Solution Manual For Modern Compressible Flow: With Historical Perspective, 3rd Edition

Preview (31 of 362 Pages)

100%

Purchase to unlock

Loading page image...

fnstructor'sSolutionsManualtoaccompanyMODERNCOMPRESSIBLEFLOWThirdEditionJohn D.Anderson,Jr.Curatorfor AerodynamicsNationalAirandSpaceMuseumandProfessorErneritusUniversity of Maryland

Loading page image...

l.l.008121.2e:#:sffiP:mr.g/m-lpRT(10X1.01x105)(1.38xLO''11ZZO1pr= p_(10)(1.01x105)RrpRT(11.0)(8314)(320)87xI0'"lm0.,0345kg-molekg/n:1.3Fromthedefinitionofenthalpy,h:e*pv:e*RTEoracalorically perfectgas,thisbebomescpT:c"T+RT,or6JFForathermally perfectgzrs,EQ.(A1)isfirstdifferentiateddh:de+ Rdtof,codT:c"dT:Rdtof,(Al)

Loading page image...

t.4sz-sr:c,tn?-.^f(a)R:1716ft-lb/slug"R"r=h= W=6oo6ftJb/slug.Rs2-sr:(6006)tn(1.687)-(t716)tn4.5(b)R:287joulelkg.K"r=h:W:too4.5jouletkg"Ksz-sr= (1004.5)tn(1.657)_287n@.1)52-sl=59.9ftJb/slue"s2-sl=3'6ioul/1.5Pr-PrW:ppz824.3=--r'RT,(1716X400)t.4=1800(400/500)0.400121.6Volume of room:(20X15XB)=2400ffStandardsealeveldensity:0.002377slug/ft3Massofair:(0.002377)e400)=IS.ZOslud

Loading page image...

1.7(a)dp=-pVdVanddp:ptdp,orCombining:dp,_L=_pVdVprdp:-"p2vdvdp4=-tpVdVpdP=-rpv2dVVp(b),,:1=-17P(1'4[l^ol"1d):7'07x1o-6m2/NY:r,py2+= -(7.07iiro{y1r.23x10)2(0.01)+=FT?.ro=1p(c)Here,&*iffbelargerbytherati,(tooo\'p"[roly:(:8.7xrof(-*9'Comment:Byincreasingthevelocityofa factorof100,the fractionalchangeindensityisincreasedbyfactorof104.This isjustanotherindicationofwhyhigh-speedflowsmustbedp:dPpr

Loading page image...

CHAPTER22-lConsideratwo-dimensionalbodyinaflow,assketchedinFigureA.Acontrol volumeisdrawnaroundthisbody, as giveninthedashedlinesinFigureA.Thecontrol volumeisboundedby:l.Theupper andlowerstreamlinesfaraboveand belowthe body (abandhi,respectively.)2.Lines perpendiculartotheflowvelocity faraheadofandbehindthe body (aiandbh, respectively)..3.Acut that surroundsandwraps the surfaceofthebody(cdefg)The entire control volumeisabcdefiria.Thelyidthofthe control volumeinthezdirection(perpendiculartothepage)isunity.StationsIiand2arcinflowandoutflowstations.respectively.Assumethatthecontour abhiisfarenoughfromthebodysuchthat thepressure isinflowvelocity urisuniformacrossai(asitwouldbeinafreestream,oratest sectionofawindtunnel.)Theoutflow velocityu2is notuniformacrossbh,becausethepresenceofthebodyhascreatedu*ukattheoutflow station.However,assumethat bothu1andu28rointhexdirection:hence,u1=coostdfltandu2:f(y).Considerthesurfaceforcesonthe controlvolumeshowninFigrreA.Theystemfromtwocontributions:l.Thepressuredistributionover the surface, abhi,-lJoutabhi

Loading page image...

