R. von Mises - Mathematical theory of compressible fluid flow (798534), страница 57
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On theother hand, construction of a compression wave along the bend involvesother Cauchy data, namely, data compatible with a compression w a v e ;corresponding to these different boundary data, there is then again one solution. If, as seen in a preceding example, a compression wave turns out to beimpossible, this simply means that one cannot prescribe initial data compatible with that shape of bend and a compression wave, just as—toquote a trivial instance—one cannot ''prescribe'' a varying value of 0 alonga straight wall.
If, finally, qi and the shape of the bend are such that bothan expansion and a compression wave are possible, it means (far from contradicting the uniqueness theorem) merely that of two quite different setsof Cauchy data along XABY, one furnishes an expansion wave, the other acompression wave. And, these lead to two different constant states.(c) Flow turning in a concave bend (see Fig. 117). Here 0ι — 0 = Δ 0 < 0.W e start with two numerical examples.1.
L e t Mi = 1.22; hence= 55.9°, Q = 94°. L e t 0i = 0°, 0 = 10°,so that Δ 0 = 0! — 0 = —10° (Fig. 117.1). Consider a forward wave (straightC~-lines inclined upstream). Then from Qi — 0 = Q — 02 , Q2 = 94° +10° = 104°, from which follow M = 1.57, a = 39.6°. This is an expansion wave which can turn the flow in the desired direction, since Q =Qi + 02 - 0i < 220.45°. In this example, i.e., for this bend and M = 1.22,a backward wave cannot be constructed since Qi + Δ 0 = 94° — 10° < 90°.2a ix22X2222x2. N o w take M i =1.57,a i=39.6°, Qi =104° (Fig.
117.2a). Here(1)F I G . 117. Flows in the same concave bend: ( 1 ) Forward expansion wave, onlysolution; (2a) Forward expansion w a v e ; (2b) Backward compression wave.18.4 M O R E E L A B O R A T EQ, a =waveto Q22305EXAMPLESBi = Q - 02, with 0! = 0° and 0 = 10°, gives Q = 114°, M = 1.92,31.3°, a forward wave. In the case shown in Fig. 117.2b a backwardcan also be constructed, the reverse of the first forward wave, leading= 94°, a = 55.9°, M = 1.21 through a compression wave.222222I t is seen that in both these cases the straight M a c h lines converge.Figure 118 shows this schematically. For the compression wave and the C(Fig.
118a), both 0 and a increase, while in the expansion case (Fig. 118b)the angle 180° — a, which the downstream-inclined C~ makes with the positive direction of the bend likewise increases. Hence in both cases thestraight Mach lines converge; they will, in general, have an envelope, anda simple wave solution exists only in a certain neighborhood of the bend,such that the envelope lies outside this neighborhood.
T h e extent of sucha neighborhood depends on the ratio of Δ 0 to the increment in the arc length,i.e., upon the curvature of the bend; in particular, no simple wave solutionexists in the case of a sharp concave corner. This case can be dealt with bymeans of a solution involving a shock (see Art. 22). Figure 118c shows suchan envelope or limit line, <£, in the case of a smooth concave bend; the limitline is cusped. Through the point Ρ passes one straight Mach line, and onit velocity, pressure, etc., are uniquely determined. Through the point Q,however, pass in general three straight Mach lines, each defining a different" f l o w " — a situation with no physical reality.+( d ) Successive simple waves.
Consider a combination of the cases considered in the two preceding subsections. Assume, e.g., a contour of theshape indicated in the first Fig. 119. T w o regions of constant state XA andDY are joined by a contour which has first positive, then negative, thenagain positive curvature along AB, BC, and CD, respectively. Assume qjknown. Again we have incomplete Cauchy data: both q and 0 are givenalong ΧΑ, but 0 alone is known along ABCDY. W e can link the regions ofconstant state I and I I by three simple waves. T h e two possible solutions ofthis kind are shown in the hodograph: (a) 12, 213, 31'.compression wave,expansion wave, compression w a v e ; (b) 14, 415, 51: expansion, compression,expansion.
T h e end velocity α is in both cases equal to qi .π4. More elaborate examples involving simple wavesW e consider here several problems of two-dimensional ducts™ A t the endof Sec. 16.4 we saw that the supersonic flow in a duct is determinedif the velocity q along an entrance section AB and the form of the twowalls is given. Carrying out the required steps (see Fig. 91) for an arbitraryvelocity distribution q along A Β and arbitrary shapes of walls is difficult,and in general approximations will be needed. However, under certain circumstances a solution involving simple waves exists, and this simplifiesthe problem.306IV.PLANESTEADYPOTENTIALFLOW(a) Uniform entrance velocity. General case. Consider a duct whoseare straight and parallel at the entrance, then begin to bend (see Fig.A t the entrance a uniform velocity q is given; hence XADAXwillregion of uniform velocity, where AD and AD are straight M a c h0walls120).be alinesF I G .
