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GasDynamics 1
Isentropic Flow with Area Variation
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GasDynamics
Simple convergent nozzle
2
If the flow is subsonic
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GasDynamics
Isentropic Mach number relations
3
+=
20
2
11 M
T
T
120
2
11
+=
M
P
P
1
1
20
2
11
+=
M
( )121
2
2
*1
1
1
21 +
+
+
+
=
M
MA
A
2
1
20
2
11
+= M
a
a
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GasDynamics
Example (Duct Flow)
4
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GasDynamics
Analysis:
5
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GasDynamics 6
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GasDynamics 7
The maximum velocity is found by converting all the thermal energy to kinetic energy. Taking
zero thermal energy to correspond to absolute zero (despite the fact that air would not be a gas at
this point) one could estimate
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GasDynamics
Sonic Properties(Let * denote a property at the sonic state)
8
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GasDynamics
Effect of Area Change
9
In the isentropic limit the mass, momentum, and energy equation for
a compressible ideal gas reduce to:
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GasDynamics 10
Substitute energy then mass into momentum:
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GasDynamics 11
equation singular when M2 = 1 if M2 = 1, one needs dA = 0 area minimum necessary to
transition from subsonic tosupersonic flow!!
can be shown area maximumnot relevant
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GasDynamics 12
Substitute from earlier-developed relations and get:
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GasDynamics
Reference Stagnation and Critical Conditions
for Isentropic Flow
13
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GasDynamics
Area versus Mach number for a calorically
perfect ideal gas
14
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GasDynamics
Choking
15
Consider mass flow rate variation with pressure difference
small pressure difference gives small velocity, small mass flow
as pressure difference grows, velocity and mass flow rate grow
velocity is limited to sonic at a particular duct location
this provides fundamental restriction on mass flow rate
can be proven rigorously that sonic condition gives maximum mass flow
rate
A flow which has a maximum mass flow rate is known as
choked flow. Flows will choke at area minima in a duct.
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GasDynamics
Isentropic Flow in a Converging Nozzle
16
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GasDynamics 17
1. Pb = Po, Pb /Po = 1. No flow occurs. Pe = Pb, Me=0.
2. Pb > P* or P*/Po < Pb /Po < 1. Flow begins to increase as
the back pressure is lowered. Pe = Pb, Me < 1.
3. Pb = P* or P*/Po = Pb /Po < 1. Flow increases to the choked
flow limit as the back pressure is lowered to the criticalpressure. Pe = Pb, Me=1.
4. Pb < P* or Pb /Po < P*/Po < 1. Flow is still choked and
does not increase as the back pressure is lowered below the
critical pressure, pressure drop from Pe to Pb occurs outsidethe nozzle. Pe = P*, Me=1.
5. Pb = 0. Results are the same as for item 4.
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GasDynamics
Choking Example
18
Given: Air with stagnation conditions Po = 200 kPa To = 500 K flows through a
throat to an exit with Mach number of 2.5. The desired mass flow is 3.0 kg/s,
Find: a) throat area, b) exit pressure, c) exit temperature, d) exit velocity, and e)
exit area.
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GasDynamics
Analysis:
19
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GasDynamics 20
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GasDynamics
Discharge Example
21
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GasDynamics 22
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GasDynamics
Isentropic Flow in a
Converging-Diverging Nozzle
23
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GasDynamics 24
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GasDynamics 25
PA = Po, or PA/Po = 1. No flow occurs. Pe = Pb, Me = 0.
Po > PB > PC > P* or P*/Po < PC/Po < PB/Po < 1. Flow begins
to increase as the back pressure is lowered. The velocity
increases in the converging section but M< 1 at the throat; thus,
the diverging section acts as a diffuser with the velocity
decreasing and pressure increasing. The flow remains subsonicthrough the nozzle. Pe = Pb and Me < 1.
Pb = PC = P* or P*/Po = Pb/Po = PC/Po and Pb is adjusted so
that M=1 at the throat. Flow increases to its maximum value at
choked conditions; velocity increases to the speed of sound atthe throat, but the converging section acts as a diffuser with
velocity decreasing and pressure increasing. Pe = Pb, Me < 1.
