**9.1 ****Normal Shock Waves**

*Shock waves *are large-amplitude waves that travel in a gas. They emanate from the wings of a supersonic aircraft, from a large explosion, from a jet engine, and ahead of the projectile in a gun barrel. They can be oblique waves or normal waves. First, we will consider the normal shock wave, as shown in Fig. 9.7. In this ﬁgure it is stationary so that a steady ﬂow exists. If *V*1 were superimposed to the left, the shock

**F****i****g****u****re 9.7 **A stationary shock wave.

would be traveling in stagnant air with velocity *V *and the *induced velocity *behindthe shock wave would be (*V _{1}*–

*V*

_{2}). The shock wave is very thin, on the order of 10

^{−4}mm, and in that short distance large pressure changes occur causing enormous energy dissipation. The continuity equation with

*A*

_{1}=*A*

_{2}is

where the areas have divided out since *A*1 = *A*2. Assuming that the three quantities *r*1, *p*1, and *V*1 before the shock wave are known, the above three equations allow us to solve for three unknowns *r*2, *p*2, and *V*2 since, for a given gas, *k *is known.

Rather than solve the above three equations simultaneously, we write them in terms of the Mach numbers M and M , and put them in more convenient forms.

First, the momentum equation (9.37), using Eq. (9.35) and *V *2 = M2 *pk */ρ, can be written as

In like manner, the energy equation (9.36), with *p *= ρ*R**T *and *V *2 = M2 *kR**T *, can be written as

If the pressure and temperature ratios from Eqs (9.38) and (9.39) are substituted into Eq. (9.40) the downstream Mach number is related to the upstream Mach num- ber by (the algebra to show this is complicated)

For air, the preceding equations simplify to

Several normal shock-ﬂow relations for air have been presented in Table D.2. The use of that table allows one to avoid using Eq. (9.44). In addition, the ratio *p*02 /*p*01 of the stagnation point pressures in front of and behind the shock wave are listed.

**F****i****g****u****re 9.8 **Flοw with shοck waves in a nοzzle.

Return to the converging-diverging nozzle and focus attention on the ﬂow below curve *C *of Fig. 9.6. If the receiver pressure decreases to *p */*p *= *a *in Fig. 9.8, a normal *r *0 shock wave would be positioned somewhere inside the nozzle as shown. If the receiver pressure decreased still further, there would be some ratio *p */*p *= *b *that would posi- *r *0 tion the shock wave at the exit plane of the nozzle. Pressure ratios *c *and *d *would result in oblique shock-wave patterns similar to those shown. Pressure ratio *e *is associated with isentropic ﬂow throughout, and pressure ratio *f *would provide an exit pressure greater than the receiver pressure resulting in a billowing out, as shown, of the exiting ﬂow, as seen on the rockets that propel satellites into space.

__EXAMPLE 9.3__

A normal shock wave travels at 600 m/s through stagnant 20°C air. Estimate the velocity induced behind the shock wave. (a) Use the equations and (b) use the normal shock-ﬂow table D.2. Refer to Fig. 9.7.

**Solution**

Superimpose a velocity of 600 m/s so that the shock wave is stationary and *V*1 = 600 m/s, as displayed in Fig. 9.7. The upstream Mach number is

(a) Using the equations, the downstream Mach number and temperature are, respectively,

Air ﬂows from a reservoir maintained at 20°C and 200 kPa absolute through a converging-diverging nozzle with a throat diameter of 6 cm and an exit diameter of 12 cm to a receiver. What receiver pressure is needed to locate a shock wave at a position where the diameter is 10 cm? Refer to Fig. 9.8.

