REG NACA-TR-1143-1953 A vector study of linearized supersonic flow applications to nonplanar problems.pdf

1、REPORT 1143A VECTOR STUDY OF LINEARIZED SUPERSONIC FLowAPPLICATIONS TO NONPLANAR PROBLEMS 1 .By JOHNC. MARTIN “SUMMARYA VeCIOr8hLdY Of h partiddi$erentid eQUdiO?L Of 8.?.eudyliarized 8uper80ni.c$OW h prwnl.ed. Qeneral expwti,which relate the velociiypotential in the stream to the cwndihlmaon the dis

2、turbing surfacea, are derived. In connection withtheze general expre+whw the c0nc4pt of the jinfite part of ,anintegral is discussed.A di.scwnin of probkm8 deali with planar bodies is givenand the condittiy for the 8olution to be unique are int%stigakd.Probkm8 CO?lC8T?Lhgnonphar 8y8tem8 are inVYBIZO

3、R small constantT= J(zg)g+% Y) zf9=-Jmr.sr acl 7l-Jl.= aP(b/2)Vp-.a0“ indicates integration over closed line 6rJsurfacefJ denotes finite part of integralTHEORY ,This report deals with the linearized partial-diilerentislequation of steadY supersonic flow. This equation is givenbyFwmetimes denotedbyTh

4、e analogous divergence of the hyperbolic gradient operatoris denoted byWh=V.Vh=-/3+ad 2W+$The following identities are needed. Let E be a voctorand# and A be scalar $mctions -of r, y, and z. Then, v.#E=#v.E+E.v# (2bVX(VXE)=V(V.E)-PE (2C)V.(vxll)=o (2d)VhVA=VhA.V# (20). Vh.#E=#Vh.E+E.Vh# (2f)VX(VhXE)

5、=Vh(V.E) -PhE (2g)VhX(VXE)=V(Vh.E) -vhE (211)Vh.(VhXE)=O (2i)These identities can be proved by direct expansion.The divergence theorem may be eqressed as$Enda=PEdv (3)Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-A VECTOR STUDY OF LRTEKMZED SUPERSO

6、MC FLOW AJ?PLICATIONS TO NONPLANAR PROBUIhlS 847where n is the normal unit vector to the element of area da. ”The vector n is eXpressedmathematically azn= ivl+ jv2+ kuwhore VI,v1,rmd V3are the direction cosines of the outwarddrrwn normal to the element of area da.A theorem more general than the dive

7、rgence theorem isgiven by (this theorem follows from the results of ref. 8, p.87)$(C, vIE.+ CMEV+ ,E,)dawhere the subscripts X, y, and z refer to components of thevector E, and Cl, Cz, and G are arbitrary constants. Notethat if Cl= CZ= Cs= 1 the preceding equation reduces toequation (3). If C,=p aci

8、/=c3=lthe preceding equation reduces toor4E”nb=P”Edwheretih= if?%l+jvs+kvIf the divergence theorem as ezprwedis applied to a volume throughout whichv.E=o(4)by equation (3)then the surface integral over the bounding surface is$ E-n da=Oprovided that no surfaces tit inside the volume of integra-tion a

9、crosswhich the normal component of E is discontinuous.Similarly, if equation (4) is applied to a volume throughoutwhich Vh.E=Othen the surface integral over the bounding surface is$ E.nb da=Oprovided that there are no surfaces inside the volume ofintegration across which E.nh is discontinuous. It is

10、 in-teresting to note, however, that surfaces exist tilde thevolume of integration across which E.n can be discontinuouswhile at the same time E.nh remains continuous. It followsthat for such a surface n and nh must satisfy the relationn.nh=O (5)Let Q(z,y,z) =0 be the equation of such Q surface. The

11、n,1n= VQIIQ2+QV+QZ2andwhere the subscripts indicate differentiation. Substitutingthe preceding expressionsfor n and n. into equation (5) yields(6)Any solution of equation (6) set equal to zero is the equationof a surface across which V.E may be discontinuous whileV.E remains continuous. The fact tha