e__6FigureAThesr:rface shear stressonab andhihas beennegfected.Also,notethatin Figwe A.thecuts cdandfgaretakenadjacenttoeachotherihence anyshe6rstess orpressuredistributiononone is'equal and oppositetothatontheother; i.e.,thesurfaceforcesoncdandfgcancel each other.Also,note thatthesurfaceon defistheewaland opposite reactiontotheshear shessandi-*pressuredistributioncreatedby theflowoverthesurfaceofthebody.Toseethis more clearly,examineFigrueB.Ontheleftisshowntheflowover thebody.The movingfluidexertspressureandshearstressdistibutionsoverthebodysurfacewhichcreatearesultantaerodynamic forceperunitspanR'on thebody.Inturn, by Newton'sttrirdlaw,thebodyexertsequal and oppositepressureandshear stressdistributions on theflow,i.e., onthepartoftheconftolsurfaceboundedbydef.Hence,the body'exertsaforce-Ronthecontolsurface,asshown on theriofFigureB.Withtheaboveinmind,thetotalsurfaceforceontheentirecontrol volumeisSurfaceforce:-lJabhi

Loading page image...

Morgover, this isthetotalforceonthecontrol volumeshownin FigureAbecausethevolumetricbodyforceisnegligible.Consider theintegralform ofthe momentum equationasgivenby,Equation(2.1l)inthetext.Theright-handsideof thisequation isphysicallythe foree on thefluidmovingthroughthecontrolvolume.For thecontolvolumeinFigureA,this forceissimplythe expression givenbyEquation(1).Hence,using Equation(2.11),withtheright-handsidegivenbyEquation (1), wehaveovlldv+'VQ)flS(pv'ds)v:-lJnns-n'abhiFlowexertspandron the surfaceofthebody,givingaresultantaerodynamicforce RFigureBAssumingsteadyflow,Equation(2)becomesR':-flrou'ds)v-lJnusa surfaceforceonthesection of.thecontrolvolumedelthatequals-R(3)

Loading page image...

Equation(3) isavector equation.consideragainthe confrol volumeinFigureA.TakethexcomponentofEquation (3), nothingthattheinflowandoutflowvelocitiesu1and u2 dreinthexdirectionandthexcomponentofR'isthe aerodynamic drag perunitspanD,:D':-#(pv'ds)'-lf(pds).Sabhi(4)InEquation(4), the last term is thecomponentofthepressureforceinthexdirection.[TheexpressionGdS),.is thexcomponentofthepressureforceexertedonthe elementalareadSofthecontrolsurface.]Recall that the boundariesofthecontrol volumeabhiare chosenfarenoughfromthebodysuchth"t pis constantalong theseboundaries. Foraconstant pressure.llro dS)*=oabhibecause,lookingalong thexdirection inFigureA,the pressureforceonabhipushingtowardthe./rightexacflybalancesthepressureforce pushingtowardtheleft.This istrue no matterwhattheshapeofabhi is,aslongasp isconstant along thesurface.Therefore,substitutingEquation(5)into(4), weobtainItlD':-(flfnV.dS)un\rsEvaluatingthe surfaceintegral inEquation (6), wenotefromFigureAthat:l.Thesections ab,hianddefaresteamlinesoftheflow.SincebydefinitionV'isparallel to thestreamlines anddSis perpendiculartothe controlsurface, alongthesesectionsVanddSareperpendicular vectors, and henceV'dS=0.Asaresult.thecontributions ofab,hiand defto theintegral inEquation(6)are zero.(s)(6)

Loading page image...

2'The cuts cd andfgare adjacenttoeachother.Themassfluxoutofoneisidenticallythemassfluxintotheother.Hence,the contributionsofcdandfg tothe integralinEquation (6)canceleach other.Asaresult' theonlycontributions to the integralinEquation(6)comefromsectionsaiandbh.These sections areorientedinthey direction.Also,the control volumehasunitdepthinthe zdirection(perpendiculartothepage).Hence,forthese sections,dS:dy(l).The integralinEquation(6)becomesflrou'ds)u:-I,pru2rdy.JJnfzdySNotethatthe minusin frontofthefirst tennontheright-handsideofEquation(7) isduetovanddsbeinginoppositedirectionsalongai (stationIisaninflowboundary);incontrast,vanddsareinthesamedirectionoverhb(station2isanuoutflow boundary),and hencethesecondtennhasapositivesign.BeforegoingfiftherwithEquation(7),considertheintegralformofthecontinuityequationforsteadyflow.Applied tothecontrol volume in FigureA,thisbecomesla1b-J,prur dy+Jnp2u2dy:0or.J,"p,u,at:JrtpzwdyMultiplyingEquation (8) byu1,whichisaconstant,weobtainJ'"o'u"dv:IJPzwurdYSubstitutingEquation (9) intoEquation (7), we havefrou'ds)u: -IJpzvuray*J,'nizdy(7)(8)(e)

