R. von Mises - Mathematical theory of compressible fluid flow (798534), страница 40
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N o w (2c) is a direct relation between theu- and p/p-values, i.e., between the u- and ^-values. If (2b) is divided by mand m replaced by p\(ui — c) or p (u — c) from (2a), a second such relationfollows. W e see therefore that there are two equations in the five variablesU\, Vi, u , v , and c. In other words, there is a simple infinity of pairs u ,vcorresponding to a given pair ui,vi , or: The end points of a shock transitionfrom a fixed initial point (ui,vi) lie on a curve in the speedgraph plane. W eshall call this the shock curve.In order to find the curve explicitly it is necessary to eliminate c, andthis is conveniently done if we start from the shock conditions in the formof Eqs. (21), which read, when expressed in terms of u and v,u2222227+2j7 — 1-(uic)(u-22c)(31)n2/=(ui-c), 7 — 12x2/+ — - — Vi = {u2-c), 7 — 1+ —-—2v.2T h e second equality is linear in c; if this is solved for c and the result substituted into the first equality, the condition on u ,v may be transformed22into(32)(vi-v)2 2-2(u2-u )V +xV2 ) 2-(^1 ) 2(u2-UiY = 0,210III.
ONE-DIMENSIONALFLOWThis is a second-order relation between the two variables v '/νχ(u — Ui) /v\\ these variables can be given parametrically by2and22(32')vi+21 '2yh2v+1'where the parameter η is the pressure ratio introduced in (23) [see the derivation of (33) and (34) below]. In graphing the relation (32) we do notconsider, of course, negative values of v. W e also disregard values of vless than vi, since v ^ Vi by (30').
T h e result is the shock diagram, Fig. 77,giving a graphical representation of all possible shock transitions. Eachcurve has a 90° corner at the point (u vi)and consists of all points (1^2,^2)which are possible end points of a shock transition starting from the staterepresented by the corner point.
T h e right-hand branch, u > u , appliesif c > Ui, and the left, u < u , if c < u (see p. 202). Each curve is symmetric with respect to the vertical line through the corner and all curvesare asymptotic to the two lines through (ν,ιβ) whose slopes, measuredfrom the u-axis, are az\^2y/(y — 1) = d b \ / l + Λ · T h e dimensionlessparameters of the shock, p /pi = η, p /pi = ξ, M[, and Mi have constant22u22xxx222u(u2;>ou;<ouFIG.
77. Shock diagram. T h e curve with corner at (ui , vi) consists of all points, v ) which are possible end points of a shock transition starting from ( w i , v\).214.6RIEMANNvalues on any ray through (u 0).211PROBLEMIn fact, if the slope of such a line is λ =hv /(u2 — Ui), then2(33)I= ^V^-= ft-2= ( I T - ) ' (ξ -I) ^12l)Mf,2Λvv\ 2/where (30) and M = w /a have been used to obtain the last equality.Using (24) and (25) we then find22222,22=y-h !j -12tt -Ι[2 ΜΊ7=22-Ί)τ-2(γ (7-2y=1)][(1 ) W7η(η +h)Gi -I)Ι2-1)Μ Γ +-I)22]W=(1 -2ΜΓ) *2T h e values of Afi for various rays through (u ,0) are indicated in theshock diagram, Fig.
77.xI t is worth noting that the curves in the shock diagram have a secondorder contact with the ± 4 5 ° lines through the corner point. N o w theselines are the images of particle lines in corresponding simple waves. Thuswe see that, for a weak shock (all ratios close to 1), the state variables of aparticle undergo approximately the same kind of changes as they do in a simplewave.A l l possible shock transitions are obtained from the shock diagram ofFig. 77 by displacing it horizontally.506. Example of a shock phenomenon. The Riemann problem51W e consider a straight tube of length Α Β = Ζ, closed at both ends, andassume that at t = 0 the fluid in the tube has constant pressure p , anddensity p , and a uniform velocity Uo directed from A to B.
If we take asthe origin the position of A at t = 0, the boundary conditions, namely,00(35a)u =uo,ν =v0at t = 0,0 < χ < I,are realized if before time t = 0 the tube with the enclosed fluid mass hasbeen moving (with velocity Uo) as one rigid body, and then the tube issuddenly stopped at t = 0. T h e stopping introduces the boundary conditions(35b)u = 0at χ = 0andχ =Ifor all t.N o w the conditions (35a) determine a unique and continuous solutionat each point within the characteristic triangle ABC, where AC and BChave slopes u + a and u — α , respectively (the slopes being measuredfrom the ί-axis with O o , the sound velocity,*corresponding to v or to p ,000000III.212O N E - D I M E N S I O N A LFLOWtL\F I G .