12、t the Mach conefrom any arbitr point satisfies equation (6) can be easilyverifbd. The equation of the envelope of theMach cones froman arbitrary line also satizk equation (6) (ref. 9, p. 106).RiNITE PART OF INTEGRALSWHICH ARISE “IN STEADY SUPERSONICPLOWIn the following sections use iz made of the co

13、ncept of thefinite part of an iniin.iteinteggal. This concept was intro-duced by Hadamard (ref. 4) and has been used by a numberof other investigators. The concept of the finite part is,however, sometimes confusing. This section was thereforeincluded in an attempt to give a realistic picture of thef

14、inite-part concept and also to present the however,for the purposes of this report such a “derivation is notneeded.The result of applying equation (4) to the vector W isWhen satisfiea equation (1) throughout the volume ofintegration, the rightihand side of equation (7) is zero;thus,(8)whenW=oEquatio

15、n (7) apped to a volume (denoted-by o,)enclosed in the forward Mach cone from the point (w,z).This vohune is bounded by tie surface given by R=R,where R is a small constant, and an arbitrary surface 81enclosed in tie forward Mach cone from the point (z,y,z).A cross section of the region of integrati

16、on is shown infigure 1. Ne that this region is analogous to the regionthat is sometimes used in calculating the potential functionsatisfying Laplaces equation (ref. 3, pp. 151153). Forregions such as the one shown in figure 1, equation (7) mqybe written as. J,0*v%f#J0?0(9)Fmum 1.Go= section of the r

17、egion of integrationueed in connectionwith eguation (9).where T represents tie area of integration when R=Rf.The integralover the area T may be reduced towhere r is given byf-=J(w-.f)*3/yyy 7#y+y9qzqf)2)2Since R is a constant, equation (10) can be written w1H V+ n,+fi daEl T )Equation (9) can now be

18、 written as(lo)(11)(12)If # is required to satisfy the linearized parti)n“3)HIf R is made smaller and smallerthe integrand of the integralover the area Tin equation (13) remains iin.iteexcept.on thesmall area close to the point (z,y,z). In anticipation of tak-ing the limit of equation (13) as R appr

19、oaches zero, the smallarea close to the point (z,y,z) is removed from the area T.The area T is divided into two parta. One part is the area ofT which is downstream of the surface given bywhere eis smallbut larger than R. This area is denoted by r.The remai there-fore, the product of l/R and these in

20、tegrals either approacheszero in at least the order of R or approaches iniinity as Rapproaches zero. Thus it follows that the integrals over theareas r and Tr have no iinite terms remaining after the limit(R+ 0) has been taken. The sum of the terms of equation(17) must be zero; thus the singularitie

21、s resulting from theintegrals over the areas r and T must cancel the singularitieswhich arise from the integral over the area S1.From the preceding considerations it follows that onemethod of evaluating the finite part of infinite integrals of thetype appearing in equation (17) is to evaluate the in

22、tegralwhen R is small but not zero and neglect the terms multi-3100G_G asHadanmrd points out (ref. 4, p. 147), these singular pointsmust be removed from the area of integration before thefinite part is t3ken. Particukw attention should be given toparagraph 92 of reference 4 since the special type of

23、 integralsdiscussed therein sometimes arises in dealing with planarproblems.Robinson (ref. 2) has shown that when using Hadamardsmethods the order of integration may be changed withoutaffecting the finite part and that it is permissible to differen-tiate under the integral sign of a multiple integra

24、l withoutconsidering the variable limits which lie along the boundarywhere the integrand is singular, provided that only the ii.nitepart is taken. Both Hadsmmd and Robinson have shownthat in differentiating an improper integral which has anintegrand that has a one-half power singularity alongvariabl

25、e limits the variable limits may be neglected providedthe finite part of the resulting integral is taken.The term %nite part” is somewhat misleading since thefinite part of an integral can be inlinite. In certain cases theintegral is in6nite even after the terms which approachiniinity as R approache

