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1EDUNEX ITB
1
18 October 2021
Fakultas Teknik Mesin dan Dirgantara
Oblique & Expansion Waves
OBLIQUE SHOCK WAVES
AE3110 Aerodynamics 1
Dr. -ing. Mochammad Agoes M. ST. MSc.Ema Amalia, ST., MT.
Pramudita Satria Palar, ST, MT, PhD
2EDUNEX ITB
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WEEK 9
Mach wave (week 1)
Review : Mach Wave and Normal Shock Wave
Source : Physics.animation.com
Normal Shock wave (week 5)
See : 4.21 & 7.04 sec
Norm
al S
W
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WEEK 9
Mach wave (week 1)
Review : Mach Wave and Normal Shock Wave
𝜇 Mach angle
𝜇 = 𝑠𝑖𝑛−11
𝑀𝑎
𝑉
𝑀∞ > 1.0 𝑀∞ > 1.0
▪ Isentropic
▪ It is only function of 𝑴∞
▪ streamline is straight
▪ Non-isentropic
▪ It is only function of 𝑴∞
▪ streamline is straight
𝑀∞ > 1.0 𝑀∞ < 1.0
𝜅 = 90𝑜
Normal Shock wave (week 5)
Supersonic Supersonic Supersonic Subsonic
What does it happen when the angle is in between ?
𝜇 ≪ 𝛽 ≪ 𝜅 𝛽
Oblique SW
If It is only function of 𝑴∞ ?
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WEEK 9
What is Oblique Shock Wave?
Oblique shockwave :
▪ Supersonic flow is turned into itself
▪ It is generated from upstream uniform supersonic turned by a
deflected surface upward through an angle θ
Important parameters of Oblique Shock wave
1. Incoming flow Mach number, 𝑴𝟏
2. Deflected surface angle, θ
3. Oblique shock wave angle, 𝜷
𝛽 Through the oblique shockwave
the supersonic flow changes to
lower supersonic
The fluid properties change in
discontinuity across the shock wave
(Non-isentropic)
5EDUNEX ITB
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WEEK 9
What is Oblique Shock Wave?
Experiment Wind tunnel Flow through test section wind tunnel is visualized
using a schlieren detecting flow density.
In Flight
Movie of
Oblique
SW
CFD
Simulation
On TSTO
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WEEK 9
Application of Oblique and Expansion wave
Design of Intake of Fighter
Analysis of wave interaction Design of supersonic airfoil
Design of supersonic Nozzle
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WEEK 9
Application of Oblique shockwave : Fighter Intake
MIG-21 FIGHTER
Fixed or movable shock cone (spike)
Please mention from which locations
oblique shockwaves may be
generated?
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WEEK 9
Application of Oblique shockwave : Fighter Intake
F-16 FIGHTER SU-27 FIGHTER
XB-70 FIGHTER
Grumman F14
FIGHTER
F-22 RAPTOR
FIGHTER
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WEEK 9
Source of Oblique shockwave
aV Subsonic aV =Sonic Supersonic aV
Mach wave
Mach Angle
▪ Oblique SW is created by disturbance which propagate
by molecular collision at speed of sound
▪ Source of propagation as a beeper.
▪ The beeper emits a sound disturbance which propagates
in all direction at the speed of sound
▪ For the beeper moving in subsonic speed, the beeper
always stays in the family of circular sound wave.
▪ For the supersonic speed of the beeper motion, it is
constantly outside the family of circular sound wave.
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WEEK 9
Source of Oblique shockwave
Oblique shock
▪ If the disturbance is stronger than small
beeper emitting sound wave such as a
wedge shown in the right figure, the wave
front become stronger than a Mach wave
▪ The strong disturbance coalesce into an
oblique shockwave at an angle β to the free
stream, where β > θ
𝜃
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WEEK 9
Oblique Shock wave .vs. Normal Shock wave
The basic phenomenon
is similar to normal
shock wave
▪ Pressure increases.
▪ Density increases.
▪ Temperature increases.
▪ Entropy increases
▪ Mach number decreases.
▪ Velocity decreases.
▪ Total pressure decreases.
▪ Total temperature remains
constant.
The differences between
them
▪ Behind the oblique shockwave flow is
deflected and be parallel to the surface
▪ Flow behind normal shockwave is
always subsonic, but behind oblique
shock mostly it is supersonic
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WEEK 9
Governing Equation for Oblique Shockwave
Assumptions:
▪ The flow is steady.
▪ The flow is adiabatic.
▪ There are no viscous effect on
the side of the control volume.
