In this section, we move to the left-hand branch of the road map given in Figure 10.3; that is, we study in detail the compressible flow through nozzles. To expedite this study, we first derive an important equation which relates Mach number to the ratio of duct area to sonic throat area.
Consider the duct shown in Figure 10.9. Assume that sonic flow exists at the throat, where the area is A*. The Mach number and the velocity at the throat are denoted by M* and u*, respectively. Since the flow is sonic at the throat, M* = 1 and u* = a*. (Note that the use of an asterisk to denote sonic conditions was introduced in Section 7.5; we continue this convention in our present discussion.) At any other section of this duct, the area, the Mach number, and the velocity are denoted by A, M, and m, respectively, as shown in Figure 10.9. Writing Equation (10.1) between A and A*, we have
p*u*A* = puA [10.26]
Since u* = a*, Equation (10.26) becomes
A p* a* p* po a* . _
A* p и po p и
where po is the stagnation density defined in Section 7.5 and is constant throughout
an isentropic flow. From Equation (8.46), we have
p*_ _ ( 2
Po VX + 1 /
Also, from Equation (8.43), we have
Also, recalling the definition of M* in Section 8.4, as well as Equation (8.48), we have
[10.30]
Squaring Equation (10.27) and substituting Equations (10.28) to (10.30), we obtain
AV = (_2_V(y~’) (l + ^_AM22,{У~П 1 + [(X – l)/2M2 A*) Vx + 1/ V 2 ) [(y+)/2]M2
[10.31]
Algebraically simplifying Equation (10.31), we have
Equation (10.32) is very important; it is called the area-Mach number relation, and it contains a striking result. “Turned inside out,” Equation (10.32) tells us that M = f (A / A*); that is, the Mach number at any location in the duct is a function of the ratio of the local duct area to the sonic throat area. Recall from our discussion of Equation (10.25) that A must be greater than or at least equal to A*; the case where A < A* is
physically not possible in an isentropic flow. Thus, in Equation (10.32), A/A* > 1. Also, Equation (10.32) yields two solutions for M at a given А/A*—a subsonic value and a supersonic value. Which value of M that actually holds in a given case depends on the pressures at the inlet and exit of the duct, as explained later. The results for А/A* as a function of M, obtained from Equation (10.32), are tabulated in Appendix A. Examining Appendix A, we note that for subsonic values of M, as M increases, AI A* decreases (i. e., the duct converges). At M = 1, A / A* = 1 in Appendix A. Finally, for supersonic values of M, as M increases, А/A* increases (i. e., the duct diverges). These trends in Appendix A are consistent with our physical discussion of convergent-divergent ducts at the end of Section 10.2. Moreover, Appendix A shows the double-valued nature of M as a function of A/A*. For example, for А/A* = 2, we have either M = 0.31 or M = 2.2.
Consider a given convergent-divergent nozzle, as sketched in Figure 10.10a. Assume that the area ratio at the inlet A,-/A* is very large and that the flow at the inlet is fed from a large gas reservoir where the gas is essentially stationary. The reservoir pressure and temperature are po and 7b, respectively. Since A,-/A* is very large, the subsonic Mach number at the inlet is very small, M ~ 0. Thus, the pressure and temperature at the inlet are essentially po and 7o, respectively. The area distribution of the nozzle A = A(x) is specified, so that A/ A* is known at every station along the nozzle. The area of the throat is denoted by A,, and the exit area is denoted by Ae. The Mach number and static pressure at the exit are denoted by Me and pe, respectively. Assume that we have an isentropic expansion of the gas through this nozzle to a supersonic Mach number Me = Me^ at the exit (the reason for the subscript 6 will be apparent later). The corresponding exit pressure is pe>6. For this expansion, the flow is sonic at the throat; hence, M = 1 and A, = A* at the throat. The flow properties through the nozzle are a function of the local area ratio A/A* and are obtained as follows:
1. The local Mach number as a function of x is obtained from Equation (10.32), or more directly from the tabulated values in Appendix A. For the specified A — A(x), we know the corresponding A/A* = fix). Then read the related subsonic Mach numbers in the convergent portion of the nozzle from the first part of Appendix A (for M < 1) and the related supersonic Mach numbers in the divergent portion of the nozzle from the second part of Appendix A (for M > 1). The Mach number distribution through the complete nozzle is thus obtained and is sketched in Figure 10.10Й.
