R. von Mises - Mathematical theory of compressible fluid flow (798534), страница 29
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For example, if one uses the value given above for μ at Τ — 518.4, the first formula (41), gives 10 μ = 4.07 slug/ft sec, while the second formula (43)yields 4.08 slug/ft sec at Τ = 581.4 (i.e., 122°F). Table I I for ν is computed from Eq. (43).A formula very similar to (43) has been found in experiments on heatconduction :79^'k"\T")\ Τ+ 225/'From (43) and (44) it would follow that the ratio μ/k, and consequently thePrandtl number also, is essentially constant, giving Ρ = 0.713 in general.150III.ONE-DIMENSIONALFLOWI I — K I N E M A T I C V I S C O S I T Y V I N FT /SECTABLE2po = 2116.2 lb/ft2= 29.92 in H gValues of ν Χ 104Temperature (deg. F )P/Po1432506886104122110201.330.1330.0661.420.1420.0711.520.1520.0761.620.1620.0811.720.1720.0861.820.1820.0911.920.1920.096On the other hand a theoretical value of P , based on the kinetic theory ofgases, is10(45)Ρ =49Τ= 0.7377—fory =1.4.οIn any case, the value 0.75 is a close approximation to the actual value of P .6.
ConclusionsW e consider first the case when the Prandtl number Ρ is 0.75. Then, asseen in Sec. 4, the solution passing through the singular points 1 and 2 inthe ν,θ-plane, the transition curve, is the straight line (36). If the value ofθ from the solution (36) is substituted into the first of Eqs. (14), we find(46)« J = ί [(γ +m axyDv -yV2v+(y -l)e],which differs only by the factor 1/y from Eq.
(15), which was derived underthe assumption k = 0. Consequently the former discussion (Sec. 3) holdsunchanged, provided that χ is replaced by x/y. In particular, the thicknessof the transition layer defined there is now(47)L = yL0(P = f ) ,where L is the expression (26).0There remains the case of a finite Ρ different from f, but constant, andthe case in which Ρ is not constant.For general Ρ the integral curves of (30) are qualitatively similar tothose shown in Fig.
54 for the case Ρ = f, except that the singular integralcurve S joining the points 1 and 2 is no longer a straight line. In particular,the sign of άθ/dv is unchanged in the various regions bounded by the isoclines λ = ± oo and λ = 0 (these t w o curves being isoclines even for nonconstant P ) .
Although the solution S cannot be given in closed form, it11.6151CONCLUSIONSis possible to find bounds for θ(ν) on S, and these are sufficient for givingsignificant estimates concerning the thickness of the transition layer.First we compute the slopes of the integral curves at 1 and 2, in the caseΡ = constant. Since Ν = D = 0 at these points, the slopes are found byapplying PHospitaPs rule to the right-hand side of (31). Thus the slopeθ ' = άθ/dv of any solution at 1 or 2 must satisfy the quadratic equation(48).Θ' = £ e' - ( - ixi - vyaa73or, using l/\/2vi = 1 +(48-,θ' +71/fMi2=-from (29),e^ + e ' ^ - i j - D - i f h - D ^ .
o .Thus, of the two values of θ ' at each of the t w o points, one is positive andone negative. A s in Fig. 54, the two negative roots, say θ[ at 1 and B at2, must give the slopes of the transition curve S at the singular points 1and 2, respectively. A more detailed study of (48') shows that θ ί <— (τ — 1)/γ< θ when Ρ < f, and that the inequalities are reversed ifΡ > f. For Ρ = f, Eq. (48') is satisfied by θ ί = θ = -(y- l)/y, whichagrees with (40).222Equation (48) also expresses the fact that any solution passing through1 or 2 in one of the possible directions determined by (48) is actually tangent to the appropriate isocline at the point (i.e., to the isocline whose λfollows by (33) from the slope θ ' ) .
I n particular, the singular solution curveS which passes through 1 and 2 is tangent to a certain isocline parabola Αχat 1 and tangent to another parabola A at 2. T h e bounds on the solutionS follow from the fact that S must lie on the concave side of Αι between 1 and22 and on the convex side of A .For example, suppose Ρ < f. Then, since θ ι < — (7 — l ) / 7 , the isocline Ai lies above its chord 12, as shown in Fig. 56a. A t any point of A ,a solution of (30) has the same slope as the tangent T, at points below Αχa less negative slope, and at points above Α a more negative slope (morevertical) or positive slope (beyond λ = ± oo). N o integral curve can leavethe point 1 in the area between Ai and T , for, geometrically, such a curvemust have a slope less negative than that of Τ near 1 whereas a solutionof (31) must have a more negative slope than T.
Along an integral curveleaving 1 above T, the tangent to the curve must rotate in a clockwisedirection and it is clear that such a curve cannot reach the point 2. Thusthe integral curve S runs below A1 as it leaves the point 1. I t cannot meetΑχ at a point Ρ between 1 and 2 (this assumption is illustrated in Fig. 56a).For at Ρ a solution has the same slope as T , but no longer the same slope2xλ?152III.ONE-DIMENSIONAL(b)FLOW^F I G . 56.
