chapter 11: steady waves in compressible flow

8/3/2019 Chapter 11: Steady Waves in Compressible Flow

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bjc 11.1 9/17/09

C

HAPTER

11

S

TEADY

W

AVES

IN

C

OMPRESSIBLE

F

LOW

11.1 O

BLIQUE

SHOCK

WAVES

The figure below shows an oblique shock wave produced when a supersonic flow

is deflected by an angle . 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 neighborhoodthe shock may be regarded as locally planar to any required level of accuracy and

the flows on either side can be regarded as uniform. With the proper orientation

of axes the flow is locally two-dimensional. Therefore it is quite general to con-

sider a straight oblique shock wave in a uniform parallel stream in two-

dimensions as shown below.

Figure 11.1 Flow geometry near a plane oblique shock wave.

Balancing mass, two components of momentum and energy across the indicated

control volume leads to the following oblique shock jump conditions.

U1

u1

v1

U2

u2v

2

1

P1

T1

M1

, , ,

2 P2 T2 M2, , ,controlvolume


2/22

Oblique shock waves

9/17/09 11.2 bjc

. (11.1)

Since is constant, and the jump conditions become,

. (11.2)

When the ideal gas law is included, the system of equations (11.2)

closes allowing all the properties of the shock to be determined. Note that, with

the exception of the additional equation, , 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 beconverted to oblique shock relations with the substitution

. (11.3)

Recall the Rankine-Hugoniot relation,

, (11.4)

1u

1

2u

2=

P1

1u

1

2+ P

2

2u

2

2+=

1u

1v

1

2u

2v

2=

h1

1

2--- u

1

2v

1

2+( )+ h

2

1

2--- u

2

2v

2

2+( )+=

u v1 v2=

1u

1

2u

2=

P1 1u1

2+ P

2 2u22

+=

v1 v2=

h1

1

2---u

1

2+ h

2

1

2---u

2

2+=

P RT=

v1

v2

=

M1

M1Sin

M2

M2Sin ( )

2

1

------

P

2

P1------ 1+

P2

P1------ 1

+

P

2

P1

------ 1+ P2

P1

------ 1

------------------------------------------------------=


3/22

Oblique shock waves

bjc 11.3 9/17/09

plotted below.

Figure 11.2 Plot of the Hugoniot relation (11.4)

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,

(11.5)

where . From the conservation of total enthalpy, for a calorically

perfect gas in steady adiabatic flow,

. (11.6)

The Prandtl relation is extremely useful for easily deriving all the various normaland oblique shock relations. The oblique shock relations generated using (11.3)

are the following.

20 40 60 80 100

2

4

6

8

10

P2

P1

2

1

1+ 1------------

1.2=

1.4=

1.66=

u1u

2a*

2 1

1+-------------

v1

2=

a*2

RT*

=

Cp

Tt

Cp

T1

2---U

2+

a2

1------------

1

2---U

2+

1+

2 1( )--------------------a*

2= = =


4/22

Oblique shock waves

9/17/09 11.4 bjc

. (11.7)

The stagnation pressure ratio across the shock is,

. (11.8)

Note that (11.8) can also be generated by the substitution (11.3).

11.1.1 EXCEPTIONALRELATIONS

One all new relation that has no normal shock counterpart is the equation for the

absolute velocity change across the shock.

. (11.9)

Exceptions to the substitution rule (11.3) are the relations involving the static and

stagnation pressure and across the wave. The reason for this is

as follows. Consider

(11.10)

Similarly

P2

P1

------2M

1

2Sin

2 1( )

1+( )------------------------------------------------------=

2

1------

u1

u2-----

1+( )M1

2Sin

2

1( )M12Si n2 2+----------------------------------------------------= =

T2

T1

------2M

1

2Sin

2 1( )( ) 1( )M

1

2Sin

2 2+( )

1+( )2M1

2Si n

2

---------------------------------------------------------------------------------------------------------------------=

M2

2Sin

2 ( )