26、s zero have been neglected.SCALARPoTBmUI#The preceding arguments show that the fite parta ofequation (17) can be equated to zero; thus,J(2(z,y,z)+f *, )1VW; . n, da=Owhere the symbol j before the integral denotes that only thetite part is to be taken. The preceding equation may besolved for the valu

27、e of the potential at the point (z,v,z);the rw.dt of this operation is given byIt should be remembered that surfaces can exist inside theProvided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-.850 RDPOIIT 114NATIONALforward Mach cone from the point (z,g,z)”

28、 acrossfurthermore, the right-hand side of.equation (20) remains iin.ite as R approaches zero. If inequation (20) R is made to approach zero and only the finiteparts of the integrals are retained, then the resulting expres-sion is(21)where 01represents the volume O.when R is equal to zero.Equation (

29、21) is equation (58) of reference 4 where /32hasbeen set equal to one. Note that the volume integral inequation (21) has the appemance of the integral for thepotential resulting from a volume distribution of sourcesin an incompressible flow.The assumption has been made that is continuousthroughout t

30、he volume 01. It is also awmmed that nosurfaces exist inside 01across which bdfb% is discontinuous.If equation (21) is applied to a volume%, which has surfacesacross which $ andfor the derivative of 4 in the directionbf nh k dkcontinuo, the sf of tintinti be removed from the volume of integration by

31、 allowing thearbitrary surface SI to envelop them (see dv (q%Equation (2h) indicates thatvhx(Vx G)= V(Vh.G)-WhG (30)Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-A VECTOR STUDY OF LINEARIZED SUPERSONIC FLOW APPLICATIONS TO NONPLANAR PROBI.iEWS 851/

32、7=0.s./zA/FImmI 3.A cross section of the region in the forward Ma however, since 0 and A are found by inte-gration only in the forward Mach cone from the point(z,y, z), equation (31) hardly seems to be a statement ofthe Hehnholtz theorem as is commonly given. The resultgiven by equation (31) waa obt

33、ained by Robinson inreference 2. HYPEllBOLICVECTORPOTEEquution (31) indicate-s that the perturbation velocityvector can be divided into two parts. One part is thegradient of a scalar function, and the other is the hyperboliccurl of a vector function. The vector-function is analogousto the common vec

34、tor potential (ref. 3, pp. 104 and 188);therefore, the vector function is refeed to as the hyperbolicvector potential. Thus, if q is the total perturbationvelocity vector, thenq=V+VhxA (32)where is the scalar potential and A is the hyperbolic vectorpotential. The part of the velocity vector which is

35、 made upof the hyperbolic curl of the vector potential is denoted by q.By direct expansion it can be shown thatVh.q=VhW$+Vh.(vhXA) =W+Vh. (VhxA)=O (33)Equation (33) indicates that tke hyperbolic divergence ofthe perturbation velocity vector is zero.The vorticity vector is given byIl=vxqTherefore, fr

36、om euation (32),(34)H=VX (VhXA) .orH= Vh(V.A)WhAFrom equations (2d) and (29), the divergence of the hyper-bolic vector potential is zero; thus,H= FhA (35)Each component of equation (35) is a partiaklifkrentialequation of the form of equation (24); thus, from equation(23) each component of equation 5

37、) has a solution given by(36)where the subscript i refers to any component of the vectorH. Since each component of A is given by equation (36),then“ (37)The velocity vector resulting from the hyperbolic vectorpotential is therefore given by.Sq=vhx do (38)orJ62 * (y q).-$ f)Hii Qu=2r q (39a)where the

38、 subscripts refer to the components of the vec.$orH. The results given by equations (39) were obtained byRobinson in reference 2.VORTEXSERBW3If the vorticity is confined to a surface S%,equation (37)becomes(40)Equation (4o) is an expression for the hyperbolic vectorpotential :ulting from a surface o

39、f vorticity. Note thus,Equations (18) and (46) can be combti”ed to yield(47)are unkno:GURE S.Two disturbing surfaces intersecting at right angk%.enclosed by the forward Mach cone from the point (z,v,z),the y=O plane, the z= O plane, and an arbitrary surfaceupstream of the disturbance (see fig. 10).