▪ There are no body forces.
Fixed Control volume approach (Eulerian)
flow is steady
Continuity equation
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WEEK 9
Governing Equation for Oblique Shockwave
▪ The flow is steady.
▪ The flow is adiabatic.
▪ There are no viscous effect on
the side of the control volume.
▪ There are no body forces.
Momentum Equations
Remember, we
have two-
dimensional flow!
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WEEK 9
Governing Equation for Oblique Shockwave
▪ The flow is steady.
▪ The flow is adiabatic.
▪ There are no viscous effect on
the side of the control volume.
▪ There are no body forces.
Momentum Equations (Tangential direction)
The pressure in b cancels that of f,
so does c and e.
The pressure in a and
d have no tangential
component
Tangential component of
the velocity is constant
across the normal shock!
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WEEK 9
Governing Equation for Oblique Shockwave
▪ The flow is steady.
▪ The flow is adiabatic.
▪ There are no viscous effect on
the side of the control volume.
▪ There are no body forces.
Momentum Equations (Normal direction)
Note that u is the
component velocity in the
normal direction
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WEEK 9
Governing Equation for Oblique Shockwave
▪ The flow is steady.
▪ The flow is adiabatic.
▪ There are no viscous effect on
the side of the control volume.
▪ There are no body forces.
Energy Equation
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WEEK 9
Governing Equation for Oblique Shockwave
Normal vs oblique shock wave
The equations are the
same!
It is only the normal
component that matters!
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WEEK 9
Governing Equation for Oblique Shockwave
Oblique SW
Mn2
Mn1 Mt
Mt
Rotated by 𝛽 deg
β
β
θβ
Flow kinematic
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WEEK 9
Governing Equation for Oblique Shockwave
β
From Normal shock wave relation
Left side kinematic
relation 𝑀𝑛1 = 𝑀1 sin 𝛽
M1
Right side
kinematic relation
𝑀2 =𝑀𝑛2
sin(𝛽 − 𝜃)
Left side Right side
𝑀22 𝑠𝑖𝑛2 𝛽 − 𝜃 =
2 + (𝛾 − 1)𝑀12 𝑠𝑖𝑛2 𝛽
2𝛾 𝑀12 𝑠𝑖𝑛2 𝛽 − (𝛾 − 1)
𝜌2𝜌1
=(𝛾 + 1)𝑀1
2 𝑠𝑖𝑛2 𝛽
2 + (𝛾 − 1)𝑀12 𝑠𝑖𝑛2 𝛽
𝑝2𝑝1
= 1 +2𝛾
(𝛾 + 1)𝑀1
2 𝑠𝑖𝑛2 𝛽 − 1
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WEEK 9
Governing Equation for Oblique Shockwave
Theta-beta-Mach number relation
Using some trigonometric
manipulation
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WEEK 9
Graph of the relation of 𝜷 − 𝜽 −𝑴
• If θ = 0, then β equals either 90◦ or μ.
• The case of β = 90◦ corresponds to a
normal shock.
• The case of β = μ corresponds to the Mach
waveType equation here.. • In both cases, the flow streamlines
• experience no deflection across the wave.
𝜽 = 𝟎
𝜷 = 𝝁
𝜷 = 𝟗𝟎
Mach wave
Normal shockwave
𝜇 = 𝑠𝑖𝑛−11
𝑀
𝜇 = 𝑠𝑖𝑛−11
2= 30𝑜
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WEEK 9
Graph of the relation of 𝜷 − 𝜽 −𝑴
• There are two straight oblique shock solutions for a given ϴ: Strong and weak shock solution.
• In nature, it is typically a weak shock that occurs.
weak shock solution
strong shock solution
𝟖𝟒. 𝟓
𝟑𝟗. 𝟓
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WEEK 9
Graph of the relation of 𝜷 − 𝜽 −𝑴
24
This is the line of M2=1, meaning that for the
vast majority of cases involving the weak
shock solution, the downstream Mach
number is supersonic M2 > 1, except for
some cases.