Let’s use the isentropic-ﬂow table D.1 and the normal shock-ﬂow table D.2. At the throat for this supersonic ﬂow M = 1. The Mach number just before the shock wave is interpolated from Table D.1 where *A*1 /*A** = 10 /6 = 2.778 to be

since the stagnation pressure does not change in the isentropic ﬂow before the shock wave so that *p*01 = 200 kPa. From just after the shock wave to the exit, isentropic ﬂow again exists so that from Table D.1 at M = 0.5078

We have introduced an imaginary throat between the shock wave and the exit of the nozzle. The exit area *A *is introduced by

Using Table D.1 at this area ratio (make sure the subsonic part of the table is used), we ﬁnd

using *p*0 *e *= *p*02 for the isentropic ﬂow after the shock wave. The exit pressure is equal to the receiver pressure for this isentropic subsonic ﬂow.

**9.2 ****Oblique Shock Waves**

Oblique shock waves form on the leading edge of a supersonic sharp-edged airfoil or in a corner, as shown in Fig. 9.9. A steady, uniform plane ﬂow exists before and after the shock wave. The oblique shock waves also form on axisymmetric projectiles.

**F****i****g****u****re 9.9 **Οblique shοck waves: (*a*) ﬂοw οver a wedge and (*b*) ﬂοw in a cοrner.

The oblique shock wave turns the ﬂow so that V_{2} is parallel to the plane surface.

Another variable, the angle through which the ﬂow turns, is introduced but the additional tangential momentum equation allows a solution. Consider the control volume of Fig. 9.10 surrounding the oblique shock wave. The velocity vector **V _{1 }**is assumed to be in the

*x*-direction and the oblique shock wave turns the ﬂow through the

*wedge angle*or

*deﬂectio*

*n angle*

*q*so that

**V**2 is parallel to the wall

*.*The oblique shock wave makes an angle of

*b*with

**V**1. The components of the velocity vectors are shown normal and tangential to the oblique shock. The tangential components of the velocity vectors do not cause ﬂuid to ﬂow into or out of the control volume, so continuity providesThe pressure forces act normal to the control volume and produce no net force tan- gential to the oblique shock. This allows the tangential momentum equation to take the form

**F****i****g****u****re 9.10 **Oblique shock-wave control volume.

since the tangential velocity terms cancel.

Observe that the tangential velocity components do not enter the three Eqs. (9.45), (9.48), and (9.49). They are the same three equations used to solve the normal shock- wave problem. So, the components *V*1*n *and *V*2*n *can be replaced with *V*1 and *V*2, respectively, of the normal shock-wave problem and a solution obtained. Table D.2 may also

To often simplify a solution, we relate the oblique shock angle *b *to the deﬂection angle *q*. This is done by using Eq. (9.45) to obtain

With this relationship the oblique shock angle *b *can be found for a given incoming Mach number and wedge angle *q*. A plot of Eq. (9.52) is useful to avoid a trial-and- error solution. It is included as Fig. 9.11. Three observations can be made by studying the ﬁgure.

**F****i****g****u****re 9.11 **Οblique shοck wave angle *b *related tο wedge angle *q *and Mach

For given Mach number M and wedge angle *q *there are two possible oblique shock angles *b*. The larger one is the “strong” oblique shock wave and the smaller one is the “weak” oblique shock wave.

• For a given wedge angle *q *there is a minimum Mach number for which there is only one oblique shock angle *b*.

• If the Mach number is less than the minimum for a particular *q*, but greater than one, the shock wave is detached as shown in Fig. 9.12. Also, for a given M1 there is a sufﬁciently large *q *that will result in a detached

shock wave.

The required pressure rise determines if a weak shock or a strong shock exists. The pressure rise is determined by ﬂow conditions.

For a detached shock wave around a blunt body or a wedge, a normal shock wave exists on the stagnation streamline; the normal shock is followed by a strong oblique shock, then a weak oblique shock, and ﬁnally a Mach wave, as shown in Fig. 9.12. The shock wave is always detached on a blunt object.