40、The result ofapplying equation (46) to this volume isThe surface SI has been taken to be the y=O plane (z posi- “tive) and the z=O plane (y negative); thus, Dntu “Wrb3r(y=o 0n8)L“AIrurtt InuInf (Xy,z)(67)-Mach coneLA- IL.t/ IxI?mmm 9.l3ione of integration for equation (55).Provided by IHSNot for Res

41、aleNo reproduction or networking permitted without license from IHS-,-,-A VECTOR STUDY OF LJNEARD SUPWRSONIC FLOW APPLICATIONS TO NONPLANAR PROBI.JIMS 857The potential function ($,T,)is chosen so that# (t,%r) =+(%W)where a is positive. Ii this case, (81)where h is given by equation (72). Substitutin

42、g equation (72) into equation (81) yields (remember that dJOb zeroupstream of the Mach cone from the v-), for S o.( ) 1 PP2 _2*9- .(z,P,e)=y 17 io+co Ppcoso o 2 (;.Qqsine-cose)(cosO+sinO) ;zx if Xpp sin 8()The potential in region is a special case tl= of equation (82). Setting o=; in equation (82) y

43、ields (in Cartesiancoordinate) ,4(Z, o-, Z)=% (1 )z+l (83)Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-,A VECWOR STUDY OF UNEARKED SUPERSONIC FLOW APPLICATIONS TO NONPLANAR PROBLEMS 871From equation (S3) the pressure-differenm coeilicient is found

44、 to be given byACP=(l-fi (84)The potential in region V is the potemtial in region IV plus the difference between the potential in region I and thepotential in region II; thus, from equations (62a), (62b), and (83), the potentiaI in region V is found to be given by2PW-?3+WW-K%-ARGG31 85)+(Z,o-,z)=;Fr

45、om equation (85) the pressure-difference coeiiicient is found to be given byAt,=-%5-2fi-llF?+Jmd (86)The potential in region VI is the potential in region Ill plus the diiferwme between the potential in region I and thepotential in region II; thus, from equations (62a), (62b), and (79), the potentia

46、l in region VI is found to be given by F/3 ;z,+(Z,O-,Z)=* -z cos-”zyx) :1, “1I ,. ,.,.1I I I I I I I I I Io2 .4 .6 .8 1.0x-FChordwiie(a) Tail of one fin.FIGURE38.Chordwise and spanwim prees distributions on a h of espect ratio 1.5 at a Maoh number of W for tailsconsisting of rectangular fins.Provide

47、d by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-2VACPTA VECTOR STUDY OF SUPDRSOMC FLOW APPLICATIONS TOzI,4-3 -2 -I -0 .2 .4 .6 .8 Lo2$T 2 - MJchGlles1t-i-xNONPIAN-4R PROBLEMS4L=:(f) ,3-w -T- -1pb %!(f) “I:.o Lox-FChofdwke873(b) Tail of tiVOh.FKJUR.EX3.-cont

48、inued.Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-. - .t 4“3 -2 -2vAI$”1-RDPORT 1143NATIONAL ADVISORYfIcoMMrrrEO FOR A13RONAUTICSr / /& /“x/ - -Mach lines2.1l/ i./ “/ / 1.4&x*o =-3k“w)._(c) a71-1 1 t ! I I 1 t ! 1 I0 .2 .4 .6.8 1.02$SpaM-se(o) Tail of four fins.FIGURE 36.-Contiued.2 - =+(*) -.-_ -2VACP=+(+)I0 . -1 I 1 I r 1 1 1 1 I I0 .2.4 % 6 .8 .OChordw&.P