Post on 01-Mar-2018
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Howtoughisthisgonna get?
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Justhavetherightattitude,youllbefine.
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Coverage
CompressibleFlow
SupersonicFlowNormalShockWaves
ObliqueShockWavesExpansionWaves
CompressibleFlowThroughNozzles,Diffusers
andWindTunnels
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CompressibleFlow
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PerfectGas
Aperfectgasisquite,discreetanddefinitelyjustagas.
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PerfectGas
RTp =pressure density
specificgasconstant
R=287J/(kgK)=1716(ftlb)/(slugR)
forair
temperature
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InternalEnergy
dvv
e
dTT
e
de
+
=
Mathematically,
But,
)()( vfTfe =Thus,
dTcdTTede v==
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Enthalpy
pveh +=
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Enthalpy
dTT
h
dpp
h
dh
+
=
Mathematically,
But,
)()( pfTfh =Thus,
dTcdTThdh p==
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SpecificHeatsPreviously,
Alternativeview:
v
vTec
=
dTcdTT
e
de v=
= dTcdTT
h
dh p=
=
p
pThc
=
constantvolume
specificheat
constantpressure
specificheat
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SpecificHeats
dTcde v= dTcdhp
=
Tce v= Tch p=
CaloricallyPerfectGas:cvandcpareassumedconstant.
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RelationBetweenSpecificHeats
pveh +=
RTTcTcvp
+=
Rcc vp +=
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RelationBetweenSpecificHeats
Rcc vp +=
pp
v
p
p
c
R
c
c
c
c+=
v
p
c
c=
pc
R+=
11
1=
Rcp
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RelationBetweenSpecificHeats
Rcc vp +=
RcR
v+=
1
1=
R
cv
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SpecificHeatsRatio
v
p
c
c
=
airfor4.1=
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FirstLawofThermodynamics
dewq =+
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Processes
AdiabaticProcess.
Oneinwhichnoheatisaddedtoortakenawayfromthesystem.
ReversibleProcess.
Oneinwhichnodissipativephenomenaoccur,thatis,wheretheeffects
ofviscosity,thermalconductivityandmassdiffusionareabsent.
IsentropicProcess.
Onethatisbothadiabaticandreversible.
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WorkForareversibleprocess,closedsystem
depdvq =
pdvw =
TheFirstLawbecomes,
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Entropy
irrev
dsT
qds +=
T
q
ds rev
=
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SecondLawofThermodynamics
T
qds
0ds
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Entropy
depdvq =
depdvTds =
depdvTds +=
vdppdvdedh ++=
vdpdhq =
vdpdhTds =
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EntropydepdvTds += vdpdhTds =
T
pdv
T
dTcds v +=
T
vdp
T
dTcds
p =
v
Rdv
T
dTc
ds v
+= p
Rdp
T
dTc
ds p
=
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v
Rdv
T
dTcds v +=
p
Rdp
T
dTcds
p =
1
2
1
2 lnlnp
pR
T
Tcs p =
1
2
1
2 lnlnv
vR
T
Tcs v +=
Entropy
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IsentropicProcessLets=0inthepreviousequations.
)1/(
1
2
1
2
1
2
=
=
T
T
p
p
Tattoothisexpressiononyourminds.
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Example
1) A perfect gas is expanded adiabatically from 5 to 1 bar
by the law pV1.2 = constant. The initial temperature is200C. Calculate the change in specific entropy. R = 287.15
J/kgK, =1.4
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Example
1) Consider a point in a flow where the velocity andtemperature are 230m/s and 375K respectively.Calculate the total enthalpy at this point.
2) An airfoil is in a freestream where P
= 0.75 atm, = 0.942 kg/m
3 and V = 325 m/s. At a point onthe airfoil surface, the pressure is 0.62 atm.Assuming isentropic flow, calculate the velocity at
the point.
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ExampleConsideraBoeing747flyingatastandardaltitudeof36,000ft.The
pressureatapointonthewingis400Ib/ft2.Assumingisentropicflow
overthewing,calculatethetemperatureatthispoint.
R.391Tandlb/ft476pft,36,000ofaltitudestandardaAt 2 ==
)1/(
=
T
T
p
p
R372476
400
91.3
4.1/4.0/)1(
=
=
=
p
p
TT
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CompressibilityMeasureoftherelativevolumechangewithpressure
p p+dp
dpd
1=
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Compressibility
T
Tdp
d
=
1
Isothermalcompressibility
Isentropiccompressibility
s
s dp
d
=
1
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Compressibility
dp
d
1=
dp
d
1=
dpd =
Wheneverafluidexperiencesachangeinpressure,
thereisacorrespondingchangeindensity.
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Compressibility
dpd =
forsolidsandliquidsissmall;
thusdissmallforeverydp
(ispracticallyconstant)
foragasinlowspeedflowmaybelarge,
butasmalldp dominates
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Compressibility
p +dp
p
p +dp
p
IncompressibleFlow
CompressibleFlow
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ContinuityEquation
=+
S
SdVdt 0VV
0=+ V
t
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MomentumEquation
+=+
S SdfSpdVSdVdVt
VVV)(V
xfx
p
Dt
Du +
= yf
y
p
Dt
Dv +
=
zfz
p
Dt
Dw +
=
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EnergyEquation
+=
++
+
S S
dVfSdVpdqSdVVedVet
VV
2
V
2
V)(V2
V2
&
)()2/( 2
VfVpqDt
VeD+=+ &
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Dontpanic,thingsdonthavetobethishard.
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OtherEquations
RTp =
Tce v=
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StagnationProperties
Propertiesthatwouldexistatapointinaflow
IF(inourimagination)thefluidelementpassingthroughthatpointwerebroughtdown
torestadiabatically.
Everypointinaflowhasbothstaticand
stagnationproperties.
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Apoint(orpoints)intheflowwhereV=0.
a)Fluidelementadiabaticallyslowdown
b)Aflowimpingesonasolidobject
V1
V2=0
Total(Stagnation)Conditions
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Thesamething
TotalPressure
StagnationPressurePitot Pressure
ReservoirPressure
ImpactPressure
HeadPressure
NosePressure
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TotalEnthalpyTotalenthalpyisconstantinasteadyadiabaticinviscid flow.
2
constant2
0
Vhh +==
Thisistheenergyequationforsteadyadiabaticinviscid flow.
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TotalTemperatureTotaltemperatureisconstantinasteadyadiabaticinviscid flow
foracaloricallyperfectgas.
