Supersonic Wind Tunnels

Return to the road map given in Figure 10.3. The material for the left and right branches is covered in Sections 10.3 and 10.4, respectively. In turn, a mating of these two branches gives birth to the fundamental aspects of supersonic wind tunnels, to be discussed in this section.

Imagine that you want to create a Mach 2.5 uniform flow in a laboratory for the purpose of testing a model of a supersonic vehicle, say, a cone. How do you do it? Clearly, we need a convergent-divergent nozzle with an area ratio Ae/A* = 2.637 (see Appendix A). Moreover, we need to establish a pressure ratio, pa/pe — 17.09, across the nozzle in order to obtain a shock-free expansion to Me = 2.5 at the exit. Your first thought might be to exhaust the nozzle directly into the laboratory, as sketched in Figure 10.16. Here, the Mach 2.5 flow passes into the surroundings as a “free jet.’’ The test model is placed in the flow downstream of the nozzle exit. In order to make certain that the free jet does not have shock or expansion waves, the nozzle exit pressure pe must equal the back pressure pB, as originally sketched in Figure 10.14c. Since the back pressure is simply that of the atmosphere surrounding the free jet, pa — pe = 1 atm. Consequently, to establish the proper isentropic expansion through the nozzle, you need a high-pressure reservoir with po — 17.09 atm at the inlet to the nozzle. In this manner, you would be able to accomplish your objective, namely, to produce a uniform stream of air at Mach 2.5 in order to test a supersonic model, as sketched in Figure 10.16.

Figure 10.l 6 Nozzle exhausting directly to the atmosphere.

In the above example, you may have a problem obtaining the high-pressure air supply at 17.09 atm. You need an air compressor or a bank of high-pressure air bottles—both of which can be expensive. It requires work, hence money, to create reservoirs of high-pressure air—the higher the pressure, the more the cost. So, can you accomplish your objective in a more efficient way, at less cost? The answer is yes, as follows. Instead of the free jet as sketched in Figure 10.16, imagine that you have a long constant-area section downstream of the nozzle exit, with a normal shock wave standing at the end of the constant-area section; this is shown in Figure 10.17. The pressure downstream of the normal shock wave is p2 = Рв = 1 atm. At M = 2.5, the static pressure ratio across the normal shock is рг/ре = 7.125. Hence, the pressure upstream of the normal shock is 0.14 atm. Since the flow is uniform in the constant-area section, this pressure is also equal to the nozzle exit pressure; that is, pe = 0.14 atm. Thus, in order to obtain the proper isentropic flow through the nozzle, which requires a pressure ratio of po/Pe — 17.09, we need a reservoir with a pressure of only 2.4 atm. This is considerably more efficient than the 17.09 atm required in Figure 10.16. Hence, we have created a uniform Mach 2.5 flow (in the constant-area duct) at a considerable reduction in cost compared with the scheme in Figure 10.16.

In Figure 10.17, the normal shock wave is acting as a diffuser, slowing the air originally at Mach 2.5 to the subsonic value of Mach 0.513 immediately behind the shock. Hence, by the addition of this “diffuser,” we can more efficiently produce our uniform Mach 2.5 flow. This illustrates one of the functions of a diffuser. However, the “normal shock diffuser” sketched in Figure 10.17 has several problems;

1. A normal shock is the strongest possible shock, hence creating the largest total pressure loss. If we could replace the normal shock in Figure 10.17 with a weaker shock, the total pressure loss would be less, and the required reservoir pressure po would be less than 2.4 atm.

2. It is extremely difficult to hold a normal shock wave stationary at the duct exit; in real life, flow unsteadiness and instabilities would cause the shock to move somewhere else and to fluctuate constantly in position. Thus, we could never be certain about the quality of the flow in the constant-area duct.

Nozzle exit Normal shock Figure 1 0*1 7 Nozzle exhausting into a constant-area duct, where a normal shock stands at the exit of the duct.

3. As soon as a test model is introduced into the constant-area section, the oblique

waves from the model would propagate downstream, causing the flow to become

two – or three-dimensional. The normal shock sketched in Figure 10.17 could not

exist in such a flow.

Hence, let us replace the normal shock in Figure 10.17 with the oblique shock diffuser shown in Figure 10.15/?. The resulting duct would appear as sketched in Figure 10.18. Examine this figure closely. We have a convergent-divergent nozzle feeding a uniform supersonic flow into the constant-area duct, which is called the test section. This flow is subsequently slowed to a low subsonic speed by means of a diffuser. This arrangement—namely, a convergent-divergent nozzle, a test section, and a convergent-divergent diffuser—is a supersonic wind tunnel. A test model, the cone in Figure 10.18, is placed in the test section, where aerodynamic measurements such as lift, drag, and pressure distribution are made. The wave system from the model propagates downstream and interacts with the multireflected shocks in the diffuser. The pressure ratio required to run the supersonic tunnel is ро/рв – This can be obtained by making po large via a high-pressure reservoir at the inlet to the nozzle or by making pB small via a vacuum source at the exit of the diffuser, or a combination of both.

