Prandtl-Meyer Expansion Waves

Oblique shock waves, as discussed in Sections 9.2 to 9.5, occur when a supersonic flow is turned into itself (see again Figure 9.1a). In contrast, when a supersonic flow is turned away from itself, an expansion wave is formed, as sketched in Figure 9.1 b. Examine this figure carefully, and review the surrounding discussion in Section 9.1 before progressing further. The purpose of the present section is to develop a theory which allows us to calculate the changes in flow properties across such expansion waves. To this stage in our discussion of oblique waves, we have completed the left – hand branch of the road map in Figure 9.5. In this section, we cover the right-hand branch.

The expansion fan in Figure 9.1 b is a continuous expansion region which can be visualized as an infinite number of Mach waves, each making the Mach angle /г [see Equation (9.1)] with the local flow direction. As sketched in Figure 9.23, the expansion fan is bounded upstream by a Mach wave which makes the angle i with respect to the upstream flow, where /xi = arcsin(l/M]). The expansion fan is bounded downstream by another Mach wave which makes the angle /x2 with respect to the downstream flow, where fi2 = arcsin(l/M2). Since the expansion through the

Figure 9.33 Prandtl-Meyer expansion.

wave takes place across a continuous succession of Mach waves, and since ds = 0 for each Mach wave, the expansion is isentropic. This is in direct contrast to flow across an oblique shock, which always experiences an entropy increase. The fact that the flow through an expansion wave is isentropic is a greatly simplifying aspect, as we will soon appreciate.

An expansion wave emanating from a sharp convex corner as sketched in Figures 9ЛЬ and 9.23 is called a centered expansion wave. Ludwig Prandtl and his student Theodor Meyer first worked out a theory for centered expansion waves in 1907-1908, and hence such waves are commonly denoted as Prandtl-Meyer expansion waves.

The problem of an expansion wave is as follows: Referring to Figure 9.23, given the upstream flow (region 1) and the deflection angle 6, calculate the downstream flow (region 2). Let us proceed.

Consider a very weak wave produced by an infinitesimally small flow deflection d6 as sketched in Figure 9.24. We consider the limit of this picture as d6 -> 0; hence, the wave is essentially a Mach wave at the angle p, to the upstream flow. The velocity ahead of the wave is V. As the flow is deflected downward through the angle d9, the velocity is increased by the infinitesimal amount dV, and hence the flow velocity behind the wave is V +dV inclined at the angle d0. Recall from the treatment of the momentum equation in Section 9.2 that any change in velocity across a wave takes place normal to the wave; the tangential component is unchanged across the wave. In Figure 9.24, the horizontal line segment A В with length V is drawn behind the wave. Also, the line segment AC is drawn to represent the new velocity V + dV behind the wave. Then line SC is normal to the wave because it represents the line along which the change in velocity occurs. Examining the geometry in Figure 9.24, from the law of sines applied to triangle ABC, we see that

However, from trigonometric identities,

From Equation (9.1), we know that ц, = arcsin(l/M). Hence, the right triangle in Figure 9.25 demonstrates that

1

Vm2 -1

Substituting Equation (9.31) into (9.30), we obtain

Equation (9.32) relates the infinitesimal change in velocity dV to the infinitesimal deflection dQ across a wave of vanishing strength. In the precise limit of a Mach wave, of course dV and hence dQ are zero. In this sense, Equation (9.32) is an approximate equation for a finite dQ, but it becomes a true equality as dQ —0. Since the expansion fan illustrated in Figures 9.1 b and 9.23 is a region of an infinite number of Mach waves, Equation (9.32) is a differential equation which precisely describes the flow inside the expansion wave.

Return to Figure 9.23. Let us integrate Equation (9.32) from region 1, where the deflection angle is zero and the Mach number is Mt, to region 2, where the deflection angle is в and the Mach number is My

[9.33]

To carry out the integral on the right-hand side of Equation (9.33), dV/V must be obtained in terms of M, as follows. From the definition of Mach number, M = V/a, we have V = Ma, or

In V — In M + lna

Differentiating Equation (9.34), we obtain

dV _ dM da

~V ~ ~M~ + ~a

From Equations (8.25) and (8.40), we have

[9.36]

Solving Equation (9.36) for a, we obtain

Differentiating Equation (9.37), we obtain

substituting Equation (9.38) into (9.35), we have

dV _ 1 dM

~V~ ~ 1 + [(y – 1)/2]M2 ~M

Equation (9.39) is a relation for dV/V strictly in terms of M—this is precisely what is desired for the integral in Equation (9.33). Hence, substituting Equation (9.39) into (9.33), we have

is called the Prandtl-Meyer function, denoted by v. Carrying out the integration, Equation (9.41) becomes

The constant of integration that would ordinarily appear in Equation (9.42) is not important, because it drops out when Equation (9.42) is used for the definite integral in Equation (9.40). For convenience, it is chosen as zero, such that v(M) = 0 when M = 1. Finally, we can now write Equation (9.40), combined with (9.41), as

where v(M) is given by Equation (9.42) for a calorically perfect gas. The Prandtl – Meyer function v is very important; it is the key to the calculation of changes across an expansion wave. Because of its importance, v is tabulated as a function of M in Appendix C. For convenience, values of д are also tabulated in Appendix C.