00 Tch p=
constant0=T
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TotalEnthalpyandTotalTemperature
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TotalPressureandTotalDensityTotalpressureandtotaldensitycanalsobedefined
inaflowsimilartohowtotalenthalpyortotaltemperatureisdefined,butthereisanadditional
requirementtotheprocessofbringingaparticleto
rest,thatis,theprocessmustalsobereversible,inotherwords,theprocesshastobeisentropic.
0p 0
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TotalPressureandTotalDensityTotalpressureandtotaldensityareconstantinanisentropicflow.
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FlowRegimes
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FlowRegimes
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FlowRegimes
SubsonicFlow
TransonicFlow
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FlowRegimes
TransonicFlow
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FlowRegimes BowShockWave
BowShock M>1
M
>1
M
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FlowRegimesSupersonicFlow
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FlowRegimesHypersonicFlow
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WhatisaShockWave?
Shockwave:Alargeamplitudecompression
wave,suchasthatproducedbyanexplosion,
causedbysupersonicmotionofabodyina
medium.
FromtheAmericanHeritageDictionary
oftheEnglishLanguage,1969
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WhatisaShockWave?
Ashockwaveisanextremelythinregion,
typicallyontheorder105 cm,acrosswhich
theflowpropertiescanchangedrastically.
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NormalShockWaves
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ObliqueShockWaves
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ObliqueShockWave
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Whatwediscussedsofar
Taketimetochewtheinformation.
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NormalShockWaves
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NSWEquations
assumptions1]Theflowissteady
2]Theflowisadiabatic
3]Therearenoviscous
effectsonthesidesofthe
controlvolume.
4]Therearenobody
forces
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NSWEquations
continuity
=+
S
SdVdt
0VV
zero
=S
SdV 0
2211 uu =
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NSWEquations
momentum
zero
+=+
S S
dfSpdVSdVdVt
VV
V)(V
zero
=S S
SpdVSdV )(
2
222
2
111 upup +=+
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NSWEquations
energy
zero zero
+=
++
+
S S
dVfSdVpdqSdVV
edV
et
VV
2
V
2
V)(V2
V2
&
zero
=
+
S S
SdVpSdVV
e2
2
22
2
22
2
11
uh
uh +=+
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NSW:5equations,5unknowns
2211 uu =2
222
2
111 upup +=+
22
2
22
2
11
uh
uh +=+
momentum
energy
continuity
22 Tch p=
222 RTp =equationofstate
enthalpy
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Whatissoundandhowdoesittravel?
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Whatissoundandhowdoesittravel?
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Whatissoundandhowdoesittravel?
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SpeedofSound
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SpeedofSound
s
pa
=
p
a=
RTa =
Thespeedofsoundina
caloricallyperfectgasisafunctionoftemperatureonly.
Atsealevel,a=340.9m/s
ora=1117ft/s.
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SpecialFormsoftheEnergyEquation
22
22
2
21
1
uhuh +=+
22
2
22
2
11
VhVh +=+
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SpecialFormsoftheEnergyEquation
22
2
22
2
11
uTcuTc pp +=+
2121
2
2
2
2
2
1
2
1 uaua +
=+
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SpecialFormsoftheEnergyEquation
2121
2
2
2
2
2
1
2
1 uaua +
=+
121
2
022
=+
aua
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SpecialFormsoftheEnergyEquation
2121
2
2
2
2
2
1
2
1 uaua +
=+
2*22
)1(2
1
21aua
+=+
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SpecialFormsoftheEnergyEquation
22
2
22
2
11 uTcuTc pp +=+
0
2
2TcuTc pp =+
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SpecialFormsoftheEnergyEquation
20
2
11 M
T
T +=
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SpecialFormsoftheEnergyEquation
)1(
20
211
+=
Mp
p
)1(1
20
211
+=
M
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SonicConditions
Similartotheideaofastagnationcondition.
Howeverinsteadofbringingaparticletorest,
itisacceleratedordeceleratedtoMach1.
Everypointinaflowhasanassociatedstatic,
stagnationandsonicproperties.
*p
* *T
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SonictoStagnationRatios
833.01
2
0
*
=+= T
T
528.01
2 )1(
0
*
=
+=
p
p
634.01
2 )1(1
0
*
=
+=
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Characteristic(reference)MachNumber
**
*
a
u
RT
uM ==
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M&M*
2*22
)1(2
1
21 a
ua
+
=+
2*2
)1(2
1
2
1
1
)/(
+=+
u
aua
2
11
)1(2
1
1
)/1( 2
*
2
+=
M
M
2
22*
)1(2
)1(
M
MM
+
+=
)1(/)1(
22*
2
+=
MM
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ExampleConsiderapointinanairflowwherethelocalMachnumber,static
pressure,andstatictemperatureare3.5,0.3atm,and180K,respectively.
Calculatethelocalvaluesofp0,T0,T*,a*,andM*atthispoint.
)1(
20
2
11
+=
Mp
p
20
2
11 M
T
T +=
)14.1(4.1
20 5.32
14.11
3.0
+=
patm9.220=p
20 5.32
14.11
180
+=
TK6210=T
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ExampleConsiderapointinanairflowwherethelocalMachnumber,static
pressure,andstatictemperatureare3.5,0.3atm,and180K,respectively.
Calculatethelocalvaluesofp0,T0,T*,a*,andM*atthispoint.
833.01
2
0
*
=+
=T
T
m/s456)5.517)(287(4.1** === RTa
833.0621
*
=T
K5.517* =T
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ExampleConsiderapointinanairflowwherethelocalMachnumber,static
pressure,andstatictemperatureare3.5,0.3atm,and180K,respectively.
Calculatethelocalvaluesofp0,T0,T*,a*,andM*atthispoint.
2
22*
)1(2
)1(
M
MM
+
+=
06.25.3)14.1(2
5.3)14.1(*
2
2
=+
+=M
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ExampleConsideranairfoil inafreestream whereM=0.6andP =1atm,as
shownbelow.Atpoint1ontheairfoil thepressureisP1=0.7545atm.
CalculatethelocalMachnumberatpoint1.Assumeisentropicflowover
theairfoil.
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ExampleConsideranairfoil inafreestream whereM=0.6andP =1atm,as
shownbelow.Atpoint1ontheairfoil thepressureisP1=0.7545atm.
CalculatethelocalMachnumberatpoint1.Assumeisentropicflowover
theairfoil.
