Introduction to Compressible Fluid Flows (Part II)

8/2/2019 Introduction to Compressible Fluid Flows (Part II)

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Compressible Fluid Flow (Part II)

Oblique shocks

Normal shocks are really just the starting point for understanding shock systems. Locally, there is alwayssome (perhaps moving) coordinate system with respect to which the shock is normal. However, it is notalways convenient to work in a moving coordinate system, especially since one has to transform back to a“hardware frame” to discuss, e.g. stagnation properties. In the hardware frame of reference, we canobserve and exploit the phenomenon of oblique shocks – shock for which the flow velocity is notperpendicular to the shock front, but rather approaches the front at an angle. The angle the shock frontmakes with the flow velocity is called the wave angle and is denoted by β; the deflection angle or wedgeangle is called δ.

Before we lay out the equations governing oblique shocks, we should consider a key question, namely:how do our various quantities of interest (total and static pressure and temperature, etc.) transform whenwe shift to a frame which is moving with respect to our original hardware frame? That is, we start with a

coordinate frame and flow situation where we know ( ) z y xV ,,r

, ( ) z y xT S ,, , etc. (for simplicity we

consider steady flow). If we boost to a frame moving at a speed U r

with respect to our original system,how do all these quantities change? In the new frame, the velocity transforms as

( ) ( ) U z y xV z y xV V rrrr

−=′′′′→ ,,,, . It is important to understand that the static thermodynamicproperties do not change . Why? The static temperature, for example, is just the temperature of a

δβ



Suppose we start with a normal shock with, as usual, upstream conditions indicated by subscript x and

downstream by y. Since we want to boost this system in the vertical direction, we indicate the normalcomponent of the total Mach number with a superscript N. Thus we consider a normal shock with

upstream Mach number N x M and downstream Mach number N

y M . Let’s look at this from a frame

moving parallel to the shock front with a speed U .

Note the static temperature is the same in the normal and boosted frames. However, the value of total

temperature differs between the normal and boosted frames; we use N TxT to denote the total temperature in

the normal frame. Note N Ty

N Tx T T = for a normal shock (since there is no heat addition or work

extraction). Note that by the reasoning above, the total temperatures in the boosted frames are given by

P

N TyTy

P

N TxTx

cU

T T

cU

T T

2

22

2

+=

+=

so it follows that TyTx T T = (which also follows from having zero energy transfer in the boosted frame).

N y

N Ty

N y

Sy N y

N y

z RT M

RT M V

γ =

γ = N x

N Tx

N x

Sx N x

N x

z RT M

RT M V

γ =γ =

β

U U

β−δ



The total temperatures cancel as well. Since we know ( ) N x

N y M f M = , the entire right hand side of this

equation can be written as a function of N x M . Furthermore, N

x M and x M are related by

β= sin x N x M M . Using some trigonometric identities, the relation between wave angle, deflection

angle, and Mach number can be expressed as

( )+β+γ −β

β=δ22cos

1sintan

2tan 2

22

x

x

M M

.

This is called the δ-β-M equation. We can use it to solve for δ as a function of β and M x.It is also possibleto invert this equation and solve explicitly for β as a function of δ and M. Before we do so, let’s take alook at the relation among these quantities. If we plot δ for a range of β’s, we find that for any given

supersonic upstream Mach number, 0=δ when ( ) x M 1sin 1−=β or wheno902 =π=β . In

between, there is a maximum value of δ when β reaches a value given by

( )

+γ

+−γ

++γ +−+γ

γ =β422

2max

2

161

21

111411

sin x x x x M M M M

(Try to reproduce this result). See the Figure 1 for a plot of δ vs β for Mach numbers up to 3.5.

Note several interesting features of this plot:

1. For any given value of deflection angle δ, there are either two, one, or zero solutions for β. The largevalue of β is called the strong shock solution and the small value is called the weak shock solution. Note

that when 0=δ , the strong shock is a normal shock ( o90=β ) and the weak shock is a Mach wave

( ( ) x M 1sin 1−=µ=β ). Whether a given flow situation results in a weak or strong shock depends, as

always, on the boundary conditions. However, for many practical applications, the weak shock solution is



Figure 1

Flow_z.xls:Oblique Shock Chart 9/27/00:9:43 AM

Oblique Shock Relations

0

5

10

15

20

25

30

35

40

0 10 20 30 40 50 60 70 80 90

Wave Angle (degrees)

W e d g e

A n g

l e ( d e g r e e s )

1.1 1.2

1.3 1.4

1.5 1.6

1.7 1.8

1.9 2

2.1 2.2

2.3 2.4

2.5 2.6

2.7 2.8

2.9 3

3.1 3.2

3.3 3.4

3.5 Weak | Strong

Sonic Line

Upstream Mach Number Mx=Strong shocks,Subsonic My

Weak shocks,Supersonic My



downstream conditions is close to the weak/strong dividing line on the δ-β-M plot. We can also plot thedownstream Mach number as a function of upstream Mach number, as shown in Figure 2. Note that the

upper limit is the Mach-wave solution ( x y M M = , ( ) x M 1sin1−

=µ=β ) and the lower limit is thefamiliar normal shock characteristic.