1( )M1

2Si n

2 2+

2M1

2Sin

2 1( )

------------------------------------------------------=

Pt2

Pt1

-------- 1+( )M

1

2Sin

2

1( )M1

2Sin

2 2+

---------------------------------------------------

1------------

1+

2M1

2Sin

2 1( )

-----------------------------------------------------

1

1------------

=

U2

U1

------- 2

1 4M

1

2Sin

2 1( ) M1

2Sin

2 1+( )

1+( )2M1

4Sin

2

--------------------------------------------------------------------------------=

Pt2

P1

Pt1

P2

Pt2

P1

--------P

t2

Pt1

--------P

t1

P1

--------P

t2

Pt1

-------- 1 1

2------------M

1

2+

1------------

= =


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Oblique shock waves

bjc 11.5 9/17/09

(11.11)

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.

11.1.2 FLOWDEFLECTIONVERSUSSHOCKANGLE

The most basic question connected with oblique shocks is: Given the free stream

Mach number, , and flow deflection, , what is the shock angle, ? The nor-

mal velocity ratio is

. (11.12)

From the velocity triangles in Figure 11.1,

. (11.13)

Now

. (11.14)

An alternative form of this relation is

. (11.15)

The shock-angle-deflection-angle relation (11.15) is plotted in Figure 11.3 for

several values of the Mach number.

Pt1

P2

--------P

t1

Pt2

--------P

t2

P2

--------P

t1

Pt2

-------- 1 1

2------------M

2

2+

1------------

= =

M1

u2

u1

-----

1( )M1

2Sin

2 2+

1+( )M1

2Sin

2

---------------------------------------------------

u2

v2

-----

v1

u1

-----= =

Tanu

1

v1

----- ; Tan ( )u

2

v2

-----= =

Tan ( ) Tan 1( )M1

2Sin2 2+

1+( )M1

2Sin

2

---------------------------------------------------

=

TanCot M

1

2Sin

2 1( )

1 1+

2

-------------

M1

2M

1

2Sin

2+

-------------------------------------------------------------------=


6/22

Oblique shock waves

9/17/09 11.6 bjc

Figure 11.3 Flow deflection versus shock angle for oblique shocks.

Corresponding points in the supersonic flow past a circular cylinder sketched

below are indicated on the contour.

0.25 0.5 0.75 1 1.25 1.5

0.2

0.4

0.6

0.8

M1

1.5=

M1

2=

M1

3=

M1

5=

2Sin 1 M1

=

strongsolution

weak

solution

1.4=

a

bc

d

M1

M1

1.5=

a

b

c

d

M>1

M


7/22

Weak oblique waves

bjc 11.7 9/17/09

At point 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 to .

the shock weakens and the deflection of the flow behind the shock increases until

a point of maximum flow deflection is reached at . The flow solution between

and is referred to as the strong solution in Figure 11.3. Notice that the Mach

number behind the shock is subsonic up to point where the Mach number just

downstream of the shock is one. Between and the flow corresponds to the

weak solutions indicated in Figure 11.3. 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 corresponding to

an infinitesimally small disturbance.

Note that as the freestream Mach number becomes large, the shock angle becomes

independent of the Mach number.

. (11.16)

11.2 WEAKOBLIQUEWAVES

In this section we will develop the differential 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

on the flow variables upstream of the wave. The sketch below depicts the case

where the flow deflection is very small . Note that is notclose to one.