)1(
20
211
+=
Mp
p
Thefreestream totalpressureis,
)14.1(4.1
20 6.0214.11
1
+=p atm276.10=p
Thisisalsothetotalpressureatpoint1because
totalpressureisconstantinanisentropicflow.
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ExampleConsideranairfoil inafreestream whereM=0.6andP =1atm,as
shownbelow.Atpoint1ontheairfoil thepressureisP1=0.7545atm.
CalculatethelocalMachnumberatpoint1.Assumeisentropicflowover
theairfoil.
)1(20
2
11
+=
Mp
p )14.1(4.12
2
14.11
7545.0
276.1
+= M 0.91=M
Atpoint1,
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Example
V1
=?
Consideranairfoil inafreestream whereM=0.6andP =1atm,as
shownbelow.Atpoint1ontheairfoil thepressureisP1=0.7545atm.
Calculatethevelocityatpoint1whenthefreestream temperatureis59oF.
Assumeisentropicflowovertheairfoil.
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ExampleConsideranairfoil inafreestream whereM=0.6andP =1atm,as
shownbelow.Atpoint1ontheairfoil thepressureisP1=0.7545atm.
Calculatethevelocityatpoint1whenthefreestream temperatureis59oF.
Assumeisentropicflowovertheairfoil.
)1(
11
=
T
T
p
p R9.4781
7545.0519
4.1/)14.1(/)1(
1
1 =
=
=
p
pTT
ft/s6.1072)9.478)(1716(4.111 === RTa
R51959460 =+=T
ft/s4.965)6.1072)(9.0(111 === aMV
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Whenisaflowcompressible?
3.0>MRuleofthumb:
Why?BecauseChuckNorrissaysso?
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Thisguydontthinkso.
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Whenisaflowcompressible?
0
2
4
6
8
10
12
14
0 0.5 1 1.5 2 2.5 3
Mach Number
Stagnat
ion
toStaticDensityRatio
Cp/Cv=1.4
1
1
20
2
11
+=
M
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Whenisaflowcompressible?
)1(1
20
2
1
1
+=
M
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Whenisaflowcompressible?
0
2
4
6
8
10
12
14
0 0.5 1 1.5 2 2.5 3
Mach Number
Stagnat
ion
to
StaticDensityRatio
Cp/Cv=1.4
0.95
1
1.05
1.1
1.15
1.2
0 0.1 0.2 0.3 0.4 0.5
Mach Number
Stagnation
to
StaticDensityRa
tio
Cp/Cv=1.4
1
1
20
211
+=
M
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CompressibilitySensitivitywith
0
5
10
15
20
25
30
0 0.5 1 1.5 2 2.5 3
Mach Number
Stagnati
on
to
StaticDensityRat
io
Cp/Cv=1.4
Cp/Cv=1.2
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NormalShockWaves
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NSW:5equations,5unknowns
2211 uu =2222
2111 upup +=+
22
2
22
2
11
uh
uh +=+
momentum
energy
continuity
22 Tch p=
222 RTp =equationofstate
enthalpy
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Prandtl Relation
21
2*uua =
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MachNumbersRelation
21
2*
uua =*
2
*
11 MM=
2
2
2*
)1(2)1(
MMM
+
+=
2/)1(
]2/)1[(1
21
2
12
2
+=
M
MM
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MachNumbersRelation
2/)1(
]2/)1[(12
1
2
12
2
+=
M
MM
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DensityRatio
2211 uu =
2*
12*
2
1
12
2
1
2
1
1
2 Ma
u
uu
u
u
u====
2
2
2*
)1(2
)1(
M
MM +
+=
2
1
21
1
2
)1(2
)1(
M
M
++=
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DensityRatio
2
1
2
1
1
2
)1(2
)1(
M
M
+
+=
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PressureRatio2
222
2
111 upup +=+
===
1
22
112111
2
22
2
1112 1)(uuuuuuuupp
=
=
=
1
22
1
1
2
2
1
2
1
1
2
1
2
11
1
12 111
u
uM
u
u
a
u
u
u
p
u
p
pp
+
+=
2
1
2
12
1
1
12
)1(
)1(21
M
MM
p
pp
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PressureRatio
( )1)1(
21
2
1
1
2
+
+= M
p
p
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TemperatureRatio
=
1
2
1
2
1
2
pp
TT
( ) 21
2
12
1
1
2
)1(
)1(211
21
M
MMT
T
+
+
++=
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( ) 2
1
2
12
1
1
2
)1(
)1(21
1
21
M
MM
T
T
+
+
++=
TemperatureRatio
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EntropyChange
WhymustM1 1?
Theequationsdescribingtherelationship
betweenupstreamanddownstream
propertiesdonotexplicitlyrestrictthevalue
fortheupstreamMachnumber.
Whathas the2ndLawofThermodynamics
gottosayaboutthis?
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EntropyChange
( )
( )
++
+
+
++=
1)1(
21ln
)1(
)1(21
1
21ln
2
1
2
1
2
12
1
MR
M
MMcs p
1
2
1
2 lnln
p
pR
T
Tcs p =
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EntropyChange
( ) ( )
++
+
+
++= 1
)1(
21ln
)1(
)1(21
1
21ln
2
121
2
12
1
MRM
MMcs
p
The2ndLawstatesthat 0sIf 11=M 12 ss = 0=sthen or
If 11>M then 0>s
Butif 11
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WhathappenstothetotalpropertiesacrossaNSW?
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TotalTemperatureChange
BecausetheflowacrossaNSWisadiabatic,
totaltemperatureisconserved.
2,01,0 TT =
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TotalPressureChange
a
a
a
apaa
p
pR
T
Tcss
1
2
1
212 lnln =
1,0
2,0
1,0
2,0
12 lnlnp
pR
T
Tcss p =
1,0
2,0
12 ln
p
pRss =
Rsse
p
p/)(
1,0
2,0 12 =
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TotalPressureChange
Rss
ep
p/)(
1,0
2,0 12
=
Since 012 ss
1,02,0 pp < Thatis,totalpressuredecreases
acrossaNSW.
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NormalShockWave
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NormalShockWave
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ExampleConsideranormalshockwaveinairwheretheupstreamflowproperties
areu1=680m/s,T1=288K,andp1=1atm.Calculatethevelocity,
temperature,andpressuredownstreamoftheshock.
