be determined.
Assumptions 1 The tank is stationary and thus the kinetic and potential energy
changes are zero. 2 There are no work interactions.
Analysis (a) We take the contents of both tanks as the system. This is a closed
system since no mass enters or leaves. Noting that the volume of the system is
constant and thus there is no boundary work, 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



Q
Change in internal, kinetic,
potential, etc. energies
 Qout  U A  U B  m(u 2  u1 ) A  m(u 2  u1 )B
TANK B
3 kg
150C
x=0.5
TANK A
2 kg
1 MPa
300C
(since W  KE = PE = 0)
The properties of steam in both tanks at the initial state are (Tables A-4 through A-6)
P1, A  1000 kPa v 1, A  0.25799 m 3 /kg

T1, A  300C u1, A  2793.7 kJ/kg
T1, B  150C  v f  0.001091, v g  0.39248 m 3 /kg

u fg  1927.4 kJ/kg
x1  0.50
 u f  631.66,
v 1, B  v f  x1v fg  0.001091  0.50  0.39248  0.001091  0.19679 m 3 /kg
u1, B  u f  x1u fg  631.66  0.50  1927.4   1595.4 kJ/kg
The total volume and total mass of the system are
V  V A  V B  m Av 1, A  m B v 1, B  (2 kg)(0.25799 m 3 /kg)  (3 kg)(0.19679 m 3 /kg)  1.106 m 3
m  m A  m B  3  2  5 kg
Now, the specific volume at the final state may be determined
v2 
V
m

1.106 m 3
 0.22127 m 3 /kg
5 kg
which fixes the final state and we can determine other properties
T2  Tsat @ 300 kPa  133.5 C

v2 v f
0.22127  0.001073

 0.3641
 x2 
3
v g  v f 0.60582  0.001073
v 2  0.22127 m /kg 
u2  u f  x2u fg  561.11  0.3641  1982.1  1282.8 kJ/kg
P2  300 kPa
(b) Substituting,
 Qout  U A  U B  m(u 2  u1 ) A  m(u 2  u1 )B
 (2 kg)(1282.8  2793.7)kJ/kg  (3 kg)(1282.8  1595.4)kJ/kg  3959 kJ
or
Qout  3959 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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4-37 allowed to mix until equilibrium is established. The temperature and... 4-46