2. Once the Mach number distribution is known, then the corresponding variation of temperature, pressure, and density can be found from Equations (8.40), (8.42), and (8.43), respectively, or more directly from Appendix A. The distributions of p/po and T/Tq are sketched in Figure 10.10c and d, respectively.
Examine the variations shown in Figure 10.10. For the isentropic expansion of a gas through a convergent-divergent nozzle, the Mach number monotonically increases from near 0 at the inlet to M = 1 at the throat, and to the supersonic value Mefi at the exit. The pressure monotonically decreases from p0 at the inlet to 0.528/?o at the throat and to the lower value рем at the exit. Similarly, the temperature monotonically
decreases from T0 at the inlet to 0.8337b at the throat and to the lower value 7′(,.f) at the exit. Again, for the isentropic flow shown in Figure 10.10, we emphasize that the distribution of M, and hence the resulting distributions of p and /’, through the nozzle depends only on the local area ratio A/A*. This is the key to the analysis of isentropic, supersonic, quasi-one-dimensional nozzle flows.
Imagine that you take a convergent-divergent nozzle, and simply place it on a table in front of you. What is going to happen? Is the air going to suddenly start flowing through the nozzle of its own accord? The answer is, of course not! Rather, by this stage in your study of aerodynamics, your intuition should tell you that we have to impose a force on the gas in order to produce any acceleration. Indeed, this is the essence of the momentum equation derived in Section 2.5. For the inviscid flows
considered here, the only mechanism to produce an accelerating force on a gas is a pressure gradient. Thus, returning to the nozzle on the table, a pressure difference must be created between the inlet and exit; only then will the gas start to flow through the nozzle. The exit pressure must be less than the inlet pressure; that is, pe < p0. Moreover, if we wish to produce the isentropic supersonic flow sketched in Figure
10.10, the pressure pjpo must be precisely the value stipulated by Appendix A for the known exit Mach number Me g; that is, pe/po = Ре, бІРо■ If the pressure ratio is different from the above isentropic value, the flow either inside or outside the nozzle will be different from that shown in Figure 10.10.
Let us examine the type of nozzle flows that occur when pe/po is not equal to the precise isentropic value for Me b, that is, when ре/Po ф Ре, б/Ро – To begin with, consider the convergent-divergent nozzle sketched in Figure 10.1 la. If pe = po, no pressure difference exists, and no flow occurs inside the nozzle. Now assume that pe is minutely reduced below pa, say, pe = 0.999p0. This small pressure difference will produce a very low-speed subsonic flow inside the nozzle—essentially a gentle wind. The local Mach number will increase slightly through the convergent portion, reaching a maximum value at the throat, as shown by curve 1 in Figure 10.1 li>. This Mach number at the throat will not be sonic; rather, it will be some small subsonic value. Downstream of the throat, the local Mach number will decrease in the divergent section, reaching a very small but finite value M,,, і at the exit. Correspondingly, the pressure in the convergent section will gradually decrease from p0 at the inlet to a minimum value at the throat, and then will gradually increase to the value pe, i at the exit. This variation is shown as curve 1 in Figure 10.1 lc. Please note that because the flow is not sonic at the throat in this case, At is not equal to A*. Recall that A*, which appears in Equation (10.32), is the sonic throat area. In the case of purely subsonic flow through a convergent-divergent nozzle, A* takes on the character of a reference area; it is not the same as the actual geometric area of the nozzle throat At. Rather, A* is the area the flow in Figure 10.11 would have if it were somehow accelerated to sonic velocity. If this did happen, the flow area would have to be decreased further than shown in Figure 10.1 la. Hence, for a purely subsonic flow A, > A*.
Assume that we further decrease the exit pressure in Figure 10.11, say, to the value pe = pe,2- The flow is now illustrated by the curves labeled 2 in Figure 10.11. The flow moves faster through the nozzle, and the maximum Mach number at the throat increases but remains less than 1. Now, let us reduce pe to the value pe = pe, з, such that the flow just reaches sonic conditions at the throat. This is shown by curve 3 in Figure 10.11. The throat Mach number is 1, and the throat pressure is 0.528p^. The flow downstream of the throat is subsonic.
