applied gas dynamics oblique shock and expansion … gas dynamics oblique shock and expansion waves...

Applied Gas Dynamics

Oblique Shock and ExpansionWaves

Ethirajan Rathakrishnan

Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c© 2010 Ethirajan Rathakrishnan 1 / 331

Introduction

The normal shock wave, a compression front normal to the flow direc-tion, was studied in some detail in Chapters 2 and 3. However, in a widevariety of physical situations, a compression wave inclined at an angleto the flow occurs. Such a wave is called an oblique shock. In deed, allnaturally occurring shocks in external flows are oblique.


In steady subsonic flows, we generally do not think in terms of wavemotion. It is usually much simpler to view the motion from a frame ofreference in which the body is stationary and the fluid flows over it. Ifthe relative speed is supersonic, the disturbance waves cannot propa-gate ahead of the immediate vicinity of the body and the wave systemtravels with the body. Thus, in the reference frame in which the body isstationary, the wave system is also stationary; then the correspondencebetween the wave system and the flow field is direct.


The normal shock wave is a special case of oblique shock waves, withshock angle β = 90◦. Also, it can be shown that superposition of auniform velocity, which is normal to the upstream flow, on the flow fieldof the normal shock will result in a flow field through an oblique shockwave. This phenomenon will be employed later in this chapter to get theoblique shock relations. Oblique shocks are usually generated when asupersonic flow is turned into itself.


The opposite of this, i.e. when a supersonic flow is turned away fromitself, results in the formation of an expansion fan. These two families ofwaves play a dominant role in all flow fields involving supersonic veloc-ities. Typical flows with oblique shock and expansion fan are illustratedin Figure 4.1.

(a) Compression corner

θ

Oblique shock Expansion fan

11 2

M2 < M1M1

M2 > M1

2

M1

(b) Expansion corner

θ

Figure 4.1Supersonic flow over compression and expansion corners.


In Figure 4.1a the flow is deflected into itself by the oblique shockformed at the compression corner, to become parallel to the solid walldownstream of the corner. All the streamlines are deflected to the sameangle θ at the shock, resulting in uniform parallel flow downstream ofshock. The angle θ is referred to as flow deflection angle. Across theshock wave, the Mach number decreases and the pressure, densityand temperature increase. The corner which turns the flow into itself iscalled compression or concave corner.


In contrast, in an expansion or convex corner, the flow is turned awayfrom itself through an expansion fan, as illustrated in Figure 4.1b. Allthe streamlines are deflected to the same angle θ after the expansionfan, resulting in uniform parallel flow downstream of the fan. Across theexpansion wave, the Mach number increases and the pressure, den-sity and temperature decrease. From Figure 4.1, it is seen that theflow turns suddenly across the shock and the turning is gradual acrossthe expansion fan and hence all flow properties through the expansionfan change smoothly, with the exception of the wall streamline whichchanges suddenly.


Oblique shock and expansion waves prevail in two- and three-dimensionalsupersonic flows, in contrast to normal shock waves, which are one-dimensional. In this chapter, we shall focus our attention only on steady,two-dimensional (plane) supersonic flows.


Oblique Shock Relations

The flow through an oblique shock is illustrated in Figure 4.2b. Theflow through a normal shock (Figure 4.1a) has been modified to resultin flow through an oblique shock, by superimposing a uniform velocityVy (parallel to the normal shock) on the flow field of the normal shock(Figure 4.2a).

βV1

Vx1

V2βV1

Vy

Vy

Vx2

V2

Normal shock Oblique shock

(a)

θθ

(b)

β

Figure 4.2Flow through an oblique shock wave.


The resultant velocity upstream of the shock is V1 =√

V 2x1 + V 2

y and

is inclined at an angle β = tan−1(Vx1/Vy ) to the shock. This angle βis called shock angle. The velocity component Vx2 is always less thanVx1; therefore, the inclination of the flow ahead of the shock and afterthe shock are different. The inclination ahead is always more than thatbehind the shock wave, i.e., the flow is turned suddenly at the shock.Because Vx1 is always more than Vx2, the turning of the flow is alwaystowards the shock.


The angle θ by which the flow turns towards the shock is called flowdeflection angle and is positive as shown in Figure 4.2. The rotationof the flow field in Figure 4.2a by an angle β results in the field shownin Figure 4.2b, with V1 in the horizontal direction. The shock in thatfield inclined at an angle β to the incoming supersonic flow is called theoblique shock.


The relations between the flow parameters upstream and downstreamof the flow field through the oblique shock, illustrated in Figure 4.2b,can be obtained from the normal shock relations in Chapter 3, since thesuperposition of uniform velocity Vy on the normal shock flow field inFigure 4.2a does not affect the flow parameters (e.g. static pressure)defined for normal shock. The only change is that in the present casethe upstream Mach number is

M1 =Resultant velocitySpeed of sound

=V1

a1


The component of the upstream Mach number M1 normal to the shockwave is

Mn1 = M1 sin β (4.1)

Thus, replacement of M1 with M1 sin β in normal shock relations givenby Eqs. (3.13), (3.16), (3.18) and (3.19) results in the following relationsfor an oblique shock.

ρ2

ρ1=

(γ + 1)M21 sin2 β

(γ − 1)M21 sin2 β + 2

(4.2)

p2

p1= 1 +

2γ

γ + 1(M2

1 sin2 β − 1) (4.3)


T2

T1=

a22

a21

= 1 +2(γ − 1)

(γ + 1)2

M21 sin2 β − 1

M21 sin2 β

(γM21 sin2 β + 1) (4.4)

s2 − s1

R= ln

8

>

>

>

>

>

:

»

1 +2γ

γ + 1(M2

1 sin2 β − 1)

–1/(γ−1)"

(γ + 1)M21 sin2 β

(γ − 1)M21 sin2 β + 2

#

−γ/(γ−1)9

>

>

>

>

>

;

= ln

8

>

>

:

p01

p02

9

>

>

;(4.5)


The normal component of Mach number behind the shock Mn2 is givenby

M2n2 =

M21 sin2 β + 2

γ−12γ

γ−1M21 sin2 β − 1

(4.6)

From the geometry of the oblique shock flow field shown in Figure 4.2,it is seen that the Mach number behind the oblique shock, M2, is relatedto Mn2 by

M2 =Mn2

sin(β − θ)(4.7)

In the above equations, M2 = V2/a2 and Mn2 = Vx2/a2. The Machnumber M2 after a shock can be obtained by combining Eqs. (4.6) and(4.7).


Numerical values of the oblique shock relations for a perfect gas, withγ = 1.4, are presented in graphical form. The same in tabular form isgiven in Table 3 of the Appendix.

It is seen from the oblique shock relations given by Eqs. (4.1)-(4.5)that the ratio of thermodynamic variables depends only on the normalcomponent of velocity (M1 sin β) ahead of the shock. But, from normalshock analysis we know that this component must be supersonic, i.e.M1 sin β ≥ 1.


This requirement imposes the restriction on the wave angle β that itcannot go below a limiting minimum value for any given M1. At thisminimum limiting value of shock angle, the shock gets degenerated toan isentropic wave (also called Mach wave) across which the change offlow properties become negligibly small. Such a weak isentropic waveis termed Mach wave. The maximum value of β is that for a normalshock, β = π/2.


Thus for a given initial Mach number M1, the possible range of waveangle is

sin−1(

1M1

)

≤ β ≤ π

2(4.8)

The limiting values of the wave angle in Eq. (4.8) are of special sig-nificance. The limiting minimum value is sin−1 ( 1

M

)

is the Mach angleµ and the maximum value π

2 corresponds to normal shock. Thus, thestrongest wave possible in a given supersonic flow is the normal shockcorresponding to the given M1. The weakest wave is the Mach wavecorresponding to the given M1.


It is essential to note that, the shock wave formation is not mandatory ina supersonic flow. For example in uniform supersonic streams such asthe flow in a supersonic wind tunnel test-section, no shocks are formedwhen the test-section is empty. Whereas, the weakest limiting isentropicwaves, namely the Mach waves are always present in all supersonicflows. Even in the empty test-section of a supersonic tunnel the Machwaves are present. But we know that the waves in a supersonic flow isdue to perturbations in the flow field.


Therefore, it is natural to ask, “in an undisturbed uniform supersonicflow why should there be Mach waves present?” The answer to thisquestion is the following. In an uniform supersonic flow such as that in awind tunnel test-section, if the test-section walls are absolutely smooththere will not be any Mach wave present in the flow. However, absolutesmooth surface is only a theoretical assumption. For instance, evensurfaces such as that of a good quality Schlieren mirror has a finish ofonly about λ/20, where λ is the wave length of light.


Thus, any practical surface is with some roughness and not absolutelysmooth. Therefore, any supersonic flow field generated by a practicaldevice is bound to posses Mach waves. In deed, the size of the gasmolecules are enough to cause Mach wave generation. Therefore, evenin a free supersonic flow without any solid confinement Mach waves willbe present.


An important feature to be inferred here is that the Mach waves, likecharacteristics will be running to the left and right in the flow field. Be-cause of this the Mach waves of opposite families prevailing in the flowfield cross each other. But being the weakest degeneration of waves,the Mach waves would continue to propagate as linear waves even afterpassing through a number of Mach waves.

In other words, the Mach waves would continue to be simple waveseven after intersecting other Mach waves. Because of this nature ofthe Mach waves, a flow region traversed by the Mach waves is simplethroughout (see Section 4.10).


Relation between β and θ

It is seen from Eq. (4.7) that, for determining M2 the flow deflectionangle θ must be known. Further, for each value of shock angle β at agiven M1 there is a corresponding flow turning angle θ. Therefore, θ canalso be expressed as a unique function of M1 and β. From Figure 4.2,we have

tan β =Vx1

Vy(4.9)

tan(β − θ) =Vx2

Vy(4.10)


Combining Eqs. (4.9) and (4.10), we get

tan(β − θ)

tan β=

Vx2

Vx1(4.11)

By continuity,Vx2

Vx1=

ρ1

ρ2

Now, substituting for ρ1/ρ2 from Eq. (4.2), we get

tan(β − θ)

tan β=

(γ − 1)M21 sin2 β + 2

(γ + 1)M21 sin2 β

(4.12)


Equation (4.12) is an implicit relation between θ and β, for a given M1.With some trigonometric manipulation, this expression can be rewrittento show the dependence of θ on Mach number M1 and shock angle β,as

tan θ = 2 cot β

M21 sin2 β − 1

M21 (γ + cos 2β) + 2

(4.13)

Equation (4.13) is called the θ–β–M relation. This relation is importantfor the analysis of oblique shocks.


The expression on the right hand side of Eq. (4.13) becomes zero atβ = π/2 and β = sin−1( 1

M1), which are the limiting values of β, defined

in Eq. (4.8). The deflection angle θ is positive in this range and musttherefore have a maximum value. The results obtained from Eq. (4.13)are plotted in Figure 4.3, for γ = 1.4. From the plot of θ–β–M (Figure4.3) curves, the following observations can be made.


30

20

M2 < 1

0

M2

=1

θ=

θ max

M2 > 1

80604020

40

2

1.4

1.6

0Wave angle β (degree)

M1 = ∞

10

5

3

4

1.2

Flo

wdefl

ection

angle

θ(d

egre

e)

10

Figure 4.3Oblique shock solution.


1 For any given supersonic Mach number M1, there is a maximumvalue of θ. Therefore, at a given M1, if θ > θmax, then no solutionis possible for a straight oblique shock wave. In such cases, theshock will be curved and detached, as shown in Figure 4.4.

M1 > 1

Detached shockDetached shock

M1 > 1 θ > θmax θ > θmax

Figure 4.4Detached shocks.


1 When θ < θmax, there are two possible solutions, for each value ofθ and M, having two different wave angles. The larger value of βis called the strong shock solution and the smaller value of β is re-ferred to as the weak shock solution. For strong shock solution theflow behind the shock becomes subsonic. For weak shock solu-tion, the flow behind the oblique shock remains supersonic, exceptfor a small range of θ slightly smaller than θmax, the zone boundedby the M2 = 1 curve and θ = θmax curve shown in Figure 4.3.


1 If θ = 0, then β = π/2, giving rise to a normal shock, or β de-creases to the limiting value µ, i.e., shock disappears and onlyMach waves prevail in the flow field. That is, when the flow turn-ing angle θ is zero, the following two solutions are possible for theshock angle β, for a given M1. (a) Either β = π/2 giving rise to anormal shock which does not cause any flow deflection, but woulddecelerate the flow to subsonic level, or (b) β = sin−1(1/M1) = µcorresponding a Mach wave, which even though inclined to the up-stream flow, would not cause any flow deflection, being the limitingcase of the weakest isentropic wave for a given M1.


A very useful form of θ–β–M relation can be obtained by rearranging Eq.(4.12) in the following manner: Dividing the numerator and denominatorof the right hand side of Eq. (4.12) by 2M2

1 sin2 β and solving, we obtain

1

M21 sin2 β

=γ + 1

2tan (β − θ)

tan β− γ − 1

2


This can be simplified further to result in

M21 sin2 β − 1 =

γ + 12

M21

sin β sin θ

cos (β − θ)(4.14)

For small deflection angles θ, Eq. (4.14) may be approximated as

M21 sin2 β − 1 ≈

γ + 12

M21 tan β

θ (4.15)

If M1 is very large, then β ≪ 1, but M1β ≫ 1 and Eq. (4.15) reduces to

β =γ + 1

2θ (4.16)


It is important to note that, oblique shocks are essentially compressionfronts across which the flow decelerates and the static pressure, statictemperature and static density jump to higher values. If the decelerationis such that the Mach number behind the shock continues to be greaterthan unity, the shock is termed weak oblique shock. If the downstreamMach number becomes less than unity then the shock is called strongoblique shock. It is essential to note that, only weak oblique shocks areusually formed in any practical flow and it calls for special arrangementto generate strong oblique shocks.

One such situation where strong oblique shocks are generated withspecial arrangements is the engine intakes of supersonic flight vehicles,where the engine has provision to control its backpressure. When thebackpressure is increased to an appropriate value, the oblique shockat the engine inlet would become a strong shock and decelerate thesupersonic flow passing through it to subsonic level.


Shock Polar

Shock polar is a graphical representation of oblique shock solutions.We have seen in Section 4.3 that, in general, for any specified value offlow turning angle θ there are two possible shock angles, giving rise tostrong and weak solutions.


The shock polar for the oblique shock geometry illustrated in Figure 4.5may be drawn as follows.

y

Vy2

Oblique shock

Vx2

θ1

xV1

V2

1 2

Figure 4.5Oblique shock in physical (xy ) plane.


An oblique shock with upstream velocity V1 in the xy-Cartesian coor-dinate system, shown in Figure 4.5, has the velocity components Vx2

and Vy2 in the downstream field, as shown. The xy-plane in Figure 4.5is called the physical plane. Let Vx1, Vy1, Vx2 and Vy2 be the x and ycomponents of flow velocity ahead of and behind the shock.


Now, let us represent the oblique shock field in a plane with Vx and Vy

as the coordinate axes, as shown in Figure 4.6. This plane is called thehodograph plane.

B

Aθ1

Vy

Vx2

Vy2

Vx1(V1)

Vx

Figure 4.6Oblique shock geometry in hodograph (VxVy ) plane.


In the hodograph plane, point A represents the flow field ahead of theshock marked as region 1 in the physical plane of Figure 4.5. Similarly,region 2 in the physical plane is represented by point B in the hodographplane. If the deflection angle θ1 in Figure 4.6 is increased, then theshock becomes stronger and, therefore, the velocity V2 decreases. Onesuch point for θ2 is shown by point C in Figure 4.7.


The loci of all such points for θ values from zero to θmax, representingall possible velocities behind the shock are given in Figure 4.7. Such alocus is defined as a shock polar.

θ1

0A

B

C

V3

V2

V1

θ2

Vx

Vy

Figure 4.7Shock polar for a given V1.