( )1
)1(
21
2
11
2 +
+= Mp
p
( ) 2
1
2
12
1
1
2
)1(
)1(21
1
21
M
MM
T
T
+
+
++=
m/s340)288)(287(4.111 === RTa
2340/680/ 111 === auM
( )12
)14.1(
)4.1(21
1
22 +
+=p
atm5.42=
p
K4862 =T
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ExampleConsideranormalshockwaveinairwheretheupstreamflowproperties
areu1=680m/s,T1=288K,andp1=1atm.Calculatethevelocity,
temperature,andpressuredownstreamoftheshock.
m/s442)486)(287(4.122 === RTa
m/s255)442(5774.0222 === aMu
2/)1(
]2/)1[(12
1
2
12
2
+=
M
MM 5774.0
2/)14.1()2(4.1
2]2/)14.1[(12
2
2 =
+=M
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Example
2=M
K3.223
N/m10x65.2 24
=
=
T
p 2.02 =M
isentropic
?pand? 22 ==T
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TheSolutionPlanGiven: 2=M
24 N/m10x65.2=p K3.223=T
Required: 22 andTp
)1(
2,0
211
+=
Mp
p
=p
p
p
p
p
p
p
p1
,01
1,0
,0
1,0
= ,0,0
1,0
1,0NSWbehindpressuretotalCompute pp
pp
)1(
2
1
1
1,0
2
11
+=
Mp
p ( )1)1(
21
21 +
+=
Mp
p
2/)1(]2/)1[(1
2
2
2
1
+=
MMM
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TheSolutionPlanGiven: 2=M
24 N/m10x65.2=p K3.223=T
Required: 22 andTp
2,0
2
11
+= MT
T
ThetotaltemperaturesinfrontofandbehindtheNSWarethesame
becausetheflowacrossaNSWisadiabatic;also,totaltemperatureand
totalpressuresremainconstantinanisentropicflow,thus,
ComputethetotaltemperatureinfrontofandbehindtheNSW.
== ,02,01,0 TTT = ,02,01,0 ppp
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TheSolutionPlanGiven: 2=M
24 N/m10x65.2=p K3.223=T
Required: 22 andTp
Computethepressureandtemperatureatpoint2using,
)1(
2
2
2
2,0
2
1,0
2
1
1
+==
Mp
p
p
p 22
2
2,0
2
,0
2
1
1 MT
T
T
T
+==
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ExampleGiven:
)1(
2,0
2
11
+=
Mp
p
2,0
2
11
+= MT
T
2=M24N/m10x65.2=p 2.02 =M
Thefreestream totalpressureandtotaltemperatureare,
K3.223=T
)14.1(4.1
2,0 22
14.11
26500
+=
p 25,0 N/m10x07.2=p
2,0 22
14.11
3.233
+=
TK9.014,0 =T
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ExampleGiven: 2=M
24 N/m10x65.2=p 2.02=M
TheMachnumberbehindtheNSWis,
K3.223=T
2/)1(
]2/)1[(12
22
1
+=
M
MM
577.02.0)2(4.1
)2(2.012
2
1 =
+=M
25
,0
N/m10x07.2=
p K9.014,0
=TComputed:
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Example
Thepressureratiosare,
Given: 2=M24 N/m10x65.2=p 2.02=MK3.223=T
25
,0
N/m10x07.2=
p K9.014,0
=TComputed: 577.01
=M
824.7
2
11
)1(
2,0 =
+=
M
p
p
253.12
11
)1(
2
1
1
1,0 =
+=
Mp
p
( ) 5.41)1(
21
21 =+
+=
Mp
p
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Example
ThetotalpressurebehindtheNSWis,
Given: 2=M24 N/m10x65.2=p 2.02=MK3.223=T
25
,0
N/m10x07.2=
p K9.014,0
=TComputed: 577.01
=M
7209.0)5.4)(824.7/1)(253.1(1
,01
1,0
,0
1,0 ===
pp
pp
pp
pp
824.7,0 =
p
p253.1
1
1,0 =p
p5.41 =
p
p
255
,0
,0
1,0
1,0 N/m10x49.1)10x07.2)(7209.0( ===
pp
pp
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Example
ThetotaltemperaturesinfrontofandbehindtheNSWarethesame,
Given: 2=M24 N/m10x65.2=p 2.02=MK3.223=T
25
,0
N/m10x07.2=p K9.014,0
=TComputed: 577.01
=M
824.7,0 =
p
p253.1
1
1,0 =p
p5.41 =
p
p 251,0 N/m10x49.1=p
K9.014,01,0 == TT
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Example
Theflowbetweenpoints1and2isisentropic,thus,thetotalpressureandthetotaltemperatureareconstant.
Given: 2=M24 N/m10x65.2=p 2.02=MK3.223=T
25
,0
N/m10x07.2=p K9.014,0 =TComputed: 577.01=M
824.7,0 =
p
p253.1
1
1,0 =p
p5.41 =
p
p 251,0 N/m10x49.1=p
K9.014,01,0 == TT
)1(
2
2
2
2,0
2
11
+=
Mp
p ( ) atm42.1)2.0(2.01149000 25.32
2
=+= pp
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Example
Theflowbetweenpoints1and2isisentropic,thus,thetotalpressureandthetotaltemperatureareconstant.
Given: 2=M24 N/m10x65.2=p 2.02=MK3.223=T
25
,0
N/m10x07.2=p K9.014,0 =TComputed: 577.01=M
824.7,0 =
p
p253.1
1
1,0 =p
p5.41 =
p
p 251,0 N/m10x49.1=p
K9.014,01,0 == TT
2
2
2
2,0
2
11 M
T
T += K399)2.0(2.01
9.4012
2
2
=+= TT
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SCRAMJETTheresultsofthepreviousexampleare,
K3992=Tatm42.12=p
,2originaltheofinstead10If == MM
K46532=Tatm7.322=p
Theseresultsdescribeanextremeenvironmentthatisverydifficultto
handleforaramjet.
Thesolutionis,DONOTslowtheflowtoM2=0.2.
Keeptheflowsupersonicallthroughout.
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SubsonicCompressibleFlow
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SubsonicCompressibleFlow)1(
20
2
11
+=
Mp
p
=
11
2)1(
02
p
pM
=
11
2 )1(02
p
paV
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SupersonicCompressibleFlow
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SupersonicCompressibleFlow
1
2
2
2,0
1
2,0
p
p
p
p
p
p=
)1(
2
2
2
2,0
2
11
+=
Mp
p
( )1)1(
21
2
1
1
2
+
+= M
p
p
2/)1(
]2/)1[(12
1
2
12
2
+=
M
MM
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SupersonicCompressibleFlow
121
)1(24
)1( 2
1
)1(
2
1
2
12
1
2,0
++
+=
M
M
Mp
p
RayleighPitot TubeFormula
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Example
APitot tubeisinsertedintoanairflowwherethestaticpressureis1atm.