**F****i****g****u****re 9.12 **Detached shock waves around (*a*) a plane, blunt object and (*b*) a wedge.

**EXAMPLE 9.5**

Air at 30°C ﬂows around a wedge with an included angle of 60° (see Fig. 9.9*a*). An oblique shock emanates from the wedge at an angle of 50°. Determine the approach velocity of the air. Also ﬁnd M and *T *.

**Solution**

From Table D.2 interpolation provides M_{2n} = 0.5176 so that

**9.3 ****Expansion Waves**

Supersonic ﬂow exits a nozzle (the pressure ratio *f *in Fig. 9.8), and billows out into a large exhaust plume. Also, supersonic ﬂow does not separate from the wall of a nozzle that expands quite rapidly, as shown in Fig. 9.8. How is this accomplished? Consider the possibility that a single ﬁnite wave, such as an oblique shock, is able to turn the ﬂow around the convex corner, as shown in Fig. 9.13*a*. From the tangential momentum equation, the tangential component of velocity must remain the same on both sides of the ﬁnite wave. For this to be true, *V*2 > *V*1. As before, this increase in velocity as the ﬂuid ﬂows through a ﬁnite wave requires an increase in entropy, a violation of the second law of thermodynamics, making a ﬁnite wave an impossibility.

A second possibility is to allow an inﬁnite fan of Mach waves, called an *expansion fan*, emanating from the corner, as shown in Fig. 9.13*b*. This is an ideal isentropic process so the second law is not violated; such a process may be approached in a real application. Let’s consider the single inﬁnitesimal Mach wave displayed in Fig. 9.14, apply our fundamental laws, and then integrate around the corner. Since the tangential velocity components are equal, the velocity triangles yield

**F****i****g****u****re 9.13 **Supersonic ﬂow around a convex corner.

The solution to this relationship is presented for air in Table D.3 to avoid a trial-and- error solution for M given the angle *q**. *If the pressure or temperature is desired, the isentropic ﬂow table can be used. The Mach waves that allow the gas to turn the corner are sometimes referred to as *expansion waves.*

Observe from Table D.3 that the expansion fan that turns the gas through the angle *q *results in M = 1 before the fan to a supersonic ﬂow after the fan. The gas

speeds up as it turns the corner and it does not separate. A slower moving subsonic ﬂow would separate from the corner and would slow down. If M = ∞ is substituted into Eq. (9.60), θ = 130.5°, which is the maximum angle through which the ﬂow could possibly turn. This shows that turning angles greater than 90° are possible, a rather surprising result.

__EXAMPLE 9.6__

Air at 150 kPa and 140°C ﬂows at M = 2 and turns a convex corner of 30°. Estimate

the Mach number, pressure, temperature, and velocity after the corner.

**Solution**

Table D.3 assumes the air is initially at M = 1. So, assume the ﬂow originates from M = 1 and turns a corner to M1 = 2 and then a second corner to M2, as shown. From Table D.3, an angle of 26.4° is required to accelerate the ﬂow from M = 1 to M = 2. Add another 30° to 26.4° and at *q *= 56.4° we ﬁnd that

M_{2} = 3.37

Using the isentropic ﬂow table D.1, the entries from the reservoir to state 1 and also to state 2 can be used to ﬁnd

__Quiz No. 1__

1. Two rocks are slammed together by a friend on one side of a lake. A listening device picks up the wave generated 0.45 s later. The distance across the lake is nearest

(A) 450 m

(B) 550 m

(C) 650 m

(D) 750 m

2. The Mach number for a projectile ﬂying at 10 000 m at 200 m/s is (A) 0.62

(B) 0.67

(C) 0.69

(D) 0.74

3. A supersonic aircraft passes 200 m overhead on a day when the temperature is 26°C. Estimate how far the aircraft is from you when you hear its sound if its Mach number is 1.68.