We know that the flow process across a shock wave is adiabatic. There-fore, from our definition of a∗ (Section 2.2), it is the same in the fieldsupstream and downstream of the shock. Hence a∗ can be convenientlyused to nondimensionalize the velocities in Figure 4.7 to obtain a shockpolar which is the locus of all possible M∗

2 for a given M∗

1 , as shown inFigure 4.8.


C

θ

1.0 Vx/a∗

θmax

N

B

Sonic circle

M∗

=1

Vy/a∗

A

D

E

β

0

Figure 4.8Dimensionless shock polar.

The advantage of using the characteristic Mach number M∗ instead oflocal Mach number M and local velocity V to plot the shock polar is that,as M → ∞, M∗ → 2.45 (see Section 2.3).


Hence, when plotted in terms of M∗, the shock polar becomes com-pact. Note that in Figure 4.8, the circle with radius M∗ = 1 is called thesonic circle; inside the circle all velocities are subsonic; outside of it, allvelocities are supersonic.


The shock polar can also be described by an analytical equation calledthe shock polar relation. The derivation of the relation is given in classictexts such as those by Shapiro (1953) or Thompson (1972) and is ofthe form

Vy

a∗

2

=(M∗

1 − Vx/a∗)2[(Vx/a∗)M∗

1 − 1]

2γ + 1

M∗2

1 −(

Vx

a∗

)

M∗

1 + 1(4.17)


The shock polars for different Mach numbers form a family of curves,as shown in Figure 4.9. Note that, for M∗

1 = 2.45 (M1 → ∞), the shockpolar is a circle.

Vx/a∗

0.41

M1 = ∞

2.45

2

4

Vy/a∗

Figure 4.9Shock polars for different Mach numbers.


Supersonic Flow Over a Wedge

From studies on inviscid flows, we know that any streamline can beregarded as a solid boundary. In our present study, we treat the su-personic flow as inviscid and, therefore, here also the streamlines canbe assumed as solid boundaries. Thus the oblique shock flow resultsalready described can be used for solving practical problems like super-sonic flow over a compression corner, as shown in Figure 4.10. For anygiven values of M1 and θ, the values of M2 and β can be determinedfrom oblique shock charts or table (oblique shock charts and table aregiven in the Appendix).


Solid wall

Oblique shock

M2

M1

βθ

Figure 4.10Supersonic flow over a compression corner.


In a similar fashion, problems like supersonic flow over symmetrical andunsymmetrical wedges (Figure 4.11) and so on also can be solved withoblique shock relations, assuming the solid surfaces of the objects asstreamlines in accordance with nonviscous (or inviscid) flow theory.

M2

M1

M2

M1

ββ

(b) Unsymmetrical wedge

β β1

(a) Symmetric wedge

M2 M2′

θ

θ θ

θ1

Figure 4.11Flow past (a) symmetrical and (b) unsymmetrical wedges.


In Figure 4.11b, the flow on each side of the wedge is determined onlyby the inclination of the surface on that side. If the shocks are attachedto the nose, the upper and lower surfaces are independent and there isno influence of wedge on the flow upstream of the shock waves.


In our discussion on shock angle β and flow turning angle θ (Section4.3), we have seen that when θ decreases to zero, β decreases to thelimiting value µ giving rise to Mach waves in the supersonic flow field(see Figure 4.12b), which is given from Eq. (4.14) as

M21 sin2 µ − 1 = 0 (4.18)

Shock wave

M1 M2

θ

(−)

(+)

P P

β

M1

(a) Oblique shock withMach wave θ → 0(b) Degeneration to (c) Left (−) and right (+) running

charateristics at an arbitrarydeflection angle θ

Mach wave

Mµ

µ

µ

M2

point in the flow

Figure 4.12Waves in a supersonic stream.


Also, the pressure, temperature and density jump across the shock(p2 − p1, T2 − T1 and ρ2 − ρ1) given by Eqs. (4.2)–(4.5) become zero.There is, in fact, no finite disturbance in the flow. The point P in Figure4.12b may be any point in the flow field. Then the angle µ is simply acharacteristic angle associated with the Mach number M by the relation

µ = sin−1(

1M

)

(4.19)

This is called Mach angle-Mach number relation.


These lines which may be drawn at any point in the flow field with incli-nation µ are called Mach lines or Mach waves. It is essential to under-stand the difference between the Mach waves and Mach lines. Machwaves are the weakest isentropic waves in a supersonic flow field andthe flow through them will experience only negligible changes of flowproperties. Thus, a flow traversed by the Mach waves do not experi-ence change of Mach number. Whereas the Mach lines, even thoughare weak isentropic waves will cause small but finite changes to theproperties of a flow passing through them.


In uniform supersonic flows, the Mach waves and Mach lines are linearand inclined at an angle given by µ = sin−1 (1/M). But in nonuniformsupersonic flows the flow Mach number M varies from point to pointand hence the Mach angle µ, being a function of the flow Mach number,varies with M and the Mach lines are curved.


In the flow field at any point P (Figure 4.12c), there are always two lineswhich are inclined at angle µ and intersect the streamline, as shown inFigure 4.12c. In a three-dimensional flow, the Mach wave is in the formof a conical surface, with vertex at P. Thus, a two-dimensional flow ofsupersonic stream is always associated with two families of Mach lines.These are represented with plus and minus sign. In Figure 4.12c, theMach lines with ‘+’ sign run to the right of the streamline when viewedthrough the flow direction and those lines with ‘−’ sign run to the left.


These Mach lines which introduce an infinitesimal, but finite changeto flow properties when a flow passes through them are also referredto as characteristics, which are not physical unlike the Mach lines andMach waves. But the mathematical concept of characteristics (taken asidentical to the Mach lines), even though not physical forms the basis forthe numerical method termed method of characteristics, used to designcontoured nozzles to generate uniform and unidirectional supersonicflows. Method of characteristics is presented in chapter 9.


At this stage it is essential to note the difference between the Machwaves, characteristics and expansion waves. Even though all these areisentropic waves, there is a distinct difference between them. Machwaves are weak isentropic waves across which the flow experiencesinsignificant change in its properties. Whereas, the expansion wavesand characteristics are isentropic waves which introduce small, but finiteproperty changes to a flow passing them.


Thus, even though we loosely state that the Mach lines and Mach wavesare isentropic waves in a supersonic flow, inclined at angle µ to thefreestream direction, in reality they are distinctly different. Mach wavesare the weakest degeneration of isentropic waves to the limiting case ofzero strength that a flow across which will not experience any changeof property. Whereas, a Mach line is a weak isentropic wave in a super-sonic flow field, causing small but finite change of properties to the flowpassing through it.


The characteristic lines play an important role in the compression andexpansion processes in the sense that it is only through these lines thatit is possible to retard or accelerate a supersonic flow isentropically.Also, this concept will be employed in designing supersonic nozzleswith Method of Characteristics in Chapter 9.


Weak Oblique Shocks

In Section 4.5, we have seen that the compression of supersonic flowwithout entropy increase is possible only through the Mach lines. Inthe present discussion on weak shocks also, it will be shown that theseweak shocks, which result when the flow deflection angle θ is small andMach number downstream of shock M2 > 1, can also compress theflow with entropy increase almost close to zero.


It is important to note that, when we discussed about flow throughoblique shocks, we considered the shock as weak when the down-stream Mach number M2 is supersonic (even though less than the up-stream Mach number M1). When the flow traversed by an oblique shockbecomes subsonic (i.e. M2 < 1), the shock is termed strong. But whenthe flow turning θ caused by a weak oblique shock is very small, then theweak shock assumes a special significance. This kind of weak shockswith both decrease of flow Mach number (M1 − M2) and flow turningangle θ are small can be regarded as isentropic compression waves.


For small values of θ, the oblique shock relations reduce to very simpleforms. For this case,

sin θ ≈ θ and cos (β − θ) ≈ cos β

Therefore, Eq. (4.14) simplifies to

M21 sin2 β − 1 ≈

γ + 12

M21 tan β

θ

Also, M2 > 1 for weak oblique shocks. Therefore, we may approximatethis weak shock with both (M1 − M2) and θ extremely small as a Machline.


Thus, the shock angle β can be regarded as almost equal to the Machangle µ. With this approximation, we can express tan β as follows.

sin β ≈ sin µ =1M

cos β =

√

1 − sin2 β

=1M

√

M2 − 1

tan β =1√

M2 − 1


Substituting for tan β is preceding equation, we get

M21 sin2 β − 1 ≈ γ + 1

2M2

1√

M21 − 1

θ (4.20)

Equation (4.20) is considered to be the basic relation for obtaining allother appropriateexpressions for weak oblique shocks since all oblique shock relationsdepend on M1 sin β, which is the component of upstream Mach numbernormal to the shock.


It is seen from Eqs. (4.3) and (4.20) that the pressure change across ashock ∆p

p1, termed the shock strength can be easily expressed as

p2 − p1

p1=

∆pp1

≈ γM21

√

M21 − 1

θ (4.21)

Equation (4.21) shows that the strength of the shock wave is propor-tional to the flow deflection angle θ.


Similarly, it can be shown that the changes in density and temperatureare also proportional to θ. But the change in entropy, on the other handis proportional to the third power of shock strength as shown below. ByEq. (4.5), we have

s2 − s1

R= ln

[

1 +2γ

γ + 1m

1/(γ−1)

(1 + m)−γ/(γ−1)

γ − 1γ + 1

m + 1

γ/(γ−1)]

(4.22)where m = (M2

1 − 1) [Note that for weak oblique shocks under consideration,i.e. for weak oblique shocks with (M1 − M2) ≪ 1 and θ very small, M2

1 sin2 β isapproximated as M2

1 ].


For values of M1 close to unity, m is small and the terms within theparentheses are like 1 + ε, with ε ≪ 1. Expanding the terms as loga-rithmic series, we get

s2 − s1

R=

2γ

(γ + 1)2

m3

3+ higher-order terms

ors2 − s1

R≈ 2γ

(γ + 1)2

(M21 − 1)3

3(4.23)


Because the entropy cannot decrease in an adiabatic flow, Eq. (4.23)stipulates that M1 ≥ 1. Thus, the increase in entropy is of third order in(M2

1 − 1). This may be written in terms of shock strength, ∆p/p, withEq. (3.16) as

s2 − s1

R≈ γ + 1

12 γ2

∆pp1

3

(4.24)

But by Eq. (4.21), the shock strength is proportional to θ and hence

∆s ∼ θ3 (4.25)


Thus, a small but finite change of pressure across a weak obliqueshock, for which there are corresponding first-order changes of den-sity and temperature, gives only a third-order change of entropy, i.e. aweak shock produces a nearly isentropic change of state.

Now, let the wave angle β for the weak shock be different from the Machangle µ by a small angle ε◦. That is,

β = µ + ε

where ε ≪ µ. Therefore, sin β = sin (µ + ε) = sin µ + ε cos µ. Also,

sin µ = 1/M1 and cot µ =√

M21 − 1.


Thus,

M1 sin β ≈ 1 + ε√

M21 − 1 (4.26)

orM2

1 sin2 β ≈ 1 + 2ε√

M21 − 1 (4.27)

From Eqs. (4.20) and (4.27), we obtain

ε =γ + 1

4M2

1

M21 − 1

θ (4.28)


That is, for a finite flow deflection angle θ, the direction of weak obliqueshock wave differs from the Mach wave direction µ by an amount ε,which is of the same order as θ.


Supersonic Compression

Compressions in a supersonic flow are not usually isentropic. Gener-ally, they take place through a shock wave and hence are nonisentropic.But there are certain cases, for which the compression process can beregarded isentropic. A compression process which can be treated asisentropic is illustrated in Figure 4.13, where the turning of the flow isachieved through large number of weak oblique shocks.


This kind of compression through a large number of weak compres-sion waves is termed continuous compression. This kind of corners arecalled continuous compression corners. Thus, the geometry of the cor-ner should have continuous smooth turning to generate large numberof weak (isentropic) compression waves.

M2

Shock

M1

θ

Figure 4.13Smooth continuous compression.


The weak oblique shocks divide the field near the wall into segments ofuniform flow. Away from the wall the weak shocks might coalesce andform a strong shock as illustrated in Figure 4.13. In Section 4.6, we haveseen that the entropy increase across a weak wave is of the order ofthird power of deflection angle θ. Let the flow turning through an angle,shown in Figure 4.13, be taking place through n weak compressionwaves, each wave turning the flow by an angle ∆θ.


The overall entropy change for this compression process is

(sn − s1) ∼ n(∆θ)3 ∼ n∆θ(∆θ)2 ∼ θ(∆θ)2

Thus, if the compression is achieved through a large number of weakcompression waves, the entropy increase can be reduced to a verylarge extent, as compared to a single shock causing the same net de-flection. When ∆θ is made vanishingly small, a smooth continuousturning of the flow as shown in Figure 4.13 is achieved. The entropy in-crease associated with such a continuous smooth compression processis vanishingly small, i.e. the compression can be treated as isentropic.


At this stage it is natural to ask, whether this kind of isentropic com-pression is only of theoretical interest or it is used in practical devicestoo? The answer to this question is that it is used in practical devicestoo. For example, in the gas turbine engines used to propel supersonicaircraft such as fighters, the freestream supersonic air stream enter-ing the engine intake needs to be decelerated to incompressible Machnumbers (of the order of 0.2) before reaching the combustion chamber,because with the present technology continuous and stable combustionis possible only at low incompressible Mach numbers.


This can be achieved by a single normal shock or even with a strongoblique shock to decelerate the supersonic stream to a subsonic Machnumber and then the subsonic stream can be decelerated further in adiffuser to reach the required incompressible Mach number before en-tering the combustion chamber. But both these decelerations will resultin a large increase of entropy and the associated large pressure loss.This kind of large increase of entropy is desirable for an efficient mixingof fuel and air in the combustion chamber, but the severe pressure losswith the nonisentropic compression through the shocks is undesirable.


We know that, the engine is used to generate thrust by reaction. Themomentum thrust produced by an engine is

Thrust = m Vj

where m is the mass flow rate of the combustion products of the fuel-airmixture burnt in the combustion chamber, expanded through the nozzleof the engine and Vj is the flow velocity at the nozzle exit. By Bernoulliprinciple it is known that, a large velocity Vj can be generated by ex-panding a gas at high stagnation pressure p0. Thus, the aim of theprocess through the engine is to achieve high p0. If possible we canuse a compressor to achieve the desired level of p0.


But carrying a compressor in a gas turbine engine is not a practicallypossible solution, mainly due to the weight penalty and the need foradditional source of energy to run the compressor. Therefore, as analternative, the high pressure required is achieved through combustionwhere liberation of thermal energy by burning fuel-air mixture results inlarge increase of total temperature T0 and the associated increase oftotal pressure p0. Now, we can notice an interesting point if we keenlyobserve the process involved. The vehicle is flying at a supersonicMach number.


Because of the skin friction, shock and expansion waves around thevehicle and other drag producing causes the vehicle encounters drag.This drag has to be compensated with thrust to maintain the supersonicflight speed. Thus, the basic work of the engine is to supply the requiredmomentum to compensate the momentum loss due to the drag. In otherwords, basically the loss caused by the drag can be viewed as loss oftotal pressure p0. Therefore, the engine must compensate the pressureloss in order to maintain the constant p0 required for the supersonicflight at the given altitude.


Instead of adding the stagnation pressure equivalent to compensate forthe pressure loss due to drag, we are doing the same thing in an in-direct manner. This is done through combustion. For performing com-bustion, the supersonic air entering the engine is decelerated to lowincompressible Mach number, fuel is mixed with the air and combus-tion is performed at such a low Mach number to increase p0 throughthe increase of stagnation temperature T0. The combustion products atlow-Mach number is accelerated through the engine nozzle to achievethe required jet velocity at the nozzle exit.


