R. von Mises - Mathematical theory of compressible fluid flow (798534), страница 28
Текст из файла (страница 28)
(28) and (29)with the second of Eqs. (21) for νι ,θι, or for v ,0 ; one finds22(W)' 'Vc = M±2M"+2[Mi/(7+2~1/y}= Mll )2*+M22[M/("71 }+ 1/y}222'From the last equality we find also(30')2yMi M222= 2 +(-T1)(ΜΧ2+M ),2as a relation between the Mach numbers M and Mtransition, which will be of importance later on.14. The complete problem2before and after the24If we do not neglect heat conduction, as in Sec.
3, a steady one-dimensional flow is determined by the two first-order differential equations (14)in the unknowns ν and Θ. One could eliminate either ν or θ to obtain onesecond-order differential equation for θ or v, respectively. T h e resultingequation, however, would not be easy to handle, and a better procedure isto eliminate χ by dividing the two equations (14), obtaining(31)dO _ p gR0dvΘ7(7-1) -o +kν +V2v-c2v - V2vFor each value of c the solutions of the first-order differential equation (31)consist of a set of o o curves in the z;,0-plane; the solutions of (31) represent, therefore, o o different ν,θ-relations. For each such solution the relations between χ and ν and between χ and θ may then be found by quadratures, using the original equations (14).T h e coefficient of the right-hand member in (31) is, of course, a dimensionless quantity.
I t is customary to use the name Praudit number for thequotient125(32)Ρ= ψ=^ Ί °Κϊ -For dry air under normal conditions, the value of Ρ varies only slightly,roughly between 0.68 and 0.77 (see Sec. 5 ) . W e then write (31) in the form(33)^ = | l ^ J pdv37X)where(34)λ_ θ/(7-V2v2v - V2v1) -θ +ν +-cW e first consider the case Ρ = constant; then a curve λ = constant in144III. ONE-DIMENSIONAL FLOWthe f,0-plane is an isocline of Eq. (31), i.e., the locus of all points at whichthe solutions of (31) have a given slope, namely, 4 Ρ λ ( γ— 1)/3γ. T h eisoclines have equations of the form:(av + 60 + cf= 2(λ +l)\where α = 2 λ + 1 , 6 = λ — 1 / ( 7 — 1 ) , and all pass through the singularpoints which make both the numerator Ν and the denominator D of λvanish.
T h e points of intersection of the isoclines D = 0, Ν= 0 are thesame as those of the isoclines D = 0, Ν + D = 0. Hence consider the twoisoclines:D = 0,λ =± o o(35)θ + 2v -\/2v = 0or(Θ + 2vf= 2vandΝ + D = 0,(36)λ =y—-17Θ+ ν -orc = 0_1θ =7 - 1(c -v).7Eliminating θ between (35) and (36), which hold simultaneously at a pointof intersection ( θ ; , v -), we findt(37)(7 +IK- -7V2^ +(7 -l)c = 0( i = 1, 2 ) .N o t e that (35), (36), and (37) are exactly the same equations as (21) and(17), holding at the beginning and the end of the special flow consideredin Sec. 3.* I t follows then, as in Sec. 3, that the τι-, ρ-, ρ-, and T-values forthe points 1 and 2 satisfy Eqs. (22), (23), and (24).
Also for M ,xM,2Eqs. (30) and (30 ) hold.rA s seen previously, (37) will have two distinct positive roots in ν if andonly if c satisfies (16). For c <however, one or both of the correspondingvalues of θ will be negative. Thus ifthere will be two distinct points of intersection, 1 and 2, with both υ and Οpositive, and they are the same points as in the preceding section. For c = f ,one point of intersection is ν =corresponds to Μν = y /2(y2+=l) , θ = 7/(7 +2θ = 0; from (28) it is seen that this pointo o .
When c = 49/48, the points 1 and 2 coincide atl ) , corresponding to Μ2=L From theresult at the end of Sec. 3 it is clear that the straight line 2v/yS =1, on* This can be easily understood since in the second E q . (14) the result is the samewhether k or άθ/dx is put equal to zero.11.4THE COMPLETE PROBLEM145θνO.IO0.200.300.400.50F I G . 52. Set of parabolas λ = constant in the velocity-temperature plane, forc = 0.7.which Μ=1, separates the points 1 and 2. In what follows we assumethat (38) holds.Figure 52 shows a full set of λ-curves; except for λ =—1 and λ=1/(7 — 1), all these curves are parabolas passing through the points 1 and2 and tangent to the θ-axis.
T h e parabola λ=ztoo,Eq. (35), is independent of c, has a vertical tangent at ν = 0 = 0, and passes through thepoint ν =(39)θ = 0. I t intersects the curveλ = 0, Ν = 0:{^ΖΓΪ"v~c)=2 vat the points 1 and 2, through which passes also the straight line λ =— 1,Eq. (36). T h e slope of the latter, — ( 7 — 1)/γ, does not depend on c, butits location does. For λ =1 / ( 7 — 1 ) , the λ-curve degenerates into a pairof vertical lines through 1 and 2, respectively.
Using this set of λ-curves,for Ρ constant, the integral curves for (31) may be found graphically (seeFig. 54 and following text).For Ρ not necessarily constant, the λ-curves are no longer isoclines of(31). The geometrical construction of the integral curves, by means of theso-called isocline method, fails. Further information about the solutions146III.ONE-DIMENSIONALFLOWmust come from the general theory of first-order differential equations. W enow assume that Ρ is, at least, a continuous function of ν and Θ, notvanishing at either of the points 1 and 2.
