535
Steam is accelerated in a nozzle from a velocity of 80 m/s. The mass flow rate, the exit velocity,
and the exit area of the nozzle are to be determined.
Assumptions
1
This is a steadyflow process since there is no change with time.
2
Potential energy
changes are negligible.
3
There are no work interactions.
Properties
From the steam tables (Table A6)
kJ/kg
3196.7
/kg
m
0.057838
C
00
4
MPa
5
1
3
1
1
1
=
=
⎭
⎬
⎫
°
=
=
h
T
P
v
and
kJ/kg
3024.2
/kg
m
0.12551
C
00
3
MPa
2
2
3
2
2
2
=
=
⎭
⎬
⎫
°
=
=
h
T
P
Analysis
(
a
)
There is only one inlet and one exit, and thus
&&&
mmm
12
=
=
. The mass flow rate of steam
is
kg/s
6.92
=
×
=
=
−
)
m
10
50
)(
m/s
80
(
/kg
m
0.057838
1
1
2
4
3
1
1
1
A
V
m
&
(
b
) We take nozzle as the system, which is a control volume since mass crosses the boundary. The
energy balance for this steadyflow system can be expressed in the rate form as
out
in
energies
etc.
potential,
kinetic,
internal,
in
change
of
Rate
(steady)
0
system
mass
and
work,
heat,
by
nsfer
energy tra
net
of
Rate
out
in
0
E
E
E
E
E
&
&
4
43
4
42
1
&
1
&
&
=
=
Δ
=
−
Ê
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
+
−
=
−
≅
Δ
≅
+
=
+
2
0)
pe
W
(since
/2)
V
+
(
)
2
/
(
2
1
2
2
1
2
out
2
2
2
out
2
1
1
V
V
h
h
m
Q
h
m
Q
V
h
m
&
&
&
&
&
&
Substituting, the exit velocity of the steam is determined to be
()
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
+
−
=
−
2
2
2
2
2
/s
m
1000
kJ/kg
1
2
m/s)
(80
3196.7
3024.2
kg/s
6.916
kJ/s
120
V
It yields
V
2
= 562.7 m/s
(
c
) The exit area of the nozzle is determined from
( )( )
2
4
m
10
15.42
−
×
=
=
=
⎯→
⎯
=
m/s
562.7
/kg
m
0.12551
kg/s
6.916
1
3
2
2
2
2
2
2
V
m
A
A
V
m
&
&
Steam
1
2
120 kJ/s
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Air is compressed at a rate of 10 L/s by a compressor. The work required per unit mass and the
power required are to be determined.
Assumptions
1
This is a steadyflow process since there is no change with time.
2
Kinetic and potential
energy changes are negligible.
3
Air is an ideal gas with constant specific heats.
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 Spring '06
 OLER
 Thermodynamics, Energy, Kinetic Energy, Heat, Mass flow rate

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