Steady waves in compressible ﬂow - Stanford...

Chapter 12

Steady waves in compressibleflow

12.1 Oblique shock waves

Figure 12.1 shows an oblique shock wave produced when a supersonic flow is deflected byan 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 begenerated 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-uniformflow. However, at each point along the shock, the change in flow properties takes place ina very thin region much thinner than the radius of curvature of the shock. If we consider asmall neighborhood of the point in question then within the small neighborhood, the shockmay 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 locallytwo-dimensional. Therefore it is quite general to consider a straight oblique shock wave ina uniform parallel stream in two-dimensions as shown below.

Balancing mass, two components of momentum and energy across the indicated controlvolume leads to the following oblique shock jump conditions.

⇢1

u1

= ⇢2

u2

P1

+ ⇢1

u1

2 = P2

+ ⇢2

u2

2

⇢1

u1

v1

= ⇢2

u2

v2

h1

+1

2

�u1

2 + v1

2

�= h

2

+1

2

�u2

2 + v1

2

�

(12.1)

Since ⇢u is constant, v1

= v2

and the jump conditions become

⇢1

u1

= ⇢2

u2

P1

+ ⇢1

u1

2 = P2

+ ⇢2

u2

2

v1

= v2

h1

+1

2u1

2 = h2

+1

2u2

2.

(12.2)

When the ideal gas law P = ⇢RT is included, the system of equations (12.2) closes allowingall the properties of the shock to be determined. Note that, with the exception of theadditional equation, v

1

= 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. Thereforealmost all of the normal shock relations can be converted to oblique shock relations withthe substitution

M1

! M1

Sin (�)

M2

! M2

Sin (� � ✓) .(12.3)

Recall the Rankine-Hugoniot relation


⇢2

⇢1

=�⇣P2

P1

+ 1⌘+⇣P2

P1

� 1⌘

�⇣P2

P1

+ 1⌘�⇣P2

P1

� 1⌘ (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 ornormal) and the associated density rise. The jump conditions for oblique shocks lead to amodified form of the very useful Prandtl relation

u1

u2

= (a⇤)2 �✓� � 1

� + 1

◆v21

(12.5)

where (a⇤)2 = �RT ⇤. From the conservation of total enthalpy, for a calorically perfect gasin steady adiabatic flow

CpTt = CpT +1

2U2 =

a2

� � 1+

1

2U2 =

� + 1

2 (� � 1)(a⇤)2. (12.6)

The Prandtl relation is extremely useful for easily deriving all the various normal andoblique shock relations. The oblique shock relations generated using (12.3) are


P2

P1

=2�M

1

2Sin2 (�)� (� � 1)

� + 1

⇢2

⇢1

=u1

u2

=(� + 1)M

1

2Sin2 (�)

(� � 1)M1

2Sin2 (�) + 2

T2

T1

=

�2�M

1

2Sin2 (�)� (� � 1)� �

(� � 1)M1

2Sin2 (�) + 2�

(� + 1)2M1

2Sin2 (�)

M2

2Sin2 (� � ✓) =(� � 1)M

1

2Sin2 (�) + 2

2�M1

2Sin2 (�)� (� � 1).

(12.7)

The stagnation pressure ratio across the shock is

Pt2

Pt1=

✓(� + 1)M

1

2Sin2 (�)

(� � 1)M1

2Sin2 (�) + 2

◆ ��1

✓� + 1

2�M1

2Sin2 (�)� (� � 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 absolutevelocity change across the shock.

✓U2

U1

◆2

= 1� 4

�M

1

2Sin2 (�)� 1� �

�M1

2Sin2 (�) + 1�

(� + 1)2M1

4Sin2 (�)(12.9)

Exceptions to the substitution rule (12.3) are the relations involving the static and stag-nation pressure, Pt2/P1

and Pt1/P2

across the wave. The reason for this is as follows.Consider

Pt2

P1

=Pt2

Pt1

Pt1

P1

=Pt2

Pt1

✓1 +

� � 1

2M

1

2

◆ ��1

. (12.10)

Similarly

Pt1

P2

=Pt1

Pt2

Pt2

P2

=Pt1

Pt2

✓1 +

� � 1

2M

2

2

◆ ��1

. (12.11)


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 Machnumber, M

1

, and flow deflection, ✓, what is the shock angle, �? The normal velocity ratiois

u2

u1

=(� � 1)M

1

2Sin2� + 2

(� + 1)M1

2Sin2�=

u2

u1

v1

v2

. (12.12)

From the velocity triangles in figure 12.1

Tan (�) =u1

v1

Tan (� � ✓) =u2

v2

.