In the deceleration process through shock/shocks at the engine intake,considerable total pressure is lost. Therefore, it would be appropri-ate and beneficial if the fuel is added to the air entering the enginewith supersonic speed and the combustion is performed at the samefreestream supersonic Mach number. But even though this is the mostsuitable and efficient situation, we are not in a position to do so. This isbecause the technology for performing stable combustion at supersonicMach number is not yet established. Many research groups in variouscountries are working on establishing combustion at supersonic Machnumbers. In deed, stable combustion at Mach number around 2 is re-ported by few advanced countries, such as USA, China, Britain, Franceand Japan.


Once the technology for supersonic combustion is established, the pres-sure lost in decelerating the supersonic air stream to the incompressibleMach number to enable combustion with the present technology can beeliminated to a large extent. This will result in a significant increase ofthe engine efficiency. In other words, the pressure loss associated withthe deceleration of supersonic or hypersonic flow entering the engineto the required incompressible Mach number for stable combustion withthe present technology can be completely eliminated if technology isdeveloped to perform stable combustion at supersonic/hypersonic Machnumbers.


Supersonic Expansion by Turning

Consider the turning of a two-dimensional supersonic flow through afinite angle at a convex corner, as illustrated in Figure 4.14. Let usassume that the flow is turned by an oblique shock at the corner, asshown in the figure.

V2n

Shock

V1n

V2

Vt1

Vt2

V1

θ

Figure 4.14Supersonic flow over a convex corner.


The flow turning shown in Figure 4.14 is possible only when the normalcomponent of velocity V2n after the shock is greater than the normalcomponent V1n ahead of the shock, since V1t and V2t on either side ofthe shock must be equal. Although this would satisfy the equations ofmotion, it would lead to a decrease of entropy across the shock. There-fore, this turning process is not physically possible. From the geometryof the flow shown in Figure 4.14, it follows that V2n must be greater thanV1n. The normal momentum equation yields

p1 + ρ1V 21n = p2 + ρ2V 2

2n


Combining this with continuity equation

ρ1V1n = ρ2V2n

we obtainp2 − p1 = ρ1V1n(V1n − V2n)

Because V2n > V1n, it follows that the pressure downstream of thecorner should be less than the pressure upstream of the corner (p2 <p1). For this, the flow should pass through an expansion fan at thecorner. Thus, the wave at the convex corner must be an expansion fan,causing the flow to accelerate. In other words, the shock wave shownat the convex corner in Figure 4.14 is a physically impossible solution.


In an expansion process, the Mach lines are divergent, as shown inFigure 4.15 and, consequently, there is a tendency to decrease thepressure, density and temperature of the flow passing through them.In other words, an expansion is isentropic throughout. It is essential tonote that, the statement “expansion is isentropic throughout” is not truealways.


To gain an insight into the expansion process, let us examine the cen-tered and continuous expansion processes illustrated in Figures 4.15aand 4.15b.

M2

µ2

P

M1

θ

µ1

(a) Centred expansion (b) Continuous (simple) expansion

M1

M2

Figure 4.15Centered and continuous expansion processes.


We know that the expansion rays in an expansion fan are isentropicwaves across which the change of pressure, temperature, density andMach number are small but finite. But when such small changes coa-lesce they can give rise to a large change. One such point where sucha large change of flow properties occurs due to the amalgamation ofthe effect due to a large number of isentropic expansion waves is pointP, which is the vertex of the centered expansion fan in Figure 4.15a.As illustrated in Figures 4.15a, the pressure at the wall suddenly dropsfrom p1 to p2 at the vertex of the expansion fan. Similarly, the tempera-ture and density also drop suddenly at point P. The Mach number at Psuddenly decreases from M1 to M2.


The entropy change across the vertex of the expansion fan is

s2 − s1 = cp lnT2

T1− R ln

p2

p1

It is seen that entropy change associated with the expansion process atpoint p is finite. Thus, the expansion process at point P is nonisentropic.Therefore, it is essential to realize that a centered expansion process inisentropic everywhere except at the vertex of the expansion fan, whereit is nonisentropic. But for the continuous expansion illustrated in Figure4.15b, there is no sudden change of flow properties.


Even at the wall surface the properties change gradually as shown inthe figure, due to the absence of any point such as P in Figure 4.15a,where all the expansion rays are concentrated. Therefore, the continu-ous expansion is isentropic everywhere.

The expansion at a corner (Figure 4.15a) occurs through a centeredwave, defined by a ‘fan’ of straight expansion lines. This centered wave,also called a Prandtl-Meyer expansion fan, is the counterpart, for a con-vex corner, of the oblique shock at a concave corner.

A typical expansion over a continuous convex turn is shown in Figure4.15b. Since the flow is isentropic, it is reversible.


The Prandtl-Meyer Expansion

Now, we are familiar with the fact that the supersonic flow passingthrough a Prandtl-Meyer expansion would experience a smooth, grad-ual change of flow properties. The Prandtl-Meyer fan consists of aninfinite number of Mach (expansion) lines, centered at the convex cor-ner. The expansion fan has a wedge-like shape with the corner as theapex, as shown in Figure 4.16. There are two Mach lines, one at anangle µ1 to the initial flow (upstream of fan) direction and the other atan angle µ2 to the final flow (downstream of fan) direction.


The change in Mach number from M1 to M2 takes place through infinitenumber of expansion lines. There is a wedge-like space with the corneras the apex, as shown in Figure 4.16, bounded by the two Mach linesgiven by µ1 and µ2.

2

Expansion fan

A

Vr

V

φ = 0

M2

M1

φθ

Vφ= a

A

P

φ = 0

1

r

µ

µ1 µ2

Figure 4.16Prandtl-Meyer expansion over a sharp corner.


As the streamlines turn smoothly across the expansion fan, the flowvelocity increases continuously and the pressure, density and tempera-ture decrease continuously. This type of flow was first studied by Meyer,a student of Prandtl in 1907. This is a turning problem in which thestreamlines are continuous and the flow is isentropic everywhere ex-cept at the vertex of the expansion fan, where the flow experiences asudden change of flow properties and the process is non-isentropic.


The argument that expansion process is isentropic holds throughout forexpansion over a continuous corner shown in Figure 4.17.

Expansion rays

M2

M1

Figure 4.17Prandtl-Meyer flow over a continuous corner.


It is important to note that the, flow process across a centered expan-sion fan is isentropic everywhere except at the vertex of the fan (point Ain Figure 4.16), where the flow process in non-isentropic because of theaccumulated effect of a large number of isentropic ways at a point. Aswe discussed earlier in this chapter, an important difference betweenthe Mach lines or expansion rays and the Mach wave is the following.Even though all are isentropic rays, across the Mach lines and expan-sion rays the properties change are small but finite. Whereas, across aMach wave the flow properties change are insignificant, i.e. negligiblysmall.


This type of flow process is important from the mathematical point ofview, since it is the only problem where an exact solution to the nonlin-ear compressible flow equation exists. In polar coordinates, the equa-tions of motion for such flow can be written as

∂

∂r(ρrVr ) +

∂

∂φ(ρVφ) = 0 (4.29)

Vr∂Vr

∂r+

Vφ

r∂Vr

∂φ−

V 2φ

r= −1

ρ

∂p∂r

(4.30)


Vr∂Vφ

∂r+

Vφ

r∂Vφ

∂φ+

VrVφ

r= −1

ρ

1r

∂p∂φ

(4.31)

Equation (4.29) is the continuity equation and Eqs. (4.30) and (4.31)are the r and φ–momentum equations, respectively.


Momentum equation of this type for inviscid flow is also called Euler’sequation. Note that, in the above equations r and φ are used as thevariables for polar coordinates instead of the conventional r and θ toavoid clash with flow deflection angle θ.


From the geometry of the flow, it is reasonable to assume that the flowproperties do not change along any radial line, i.e. the flow propertiesvary only with φ and are independent of r . This is a reasonable (alsovalid) assumption and considerably simplifies the equations.


By the above assumption, the derivatives of flow properties with respectto r are zero and so Eqs. (4.29)-(4.31) are no more partial. i.e.

∂Vr

∂r=

∂p∂r

=∂Vφ

∂r= 0

Equations (4.29)-(4.31) now become

ρVr +d

dφ(ρVφ) = 0 (4.32)

Vφ =dVr

dφ(4.33)

Vφ

(

dVφ

dφ+ Vr

)

= −1ρ

dpdφ

(4.34)


But dp/dφ can be expressed as

dpdφ

=dpdρ

dρ

dφ= a2 dρ

dφ

Using this equation and relation (4.32), Eq. (4.34) can be rewritten as

(

Vr +dVφ

dφ

)

(

1 −V 2

φ

a2

)

= 0 (4.35)

This is the governing equation for the Prandtl-Meyer flow. First, let theterm in the first set of parentheses in Eq. (4.35) be zero, i.e.

Vr +dVφ

dφ= 0


Therefore, from Eq. (4.34), we get

dpdφ

= 0

This implies that the pressure is constant throughout the flow field andso there can be no expansion. But the basic geometry considered is anexpansion field and, therefore, the above solution is impossible.


Hence, from Eq. (4.35) we have

1 −V 2

φ

a2 = 0

i.e.Vφ = a (4.36)

The velocity is constant along the radial lines and is equal to the localspeed of sound. This means that the radial lines must correspond toMach lines. If the Mach lines are assumed to be identical to Machwaves, Mach angle is given by

sin µ =Vφ

V=

aV

=1M

Hence, all rays in an expansion field are Mach lines and the entire flowis isentropic.


Velocity Components Vr and Vφ

By compressible Bernoulli’s equation, we have

γ

γ − 1pρ

+V 2

2=

γ

γ − 1p0

ρ0=

V 2max

2

where p0 and ρ0 are the stagnation pressure and density, respectively.The resultant velocity V and the φ-component of velocity Vφ are givenby

V 2 = V 2r + V 2

φ

V 2φ = a2 =

γpρ


With the above two relations, Bernoulli’s equation may be rewritten as

V 2r +

γ + 1γ − 1

V 2φ = V 2

max (4.37)

For the present flow, the velocity and pressure are constants along theradial lines and, therefore, Eq. (4.30) reduces to

dVr

dφ= Vφ

Substituting the above expression for Vφ into Eq. (4.37), we get

dVr

dφ= Vmax

√

γ − 1γ + 1

√

1 −(

Vr

Vmax

)2


Integration of the above equation by separation of variables gives

φ

√

γ − 1γ + 1

= arc sin(

Vr

Vmax

)

+ constant (4.38)

From Figure 4.16, when φ = 0, constant = 0. Therefore, Eq. (4.38)yields

Vr = Vmax sin

(

φ

√

γ − 1γ + 1

)

(4.39)

From Eqs. (4.39) and (4.33), we obtain

Vφ = Vmax

√

γ − 1γ + 1

cos

(

φ

√

γ − 1γ + 1

)

(4.40)


For φ = 0, Eq. (4.40) gives

Vφ = Vmax

√

γ − 1γ + 1

Substituting for Vmax from Bernoulli’s equation, we get

Vφ =

√

2γ

γ + 1p0

ρ0=

√

2γ + 1

a0 = a∗

That is, at the beginning of the fan the φ-component of velocity corre-sponds to the speed of sound at sonic condition.


We can also express the pressure at any point in the Prandtl-Meyerflow in terms of φ as follows. With the resultant velocity V 2 = V 2

r + V 2φ ,

Bernoulli’s equation can be expressed as

V 2r + V 2

φ = V 2max

[

1 −(

pp0

)γ−1

γ

]


Replacing Vr and Vφ by Eqs. (4.39) and (4.40), we get

Vmax sin

(

φ

√

γ − 1γ + 1

)

2

+

Vmax

√

γ − 1γ + 1

cos

(

φ

√

γ − 1γ + 1

)

2

=

V 2max

[

1 −(

pp0

)γ−1

γ

]

sin2

(

φ

√

γ − 1γ + 1

)

+γ − 1γ + 1

cos2

(

φ

√

γ − 1γ + 1

)

=

1 −(

pp0

)γ−1

γ


1 − cos2

(

φ

√

γ − 1γ + 1

)

+γ − 1γ + 1

cos2

(

φ

√

γ − 1γ + 1

)

=

(

pp0

)γ−1

γ

cos2

(

φ

√

γ − 1γ + 1

)

γ − 1γ + 1

− 1

= −(

pp0

)γ−1

γ

cos2

(

φ

√

γ − 1γ + 1

)

2γ + 1

=

(

pp0

)γ−1

γ

1γ + 1

1 + cos

(

2 φ

√

γ − 1γ + 1

)

=

(

pp0

)γ−1

γ


or(

pp0

)γ−1

γ

=1

γ + 1

1 + cos

(

2φ

√

γ − 1γ + 1

)

(4.41)

From Eq. (4.41), it is seen that the flow can be turned by an angle φmax

where the pressure p will become zero. This condition results in

φmax =π

2

√

γ + 1γ − 1

(4.42)


For this situation,Vr = Vmax, Vφ = 0

For a perfect gas, such as air with γ = 1.4, the maximum turning angledue to expansion is

φmax = 220.5◦

Thus, the limiting maximum flow turning angle associated with Prandtl-Meyer expansion is

θ = 130.5◦


From the foregoing discussion, it is seen that the beginning of the Prandtl-Meyer expansion is marked by φ = 0, where the radial component offlow velocity is zero and the expansion can go to the maximum extentwhere the static pressure of the flow goes to zero and the φ-componentof the flow velocity also vanishes. The value of the expansion and flowdeflection angles are 220.5◦ and 130.5◦, respectively.


The expansion described can be shown graphically as shown in Figure4.18. The normal direction to the hodograph curve in the figure givesthe Mach line direction in the physical plane.

0

p2 = 0

VVmax

Vmax

a∗

PV

130.5

◦

A

Vr 0.2 0.6 0.8

Vmax

θθ

V

Vφφ

Figure 4.18Expansion around a corner.


The Prandtl-Meyer Function

It is known from basic studies on fluid flows that, a flow which preservesits own geometry in space or time or both is called a self-similar flow.In the simplest cases of flows, such motions are described by a singleindependent variable, referred to as similarity variable. The Prandtl-Meyer function is such a similarity variable.


From the geometry of flow in Figure 4.16, θ = φ + µ − π/2, whereθ is the isentropic turning angle. Now, we define the quantity ν suchthat ν = ±θ, where ‘+’ holds across a right-running characteristic and‘−’ across a left-running characteristic. This function ν is called thePrandtl-Meyer function. From the above definition,

ν = φ + µ − π/2 (4.43)

orφ = π/2 + ν − µ (4.44)


From the flow geometry of Figure 4.16

tan µ =Vφ

Vr

Substituting for Vφ and Vr from Eqs. (4.39) and (4.40), we get

tan µ =

√

γ − 1γ + 1

cot

(

φ

√

γ − 1γ + 1

)

(4.45a)

or

cot µ =

√

γ + 1γ − 1

tan

(

φ

√

γ − 1γ + 1

)

(4.45b)

However,cot µ =

√

M2 − 1 (4.46)


With this the above equation can be expressed as√

γ − 1γ + 1

(M2 − 1) = tan

(

φ

√

γ − 1γ + 1

)

or

φ =

√

γ + 1γ − 1

arc tan

√

γ − 1γ + 1

(M2 − 1)

Substitution of φ from Eq. (4.44) into the above equation results in

ν =

√

γ + 1γ − 1

arc tan

√

γ − 1γ + 1

(M2 − 1) + µ − π

2

From Eq. (4.46), we have the Mach angle

µ = arc cot√

(M2 − 1)


Also,

arc cot√

(M2) = π/2 − arc tan√

(M2 − 1)

Using the above relations, we obtain the relation for the Prandtl-Meyerfunction in terms of the Mach number M1 just upstream of the expansionfan as

ν =

√

γ + 1γ − 1

arc tan

√

γ − 1γ + 1

(M21 − 1) − arc tan

√

(M21 − 1) (4.47)

Equation (4.47), expressing the Prandtl-Meyer function ν in terms of theMach number, is a very important result of supersonic flow. From thisrelation, it is seen that for a given M1, there is a fixed ν. Physically, ν isthe flow inclination from M1 = 1 (i.e. φ = 0) line.