Upon comparing Figures 10.10 and 10.11, we are struck by an important physical difference. For a given nozzle shape, there is only one allowable isentropic flow solution for the supersonic case shown in Figure 10.10. In contrast, there are an infinite number of possible isentropic subsonic solutions, each one corresponding to some value of pe, where po > pe > Ре. з – Only three solutions of this infinite set of solutions are sketched in Figure 10.11. Hence, the key factors for the analysis of purely subsonic flow in a convergent-divergent nozzle are both А/A* and pe/Po-
Consider the mass flow through the convergent-divergent nozzle in Figure 10.11. As the exit pressure is decreased, the flow velocity in the throat increases; hence, the
Figure 10.11 Isentropic subsonic nozzle flow.
|
mass flow increases. The mass flow can be calculated by evaluating Equation (10.1) at the throat; that is, m — p, utAt. As pe decreases, u, increases and p, decreases. However, the percentage increase in ut is much greater than the decrease in p,. As a result, m increases, as sketched in Figure 10.12. When pe — pe^, sonic flow is achieved at the throat, and m = p*u*A* = p*u*A,. Now, if pe is further reduced below pe>3, the conditions at the throat take on a new behavior; they remain unchanged. From our discussion in Section 10.2, the Mach number at the throat cannot exceed 1; hence, as pe is further reduced, M will remain equal to 1 at the throat. Consequently, the mass flow will remain constant as pe is reduced below pe 3, as shown in Figure 10.12. In a sense, the flow at the throat, as well as upstream of the throat, becomes “frozen.” Once the flow becomes sonic at the throat, disturbances cannot work their way upstream of the throat. Hence, the flow in the convergent section of the nozzle no longer communicates with the exit pressure and has no way of knowing that the
Figure 10.12 Variation of mass flow with exit
pressure; illustration of choked flow.
|
exit pressure is continuing to decrease. This situation—when the flow goes sonic at the throat, and the mass flow remains constant no matter how low pe is reduced—is called choked flow. It is a vital aspect of the compressible flow through ducts, and we consider it further in our subsequent discussions.
Return to the subsonic nozzle flows sketched in Figure 10.11. Question: What happens in the duct when pe is reduced below p,..fl In the convergent portion, as described above, nothing happens. The flow properties remain fixed at the conditions shown by curve 3 in the convergent section of the duct (the left side of Figure 10.11 b and c). However, a lot happens in the divergent section of the duct. As the exit pressure is reduced below pe 3, a region of supersonic flow appears downstream of the throat. However, the exit pressure is too high to allow an isentropic supersonic flow throughout the entire divergent section. Instead, for pe less than pe,3 but substantially higher than the fully isentropic value pe$ (see Figure 10.10c), a normal shock wave is formed downstream of the throat. This situation is sketched in Figure 10.13.
In Figure 10.13, the exit pressure has been reduced to pe^, where рсл < pe,3, but where pe 4 is also substantially higher than pe 6. Here we observe a normal shock wave standing inside the nozzle at a distance d downstream of the throat. Between the throat and the normal shock wave, the flow is given by the supersonic isentropic solution, as shown in Figure 10.13Z? and c. Behind the shock wave, the flow is subsonic. This subsonic flow sees the divergent duct and isentropically slows down further as it moves to the exit. Correspondingly, the pressure experiences a discontinuous increase across the shock wave and then is further increased as the flow slows down toward the exit. The flow on both the left and right sides of the shock wave is isentropic; however, the entropy increases across the shock wave. Hence, the flow on the left side of the shock wave is isentropic with one value of entropy, V|, and the flow on the right side of the shock wave is isentropic with another value of entropy S2, where 52 > ■Si – The location of the shock wave inside the nozzle, given by d in Figure 10.13a, is determined by the requirement that the increase in static pressure across the wave plus that in the divergent portion of the subsonic flow behind the shock be just right to achieve pe 4 at the exit. As pe is further reduced, the normal
Normal shock wave
Figure 10.13 Su personic nozzle flow with a normal shock inside the nozzle.
|
shock wave moves downstream, closer to the nozzle exit. At a certain value of exit pressure, pe — pe 5, the normal shock stands precisely at the exit. This is sketched in Figure 10.14a to c. At this stage, when pe — ре $, the flow through the entire nozzle, except precisely at the exit, is isentropic.