T h e points 1 and 2 are thensingular points for the differential equation (31). If we analyze each ofthese singular points by considering the linear terms in the Taylor expansions of the numerator and denominator of (31) about that point, we findthat 1 (supposing V\ >v ) is a nodal point, and 2 a saddle point. Thus2exactly t w o integral curves pass through point 2, while an infinite numberof solutions pass through point 1, all of them except one having a commontangent at 1. T h e two pairs of directions at 1 and 2 can be computedeasily, as will be seen in Sec.
6. One and only one integral curve of (30)passes through both singular points, and this solution represents a transition between the two states v ,θι and v , θ . This corresponds to thex22solution which was considered in Sec. 3; there k =0, so that Ρ=oo,and the complete solution is represented in the υ,θ-plane by the parabolaλ =0 (depending on c ) . Furthermore, in the shaded region of Fig. 53,between the parabolas (35) and (39), we have Ν > 0 and D < 0. Consequently άθ/dxand dv/dx as determined by (14) are positive and negative,respectively, in this region, the solutions of (31) have negative slopesthere, and the direction of the negative y-axis corresponds to increasing x.I t will be shown in Sec.
6 that the transition curve lies in this region.T h e integral curves in Fig. 54 have been found under the assumptionthat Ρ= constant =f. Thus, from (33), the slope of an integral curve is(7 — 1)λ/7 as it crosses any λ-curve. In particular, the slope of an integralθF I G . 53. T h e curves λ = 0, — 1, ± »in the υ,θ-plane for c = 0.7.11.5NUMERICAL DATA147F I G . 54. T h e set of integral curves corresponding to F i g . 52 for Prandtl numberΡ = Y±.
T h e transition curve from 1 to 2 is a straight line.curve at any point of the straight lineX = — 1 is — (7 — l)/7, which is alsothe slope of the straight line λ = — 1, Eq. (36). In the case Ρ = f, therefore, the transition from 1 to 2 occurs along the straight line (36). W h e nthis solution is expressed, by means of (13), in terms of the given variablesu and T one findsy2(40)%: + — g R TΔ7 — 12= %r + c T2p= C2= constant.This relation, valid only for Ρ = J, was found by R. Becker in 1922.Finally, Fig. 55 shows the solutions* determining the transition from 1 to2 for the three values Ρ =Ρ = f, and Ρ = 1. These are found by computing the directions at the singular points 1 and 2, and then by graphicalintegration from a diagram such as Fig.
52. T h e interpretation of these results will be given in Sec. 6.65. Numerical dataT h e conclusions to be drawn in the following section depend, to a certainextent, upon the numerical values of the physical constants which appearin the equations. Therefore we begin by establishing a set of standard valuesof constants, referring to the conditions of air at room temperature under* T h e ordinates have been magnified 5:1.148III.ONE-DIMENSIONALFLOWM,«2F I G .
55. The transition curves for Ρ = J£,and 1, with bounding parabolas.a pressure of 1 atmosphere. For the convenience of the reader therefollows a list of the numerical values of constants employed in this book.(a) General constant:g = 32.17 ft/sec= 980.7 cm/sec .22(b) D r y air constants (diatomic, perfect g a s ) :7= I = 1.4;gR7 - 1C p=y C vR = 53.33 ft/°F = 29.26 m/°C;= 4289 ft /sec °F = 0.717 Χ 10 cm /sec ° C ;22722= 6005 ft /sec ° F = 1.004 Χ 10 cm /sec ° C .22722(c) Assumed standard conditions:Temperature = 59°F = 15°C, corresponding to the absolutetemperatureΤ = 518.4 in Fahrenheit degrees = 288 in Celsius degrees;ρ= 0.00238 slug/ftV = gRpT3= 0.001226 g/cm ;= 2116 lb/ft32= 1.013 Χ 10 dynes/crh ;61116 ft/sec = 340.1 m/sec.211.5NUMERICAL149DATA(d) Viscosity (experimental values):μ= 3.73 Χ 10" slug/ft sec = 0.000179 g/cm sec;ν= ίί = 0.000157 ft /sec = 0.146 cm /sec.722Ρ(e) H e a t conductivity (experimental values):k(f)= 0.00314 lb/sec ° F = 2.52 Χ 10 dynes/sec °C.3Prandtl number:Ρ= £*ίί = 0.713.ΚιT h e coefficients of viscosity and heat conductivity depend, in general, onthe state variables ρ, ρ, T.
Experimental evidence indicates that μ and k(but not v) can be considered as functions of temperature alone. A n olderformula of Lord Rayleigh gives μ proportional to the f power of Τ in therange between the freezing and boiling points of water:7'/φ'\0.75( 4 I) * , - ( ; ,).from which one derives«ί - ( ί Γ · ? - ( ί Γ · ί - ^ Γ ·A more recent investigation gives, within the limits 32°F t o 950°F,8(43)Κilμ"}= (TLY\T"Jχ("T\ Tf+™A+ 223.2/1when V and T" are absolute temperatures in Fahrenheit degrees.
![](https://s.studizba.com/z.php?f=/templates/si/i/noavatar.png&w=100&h=100&t=1"){kind=avatar}
