Supersonic - BANAL 2013

7/25/2019 Supersonic - BANAL 2013

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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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Rcc vp +=

pp

v

p

p

c

R

c

c

c

c+=

v

p

c

c=

pc

R+=

11

1=

Rcp


17/282


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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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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22

2

22

2

11

uTcuTc pp +=+

2121

2

2

2

2

2

1

2

1 uaua +

=+


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2121

2

2

2

2

2

1

2

1 uaua +

=+

121

2

022

=+

aua


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2121

2

2

2

2

2

1

2

1 uaua +

=+

2*22

)1(2

1

21aua

+=+


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22

2

22

2

11 uTcuTc pp +=+

0

2

2TcuTc pp =+


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20

2

11 M

T

T +=


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)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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833.01

2

0

*

=+

=T

T

m/s456)5.517)(287(4.1** === RTa

833.0621

*

=T

K5.517* =T


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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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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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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


Calculatethevelocityatpoint1whenthefreestream temperatureis59oF.

Assumeisentropicflowovertheairfoil.


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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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100/282

Whenisaflowcompressible?

3.0>MRuleofthumb:

Why?BecauseChuckNorrissaysso?


101/282

Thisguydontthinkso.


102/282


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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)1(1

20

2

1

1

+=

M


104/282


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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107/282

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 =


110/282

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


111/282

MachNumbersRelation

2/)1(

]2/)1[(12

1

2

12

2

+=

M

MM


112/282

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


115/282

PressureRatio

( )1)1(

21

2

1

1

2

+

+= M

p

p


116/282

TemperatureRatio

=

1

2

1

2

1

2

pp

TT

( ) 21

2

12

1

1

2

)1(

)1(211

21

M

MMT

T

+

+

++=


117/282

( ) 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?


119/282

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 =


120/282

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?


122/282

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 =


124/282

TotalPressureChange

Rss

ep

p/)(

1,0

2,0 12

=

Since 012 ss

1,02,0 pp < Thatis,totalpressuredecreases

acrossaNSW.


125/282

NormalShockWave


126/282

NormalShockWave


127/282

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


128/282

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


129/282

Example

2=M

K3.223

N/m10x65.2 24

=

=

T

p 2.02 =M

isentropic

?pand? 22 ==T


130/282

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


131/282


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


132/282


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

+==


133/282

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


134/282

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:


135/282

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


136/282

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


137/282

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


138/282

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


139/282

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


140/282

SCRAMJETTheresultsofthepreviousexampleare,

K3992=Tatm42.12=p

,2originaltheofinstead10If == MM

K46532=Tatm7.322=p

Theseresultsdescribeanextremeenvironmentthatisverydifficultto

handleforaramjet.

Thesolutionis,DONOTslowtheflowtoM2=0.2.

Keeptheflowsupersonicallthroughout.


141/282


142/282

SubsonicCompressibleFlow


143/282

SubsonicCompressibleFlow)1(

20

2

11

+=

Mp

p

=

11

2)1(

02

p

pM

=

11

2 )1(02

p

paV


144/282

SupersonicCompressibleFlow


145/282


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


146/282


121

)1(24

)1( 2

1

)1(

2

1

2

12

1

2,0

++

+=

M

M

Mp

p

RayleighPitot TubeFormula


147/282

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


148/282

Example



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


149/282

Example



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


150/282

Example



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


151/282

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.


152/282

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


153/282

Whatwediscussed


154/282


155/282

ObliqueShockWave


156/282

ExpansionWave


157/282

PropagationofDisturbance


158/282



159/282


Thephysicalgenerationofwavesina

supersonicflowisduetothepropagationof

informationviamolecularcollisionsanddue

tothefactthatsuchpropagationcannot

workitswayintocertainregionsofthesupersonicflow.


160/282

Whydoesithavetobeoblique?


161/282

MachWave

Mv

a

Vt

at 1sin ===

M

1sin 1=

Machangle


162/282

OSWvs MachWave


163/282

ObliqueShockWaves


164/282

ObliqueShockWaves

waveangle

deflectionangle

OSW


165/282

ContinuityEquation

controlvolume

=+

S

SdVdt

0VV

zero

2211 uu =

=S

SdV 0


166/282

MomentumEquation

controlvolume

zero

21 ww =

+=+

S S

dfSpdVSdVdVt

VV

V)(V

zero

=S S SpdVSdV )(

=S S

SpdwSdV tangential)()(tangentialcomponent


167/282

MomentumEquation

controlvolume

zero

2

222

2

111 upup +=+

+=+

S S

dfSpdVSdVdVt

VV

V)(V

zero

=S SSpdVSdV )(

=S S

SpduSdV normal)()(normalcomponent


168/282

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


169/282

Summary

continuity

momentum

energy

2211 uu =

21 ww =2

222

2

111 upup +=+

22

2

22

2

11

uh

uh +=+


170/282

Summary

TheseequationsaresimilartotheNSWequations.

