notes07 oblique shock waves

7/30/2019 Notes07 Oblique Shock Waves

1/26

Oblique Shock Waves

Text Chapter 9

The next topic to be uncovered in this class is the problem ofoblique shock waves. The motivation comes again from the quasi-

one-dimensional nozzle flows that is shown below:

We considered the physical and theoretical reasons that normal

shock waves form, summarized partly by the figure below:

1


2/26

This idea can be further expanded by considering the following

figure:

in which we see the physical pattern that might occur if

disturbances propagate in subsonic and supersonic flows.

Relative to figure (b) we note that the lines extending from the

disturbance outward at angles define two regions of the flow.

Zone of action The region of the flow affected by

disturbances.

Zone of silence The region of the flow not affected by

disturbances.

The line separating the two regions defines theMach Wave and is

found from the equation:

V

a

Vt

at 1sin === (9.1)

2


3/26

or

1sin 1=

TheMach Wave is an infinitesimally weak shock wave. An actual

oblique shock wave is formed from many such waves coalescing

into finite disturbances, similar to the normal shock wave. The

figure below relates the two types of waves:

The figure presents an interesting but somewhat confusing picture

of oblique shock waves, since the Mach waves behind a shock

wave have a much larger angle than those before. The Mach wave

in the figure is drawn for conditions upstream of the shock wave. It

is also important to note that Mach waves travel relative to the

current flow conditions and the Mach angle is determined relativeto those conditions not necessarily the horizontal.

3


4/26

A more accurate picture of shock and Mach waves is given below:

It is also important to note the flow direction as it is significant that

it follows tangent to the surface.

It is curious in a way that the flow no longer remains horizontal in

a two-dimensional shock wave, and is indicative of the fact that

shock waves form for two reasons.

1. high pressure downstream

2. flow direction changes

However, it should be noted that it is more than a simple flow

direction change that causes a shock wave, rather, it is flow turning

into itselfas shown in the left figure on the next page:

4


5/26

Flow turning away from itselfresults in an expansion wave.

Properties change across an oblique shock wave in manner similarto normal shock waves, that is:

M: M1>1 M2?

p: p1


6/26

The derivation of the equations is accomplished by again applying

the governing equations, this time to the control volume shown

below:

Noteworthy is that velocity is decomposed into normal and

tangential components. The figure also introduces the angles:

- shock wave angle

- flow deflection angle ( in older texts /NACAreports)

6


7/26

Apply the continuity equation

0 =S

dSnVr

(7.1)

b,c,e & f

lie on a streamline

2211

2211 0000

uu

AuAu

=

+++++(9.2)

Momentum equation:

( ) =SS

dSnpVdSnV rr

(7.7)

Tangential Direction:

( ) ( ) =SS

dSnpwdSnV tanr

Gives:

( ) ( ) ( )cebfbf AppAppwAuwAu +=+++++ 0000 22221111

222111 wuwu =

Use (9.2) 21 ww = (9.4)

Therefore the tangential velocity component does not change

across the oblique shock wave.

7


8/26


9/26

We need next to consider the velocity:

(9.4)22

21

22

22

21

21

22

21 uuwuwuVV =++=

So finally:

22

22

2

21

1u

hu

h +=+ (9.9)

Then for a oblique shock wave the equations become:

2211 uu = (9.2)

21 ww = (9.4)

2222

2111 upup +=+ (9.6)

ohu

hu

h =+=+22

22

2

21

1 (9.9)

So that, with the exception of Eq. (9.4), the oblique shock jump

relations are identical to the normal shock relations, provided that

we use the normal velocity to define the conditions. That is:

sin11 MMn = (9.10)

9


10/26

Oblique Shock Relations

So that Equations (8.50), (8.52), (8.57) and (8.59) become:

+

=

2

1

2

11

2

2

2

1

1

2

n

n

nM

M

M(9.11)

( )

( ) 2

2

12

1

1

12

1

n

nM

M

+

+

=

(9.12)

( )( )1

1

21 2

1

2

1

++=

nM

p

p

(9.13)

2

1

1

2

1

2

pp

TT = (9.14)

and finally from the geometry we define:

( )=

sin

22

nMM (9.15)

However, this is not quite enough to define everything we need for

an oblique shock analysis. We still need a way to determine how

the shock angle relates to the surface flow angle.

10


11/26

--M Relation

Returning back to the oblique shock relation:

1

1tanw

u= (9.16)

( )2

2tanw

u= (9.17)

Eq. (9.4)( )

2

1

1

2

tan

tan

==

u

u(9.18)

11


12/26

Eq. (9.12)( ) ( )

( )

22

1

221

sin1

sin12

tan

tan

M

M

+

+=

(9.19)

Or after significant manipulation:

( ) 22cos

1sincot2tan

21

221

++

=

M

M(9.20)

--M Relation

12


13/26

13


14/26

14


15/26

Observations on --M Equation

1. max Given M1 there is a maximum deflection angle thatcan occur, i.e., only detached shocks are possible.

2. Given M1 and


16/26

16


17/26

17


18/26

Example 1 - A uniform supersonic stream with M1=3.0, p1=1atm, and

T1=288K encounters a compression corner which deflects the stream by an

angle =20. Calculate the shock wave angle, andp2, T2, M2, p02 and T02behind the shock wave.

18


19/26

Example 2 The flow deflection angle of Example 1 is increased to =30.Calculate the pressure and Mach number behind the wave, and compare

these results with those of Example 1.

19


20/26

Example 3 The free-stream Mach number in Example 1 is increased to 5.

Calculate the pressure and Mach number behind the wave, and compare

these results with those of Example 1.

20


21/26

Example 4 Consider a Mach 2.8 supersonic flow over a compression

corner with a deflection angle of 15. If the deflection angle is doubled to30, what is the increase in shock strength (as measured by the pressureratio)? Is it also doubled?

21


22/26

Example 5 - Consider a compression corner with a deflection angle of 28.Calculate the shock strengths whenM1=3.0 and whenM1 is doubled to 6. Isthe shock strength also doubled?

22


23/26

Example 6 Consider a Mach 4 flow over a compression corner with a

deflection angle of 32. Calculate the oblique shock wave angle for the weakshock case using (a) the --Mchart, and (b) the --Mequation. Compare

the results from the two sets of calculations.

23


24/26

Example 7 A 10 half-angle wedge is placed in a mystery flow ofunknown Mach number. Using a Schlieren system, the shock wave angle is

measured as 44. What is the free-stream Mach number?

24


25/26

Example 8 Consider a 15 half-angle wedge at zero angle of attack.Calculate the pressure coefficient on the wedge surface in a Mach 3 flow or

air.

25


26/26

Example 9 Consider a 15 half-angle wedge at zero angle of attack in aMach 3 flow of air. Calculate the drag coefficient. Assume that the pressure

exerted over the base of the wedge, the base pressure, is equal to that of the

free-stream pressure.

notes07 oblique shock waves

Documents