T h e two isoclines bounding a transition curve when Ρ is constant; shownforΡ<as Αχ, and can only continue beyond Ρ in a manner similar to the solutionsleaving 1 above T. Thus S runs below Αχ between 1 and 2.Since θ2>— ( τ — l)/y(which is the slope of the chord joining 1 andlies above this chord (but of course below A ι , since2 ) , the isocline A2θ ί < θ ) , as in Fig. 56b. A n analogous argument shows that S must run2above A2between 1 and 2, since a solution at a point of A ,different from21, has the slope θ and continues below A22with less negative slope, so thatit cannot reach the point 2. Thus the t w o isoclines Αχ and A2bound acrescent-shaped area within which S must lie between 1 and 2.
For Ρ >f,these t w o isoclines lie below the chord joining the points 1 and 2. If Ρ=f, then Αχ,Ρ =A,2and S coincide with the chord, while if k =0 (so thatoo) all coincide with the isocline λ = 0.N o w let a and β be the λ-values corresponding to the isoclines Αχ and A2with a ^ β, e.g., if Ρ < f, then a corresponds to Aand β to Αχ, and let2θ (ν)αand θβ(ν) denote the ordinates on these two isoclines and θ(ν)ordinate on S.
Then the fact that S runs between Αχ and A2theis expressedby the inequalities(49)θ (ν)α= θ(ν)=θ (ν),βI n the particular cases Ρ = f and Ρ =v2 =ν=νχ.oo, equality holds throughout in(49).Next we introduce the abbreviationsθ„ =(50)\f2v-2v^_Θ» = (y - l)(v - V2^+c);i.e., θ „ and θο are ordinates on the two isoclines λ =± oo and λ = 0 respectively. Then the first differential equation (14) may be written as11.6153CONCLUSIONSso that (49) leads to(51)θ-αΘΜg θ, -5Ξθαό πι dxAlso, Eq.
(34) may be written asλ =θ -Oo(Θ- Θ j (from which it follows that the ordinate θ-7λiyon the isocline corresponding toany value λ satisfiesr\σλr\~~ WooOoo—1 -O—(70-1)λ'Consequently, (51) may be written as(52)-9001 -~ °< ί ϋ ^ . < _( - \)a ~ 3 m dx ~θ7° ° ~~ °1 - (7 θθ1)0'This result may be compared with the differential equation (15) determining ν in the case k = 0, Ρ = α>, discussed in Sec. 3, namely,(53)^ ? = - ( θ . - θ , ) .3 m dxI t is seen that the upper and lower limits for dv/dx, as determined by theequality signs in (52), lead to the same differential equation as (53), exceptthat χ is to be replaced by :r/[l — (7 — I)a] and x/[l — (7 — 1)β], respectively.
Consequently, if we assume μ constant, the thickness L of the transition layer must satisfy11(54)Lo[l -(7 -1 ) « ] ^ L £ Lo[l -(7 -1)0],where L is the expression (26). Here a and β are the λ-values for the twoisoclines Αχ and A tangent to S at 1 and 2 respectively, with a ^ β. T odetermine the exact limits in (54), it is merely necessary to compute theslopes O i andΘ 2 of the solution θ(ν) at the points 1 and 2, i.e., the negative root of (48 ) at each of these two points.
Then a and β are the (negative) values02r3747— IPθί,37Θ4 7 —2I Phere a is the second expression if Ρ < f and the first if Ρ > f. When Ρ =f, we find θ ί = θ = — ( 7 — 1)/τ, as mentioned above; in this case bothbrackets in (54) reduce to 7, in agreement with (47).A similar argument can be applied to the case where μ is not constant,but a given increasing function of T. I t follows from (52) that the estimate2154III. ONE-DIMENSIONALFLOW(54) still holds, provided that L on the right is computed for the maximumvalue of μ (the value at point 2) while L on the left is computed for theminimum value of μ (the value at point 1).00Finally there remains the case of Ρ not constant, to which the precedingtheory can be adapted.
T h e result is that if the expressions in (55) arenow computed for the maximum and minimum values of Ρ in the shadedregion of Fig. 53, and the two extreme values taken for α and β, then theestimate (54) still holds. T h e transition curve lies between the parabolasλ = a and λ = β. A n y variation in μ is treated as in the last paragraph.ExampleSuppose the initial Mach number is Μ ι = 2. Then from (30') we haveΜ2 = £ for y = 1.4, while from (29) we find"1 = ZUJ1- 7 = -1 = —= 0.179,2= 2.143.Thus the equations (48) for the slopes at the two endpoints are01_4P-3 X T 4Θ( + 0.071θί +.0.821, _θ2~4P θί +0.857Ϊ21.143-θί -T h e negative roots areθί =-0.43,θ2=-0.22for Ρ = ±θί =-0.20,θ2=-0.34for Ρ = 1.T h e limits for the thickness of the transition layer are then1.46 L0^ L ^ 1.88 L01.21 Lo ^ L ^ 1.35 Ufor Ρ =}for Ρ = 1.T h e quantity L was found, for a numerical example in Sec.
3, t o be of theorder of less than one-tenth mm.T h e situation in the ν,θ-plane is sketched in Fig. 55 (see end of Sec. 4)with the ordi nates θ magnified in the ratio 5 : 1 . T h e broken lines are theintegral curves for Ρ = J, Ρ = J, and Ρ = 1; the solid lines are the isoclines for the four slopes θ ' found above (and for λ = 0 ) .012.1155GENERAL EQUATIONSArticle 12Nonsteady Flow of an Ideal Fluid121. General equationsW e consider an ideal fluid: a perfect gas with μ = k = 0; that is, viscosity and heat conduction are neglected.
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