(12.13)

Now

Tan (� � ✓) = Tan (�)

✓(� � 1)M

1

2Sin2 (�) + 2

(� + 1)M1

2Sin2 (�)

◆. (12.14)

An alternative form of this relation is

Tan (✓) = Cot (�)

0

@ M1

2Sin2 (�)� 1

1 +⇣�+1

2

⌘M

1

2 �M1

2Sin2 (�)

1

A . (12.15)

The shock-angle-deflection-angle relation (12.15) is plotted in figure 12.3 for several valuesof the Mach number.

Corresponding points in the supersonic flow past a circular cylinder sketched below areindicated on the M

1

= 1.5 contour. At point a the flow is perpendicular to the shock waveand the properties of the flow are governed by the normal shock relations. In moving frompoint a to b the shock weakens and the deflection of the flow behind the shock increasesuntil a point of maximum flow deflection is reached at b . The flow solution between aand b is referred to as the strong solution in figure 12.3. Notice that the Mach numberbehind the shock is subsonic up to point c where the Mach number just downstream of


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 indicatedin figure 12.3. If one continued along the shock to very large distances from the spherethe shock will have a more and more oblique angle eventually reaching the Mach angle� ! µ = Sin�1 (1/M

1

) 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 inde-pendent of the Mach number.


limM

1

!1Tan (✓) =

Cos (�)Sin (�)⇣�+1

2

⌘� Sin2�

(12.16)

12.2 Weak oblique waves

In this section we will develop the di↵erential equations that govern weak waves generatedby a small disturbance. The theory will be based on infinitesimal changes in the flow andfor this reason it is convenient to drop the subscript 010 on the flow variables upstreamof the wave. The sketch below depicts the case where the flow deflection is very smalld✓ << 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 hori-zontal axis of the figure. For a weak disturbance, the shock angle is very close to the Machangle Sin (µ) = 1/M . Let

Sin (�) =1

M+ " (12.17)

and make the approximation

M2Sin2 (�) ⇠= 1 + 2M". (12.18)

Using (12.18) we can also develop the approximation

Cot (�) ⇠=�M2 � 1

�1/2✓1� M3

M2 � 1"

◆. (12.19)

Using (12.18) and (12.19) the (�, ✓) relation (12.15) can be expanded to yield


Tan (d✓) ⇠= d✓ ⇠=4

� + 1

�M2 � 1

�1/2

M". (12.20)

The velocity change across the shock (??) is expanded as

✓U2

� U1

U1

+ 1

◆2

= 1� 4

⇣M2

�1

M + "�2 � 1

⌘⇣�M2

�1

M + "�2

+ 1⌘

(� + 1)2M4

�1

M + "�2

. (12.21)

Retaining only terms of order " the fractional velocity change due to the small deflectionis

✓dU

U+ 1

◆2

= 1� 8

(� + 1)M". (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

(M2 � 1)1/2d✓ (12.24)

or

dU

U= � 1

(M2 � 1)1/2d✓ (12.25)

where d✓ is measured in radians. Other small deflection relations are

dP

P=

�M2

(M2 � 1)1/2d✓

d⇢

⇢=

M2

(M2 � 1)1/2d✓

dT

T=

(� � 1)M2

(M2 � 1)1/2d✓

(12.26)


and

dPt

Pt= � 2�

3(� + 1)2�M2Sin2� � 1

�3

= � 16�M3

3(� + 1)2"3 = �ds

R(12.27)

or using (12.20)

dPt

Pt= � � (� + 1)M6

12(M2 � 1)3/2(d✓)3 = �ds

R. (12.28)

Note that the entropy change across a weak oblique shock wave is extremely small; thewave is nearly isentropic. The Mach number is determined from

dM2

M2

=dU2

U2

� dT

T= � 2

(M2 � 1)1/2d✓ � (� � 1)M2

(M2 � 1)1/2d✓. (12.29)