If the wall inclination is θ, then ν2 can be obtained by adding θ to ν1, i.e.ν2 = ν1 + θ. As M1 varies from 1 to ∞, ν increases from 0 to νmax. Whenφ = 0, M1 = 1 and ν = 0. When φ1 = 220.5◦ or θmax = 130.5◦, M2 = ∞and ν = νmax. This maximum for ν is

νmax =π

2

(√

γ + 1γ − 1

− 1

)

(4.48)


For air with γ = 1.4, νmax = 130.5◦ for M1 = ∞. The Prandtl-Meyer func-tion is tabulated in the isentropic flow tables (Table 1) in the Appendix.


Example 4.1 Isentropic expansion:

A Mach 2 air stream passes over a 10◦ expansion corner. Find theMach number of the flow downstream of the expansion fan.

Solution

Given, M1 = 2 and θ = 10◦. From isentropic table, for M1 = 2, thePrandtl-Meyer function is

ν1 = 26.38◦


Therefore, the value of the Prandtl-Meyer function ν2, downstream ofthe expansion fan which turns the flow by 10◦ is

ν2 = ν1 + θ = 26.38◦ + 10◦ = 36.38◦

Again, from isentropic table, for ν2 = 36.38◦, we get the Mach numberM2 downstream of the expansion fan as

M2 = 2.386


Compression

The supersonic flow over a continuous compression corner, discussedin Section 4.7, resulting in isentropic compression of flow is similar tothe Prandtl-Meyer expansion except that compression results in isen-tropic declaration of flow and expansion accelerates the flow isentrop-ically. In expansion, all Mach lines diverge, whereas in compression,all Mach lines converge. The purpose of Mach lines is to change thedirection of the streamlines in such a way that they become parallel tothe solid boundary, as shown in Figure 4.13.


For isentropic compression also, the Prandtl-Meyer function can beused, but this is restricted to the flow near the wall only. This is be-cause as illustrated in Figure 4.13, the isentropic compression wavesemanating from the continuous compression corners, progressively ap-proach each other in the flow field away from the wall and ultimatelycoalesce to form a shock. As we know, the process across a shockis non-isentropic. Therefore, only in the proximity of the wall of a con-tinuos compression, where the weak compressions waves generatedretain their individual identity, the waves can be approximated as isen-tropic compression waves.


Example 4.2 Isentropic compression:

A Mach 2 air stream passes over a 10◦ isentropic compression corner.Find the downstream Mach number of the flow following the 10◦ turn.

Solution

Given, M1 = 2, θ = 10◦. From isentropic table, for M1 = 2,

ν1 = 26.38◦


After the compression, the Prandtl-Meyer function becomes

ν2 = ν1 − θ = 16.38◦

Again, from isentropic table, for ν2 = 16.38◦, the corresponding Machnumber is

M2 = 1.652


Instead of continuous change, if the compression is caused by onlyone kink at the wall of about 10◦, the compression cannot be isentropicand the turning of the flow will take place through a shock wave asshown in Figure E4.2. Only when the compression is continuous, theprocess can be treated as isentropic. However, for small values of θ,the compression through a single compression wave such as the oneshown in Figure E4.2, the process can still be treated as isentropic andreasonably accurate results can be obtained.


Flow turning angle θ ≤ 5◦ may be taken as the limit for considering thecompression to be isentropic.

Shock

M2

10◦

M1

Figure E4.2


From the above discussion it is clear that, the relation between thePrandtl-Meyer function and the flow turning angle may be expressedas

νn = νn−1 + |θn − θn−1| (expansion) (4.49a)

νn = νn−1 − |θn − θn−1| (compression) (4.49b)


Simple and Nonsimple Regions

The waves causing isentropic expansion and compression discussedin Section 4.9 are called simple waves. A simple wave is a straightMach line, with constant flow conditions, and is governed by the simplerelations [Eq. (4.49)] between the flow direction and the Prandtl-Meyerfunction.


A supersonic flow field with simple and nonsimple regions is shown inFigure 4.19. A supersonic expansion or compression zone with Machlines which are straight is called a simple region. Equations (4.49a) and(4.49b) which govern the simple region are not applicable to nonsimpleregions. Such a region may be treated by the method of characteristicsis discussed in details in Chapter 8.

Nonsimple region

Simple region

Uniform flow

Figure 4.19Simple and nonsimple regions in supersonic flow.


The Mach lines which are straight in the simple region become curved inthe nonsimple region, after intersection with other Mach lines. However,the wave segments between adjacent cross-over points may be treatedas linear, without introducing significant error to the calculated results.


Reflection and Intersection of Shocks and Expansion Waves

When an oblique shock is intercepted by a solid wall, it is reflected. Apossible flow field is as shown in Figure 4.20. If the incident shock issufficiently weak, the reflection would be regular and could be treatedaccording to the linear theory, giving just a reflected shock of the samestrength as the incident shock. For a shock which is not necessarilyweak, the incident shock deflects the flow through an angle θ towardsthe wall. A second reflected shock of opposite family is required to turnthe flow back again by an amount θ, to satisfy the constraint of thewall. That is, the flow passing through the incident shock is deflectedby an angle θ towards the wall, as shown in Figure 4.20.


1

32Incident shock

Shock

M2

p2 p3

p3

p1

β

Streamline

Wall

Reflected shock

β′

p1

M1

M3

Figure 4.20Shock reflection from a rigid wall.

Therefore, the reflected shock must deflect the flow in the oppositedirection to the same angle θ, rendering the flow to travel parallel to thesolid wall, in accordance with the flow physics; that the streamline overa solid wall has to be parallel to the wall surface. Therefore, the Machnumber M2 and wave angle of the reflected shock are dictated by theflow turning angle θ.Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c© 2010 Ethirajan Rathakrishnan 134 / 331

Although the flow deflection θ, caused by the incident and reflectedshocks, are equal in magnitude, the pressure ratios of the incident andreflected shocks are not equal, since M2 < M1. The pressure distribu-tion along a streamline and along the wall are as shown in Figure 4.20.The strength of the reflected shock wave can be defined by the overallpressure ratio,

p3

p1=

p3

p2

p2

p1

But p3/p2 and p2/p1 are the strengths of reflected and incident shocks,respectively. Therefore, the strength of reflection is given by the productof individual shock strengths.


The reflection from a rigid surface, in general, is not specular, i.e. thereflected shock inclination (β′) is not equal to the incident shock incli-nation (β). Now there exists one of the following two possibilities; eitherβ > β′ or β < β′. These two cases are opposite, and the net resultdepends on the particular values of M1 and θ. These results cannotbe written explicitly in general form but may easily be obtained, for theparticular cases with specified values of M1 and θ, from oblique shockcharts. For high Mach numbers (M1), β > β′, whereas for low Machnumbers, β < β′.


It is important to recall that, the reflection of a wave from a surfaceis governed by the surface roughness. From elementary physics weknow that reflection from a surface could be either specular or diffuse.Specular reflection is that in which the reflection is like a mirror image.That is, the wave angles of the incident and reflected waves will bethe same. For this the surface should be absolutely smooth. But weknow that absolute smoothness or zero roughness for a surface is onlyan imaginary situation. In reality all physical surfaces are with someroughness.


However, the roughness of a surface can vary depending upon the man-ufacturing technique used. For example the surface finish of a bestquality mirror is about λ/20, where λ is the wave length of light. There-fore, any physical surface is rough enough to cause diffuse reflection.Thus, specular reflection is essentially an assumption made to obtainsimplified solutions in most of the practical problems.


When an oblique shock is intercepted by another oblique shock of thesame strength but of opposite family, the possible flow field will looklike the one shown in Figure 4.21. Viewing in the flow direction, theshocks running to the left are named as one family (left-running family)and the shocks running to the right are called (right-running family) theopposite family.


M2

M2

M1

M3

M3

Figure 4.21Interaction of oblique shocks of opposite families with equal strength.

The shocks ‘pass through’ each other, but are slightly ‘bent’ in the pro-cess. The flow downstream of the shock system is parallel to the initialflow.


When two shocks of unequal strength intersect, a new flow geometryis established, as shown in Figure 4.22. The flow field is divided intotwo portions by the streamline through the intersection point. The twoportions experience different changes while passing through the shockwave system. The overall results must be such that, after crossing thewave system the two portions have the same pressure and same flowdirection, i.e., p3 = p3′ and θ3 = θ3′ . The flow downstream of thereflected shocks (zone 3) need not be in the freestream direction. Thesetwo requirements determine the final direction δ and the final pressurep3.


M1

M2′ , θ2

′

δ

M2, θ2 M3, θ3

M3′ , θ3′

Slipstreamp3′

p3

Figure 4.22Intersection of oblique shocks of opposite families with different strengths.


The streamline shown with dashed line, having two flow fields of differ-ent parameters (T and ρ) on either side of it, is called contact surface.The contact surface may also be idealized as a surface of discontinu-ity. The contact surface can either be stationary or moving. Unlike theshock wave, there is no flow of matter across the contact surface. Inliterature, we can find this contact surface being referred to by differ-ent names, e.g. material boundary, entropy discontinuity, slipstream orslip surface, vortex sheet, and tangential discontinuity. It is essential tonote that the contact surface is a fluid boundary across which there isno mass transport.


Further, the surface can tolerate temperature and density gradients, butnot pressure gradient. In other words, the temperature and density oneither side of the slipstream can be different but the pressure on bothsides must be equal. This may also be stated as follows. The con-tact surface can tolerate thermal and concentration imbalance but notpressure imbalance.


Example 4.3

An oblique from a wedge in a Mach 2 air stream is reflected from a flatsurface. If the flow turning caused by the shock is 11◦, determine (a) thewave angle of the reflected shock, (b) the Mach number downstream ofthe reflected shock and (c) the strength of reflection.


Solution

(a) For the incident shock, M1 = 2 and θ = 11◦. For this Mach numberand θ, from oblique shock table, we get

β1 = 40.42◦, M2 = 1.6,p2

p1= 1.795

For M2 = 1.6 and θ = 11◦, from oblique chock chart I, we have

β2 = 52.88◦


The normal component of Mach number upstream of the reflected shockis

M2n = M2 sin β2 = 1.6 × sin 52.88

= 1.276

For M2n = 1.276, from normal shock table,

M3n = 0.808,p3

p2= 1.732


(b) The Mach number downstream of the reflected shock is

M3 =M3n

sin (β2 − θ2)

=0.808

sin (52.88 − 11)

= 1.213

(c) The strength of reflection is

p3

p1=

p3

p2

p2

p1

= 1.732 × 1.795

= 3.11


Intersection of shocks of same family

When a shock intersects another shock of the same family, the shockscannot pass through as in the case of intersection of shocks of oppositefamily. The shocks will coalesce to form a single stronger shock, asshown in Figure 4.23, where shocks of the same family are producedby successive corners in the same wall.

O

Contact surface

M

N

Weak reflection

A BC

Figure 4.23Intersection waves of same family.


If the second shock BO is much weaker than the first one AO, then OCwill be a compression wave. This intersection may also be describedas follows: the second shock is partly transmitted along OM, thus aug-menting the strength of the first one and partly reflected along OC.

Shocks are referred to as left-running and right-running depending onwhether they run to the left or right when viewed in the flow direction.Left-running waves constitute one family and the right-running wavesconstitute the opposite family.


Example 4.4

For the flow field shown in Figure E4.4, determine βr with respect to theflow direction in zone 2, M2 and M3 if M1 = 2.0 and βi = 40◦.

M1

Refelected shock

M3

βr

M2

βi

Incident shock

Figure E4.4


Solution

From oblique shock chart 1, for M1 = 2.0 and βi = 40◦, we getθ = 10.5◦. This θ corresponds to the angle through which the flow isturned by the incident shock wave. Also, the angle through which theflow is turned back after the reflected wave should be 10.5◦, to satisfythe streamline concept that the flow should be parallel to the surfaceover which it flows. From oblique shock chart 2, we have


M2 = 1.63 for M1 = 2.0, θ = 10.5◦

M3 = 1.3 for M2 = 1.63, θ = 10.5◦

For M2 = 1.63 and θ = 10.5◦, from oblique shock chart 1, we get thecorresponding shock angle as β = 50.2◦.

The angle between the flow directions of zone 2 and the reflected waveis

βr = 50.2 − 10.5 = 39.7◦


Example 4.5

A Mach 4.0 air stream at 105 Pa is turned abruptly by a wall into theflow with a turning angle of 20◦, as shown in Figure E4.5. If the shockis reflected by another wall, determine the Mach number and pressuredownstream of the reflected shock.

1

θ

θM3

2

3M1

M2

θ

Figure E4.5


Solution

For M1 = 4.0 and θ = 20◦, from the oblique chart, the shock angle is

β12 = 32.5◦

Hence,M1n = M1 sin β = 2.149

From normal shock table, for M1n = 2.149, we get

M2n = 0.554,p2

p1= 5.226


Therefore,

M2 =Mn2

sin (β − θ)= 2.56

Now, for M2 = 2.56 and θ = 20◦, from oblique shock chart,

β23 = 42.11◦

M ′

2n = M2 sin β23 = 1.72

For M ′

2n = 1.72, normal shock table gives

M3n = 0.64,p3

p2= 3.285


Hence, the Mach number downstream of the reflected shock is

M3 =0.64

sin (42.11 − 20)

= 1.7

Thus,

p3

p1=

p3

p2

p2

p1= 3.285 × 5.226

= 17.17

p3 = 1.717MPa

Note: Problems involving oblique shocks can also be solved using theoblique shock tables instead of oblique shock charts.


Wave Reflection from a Free Boundary

Reflection of shock from a solid wall is shown in Figure 4.20 and basedon the analysis of that figure, we also know that the reflection from asolid boundary, though generally is not specular, is a like reflection. Thatis, an incident shock will be reflected as a shock and an incident expan-sion wave will be reflected as an expansion wave by a solid boundary.However, when the boundary is a free boundary (such as a fluid bound-ary) the reflection will not be a like reflection.


The wave patterns shown emanating from the nozzle exit in Figures2.11b and 2.11c experience such reflection from a free boundary. Al-though they are not inherently quasi-one-dimensional flows, the wavepattern shown is frequently encountered in the study of nozzle flows.

The gas jet emanating from a nozzle and propagating into the surround-ing still atmosphere (at pressure pa) has a boundary surface which in-terfaces with the surroundings still atmosphere. As in the case of theslipstream discussed in Section 4.11, the pressure across this bound-ary must be preserved, i.e. the pressure just below the jet boundarymust be equal to the pressure pa of the atmosphere in which the jetpropagates, along its complete length. Therefore, the waves must re-flect from the jet boundary in such a manner to preserve the pressureequilibrium all along the jet boundary. The free boundary, unlike a solidboundary, can change its shape and direction.


Examine the reflection of an oblique shock from a free boundary, asshown in Figure 4.24. The pressure p1 in region 1 is equal to the sur-rounding atmosphere. The pressure in the region downstream of theincident shock is p2, which is higher than p1 (p2 > p1). At the edge ofthe jet boundary, shown by the dashed line in Figure 4.24, the pressuremust always be p1. Therefore, when the incident shock impinges onthe boundary, it must be reflected in such a manner as to result in p1,in region 3 behind the reflected wave. Hence, we have p3 = p1 andp1 < p2, i.e. p3 < p2. This situation demands that the reflected wavemust be an expansion wave, as shown in Figure 4.24.


The flow in turn is deflected upward by both the incident shock and thereflected expansion fan, resulting in the upward deflection of the freeboundary, as shown in Figure 4.24.

Incident shock

12

3

Constant p1

p1

Reflected expansion fan

Figure 4.24Shock wave reflection from a free boundary.