118. Convergent Mach lines in concave bend: (a) Backward compressionwave; (b) Forward expansion wave; (c) Envelope with cusp—three straight characteristics through Q, one through P.18.4 M O R E E L A B O R A T E307EXAMPLESF I G .
119. Successive simple waves in physical plane, and hodograph plane.whose slopes are given, since q = qx = q . Adjacent to the straight characteristic AD is a simple forward wave with straight Mach lines C~~, andadjacent to AD a backward wave with straight Mach lines C . The crosscharacteristics Ci and CD issuing from D are both curved; they intersectthe walls at the points Ε and Ε respectively.
Assume first that the wallsare still curved at these points. Since we know q at D> we know the constant2ξ of the forward wave ADE, and since 0 is known along the streamlineAE, the g-values follow there. Therefore we know q along each of thestraight C~ in ADE, and this determines Ct (see end of Sec. 16.4). Acorresponding graphical procedure would be the following.
W e proceed fromD in continuation of the direction AD to the intersection with a straightC~ close to D; at this point of intersection we again know q and 0, hence thedirection of a short segment of the line, and we can go on. Similar considerations hold for the wave ADE.A0+penetrationbackwardwaveF I G . 120. Duct with uniform entrance v e l o c i t y : general case.308IV. P L A N E S T E A D Y P O T E N T I A LFLOWN o w if, as assumed, the walls at Ε and Ε are curved, there are in general no other simple waves, and we are again faced with the general problem.Knowledge of q (and compatible Θ) along the characteristics ED, ED determines the flow in the characteristic quadrangle EDEF, the "penetration zone".
N e x t , the flow may be found in the triangular region limitedby EF, by the characteristic which continues EF, and by a piece of theupper wall along which θ is given, and similarly for the lower wall. This isthe type of " m i x e d " problem discussed in Arts. 10 and 16 where values aregiven along two arcs, one of which is characteristic. A s the next step we haveto solve again a characteristic boundary-value problem, etc.(b) Short bend between straight walls.
T h e situation becomes simpler if,after a short bend starting at A, the wall again becomes rectilinear, so thatΕ and Ε lie again on straight walls (see Fig. 121). U p to ADA there isuniform flow K ; then, as before, two simple waves follow, the forwardwave W and the backward wave Wj · Again the cross-characteristics DCand DC can be found. So far it is the same as before. N o w , however, sincethe wall beyond Β is straight and the C"-characteristic BC is also straight,0AmaximumcircleF I G . 121.
Short bend bet ween straight walls in physical plane, and hodograph plane.18.4 M O R E E L A B O R A T E309EXAMPLESa region of constant flow is determined in the triangle BCE, where CE isa straight C of known direction. Everything is similar in BCE. Therefollows, next, a penetration zone, CDCF, where CF and CF are characteristics. I n this zone there is general flow, determined by the characteristicboundary-value problem. N o w , however, adjacent to the triangle BCE ofconstant state, there is again a simple wave, W , with straight C -lines,and similarly there is a forward wave Wc adjacent to CE.
Since q and Θare known at C, we find the constant 2η of W by Q + θ = 2η.++ccc0However the wave W with the cross-characteristic CF is not completelyknown, unless the penetration problem has been solved. Nevertheless thevelocity q at F is known without the penetration problem being solved,as can be seen by considering the hodograph (Fig. 121). First the vectorqo = O'D' is located. From D' start the epicycloids D'C and D'Climitedby the radii O'C and O'C' which are parallel to BE and BE, respectively;these arcs D'C and U C' are images of the waves W and Wx , and thepoints C, C' represent the constant flows in BCE and BCE. T h e imagesof the waves W and Wc are the epicycloidal arcs CF' and C'F', whichintersect at F', and determine the vector O'F= q3.c3Ac1In W the correspondence between θ and q is known but this is notenough.
If, however, the shapes of the characteristics CF and CF areknown (after the penetration problem has been solved), this determinesthe velocity distribution along CF, which is a cross-characteristic for W .Then, the velocity distribution everywhere in the " q u a d r i l a t e r a l "Wis known, in particular on the cross-characteristic issuing from Ε whichlimits W . A t F there again begins a uniform flow, etc.
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