Chapter 12
Steady waves in compressible
flow
12.1
Oblique shock waves
Figure 12.1 shows an oblique shock wave produced when a supersonic flow is deflected by
an angle ✓.
Figure 12.1: Flow geometry near a plane oblique shock wave.
We can think of the deflection as caused by a planar ramp at this angle although it could be
generated by the blockage produced by a solid body placed some distance away in the flow.
In general, a 3-D shock wave will be curved, and will separate two regions of non-uniform
flow. However, at each point along the shock, the change in flow properties takes place in
a very thin region much thinner than the radius of curvature of the shock. If we consider a
small neighborhood of the point in question then within the small neighborhood, the shock
may be regarded as locally planar to any required level of accuracy and the flows on either
12-1
CHAPTER 12. STEADY WAVES IN COMPRESSIBLE FLOW
12-2
side can be regarded as uniform. With the proper orientation of axes the flow is locally
two-dimensional. Therefore it is quite general to consider a straight oblique shock wave in
a uniform parallel stream in two-dimensions as shown below.
Balancing mass, two components of momentum and energy across the indicated control
volume leads to the following oblique shock jump conditions.
⇢1 u1 = ⇢2 u2
P1 + ⇢1 u1 2 = P2 + ⇢2 u2 2
(12.1)
⇢1 u1 v1 = ⇢2 u2 v2
h1 +
1
1
u 1 2 + v 1 2 = h2 +
u2 2 + v 1 2
2
2
Since ⇢u is constant, v1 = v2 and the jump conditions become
⇢1 u1 = ⇢2 u2
P1 + ⇢1 u1 2 = P2 + ⇢2 u2 2
(12.2)
v1 = v2
1
1
h1 + u 1 2 = h 2 + u 2 2 .
2
2
When the ideal gas law P = ⇢RT is included, the system of equations (12.2) closes allowing
all the properties of the shock to be determined. Note that, with the exception of the
additional equation, v1 = v2 , the system is identical to the normal shock jump conditions.
The oblique shock acts like a normal shock to the flow perpendicular to it. Therefore
almost all of the normal shock relations can be converted to oblique shock relations with
the substitution
M1 ! M1 Sin ( )
M2 ! M2 Sin (
Recall the Rankine-Hugoniot relation
(12.3)
✓) .
CHAPTER 12. STEADY WAVES IN COMPRESSIBLE FLOW
⇢2
=
⇢1
⇣
⇣
P2
P1
P2
P1
⌘ ⇣
+ 1 + PP21
⌘ ⇣
P2
+1
P1
1
1
⌘
⌘
12-3
(12.4)
plotted in figure 12.2.
Figure 12.2: Plot of the Hugoniot relation (12.4)
This shows the close relationship between the pressure rise across the wave (oblique or
normal) and the associated density rise. The jump conditions for oblique shocks lead to a
modified form of the very useful Prandtl relation
⇤ 2
u1 u2 = (a )
✓
1
+1
◆
v12
(12.5)
where (a⇤ )2 = RT ⇤ . From the conservation of total enthalpy, for a calorically perfect gas
in steady adiabatic flow
1
C p Tt = C p T + U 2 =
2
a2
1
+1
+ U2 =
(a⇤ )2 .
1 2
2(
1)
(12.6)
The Prandtl relation is extremely useful for easily deriving all the various normal and
oblique shock relations. The oblique shock relations generated using (12.3) are
CHAPTER 12. STEADY WAVES IN COMPRESSIBLE FLOW
P2
2 M1 2 Sin2 ( )
=
P1
+1
(
12-4
1)
⇢2
u1
( + 1) M1 2 Sin2 ( )
=
=
⇢1
u2
(
1) M1 2 Sin2 ( ) + 2
(12.7)
2 M1 2 Sin2 ( ) (
1) (
1) M1 2 Sin2 ( ) + 2
T2
=
T1
( + 1)2 M1 2 Sin2 ( )
M2 2 Sin2 (
✓) =
(
1) M1 2 Sin2 ( ) + 2
.
2 M1 2 Sin2 ( ) (
1)
The stagnation pressure ratio across the shock is
Pt2
=
Pt1
✓
( + 1) M1 2 Sin2 ( )
(
1) M1 2 Sin2 ( ) + 2
◆
1
✓
+1
2
2
2 M1 Sin ( )
(
1)
◆
1
1
.
(12.8)
Note that (12.8) can also be generated by the substitution (12.3).
12.1.1
Exceptional relations
One all new relation that has no normal shock counterpart is the equation for the absolute
velocity change across the shock.
✓
U2
U1
◆2
=1
4
M1 2 Sin2 ( )
M1 2 Sin2 ( ) + 1
1
( + 1)2 M1 4 Sin2 ( )
(12.9)
Exceptions to the substitution rule (12.3) are the relations involving the static and stagnation pressure, Pt2 /P1 and Pt1 /P2 across the wave. The reason for this is as follows.
Consider
✓
Pt2
Pt2 Pt1
Pt2
=
=
1+
P1
Pt1 P1
Pt1
1
2
M1
2
◆
M2
2
◆
1
.
(12.10)
.
(12.11)
Similarly
✓
Pt1
Pt1 Pt2
Pt1
=
=
1+
P2
Pt2 P2
Pt2
1
2
1
CHAPTER 12. STEADY WAVES IN COMPRESSIBLE FLOW
12-5
The stagnation to static pressure ratio in each region depends on the full Mach number,
not just the Mach number perpendicular to the shock wave.
12.1.2
Flow deflection versus shock angle
The most basic question connected with oblique shocks is: given the free stream Mach
number, M1 , and flow deflection, ✓, what is the shock angle, ? The normal velocity ratio
is
u2
(
1) M1 2 Sin2 + 2
u2 v 1
=
=
.
2
2
u1
u1 v 2
( + 1) M1 Sin
(12.12)
From the velocity triangles in figure 12.1
T an ( ) =
u1
v1
(12.13)
u2
✓) =
.
v2
T an (
Now
T an (
✓) = T an ( )
✓
(
1) M1 2 Sin2 ( ) + 2
( + 1) M1 2 Sin2 ( )
◆
.
(12.14)
An alternative form of this relation is
0
T an (✓) = Cot ( ) @
2
1+
⇣
M1
⌘
+1
2
Sin2 (
M1
2
)
M1
1
2
Sin2 (
)
1
A.
(12.15)
The shock-angle-deflection-angle relation (12.15) is plotted in figure 12.3 for several values
of the Mach number.
Corresponding points in the supersonic flow past a circular cylinder sketched below are
indicated on the M1 = 1.5 contour. At point a the flow is perpendicular to the shock wave
and the properties of the flow are governed by the normal shock relations. In moving from
point a to b the shock weakens and the deflection of the flow behind the shock increases
until a point of maximum flow deflection is reached at b . The flow solution between a
and b is referred to as the strong solution in figure 12.3. Notice that the Mach number
behind the shock is subsonic up to point c where the Mach number just downstream of
CHAPTER 12. STEADY WAVES IN COMPRESSIBLE FLOW
12-6
Figure 12.3: Flow deflection versus shock angle for oblique shocks.
the shock is one. Between c and d the flow corresponds to the weak solutions indicated
in figure 12.3. If one continued along the shock to very large distances from the sphere
the shock will have a more and more oblique angle eventually reaching the Mach angle
! µ = Sin 1 (1/M1 ) corresponding to an infinitesimally small disturbance.
Figure 12.4: Supersonic flow past a cylinder with shock structure shown.
Note that as the free-stream Mach number becomes large, the shock angle becomes independent of the Mach number.
CHAPTER 12. STEADY WAVES IN COMPRESSIBLE FLOW
Cos ( ) Sin ( )
⌘
lim T an (✓) = ⇣
+1
M1 !1
Sin2
2
12.2
12-7
(12.16)
Weak oblique waves
In this section we will develop the di↵erential equations that govern weak waves generated
by a small disturbance. The theory will be based on infinitesimal changes in the flow and
for this reason it is convenient to drop the subscript 0 10 on the flow variables upstream
of the wave. The sketch below depicts the case where the flow deflection is very small
d✓ << 1. Note that M is not close to one.
Figure 12.5: Small deflection in supersonic flow.
In terms of figure 12.3 we are looking at the behavior of weak solutions close to the horizontal axis of the figure. For a weak disturbance, the shock angle is very close to the Mach
angle Sin (µ) = 1/M . Let
Sin ( ) =
1
+"
M
(12.17)
and make the approximation
M 2 Sin2 ( ) ⇠
= 1 + 2M ".
(12.18)
Using (12.18) we can also develop the approximation
Cot ( ) ⇠
= M2
1
1/2
✓
1
◆
M3
" .
M2 1
Using (12.18) and (12.19) the ( , ✓) relation (12.15) can be expanded to yield
(12.19)
CHAPTER 12. STEADY WAVES IN COMPRESSIBLE FLOW
M2 1
4
+1
M
T an (d✓) ⇠
=
= d✓ ⇠
12-8
1/2
".
(12.20)
The velocity change across the shock (??) is expanded as
✓
U2
U1
U1
+1
◆2
=1
4
⇣
1
M
M2
+"
⌘⇣
1
M2
2
( + 1)2 M 4
1
M
1
M
+"
+"
2
+1
2
⌘
.
(12.21)
Retaining only terms of order " the fractional velocity change due to the small deflection
is
✓
dU
+1
U
◆2
8
".
( + 1) M
=1
(12.22)
Equation (12.22) is approximated as
dU
=
U
4
".
( + 1) M
(12.23)
Write (12.23) in terms of the deflection angle
dU
=
U
4
"=
( + 1) M
4
( + 1) M
✓
+1
4
◆
M
(M 2
1)1/2
d✓
(12.24)
or
dU
=
U
1
(M 2
1)1/2
d✓
(12.25)
where d✓ is measured in radians. Other small deflection relations are
dP
M2
=
d✓
P
(M 2 1)1/2
d⇢
M2
=
d✓
⇢
(M 2 1)1/2
dT
(
1) M 2
=
d✓
T
(M 2 1)1/2
(12.26)
CHAPTER 12. STEADY WAVES IN COMPRESSIBLE FLOW
12-9
and
dPt
=
Pt
2
2
2
2 M Sin
3( + 1)
3
1
16 M 3 3
" =
3( + 1)2
=
ds
R
(12.27)
or using (12.20)
dPt
=
Pt
( + 1) M 6
12(M 2
1)
3/2
ds
.
R
(d✓)3 =
(12.28)
Note that the entropy change across a weak oblique shock wave is extremely small; the
wave is nearly isentropic. The Mach number is determined from
dM 2
dU 2
=
M2
U2
dT
=
T
2
(M 2
1)
(
d✓
1/2
(M 2
1) M 2
1)1/2
d✓.
(12.29)
Adding terms
dM 2
M2
=
2
⇣
1+
1
2
(M 2
M2
⌘
1)1/2
d✓.
(12.30)
Eliminate d✓ between (12.25) and (12.30) to get an integrable equation relating velocity
and Mach number changes.
dU 2
1
dM 2
⌘
=⇣
2
2
U
1 + 21M2 M
(12.31)
The weak oblique shock relations (12.26) are, in terms of the velocity.
dP
=
P
dT
=
T
d⇢
=
⇢
M2
2
✓
M2
2
dU 2
U2
◆
1
dU 2
U2
M2
dU 2
U2
(12.32)
CHAPTER 12. STEADY WAVES IN COMPRESSIBLE FLOW
12-10
These last relations are precisely the same ones we developed for one dimensional flow
with area change in the absence of wall friction and heat transfer in chapter 9. From that
development we had
M 2 dU 2
=
2
U2
1
⇣
1
2 1+
M2
1
2
M2
⌘
dA
A
dM 2
=
M2
dA
.
A
(12.33)
If we eliminate dA/A between these two relations, the result is
dU 2
1
dM 2
⇣
⌘
=
2
U2
1 + 21M2 M
(12.34)
which we just derived in the context of weak oblique shocks.
12.3
The Prandtl-Meyer expansion
The upshot of all this is that
dU 2
=
U2
2
(M 2
1)1/2
d✓
(12.35)
is actually a general relationship valid for steady, isentropic flow. In particular it can be
applied to negative values of d✓. Consider flow over a corner.
Figure 12.6: Supersonic flow over a corner.
CHAPTER 12. STEADY WAVES IN COMPRESSIBLE FLOW
12-11
Express the angle in terms of the Mach number.
d✓ =
M2
⇣
2 1+
1
1/2
1
M2
2
⌘
dM 2
M2
(12.36)
Now integrate the angle between the initial and final Mach numbers.
Z
✓
0
d✓ =
0
Z
M2
M1
M2
⇣
2 1+
1
1/2
1
M2
2
⌘
dM 2
M2
(12.37)
Let ! be the angle change beginning at the reference mach number M1 = 1. The integral
(12.37) is
! (M ) =
✓
+1
1
◆1/2
T an
1
✓
1
+1
◆1/2
M2
1
1/2
!
T an
1
M2
1
1/2
.
(12.38)
This expression provides a unique relationship between the local Mach number and the
angle required to accelerate the flow to that Mach number beginning at Mach one. The
straight lines in figure 12.5 are called characteristics and represent particular values of the
flow deflection. According to (12.38) the Mach number is the same at every point on a
given characteristic. This flow is called a Prandtl-Meyer expansion and (12.38) is called
the Prandtl-Meyer function, plotted below for several values of .
Figure 12.7: Prandtl-Meyer function for several values of .
CHAPTER 12. STEADY WAVES IN COMPRESSIBLE FLOW
Note that for a given
12-12
there is a limiting angle at M 2 ! 1.
!max
⇡
=
2
✓
+1
1
◆1/2
1
!
(12.39)
⇣⇡ ⌘
For
= 1.4, !max = 1.45
. The expansion angle can be greater than90 . If the
2
deflection is larger than this angle there will be a vacuum between !max and the wall.
12.3.1
Example - supersonic flow over a bump
Air flows past p
a symmetric 2-D bump at a Mach number of 3. The aspect ratio of the
bump is a/b = 3.
Figure 12.8: Supersonic flow over a bump.
Determine the drag coefficient of the bump assuming zero wall friction.
Cd =
Drag f orce per unit span
2
1
2 ⇢1 U1 b
(12.40)
Solution
The ramp angle is 30 producing a 52 oblique shock with pressure ratio
P2
= 6.356.
P1
(12.41)
The expansion angle is 60 producing a Mach number
M2 = 1.406
(12.42)
! = 9.16 .
The stagnation pressure is constant through the expansion wave and so the pressure ratio
over the downstream face is
CHAPTER 12. STEADY WAVES IN COMPRESSIBLE FLOW
12-13
M3 = 4.268
(12.43)
! = 69.16
and
0
1+
⇣
P3 @
⇣
=
P2
1+
1
2
1
2
⌘
⌘
M2 2
M3 2
1
A
1
=
1 + 0.2(1.406)2
1 + 0.2(4.268)2
!3.5
=
✓
1.395
4.643
◆3.5
= 0.0149
(12.44)
and
P3
P3 P2
=
= 0.0149 ⇥ 6.356 = 0.0945.
P1
P2 P1
(12.45)
The drag coefficient becomes
Cd =
12.4
P2 2b Sin (30) P3 2b Sin (30)
6.356 0.0945
=
= 0.994.
2
1.4
2 (9)
2 M 1 P1 b
(12.46)
Problems
Problem 1 - Use the oblique shock jump conditions (12.2) to derive the oblique shock
Prandtl relation (12.5).
Problem 2 - Consider the supersonic flow past a bump discussed in the example above.
Carefully sketch the flow putting in the shock waves as well as the leading and trailing
characteristics of the expansion.
Problem 3 - Consider a streamline in compressible flow past a 2-D ramp with a very small
ramp angle.
Figure 12.9: Supersonic flow past a 2-D ramp.
CHAPTER 12. STEADY WAVES IN COMPRESSIBLE FLOW
12-14
Determine the ratio of the heights of the streamline above the wall before and after the
oblique shock in terms of M1 and d✓ find the unknown coefficient in (12.47).
H2
= 1 + (???) d✓
H2
(12.47)
Pay careful attention to signs.
Problem 4 - Consider a body in subsonic flow. As the free-stream Mach number is
increased there is a critical value, Mc , such that there is a point somewhere along the body
where the flow speed outside the boundary layer reaches the speed of sound. Figure 12.10
illustrates this phenomena for flow over a projectile.
Figure 12.10: Projectile in high subsonic flow.
In this figure 12.10 the critical Mach number is somewhere between 0.840 and 0.885 as
evidenced by the weak shocks that appear toward the back of the projectile in the middle
picture. The local pressure in the neighborhood of the body is expressed in terms of the
pressure coefficient.
CP =
P P1
2
1
2 ⇢1 U1
(12.48)
Show that the value of the pressure coefficient at the point where sonic speed occurs is
CHAPTER 12. STEADY WAVES IN COMPRESSIBLE FLOW
C Pc =
✓
1+
1
2
M1c 2
+1
2
◆
2 M1c
1
1
.
2
12-15
(12.49)
State any assumptions needed to solve the problem.
Problem 5 - Consider frictionless (no wall friction) supersonic flow over a flat plate of
chord C at a small angle of attack as shown in figure 12.11.
Figure 12.11: Supersonic flow past a flat plate at a small angle of attack.
The circulation about the plate is defined as
=
I
U ds.
(12.50)
where the integration is along any contour surrounding the plate.
1) Show that, to a good approximation, the circulation is given by
=
2U1 C
M1 2
1
1/2
↵
(12.51)
where the integration is clockwise around the plate.
2) Show that, to the same approximation, Lif tperunitspan = ⇢1 U1 .
Problem 6 - Consider frictionless (no wall friction) flow of air at M = 2 over a flat plate
of chord C at 5 angle of attack as shown in figure 12.12.
Figure 12.12: Supersonic flow over a flat plate at 5 angle of attack.
CHAPTER 12. STEADY WAVES IN COMPRESSIBLE FLOW
12-16
Evaluate the drag coefficient of the plate. Compare with the value obtained using a weak
wave approximation.
Problem 7 - Consider frictionless (no wall friction) supersonic flow of Air over a flat plate
of chord C at an angle of attack of 15 degrees as shown in figure 12.13.
Figure 12.13: Supersonic flow over a flat plate at 15 angle of attack.
Determine the lift coefficient
CL =
L
2
1
2 ⇢1 U1 C
(12.52)
where L is the lift force per unit span.
Problem 8 - Figure 12.14 shows a symmetrical, diamond shaped airfoil at a 5 angle of
attack in a supersonic flow of air.
Figure 12.14: Supersonic flow past a diamond shaped airfoil.
Determine the lift and drag coefficients of the airfoil.
CL =
Lif t per unit span
2
1
2 ⇢1 U1 C
Drag per unit span
CD =
2
1
2 ⇢1 U1 C
(12.53)
CHAPTER 12. STEADY WAVES IN COMPRESSIBLE FLOW
12-17
What happens to the flow over the airfoil if the free-stream Mach number is decreased to
1.5? Compare your result with the lift and drag of a thin flat plate at 5 angle of attack
and free-stream Mach number of 3.
Problem 9 - The figure below shows supersonic flow of Air over a 30 wedge followed by
a 10 wedge. The free stream Mach number is 3.
Figure 12.15: Supersonic over a wedge with a shoulder.
1) Determine M2 , M3 and the included angle of the expansion fan, .
2) Suppose the flow was turned through a single 10 wedge instead of the combination
shown above. Would the stagnation pressure after the turn be higher or lower than in the
case shown? Why?
Problem 10 - Figure 12.16 shows a smooth compression of a supersonic flow of air by a
concave surface. The free-stream Mach number is 1.96. The weak oblique shock at the
nose produces a Mach number of 1.932 at station 1. From station 1 to station 2 the flow
is turned 20 degrees.
Figure 12.16: Supersonic flow compressed by a concave surface.
1) Determine the Mach number at station 2.
2) Determine the pressure ratio P2 /P1 .
3) State any assumptions used.
CHAPTER 12. STEADY WAVES IN COMPRESSIBLE FLOW
12-18
Problem 11 - Figure 12.17 shows supersonic flow of air in a channel or duct at a Mach
number of three. The flow produces an oblique shock o↵ a ramp at an angle of 16 degrees.
The shock reflects o↵ the upper surface of the wind tunnel as shown below. Beyond the
ramp the channel height is the same at the height ahead of the ramp.
Figure 12.17: Mach 3 flow in a duct with a ramp.
1) Determine the Mach number in region 2.
2) Determine the Mach number in region 3.
3) Describe qualitatively how Pt and Tt vary between regions 1, 2 and 3.
4) Suppose the channel height is 10 cm . Precisely locate the shock reflections on the upper
and lower walls.
5) Suppose the walls were lengthened. At roughly what point would the Mach number
tend to one?
Problem 12 - Figure 12.18 shows supersonic flow of air turned through an angle of 30 .
The free stream Mach number is 3.
Figure 12.18: Supersonic flow turned 30 .
In case (a) the turning
turning is accomplished
pressure change in each
merit of one design over
is accomplished by a single 30 wedge whereas in case (b) the
by two 15 degree wedges in tandem. Determine the stagnation
case, (Pt2 /Pt1 )|(a) and (Pt3 /Pt1 )|(b) and comment on the relative
the other.
Problem 13 - Figure 12.19 shows the flow of helium from a supersonic over-expanded
round jet. If we restrict our attention to a small region near the intersection of the first
two oblique shocks and the so-called Mach disc as shown in the blow-up, then we can use
CHAPTER 12. STEADY WAVES IN COMPRESSIBLE FLOW
12-19
oblique shock theory to determine the flow properties near the shock intersection (despite
the generally non-uniform 3-D nature of the rest of the flow). The shock angles with respect
to the horizontal measured from the image are as shown.
Figure 12.19: Supersonic flow from an over expanded round jet.
1) Determine the jet exit Mach number. Hint, you will need to select a Mach number that
balances the pressures in regions 2 and 4 with a dividing streamline that is very nearly
horizontal as shown in the picture.
2) Determine the Mach number in region 2.
3) Determine the flow angles and Mach numbers in regions 3 and 4.
4) Determine P2 /P1 and P4 /P1 . How well do the static pressures match across the dividing
streamline (dashed line) between regions 2 and 4?
Problem 14 - Figure 12.20 shows the reflection of an expansion wave from the upper wall
of a 2-D, adiabatic, inviscid channel flow. The gas is helium at an incoming Mach number,
M1 = 1.5 and the deflection angle is 20 . The flow is turned to horizontal by the lower
wall which is designed to follow a streamline producing no reflected wave. Determine M2 ,
M3 and H/h.
Figure 12.20: Supersonic flow in an expansion.