R. von Mises - Mathematical theory of compressible fluid flow (798534), страница 78

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T h e angle 0 is easilyeliminated between these last two equations to give a single equation for p.Once ρ has been determined, the second equation in (16) yields r i , and thesecond in (17) r ; the first in each pair will then give the common inclina­2tion 0 of the streamlines between CD and CE.I n addition the densityratios p /pi and p /p across the two reflected shocks can be determined from342(22.23), since the corresponding pressure ratios p/ρι and p/p2are known.Hence the two densities p and P4 below and above the discontinuity line CF3can be computed. Finally the M a c h numbers on either side of this line canbe found from (22.22).In practice, the common values of 0 and ρ in (26) and (27) are found in23.6I N T E R S E C T I O N OF T W O421SHOCKSthe same way as for the example discussed in Sec. 3. A first approximationto 0, which is accurate when all four shocks are very weak, is obtained byusing the results of Sec.

1. Thus although the theorem at the end of thatsection does not apply here (the transitions concerned being of differentkinds) we can adapt the theory to the present example.W e note that if F does not depend on 0 explicitly then neither does 'Fnor F', and these two derivatives differ only in sign: F = — F', see Eqs.(2) and ( 4 ) .

Consider the flow below C. Then for the transition of positivedeflection across the shock CA, we have according to Eq. ( 3 ) :fFi -(28)F=0-F^0i +^Fo0i2+O(0i ),3and for the second transition, of negative deflection, across the shock CD,Eq. ( 1 ) gives(29)= F[{fl - 0i) + \Fi{fl -Fz0 i ) + 0 ( 0 - 0O .23N o w in (28) we may replace F by F' and F " successively to obtainF[=F'o -Fi=Fo +Fo0i +O(0i ),2O(0i).Hence on substituting into (29) and adding (28) to the result we find(30)F3-F0=200 +-Fo(6±Fo(0 -20O+20(0 -θ,0 ) ,3X1and similarly(31)Ft -F=0-F'o(e-20 ) +2iFo(6-20 )22+0(0 -0 , 0 ) .223In particular we may take F = p.

Then on equating the right-hand sidesof (30) and (31) we see that for the pressure to be the same above and be­low the discontinuity line CF, the inclination of that line must satisfyθ - 20! = - (0 - 20 ) or2(32)0 =0i +0 ,2to the second order in the deflections 0 , 0 . Another way of stating thisresult is to say that the deflections caused by diagonally opposite shocks areapproximately equal.A s an example we take M = 3, 0 = 20° and 0 = — 1 0 ° with y = 7/5.Then in the usual way we find for weak shocks CA and CB: Mi = 1.9941,/p= 3.7713 and M = 2.5050, p /p = 2.0545.

W i t h the first approxi­mation 0 = 10° we obtain p/ρι = 1.7050 across CD and p/p = 3.2158across CE. ThusXX0VlQ2222020 =10°:Vo6.43016.6069across CD,across CE.422V. 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 SThis indicates that the correct value of θ is somewhat smaller. W e thereforetry_ οΛίη'V — 6.5400Vo ~ 6.4946across CD,across CE.B y linear interpolation we obtain the second approximation θ = 9°44'and this gives p/p = 6.52, correct to t w o decimal places, across both CDand CE.

T h e corresponding density ratios are p/po = 3.56 and p/po =3.61 so that034P3which differs very little from unity. Similarly the difference between M =1.63 and M = 1.66 is very slight, so that the discontinuity is insignificanteven though we started with large deflections θχ, θ . These values wereused in constructing Fig. 163; the apparently equal spacing of the stream­lines throughout DC Ε illustrates the weakness of the discontinuity CF.342Finally, we must consider the conditions under which the present solu­tion is valid. If the t w o shocks AC, BC have equal strengths, so that theyalso have equal and opposite inclinations, then the solution will be sym­metric about the horizontal line through C.