CalculatetheflowMachnumberwhenthePitot tubemeasures(a)1.276
atm,(b)2.714atm,(c)12.06atm.
First,determinethetotalpressurethatdividessubsonicandsupersonicflow.
)1(
20
2
11
+=
Mp
pppp 893.11
2
14.11
)14.1(4.1
2
0 =
+=
subsonic.isflowtheatm,893.1When 0p
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Example
APitot tubeisinsertedintoanairflowwherethestaticpressureis1atm.
CalculatetheflowMachnumberwhenthePitot tubemeasures(a)1.276
atm,(b)2.714atm,(c)12.06atm.
(a)Flowissubsonic
=
11
2)1(
02
p
pM
=
11
276.1
14.1
2 4.1)14.1(
M
6.01=M
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Example
APitot tubeisinsertedintoanairflowwherethestaticpressureis1atm.
CalculatetheflowMachnumberwhenthePitot tubemeasures(a)1.276
atm,(b)2.714atm,(c)12.06atm.
(b)Flowissupersonic
3.11=M1
21
)1(24
)1( 2
1
)1(
2
1
2
1
2
1
2,0
+
+
+=
M
M
M
p
p
1
714.2
4.2
8.2)4.0(
8.06.5
76.5 2
1
5.3
2
1
2
1
1
2,0 =+
=
M
M
M
p
p
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Example
APitot tubeisinsertedintoanairflowwherethestaticpressureis1atm.
CalculatetheflowMachnumberwhenthePitot tubemeasures(a)1.276
atm,(b)2.714atm,(c)12.06atm.
(c)Flowissupersonic
0.31=M1
21
)1(24
)1( 2
1
)1(
2
1
2
1
2
1
2,0
+
+
+=
M
M
M
p
p
1
06.12
4.2
8.2)4.0(
8.06.5
76.5 2
1
5.3
2
1
2
1
1
2,0 =+
=
M
M
M
p
p
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Example
Consider a hypersonic missile
flying at Mach 8 at analtitude of 20,000 ft, where
the pressure is 973.3 Ib/ft2.
The nose of the missile is
blunt and is shaped like that
shown below. Calculate the
pressure at the stagnation
point on the nose.
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Example
Given: 8=M 2lb/ft3.973=p
Required: ?0=p
4.2
8.2)4.0(
8.06.5
76.5 2
5.3
2
2
0
+
=
M
M
M
p
p
4.2
)8(8.2)4.0(
8.0)8(6.5
)8(76.5 25.3
2
2
0
+
= pp
24
0 lb/ft10x07.8)87.82)(3.973( ==p
atm1.380=p
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Whatwediscussed
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ObliqueShockWave
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ExpansionWave
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PropagationofDisturbance
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PropagationofDisturbance
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PropagationofDisturbance
Thephysicalgenerationofwavesina
supersonicflowisduetothepropagationof
informationviamolecularcollisionsanddue
tothefactthatsuchpropagationcannot
workitswayintocertainregionsofthesupersonicflow.
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Whydoesithavetobeoblique?
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MachWave
Mv
a
Vt
at 1sin ===
M
1sin 1=
Machangle
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OSWvs MachWave
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ObliqueShockWaves
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ObliqueShockWaves
waveangle
deflectionangle
OSW
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ContinuityEquation
controlvolume
=+
S
SdVdt
0VV
zero
2211 uu =
=S
SdV 0
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MomentumEquation
controlvolume
zero
21 ww =
+=+
S S
dfSpdVSdVdVt
VV
V)(V
zero
=S S SpdVSdV )(
=S S
SpdwSdV tangential)()(tangentialcomponent
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MomentumEquation
controlvolume
zero
2
222
2
111 upup +=+
+=+
S S
dfSpdVSdVdVt
VV
V)(V
zero
=S SSpdVSdV )(
=S S
SpduSdV normal)()(normalcomponent
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EnergyEquation
controlvolume
zero
22
2
22
2
11
uh
uh +=+
zero
+=
++
+
S S
dVfSdVpdqSdVV
edV
et
VV
2
V
2
V)(V2
V2
&
zero
=
+
S S
SdVpSdVV
e2
2
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Summary
continuity
momentum
energy
2211 uu =
21 ww =2
222
2
111 upup +=+
22
2
22
2
11
uh
uh +=+
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Summary
TheseequationsaresimilartotheNSWequations.
2211 uu = 21 ww = 2
222
2
111 upup +=+22
2
22
2
11
uh
uh +=+
TheonlydifferenceisthatuhereisnotthetotalvelocityasinaNSW,butratherthenormalvelocity
oftheOSW.
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Summary
Hencesimilarresultscanbeexpected.
2211 uu = 21 ww = 2
222
2
111 upup +=+22
2
22
2
11
uh
uh +=+
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OSWEquations
2/)1(
]2/)1[(12
1,
2
1,2
2,
+=
n
n
nM
MM
2
1,
2
1,
1
2
)1(2
)1(
n
n
M
M
+
+=
( )1)1(
21
2
1,
1
2 +
+= nM
p
p
( ) 2
1,
2
1,2
1,
1
2
)1(
)1(21
1
21
n
n
nM
MM
T
T
+
+
++=
sin11, MMn =
)sin(
2,
2
= nM
M
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MRelation
2)2cos(
1sincot2tan
2
1
22
1
++
=
M
M
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MRelation
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MRelation
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Whatdoesthegraphortheequationsay?
Thereexistsamaximumdeflection
angleforeveryupstreamMachnumber.
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Whatdoesthegraphortheequationsay?
Thereexistsamaximumdeflection
angleforeveryupstreamMachnumber.
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Whatdoesthegraphortheequationsay?
Thereisaweakshockandastrongshock
solutioncorrespondingtothetwovaluesof
thewaveangle.
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Whatdoesthegraphortheequationsay?
If =0, then =90o or =. These
correspond to a NSW and a Mach
wave. In both cases, there is no flow
deflection.
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Whatdoesthegraphortheequationsay?
Ingeneral,forattachedshockswithafixed
deflectionangle,thewaveangledecreases
astheupstreamMachnumberincreases
andtheshockwavebecomesstronger.The
reverseisalsotrue.
Wh d h h h i ?
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Whatdoesthegraphortheequationsay?
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Wh t d th h th ti ?
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Whatdoesthegraphortheequationsay?