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GasDynamics 26
PC > Pb > PE or PE/Po < Pb/Po < PC/Po < 1. The fluid that achieved sonic velocity
at the throat continues to accelerate to supersonic velocities in the diverging section
as the pressure drops. This acceleration comes to a sudden stop, however, as a
normal shock develops at a section between the throat and the exit plane. The flowacross the shock is highly irreversible. The normal shock moves downstream away
from the throat as Pb is decreased and approaches the nozzle exit plane as Pbapproaches PE. When Pb = PE, the normal shock forms at the exit plane of the
nozzle. The flow is supersonic through the entire diverging section in this case,
and it can be approximated as isentropic. However, the fluid velocity drops to
subsonic levels just before leaving the nozzle as it crosses the normal shock.
PE > Pb > 0 or 0 < Pb/Po < PE/Po < 1. The flow in the diverging section is
supersonic, and the fluids expand to PF at the nozzle exit with no normal shock
forming within the nozzle. Thus the flow through the nozzle can be approximated
as isentropic. When Pb = PF, no shocks occur within or outside the nozzle. When
Pb < PF, irreversible mixing and expansion waves occur downstream of the exitplane or the nozzle. When Pb > PF, however, the pressure of the fluid increases
from PF to Pb irreversibly in the wake or the nozzle exit, creating what are called
oblique shocks.
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GasDynamics
Example : Converging-Diverging Area
27
A converging-diverging nozzle has an exit-area-to-throat area ratio of 2.Air enters this nozzle with a stagnation pressure of 1000 kPa and a
stagnation temperature of 500 K. The throat area is 8 cm2. Determine the
mass flow rate, exit pressure, exit temperature, exit Mach number, and exit
velocity for the following conditions:
Sonic velocity at the throat, diverging section acting as a nozzle.Sonic velocity at the throat, diverging section acting as a diffuser.
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GasDynamics 28
For A/A* = 2, Table A-32 yields two Mach numbers, one > 1 and one < 1.When the diverging section acts as a supersonic nozzle, we use the value for M> 1.
Then, for AE/A* = 2.0, ME = 2.197, PE/Po = 0.0939, and TE/To = 0.5089,
P P kPa kPa
T T K K
E o
E o
= = =
= = =
0 0939 0 0939 1000 939
08333 0 5089 254 5
. . ( ) .
. . (500 ) .
C kRT
kJ
kg K
K
m
s
kJkg
m
s
E E=
=
=
14 0 287 254 5
1000
319 7
2
2
. ( . )( . )
.
Analysis:
r
V M Cm
s
m
sE E E= = =2197 319 7 702 5. ( . ) .
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GasDynamics
The mass flow rate can be calculated at any known cross-sectional area where the
properties are known. It normally is best to use the throat conditions. Since the flow
has sonic conditions at the throat, Mt = 1, and
*0.8333
0.8333 0.8333(500 ) 416.6
T Tt
T To o
T T K K t o
= =
= = =
21000
21.4(0.287 )(416.6 )
409.2
V C kRT t t t
m
kJ sKkJkg K
kg
m
s
= =
=
=
*0.528
0.528 0.528(1000 ) 528
P Pt
P Po o
P P kPa kPat o
= =
= = =
tP
RT
kPa
kJ
kg KK
kJ
m kPa
kg
m
= = =
=
**
*
(528 )
( . )( . )
.
0 287 416 6
4 416
3
3
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GasDynamics
&
. (8 )( . )( )
.
m AV
kg
mcm
m
s
m
cm
kg
s
t t t=
=
=
4 416 409 2100
1446
3
22
2
When the diverging section acts as a diffuser, we use M< 1. Then, for
AE /A* = 2.0, ME = 0.308, PE /Po = 0.936, and TE /To = 0.9812,
P P kPa kPa
T T K K
E o
E o
= = =
= = =
0 0939 0 936 1000 936
08333 0 9812 490 6
. . ( )
. . (500 ) .
C kRT
kJ
kg K
K
m
s
kJkg
m
s
E E=
=
=
14 0287 490 6
1000
444 0
2
2
. ( . )( . )
.
Since M= 1 at the throat, the mass flow rate is the same as that in the first
part because the nozzle is choked.
r
V M Cm
s
m
sE E E
= = =0 308 444 0 136 7. ( . ) .
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