Reflection of an expansion fan from a free boundary is illustrated inFigure 4.25. The rays of the expansion fan are reflected from the freeboundary as compression waves. These waves coalesce, forming ashock wave, as shown. The wave interaction shown in Figure 4.25 canbe analyzed by the method of characteristics presented in Chapter 9.

p3 = p1 > p2

1

2

3

Constant p1

p1

Figure 4.25Reflection of an expansion fan from a free boundary.


From the above discussions, the following two observations can bemade.

1 The reflection of an incident wave from a solid boundary is termedlike reflection, i.e., a shock wave reflects as a shock and an expan-sion wave reflects as an expansion wave.

2 The reflection of an incident wave from a free boundary is calledunlike reflection (opposite manner), i.e. a shock wave reflects asan expansion wave and an expansion wave reflects as a shock (acompression wave).


Consider the overexpanded nozzle flow illustrated in Figure 2.11(b).The flow pattern downstream of the nozzle exit will appear as shownin Figure 4.26a. The various reflected waves form a diamond patternthroughout supersonic region of the exhaust jet. Shadowgraph picturesof a Mach 1.8 jet from a straight convergent-divergent nozzle of semi-divergence angle 7◦, at overexpanded (NPR 4), correctly expanded(NPR 5.24) and underexpanded (NPR 7) states are shown in Figures4.26b, c and d, respectively.


Free boundary

M > 1pe < pa

pa

Figure 4.26 (a) Diamond wave pattern in the exhaust from a supersonicnozzle.


Figure 4.26 (b) Shadowgraph picture of overexpanded (NPR 4) Mach 1.8 jet.


Figure 4.26 (c) Shadowgraph picture of correctly expanded (NPR 5.24) Mach1.8 jet.


Figure 4.26 (d) Shadowgraph picture of underexpanded (NPR 7) Mach 1.8jet.


At NPR 4 the Mach 1.8 jet is overexpanded with nozzle exit pressurepe less than the backpressure pa, which is the pressure of the ambi-ent atmosphere to which the jet is discharged. Therefore, there is anoblique shock at the nozzle exit to increase pe to come to equilibriumwith pa. This oblique shock formed at the nozzle exit is in the form of anaxisymmetric truncated-cone. The shock on impingement with the jetboundary reflects as compression waves as seen in Figure 4.26b, asdiscussed above for Figure 4.26a.


These compression and expansion waves experience back and forthreflections till the flow ceases to become supersonic. Thus, up to somedistance downstream of the nozzle exit, waves prevail in the jet field, asseen in Figure 4.26b. The compression and expansion waves exhibit apattern resembling diamond shape.


At NPR 5.24, Mach 1.8 is correctly expanded with pe = pa. For thiscase, we may tend to think that there will not be any wave in the jet field.But it is not correct. This is because, even though there is pressureequilibrium at the nozzle exit, the supersonic jet issuing from the nozzlefind a large space to relax. This relaxation effect forces the flow todeflect away from the nozzle axis. We know that supersonic flows arewave dominated and any change of flow property or orientation will takeplace only through waves. In accordance with this flow physics, therelaxation effect causes the formation of an expansion fan at the nozzleexit, as seen in Figure 4.26c [Rathakrishnan, 2008].


The waves of the expansion fan reflect back as compression waves fromthe jet boundary, coalesce and become stronger compression wavesand cross each other at the jet axis. These waves on reaching the jetboundary reflect back as expansion waves and the process continues.It is seen in Figure 4.26c that diamond pattern of waves prevail in thecorrectly expanded jet also, indicating that the correctly expanded freejet is also wave dominated.


Mach 1.8 jet a NPR 7 is underexpanded with pe > pa. Therefore, toestablish pressure equilibrium pe should be reduced to become equalto pa. This is caused by the expansion fan positioned at the nozzleexit, as seen in Figure 4.26d. The expansion waves of the fan reflectback as compression waves and the back and forth reflection of thesecompression and expansion waves establish diamond pattern in thenear field downstream of the nozzle exit. Now we may have a doubt;whether there is any difference between the expansion fan at the nozzleexit for correctly expanded and underexpanded states?


The answer to this question is the that there is a distinct difference be-tween these two expansion fans. For the correctly expanded jet, theexpansion fan is only due to the relaxation process. But for the under-expanded jet the expansion is due to the combined effect of pressureinequality and relaxation process. Therefore, the expansion fan for theunderexpanded jet is significantly stronger than that of the correctly ex-panded jet.


In fact, even for the overexpanded jet the flow at the nozzle exit wouldpass through the oblique shock caused by the overexpansion level andan expansion fan caused by the relaxation effect. An elaborate dis-cussion giving a deeper isight into the flow process of correctly andincorrectly expanded free jets with qualitative and quantitative measureof the flow is given in Chapter 12 on jets.


Example 4.6

Consider a two-dimensional duct carrying a perfect gas with uniformconditions of p1 = 1 atm and M1 = 2.0. Design a 10 turning elbow toachieve a uniform downstream state 2 for each of the following cases:(a) M2 > M1, s2 = s1, (b) M2 < M1, s2 = s1 (c) M2 < M1, s2 > s1, (d)M2 = M1, s2 = s1. For each case find the numerical values for M, pand duct area (compared to that of state 1).


Solution

(a) Given, M2 > M1, s2 = s1. From isentropic table, for M1 = 2.0,

ν1 = 26.38◦

Hence,ν2 = ν1 + |∆θ| = 26.38 + 10 = 36.38◦

For ν2 = 36.38◦, from isentropic table,

M2 = 2.387


For M1 = 2.0, again from isentropic table,

p1

p0= 0.1278,

A∗

A1= 0.5924

For M2 = 2.387,

p2

p0= 0.06948,

A∗

A2= 0.4236

Therefore,

p2 = 0.06948 p0

= 0.06948 × p1

0.1278

= 0.544 p1


d22

d21

=A2

A1

=A2

A∗

A∗

A1= 1.40

d2 =√

1.4 d1


(b) Given, M2 < M1, s2 = s1. It is an isentropic compression, therefore,

ν2 = ν1 − |∆θ| = 16.38◦

From isentropic table, for ν2 = 16.38◦,

M2 = 1.655

p2

p0= 0.2168

A∗

A2= 0.7713


Therefore,

p2 = 0.2168 p0

= 0.2168 × p1

0.1278

= 1.696 p1

d22

d21

=A2

A∗

A∗

A1

=0.59240.7713

= 0.768

d2 =√

0.768 d1


(c) Given, M2 < M1, s2 > s1.

Unlike cases (a) and (b), case (c) is nonisentropic since s2 > s1. There-fore, there should be a shock in the duct to decelerate the flow from M1

to M2. From oblique shock chart 1, for M1 = 2.0 and θ = 10◦, β = 39.5◦.Thus,

M1n = M1 sin β = 1.27

From normal shock table, at M1n = 1.27, we have

M2n = 0.8016,p2

p1= 1.715

ρ2

ρ1= 1.463,

a2

a1= 1.083


Hence,

M2 =M2n

sin (β − θ)

= 1.628

p2 = 1.715 p1

By continuity,ρ2A2V2 = ρ1A1V1


Therefore,

d22

d21

=A2

A1=

ρ1V1

ρ2V2

=ρ1

ρ2

M1

M2

a1

a2= 0.775

d2 =√

0.775 d1

(d) Given, M2 = M1, s2 = s1.

Thus, there is no change in area and flow properties from 1 to 2.


Example 4.7

A Mach 2 uniform air stream at p1 = 800 kPa and temperature 270 Kexpands through two convex corners of 10◦ each, as shown in FigureE4.7. Determine the Mach number M3, downstream of the second fanand p2, T2 and the angle of the angle of the second expansion fan.

M1 M2

M3

Figure E4.7


Solution

From isentropic table, for M1 = 2.0, we get

ν1 = 26.38◦

Therefore, the Prandtl-Meyer function after the first fan is

ν2 = ν1 + 10◦ = 36.38◦

From isentropic table, for ν2 = 36.38◦, M2 = 2.38. The Prandtl-Meyerfunction after the second fan is

ν3 = ν2 + 10 = 46.38◦


For ν3 = 46.38◦, from isentropic table, we get M3 = 2.83 and

M3 = 2.83

Also,

p1

p01= 0.1278,

T1

T01= 0.5556 at M1 = 2.0

p2

p02= 0.0706,

T2

T02= 0.4688 at M2 = 2.38


But for isentropic flow, p02 = p01, T02 = T01. Therefore,

p2 =0.07060.1278

× p1

=0.07060.1278

× 800

= 441.94 kPa

T2 =0.46880.5556

× 270

= 227.82 K


After the second fan, following the same procedure as above and usingisentropic table, we get

p3 = 22.03 kPa

T3 = 186.8 K

The fan angle for the second fan is

µ23 = θ + µ2 − µ3

= 10 + 24.85 − 20.69

= 14.16◦


Detached Shocks

Detached shock is a compression front made up of approximately anormal shock followed by an infinite number of strong and weak obliqueshocks of varying strength, which finally degenerates to a Mach wave.This is the shock which results when the wall deflection angle θ isgreater than θmax (Section 4.3). We know from the discussion of Section4.3 that, there is no rigorous analytical procedure available for analyzingproblems in which the deflection angles are more than θmax.


But experimentally, it has been observed that a flow with θ > θmax willhave a configuration as shown in Figure 4.27a, with the shock waveslightly upstream of the nose even for a sharp nosed body. The shapeof the detached shock and its detachment distance depend on the ge-ometry of the object facing the flow and the Mach number M1.


M > 1

M > 1

M > 1

Shock

M < 1M < 1

M > 1

Sonic line

M1 θ/2

θ/2

(a)

(b)Figure 4.27

Detached shock waves, (a) schematic sketches and (b) detached shock atthe nose of a body at Mach 7.Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c© 2010 Ethirajan Rathakrishnan 192 / 331

For an object with a blunt-nose, as shown in Figure 4.27b, the shockwave is detached at all supersonic and hypersonic Mach numbers. There-fore, even a streamlined body like a cone is a ‘blunt-nosed’ body as faras the oncoming flow is concerned, when θ > θmax.

From Figure 4.27, it is seen that the shock portion at the nose of theobject can be approximated to a normal shock. Therefore, immediatelybehind it there will be subsonic flow. Hence, the wedge portion of thesharp nosed body gives rise to acceleration of the flow from subsonicto transonic before the flow reached the shoulder.


Because of this acceleration to transonic Mach number, the flow en-counters an expansion fan at the shoulder, in order to become parallelto the body downstream of the shoulder, in accordance with the stream-line concept. Thus, the flow traversed by the expansion fan becomessupersonic. Therefore, there will be a sonic line, which will emanatefrom shoulder of the wedge and the blunt nosed body, as shown in Fig-ure 4.27a. For blunt nosed bodies, it is difficult to determine the exactposition of the sonic line.


For a given wedge angle θ, when M1 is high enough the shock is at-tached to the nose. As M1 decreases, the shock angle increases; withfurther decrease of Mach number, a value is reached for which the flowthrough the shock becomes subsonic. The shoulder of the body, behindthe nose, now has an effect on the whole shock, which may be curved,even though attached. These conditions correspond to the region be-tween the lines with M2 = 1 and θ = θmax in Figure 4.3. At the Machnumber corresponding to θmax, the shock wave begins to detach. Thisis called the shock detachment Mach number. With further decrease ofM1, the detached shock moves upstream of the nose.


The analysis of the flow field associated with detached shock becomesvery difficult because of the transonic flow, which prevails behind theshock. In this case, we mainly look for the shape of the shock, the de-tachment distance, and the shape of the sonic line. But such a solutiondoes not exist. The approximation we usually make is that the sonic lineis linear.

The strength of the detached shock is the maximum near the stagna-tion streamline, where it is approximated as a normal shock and then itcontinuously decreases by becoming oblique until finally it becomes aMach line, far away from the object.


It is interesting to note that, the detached shock formed ahead of anobject is a special compression front consisting of infinite number ofcompression wave segments, beginning from nearly normal shock nearthe stagnation streamline, strong oblique shock segments above thatfollowed by weak oblique shock segments and finally terminating into aMach line, far away from the object.


From our discussions on detached shock it is evident that, higher thefreestream Mach number, closer will be the curved shock to the apex(for conical nose) and nose tip for blunt nosed body. The density nearthe nose tip will be higher than the rest of the nose portion because theshock is normal near the stagnation point. The flow behind this normalportion of the shock will be subsonic.


Hence the wedge portion gives rise to acceleration of the flow fromsubsonic to supersonic. Therefore, there will be a sonic line, which willemanate from the shoulder of the wedge. For blunt nosed bodies deter-mination of the sonic line location is difficult. The problem of detachedbecomes complicated owing to the presence of sonic flow zone. If atall a solution to the flow exists, it should give the shape of the shock,shock detachment distance and the shape of the sonic line. But sucha solution does not exist. One approximation usually made is that thatthe sonic line is linear.


The shock strength is maximum near the stagnation point and then itdecreases continuously by becoming oblique, until finally it becomes aMach line. Because the strength decreases continuously, the entropychange also continuously decreases and hence the flow behind the de-tached shock is rotational.

For the conical nose, there is an apex. Therefore, when θ is decreasedbelow θmax, the flow detachment angle for the given freestream Machnumber, there will be two attached oblique shocks.


Also when the freestream Mach number for a conical nosed body (Fig-ure 4.27a), the shock will become attached. But for blunt nosed bodied(Figure 4.27), the shock is never attached. For these bodies, even forM∞ = ∞, the shock is detached. This is because the leading edgeangle is 90◦. For blunt nosed bodies there is always a detached shockfor any Mach number. But for sharp nosed bodies, provided θ < 45◦,there is a shock detachment Mach number blow which only the shockwill be detached. That is, when the Mach number is continuously de-creased, the shock, which is oblique so far, gets detached. For θ > 45◦,the shock detachment Mach number is extremely low.


Mach Reflection

A close look at the detached shock field will reveal that the complica-tions associated with such a flow field are due to the appearance ofsubsonic regions in the flow. Similar complications leading to a condi-tion where no solution with simple oblique shock wave is possible willarise in a flow field with shock reflections. Intersection of normal shockand the right-running oblique shock gives rise to a reflected left-runningoblique shock, in order to bring the flow into the original direction, asillustrated in Figure 4.28, this kind of reflection is termed Mach reflec-tion. The left-running shock must have lesser strength compared to theright-running shock because of the flow deflection angle θ involved inthe compression process associated with the shock, but M1 > M2.


M1

Slipstream

M2

M1 > M2 > M3

M3

M3

M1

Incident shock Reflected shock

Figure 4.28Mach reflection.


It may so happen that M2 is less than the shock detachment Mach num-ber for the wall deflection required; in such a case, the entire pictureof the flow field changes, all the shocks become curved and the flowbehind the shock system need not be parallel to the wall as shown inFigure 4.29. Some other phenomenon might take place later on to bringthe flow parallel to the wall.

M1

M2

M1

M3

M2

Slipstream

Figure 4.29 Flow past a shock system.


A similar phenomenon also occurs when two oblique shocks of oppositefamily intersect with a normal shock bridging them, as shown in Figure4.30.

Oblique shock

M2

M3

M2Normal shock

Oblique shock

SlipstreamM1

Figure 4.30 Intersection of oblique shocks.


From our discussions on shocks we know that, in the case of normallyoccurring oblique shocks, the strong shock solution is physically impos-sible. But a detached shock even tough naturally occurs over blunt-nosed bodies and over sharp-nosed bodies with nose angle larger thanthe shock detachment angle at a given supersonic (hypersonic) Machnumber, posses both strong and weak oblique shocks at some locationof its shape. Thus, detached shock is a special case of among the nat-urally occurring shocks, consisting of strong oblique shock. Therefore,we can state that all naturally occurring oblique shocks are weak whenthey are just a single shock, but when they appear as a set of obliqueshocks, as in the case of detached shock, the combination can have aset of strong and weak oblique shocks.