𝑀2 > 1
𝑀2 < 1
𝑴𝟐 = 𝟏
weak shock solution
strong shock solution
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WEEK 9
Graph of the relation of 𝜷 − 𝜽 −𝑴
ϴmax is the maximum theta for a given Mach number
Maximum theta
𝜽𝒎𝒔𝒙 = 𝟐𝟑
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WEEK 9
Shock Polar
Graphical representation of Oblique shock properties
𝑉𝑥
𝑉𝑦
𝑉𝑥 = 𝑉1
𝑉𝑦 ≠ 0
𝑉𝑦 = 0
𝑉𝑥 = 𝑉2
upstream
downstream
𝑉𝑥/𝑎∗
𝑉𝑦/𝑎∗
▪ For θ = 0, there are two points A and E. Point
A correspond to Mach line and point E is
normal shockwave
▪ For a given θ, two solutions : Points B and D
Point B weak SW 𝑉2 > 1.0 and point D
strong SW
▪ Point C is the tangent of the shock polar →
max deflection angle 𝜃𝑚𝑎𝑥
Look at the curve A-B-C-D-E
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WEEK 9
Example 1
Flow through Ramp
Air flowing at Mach number 2.0 with temperature of 300K and pressure of 1
atm is compressed by turning through a ramp with angle of 10o. Ratio of
specific heat of air is 1.4. For each of the two possible solutions calculate:
a) The shock angle, β
b) The Mach number downstream of the shock wave
c) The temperature and pressure behind shockwave
d) The change of entropy
e) What is the maximum deflection angle if the shock remains attached?
Θ=10o
M= 2.0β = ?𝑇 = 300 𝐾
𝑝 = 1 𝑎𝑡𝑚
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WEEK 9
Solution of example 1
Mach number Downstream
𝑀22 𝑠𝑖𝑛2 𝛽 − 𝜃 =
2 + (𝛾 − 1)𝑀12 𝑠𝑖𝑛2 𝛽
2𝛾 𝑀12 𝑠𝑖𝑛2 𝛽 − (𝛾 − 1)
Solution of weak shockwave : M2=1.64 (supersonic)
Solution of strong shockwave : M2=0.60 (subsonic)
Temperature and pressure downstream
𝜌2𝜌1
=(𝛾 + 1)𝑀1
2 𝑠𝑖𝑛2 𝛽
2 + (𝛾 − 1)𝑀12 𝑠𝑖𝑛2 𝛽
𝑝2𝑝1
= 1 +2𝛾
(𝛾 + 1)𝑀1
2 𝑠𝑖𝑛2 𝛽 − 1
𝜌2
𝜌1= 2.65
𝑝2𝑝1
= 4.44𝑇2𝑇1
= 1.68
𝑝2 = 1.72 𝑎𝑡𝑚
𝑇2 = 351 𝐾
weak SW
𝑝2 = 4.46 𝑎𝑡𝑚
strong SW
𝑇2 = 504 𝐾
𝜌2
𝜌1= 1.46
𝑝2𝑝1
= 1.7𝑇2𝑇1
= 1.17
𝛽 = 39.31
𝛽 = 83.7
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WEEK 9
Solution of example 1
Entropy change
∆𝑠 = 𝑠2 − 𝑠1 = 𝑐𝑝 ln𝑇02𝑇01
− 𝑅 ln𝑝02𝑝01
𝑝02
𝑝01=
𝑝2
𝑝1
𝑇1
𝑇2
𝛾
𝛾−1=
𝑝2
𝑝2
(2−1
𝛾) 𝜌2
𝜌1
𝛾
𝛾−1
𝑃𝑅 =𝑝𝑜2
𝑝01= 0.98 𝑜𝑟 0.73
weak strong
Δs =5.98 J/Kg.K, or Δs =86.42 J/Kg.KOblique shockwave calculator
http://www.dept.aoe.vt.edu/~devenpor/aoe3114/calc.html
https://www.engineering.com/calculators/oblique_flow_relations.htm
Θ=10o
M= 2.0𝑇 = 300 𝐾
𝑝 = 1 𝑎𝑡𝑚
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WEEK 9
Solution of example 1
For the same case and what does it happen if 𝜃 = 23o
Θ=23o
M= 2.0𝑇 = 300 𝐾
𝑝 = 1 𝑎𝑡𝑚
𝜃 > 𝜃𝑚𝑎𝑥
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WEEK 9
Example 1 (cont’d)
Flow through Ramp
It is still the same freestream condition, but the ramp is carried out in two
steps with a half angle of the previous problem ( 5 deg for one step),
Compare pressure recovery this problem to the previous problem for weak
solution
Θ1=5o
M= 2.0β = ?𝑇 = 300 𝐾
𝑝 = 1 𝑎𝑡𝑚
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WEEK 9
Example 1 (cont’d)
Comparison between Single and double ramp
Single Ramp Double Ramps
Deflected angle 10 deg 5 + 5 deg
Tan(θ) 0.176 2 x 0.0875 = 0.175
Oblique SW angle 39.31 deg 34.3 deg ; 37.95
Rear Mach number 1.65 1.82 → 1.648
Pressure Recovery 0.985 0.997 * 0.998 =0.995
Case : incoming Mach number = 2.0
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WEEK 9
Reflected Shock Wave
The reflected shock wave occurs when an incident wave meets a
wall/ surface and the flow deflected and turned into itself
M1
Deflected SW
12 3M2
M3