The main source of total pressure loss in a supersonic wind tunnel is the diffuser. How does the oblique shock diffuser in Figure 10.18 compare with the hypothetical normal shock diffuser in Figure 10.17? Is the total pressure loss across all the reflected oblique shocks in Figure 10.18 greater or less than across the single normal shock wave in Figure 10.17? This is an important question, since the smaller the total pressure loss in the diffuser, the smaller is the pressure ratio ро/рв required to run the supersonic tunnel. There is no pat answer to this question. However, it is usually true that progressively reducing the velocity of a supersonic flow through a series of oblique shocks to a low supersonic value, and then further reducing the flow to subsonic speeds across a weak normal shock, results in a smaller total pressure loss than simply reducing the flow to subsonic speeds across a single, strong normal shock wave at the initially high supersonic Mach number. This trend is illustrated by Example 9.4. Therefore, the oblique shock diffuser shown in Figures 10.15/? and 10.18 is usually more efficient than the simple normal shock diffuser shown in Figure 10.17. This is not always true, however, because in an actual real-life oblique shock diffuser, the

shock waves interact with the boundary layers on the walls, causing local thickening and even possible separation of the boundary layers. This creates an additional total pressure loss. Moreover, the simple aspect of skin friction exerted on the surface generates a total pressure loss. Hence, actual oblique shock diffusers may have efficiencies greater or less than a hypothetical normal shock diffuser. Nevertheless, virtually all supersonic wind tunnels use oblique shock diffusers qualitatively similar to that shown in Figure 10.18.

Notice that the supersonic wind tunnel shown in Figure 10.18 has two throats: the nozzle throat with area At i is called the first throat, and the diffuser throat with area A, 2 is called the second throat. The mass flow through the nozzle can be expressed as m — pu A evaluated at the first throat. This station is denoted as station 1 in Figure 10.18, and hence the mass flow through the nozzle is mi — pxuAt = p*a*Att j. In turn, the mass flow through the diffuser can be expressed as m = pu A evaluated at station 2, namely, m2 = p2u2A,2- For steady flow through the wind tunnel, m = rhj. Hence,

раА, л = р2и2А, л

Since the thermodynamic state of the gas is irreversibly changed in going through the shock waves created by the test model and generated in the diffuser, clearly p2 and possibly u2 are different from p* and a*, respectively. Hence, from Equation (10.33), the second throat must have a different area from the first throat; that is, Лі2 ф A, |.

Question: How does A, 2 differ from A, ? Let us assume that sonic flow occurs at both stations 1 and 2 in Figure 10.18. Thus, Equation (10.33) can be written as

A,,2 = /Offlf Af, 1 P2a2

Recall from Section 8.4 that a* is constant for an adiabatic flow. Also, recall that the flow across shock waves is adiabatic (but not isentropic). Hence, the flow throughout the wind tunnel sketched in Figure 10.18 is adiabatic, and therefore a* = a*2. In turn, Equation (10.34) becomes

At,2 _

At, і P2

Recall from Section 8.4 that T* is also constant throughout the adiabatic flow of a calorically perfect gas. Hence, from the equation of state,

РІ = p*/RT* = p p2 p2/RTf P2

Substituting Equation (10.36) into (10.35), we have

At,2 _ p

Аг, 1 P2

From Equation (8.45), we have

Substituting the above into Equation (10.37), we obtain

Examining Figure 10.18, the total pressure always decreases across shock waves; therefore, po,2 < Po. i – In turn, from Equation (10.38), A,2 > А, л. Thus, the second throat must always be larger than the first throat. Only in the case of an ideal isentropic diffuser, where p0 — constant, would A,2 = A, and we have already discussed the impossibility of such an ideal diffuser.

Equation (10.38) is a useful relation to size the second throat relative to the first throat if we know the total pressure ratio across the tunnel. In the absence of such information, for the preliminary design of supersonic wind tunnels, the total pressure ratio across a normal shock is assumed.

For a given wind tunnel, if At 2 is less than the value given by Equation (10.38), the diffuser will “choke”; that is, the diffuser cannot pass the mass flow coming from the isentropic, supersonic expansion through the nozzle. In this case, nature adjusts the flow through the wind tunnel by creating shock waves in the nozzle, which in turn reduce the Mach number in the test section, producing weaker shocks in the diffuser with an attendant overall reduction in the total pressure loss; that is, nature adjusts the total pressure loss such that po. i/Au = Рол/Рв satisfies Equation (10.38). Sometimes this adjustment is so severe that a normal shock stands inside the nozzle, and the flow through the test section and diffuser is totally subsonic. Obviously, this choked situation is not desirable because we no longer have uniform flow at the desired Mach number in the test section. In such a case, the supersonic wind tunnel is said to be unstarted. The only way to rectify this situation is to make Al2/A, a large enough so that the diffuser can pass the mass flow from the isentropic expansion in the nozzle, that is, so that Equation (10.38) is satisfied along with a shock-free isentropic nozzle expansion.

As a general concluding comment, the basic concepts and relations discussed in this chapter are not limited to nozzles, diffusers, and supersonic wind tunnels. Rather, we have been discussing quasi-one-dimensional flow, which can be applied in many applications involving flow in a duct. For example, inlets on jet engines, which diffuse the flow to lower speeds before entering the engine compressor, obey the same principles. Also, a rocket engine is basically a supersonic nozzle designed to optimize the thrust from the expanded jet. The applications of the ideas presented in this chapter are numerous, and you should make certain that you understand these ideas before progressing further.

In Section 1.2, we subdivided aerodynamics into external and internal flows. You are reminded that the material in this chapter deals exclusively with internal flows.

Example 10.4 I For the preliminary design of a Mach 2 supersonic wind tunnel, calculate the ratio of the diffuser throat area to the nozzle throat area.

Solution

Assuming a normal shock wave at the entrance of the diffuser (for starting), from Appendix B, po. i/Poj = 0.7209 for M = 2.0. Hence, from Equation (10.38),

d^ = m = _L_ =| 1.387

At, і Рол 0.7209 —’——-