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530
Air is accelerated in a nozzle from 45 m/s to 180 m/s. The mass flow rate, the exit
temperature, 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
Air is
an ideal gas with constant specific heats.
3
Potential energy changes are negligible.
4
The
device is adiabatic and thus heat transfer is negligible.
5
There are no work interactions.
Properties
The gas constant of air is 0.287
kPa.m
3
/kg.K (Table A1). The specific heat
of air at the anticipated average temperature
of 450 K is
c
p
= 1.02 kJ/kg.
°
C (Table A2).
Analysis
(
a
) There is only one inlet and one
exit, and thus
±±±
mmm
12
==
. Using the ideal
gas relation, the specific volume and the
mass flow rate of air are determined to be
/kg
m
0.4525
kPa
300
)
K
473
)(
K
/kg
m
kPa
0.287
(
3
3
1
1
1
=
⋅
⋅
=
=
P
RT
v
kg/s
1.094
=
=
=
)
m/s
45
)(
m
0.0110
(
/kg
m
0.4525
1
1
2
3
1
1
1
V
A
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
±
±
²
²³
²
²´
µ
±
µ
±
±
=
=
Δ
=
−
Ê
()
2
0
2
0
0)
pe
W
(since
/2)
+
(
)
2
/
(
2
1
2
2
1
2
,
2
1
2
2
1
2
2
2
2
2
1
1
V
V
T
T
c
V
V
h
h
Q
V
h
m
V
h
m
ave
p
−
+
−
=
⎯→
⎯
−
+
−
=
≅
Δ
≅
≅
=
+
±
±
±
±
Substituting,
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
+
−
⋅
=
2
2
2
2
2
/s
m
1000
kJ/kg
1
2
)
m/s
45
(
)
m/s
180
(
)
C
200
)(
K
kJ/kg
1.02
(
0
D
T
It yields
T
2
=
185.2
°
C
(
c
)
The specific volume of air at the nozzle exit is
/kg
m
1.315
kPa
100
)
K
273
185.2
)(
K
/kg
m
kPa
0.287
(
3
3
2
2
2
=
+
⋅
⋅
=
=
P
RT
m/s
180
/kg
m
1.315
1
kg/s
.094
1
1
2
3
2
2
2
A
V
A
m
=
⎯
=
±
→
A
2
= 0.00799 m
2
=
79.9 cm
2
AIR
P
2
= 100
kPa
P
1
= 300
kPa
T
1
=
200 C
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View Full Document 534
Heat is lost from the steam flowing in a nozzle. The velocity and the volume flow
rate at the nozzle exit are to be determined.
Assumptions
1
This is a steadyflow
process since there is no change with
time.
2
Potential energy change is
negligible.
3
There are no work
interactions.
Analysis
We take the steam 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
Energy balance:
0)
pe
since
2
2
0
out
2
2
2
2
1
1
out
in
energies
etc.
potential,
kinetic,
internal,
in
change
of
Rate
(steady)
0
system
mass
and
work,
heat,
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This note was uploaded on 01/03/2011 for the course MEEN 315 taught by Professor Ramussen during the Fall '07 term at Texas A&M.
 Fall '07
 RAMUSSEN

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