At this stage, it is important to realize that, as we saw in Section 4.11,the reflection of wave from a free boundary is unlike. In accordancewith this, the reflection of the right-running shock from the intersectionpoint of the normal shock in Figure 4.28 should be an expansion fan.But that is not the case in this reflection. This is because, even thoughthe normal shock can be regarded as a free boundary, it is adjacent toa solid wall. Because of this proximity the shock could able to resist theinertia of the impinging shock without deflecting to an extent to result ina kink at the point of oblique shock incidence.


Due to the absence of any kink, as shown in Figure 4.24, the shockexperiences like reflection and gets reflected as a shock. But it is es-sential to note that, the shape of the normal shock near the wall, shownas linear is an approximated idealization. In actual flows, the normalshock will assume a shape which is non-linear as in Figure 4.29. Butit is not possible to predict the exact non-linear shape of the shock ina problem of this kind. The only means to identify the wave shape isthrough optical flow visualization, such as Schlieren or shadowgraph.


Example 4.8

For the flow field shown in Figure E4.8, find the flow Mach number atzones 2, 3 and 4. Assume the slipstream deflection to be negligible.

1

M1 = 3

β4

β2

β3

2

3

4

10◦

10◦

Figure E4.8


Solution

From oblique shock chart 1, for M1 = 3.0 and θ2 = 10◦, we have

β2 = 27.5◦

Therefore,

M1n2 = 3 sin 27.5◦

= 1.38

From normal shock table, for M1n2 = 1.38, we have

p2

p1= 2.055, M2n1 = 0.7483


Therefore,

M2 =M2n1

sin (β2 − θ2)

= 2.49

For M2 = 2.49 and θ3 = 10◦, from oblique shock chart,

β3 = 32◦

Therefore,

M2n3 = 2.49 sin 32◦

= 1.32


From normal shock table, for M2n3 = 1.32, we get

p3/p2 = 1.866 and M3n2 = 0.7760

Therefore,

M3 =0.7760

sin (32 − 10)

= 2.07

p3

p1=

p3

p2× p2

p1

= 1.866 × 2.055

= 3.835


With no slipstream deflection,

θ4 = 20◦

From oblique shock chart 1, for M1 = 3.0 and θ4 = 20◦, we get

β4 = 37.5◦

Therefore,

M1n4 = 3 sin 37.5◦

= 1.83


From normal shock table, for M1n4 = 1.83,

p4

p1= 3.74, Mn4 = 0.6099

Therefore,

M4 =0.6099

sin (37.5 − 20)

= 2.03


Example 4.9

Find the slipstream angle δs for the flow field shown in Figure E4.9,using the oblique shock charts.

42

3

1

10◦

M∞ = 2.0

5◦

β1

δsSlipstream

β2

Figure E4.9


Solution

From oblique shock chart 1, for M∞ = 2.0 and θ1 = 10◦, from obliqueshock chart I, we get

β1 = 39◦

For M∞ = 2.0 and θ1 = 10◦, from oblique shock chart II, we get

M1 = 1.665,p1

p∞

= 1.686

For M∞ = 2.0 and θ2 = 5◦, from oblique shock chart I, we get

β2 = 34.5◦


For M∞ = 2.0 and θ2 = 5◦, from oblique shock chart II, we get

M2 = 1.805 andp2

p∞

= 1.323

Because of the slipstream, the properties in the regions (3) and (4) haveto be calculated by trial and error only.

Trial 1

Let the downstream flow be parallel to upstream flow. For θ3 = 10◦ andM1 = 1.665, from oblique shock chart 1,

β3 = 48.5◦


Therefore,

M1n3 = M1 sin β3

= 1.665 × sin 48.5◦

= 1.247

For M1n3 = 1.247, from normal shock table,

p3/p1 = 1.656


Therefore,

p3

p∞

=p3

p1× p1

p∞

= 1.656 × 1.686

= 2.79

For θ4 = 5◦ and M2 = 1.805,

β4 = 38◦


Therefore,

M2n4 = M2 sin β4

= 1.805 × sin 38◦

= 1.111


p4

p2= 1.271


Therefore,

p4

p∞

=p4

p2× p2

p∞

= 1.271 × 1, 323

= 1.68

Because p3/p∞ > p4/p∞, the slipstream has to be deflected down-wards so that the two pressures become equal.


Trial 2

Let p3

p∞

=p4

p∞

= 2.25

Therefore,

p3

p1=

p3

p∞

× p∞

p1

= 2.25 × 11.686

= 1.334

For p3/p1 = 1.334, from normal shock table,

M1n3 = 1.135


Therefore,

β3 = sin−1(

M1n3

M1

)

= 43◦

For M1 = 1.665 and β3 = 43◦, from oblique shock chart I,

θ3 = 6.5◦

i.e. 3.5◦ below freestream direction. Similarly,p4

p2=

p4

p∞

× p∞

p2

= 2.25 × 11.323

= 1.7


For p4/p2 = 1.7, from normal shock table,

M2n4 = 1.27

Therefore,

β4 = sin−1(

M2n4

M2

)

= sin−1(

1.271.805

)

= 44.7◦

For M2 = 1.805 and β4 = 44.7◦, from oblique shock chart II,

θ4 = 11◦

i.e. 6◦ below the freestream direction.


Trial 3

With downstream flow 4.5◦ below upstream flow,

For θ3 = 5.5◦ and M1 = 1.665, from oblique shock chart I,

β3 = 42.5◦

Therefore,

M1n3 = M1 sin 42.5 = 1.665 × sin 42.5

= 1.125



p3

p1= 1.31

Therefore,

p3

p∞

=p3

p1× p1

p∞

= 1.31 × 1.686

= 2.21

For θ4 = 9.5 and M2 = 1.805, from oblique shock chart I.

β4 = 43.5◦


Therefore,

M2n4 = M2 sin β4

= 1.805 × sin 43.5◦

= 1.242


For M2n4 = 1.242, from normal shock table,

p4

p2= 1.63

Thus,

p4

p∞

=p4

p2× p2

p∞

= 1.63 × 1.323

= 2.16


In trial 3, it is seen that pressures p3 and p4 are nearly equal, i.e. theassumed slipstream deflection (δs) of 4.5◦ is the correct value.

Note: This problem could have been solved using oblique shock tablestoo. But to practice the use oblique shock charts, this example has beensolved with charts.


Shock-Expansion Theory

The shock and expansion waves discussed in this chapter are the basisfor analyzing large number of two-dimensional, supersonic flow prob-lems by simply ‘patching’ together appropriate combinations of two ormore solutions. That is, the aerodynamic forces acting on a body presentin a supersonic flow are governed by the shock and expansion wavesformed at the surface of the body. This can be easily seen from thebasic fact that the aerodynamic forces acting on a body depend on thepressure distribution around it and in supersonic flow, the pressure dis-tribution over an object depends on the wave pattern on it, as shown inFigure 4.31.


4

p4

12

2’

2

(c)(b)

3

3

p1

l

p1

p4

p1

p2

α0

p3

M1

p2

p3

p2′

(a)

2ǫ t

Figure 4.31Wave pattern over objects.


Consider the two-dimensional diamond aerofoil kept at zero angle ofattach in a uniform supersonic flow, as shown in Figure 4.31(a). Thesupersonic flow at M1 is first compressed and deflected through an an-gle ε by the oblique shock wave at the leading edge, forcing the flowto travel parallel to the wedge surface. At the shoulder located at mid-chord, the flow is expanded through an angle 2ε by the expansion fan.


At the trailing edge, the flow is again deflected through an angle ε, inorder to bring it back to the original direction. Therefore, the surfacepressures on the wedge segments ahead and after the shoulder, willbe at a constant level over each segment for supersonic flow, accordingto oblique shock and the Prandtl-Meyer expansion theory.


On the diamond aerofoil, at zero angle of attack, the lift is zero becausethe pressure distributions on the top and bottom surfaces are the same.Therefore, the only aerodynamic force acting on the diamond aerofoil isdue to the higher-pressure on the forward face and lower-pressure onthe rearward face. The drag per unit span is given by

D = 2(p2 l sin ε − p3 l sin ε) = 2 (p2 − p3) (t/2)

i.e.D = (p2 − p3) t (4.50)


Equation (4.50) gives an expression for drag experienced by a two-dimensional diamond aerofoil, kept at zero angle of attack in an inviscidflow. This is in contrast with the familiar result from studies on sub-sonic flow that, for two-dimensional inviscid flow over a wing of infinitespan at a subsonic velocity, the drag force acting on the wing is zero–atheoretical result called D’Alemberts paradox. In contrast with this, forsupersonic flow, drag exists even in the idealized, nonviscous fluid.


This new component of drag encountered when the flow is supersonicis called wave drag, and is fundamentally different from the skin-frictiondrag and separation drag which are associated with boundary layer ina viscous fluid. The wave drag is related to loss of total pressure andincrease of entropy across the oblique shock waves generated by theaerofoil.


For the flat plate at an angle of attack α0 in a uniform supersonic flow,shown in Figure 4.31(c), from the uniform pressure on the top and bot-tom sides, the lift and drag are computed very easily, with the followingequations.

L = (p2′ − p2) c cos α0

(4.51)

D = (p2′ − p2) c sin α0

where c is the chord.


Example 4.10

A flat plate is kept at 15◦ angle of attack to a Mach 2.4 air stream,as shown in Figure E4.10. Solve the flow field around the plate anddetermine the inclination of slipstream to the freestream direction usingshock-expansion theory.

15◦ Slipstream

1

2’

3’

3

M1 = 2.42

Figure E4.10


Solution

Using the shock and expansion wave properties, Table 4.1 can be formed.

Table 4.1

Region M ν µ p/p01 T/T01

1 2.4 36.8◦ 24.6◦ 0.0684 0.4652 3.11 51.8◦ 18.8◦ 0.0231 0.3413 2.33 35.0◦ 25.4◦ 0.0675 0.4802′ 1.80 20.7◦ 33.8◦ 0.1629 0.6073′ 2.36 35.7◦ 25.1◦ 0.0679 0.473

Table 4.1 lists the flow properties around the flat plate. Slip-surfaceinclination relative to freestream is negligibly small. The velocity jumpacross the slip-surface is found to be 1 m/s.


Example 4.11

Determine the flow field around a symmetric double wedge of 20◦ in-cluded angle kept at 15◦ angle of attack to a Mach 2.4 air stream ata stagnation temperature of 300 K, shown in Figure E4.11, by shock-expansion theory.

3’

1

M1 = 2.4

4’

3

4

15◦

Slipstream

2

2’

20◦

Figure E4.11


Solution

The flow properties are as given in Table 4.2 below.

Table 4.2

Region M ν µ p/p01 T/T01

1 2.40 36.8◦ 24.6◦ 0.0684 0.4652 2.62 41.8◦ 22.5◦ 0.0486 0.4213 3.71 61.8◦ 15.6◦ 0.0098 0.2674 2.00 26.5◦ 30.0◦ 0.0707 0.5552′ 1.31 6.5◦ 49.7◦ 0.2736 0.7453′ 2.00 26.5◦ 30.0◦ 0.0986 0.5554′ 2.23 32.5◦ 26.6◦ 0.0689 0.501


From the above values, we get

a4 = 258.7 m/s, V4 = 517.4 m/s

a4′ = 245.8 m/s, V4′ = 548.1 m/s

The slipstream surface is inclined at ≈ 1◦ upwardrelative to the freestream.

The velocity jump across the slipstream surface is = V4′ −V4 = 548.1−517.4 = 30.7 m/s.


Example 4.12

For the flow over the half-diamond wedge shown in Figure E4.12, findthe inclinations of shocks and expansion waves and the pressure coef-ficient over the wedge surfaces 2 and 3.

M1 = 1.8β1

µ31 2

3

415◦

Figure E4.12


Solution

Region 2

From oblique shock chart 1, for M1 = 1.8 and θ = 15◦,

β1 = 51.5◦

By Eq. (4.3), we have

p2

p1= 1 +

2γ

γ + 1(M2

1 sin2 β − 1)

= 2.149


The pressure coefficient over surface 2 is

Cp2 =p2 − p1

q1=

2 (p2/p1 − 1)

γM21

= 0.506

By Eq. (4.7),

M2 =Mn2

sin (β1 − θ)

From normal shock table, for M1n = M1 sin β1 = 1.4, we have

M2n = 0.7397


Therefore,

M2 =M2n

sin (β1 − θ1)

=0.7397

sin (51.5◦ − 15◦)

= 1.24

From isentropic table, for M2 = 1.24, we have

ν2 = 4.569◦ and µ2 = 53.751◦


Therefore,

ν3 = ν2 + θ3 = 4.569 + 30

= 34.569◦

For ν3 = 34.569◦, from isentropic table,

M3 = 2.315 and µ3 = 25.6◦

Angles µ2 and µ3 are the inclination angles of the first and the lastrays of the expansion fan, respectively. Thus, the mean angle of theexpansion fan becomes

µ3 =µ2 + µ3

2

= 39.68◦


Region 3

For M1 = 1.8, from isentropic table, we have

p1

p01= 0.1740

For M3 = 2.315, from isentropic table,

p3

p03= 0.0780, also p02 = p03

The dynamic pressure of the freestream flow is

q1 =12ρ1V 2

1 =γ p1M2

1

2


Therefore,

q1

p01=

γM21

2× p1

p01

=γM2

1

2× 0.1740

= 0.3946

p1

q1=

p1

p01× p01

q1

=0.17400.3946

= 0.441


p2

q1= Cp2 +

p1

q1

= 0.506 + 0.441

= 0.947


From the normal shock table, for M1n = 1.4, we have

p02

p01= 0.9582

Thus,

p3

q1=

p3

p02× p02

p01× p01

q1

= 0.0780 × 0.9582 × 10.3946

= 0.1894


The pressure coefficient over wedge surface 3 is

Cp3 =p3 − p1

q1

= 0.1894 − 0.441

= −0.2516

Note: It is important to note that the solutions for problems involvingoblique shock, obtained using oblique shock chart or table are onlyclose to the correct results and are not 100 percent accurate. For accu-rate results we have to use the appropriate relations directly. However,the use of chart and table results in considerable saving in time andalso the resulting accuracies are good enough for any application.


Thin Aerofoil Theory

In Section 4.14 we saw that the shock-expansion theory gives a simplemethod for computing lift and drag acting over a body kept in a su-personic stream. This theory is applicable as long as the shocks areattached. This theory may be further simplified by approximating it byusing the approximate relations for the weak shocks and expansion,when the aerofoil is thin and is kept at a small angle of attack, i.e. ifthe flow inclinations are small. This approximation will result in simpleanalytical expressions for lift and drag.


From our studies on weak oblique shocks in Section 4.6, we know thatthe basic approximate expression (Eq. 4.21) for calculating pressurechange across a weak shock is

∆pp1

≈ γM21

√

M21 − 1

∆θ

Because the wave is weak, the pressure p behind the shock will not besignificantly different from p1, nor will the Mach number M behind theshock be appreciably different from the freestream Mach number M1.


Therefore, we can express the above relation for pressure change acrossa weak shock, without any reference to the freestream state (i.e. withoutsubscript 1 to the pressure and Mach number) as

∆pp

≈ γM2√

M2 − 1∆θ

Now, assuming all direction changes to the freestream direction to bezero and freestream pressure to be p1, we can write

p − p1

p1=

γM21

√

M21 − 1

(θ − 0)

where θ is the local flow inclination relative to the freestream direction.


The pressure coefficient Cp is defined as

Cp =p − p1

q1

where p is the local static pressure and p1 and q1 are the freestreamstatic pressure and dynamic pressure, respectively. In terms of freestreamMach number M1, the pressure coefficient Cp can be expressed as

Cp =p − p1

q1=

2γM2

1

p − p1

p1


Substituting the expression for (p − p1)/p1 in terms of θ and M1, we get

Cp =2θ

√

M21 − 1

(4.52)

The above equation, which states that the pressure coefficient is pro-portional to the local flow direction, is the basic relation for thin aerofoiltheory.


