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HIGH-SPEED AERODYNAMICSMACE 31321

Lecture 5Normal Shock II

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OBJECTIVE OF THIS LECTURE

• To examine the changes in flow properties across a normal shock wave

• To understand how velocity can be measured in a compressible flow

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NORMAL SHOCK WAVES• Flow conditions before & after the shock wave

p1

T1

M1

u1

p0,1

h0,1

T0,1

s1

Given conditions ahead of the wave

p2

T2

M2

u2

p0,2

h0,2

T0,2

s2

Unknown conditions behind the wave x

Assumptions: –The flow is steady and 1D.

–The flow is adiabatic.

–There are no viscous effects on the sides of the CV

The basic relations–Continuity:

–Momentum:

–Energy:

–Equation of state:

222

211 2

121 uTCuTC pp +=+

222 RTp ρ=

2211 uu ρρ =2222

2111 upup ρρ +=+

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NORMAL SHOCK RELATIONS

M1 M2

( )( ) 2

1

21

2

1

1

2

121

MM

uu

−++

==γ

γρρ

( ) ( )( ) ⎥

⎦

⎤⎢⎣

⎡+−+

⎥⎦

⎤⎢⎣

⎡−

++= 2

1

212

11

2

1121

121

MMM

TT

γγ

γγ

( )11

21 21

1

2 −+

+= Mpp

γγ

( )

( )121

1211

21

21

22

−−

−+=

γγ

γ

M

MM

See Normal Shock Property Table

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NORMAL SHOCK WAVES• Flow conditions before & after the shock wave

p1

T1

M1

u1

p0,1

h0,1

T0,1

s1

Given conditions ahead of the wave

p2

T2

M2

u2

p0,2

h0,2

T0,2

s2

Unknown conditions behind the wave x

M1

M2<1

p1 p2

T1

T2

What about p0, T0 and s?

ρ1

ρ2

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NORMAL SHOCK WAVES• Flow conditions before & after the shock wave

Fluid element with actual p1, T1, M1 , s1

Imaginary state 1a where the fluid element has been brought to rest isentropically. P0,1, T0,1, s1

Fluid element with actual p2, T2, M2 , s2

Imaginary state 2a where the fluid element has been brought to rest isentropically. P0,2, T0,2, s2

M1>1 M2<1

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• Entropy change across the shock

• Substituting

• Gives

CHANGES IN ENTROPY ACROSS A NORMAL SHOCK

M1 M2

( ) ( )( ) ⎥

⎦

⎤⎢⎣

⎡+−+

⎥⎦

⎤⎢⎣

⎡−

++= 2

1

212

11

2

1121

121

MMM

TT

γγ

γγ( )1

121 2

11

2 −+

+= Mpp

γγ

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛=−

1

2

1

212 lnln

ppR

TTCss p

( ) ( )( ) ( )⎥

⎦

⎤⎢⎣

⎡−

++−

⎭⎬⎫

⎩⎨⎧

+−+

⎥⎦

⎤⎢⎣

⎡−

++=− 1

121ln

1121

121ln 2

121

212

112 MRM

MMCss p γγ

γγ

γγ

&

⎩⎨⎧

>>==

=−1,01,0

1

112 M

Mss The entropy is caused by strong

viscous dissipation within the shock.

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• From the energy equation

• Entropy change across the shock

CHANGES IN P0 &T0 ACROSS A NORMAL SHOCK

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟

⎟⎠

⎞⎜⎜⎝

⎛=−=−

1,0

2,0

1,0

2.01212 lnln

pp

RTTCssss paa

The total temperature is constant across a normal shock wave.

222

211 2

121 uTCuTC pp +=+ 2,01,0 TCTC pp = 2,01,0 TT =

2,01,0 TT =⎟⎟⎠

⎞⎜⎜⎝

⎛−=−

1,0

2,012 ln

pp

Rss ( ) 1/

1,0

2,0 12 <= −− Rssepp

The total pressure decreases across a normal shock wave.

The lower the total pressure loss, the more efficient is the flow process.

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NORMAL SHOCK WAVES• Variations of flow properties across a normal shock

x

M1

M2<1

p1p2

T1T2

ρ1ρ2

p1

T1

M1

u1

p0,1

T0,1

s1

p2

T2

M2

u2

p0,2

T0,2

s2

p0,1

p0,2

T0,1T0,2

s1s2

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• Low-speed incompressible flow

VELOCITY MEASUREMENT

( )ρ

101

2 ppu −=

ρρ0

211

2pup

=+From Bernoulli Equation:

u1

Pitot-Static probe

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• Subsonic compressible flow – No shock waves and the flow is regarded as isentropic

VELOCITY MEASUREMENT

121

1

1,0

211

−

⎥⎦⎤

⎢⎣⎡ −+=

γγ

γ Mpp

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛−

=

−

11

21

1

1,021

γγ

γ pp

M

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛−

=

−

11

21

1

1,0212

1

γγ

γ ppau p0,1 will be measured by a Pitot probe.

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• Supersonic flow – A shock wave forms in front of the probe – The total pressure read by the probe is not equal to that of

incoming flow, i.e. p0,1≠ p0,2 .

VELOCITY MEASUREMENT

p01

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• Rayleigh Pitot-tube formula

VELOCITY MEASUREMENT

1

2

2

2,0

1

2,0

pp

pp

pp

=12

22

2,0

211

−

⎥⎦⎤

⎢⎣⎡ −+=

γγ

γ Mpp

( )11

21 21

1

2 −+

+= Mpp

γγ

( )

( )121

1211

21

21

22

−−

−+=

γγ

γ

M

MM

( )( ) ⎥

⎦

⎤⎢⎣

⎡+

+−⎥⎦

⎤⎢⎣

⎡

−−−

=−

121

1241 2

11

21

21

2

1

2,0

γγγ

γγγ γ

γ

MM

Mp

p

Tabulated versus M1 in Shock Wave Properties Table

(Normal shock relation)

(Isentropic relation)

(Normal shock relation)

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QUESTION

• A Pitot tube is inserted into an airflow where the static pressure is 1atm. Calculate the flow Mach number when the Pitot measures

a) 1.276atmb) 2.714atm

• Hints: What is p0/p when M1=1?

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• Determine firstly if the incoming flow is likely to be subsonic or supersonic.

– If M1=1, we have

– For Reading 1 , hence M1<1. From

Table A, M1 = 0.6.

– For Reading 2 , hence M1>1. From

Table B, M1 = 1.3.

SOLUTION

893.11

2,0 =p

p

893.1714.21

2,0 >=p

p

893.1276.11

2,0 <=p

p

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REFERENCES

• In “Fundamentals of Aerodynamics” by Andersons, 2nd edition.– Chapter 8, p.443-450.