R. von Mises - Mathematical theory of compressible fluid flow (798534), страница 75
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Boundary conditions for which no continuous solution exists.404V. I N T E G R A T I O N T H E O R Y A N DSHOCKSPi to the <? -axis. I t is clear, however, that if Pi lies too near to the point ofintersection, /, of the polar with the sonic circle (i.e., δ is too close to δ ) ,then the r -characteristic through Pi will not reach the g -axis and thepoint Pt will not exist. On the other hand the r~-characteristic through Piwill alwaj's intersect the g -axis. For it is easily proved that this propertyholds for any sonic or supersonic point on the upper half of the circularlimit polar A , corresponding to q —> q . Hence it holds for any sonic orsupersonic point lying above the </ -axis and within A .x0+8 θ ηzx0mxW e therefore select the arc PiP 2 of the r~-characteristic correspondingto the backward wave.
T h e first of Fig. 155 shows the centered rarefaction wave between the characteristic DS (perpendicular to the tangent atPi) and the characteristic DW (perpendicular to the tangent at P ). Allparticles cross DW horizontally and here the curved streamlines turn intostraight lines parallel to the x-axis.2In Sec. 22.3 we found that the angle between a nonzero shock and thestreamline behind it is smaller than the Mach angle downstream. Hencethe shock line AS, and the characteristic DS eventually intersect, as indicated in the figure.
Beyond the point S the solution outlined above is nolonger valid, and the character of the flow is different. Since such a changecan only take place across a characteristic, the flow pattern is bounded bythe cross-characteristic ST of the simple wave, and its rectilinear extensionTU.T h e problem of determining the solution beyond VSTUis more difficult.yo'F I G .
155. Supersonic flow along a partly inclined wall.23.2FLOW ALONG A P A R T I A L L Y I N C L I N E DWALL405I t can be anticipated that a curved extension of the shock line AS willappear, but neither its shape nor the flow pattern behind it can be expressedin terms of known functions. In such cases one has to resort to numericalmethods of integration based, for instance, on the use of characteristics ofthe equations of ideal (nonpotential) fluid flow and the shock conditions.T h e unknown shock line is determined step by step along w ith the flow oneither side. T h e extensive computations involved can be carried out on highspeed computing machines. ( T h e basic theory of such flows will be givenin Sees. 24.1 and 24.2; see also the discussion of Riemann's problem in Sees.T3315.4 and15.5.)T h e full solution in the region VSTUDAVis shown in Fig.
155. T h eregion is divided into four subregions with uniform horizontal flows in thetriangles AVS and DTU, uniform inclined flow in the triangle ASD, anda centered simple wave in SDT.Analytically the flow regions are determined as follows. T h e inclinationσ of the shock AS to the rr-axis is found by replacing the subscript 1 by 0in (22.32) and determining the. larger positive root r = cot σ of the cubicequation in r which results on setting e equal to tan δ . Once σ is found,the state 1 within the triangle ASD is determined from Eqs.
(22.21) and(22.22), with the subscripts 1, 2 replaced by 0, 1. Thus with M=Mo sin σ we compute M, pi/po , and pi/p . T h e Mach number Mi behindthe shock is then given by M i = Mcosec σ = Mcosec ( σ — δ ) .T h e simple wave extends from the line DS, making an angle arc sin ( 1 / M i )with the direction AD, to the line DW, making an angle arc sin ( 1 / M )with DU, where M is the M a c h number for which000000()n00lni nχ0ln022Q2=Qi +δ .0T h e equation of the streamlines within the wave [see (18.9)] isconstant;where r, φ are polar coordinates with origin at D. T h e state 2 in the triangle DTU is known once M has been determined.T h e distance of S from D is given by DS/sin σχ = //sin (αι — σ ι ) ; theequation of the cross-characteristic ST of the simple wave [see (18.11)] istherefore2(10)=(DS)2sinοίι +since φ is measured from the direction DU.δο — 2η fCOSοίι +δ0— 2ηhT h e distance DT is determined406V.
I N T E G R A T I O N T H E O R Y A N D S H O C K Sfrom (10) by setting φ = a and solving for r. Finally the line TU has slope2— arc sin ( 1 / M ) .2A s an example we consider the case M0= 2, δ =010°. In Sec. 22.7 wefound that for this Mach number the weak shock was inclined at an angleof 39°19'.
T h e corresponding value of M0 nwas 1.2671 and behind the shockwe had= 0.8032,MmPi =pi = 1.7066p ,01.4584p .0T h e corresponding value of M i was 1.6405.For this value of M i we find Qi =to which corresponds Mtion of DSto ADx8= 0.2215, p /p28=2is therefore 37°34', and that of DWFor these values of M i and Mp /p106.0581°, so that Q116.0581°,= 1.9884 (see Table V in Sec. 18.2). T h e inclina2to DUis 30°12'.we have from Eqs. (8.15) and (8.16):2= 0.1301, pi/p.