(A) 245 m

(B) 275 m

(C) 315 m

(D) 335 m

4. A converging nozzle with exit area of 10 cm2 is attached to a reservoir maintained at 250 kPa absolute and 20°C. If the receiver pressure is maintained at 150 kPa absolute, the mass ﬂux is nearest

(A) 0.584 kg/s

(B) 0.502 kg/s

(C) 0.428 kg/s

(D) 0.386 kg/s

5. Air ﬂows through a converging-diverging nozzle attached from a reservoir maintained at 400 kPa absolute and 20°C to a receiver. If the throat and exit diameters are 10 and 24 cm, respectively, the receiver pressure that will just result in supersonic ﬂow throughout is nearest

(A) 6.8 kPa

(B) 9.2 kPa

(C) 16.4 kPa

(D) 28.2 kPa

6. The temperature, pressure, and velocity before a normal shock wave in air are 18°C, 100 kPa absolute, and 600 m/s, respectively. The velocity after the shock wave is nearest

(A) 212 m/s

(B) 249 m/s

(C) 262 m/s

(D) 285 m/s

7. A large explosion occurs on the earth’s surface producing a shock wave that travels radially outward. At a particular location the Mach number of the wave is 2.0. Determine the induced velocity behind the shock wave if *T*1 = 15°C.

(A) 502 m/s

(B) 425 m/s

(C) 368 m/s

(D) 255 m/s

8. Air ﬂows from a reservoir through a nozzle into a receiver. The reservoir is maintained at 400 kPa absolute and 20°C. The nozzle has a 10-cm-diameter throat and a 20-cm-diameter exit. The receiver pressure needed to locate a shock wave at the exit is nearest

(A) 150 kPa

(B) 140 kPa

(C) 130 kPa

(D) 120 kPa

(A) 164 kPa

(B) 181 kPa

(C) 192 kPa

(D) 204 kPa

10. An airﬂow with a Mach number of 2.4 turns a convex corner of 40°. If the temperature and pressure are 5°C and 60 kPa absolute, respectively, the Mach number after the corner is nearest

(A) 5

(B) 3

(C) 2

(D) 1

__Quiz No. 2__

1. An underwater animal generates a signal that travels through water until it hits an object and then echoes back to the animal 0.46 s later. How far is the animal from the object?

2. A bolt of lightning lights up the sky and 1.5 s later you hear the thunder.

How far did the lightning strike from your position?

3. A supersonic aircraft passes 200 m overhead on a day when the temperature is 26°C. Estimate how far the aircraft is from you when you hear its sound if its Mach number is 3.49.

4. A small-amplitude wave travels through the 15°C atmosphere creating a pressure rise of 5 Pa. Estimate the temperature rise across the wave and the induced velocity behind the wave.

5. A converging nozzle with exit area of 10 cm2 is attached to a reservoir maintained at 250 kPa absolute and 20°C. Calculate the mass ﬂux if the receiver pressure is maintained at 100 kPa absolute.

6. Air ﬂows from a converging-diverging nozzle from a reservoir maintained at 400 kPa absolute and 20°C through a 12-cm-diameter throat. At what diameter in the diverging section will M = 2?

7. The temperature and pressure before a normal shock wave in air are 20°C and 400 kPa absolute, respectively. The Mach number after the shock wave is 0.5. Calculate the pressure and velocity after the shock.

8. Air ﬂows from a reservoir maintained at 400 kPa absolute and 20°C out a nozzle with a 10-cm-diameter throat and a 20-cm-diameter exit into a

receiver. Estimate the receiver pressure needed to locate a shock wave at a diameter of 16 cm.

9. A supersonic airﬂow changes direction 20° due to a sudden corner (see Fig. 9.9*b*). If *T*1 = 40°C, *p*1 = 60 kPa absolute, and *V*1 = 900 m/s, calculate *p *and *V* assuming a strong shock.

10. An airﬂow with M = 3.6 is desired by turning a 20°C-supersonic ﬂow with a Mach number of 1.8 around a convex corner. If the upstream pressure is 40 kPa absolute, what angle should the corner possess? What is the velocity after the corner?