1
Thermodynamics
)
(
)
(
0
=
PE
=
KE
=
since
)
(
1
2
in
pw,
1
2
out
b,
in
pw,
in
e,
h
h
m
W
t
I
Q
h
h
m
W
U
W
W

=
+
∆

=
+
∆
=
+
V
(
29
kg
4.731
/kg
m
0.001057
m
0.005
kJ/kg
1593.6
1
.
2213
5
.
0
01
.
487
5
.
0
kPa
175
/kg
m
0.001057
kJ/kg
487.01
liquid
sat.
kPa
175
3
3
1
1
2
2
2
2
3
kPa
175
@
1
kPa
175
@
1
1
=
=
=
=
×
+
=
+
=
=
=
=
=
=
=
=
v
V
v
v
m
h
x
h
h
x
P
h
h
P
fg
f
f
f
V
223.9
=
×
=
=
∆

=
+
∆
kJ/s
1
VA
1000
s)
60
A)(45
(8
kJ
4835
kJ
4835
kg
487.01)kJ/
kg)(1593.6
(4.731
kJ)
(400
V
V
V
t
I
t
I
H
2
O
P
= const.
W
pw
W
e
v
P
1
2
539
A cylinder is initially filled with saturated liquid water at a specified pressure. The water is heated electrically as it is stirred by a paddlewheel at constant
pressure.
The voltage of the current source is to be determined, and the process is to be shown on a P
v
diagram.
Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero.
2 The cylinder is wellinsulated and thus heat transfer is
negligible.
3 The thermal energy stored in the cylinder itself is negligible. 4 The compression or expansion process is quasiequilibrium.
THIS IS AN EXAMPLE OF A CLOSED STATIONARY SYSTEM ENERGY ANALYSIS:
since
∆
U
+
W
b
=
∆
H
during a constant pressure quasiequilibrium process. The properties of water are (Tables A4 through A6)
Substituting,
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2
Thermodynamics
m
m
m
1
2
=
=
/kg
m
0.4525
kPa
300
)
K
473
)(
K
/kg
m
kPa
0.287
(
3
3
1
1
1
=
⋅
⋅
=
=
P
RT
v
kg/s
0.5304
V
=
=
=
)
m/s
30
)(
m
0.008
(
/kg
m
0.4525
1
1
2
3
1
1
1
A
v
m
E
E
E
E
E
in
out
in
out

=
=
=
Rate of net energy transfer
by heat, work, and mass
system
(steady)
Rate of change in internal, kinetic,
potential, etc. energies
T
T
T
T
T
T
T
T TT
T
T TT
∆
0
0
631
Air is accelerated in a nozzle from 30 m/s to 180 m/s. The mass flow rate,
the exit temperature, and the exit area of the nozzle are to be determined.
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 Summer '08
 Sherif
 Thermodynamics, Energy, Heat, kPa, potential energy changes

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