This dividing streamline maybe replaced b y a fixed wall, and then we have the oblique reflection prob­lem of the last section. Thus, in this case, it is sufficient that the strengthη of the shock AC and its inclination ω to the horizontal determine a pointin Fig. 161 lying on or below the curve. W e have seen that when the pointlies below the curve there are t w o possible flow patterns. N o w consider thecase of unequal shocks. Their strengths and corresponding inclinationsdetermine t w o distinct points in Fig. 161. A continuity argument similar t othe one given in Sec.

3 shows that if one of these points lies below thecurve and the second lies sufficiently close to the first, then there areagain two solutions of the present kind. This may be seen by consideringthe curves in the 0,p-plane given by (26) and (27) when η and r are al­lowed to vary (see the curve ρ = p in Fig. 157). T h e same figure also showsthat whenever there is one such flow pattern there is in general a second.One of these corresponds to a pair of strong reflected shocks, and the otherto a pair of weak reflected shocks.

T h e approximation (32) applies t o thelatter. I n particular we can always obtain these flow patterns for suffi­ciently weak shocks AC, BC.24This gives an indication of some conditions under which our solution isvalid. On the other hand it is definitely not valid when either of the shocksAC, BC is the strong shock for the deflection it produces. F o r the flowbehind a strong shock is always subsonic (see Sec. 22.7) whereas the flowsin front of the reflected shocks must be supersonic.24.1A D I A B A T I C F L O W OF I N V I S C T DFLUID423Article 24Nonisentropic Flow1. Strictly adiabatic flow of an inviscid fluidIn all the examples discussed in the preceding article, solutions could begiven in terms of regions of uniform flow or simple waves, separated bystraight shock lines or straight characteristics. I t is clear that problemswith more general boundary conditions cannot be solved in this way.

Forinstance, in the example discussed in Sec. 23.2 the solution was obtainedin a limited region only. T h e state of the fluid on the boundary of thisregion forms a set of boundary conditions on the flow beyond, for whichwe anticipated that a curved shock line would be required. In this sectionwe shall investigate the nature of the inviscid flow behind a curved shockline.In Arts.

16 through 21 the theory of steady plane inviscid flow was de­veloped under the assumptions that the fluid was elastic and the motionirrotational. Both these assumptions are realized when the motion is isen­tropic and the Bernoulli function Η is constant throughout the flow. Suchis the case for the strictly adiabatic flow behind a straight shock when theincident flow is uniform. For the total head Η is the same for all particlesbehind the shock by (22.19), and since the values of p ,pare the same atall points of the shock, see Eq. (22.17), so also is the entropy. However, ifthe shock is curved the flow after it will not be isentropic since the valuesof p , P2 depend, for each particle, on the slope of the shock line at the pointwhere the particle reaches the transition, see Eq.

(22.17). On the other handthe total head is still constant behind the shock. There is no longer anover-all relation between ρ and ρ after the shock since different particleshave different values of entropy. T h e condition of strictly adiabatic flowfor the region behind the shock leads, as was seen in Sec. 1.5, only to thecondition222(ί)ξ(= ο,dtwhere S, the entropy or any given function of the entropy, is a known func­tion of ρ and p.T o study the inviscid flow behind a shock—or any inviscid flow forwhich ρ is a given function of ρ only for each particle—we must first findan equation to replace (6.17). This equation was obtained from Newton'sequation (1.1) by writing grad Ρ for (grad p)/p and using certain vectoridentities.

This can only be done for an elastic fluid and we therefore seek424V. 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 Sanother way of transforming (grad p)/p. According to Eqs. (2.11) and(2.23), which relate the three functions / ( p , p), C/(p, p), and S ( p , p), wehavedl = dU-* dp+ ^tΡ1= TdS+Ρ-V.dΡHence for the actual changes occurring in the flow, we have- grad ρ = grad I — Τ grad S,Ρwhich provides the necessary transformation.In Eq.

(6.17) grad Ρ must now be replaced by grad I — Τ grad S. T h eequation then reads(2)^7 +dt(c u r lq X q) = Γ grad S -grad gH,where(20Η = 1.For steady flow this reduces to(3)curl q X q = Τ grad S -grad gH.A0N o t e that in the derivation of this equation we have not used the specify­ing condition ( 1 ) .

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