Example
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Example
Consider a supersonic flow with M = 2, p = 1 atm, and T = 288 K. This flow
is deflected at a compression corner through 20o. Calculate M, p, T, p0,
and T0behind the resulting oblique shock wave.
M = 2
p = 1 atm
T = 288 K
M = ?
p = ?, p0= ?
T = ?, T0= ?
20o
Example
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Example
606.14.53sin2sin11, === MMn
=== 4.53,20and2For 1 M2)2cos(
1sincot2tan
2
1
22
1
++
=
M
M
( ) 82.21)1(
21
2
1,
1
2 =+
+= nMp
p
( ) 388.1)1(
)1(21
1
21
2
1,
2
1,2
1,
1
2 =+
+
++=
n
n
nM
MM
T
T
6684.02/)1(
]2/)1[(12
1,
2
1,2
2, =
+=
n
n
nM
MM
atm82.2)1(82.211
22 ==
= p
p
pp
K7.399)288(388.111
22 ==
= T
T
TT
21.1)204.53sin(
6684.0
)sin(
2,
2 =
=
=
nMM
Example
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Example
atm82.22 =p K7.3992 =T21.12 =M
457.2211
)1(
22
2
2,0=
+=
Mp
p
8.12
11
2
1
1
1,0 =+= MT
T 11
1,01,02,0 T
T
TTT
==
K4.518)288(8.12,0 ==T
2
2
2,0
2,0 pp
pp
=
atm7)82.2(457.22,0 ==p
Example
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Example
Consider an oblique shock wave with a wave angle of 30o. The upstream
flow Mach number is 2.4. Calculate the deflection angle of the flow, the
pressure and temperature ratios across the shock wave, and the Mach
number behind the wave.
M1= 2.4
M2= ?
= ?
= 30o
p2/p1=?
T2/T1=?
Example
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Example
=
++
= 5.6
2)2cos(
1sincot2tan
21
22
11
M
M
Given: M1= 2.4 = 30o
2.130sin4.2sin11, === MMn
( ) 513.11)1(
21
2
1,
1
2 =+
+= nMp
p
( ) 128.1)1(
)1(21
1
21
2
1,
2
1,2
1,
1
2 =+
+
++=
n
n
nM
MM
T
T
11.2
)5.630sin(
8422.0
)sin(
2,
2 =
=
=
nMM
8422.02/)1(
]2/)1[(12
1,
2
1,2
2, =
+=
n
n
nM
MM
What does the example tell us?
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Whatdoestheexampletellus?
Thewaveisweak.
Producedonly51%increaseinpressure.
Thisisbecausedeflectionangleissmall.Also,waveangleissmall;closetoMachwaveangleof
=== 306.244.2sinsin 1111M
Only2propertiesneedtobespecifiedtocompletely
describe(solve)agivenOSW.
Inthisexample,thesepropertieswereMand.Inthefirstexample,itwasMand.
Example
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ExampleConsider an oblique shock wave with = 35o and a pressure ratio
p2/p1= 3. Calculate the upstream Mach number.
M1= ?
= 35o
p2/p1=3
Example
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pConsider an oblique shock wave with = 35o and a pressure ratio
p2/p1= 3. Calculate the upstream Mach number.
( ) 31)1( 21 2
1,
1
2=++=
nMpp
6475.112
)1(1
1
21, =+
+
=
p
pMn
87.235sin6475.1
sinsin 1,111, ====
nn MMMM
Example
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p
Consider a Mach 3 flow. It is desired to slow this flow to a subsonic speed.
Consider two separate ways of achieving this:
(1) the flow is slowed by passing directly through a normal shock wave;
(2) the flow first passes through an oblique shock with a 40 wave angle,
and then subsequently through a normal shock.
Calculate the ratio of the final total pressure values for the two cases.
Comment on the significance of the result.
Example
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p
?1,0
2,0 =p
p
?1,0
3,0 =p
p
CaseI
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1
1,0
2
2,0
1
2
1,0
2,0
p
p
p
p
p
p
p
p
=
( ) 3333.101)1(
21
2
1
1
2 =+
+= Mp
p
1672.12
11
)1(
2
2
2
2,0 =
+=
Mp
p
4752.02/)1(
]2/)1[(1
21
2
1
2
=
+=
M
MM
7327.362
11
)1(
2
1
1
1,0
=
+=
Mp
p
( ) 7327.361672.13333.101,0
2,0 =p
p
3283.01,0
2,0 =p
p
CaseII
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1
1,0
2
2,0
1
2
1,0
2,0
p
p
p
p
p
p
p
p
=
( ) 1717.41)1(
21
2
1,
1
2 =+
+= nMp
p
2658.12
11
)1(
2
2
2
2,0 =
+=
Mp
p
5902.02/)1(
]2/)1[(1
21,
2
1,
2, =
+=
n
n
n M
M
M
72
11
)1(
2
1,
1
1,0
=
+=
nMp
p
( )72658.11717.41,0
2,0 =p
p
7544.0
1,0
2,0 =
p
p
9284.140sin3sin11, === MMn
CaseII
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2
2,0
3
3,0
2
3
2,0
3,0
p
p
p
p
p
p
p
p
=
( ) 089.41)1(
21
2
2
2
3 =+
+= Mp
p
2692.12
11
)1(
2
3
3
3,0 =
+=
Mp
p
5937.02/)1(
]2/)1[(1
22
2
23 =
+=
M
MM
805.62
11
)1(
2
2
2
2,0
=
+=
Mp
p
( ) 805.62692.1089.42,0
3,0 =p
p
7626.0
2,0
3,0 =p
p
91.1)2240sin(
5902.0
)sin(
2,
2 =
=
=
nMM
CaseII
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58.0)7626.0)(7544.0(2,0
3,0
1,0
2,0
1,0
3,0 ===p
p
p
p
p
p
33.0
ICASE1,0
2,0=
pp 58.0
IICASE1,0
3,0=
pp
76.1
IICASE1,0
3,0
ICASE1,0
2,0 =
p
p
p
p
Case II is the more efficient flow with less reduction in total pressure.
Application Design of supersonic inlets for jet engines.
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Scramjet
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Scramjet
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Scramjet
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Supersonicflowoverwedgesandcones
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(1)theshockwaveontheconeisweaker,
(2)theconesurfacepressureisless,and
(3)thestreamlinesabovetheconesurfacearecurved
ratherthanstraight.