Adding terms

dM2

M2

= �2

⇣1 + ��1

2

M2

⌘

(M2 � 1)1/2d✓. (12.30)

Eliminate d✓ between (12.25) and (12.30) to get an integrable equation relating velocityand Mach number changes.

dU2

U2

=1⇣

1 + ��1

2

M2

⌘ dM2

M2

(12.31)

The weak oblique shock relations (12.26) are, in terms of the velocity.

dP

P= ��

2M2

dU2

U2

dT

T= �

✓� � 1

2

◆M2

dU2

U2

d⇢

⇢= �M2

dU2

U2

(12.32)


These last relations are precisely the same ones we developed for one dimensional flowwith area change in the absence of wall friction and heat transfer in chapter 9. From thatdevelopment we had

1�M2

2

dU2

U2

= �dA

A

1�M2

2⇣1 + ��1

2

M2

⌘ dM2

M2

= �dA

A.

(12.33)

If we eliminate dA/A between these two relations, the result is

dU2

U2

=1⇣

1 + ��1

2

M2

⌘ dM2

M2

(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

dU2

U2

= � 2

(M2 � 1)1/2d✓ (12.35)

is actually a general relationship valid for steady, isentropic flow. In particular it can beapplied to negative values of d✓. Consider flow over a corner.

Figure 12.6: Supersonic flow over a corner.


Express the angle in terms of the Mach number.

d✓ = ��M2 � 1

�1/2

2⇣1 + ��1

2

M2

⌘ dM2

M2

(12.36)

Now integrate the angle between the initial and final Mach numbers.

Z ✓

0

d✓0 = �Z M

2

M1

�M2 � 1

�1/2

2⇣1 + ��1

2

M2

⌘ dM2

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

Tan�1

✓� � 1

� + 1

◆1/2�

M2 � 1�1/2

!� Tan�1

�M2 � 1

�1/2

. (12.38)

This expression provides a unique relationship between the local Mach number and theangle required to accelerate the flow to that Mach number beginning at Mach one. Thestraight lines in figure 12.5 are called characteristics and represent particular values of theflow deflection. According to (12.38) the Mach number is the same at every point on agiven characteristic. This flow is called a Prandtl-Meyer expansion and (12.38) is calledthe Prandtl-Meyer function, plotted below for several values of �.

Figure 12.7: Prandtl-Meyer function for several values of �.


Note that for a given � there is a limiting angle at M2 ! 1.

!max

=⇡

2

✓� + 1

� � 1

◆1/2

� 1

!(12.39)

For � = 1.4, !max

= 1.45⇣⇡2

⌘. The expansion angle can be greater than90�. If the

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 a symmetric 2-D bump at a Mach number of 3. The aspect ratio of thebump is a/b =

p3.

Figure 12.8: Supersonic flow over a bump.

Determine the drag coe�cient of the bump assuming zero wall friction.

Cd =Drag force per unit span

1

2

⇢1

U1

2b(12.40)

Solution

The ramp angle is 30� producing a 52� oblique shock with pressure ratio

P2

P1

= 6.356. (12.41)

The expansion angle is 60� producing a Mach number

M2

= 1.406

! = 9.16�.(12.42)

The stagnation pressure is constant through the expansion wave and so the pressure ratioover the downstream face is


M3

= 4.268

! = 69.16�(12.43)

and

P3

P2

=

0

@1 +

⇣��1

2

⌘M

2

2

1 +⇣��1

2

⌘M

3

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

P1

=P3

P2

P2

P1

= 0.0149⇥ 6.356 = 0.0945. (12.45)

The drag coe�cient becomes

Cd =P2

2b Sin (30)� P3

2b Sin (30)�2

M1

2P1

b=

6.356� 0.09451.42

(9)= 0.994. (12.46)

12.4 Problems

Problem 1 - Use the oblique shock jump conditions (12.2) to derive the oblique shockPrandtl 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 trailingcharacteristics of the expansion.

Problem 3 - Consider a streamline in compressible flow past a 2-D ramp with a very smallramp angle.

Figure 12.9: Supersonic flow past a 2-D ramp.