Though not as neat as the δ-β-M equation, there is an explicit solution for β in terms of δ and M x,basically by solving the appropriate cubic equation. This can be very useful for some applications, and ismuch quicker than the iterative inversion technique with which the ignorant are cursed. Here we quotedirectly from the technical note by V. Mascitti 1:

The equations governing oblique shock relations may be written in the form

0sinsinsin 246 =+β+β+β d cbwhere

δγ −+−= 22

2

sin2

x

x

M M

b

( ) δ−γ ++γ ++= 22

2

4

2

sin14112 x x

x

M M M

c

4

2cos

x M d

δ−=

and

β= shock-wave angle

x M = free-stream Mach number

δ = deflection angleγ = ratio of specific heats



Figure 2

Flow_z.xls:Oblique Mx vs My w delta 9/27/00:9:42 AM

My vs Mx for Oblique Shocks

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5

Mx

My

Weak | Strong beta=20

beta=25 beta=30beta=35 beta=40beta=45 beta=50beta=55 beta=60beta=65 beta=70beta=75 beta=80beta=85 Normal Shockdelta=0 delta=5delta=10 delta=15delta=20 delta=25delta=30 delta=35



So with this cubic equation solution and the δ-β-M equation, we can solve for either angle withoutresorting to iterative code.

The ratios of pressures and temperatures across the shock can be easily determined using normal shock

relations, once N x M and N

y M have been found. It is useful to recall that

N Tx

N Ty

TxTy

T T

T T

==

Note in particular that

( )( ) N y

N x

N Tx

N Ty

Tx

Ty

M G M G

PP

PP ==

despite the fact that N TxTx PP ≠ and N

TyTy PP ≠ , since the entropy rise across the shock can be expressed

in terms of either normal or oblique total conditions:

N Tx

N Ty

Tx

Ty

N Tx

N Ty

N Tx

N Ty x y

Tx

Ty

Tx

Ty

P

P

P

P

P

P

T

T

R

ss

P

P

T

T

=⇒

−−γ γ

=−

=−−γ γ

lnln1lnln1

However, be cautious about applying relations derived for normal shocks. For example,

( )( )

( )( ) y

x N y

N x

M G M G

M G M G

≠

Since the flow changes direction across an oblique shock, one cannot simply assume that the oblique-



The static property ratios can be determined as follows. The relation zT T ST = , where 22

11 M z −γ += ,

is always true for perfect gases. Hence we can find SxSy T T from either

N y

N x

N x

N Tx

N y

N Ty

Sx

Sy

z z

zT

zT

T

T == or y

x

xTx

yTy

Sx

Sy

z z

zT

zT

T

T == .

The static pressure ratio can be handled similarly, except that the total pressure ratio is different from 1:

( )( )

1

1

1 −γ γ

−γ γ

−γ γ

== N y

N

x N

Tx

N

Ty N x

N Tx

N

y

N

Ty

Sx

Sy

z z

P

P

zP

zP

P

Por

1

1

1 −γ γ

−γ γ

−γ γ

== y

x

Tx

Ty

xTx

yTy

Sx

Sy

z z

P

P

zP

zP

P

P.

Figure 3 shows the total pressure ratio versus Mach number for oblique shocks, with lines of constant δsuperimposed.

Supersonic aircraft designers often exploit the properties of oblique shocks to achieve good inlet pressurerecovery at high flight Mach numbers. This chart makes clear why. The advantages of an oblique shock inlet are analogous to the advantages of a swept wing – by allowing the freestream air to pass through theshock at an oblique angle, the strength of the shock is reduced because only the normal component of theflow is compressed. There is often a tradeoff with inlet weight and complexity, as ramped inlets frequentlyrequire variable geometry to achieve acceptable performance at all angles of attack and sideslip, across arange of Mach numbers. In addition, the flow coming out of an oblique shock is almost always stillsupersonic, often necessitating more ramps to generate oblique shocks. A final normal shock is almostalways a part of the shock system as well. Good pressure recovery in high Mach number flight requiresthat the flow be slowed enough to enter the terminal normal shock at a low supersonic Mach number.