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ShockReflections
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MachReflection
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IntersectionofLeftandRightRunningWaves
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IntersectionofLeftandRightRunningWaves
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Example
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Consider an oblique shock wave generated by a compression corner with
a 10 deflection angle. The Mach number of the flow ahead of the corner
is 3.6; the flow pressure and temperature are standard sea levelconditions. The oblique shock wave subsequently impinges on a straight
wall opposite the compression corner. Calculate the angle of the
reflected shock wave relative to the straight wall. Also, obtain the
pressure, temperature, and Mach number behind the reflected wave.
Example
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6.31=M
=10
2
1 lb/ft2116=p
R5191=T?=
Example
2)2cos(
1sincot2tan
2
1
22
1
++
=
M
M.24,10and63For 11 === .M
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)(1
( ) 32.21)1( 21 2
1,
1
2=++=
nMpp
( ) 294.1)1(
)1(21
1
21
2
1,
2
1,2
1,
1
2 =+
+
++=
n
n
nM
MM
T
T
96.2)1024sin(
7157.0
)sin( 1
2,
2 =
=
=
nMM
7175.02/)1(
]2/)1[(12
1,
2
1,
2, =
+
=
n
n
nM
M
M
464.124sin6.3sin 111, === MMn
Example=== 3.17103.272 .3.27,10and96.2For 21 === M
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( ) 991.11)1( 21 2
2,
2
3=++=
nMpp
( ) 229.1)1(
)1(21
1
21
2
2,
2
2,2
2,
2
3 =+
+
++=
n
n
nM
MM
T
T
55.2)103.27sin(
7572.0
)sin( 2
3,
3 =
=
=
nMM
7572.02/)1(
]2/)1[(12
2,
2
2,
3, =
+
=
n
n
nM
M
M
358.13.27sin96.2sin 222, === MMn
Example
991.13 =p
p229.13 =
T
T32.22 =
p
p294.12 =
T
T
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2p 2T1p 1T
3
1
1
2
2
33 lb/ft9774)2116)(32.2)(991.1( === p
p
p
p
pp
R825)519)(294.1)(229.1(11
2
2
33 === T
T
T
T
TT
Example
R825T
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6.31=M
=10
2
1 lb/ft2116=p
R5191=T
2
2 lb/ft4909=p
R6.6712 =T
3
3 lb/ft9774=p
R8253=T
= 3.17
55.23=M
96.22 =M
BluntBody
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BluntBody
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LudwigPrandtl Theodor Meyer
PrandtlMeyerExpansion
(centered expansionwaves)
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PrandtlMeyerExpansion
dV
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V
dVMd 12 =
ThePrandtlMeyerFunction
+
=2
2
2
]2/)1[(1
1M
M
dM
M
M
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1tan)1(
1
1tan
1
1)( 2121
+
+= MMM
)()( 12 MM =
+
=M
dM
M
MM
2
2
]2/)1[(1
1)(
+1
]2/)1[(1M
MM
Computingdownstreamproperties
1 Compute v(M )
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1. Compute v(M1).
2. Compute v(M2) = v(M1) +.3. Obtain M2 corresponding to v(M2).
4. Use appropriate isentropic equations
to relate upstream and downstream
properties.
ExampleAsupersonicflowwithM1=1.5,P1=1atm,andT1=288Kisexpanded
aroundasharpcornerthroughadeflectionangleof15o.CalculateM2,
d h l h h f d d d h
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P2,T2,P0,2,T0,2andtheanglesthattheforwardandrearwardMach
linesmakewithrespecttotheupstreamflowdirection.
M1=1.5
P1=1atm
T1=288K
15o
Example
1 Compute v(M )
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1. Compute v(M1).
v(1.5) = 11.91o2. Compute v(M2) = v(M1) +.
v(M2) = 11.91o + 15 = 26.91o
3. Obtain M2 corresponding to v(M2).
M2 = 2.0 (rounding to nearest entry in table)
Example
4. Use appropriate isentropic equations to
relate upstream and downstream
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relate upstream and downstream
properties.
45.1T
Tand671.3
p
p,5.1MFor
1
0,1
1
0,1
1 ===
8.1TTand824.7
pp,0.2MFor
2
0,2
2
0,2
2 ===
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Example
4. Use appropriate isentropic equations to
relate upstream and downstream
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relate upstream and downstream
properties.0,20,10,20,1 TTandpp,isentropicisflowtheSince ==
atm0.469atm)1)(671.3)(1(824.7
1p
p
p
p
p
p
pp 1
1
0,1
0,1
0,2
0,2
22 ===
K232)288)(45.1)(1(8.1
1T
T
T
T
T
T
TT 1
1
0,1
0,1
0,2
0,2
22 ===
Example
4. Use appropriate isentropic equations to
relate upstream and downstream
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relate upstream and downstream
properties.
===
===
1515300.2sin:lineMachrearwardofAngle
81.415.1sinsin
:lineMachforwardofAngle
11
2
111
1
1
1
M
ShockExpansionTheory:FlatPlate
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sin)('
cos)('
)('
23
23
23
cppD
cppL
cppR
=
=
=
ShockExpansionTheory:DiamondShapedAirfoil
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Schlieren Photograph SAEP Logo
ShockExpansionTheory:DiamondShapedAirfoil
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tppD
tpplplpD
)('
)2/)((2)sinsin(2'
32
3232
=
==
Example
Calculatetheliftanddragcoefficientsforaflatplateata50o angleof
attackinaMach3flow.
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M=3.0
= =50o
Example== 76.49)0.3()( 1 M
=+=+= 76.54576.49)()( 12 MM
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27.32=M
73.361
1,0 =p
p
55
2
2,0 =p
p
668.055
73.36
2
2,0
1
1,0
1
2 ===p
p
p
p
p
p
Example=== 23.1,5and3For 1 M
177.11.23sin3sin11, === MMn
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458.11
3
=p
p
cos2
)2/(
''
1
2
1
3
2
1
2
111
===
p
p
p
p
McMp
L
Sq
LCL
cos)(' 23 cppL =
( ) 125.05cos668.0458.1)3)(4.1(
22
==LC
Example
i2' 23
ppD
C
sin)(' 23 cppD =
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sin1
2
1
3
2
11
==
p
p
p
p
MSq
CD
( ) 011.05sin668.0458.1)3)(4.1(
22
==DC
tan=L
D
C
C
011.05tan125.0tan === LD CC
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Roadmap
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QuasiOneDimensionalFlow
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GoverningEquations
222111 AuAu =
Continuity
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2
2
22221
2
1111
2
1
AuAppdAAuAp
A
A
+=++
22
2
2
2
2
1
1
uh
uh +=+
Momentum
Energy
22222 and TchRTp p==Foracaloricallyperfectgas
Differentialforms
Continuity
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0)( =uAd
ududp =
0=+ ududh
Momentum(EulersEquation)
Energy
AreaVelocityRelation
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( )u
duM
A
dA12 =
SubsonictoSupersonic
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zzle?