2211 uu = 21 ww = 2

222

2

111 upup +=+22

2

22

2

11

uh

uh +=+

TheonlydifferenceisthatuhereisnotthetotalvelocityasinaNSW,butratherthenormalvelocity

oftheOSW.


171/282

Summary

Hencesimilarresultscanbeexpected.

2211 uu = 21 ww = 2

222

2

111 upup +=+22

2

22

2

11

uh

uh +=+


172/282

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


173/282

MRelation

2)2cos(

1sincot2tan

2

1

22

1

++

=

M

M


174/282

MRelation


175/282

MRelation


176/282

Whatdoesthegraphortheequationsay?

Thereexistsamaximumdeflection

angleforeveryupstreamMachnumber.


177/282


Thereexistsamaximumdeflection

angleforeveryupstreamMachnumber.


178/282


Thereisaweakshockandastrongshock

solutioncorrespondingtothetwovaluesof

thewaveangle.


179/282


If =0, then =90o or =. These

correspond to a NSW and a Mach

wave. In both cases, there is no flow

deflection.


180/282


Ingeneral,forattachedshockswithafixed

deflectionangle,thewaveangledecreases

astheupstreamMachnumberincreases

andtheshockwavebecomesstronger.The

reverseisalsotrue.

Wh d h h h i ?


181/282



182/282

Wh t d th h th ti ?


183/282


Example


184/282

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


185/282

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


186/282

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


187/282

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


188/282

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?


189/282

Whatdoestheexampletellus?

Thewaveisweak.

Producedonly51%increaseinpressure.

Thisisbecausedeflectionangleissmall.Also,waveangleissmall;closetoMachwaveangleof

=== 306.244.2sinsin 1111M

Only2propertiesneedtobespecifiedtocompletely

describe(solve)agivenOSW.

Inthisexample,thesepropertieswereMand.Inthefirstexample,itwasMand.

Example


190/282

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


191/282

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


192/282

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


193/282

p

?1,0

2,0 =p

p

?1,0

3,0 =p

p

CaseI


194/282

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


195/282

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


196/282

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


197/282

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.


198/282

Scramjet


199/282

Scramjet


200/282

Scramjet


201/282


202/282


203/282


204/282


205/282


206/282

Supersonicflowoverwedgesandcones


207/282

(1)theshockwaveontheconeisweaker,

(2)theconesurfacepressureisless,and

(3)thestreamlinesabovetheconesurfacearecurved

ratherthanstraight.


208/282

ShockReflections


209/282

MachReflection


210/282

IntersectionofLeftandRightRunningWaves


211/282

IntersectionofLeftandRightRunningWaves


212/282

Example


213/282

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


214/282

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


215/282

)(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


216/282

( ) 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


217/282

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


218/282

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


219/282

BluntBody


220/282


221/282


222/282

LudwigPrandtl Theodor Meyer

PrandtlMeyerExpansion

(centered expansionwaves)


223/282

PrandtlMeyerExpansion

dV


224/282

V

dVMd 12 =

ThePrandtlMeyerFunction

+

=2

2

2

]2/)1[(1

1M

M

dM

M

M


225/282

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 )


226/282

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


227/282

P2,T2,P0,2,T0,2andtheanglesthattheforwardandrearwardMach

linesmakewithrespecttotheupstreamflowdirection.

M1=1.5

P1=1atm

T1=288K

15o

Example

1 Compute v(M )


228/282

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


229/282


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 ===


230/282

Example




231/282


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




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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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)1/()1(

2

2

2

* 2

11

1

21 +

+

+=

M

MA

A



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528.012

)1(

0

*

=

+=

pp



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833.012

0

*

=+

=T

T


Th di t ib ti f M d


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ThedistributionofM,andtheresultingdistributionof

pandT,dependonlyonthe

localarearatioA/A*.


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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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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Supersonic - BANAL 2013

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