Application of thin aerofoil theory

Applying the thin aerofoil theory relation, Eq. (4.52), for the flat plateshown in Figure 4.31c at a small angle of attack α0, the Cp on the upperand lower surfaces of the plate can be expressed as

Cp = ∓ 2α0√

M21 − 1

(4.53)

where the minus sign is for Cp on the upper surface and the plus signis for Cp on the lower surface.


The lift and drag coefficients are respectively given by

CL =(pl − pu) c cos α0

q1c= (Cpl − Cpu) cos α0

CD =(pl − pu) c sin α0

q1c= (Cpl − Cpu) sin α0

In the above expressions for CL and CD, cos α0 ≈ 1 and sin α0 ≈ α0,since α0 is small and the subscripts l and u refer to the lower and uppersurfaces, respectively and c is the chord.


Therefore,

CL = (Cpl − Cpu)

CD = (Cpl − Cpu)α0

Using Eq. (4.53), the CL and CD of the flat plate at a small angle ofattack may be expressed as

CL =4α0

√

M21 − 1

(4.54)

CD =4α2

0√

M21 − 1


Now, consider the diamond section aerofoil shown in Figure 4.31a, withnose angle 2ε, at zero angle of attack. The pressure coefficient Cp onthe front and rear faces are given by

Cp = ± 2ε√

M21 − 1

where the + sign is for the front face where the pressure p2 is higherthan p1 and the − sign is for the rear face with pressure p3 less than p1.This can be rewritten in terms of pressure difference to give

p2 − p3 =4ε

√

M21 − 1

q1


Therefore, the drag is given by

D = (p2 − p3)t = (p2 − p3) ε c

D =4ε2

√

M21 − 1

q1c

where q1 is the freestream dynamics pressure and c is the chord of theaerofoil.

In terms of the drag coefficient, the above drag equation becomes

CD =D

q1c=

4ε2√

M21 − 1

(4.55a)

or

CD =4

√

M21 − 1

(

tc

)2

(4.55b)


In the above two applications, the thin aerofoil theory was used for spe-cific profiles to get expressions for CL and CD. A general result ap-plicable to any thin aerofoil may be obtained as follows. Consider acambered aerofoil with finite thickness at a small angle of attack treatedby linear resolution into three components, each of which contributingto lift and drag, as shown in Figure 4.32.

α0

yu(x)

yc(x) h(x)yl(x)

y

Angle of attack

c

Camber Thickness

yc(x)

x

Figure 4.32Linear resolution of aerofoil into angle of attack, camber and thickness.


By thin aerofoil theory, the Cp on the upper and lower surfaces areobtained as

Cpu =2

√

M21 − 1

(

dyu

dx

)

(4.56)

Cpl =2

√

M21 − 1

(

−dyl

dx

)

where yu and yl are the upper and lower profiles of the aerofoil. Theprofile may be resolved into a symmetrical thickness distribution h(x)and a camber line of zero thickness yc(x).


Thus, we havedyu

dx=

dyc

dx+

dhdx

= −α(x) +dhdx

(4.57a)

dyl

dx=

dyc

dx− dh

dx= −α(x) − dh

dx(4.57b)

where α(x) = α0 + αc(x) is the local angle of attack of the camber lineand α0 is the angle of attack of the freestream and αc is the angle attackdue to the camber.


The lift and drag are given by

L = q1

∫ c

0(Cpl − Cpu) dx (4.58a)

D = q1

∫ c

0

[

Cpl

(

−dyl

dx

)

+ Cpu

(

dyu

dx

)]

dx (4.58b)


Substituting Eqs. (4.56) and (4.57) into Eqs. (4.58a) and (4.58b), weget

L =2q1

√

M21 − 1

∫ c

0

(

−2dyc

dx

)

dx

=4q1

√

M21 − 1

∫ c

0α(x) dx

D =2q1

√

M21 − 1

∫ c

0

[

(

dyl

dx

)2

+

(

dyu

dx

)2]

dx

=4q1

√

M21 − 1

∫ c

0

[

α(x)2 +

(

dhdx

)2]

dx


The integrals may be replaced by average values, e.g.

α =1c

∫ c

0α(x) dx

Also, noting that by definition αc = 0, we get

α = α0 + αc

= α0 + αc

= α0


Similarly,

α2 = (α0 + αc)2

= α20 + 2α0αc + α2

c

= α20 + α2

c

Using the above averages in the lift and drag expressions, we obtainthe lift and drag coefficients as


CL =4α

√

M21 − 1

=4α0

√

M21 − 1

(4.59a)

CD =4

√

M21 − 1

[

(

dhdx

)2

+ α2(x)

]

CD =4

√

M21 − 1

[

(

dhdx

)2

+ α20 + α2

c(x)

]

(4.59b)

Equations (4.59) give the general expressions for lift and drag coeffi-cients of a thin aerofoil in a supersonic flow. In thin aerofoil theory, thedrag is split into drag due to lift, drag due to camber and drag due tothickness, as given by Eq. (4.59b). But the lift coefficient depends onlyon the mean angle of attack.Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c© 2010 Ethirajan Rathakrishnan 270 / 331

Example 4.13

A supersonic, circular arc aerofoil, shown in Figure E4.13, has chord cand thickness-to-chord ratio of 0.12. Determine the lift and drag coeffi-cients of the aerofoil in terms of the angle of attack, α.

x0

y

K

t

R

c/2c/2

Figure E4.13


Solution

Let 0 the origin of the xy-coordinate system be at the leading edge ofthe aerofoil. Now, the equation of the circular arc is given by

(

x − c2

)2+ (y + K )2 = R2 (i)

Substituting x = 0, y = 0 in Eq. (i), we get

c2

4+ K 2 = R2

Also,R − K = t


Simplifying the above two equations, we can write

R =18

c2

t

[

1 +

(

2tc

)2]

K =18

c2

t

[

1 −(

2tc

)2]

Differentiating Eq. (i), the slope dy/dx can be obtained as

dydx

=c2

(

1 − 2 xc

)

y + K


For small y , this can be approximated as

dydx

≈(

dydx

)

y=0=

4 tc

(

1 − 2xc

)

(1 − (2t/c)2)

= 0.509(

1 − 2xc

)

For the aerofoil considered,

dyu

dx=

dyc

dx,

dyl

dx=

dyc

dx,

dhdx

= 0


Therefore, when the aerofoil is at an angle of attack α, we have

dyc

dx= −α(x) +

dhdx

= −α(x) + 0.509(

1 − 2xc

)

The coefficient of lift CL is given by Eq. (4.58a), in the form

Lq1c

=1c

∫ c

0(Cpl − Cpu)dx

Substituting Eqs. (4.56) and (4.57) into the above equation, we get

CL =1c

∫ c

0− 4√

M21 − 1

(

dyc

dx

)

dx

= − 4

c√

M21 − 1

∫ c

0

−α(x) + 0.509

(

1 − 2xc

)

dx


On integrating, we obtain the lift coefficient as

CL =4α

√

M21 − 1

This result of CL implies that the lift goes to zero when the angle ofattack is zero. The drag coefficient is given by


CD =1c

∫ c

0

[

Cpl

(

−dyl

dx

)

dx + Cpu

(

dyu

dx

)]

dx

=2

c√

M21 − 1

∫ c

0

[

(

dyl

dx

)2

+

(

dyu

dx

)2]

dx

=4

c√

M21 − 1

∫ c

0

(

dyc

dx

)2

dx

=4

c√

M21 − 1

∫ c

0

[

−α(x) + 0.509(

1 − 2xc

)]2

dx


=4

c√

M21 − 1

×

∫ c

0

[

α2(x) + 0.5092(

1 − 4xc

+4x2

c2

)

− 2α(x) × 0.509(

1 − 2xc

)]

dx

=4α2(x)√

M21 − 1

+0.3451√

M21 − 1


Summary

In this chapter we have discussed the flow processes through obliqueshock and expansion waves. A shock wave which is inclined at an angleto the flow direction is called an oblique shock wave. Oblique shocksusually occur when a supersonic flow is turned into itself. The oppo-site of this, namely, when a supersonic flow is turned away from itself,results in the formation of an expansion fan. Oblique shock and expan-sion waves prevail in two- and three-dimensional supersonic flows, incontrast to normal shock waves, which are quasi-one-dimensional.


The density, pressure and temperature ratios across an oblique shockwave are given by

ρ2

ρ1=

(γ + 1)M21 sin2 β

(γ − 1)M21 sin2 β + 2

p2

p1= 1 +

2γ

γ + 1(M2

1 sin2 β − 1)

T2

T1= 1 +

2(γ − 1)

(γ + 1)2

M21 sin2 β − 1

M21 sin2 β

(γM21 sin2 β + 1)

where subscripts 1 and 2 refer to the conditions ahead of and behindthe oblique shock and β is the shock angle.


The Mach number behind an oblique shock M2 is given by

M2 =Mn2

sin(β − θ)

where Mn2 is the normal component of Mach number behind the shockand θ is the flow turning angle.

The maximum and minimum values of shock angle correspond to thosefor normal shock; β = π/2 and Mach wave, µ. Thus, the possible rangeof β is

sin−1(

1M1

)

≤ β ≤ π

2


The relation between the Mach number, shock angle and flow turningangle is given by

tan θ = 2 cot β

M21 sin2 β − 1

M21 (γ + cos 2β) + 2

This is known as the θ − β − M relation.The graphical representation of oblique shock properties is known asshock polar.


A two-dimensional flow of supersonic stream is always associated withtwo families of Mach lines. The Mach lines with (+) sign run to theright of the streamline when viewed through the flow direction are calledright-running characteristics and those Mach lines with (−) sign run tothe left are called left-running characteristics.


Supersonic flow expansion around a convex corner, involving a smooth,gradual change in flow properties is known as Prandtl-Meyer expansion.The expansion fan or the Prandtl-Meyer fan consists of an infinite num-ber of expansion waves, centered at the convex corner. All rays in anexpansion fan are isentropic lines and the entire flow (except at the apexof the expansion fan) is isentropic.

The maximum turning of an expanding flow corresponds to the situationwhere the pressure p goes to zero. This corresponds to a flow turningangle of θ = 130.5◦.


The Prandtl-Meyer expansion is a self-similar motion and the Prandtl-Meyer function is a similarity parameter. The Prandtl-Meyer function interms of Mach number is given by

ν =

√

γ + 1γ − 1

arc tan

√

γ − 1γ + 1

(M21 − 1) − arc tan

√

(M21 − 1)

The waves causing isentropic expansion and compression are calledsimple waves. Zones of supersonic expansion or compression withMach lines which are straight are called simple regions. Zones withcurved Mach lines are called non-simple regions.


An incident shock gets reflected as a shock from a solid boundary. Thiskind of reflection is called like reflection. On the other hand, an incidentshock gets reflected as an expansion fan and the expansion fan getsreflected as compression waves from a free boundary. This kind ofreflection is called unlike reflection.

For supersonic flows, drag exists even in the idealized, nonviscous fluid.This new component of drag encountered when the flow is supersonicis called wave drag and it is fundamentally different from the skin-frictiondrag and separation drag which are associated with boundary layers ina viscous fluid.


Thin aerofoil theory gives an expression for the pressure coefficient as

Cp =2θ

√

M21 − 1

It states that, the pressure coefficient is proportional to the local flowdirection.

For a flat plate at a small angle of attack α0, the lift and drag coefficientsmay be expressed as

CL =4α0

√

M21 − 1

CD =4α2

0√

M21 − 1


The general expressions of lift and drag coefficients of a thin aerofoil insupersonic flow may be written as

CL =4α0

√

M21 − 1

CD =4

√

M21 − 1

[

(

dhdx

)2

+ α20 + α2

c(x)

]

where h(x) gives the symmetrical thickness distribution, α0 is the freestreamangle of attack and αc(x) is the angle of attack due to camber. From theabove relation for CD, it is seen that the drag is split into drag due to lift,drag due to camber and drag due to thickness. But the lift coefficientdepends only on the mean angle of attack.


From the discussions of shock and expansion waves in this chapter, it isclear that any problem in supersonic flow, in principle, can be analyzedwith the relations developed for the oblique shock and expansion fans.However, it must be realized that all relations we have developed are forflow fields with simple regions. For the nonsimple regions with nonlinearwave net there is no exact analytical approach developed so far. Butwhen the wave net pattern is made with tiny segments, the relationsdeveloped for linear waves can be applied to nonlinear wave segmentsin the nonsimple region, without introducing significant error.


This kind of approach is adopted in Chapter 8 while designing con-toured nozzles to generate uniform and unidirectional supersonic flows.Experimental results with such contoured nozzles prove that, assum-ing the nonlinear waves to be straight waves within a small wave netdoes not introduce any significant error. It is essential to realize that, inmethod of Characteristics the above approximation is made for expan-sion waves which are isentropic.


When the wave net involves crossing of shocks or compression waves,this assumption is bound to introduce significant errors in our calcula-tions. The development of a theory involving nonsimple regions withshocks or compression waves is still an open question.


Exercise Problems

4.1 A Mach 2 uniform air flow passes over a wedge. An oblique shock,making an angle of 40◦ with the flow direction, is attached to the wedge.If the static pressure and temperature in the freestream are 50 kPa and0◦C, determine the static pressure and temperature behind the wave,the Mach number of the flow passing over the wedge and the wedgeangle.

[Ans. 88.75 kPa, 323.5 K, 1.61 and 21.14◦]

4.2 A Mach 2.0 air stream is isentropically deflected by 5◦ in the clock-wise direction. If the pressure and temperature before deflection are 98kPa and 97◦, determine the Mach number, pressure, temperature anddensity of the deflected flow .

[Ans. M = 2.18, p = 74 kPa, T = 276.8 K, ρ = 0.9315 kg/m3]


4.3 A Mach 3 axisymmetric wind tunnel nozzle is run by an air reservoirat 7 MPa and the nozzle exhausts to a surrounding at 1 atm. During theoperation, the pressure in the supply reservoir decreases. (a) At whatsupply pressure an oblique shock wave will first appear at the nozzleexit? (b) if a shock-free test-region extending one diameter in lengthis required, what should be the minimum supply pressure for obtainingthis test region? and (c) what is the minimum supply pressure for whicha normal shock will appear at the nozzle exit?

[Ans. (a) ≤ 3.725 MPa, (b) 2.33 MPa, (c) 360 kPa]


4.4 Air flows over a symmetrical wedge of semi-vertex angle 15◦ atMach number 2.0. Determine (a) wave angle with respect to the freestreamdirection, (b) pressure ratio across the wave, (c) temperature ratio acrossthe wave, (d) density ratio across the wave and (e) Mach number down-stream of shock, assuming the shock at the nose be strong and weak.

[Ans. Strong shock solution:(a) 79.8◦, (b) 4.355, (c) 1.662, (d) 2.62, (e) 0.646Weak shock solution:(a) 45.34◦, (b) 2.186, (c) 1.268, (d) 1.729, (e) 1.448]


4.5 An underexpanded, two-dimensional, supersonic nozzle exhaustsinto a region where p = 0.75 atm. Flow at nozzle exit plane is uniform,with p = 1.6 atm and M = 2.0. Calculate the flow direction and Machnumber after the initial expansion.

[Ans. Flow turning angle = 12.275◦, M = 2.48]

4.6 (a) Compute the maximum deflection angles for which the obliqueshock remains attached to the wedge when M1 = 2.0 and 3.0. (b)Compute the minimum values of Mach number M1 for which the obliqueshock remains attached to the wedge for flow deflection angles of θ =15◦, 25◦ and 40◦.

