gas dynamics esa 341 chapter 3
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Gas Dynamics ESA 341 Chapter 3. Dr Kamarul Arifin B. Ahmad PPK Aeroangkasa. Normal shock waves. Definition of shock wave Formation of normal shock wave Governing equations Shock in the nozzle. Shock wave. V. P. T. Definition of shock wave. - PowerPoint PPT PresentationTRANSCRIPT
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Gas DynamicsESA 341Chapter 3
Dr Kamarul Arifin B. Ahmad
PPK Aeroangkasa
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Normal shock waves
Definition of shock wave Formation of normal shock wave Governing equations Shock in the nozzle
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Definition of shock waveShock wave is a very thin region in a flow where a supersonic flow is decelerated to subsonic flow. The process is adiabatic but non-isentropic.
Shock wave
V
P
T
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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 dV causing 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.
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Types of Shock Waves:
Normal shock wave - easiest to analyze
Oblique shock wave - will be analyzed based on normal shock relations
Curved shock wave - difficult & will not be analyzed in this class
- The flow across a shock wave is adiabatic but not isentropic (because it is irreversible). So:
0201
0201
PP
TT
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Governing Equations
1
1
1
1
T
P
V
2
2
2
2
T
P
VConservation of mass:
Conservation of momentum:
Rearranging:
Combining:
AVAV 2211
122221
121121
1221
VVVPP
VVVPP
VVmAPP
2
2121
22
1
212121
PP
VVV
PPVVV
21
22
2121
11VVPP
Conservation of energy:
Change of variable:
0
22
2
21
1 22Tc
VTc
VTc ppp
2
2
1
121
22 1
2
PP
VV
combine
22
2
221
1
1
1
2
1
2V
PV
P
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Continued:
Multiplied by 2/p1:
Rearranging:
2
2
1
1
2121 1
211
PP
PP
1
2
1
2
1
2
1
2
1
211
P
P
P
P
1
2
1
2
1
2
11
111
P
P
1
2
1
2
1
2
11
111
PP
PP
or
Governing Equations cont.
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1
2
1
2
2
1
1
2
11
111
PP
PP
V
V
2
1
1
2
1
2
P
P
T
T
2
1
1
2
1
2
11
11
PP
PP
T
T
Governing Equations cont.
From conservation of mass:
From equation of state:
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Governing Equations cont.
2211 VV
222
211
2
2222
2111
1221
11 MPMP
Pa
VPVP
VVmAPP
22
21
1
2
22
2
21
1
21
1
21
1
22
M
M
T
T
Vh
Vh
C
O
M
BINE
Conservation of mass
Conservation of momentum
Conservation of energy
02
21
1
)2
11(
1
)2
11(
2
11
12
11
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
MM
MMMMMM
M
MM
M
MM
MM
MM
M
M
RTMRT
PRTM
RT
P
VV
Expanding the equations
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Governing Equations cont.
12
212
1
21
2
M
MM
Solution:
Mach number cannot be negative. So, only the positive value is realistic.
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Governing Equations cont.
1
1
1
2
1
1
121
11
22
11
21
1
21
1
21
1
2
22
21
1
2
21
2
21
21
1
2
22
21
1
2
M
P
P
M
M
P
P
M
MM
T
T
M
M
T
T
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
M
M
M
MM
MM
M
T
T
M
M
V
V
Temp. ratio
Pres. ratio
Dens. ratio
Simplifying:
1
2
3
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Stagnation pressures:
Other relations:
1
12
21
1
21
1 21
1
21
22
01
02
1
2
01
1
2
02
01
02
M
M
M
P
P
P
P
P
P
P
P
P
P
2
02
02
01
2
01
1
01
01
02
1
02
P
P
P
P
P
P
P
P
P
P
P
P
Governing Equations cont.
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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
P
PR
T
Tcss p
01
02
01
020102 lnln
P
PR
T
Tcss p
0201 TT
1ln 1201
020102
ss
P
PRss
1exp 12
01
02
R
ss
P
P Total pressure decreases across shock wave !
Governing Equations cont.
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Group Exercises 3
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.
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0sM
11 M 12 M
01
01
1
1
1
T
P
T
P
0102
0102
12
12
12
TT
PP
TT
PP
1M 2M1
2
P
P
1
2
T
T
1
2
1
2
a
a
01
02
P
P
1
02
P
P
Stationary Normal Shock Wave Table – Appendix C: