R. von Mises - Mathematical theory of compressible fluid flow (798534), страница 56
Текст из файла (страница 56)
T h e simplest type of solution compatible with these conditions is one in which the flow along AYis also uniform and the two states of uniform flow, I and I I , are linked bya single centered simple wave I I I . W i t h these data there exist in principle,as we shall see, two different simple wave solutions I I I , one a forward waveand one a backward wave, joining I and I I . T h e numerical data, however,may be such that in a particular case only one solution, or perhaps none,exists.22This can easily be seen both arithmetically and geometrically. Since qiand 0i are known along XA, in particular at A, and a scale factor a or qor q is given, Mi and ai = arc sin 1/Mi are known, and Q(qi) follows.
T obe specific we consider a backward wave. There is at A a C i which makesthe angle « i with XA. N e x t , Q(qi) + 0i = 2η determines η, and this holdsthroughout the wave.On the other hand, the angle a , valid along A Y, depends upon q which8m+22t18.3 E X A M P L E S O F S I M P L E299WAVESwe do not know; but if a backward wave I I I joins the two regions I and I I ,thenQi + 0i = Q +202 ,and, for the convex cornerΔ0 = 0! - 0 > 0,2where Δ0 denotes the deflection angle.
Since, according to Table I V , Q isalways between 90° and 220.45°, we find the conditionQ = Qi + Δ0 < 220.45°.2Similarly, in the case of a forward wave, Q — 0i = Q — 02 must hold, and,thereforexQ = Qi -2Δ0 > 90°.2If both inequalities are satisfied, we have two different solutions, which weshall characterize presently. It is now clear that the given data may alsobe such that only one, or neither of these solutions exists.In geometrical terms, from a point P i with coordinates (qi ,0i) in theepicycloidal ring one can reach a point with given 0 < 0i (i.e., on a givenradius) by moving along the Γ~ through P i until the radius in question isreached, provided this happens before being stopped at the maximumcircle. Or one can move along the Γ through Pi until the radius in question is reached, provided this happens before meeting the sonic circle.2+Numerical illustrations (for κ = 1.4, h = \/6)· The above inequalitiesmay be visualized by plotting in a A0,M-plane the curves Q = 90° + Δ0and Q = ft X 90° - Δ0 which intersect for Δ0 « 65°, Μ « 3.94.
Thesecurves together with the line Μ = 1 and the two asymptotes delimit fourregions corresponding to the above-mentioned four possibilities (Fig. 112).1. Assume, e.g., 0i = 0°, and Δ0 = 0 - 0 = 38°, M = 1.5. Fromtrigonometric tables or from Table V , « i = 41.8°, σ = 24.6°, Qi = hai += 2η = 101.9° = - 3 8 ° + Q ; thus, Q = 139.9°, which gives a = 19.5°,Μ2 = 3.
Hence everything is determined. The streamlines in the angularspace between Ci" and Ci are given explicitly by Eq. (9) with 2η = φι =101.9°. Here q and Μ increase through the wave while ρ, ρ, T, and a decrease. Such a wave is called an expansion wave (see Fig. 113.1). For thesenumerical data, Qi — Δ0 = 101.9° — 38° < 90°; hence, no second solutionexists. In fact, the point P i with coordinates Δ0 = 38°, Μ = 1.5 lies, inFig. 112, in the region where only a backward wave exists.2.
Consider the same corner, 0i = 0°, 0 = —38°, but now let M = 3, sothat ai = 19.5°. Then, σι = 49.1°, Qi = « ι + hai = 139.9° (equal to the28X2xχα ι2222x300IV.PLANESTEADYPOTENTIALFLOW65Μ432(52 0 4 060Ϊ00120780I30«(h-I)90Δ0 (degrees)F I G . 112. Regions of two, one or no simple wave solutions.previous Q ).
(a) Q - ΑΘ = 139.9° - 38° = 101.9° = Q ; M = 1.5,a = 41.8°. This forward wave is the reverse of the backward wave considered before. Here Q and Μ decrease through the wave, while pressure,density, and Mach angle increase through this compression wave (see Fig.113.2a). (b) For these same data: θι = 0°, Mi = 3, (αϊ = 19.5°,Qi = 139.9°), ΑΘ = 38°, a second solution, an expansion wave, also existssince Q + ΑΘ = 139.9° + 38° = 177.9° = Q < 220.5°. To this Q corresponds an Μ greater than 6, M = 6.4, with an a close to 9° (see Fig. 113.2b).In this case, therefore, we have two different simple wave solutions of the typeunder consideration (see in Fig. 112, the point P with coordinates ΑΘ = 38°,Μ = 3). More data are necessary to determine which of these two solutionswill materialize.2x222l222223.
Next, if M = 3, ΑΘ = 85°, then & - ΑΘ = 139.9° - 85° < 90°, andfor these initial data no compression-wave solution exists. Also Qi + ΑΘ =139.9° + 85° = 224.9° > 220.5°. Hence no expansion-wave solution existseither. In fact, the point P in Fig.