Figure 11.4 Small deflection in supersonic flow

a

a b

b a

b

c

c d

Sin1

1 M1

( )=

Tan( )M

1

limCos Sin

1+

2-------------

Sin2----------------------------------------=

1

d 1 M

d

U U+dU

M M+dM


8/22

Weak oblique waves

9/17/09 11.8 bjc

In terms of Figure 11.3 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

. Let

(11.17)

and make the approximation

. (11.18)

Using (11.18) we can also develop the approximation

. (11.19)

Using (11.18) and (11.19) the relation (11.15) can be expanded to yield

. (11.20)

The velocity change across the shock (11.9) is expanded as

. (11.21)

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

deflection is

. (11.22)

Equation (11.22) is approximated as

. (11.23)

Sin 1 M=

Sin1

M----- +=

M2Sin

2 1 2M+

Cot M2 1( )1 2 1 M3

M2

1

-----------------

,( )

Tan d( ) d 4 1+-------------

M2

1( )1 2

M-------------------------------

U2

U1

U1

--------------------- 1+ 2

1 4

M2 1

M----- +

2 1 M2 1

M----- +

2 1+

1+( )2M4 1M----- +

2---------------------------------------------------------------------------------------------------=

dUU------- 1+ 2

18

1+( )M----------------------- =

dU

U-------

4

1+( )M----------------------- =


9/22

Weak oblique waves

bjc 11.9 9/17/09

Write (11.23) in terms of the deflection angle

(11.24)

or

(11.25)

where is measured in radians. Other small deflection relations are

(11.26)

and

(11.27)

or

. (11.28)

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,

(11.29)

dU

U-------

4

1+( )M-----------------------

4

1+( )M-----------------------

1+

4-------------

M

M2

1( )1 2

-------------------------------d= =

dU

U-------

1

M2

1( )1 2

-------------------------------d=

d

dP

P-------

M2

M2

1( )1 2-------------------------------d=

d

------

M2

M2

1( )1 2

-------------------------------d=

dT

T-------

1( )M2

M2

1( )1 2

-------------------------------d=

dPt

Pt

--------2

3 1+( )2------------------------ M

2Sin

2 1( )

3

16M3

3 1+( )2------------------------

3

ds

R-----= = =

dPt

Pt

-------- 1+( )M6

12 M2

1

( )

3 2------------------------------------- d( )3 ds

R-----= =

dM2

M2

-----------dU

2

U2

----------dT

T-------

2

M2

1( )1 2

-------------------------------d 1( )M2

M2

1( )1 2

-------------------------------d= =


10/22

Weak oblique waves

9/17/09 11.10 bjc

or

. (11.30)

Eliminate between (11.25) and (11.30)to get an integrable equation relating

velocity and Mach number changes.

(11.31)

The weak oblique shock relations (11.26) are, in terms of the velocity

(11.32)

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,

(11.33)

If we eliminate between these two relations, the result is,

dM2

M2

----------- 2

1 1

2------------

M+2

M2 1( )1 2

------------------------------------------d=

d

dU2

U2

----------1

1 1

2------------

M+2

------------------------------------------

dM2

M2

-----------=

dP

P-------

M2

2-----------

dU2

U2

----------=

dT

T-------

1( )M2

2-------------------------

dU2

U2

----------=

d M2

2( ) dU2 U2( )=

1 M2

2-----------------

dU2

U2

----------dA

A-------=

1 M2

2 1 12

------------ M+2

---------------------------------------------dM

2

M2

-----------dA

A

-------=

dA A


11/22

The Prandtl-Meyer expansion

bjc 11.11 9/17/09

, (11.34)

which we just derived in the context of weak oblique shocks.

11.3 THE PRANDTL-MEYEREXPANSION

The upshot of all this is that

(11.35)

is actually a general relationship valid for steady, isentropic flow. In particular itcan be applied to negative values of . Consider flow over a corner

Figure 11.5 Supersonic flow over a corner

Express the angle in terms of the Mach number.

. (11.36)

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

dU2

U2

----------1

1 1

2------------

M+2

------------------------------------------

dM2

M2

-----------=

dU2

U2

----------2

M2

1( )1 2

-------------------------------d=

d

M1

M2

dM

21( )

1 2

2 1 1

2------------

M+2

---------------------------------------------dM

2

M2-----------=


12/22


9/17/09 11.12 bjc

. (11.37)

Let be the angle change beginning at the reference mach number .