Determine the ratio of the heights of the streamline above the wall before and after theoblique shock in terms of M

1

and d✓ find the unknown coe�cient in (12.47).

H2

H2

= 1 + (???) d✓ (12.47)

Pay careful attention to signs.

Problem 4 - Consider a body in subsonic flow. As the free-stream Mach number isincreased there is a critical value, Mc, such that there is a point somewhere along the bodywhere the flow speed outside the boundary layer reaches the speed of sound. Figure 12.10illustrates 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 asevidenced by the weak shocks that appear toward the back of the projectile in the middlepicture. The local pressure in the neighborhood of the body is expressed in terms of thepressure coe�cient.

CP =P � P11

2

⇢1U12

(12.48)

Show that the value of the pressure coe�cient at the point where sonic speed occurs is


CPc =

✓1+

��1

2

M1c2

�+1

2

◆ ��1

� 1

�2

M1c2

. (12.49)

State any assumptions needed to solve the problem.

Problem 5 - Consider frictionless (no wall friction) supersonic flow over a flat plate ofchord 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

� =

IUds. (12.50)

where the integration is along any contour surrounding the plate.

1) Show that, to a good approximation, the circulation is given by

� =2U

1

C�M

1

2 � 1�1/2

↵ (12.51)

where the integration is clockwise around the plate.

2) Show that, to the same approximation, Liftperunitspan = ⇢1

U1

�.

Problem 6 - Consider frictionless (no wall friction) flow of air at M = 2 over a flat plateof 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.


Evaluate the drag coe�cient of the plate. Compare with the value obtained using a weakwave approximation.

Problem 7 - Consider frictionless (no wall friction) supersonic flow of Air over a flat plateof 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 coe�cient

CL =L

1

2

⇢1U12C

(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 ofattack in a supersonic flow of air.

Figure 12.14: Supersonic flow past a diamond shaped airfoil.

Determine the lift and drag coe�cients of the airfoil.

CL =Lift per unit span

1

2

⇢1U12C

CD =Drag per unit span

1

2

⇢1U12C

(12.53)


What happens to the flow over the airfoil if the free-stream Mach number is decreased to1.5? Compare your result with the lift and drag of a thin flat plate at 5� angle of attackand free-stream Mach number of 3.

Problem 9 - The figure below shows supersonic flow of Air over a 30� wedge followed bya 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 combinationshown above. Would the stagnation pressure after the turn be higher or lower than in thecase shown? Why?

Problem 10 - Figure 12.16 shows a smooth compression of a supersonic flow of air by aconcave surface. The free-stream Mach number is 1.96. The weak oblique shock at thenose produces a Mach number of 1.932 at station 1. From station 1 to station 2 the flowis 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.


Problem 11 - Figure 12.17 shows supersonic flow of air in a channel or duct at a Machnumber 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 theramp 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.


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 upperand lower walls.

5) Suppose the walls were lengthened. At roughly what point would the Mach numbertend 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 is accomplished by a single 30� wedge whereas in case (b) theturning is accomplished by two 15� degree wedges in tandem. Determine the stagnationpressure change in each case,(Pt2/Pt1)|

(a) and (Pt3/Pt1)|(b) and comment on the relative

merit of one design over the other.

Problem 13 - Figure 12.19 shows the flow of helium from a supersonic over-expandedround jet. If we restrict our attention to a small region near the intersection of the firsttwo oblique shocks and the so-called Mach disc as shown in the blow-up, then we can use


oblique shock theory to determine the flow properties near the shock intersection (despitethe generally non-uniform 3-D nature of the rest of the flow). The shock angles with respectto 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 thatbalances the pressures in regions 2 and 4 with a dividing streamline that is very nearlyhorizontal as shown in the picture.


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 dividingstreamline (dashed line) between regions 2 and 4?

Problem 14 - Figure 12.20 shows the reflection of an expansion wave from the upper wallof a 2-D, adiabatic, inviscid channel flow. The gas is helium at an incoming Mach number,M

1

= 1.5 and the deflection angle is 20�. The flow is turned to horizontal by the lowerwall which is designed to follow a streamline producing no reflected wave. Determine M

2

,M

3

and H/h.

Figure 12.20: Supersonic flow in an expansion.

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