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W
hatisano
z
deLavalNozzle
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Nozzle
Thisisarocketnozzle.
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AreaMachRelation
)1/()1(
2
2
2
* 2
11
1
21 +
+
+=
M
MA
A
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Misafunctionoflocaltothroatarearatio:M=f(A/A*)
Localtothroatarearatio,A/A* 1
TherearetwoMsforeachA/A*,asubsonicandasupersonicvalue.
ForM1,asMincreasesA/A*alsoincreases(divergentduct).ForM=1,A/A*=1.
IsentropicSupersonicFlow
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IsentropicSupersonicFlow
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)1/()1(
2
2
2
* 2
11
1
21 +
+
+=
M
MA
A
IsentropicSupersonicFlow
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528.012
)1(
0
*
=
+=
pp
IsentropicSupersonicFlow
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833.012
0
*
=+
=T
T
IsentropicSupersonicFlow
Th di t ib ti f M d
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ThedistributionofM,andtheresultingdistributionof
pandT,dependonlyonthe
localarearatioA/A*.
IsentropicSupersonicFlow
Foranisentropic
i fl t h
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supersonicflowtohappen,thepressuredifference
betweentheinletandexit
hastobejustrightforthe
geometryoftheduct.
IsentropicSubsonicFlow
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Whathappenswhenthereisan
inletexitpressuremismatch?
IsentropicSubsonicFlowFreezesatchokedflow.
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For1and2,A* isareferenceareanotequaltoAt.ForsubsonicflowA
*
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Theresaninfinitenumberofisentropicsubsonicsolutionandonlyoneisentropic
supersonicsolution.
528.01
2)1(
0
*
=
+=
p
p
ChokedFlow
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uAm =&
Chokedthroat
Aconditioninaconvergent
divergentductwhereinsonic
condition has been achieved ath i f i i h
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conditionhasbeenachievedatthesectionofminimumarea,the
throat,andinformationisno
longerpropagatedfromthe
convergentportiontothe
divergentportionoftheduct.
IsentropicSupersonicFlowwithNSW
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IsentropicSupersonicFlowwithNSW
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IsentropicSupersonicFlowwithNSW
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NSWmovesfurtheraftupon
exitpressure
decrease
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DiffuserFlow
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IdealVSRealDiffuser
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Nozzleexhaustingtoatmosphere
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Nozzleexhaustingintoaconstantareaduct
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SupersonicWindTunnel
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cWindTunnel
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Supersonic
ExampleConsidertheisentropicsupersonicflowthroughaconvergentdivergent
nozzlewithanexittothroatarearatioof10.25.Thereservoirpressure
andtemperatureare5atm and600R,respectively.CalculateM,p,andT
atthenozzleexit.
25.101
121 )1/()1(2
2
2
*=
+=
+
e MA
95.3=eM
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25.102
11
2*
++
e
e
MMA
e
1422
11
)1/(
20,0 =
+==
e
ee
eM
p
p
p
p
12.42
11
20,0 =
+== eee
eM
T
T
T
T
atm035.0142/500 ===e
ep
ppp
R6.14512.4/60000 ===e
eT
TTT
ExampleConsidertheisentropicflowthroughaconvergentdivergentnozzlewith
anexittothroatarearatioof2.Thereservoirpressureandtemperature
are1atm and288K,respectively.CalculatetheMachnumber,pressure,
andtemperatureatboththethroatandtheexitforthecaseswhere
(a) theflowissupersonicattheexit,and(b) the flow is subsonic throughout the entire nozzle except at the throat
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Given:
(b)theflowissubsonicthroughouttheentirenozzleexceptatthethroat,
whereM=1.
2*=
A
Ae atm10 =p K2880 =T
Example:(a)1* =M
atm528.0)1(528.00
**
==== p
p
ppt
Atthethroat,
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K240)288(833.00
** ====
T
TTTt
Example:(a)
22
11
1
21)1/()1(
2
2
2
* =
+
+=
+
e
e
e MMA
A2.2=eM
( ) 5.3200 pp
Attheexit,
093506910/10p
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( ) 69.102.01 5.320,0 =+== eee
eM
p
p
p
p
968.1211 20,0 =
+== e
ee
e MTT
TT
atm0935.069.10/100 ===e
ep
ppp
K146968.1/28800 ===e
eTTTT
Example:(b)1* ==MMt
atm528.0)1(528.00
**
==== p
p
ppt
Atthethroat,
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K240)288(833.00
** ====
T
TTTt
Example:(b)
22
11
1
21)1/()1(
2
2
2
* =
+
+=
+
e
e
e MMA
A3.0=eM
( ) 0641201 5.3200 e Mpp
Attheexit,
atm9400641/10p
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( ) 064.12.01 20,0 =+== eee
eM
p
p
p
p
018.1211 20,0 =
+== e
ee
e MTT
T
T
atm94.0064.1/100 ===e
ep
ppp
K9.282018.1/28800 ===e
eTTTT
ExampleConsider the isentropic flow through a convergentdivergent nozzle
with an exittothroat area ratio of 2. The reservoir pressure and
temperature are 1 atm and 288 K, respectively. The exit pressure is
0.973 atm. Calculate the Mach number at both the throat and the exit
for the cases where(a) the flow is supersonic at the exit, and
(b) h fl i b i h h h i l h
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Given:
(b) the flow is subsonic throughout the entire nozzle except at the
throat, where M = 1.
2* =A
Aeatm10=p K288
0=T atm973.0=
ep
Example:
028.1
973.0
10 ==ep
p
Frombefore,theexitpressurecorrespondingtoasubsonicflow
throughoutthenozzle(exceptatthethroat), atm94.0=ep
5.3/1
482.1)964.2(5.0**
===
A
A
A
A
A
A e
e
tt
440=M
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2.015 0 =
=
e
ep
pM
964.22
111
21 )1/()1(
2
2
2
* =
+
+=
+
e
e
e MMA
A
44.0=tM
ExampleFor the preliminary design of a Mach 2 supersonic wind tunnel,
calculate the ratio of the diffuser throat area to the nozzle throat area.
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