112 with ΑΘ = 85°, Μ = 3 is in theright-hand region where no solution exists. If we use, as on previous occasions, the term cavitation or vacuum for a state with ρ = ρ = 0, q = q ,we can say that a zone of cavitation will exist between the Ct which makesthe angle - 8 0 . 6 ° ( = 139.9° - 220.5°) with the z-axis and the given secondwall which makes the angle —85° with the rc-axis.x3m(b) Flow around a convex bend. The boundary is now a convex polygon ora continuous bend. W e consider first the latter case.
Neither presents anessentially new problem compared to case (a). Again, we assume the oncoming flow as uniform. At the point where the bend begins <?i ,0i aregiven, as well as a scale constant, a or a , of the oncoming uniform flow.The bend extends between two straight walls XA and BY (see Fig. 114),8t18.3 E X A M P L E S O F S I M P L EWAVES301and the geometric shape of the whole wall XABYis given; since this wallis a streamline, 0 is given along it. Again 2η = Q(qi) + 0i = Q f e ) + 02determines q at Β for an expansion-wave solution, provided Q(qi) + Δ0 <220.45°, where Δ0 = θι - 0 > 0 ° ; likewise if Q(qi) - Δ0 > 90°, a compression wave can link the two constant states.
T h e first Mach line, say Ctin the case of the expansion wave, is determined by the angle αχ = a(qi)which it makes with XA at A. T h e last, Ct, makes the angle a =a(q )with BY. A t each intermediate point of the bend the angle 0 is known and,always by means of 2η = 0 + Q(q), the corresponding q, M, and a follow.2222If we prefer a geometric procedure, then in an epicycloidal diagram constructed as in Art. 16 we identify the radius of the inner circle with a(thus determining the scale); next, we look in the diagram for the pointA' with the given polar coordinates qifli . Through A' goes a Γο~ in thedirection of decreasing 0 toward the maximum circle, and a Γο" (not shownin figure) in the direction of increasing 0 toward the sonic circle.
Considerfirst the Γο~ . I n order to find graphically the a for an arbitrary point Ρ ofthe bend, i.e., the direction of Ct, we draw the tangent to the bend attF I G . 113. Simple waves turning the same convex corner: (1) Mi = 1.5, backwardexpansion wave, only solution.
(2a) Mi = 3, forward compression waves; (2b) Μ = 3,backward expansion wave.302IV. P L A N E S T E A D Y P O T E N T I A LFLOWΡ and a radius parallel to that tangent through 0', the origin in the hodograph plane. This parallel intersects ΤΌ~ at a point P and q is equal to thefdistance O'P'.PThe tangent at P' to Γο~ is perpendicular to the directionof Ct (since Ct170-±In exactly the same way a compression solution may be constructed usingthe Γο" through A'. Of course, it can be decided beforehand by the abovementioned criteria whether two solutions exist or one or none.
W e note thatin both cases the straight Mach lines diverge: as we proceed from A toward B, the angle θ decreases. In the first case, the C -lines point fromthe bend downstream, and the angle a which they make with the positive(downstream) direction of the bend decreases (see Fig. 115a); in the secondcase (Fig. 115b), the C~-lines make with the negative (up-stream) directionan increasing angle a, and hence, with the positive direction a decreasingangle.The approach is the same for a convex polygon. If the oncoming uniformflow is completely given, several solutions may exist for which ABCD (seeFig. 116) is a streamline and which are such that three different uniformstates, Κι, K , K are linked by centered waves, W\ and W .It is now easy to realize the connection and complete agreement betweenour present results and the uniqueness theorems for hyperbolic boundaryvalue problems (Arts.
10 and 16). Let us reason in terms of the continuousbend (Fig. 114), where the position is perhaps clearer than for the polygon or+2z2YF I G . 114. Simple wave turning a bend.18.3 E X A M P L E S O F S I M P L EWAVES303corner. Consider, e.g., a case with t w o solutions. T h e situation is the following.
A s long as we know only the shape of the streamline XABYand the qion XA, we do not have Cauchy data; to have Cauchy data we must know qalong the whole noncharacteristic curve XABY. This may be prescribed inmany ways. T h e type of solution we are considering here, which consists ofa simple wave joining t w o constant states, is just one such solution, thougha particularly simple one.
Thus, if we decide to look for a solution which isa simple wave along AB, in particular an expansion wave, we must giveq along AB in a way compatible with an expansion wave. In other words,we have to prescribe q along AB so that we are led from the given qi, 0ito the given 0 along a characteristic Γί". This is actually achieved b y theabove-explained construction, which provides q all along XABY.
Thus, ina sense, the simple-wave construction serves to supplement the boundary2YF I G . 115. T w o possibilities of simple wave between constant states for the samebend: (a) Backward expansion wave; ( b ) Forward compression w a v e .F I G . 116. Flow along a convex polygon.304IV.PLANESTEADYPOTENTIALFLOWdata for the Cauchy problem; the situation is obscured by the fact thatin a simple wave the distribution of q everywhere in the wave region followsin an immediate way from the distribution along the curve AB.
![](https://s.studizba.com/z.php?f=/templates/si/i/noavatar.png&w=100&h=100&t=1"){kind=avatar}
