normal shocks - sky.kiau.ac.irsky.kiau.ac.ir/~mostafa.khosravy/mycourses/gas_dynamics_files/4... ·...
TRANSCRIPT
Gas Dynamics 1
Normal Shocks
Gas Dynamics
Definition of shock waveShock wave is a very thin region in aflow where a supersonic flow isdecelerated to subsonic flow. The processis adiabatic but non-isentropic.
Shock wave
V
P
T
2
Gas Dynamics
Formation of Shock WaveA piston in a tube is given a small constant velocity increment to the right magnitude dV, a sound wave travel ahead of the piston.
A second increment of velocity dVcausing a second wave to move into the compressed gas behind the first wave.
As the second wave move into a gas that is already moving (into a compressed gas having a slightly elevated temperature), the second waves travels with a greater velocity.
The wave next to the piston tend to overtake those father down the tube. As time passes, the compression wave steepens.
3
Gas Dynamics
Types of Shock Waves:
Normal shock wave- easiest to analyze
Oblique shock wave- will be analyzed
based on normalshock relations
Curved shock wave- difficult & will
not be analyzedin this class
- The flow across a shock wave is adiabatic butnot isentropic (because it is irreversible). So:
0201
0201
PPTT
≠=
β
4
Gas Dynamics
Types of Shock Waves:
Normal shock wave- easiest to analyze
Oblique shock wave- will be analyzed
based on normalshock relations
Curved shock wave- difficult & will
not be analyzedin this class
- The flow across a shock wave is adiabatic butnot isentropic (because it is irreversible). So:
0201
0201
PPTT
≠=
β
Gas Dynamics
Types of Shock Waves:
Normal shock wave- easiest to analyze
Oblique shock wave- will be analyzed
based on normalshock relations
Curved shock wave- difficult & will
not be analyzedin this class
- The flow across a shock wave is adiabatic butnot isentropic (because it is irreversible). So:
0201
0201
PPTT
≠=
β
6
Gas Dynamics
Governing Equations
1
1
1
1
ρTPV
2
2
2
2
ρTPV
Conservation of mass:
Conservation of momentum:
Rearranging:
Combining:
AVAV 2211 ρρ =
( ) ( )( )( )122221
121121
1221
VVVPPVVVPPVVmAPP
−=−−=−−=−
ρρ
&
2
2121
22
1
212121
ρ
ρPPVVV
PPVVV
−=−
−=−
( ) 21
22
2121
11 VVPP −=⎟⎟⎠
⎞⎜⎜⎝
⎛+−ρρ
Conservation of energy:
Change of variable:0
22
2
21
1 22TcVTcVTc ppp =+=+
⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟⎠
⎞⎜⎜⎝
⎛−
=−2
2
1
121
22 1
2ρργ
γ PPVV
combine
22
2
221
1
1
12
12 VPVP +⎟⎟⎠
⎞⎜⎜⎝
⎛−
=+⎟⎟⎠
⎞⎜⎜⎝
⎛− ργ
γργ
γ
7
Gas Dynamics
Continued:
Multiplied by ρ2/p1:
Rearranging:
( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟⎠
⎞⎜⎜⎝
⎛−
=⎟⎟⎠
⎞⎜⎜⎝
⎛+−
2
2
1
1
2121 1
211ρργ
γρρ
PPPP
⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟⎠
⎞⎜⎜⎝
⎛−
=⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟⎠
⎞⎜⎜⎝
⎛−
1
2
1
2
1
2
1
2
1211
PP
PP
ρρ
γγ
ρρ
⎥⎦
⎤⎢⎣
⎡−⎟⎟⎠
⎞⎜⎜⎝⎛
−+
⎥⎦
⎤⎢⎣
⎡−⎟⎟⎠
⎞⎜⎜⎝⎛
−+
=
1
2
1
2
1
2
11
111
ρρ
γγ
ρρ
γγ
PP
⎥⎦
⎤⎢⎣
⎡+⎟⎟⎠
⎞⎜⎜⎝⎛
−+
⎥⎦
⎤⎢⎣
⎡+⎟⎟⎠
⎞⎜⎜⎝⎛
−+
=
1
2
1
2
1
2
11
111
PP
PP
γγ
γγ
ρρor
Governing Equations cont.
8
Gas Dynamics
⎥⎦
⎤⎢⎣
⎡+⎟⎟⎠
⎞⎜⎜⎝⎛
−+
⎥⎦
⎤⎢⎣
⎡+⎟⎟⎠
⎞⎜⎜⎝⎛
−+
==
1
2
1
2
2
1
1
2
11
111
PP
PP
VV
γγ
γγ
ρρ
2
1
1
2
1
2
ρρ
PP
TT =
⎥⎦
⎤⎢⎣
⎡+⎟⎟⎠
⎞⎜⎜⎝⎛
−+
⎥⎦
⎤⎢⎣
⎡+⎟⎟⎠
⎞⎜⎜⎝⎛
−+
=
2
1
1
2
1
2
1111
PPPP
TT
γγ
γγ
Governing Equations cont.
From conservation of mass:
From equation of state:
9
Gas Dynamics
Governing Equations cont.