The integral (11.37) is,

. (11.38)

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 11.5 are called characteristics and repre-

sent particular values of the flow deflection. According to (11.38) the Mach

number is the same at every point on a given characteristic. This flow is called a

Prandtl-Meyer expansion and (11.38) is called the Prandtl-Meyer function, plot-

ted below for several values of .

Note that for a given there is a limiting angle at .

d

0

M2

1( )1 2

2 1 1

2------------

M+2

---------------------------------------------dM

2

M2

-----------

M1

M2

=

M1

1=

M( ) 1+ 1-------------

1 2 Tan 1 1 1+-------------

1 2 M2 1( )1 2

Tan 1 M2 1( )

1 2=

5 10 15 20

0.5

1

1.5

2

2.5

3

M

M( ) 1.2=

1.4=

1.66=

M2


13/22


bjc 11.13 9/17/09

(11.39)

For , . The expansion angle can be greater than

If the deflection is larger than this angle there will be a vacuum between

and the wall.

11.3.1 EXAMPLE - SUPERSONICFLOWOVERABUMP

Air flows past a symmetric 2-D bump at a Mach number of 3. The aspect ratio

of the bump is .

Determine the drag coefficient of the bump assuming zero wall friction,

Solution

The ramp angle is producing a oblique shock with pressure ratio,

(11.40)

and downstream Mach number and Prandtl-Meyer function

. (11.41)

The expansion angle is producing a Mach number,

. (11.42)

max

2---

1+

1-------------

1 2 1 =

1.4= max

1.45

2

---

= 90

max

a b 3=

2a

bM1 3= 2 3

Cd

Drag Force per unit span

12---

1U

12b

-------------------------------------------------------------=

30 52

P2

P1

6.356=

M2

1.406 ; 9.16= =

60

M3

4.268 ; 69.16= =


14/22

Problems

9/17/09 11.14 bjc

The stagnation pressure is constant through the expansion wave and so the

pressure ratio over the downstream face is

(11.43)

and

(11.44)

The drag coefficient becomes,

11.4 PROBLEMS

Problem 1 - Use the oblique shock jump conditions (11.2) to derive the oblique

shock Prandtl relation (11.5).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.

P3

P2

------

1 1

2------------

M2

2+

1 1

2------------M

3

2+

-------------------------------------

1------------

1 0.2 1.406( )2+

1 0.2 4.268( )2+---------------------------------------

3.5 1.3954.643-------------

3.5

0.0149= = = =

P3

P1

------P

3

P2

------P

2

P1

------ 0.0149( ) 6.356( ) 0.0945= = =

Cd

P22bSin 30( ) P32bSin 30( )

2---M

1

2P

1b

-------------------------------------------------------------------------6.356 0.0945

1.4

2------- 9( )

------------------------------------ 0.994= = =

d

M1

H1

H2


15/22

Problems

bjc 11.15 9/17/09

Determine the ratio of the heights of the streamline above the wall before and after

the oblique shock in terms of and , ie, find the unknown coefficient in,

. (11.45)

Pay careful attention to signs.

Problem 4 - Consider a body in subsonic flow. As the freestream Mach number

is increased there is a critical value, , such that there is a point somewhere

along the body where the flow speed outside the boundary layer reaches the speed

of sound. Figure 223 (below) from VanDykes book illustrates this phenomena for

flow over a projectile.

1

H2

H1

------- 1 ???( )d+=

Mc


16/22

Problems

9/17/09 11.16 bjc

In this figure 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.

(11.46)

Show that the value of the pressure coefficient at the point where sonic speed

occurs is,

(11.47)

State any assumptions needed to solve the problem.

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

plate of chord at a small angle of attack as shown below.

The circulation about the plate is defined as

(11.48)

where the integration is along any contour surrounding the plate.

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

(11.49)

CP

P P( )12---U

2 =

CPc

2 1( )Mc2

+

1+---------------------------------------

1------------

1

2---Mc

2

=

C

M1>1

C

Uds=

2U

1C

M1

21( )

1 2-------------------------------=


17/22

Problems

bjc 11.17 9/17/09

where the integration is clockwise around the plate.