78. Riemann's problem.po). This solution is represented in Fig. 78 by the set of parallel particlelines of slope u . T h e figure is drawn under the assumption that u is super00sonic, so that u =b a are both positive. In this case it is immediately evi00dent that this continuous solution is inconsistent with the boundary conditions at χ = I. Thus a flow pattern which includes a shock line must besought.In the speedgraph, Fig. 79, the point P0with coordinatesUQ,VQrepresents the whole region of the physical plane in which u and ν remain constant.
So long as the laws of continuous motion hold, an adjacent regioncan only be mapped into a curve passing through P , and we saw in Sec.013.1 that such a curve must necessarily be a characteristic, and the corresponding flow a simple wave. T h e + 4 5 ° line P Pi0abscissa uending at a point with«= 0 and ordinate v < v supplies in fact an adequate solutionxx0for the region adjoining A C on the left. Figure 79 shows, in the χ,ί-plane,νFIG.
79. Solution of the Riemann problem up to the time of interaction of shockand simple wave.14.6a centered simple wavecharacteristics AC213RIEMANN PROBLEM(rarefaction wave) with center A between the(parallel to MC)and AD(parallel to MD').construction see Sec. 12.5.) All particles arrive at the line AD(For thewith zero velocity. Here the curved particle lines turn into straight lines parallel to the/-axis.
In mechanical terms: a steadily increasing region of fluid at restestablishes itself at the left end of the tube.N o similar, continuous solution exists for the right side of the tube. Herethe transition from velocity u to zero velocity takes place by means of a0shock. T o find the end point Pof the shock transition we have to draw2the shock curve with corner at Ρ ο up to the point where it meets the y-axis.T h e shock front BE in the physical plane is, by definition, the line of slope cmeasured from the ί-axis.
Since the shock diagram gives the value of M[\u .—0=c \/ao for each point, and Uo and do are known, we can compute c andthen plot the shock line BE along which the particle velocities jump fromu to 0. T h e solution up to the time to, the ordinate of E0yis shown in Fig.79. Here Ε is the point of intersection of the shock line and the loweststraight characteristic of the simple wave centered at A.T h e solutionconsists of four parts: fluid at rest to the left of AD and to the right of BE,uniform flow with u = u , ν = v within the triangle AEB,00centered simple wave between A Ε andand finally theAD.Analytically, the four regions are determined as follows. First, the simple wave extends from the ray of slope u +0a = u + (y — l)*>o/2, along00which the velocity is u , to the ray of slope0a i= (-y\) /2Vl=(y -Uo)/2l)(v:Qalong which the velocity is zero.* T h a t vi = vo — u may be seen by comput0ing the ordinate of P i in Fig.
79. Accordingly, the equation of the particlelines [see Eq. (13.8)] is(36)Second, the two unknown parameters of the shock transition, the slope cof BE and the ordinate v of P22, can be found from the two equations (31),where u and 0 are the values of Ui and u , respectively. Choosing the nega02tive value of c we find from these equationsc(37)* We assume UQ < VQ . Otherwise, a cavity forms at the left end of the tube (see alsoend of Sec.
13.3.)214III.ONE-DIMENSIONALFLOWνtuθF I G . 80. T h e only possible continuous change to zero velocity.where Mo = Uo/do. Finally, the ordinate of the point Ε in Fig. 79, thatis, the time to up to which the solution has been constructed, is the ordinateof the point of intersection of AC: χ = (uo + ao)t and BE: χ = I + ct> andis to = l/(uo + ao — c ) .A l l this holds whether M is greater or smaller than 1, since it is also truein the subsonic case, when the point C of Fig.
78 falls between the verticallines through A and B, that no continuous solution exists which satisfiesthe condition u = 0 at χ = I for all t. In fact, if there were a continuoussolution to the right of BC, it would have to be a simple wave since itwould be adjacent to the. region of constant u v within the triangle ABC.T h e map of this simple wave in the speedgraph plane, Fig. 80, must lieon the —45° line through P in the direction of decreasing u, tangent at Ρ οto the shock curve P0P2 of the preceding figure. This corresponds to caseI V of Fig. 67, that of a backward compression wave. But we have seen thatin this case the straight characteristics in the χ,^-plane converge to the left,so that the particle lines have the general shape shown in the first Fig. 80.T h a t is, the only possible continuous change precludes from the start avertical particle line through B.0y0Article 15Further Shock Problems1.
Behavior of a shock at the end of a tube or a wall (shock reflection)52In this section we discuss another example, on the basis of the theorydeveloped in the preceding article. Supplements and modifications of thetheory will be introduced below.Assume that the tube in which the flow takes place is closed at its rightend, and that a shock front moves toward this end with velocity c. Thisshock is represented in the z,£-plane (Fig. 81) by the straight line AB15.1SHOCK215REFLECTIONwhose slope is c measured from the ί-axis. If we suppose the fluid in thetube to be at rest before the shock reaches it, the particle lines below ABare vertical straight lines which continue above A Β in some other direction(see the example of a shock considered in Sec.
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