oblique & expansion waves

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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

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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)

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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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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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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







Momentum Equations

Remember, we

have two-

dimensional flow!

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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







Momentum Equations (Normal direction)

Note that u is the

component velocity in the

normal direction

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Energy Equation

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Normal vs oblique shock wave

The equations are the

same!

It is only the normal

component that matters!

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WEEK 9


Oblique SW

Mn2

Mn1 Mt

Mt

Rotated by 𝛽 deg

β

β

θβ

Flow kinematic

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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


Theta-beta-Mach number relation

Using some trigonometric

manipulation

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WEEK 9

Graph of the relation of 𝜷 − 𝜽 −𝑴

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WEEK 9


• 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


𝑀


2= 30𝑜

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• 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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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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ϴmax is the maximum theta for a given Mach number

Maximum theta

𝜽𝒎𝒔𝒙 = 𝟐𝟑

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WEEK 9

Detached Shockwave

max max

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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

𝟖𝟑. 𝟕

𝟑𝟗. 𝟑𝟏

θmax=23o

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WEEK 9


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


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


http://www.dept.aoe.vt.edu/~devenpor/aoe3114/calc.html

https://www.engineering.com/calculators/oblique_flow_relations.htm

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For the same case and what does it happen if 𝜃 = 23o

Θ=23o

M= 2.0𝑇 = 300 𝐾


𝜃 > 𝜃𝑚𝑎𝑥

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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


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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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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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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

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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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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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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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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

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WEEK 2

oblique & expansion waves

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