𝜽𝟏𝜷𝟏
𝜽𝟐
𝜷𝟐
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WEEK 9
Reflected Shock Wave
The reflected shock wave occurs when an incident wave meets a
wall/ surface and the flow deflected and turned into itself
=10 deg
M= 2.0
Deflected SW
12 3
𝜽𝟐
𝜽𝟏𝜷𝟏
Solution of weak shockwave : M1=2.0 (supersonic)
𝛽1 = 39.31𝜃1 = 10 𝑀1 = 1.64→𝑀1 = 2.0
𝛽1 = 49.4𝜃2 = 10 𝑀1 = 1.28→𝑀2 = 1.64
𝜌2
𝜌1= 1.46
𝑝2𝑝1
= 1.7𝑇2𝑇1
= 1.17
𝜌2
𝜌1= 1.42
𝑝2𝑝1
= 1.64𝑇2𝑇1
= 1.15
𝜃
38EDUNEX ITB
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WEEK 9
Shock Wave Interactions with zero slip line
M1 12 3M2
M3
𝜽𝟏𝜷𝟏
𝜷𝟐
𝜽𝟏𝜷𝟏
The shock wave from the corner of the lower surface will meet the
shock wave form the corner of the upper surface producing the
interaction in the flow field in point E. Two reflected shock waves
are generated in downstream in regions 2-3 and region 4-5.
Regions 3 and 5 are one region separated by a slip line (virtual
line) where along its line the static pressure is same.
E
A
B
C
D4
5
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WEEK 9
1 2
3 4
4’
Shock Wave Interactions with non-zero slip line
Shock wave interaction with different deflected angles
▪ The shock wave from the corner of the lower surface will meet the shock wave form the
corner of the upper surface producing the interaction in the flow field in point E.
▪ Two reflected shock waves are generated in downstream in regions 3-4 and region 2-4’.
▪ The direction of streamline of flow downstream is determined by the equilibrium of the
static pressures in regions 4 and 4’ .
▪ The virtual boundary line between the regions 4 and 4’ is called Slip line
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WEEK 9
Shock Wave Interactions with non-zero slipline
Shock wave interaction with different deflected anglesM_1 2.2
Beta_12 45deg
From Graph Theta-Mach-
Beta
Theta 18deg
M_2 1.48
P2/P1 2.67
Based on M_2 , the reflected shock wave
has maximum deflected angle (θmax) = 11.7
and what will happen?
1. A regular wave reflection is insufficient to strengthen the flow
2. A Mach reflection is generated at the upper wall and meets the
incident shock wave around the middle domain.
3. Detached SW generates at the wall
4. Slip line is formed at region 3-4
5. Behind normal SW and slip line is subsonic
6. The static pressures at the up and below slip line should be equal
And how to calculate Mach number in region 3 and 4.
Answer the calculation process is iteratively till the static
pressure in region 3 is equal to one in the region 4
𝑝4
𝑝1=
𝑝3
𝑝1=
𝑝3
𝑝2
𝑝2
𝑝1= 2.67
𝑝3
𝑝2
p2/p1 = 2.67
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WEEK 9
Shock Wave Interactions with non-zero slip line
Shock wave interaction with different deflected angles
Theta_14 Beta_14 M4 p4/p1 theta_23 Beta_23 M3 P3/p2 p3/p1
11 84 0.58 5.3 7 81 0.754 1.4 3.74
10 84.6 0.57 5.3 8 79 0.76 1.5 4.01
8 86 0.56 5.4 10 75 0.8 1.7 4.54
6 86.5 0.555 5.5 12 67 0.93 2.1 5.6
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WEEK 9
Home work
1. Spike of a fighter at the flight condition shown below
▪ Determine properties in zones 2 and 3 including
Mach number, pressure, temperature, density and
Total pressure
▪ The change in entropy s3-s1
▪ What does happen in regions 2 and 3 when
a. Freestream Mach number is increased or
decreased
b. flow deflection angle is greater
that its 𝜃𝑚𝑎𝑥
Gives answer qualitatively
K
kPa
kPa
7.216
4.19
8.151
0.2
1
1
1
1
=
=
=
=
T
P
P
M
t
1 2
3