Figure 3

Flow_z.xls:PR vs Mx 9/27/00:9:43 AM

Total Pressure Ratio for Oblique Shock

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1 1.5 2 2.5 3 3.5

Mx

P T y / P T x

delta=5

delta=10delta=15

delta=20

delta=25

delta=30

delta=35

Normal Shock



Prandtl-Meyer Expansion Waves

As a complement to oblique shocks, we now consider the so-called Prandtl-Meyer expansion fan. We haveseen that when a supersonic flow enters a concave corner, an oblique shock can result if the Mach numberis high enough, and a detached shock will form otherwise.

If, on the other hand, the corner is convex, instead of a shock that slows and compresses the flow, weobserve a flow that is accelerated and expanded. The flow disturbance associated with the corner is calleda Prandtl-Meyer expansion fan. In contrast with a compact shock wave with a few mean-free-path’sthickness, the Prandtl-Meyer fan extends over a finite angle, and is composed of Mach waves emanatingfrom the corner. The downstream Mach number is greater than the upstream Mach number, and is a

function of the upstream Mach number and the deflection angle.

Strictly speaking, the structure depicted above is a “centered” Prandtl-Meyer fan – a general fan can occuroutside any concave surface in supersonic flow. However, the essential features of external supersonicexpansion can be discussed in the context of centered fans.

Unlike the oblique shock, the PM fan is an extended object. We find that it is necessary to analyze theflow on a differential basis, that is, we analyze an infinitesimal deflection and derive a differentialequation that can be integrated to give the expansion as a function of deflection angle. We proceed toderive the relation governing small angle changes across a Mach wave. Just as in the case of an oblique

shock, we note that the transverse component of velocity is the same on both sides of the Mach wave –only the normal component of velocity changes. In the following we let the infinitesimal deflection be

θd .

Oblique shock compression Mach wave Prandtl-Meyer expansion



Thus we find

θ=µ d V dV

tan1

Now the wave angle µ is related to the Mach number by ( ) M 1sin 1−=µ .

Hence 11tan 2 −=µ M .

It remains to express the ratio V dV in terms of Mach number. Since the total temperature does notchange across the Mach wave, we write

22

11 M RT

M V T −γ +γ

=

Differentiating this expression gives the relation

22

111

M M dM

V dV

−γ +=

Combining this with the result above gives the Prandtl-Meyer relation

21

2 1 M dM d

−=θ

12 − M M

1

µ



Figure 4

Flowfunc.xls:PM nu 9/27/00:9:45 AM

Prandtl-Meyer Function ν (M)

0

20

40

60

80

100

120

0 2 4 6 8 10 12 14 16 18 20

M

ν (M)



Note a key difference between Prandtl-Meyer expansions and the shock process: while shocks increaseentropy and result in a loss of total pressure, expansions do not. The total pressure along a streamlinepassing through a Prandtl-Meyer fan does not change. Along with a constant total temperature, thismeans that the static pressure and temperature can be determined simply from the upstream anddownstream Mach numbers:

2

1

1

2

z z

T T

S

S = and

1

2

1

1

2

−γ γ

=

z z

PP

S

S

The variation of static properties through a fan can be summarized by the ratio to sonic conditionproperties, that is, by plotting *

SS T T and *SS PP . The variation of these ratios with Mach number is

shown in Figure 5. Since the Prandtl-Meyer process is isentropic, it is simply a vertical line on a T-sdiagram.

The Diamond Airfoil

Our basic toolkit of compressible flow is nearly complete. As an example of the type of problem we arenow equipped to solve, consider a so-called diamond airfoil immersed in a supersonic flowfield. This is a2d airfoil with the profile of a thin rhombus with chord length c and thickness t. If the Mach number ishigh enough that all shocks remain attached, the Mach numbers, pressures, and temperatures can all becalculated for the inviscid part of the flowfield. In particular the static pressure distribution over the airfoilsurface can be calculated. Predictions for lift and drag per unit span can be assembled, simply fromoblique-shock and Prandtl-Meyer expansion theory.

M0

Obliqueshock

PM FanObliqueshock



Figure 5

Flowfunc.xls:PM Ps and Ts 9/27/00:9:46 AM

Static T and P ratios for Prandtl-Meyer Fan

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.00 1.50 2.00 2.50 3.00 3.50 4.00 4.50 5.00

M

Ts/Ts*

Ps/Ps*

Introduction to Compressible Fluid Flows (Part II)

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