[Ans. (a) 22◦, 34◦, (b) 1.65, 2.11, 4.45]


4.7 An oblique shock wave is incident on a solid boundary as shown inFigure P4.7.

M1 = 3.5M

2

45◦

θ

Figure P4.7

The boundary is to be turned through such an angle that there will beno reflected wave. Determine the angle θ and the flow Mach angle M.

[Ans. 28.158◦, 1.78]


4.8 Air flows over a frictionless surface having a sharp corner. Theflow angle and Mach number downstream of the corner are −60◦ and4.0, respectively. Calculate the upstream Mach number for flow turningangle of (a) 15◦ clockwise, (b) 30◦ clockwise, (c) 60◦ clockwise and (d)15◦ counter-clockwise.

[Ans. (a) 1.802, (b) 2.36, (c) 4.0, (d) For this case ν is negative, whichis not physically possible. Flow can exist only up to |∆θ| = 65.785, forwhich ν1 = 0 and M1 = 1.0]


4.9 A steady supersonic flow expands from Mach number M1 = 2.0 andpressure p1 to pressure p2 = p1/2 through a centered expansion. Findthe Mach number M2 and flow direction θ2, after the expansion.

[Ans. 2.444, 11.43◦]

4.10 (a) A nozzle is designed to generate Mach 3 uniform parallel flowof air. The stagnation pressure of the air supply reservoir p0 = 7 MPaand the nozzle exhausts to sea level atmosphere. Calculate the flowangle at the exit lip of the nozzle if the atmospheric pressure pe = 1.0atm. (b) Determine the stagnation pressure of the air supply for whichthe flow angle at the exit lip would be zero.

[Ans. (a) 7.64◦, (b) 3.725 MPa]


4.11 Air at p1 = 30 kPa, T1 = 350 K and M1 = 1.5 is to be expandedisentropically to 13 kPa. Determine (a) the flow deflection angle re-quired, (b) final Mach number and (c) the temperature of air after ex-pansion.

[Ans. (a) 15.85◦, (b) 2.05, (c) 275.7 K]


4.12 Mach 2 air stream flows over three sharp corners in succession,having clockwise turning angles of 5, 10 and 15◦, respectively. (a) Cal-culate the Mach number and flow angle after each of the three corners.(b) Find the expansion fan angles and the streamline distances from thesolid boundary. (Take freestream streamline distance d1 from the wallas unity.)

[Ans. (a) 2.0, 30◦, 2.187, 27.2◦, 2.6, 22.62◦, (b) Fan angles: 7.8◦,

14.6◦, 20.34◦, Distances:d2

d1= 1.173,

d3

d1= 1.716,

d4

d1= 3.562]


4.13 A supersonic inlet is to be designed to handle air at Mach 2.4 withstatic pressure and temperature of 50 kPa and 280 K, as shown in Fig-ure P4.13. (a) Determine the diffuser inlet area Ai , if the device has tohandle 20 kg/s of air and (b) if the diffuser has to further decelerate theflow behind the normal shock so that the velocity entering the compres-sor will not exceed 30 m/s. Assuming isentropic flow behind the normalshock, determine the area Ae required and the static pressure pe there.

21

M1 = 2.43 Ae

14◦

Figure P4.13

[Ans. (a) Ai = 0.03126 m2, (b) Ae = 0.239 m2, pe = 482 kPa]


4.14 A supersonic inlet is to be designed to operate at Mach 3.0. Twopossibilities are considered, as shown in Figure P4.14. In one, the com-pression and deceleration of the flow takes place through a single nor-mal shock (Figure P4.14a); in the other, a wedge-shaped diffuser (Fig-ure P4.14b) is used and the deceleration is through two weak obliqueshocks followed by a normal shock wave. The wedge turning angles are8◦ each. Compare the loss in stagnation pressure for the two cases.

M1M1

(a) (b)

M2 M4

M3

M2

Figure P4.14 (a) Normal shock diffuser, (b) wedge-shaped diffuser.

[

Ans. (a)p02

p01= 0.3283; (b)

p04

p01= 0.5803

]


4.15 A two-dimensional flat plate is kept at a positive angle of attack ina Mach 2.0 air stream. Below the plate, an oblique shock wave starts atthe leading edge, making an angle of 42◦ with the freestream direction.On the upper side, an expansion fan is positioned at the leading edge.Find (a) the angle of attack of the plate, (b) the pressure on the lowerand upper surfaces of the plate and (c) the pressure at the trailing edgeafter the flow leaves the plate, if the freestream static pressure is 1 atm.

[Ans. (a) 12.3◦, (b) 1.928 atm and 0.473 atm, (c) 1.0 atm]


4.16 High-pressure air maintained at a constant stagnation state of 7MPa and 300 K runs a Mach 3 blowdown wind tunnel. A symmetri-cal wedge having a semi-angle α = 15◦ is placed in the test-section.Calculate the following flow properties on the face of the wedge: (a)static pressure, density and temperature, (b) stagnation pressure, (c)flow velocity and flow Mach number.

[Ans. (a) 536.93 kPa, 12.58 kg/m3, 148.7 K, (b) 6.265 MPa, (c) 552.3m/s, 2.26]


4.17 The two-dimensional aerofoil shown in Figure P4.17 is traveling ata Mach number of 3 and at an angle of attack of 2◦. The thickness tochord ratio of the aerofoil is 0.1 and the maximum thickness occurs at30 percent of the chord downstream from the leading edge. Using thelinearized theory, show that the moment coefficient about the aerody-namic centre is −0.035, the centre of pressure is at 1.217c and the dragcoefficient is 0.0354. Show also that the angle of zero lift is 0◦.

0.7c0.3c

0.1cM = 3.0

Figure P4.17


4.18 For the flat plate shown in Figure P4.18, calculate the flow Machnumbers at zones 2, 2′, 3 and 3′, assuming the slipstream deflection tobe negligible.

12◦

1′

3′2′

1

3

2M1 = 3

Figure P4.18

[Ans. M2 = 3.71, M3 = 2.726, M2′ = 2.4, M3′ = 2.95]


4.19 For the double-wedge shown in Figure P4.19, calculate the flowMach numbers at different flow zones and the slipstream.

4′

M1 = 3 3

4

1

2

3′

2′

10◦

12◦

Figure P4.19

[Ans. M2 = 3.105, M3 = 4.493, M4 = 2.58, M2′ = 1.91, M3′ = 2.71,M4′ = 2.806. Comparing p4/p01 and p4′/p01 it can be seen that theslipstream is very weak]


4.20 A two-dimensional wedge shown in Figure P4.20 moves throughthe atmosphere at sea-level, at zero angle of attack with M∞ = 3.0.Calculate CL and CD using shock-expansion theory.

c10

◦M∞

Figure P4.20

[Ans. CL = −0.0389, CD = 0.02266]


4.21 For a Prandtl-Meyer expansion, the upstream Mach number is 2and the pressure ratio across the fan is 0.5. Determine the angles of thefront and end Mach lines of the expansion fan relative to the freestream.

[Ans. 30◦, 12.86◦]

4.22 Calculate the ratios of static and total pressures across the shockwave emanating from the leading edge of a wedge of 5◦ half-angle flyingat Mach 2.2.

[Ans. 1.34, 0.99726]


4.23 An uniform supersonic flow of air at Mach 3.0 and p1 = 0.05 atmpasses over a cone of semi-vertex angle 8◦ kept in line with the flow.Determine the shock angle and the static pressure at the cone surface,just behind the shock.

[Ans. 25.61◦, 9.1 kPa]

4.24 A supersonic stream of air at Mach 3 and 1 atm passes througha sudden convex and a sudden concave corners of turning angle 15◦

each. Determine the Mach number and pressure of the flow down-stream of the concave corner.

[Ans. 2.7, 1.015 atm]


4.25 A flat plate wing of chord 1 m experiences a lift of 10200 N permeter of width. If the flow Mach number and pressure are 1.6 and 25kPa, respectively. Determine the angle of attack and the aerodynamicefficiency (CL/CD) of the wing.

[Ans. 4◦, 14]

4.26 For an oblique shock wave with a wave angle of 33◦ and upstreamMach number 2.4, calculate the flow deflection angle θ, the pressureand temperature ratio across the shock wave and the Mach numberbehind the wave.

[Ans. 10◦, 1.8354, 1.1972, 2.0]


4.27 Show that the pressure difference across a oblique shock wavewith wave angle β may be expressed in the form

p2 − p112ρ1u2

1

=4

γ + 1

(

sin2 β − 1M2

1

)

where subscripts 1 and 2 refer to states upstream and downstream ofthe shock.

4.28 Air flow with Mach number 3.0 and pressure 1 atm passes over acompression corner. If the pressure downstream of the corner is 5 atm,determine the flow turning angle.

[Ans. 25.5◦]


4.29 A Mach 2 air stream passes over a 10◦ compression corner. Theoblique shock from the corner is reflected from a flat wall which is paral-lel to the freestream, as shown in Figure P4.29. Compute the angle ofthe reflected shock wave relative to the flat wall and the Mach numberdownstream of the reflected shock.

10◦

βi

M1

βr

Figure P4.29

[Ans. 39.5◦, 1.28]


4.30 Mach 2 air stream passes over two compression corners of angles7◦ and θ, as shown in Figure P4.30. Determine the value of θ up towhich the second shock will remain attached.

3M2M1 = 2

M3

21

θ7◦

Figure P4.30

[Ans. 18◦]


4.31 Mach 2 air flow is compressed by turning it through 15◦. For eachof the possible solutions calculate the (a) shock angle, (b) Mach num-ber downstream of the shock and (c) change of entropy. What is themaximum deflection angle up to which the shock will remain attached?

[Ans. Weak solution: (a) 45.34◦, (b) 1.45, (c) 13.73 J/(kg K), Strongsolution: (a) 79.83◦, (b) 0.64, (c) 88.25 J/(kg K), 22.97◦]

4.32 Mach 3 air stream at 1 atm and 200 K is deflected at a compressioncorner through 10◦. Calculate the Mach number, static and stagnationpressures and temperatures downstream of the corner.

[Ans. 2.5, 2.06 atm, 35.4 atm, 248.36 K, 560 K]


4.33 Determine the wave angle and Mach number behind and the pres-sure ratio across the oblique shock in air with M1 = 3.0 and θ = 10◦,treating the shock as (a) weak and (b) strong.

[Ans. (a) Weak solution: β = 27.38◦, 2.5, 2.05, (b) Strong solution:β = 86.41◦, 0.489, 10.292]


4.34 Compare the pressure loss experienced by the (a) one-shock and(b) two-shock spikes shown in Figures P4.34a and b.

31 2

M1 = 2.2

(a)

12◦M1 = 2.2

31

(b)

2

6◦

12◦

Figure P4.34(a) one-shock spike, (b) two-shock spike diffuser.

[Ans. 27.3 percent, 17.2 percent]


4.35 An oblique shock created by the flow of air over a sharp corner,as shown in Figure P4.35, is with wave angle 30◦. If the Mach num-ber upstream of the incident wave is 2.4, determine the Mach numberupstream and downstream of the reflected shock wave.

M2

β12 3

2β12

M1

1

φ

M3

θ = 10◦

Figure P4.35

[Ans. 2.09, 1.85]


4.36 For the flow field shown in Figure P4.36, determine the Mach num-ber and pressure in region 3.

12

3

10◦

M1 = 2.4

p1 = 101 kPa

Figure P4.36

[Ans. 1.64, 316.43 kPa]


4.37 An oblique shock gets reflected from a solid wall, as shown inFigure P4.37. Calculate the Mach number and pressure downstream ofthe reflected shock. Also, calculate the angle φ made by the reflectedshock with the wall. Assume the medium to be air.

M2

p1 = 100 kPa

30◦ φ

M3

M1 = 2.2

Figure P4.37

[Ans. 1.83, 164.7 kPa, 29◦]


4.38 If the boundary of the overexpanded jet issuing from a nozzle with

area ratioAe

At= 2.0 and stagnation pressure 300 kPa is as shown in

Figure P4.38, determine the backpressure pb.

Jet boundary

Air

e5◦

1

2

Figure P4.38

[Ans. 37.59 kPa]


4.39 An incident shock wave with β = 30◦ impinges on a straight wall.If the flow upstream of the shock is with M1 = 2.8, p1 = 1 atm and T1

= 300 K, calculate the pressure, temperature, Mach number and thestagnation pressure downstream of the reflected shock wave.

[Ans. 3.83 atm, 441.5 K, 1.93, 25.5 atm]

4.40 Mach number 1.46 air stream is expanded through an angle of5◦. Determine the Mach number of the expanded flow. Calculate thepressure coefficient Cp = (p2 − p1) /(1

2 ρ1 V12), where subscript 1 refer

to the initial condition. Compare this Cp with that predicted by linearized(Ackeret’s) theory.

[Ans. 1.63, − 0.1474, − 0.1645]


4.41 Supersonic air stream at Mach 2.2 and 55 kPa is turned by a con-tinuous compression corner by 10◦. Determine the Mach number andpressure after the corner.

[Ans. 1.83, 97.76 kPa]

4.42 Mach 2.8 air stream at a pressure of 101 kPa is turned by anexpansion fan through 4◦. Compute the Mach number and pressureof the turned flow by (a) using Prandtl-Meyer expansion theory and (b)using small-perturbation theory.

[Ans. (a) 3.0, 74.65 kPa, (b) 3.03, 71.36 kPa]


4.43 Air at Mach 2.82 and at a pressure of 50 kPa flows along a smoothconvex wall which gradually turns the flow by 9◦. Calculate the Machnumber and pressure of the flow downstream of the wall.

[Ans. 3.3, 24.5 kPa]

4.44 Air flow at Mach 2.8 passes over a 16◦ compression corner. Whatshould be the angle of expansion corner required to bring back the flowto Mach 2.8?

[Ans. 17.73◦]


4.45 A Laval nozzle of area ratio 1.4 operating under underexpandedcondition exhibits a wave pattern as shown in Figure P4.45. If the stag-nation pressure of the air is 490 kPa and the backpressure is 25 kPa,calculate the Mach number and the angle φ, assuming γ = 1.4.

Jet boundary

e

Air

5◦

φ

M1

Figure P4.45

[Ans. 2.59, 68.67◦]


4.46 A flat plate aerofoil of 1.4 m chord is placed at an angle of attack of4◦ in Mach 2 air stream at 101 kPa. Calculate CL and CD experiencedby the aerofoil.

[Ans. 0.1616, 0.0113]


4.47 A symmetric wedge is kept in a supersonic stream of Mach num-ber 2.5 and pressure 70 kPa, as shown in Figure P4.47. Estimate thepressure at point A.

(b)

A

(a)

10◦10◦ A

Figure P4.47

[Ans. (a) 96.33 kPa, (b) 597.1 kPa]


4.48 Calculate the lift and drag coefficient experienced by a flat platekept at an angle of attack of 5◦ to an air stream at Mach 2.3 and pres-sure 101 kPa, using (a) shock-expansion theory and (b) Ackeret’s the-ory.

[Ans. (a) 0.1735, 0.0152, (b) 0.1685, 0.0147]

4.49 Calculate the CL and CD for a half-wedge of wedge angle 5◦ keptin an air stream at Mach 2 and 101 kPa at (a) 0◦ angle of attack, (b) at3◦ angle of attack.

[Ans. (a) CL = − 0.054, CD = 0.00497, (b) CL = 0.1778, CD =0.01554]


4.50 Mach 3 air at 101 kPa and −30◦C flows over a symmetrical wedgeof semi-vertex angle 5◦, leading to the generation of oblique shocks.The right-running shock impinges on a wall which “turns away from theflow” by 5◦ exactly at the point where the oblique shock wave impingeson it, i.e. the wall has a 5◦ sharp convex corner at the point where theincident shock impinges on it. Sketch the flow pattern. If the leadingedge of the wedge is 1 m above the wall how far behind this leadingedge would the change in wall angle has to occur?

[Ans. 2.34 m]


applied gas dynamics oblique shock and expansion … gas dynamics oblique shock and expansion waves...

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