To this stage in our discussion, we have dealt with pe, which is the pressure right at the nozzle exit. In Figures 10.10, 10.11, 10.13, and 10.14a to c, we have not been concerned with the flow downstream of the nozzle exit. Now imagine that the nozzle in Figure 10.14a exhausts directly into a region of surrounding gas downstream of the exit. These surroundings could be, for example, the atmosphere. In any case, the pressure of the surroundings downstream of the exit is defined as the back pressure, denoted by pB. When the flow at the nozzle exit is subsonic, the exit pressure must equal the back pressure, pe = pB, because a pressure discontinuity cannot be maintained in a steady subsonic flow. That is, when the exit flow is subsonic, the surrounding back pressure is impressed on the exit flow. Hence, in Figure 10.II, Рв = Pe. і for curve 1, pB — Pe.2 for curve 2, and pB — pej for curve 3. For the same reason, pB = pe_4 in Figure 10.13, and pB = pe^ in Figure 10.14. Hence, in discussing these figures, instead of stating that we reduced the exit pressure pe and
observed the consequences, we could just as well have stated that we reduced the back pressure pB. It would have amounted to the same thing.
For the remainder of our discussion in this section, let us now imagine that we have control over pв and that we are going to continue to decrease pB. Consider the case when the back pressure is reduced below pe 5. Whence < pB < pe 5, the back pressure is still above the isentropic pressure at the nozzle exit. Hence, in flowing out to the surroundings, the jet of gas from the nozzle must somehow be compressed such that its pressure is compatible with pB. This compression takes place across oblique shock waves attached to the exit, as shown in Figure 10.14d. When pB is reduced to the value such that pB = pee, there is no mismatch of the exit pressure and the back pressure; the nozzle jet exhausts smoothly into the surroundings without passing through any waves. This is shown in Figure 10.14-е. Finally, as pB is reduced below pe (, the jet of gas from the nozzle must expand further in order to match the lower back pressure. This expansion takes place across centered expansion waves attached to the exit, as shown in Figure 10.14/.
When the situation in Figure 10. 14й? exists, the nozzle is said to be overexpanded, because the pressure at the exit has expanded below the back pressure, p,,^ < pB.
That is, the nozzle expansion has gone too far, and the jet must pass through oblique shocks in order to come back up to the higher back pressure. Conversely, when the situation in Figure 10.14/ exists, the nozzle is said to be underexpanded, because the exit pressure is higher than the back pressure, pe (, > pB, and hence the flow is capable of additional expansion after leaving the nozzle.
Surveying Figures 10.10 through 10.14, note that the purely isentropic supersonic flow originally illustrated in Figure 10.10 exists throughout the nozzle for all cases when pB < Pe, 5- For example, in Figure 10.14a, the isentropic supersonic flow solution holds throughout the nozzle except right at the exit, where a normal shock exists. In Figure 10.14d to /, the flow through the entire nozzle, including at the exit plane, is given by the isentropic supersonic flow solution.
Keep in mind that our entire discussion of nozzle flows in this section is predicated on having a duct of given shape. We assume that A = A(x) is prescribed. When this is the case, the quasi-one-dimensional theory of this chapter gives a reasonable prediction of the flow inside the duct, where the results are interpreted as mean properties averaged over each cross section. This theory does not tell us how to design the contour of the nozzle. In reality, if the walls of the nozzle are not curved just right, then oblique shocks occur inside the nozzle. To obtain the proper contour for a supersonic nozzle so that it produces isentropic shock-free flow inside the nozzle, we must account for the three-dimensionality of the actual flow. This is one purpose of the method of characteristics, a technique for analyzing two – and three-dimensional supersonic flow. A brief introduction to the method of characteristics is given in Chapter 13.
Consider the isentropic supersonic flow through a convergent-divergent nozzle with an exit- | Example 1 0.1 to-throat area ratio of 10.25. The reservoir pressure and temperature are 5 atm and 600°R, respectively. Calculate M, p, and T at the nozzle exit.
Solution
From the supersonic portion of Appendix A, for AJ A* = 10.25,
Also,
Thus,
Te = 0.24277o = 0.2427(600) = 145.6°R
Consider the isentropic flow through a convergent-divergent nozzle with an exit-to-throat area | Example 10.2 ratio of 2. The reservoir pressure and temperature are 1 atm and 288 K, respectively. Calculate the Mach number, pressure, and temperature at both the throat and the exit for the cases where
Te 1
Te = —To = ———— (288) =
To 1.968
(b) At the throat, the flow is still sonic. Hence, from above, M, = 1.0, p, = 0.528 atm, and T, = 240 K. However, at all other locations in the nozzle, the flow is subsonic. At the exit, where Ae/A* = 2, from the subsonic portion of Appendix A,
Te 1
Te = —T0 =————- (288) =
T0 1.018
From the subsonic portion of Appendix A, for p0/p,: = 1.028, we have
– = — — = 0.5(2.964) = 1.482 * AeA*
From the subsonic portion of Appendix A, for A,/A* = 1.482, we have