How do the above results solve the problem stated in Figure 9.23; that is how can we obtain the properties in region 2 from the known properties in region 1 and the known deflection angle 91 The answer is straightforward:

1. For the given M1, obtain v{M) from Appendix C.

2. Calculate v(M2) from Equation (9.43), using the known 9 and the value of v{M ) obtained in step 1.

3. Obtain M2 from Appendix C corresponding to the value of v(M2) from step 2.

4. The expansion wave is isentropic; hence, p0 and 7b are constant through the wave. That is, Tq 2 = To, і and /?0,2 = Po. i – From Equation (8.40), we have

T2 T2/Tq2 _ 1 + [(7 — 1)/2]M)2

T, T/Tq 1 + [(y – )/2]Mj

From Equation (8.42), we have

Pi = P2/P0 = /1+[(к – 1)/2]А79Гу/(у Pi Pi/Pa \+[(y – Y)/2Ml)

Since we know both M and M2, as well as T and p, Equations (9.44) and (9.45) allow the calculation of T2 and p2 downstream of the expansion wave.

(a) Scramjet powered air-to-surface-missile concept

(/?) Scramjet powered strike/reconnaissance vehicle concept

(c) Scramjet powered space access vehicle concept Figure 9.26 Computer-generated images of possible future SCRAMjet-powered hypersonic vehicles. ICourtesy of James Weber, United States Air Force.)

Example 9.8 | In the preceding discussion on SCRAMjet engines, an isentropic compression wave was men­tioned as one of the possible compression mechanisms. Consider the isentropic compression surface sketched in Figure 9.32a. The Mach number and pressure upstream of the wave are M = 10 and pi = 1 atm, respectively. The flow is turned through a total angle of 15°. Calculate the Mach number and pressure in region 2 behind the compression wave.

Solution

From Appendix C, for M = 10, Ui = 102.3°. In Region 2,

u2 = У, – в = 102.3 – 15 = 87.3°

From Appendix C for v2 = 87.3°, we have (closest entry)

From Appendix A, for M = 10, рол/Рi = 0.4244 x 105 and for M2 = 6.4, po. i! Pi = 0.2355 x 104. Since the flow is isentropic, p0,2 = Рол, and hence

Consider the flow over a compression comer with the same upstream conditions of M{ = 10 and р = 1 atm as in Example 9.8, and the same turning angle of 15°, except in this case the comer is sharp and the compression takes place through an oblique shock wave as sketched in Figure 9.32b. Calculate the downstream Mach number, static pressure, and total pressure in region 2. Compare the results with those obtained in Example 9.8, and comment on the significance of the comparison.

Solution

From Figure 9.7 for M = 10 and в = 15°, the wave angle is /5 = 20°. The component of the upstream Mach number perpendicular to the wave is

M„,1 = Mi sin p = (10) sin 20° = 34.2

From Appendix В for M„,i = 3.42, we have (nearest entry), p2/P = 13.32, рол/Рол = 0.2322, and M„,2 = 0.4552. Hence

M„ 2 0.4552

M2 =——– —— =—————— = 5.22

sin(/3 — (9) sin(20 — 15) L _. J

Pi = (Pi/Pi)P = 13.32(1) =

The total pressure in region 1 can be obtained from Appendix A as follows. For M = 10, Poa/Pi = 0.4244 x 105. Hence, the total pressure in region 2 is

9.85 x 103 atm

Рол = (pi) = (0.2322X0.4244 x 105)(1) =

As a check, we can calculate p02 as follows. (This check also alerts us to the error incurred when we round to the nearest entry in the tables.) From Appendix A for M2 = 5.22, Рол! Рг = 0.6661 x 103 (nearest entry). Hence,

Po,2 = (P2) = (0.6661 x 103)(13.32) = 8.87 x 103 atm

Note this answer is 10 percent lower than that obtained above, which is simply due to our rounding to the nearest entry in the tables. The error incurred by taking the nearest entry is exacerbated by the very high Mach numbers in this example. Much better accuracy can be obtained by properly interpolating between table entries.

Comparing the results from this example and Example 9.8, we clearly see that the isentropic compression is a more efficient compression process, yielding a down­stream Mach number and pressure that are both considerably higher than in the case of the shock wave. The inefficiency of the shock wave is measured by the loss of total pressure across the shock; total pressure drops by about 77 percent across the shock. This emphasizes why designers of supersonic and hypersonic inlets would love to have the compression process carried out via isentropic compression waves. However, as noted in our discussion on SCRAMjets, it is very difficult to achieve such a compression in real life; the contour of the compression surface must be quite precise, and it is a point design for the given upsteam Mach number. At off-design Mach numbers, even the best-designed compression contour will result in shocks.