2) Show that, to a good approximation,

Problem 6 - Consider frictionless (no wall friction) flow of air at over a

flat plate of chord at angle of attack as shown below.

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 at an angle of attack of 15 degrees as shown below.

Determine the lift coefficient

where L is the lift force per unit span.

Lift per unit span 1U

1=

2=

C 5

M=2

=5

C

C

M1=2

= 15

C

CL

L1

2---U

2C

=


18/22

Problems

9/17/09 11.18 bjc

Problem 8 - The figure below shows a symmetrical, diamond shaped airfoil at a

angle of attack in a supersonic flow of air.

Determine the lift and drag coefficients of the airfoil.

(11.50)

What happens to the flow over the airfoil if the freestream Mach number is

decreased to 1.5? Compare your result with the lift and drag of a thin flat plate at

angle of attack and freestream Mach number of 3.

Problem 9 - The figure below shows supersonic flow of Air over a wedge

followed by a wedge. The free stream Mach number is 3.

1) Determine , and the included angle of the expansion fan, .

5

M=3

25

5

C

CL

Lift per unit span1

2---U

2C

----------------------------------------- ; CD

Drag per unit span1

2---U

2C

---------------------------------------------= =

5

3010

M1 = 3M2

M3

30

10

M2

M3


19/22

Problems

bjc 11.19 9/17/09

2) Suppose the flow was turned through a single wedge instead of the com-

bination shown above. Would the stagnation pressure after the turn be higher or

lower than in the case shown? Why?

Problem 10 - The photo below (page 137 in Van Dyke) shows a smooth com-

pression of a supersonic flow of Air by a concave surface. The freestreamMach 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.

1) Determine the Mach number at station 2

2) Determine the pressure ratio .

3) State any assumptions used.

Problem 11 - Supersonic flow of Air in a wind tunnel at a Mach number of

three produces an oblique shock off a ramp at an angle of 16 degrees. The

shock reflects off the upper surface of the wind tunnel as shown below.

10

1

2

P2

P1

M1

3=

16

231


20/22

Problems

9/17/09 11.20 bjc

1) Determine the Mach number in region 2

2) Determine the Mach number in region 3

3) Describe qualitatively how and vary between regions 1, 2 and 3.

4) Suppose the channel height is . Precisely locate the shock reflectionson the upper and lower walls.

5) Suppose the walls were lengthened. At roughly what point would the Mach

number tend to one?

Problem 12 - The figure below shows supersonic flow of air turned through an

angle of . The free stream Mach number is 3.

In case (a) the turning is accomplished by a single wedge whereas in case

(b) the turning is accomplished by two degree wedges in tandem. Determine

the stagnation pressure change in each case, and

and comment on the relative merit of one design over the other.

Problem 13 - The figure below 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 oblique shock theory to determine the flow prop-

erties near the shock intersection (despite the generally non-uniform nature of the

rest of the flow). The shock angles with respect to the horizontal measured from

the image are as shown.

Pt

Tt

10 cm

30

M1 = 3M2

30

M1 = 3

M2

15

15

M3

(a) (b)

30

15

Pt2 Pt1 a( ) Pt3 Pt1 b( )


21/22

Problems

bjc 11.21 9/17/09

1) Determine the jet exit Mach number. Hint, you will need to select a Mach num-

ber that balances the pressures in regions 2 and 4 with a dividing streamline thatis 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 and . How well do the static pressures match

across the dividing streamline (dashed line) between regions 2 and 4?

Problem 14 - The figure below 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, and the deflection angle is . The flow is

turned to horizontal

by the lower wall which is designed to follow a streamline producing no

reflected wave. Determine , and .

37

1

40

2

34

P2

P1

P4

P1

M1

1.5= 20

M1

M2M3

h

H

M2

M3

H h


22/22

Problems

chapter 11: steady waves in compressible flow

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