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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.