= 0.3407, and p /p2= 0.2330 (see8Table I in A r t . 8 ) . * Hence= 0.5875pi =p2p = 0.683921.0026p ,0= 0.9975po.PlThis example was used in constructing Fig. 155.I t should be noticed that M , p , p differ from M , po, p by well under222001 per cent. This is in agreement with the result obtained in the last sectionconcerning successive transitions through shocks with positive deflectionand backward waves. In the notation of that section 0 = 0 = 0, Δ = 10°.X2T h e above arguments hold whether δ is greater or smaller than a (pro0vided δ ^δ08 ο η0for the given M ) .
I t remains to be checked that even0when the point D lies below the Mach line AC in Fig. 154 (as it does inthe example given) no continuous solution exists which satisfies the boundary conditions ( 9 ) . T h e argument is completely analogous to that given atthe end of Sec. 14.6 for the corresponding one-dimensional case.3. Supersonic flow past a straight line profile: contact discontinuity34In this example the fluid is supposed to be moving horizontally acrossthe whole 2/-axis at constant pressure p and density p , and with supersonic speed q :000a = a,0q = q > a,QQon χ = 0 for all y.6 = 0T h e presence of the straight line profile ABin Fig.
156, of length I andinclination δ , introduces the additional boundary condition00 = δ >Οοony = χ tan δ ,0for0 < χ < I cos δ .0* This accuracy, which is not given by Tables I and V , is required for Sec. 3, wherethe present example is used.23.3FLOW PAST A STRAIGHT L I N E407PROFILEThis problem has features in common with the last one, and in particularit may be seen that there is no continuous solution on the upper side of theprofile. As before we assume that δ ύ δ η for the given M , and use astraight weak shock A Si to deflect the stream abruptly into the directionAB. Conditions before and after the shock are represented by hodographpoints P and P i , Fig. 156, with P P i an arc of the shock polar with cornerat P .
T h e deflection on the lower side of the profile is effected by meansof a simple wave centered at A , and in order not to violate the boundaryconditions on the y-axis, this must be a forward wave. T h e image of thiswave is the arc P 0 P 2 of the r - e p i c y c l o i d through P , and the lines AW2,AS* are perpendicular to the tangents of this epicycloid at P and P ,respectively. I t is of course assumed that δ is sufficiently small for theepicycloid to meet 0'P\ (see Sec. 18.3). T h e uniform flow on the lowerside of the profile is then represented by the point P .0080000+00202T h e flow above the profile may be joined to the flow below by asyF I G .
156. Supersonic flow about an inclined straight line profile.408V. I N T E G R A T I O N T H E O R Y A N D S H O C K Ssuming that a straight shock forms along an appropriate line BS and a centered backward wave SiBWi beyond BSi, the (^-characteristic through B.T h e uniform state of particles after passing through this simple wave isrepresented b y a point P lying on the r~-characteristic through P i , andthat after passing through the shock b y a point P lying on the shock polarwith corner at P . T h e two points P , Pa must satisfy t w o conditions: theymust lie on the same ray through the origin O', in order that the directionof motion of all particles be the same behond W BS , and secondly theirpressure values ρ must be the same. W e now restrict the discussion tocases where P lies on or outside the sonic circle, so that the flow behindthe shock BS is sonic or supersonic.
Under suitable conditions it is possibleto satisfy the t w o requirements above by choosing the inclination of theshock line BS and the extent of the wave SiBWi appropriately.23423X2422T h e values of the density, however, will not necessarily be the same forparticles coming from above and below the profile. Thus the dividingstreamline BC through Β will be a discontinuity line analogous to the onedescribed in Sec.
15.2 for the one-dimensional case. As there, it will be calleda contact discontinuity. I t is a characteristic of the ideal fluid equations(see Sec. 9.6).As we noted in Sec. 2, the shock line A Si meets the characteristic throughΒ above the profile. Similarly the characteristic AS intersects the shockline through Β below the profile.
Thus, as in the preceding example, the flowregions are limited, namely by the cross-characteristics $ιΤΊ , S T ° f thesimple waves, and their extensions TiU and S CU. T h e problem of determining the flow pattern outside the region V1S1T1UCS T Vis more difficult and as before numerical methods of integration must be employed.T h e flow patterns up to the states 1 and 2 may be computed as in thelast section. I t is then easy, in principle, to determine the rest of the flowpattern. L e t — δ ( t o be found) be the inclination of the straight dividingstreamline BC to the profile AB.
Then, according to (18.3) which governsthe transition through the backward wave, and (8.16):222222235δQz - Qi,(11)Moreover b y Eqs. (22.32) and (22.21), governing the shock transitionacross BS ,2δ(12)— arc tan(CT2+ d)r2(1 - cW + (1 - d)923.3FLOWPASTAS T R A I G H TL I N Ewhere r (negative) is the cotangent of the angle ABS .2Here c, d must be2computed from (22.31) with Mreplaced by Mx2409PROFILE.
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