2211 VV ρρ =
( ) ( )
( ) ( )222
211
2
2222
2111
1221
11 MPMP
Pa
VPVPVVmAPP
γγρ
γ
ρρ
+=+
=
+=+
−=− &
( )
( ) ⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−+
−+=⎟⎟⎠
⎞⎜⎜⎝
⎛
+=+
22
21
1
2
22
2
21
1
211
211
22
M
M
TT
VhVh
γ
γ
C
O
M
BINE
Conservation of mass
Conservation of momentum
Conservation of energy ( ) ( )
( )( ) ( )( ) 02
21
1
)2
11(
1
)2
11(
211
1211
1
21
22
21
22
21
22
41
42
222
22
22
221
21
21
222
2
2212
1
1
222
211
1
1
2211
=−+
−−−−
+
−+=
+
−+
−++
=−++
=
=
MMMMMMMM
M
MM
M
MM
MM
MMMM
RTMRTPRTM
RTP
VV
γγ
γ
γ
γ
γ
γγ
γγ
γγ
ρρ
Expanding the equations
10
Gas Dynamics
Governing Equations cont.
( )( )12
212
1
21
2 −−+−±=
γγγM
MM
Solution:
Mach number cannot be negative. So, only the positive value is realistic.
11
Gas Dynamics
Governing Equations cont.
( )
( )
( )
( )( )
( )( )
11
12
11
121
11
22
11
211
211
21
1
2
22
21
1
2
21
2
21
21
1
2
22
21
1
2
+−−
+=
++=
⎟⎟⎠
⎞⎜⎜⎝
⎛−+
⎟⎟⎠⎞
⎜⎜⎝⎛ −
−⎟⎠⎞⎜
⎝⎛ −+
=⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−+
−+=⎟⎟
⎠
⎞⎜⎜⎝
⎛
γγ
γγ
γγ
γγ
γγγ
γ
γ
MPP
MM
PP
M
MM
TT
M
M
TT
( )( )
( )
( )( )
2)1()1(
121
11
22
11
1221
21
21
1
2
21
2
21
21
21
21
1
1
2
2
1
2
1
2
1
1
2
+−+=
⎟⎟⎠
⎞⎜⎜⎝
⎛−+
⎟⎟⎠⎞
⎜⎜⎝⎛ −
−⎟⎠⎞⎜
⎝⎛ −+
−−+−
=
==
MM
M
MM
MMM
TT
MM
VV
γγ
ρρ
γγ
γγγ
γγγρ
ρ
ρρ
Temp. ratio
Pres. ratio
Dens. ratio
Simplifying:
1
23
12
Gas Dynamics
Stagnation pressures:
Other relations:
( )⎭⎬⎫
⎩⎨⎧
+−−
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
−+
−+=
=
−
112
211
211 2
1
1
21
22
01
02
1
2
01
1
2
02
01
02
γγγ
γ
γ γγ
M
M
M
PP
PP
PP
PP
PP
2
02
02
01
2
01
1
01
01
02
1
02
PP
PP
PP
PP
PP
PP
=
=
Governing Equations cont.
13
Gas Dynamics
Entropy change:
But, S02=S2 and S01=S1 because the flow is all isentropic before and after shockwave.
So, when applied to stagnation points:
But, flow across the shock wave is adiabatic & non-isentropic:
And the stagnation entropy is equal to the static entropy:
So:
Shock wave
1 2
⎥⎦
⎤⎢⎣
⎡−⎥
⎦
⎤⎢⎣
⎡=−
1
2
1
212 lnln
PPR
TTcss p
⎥⎦
⎤⎢⎣
⎡−⎥
⎦
⎤⎢⎣
⎡=−
01
02
01
020102 lnln
PPR
TTcss p
0201 TT =
1ln 1201
020102 >−=⎥
⎦
⎤⎢⎣
⎡−=− ss
PPRss
( ) 1exp 12
01
02 <−−=Rss
PP Total pressure decreases across shock wave !
Governing Equations cont.
14
Gas Dynamics 15
Gas Dynamics 16
Gas Dynamics 17
Gas Dynamics 18
Gas Dynamics
Home Works
1. Consider a normal shock wave in air where the upstream flow properties are u1=680m/s, T1=288K, and p1=1 atm. Calculate the velocity, temperature, and pressure downstream of the shock.
2. A stream of air travelling at 500 m/s with a static pressure of 75 kPa and a static temperature of 150C undergoes a normal shock wave. Determine the static temperature, pressure and the stagnation pressure, temperature and the air velocity after the shock wave.
3. Air has a temperature and pressure of 3000K and 2 bars absolute respectively. It is flowing with a velocity of 868m/s and enters a normal shock. Determine the density before and after the shock.
19
Gas Dynamics 0=sM
11 >M 12 <M
01
01
1
1
1
TPT
Pρ
0102
0102
12
12
12
TTPPTT
PP
=<>>>ρρ
1M 2M1
2
PP
1
2
TT
1
2
ρρ
1
2
aa
01
02
PP
1
02
PP
Stationary Normal Shock Wave Table – Appendix C:
20
Gas Dynamics
Normal Shock Wave Table
21