notes07 oblique shock waves
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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:
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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)
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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.
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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:
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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
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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)
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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.
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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)
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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.
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--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)
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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
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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
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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.
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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.
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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.
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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?
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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?
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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.
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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?
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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.
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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.