Loading page image...

flfon'ds)u=-IJpzuz(ur-uz)dySSubstitutingEquation (10) intoEquation (6)yieldso':Jrtpzuz(ur-uz)dy(10)(l1)(r2)Equation(l1)is the desired resultof thissection;itexpressesthedragofabody intermsoftheknownfreestreamvelocityu1 andtheflow-fieldpropertiesp2andu2,&croSSa verticalstation downstreamofthebody.Thesedownstearnpropertiescan bemeasuredinawindtunnel,andthedragperunitspanofthebodyD'canbe obtainedbyevaluating the integratinEquation(ll)numerically,using themeasured dataforp2andu2asafunction of y.Examine Equation (11)moreclosely.The quantrtyur-uzis thevelocitydecrement atagiveny location.Thatis,becauseofthe dragon thebody, thereisawake thattrailsdownstreamofthebody.In thiswake, thereisalossin flowvelocifur-uz.Thequantiy mwissimplythemassflux;whenmultiplied byot-u2,itgivesthedecrementinmomentum.Therefore,theintegralinEquation(11) is physicallythedecrementinmomentumflowthatexistsacrossthe--W6ke,andfrom Equation(11),this #akemomentumdecrementisequalto the dragonthe body.Forincompressibleflow,p:constantandisknown.For thiscase,Equation(11)becomes.D'=oJJuz(ur-uz)dyEquation(12)is theanswertothequestions posedatthebeginning ofthis section.Itshowshowameasurementof thevelocity distributionacrossthewakeofabodycanyieldthedrag.Thesevelocity distributionsareconventionallymeaswedwithaPitotrake:

Loading page image...

''.t.tlJ.::'rit'--!i,l:.l;il::11..--,,..::,r_..il":.iUpTerf"("\2.2ia-ILo,uet-Wa//f2(r)Denote thepressuredistributionsonthe upperandjowerwallsbypr(")and.p,(x)respectively.Thewallsarecloseenoughtothemodblsuchthatpudrdpz tr€not necessarilyequaltop-.Assumethatfacesaiandbharefarenough upstreamanddownstreamofthemodelsuchthatP:P-andv:0andaiandbh.eIIIcl?,,IIiiuttt.y-component ofEq.(2.11)inthetext:L:-#rpiat;v-lJrnaSlvsabhi++Thefustintegral=0overallsurfaces,eitherbecauseV'ds =0 orbecausev =0.Hence.bhL':-JJtnaSlv:-tJPudx-Jn,d*1abhiaiMinussignbecausey-component isindownwarddirection.Note:Intheabove,the integrals over iaandbhcancelbecausep =p.onbothfaces.HenceL':hbc.'fJp,o*-Jpuoxi-a

Loading page image...

CHAPTER33.1FromTableA.1,forM:0.7;pJp=l.3BTandTo/T:1.09g.Hence.r^\Po=pl?J:o.e (1.01xto-5y11.387y:Notethat P*/pocanbe obtainedfromTableA.lasthe valueofp/poforM: l.Hence,fromTableA.lforM=1;po/p*:1.893andTolT*:1.2.Thus.o*:n[&-)fa'.l:0.e(r.0r*ro)tr.s;7)/rr8e3:-'\p/(p"/r*:rt+)[+l:250(t.os8)/1.2:ET&m\1./\lolu*=.'ffi:^l@:bog2'A;d1.26x10"N/m'(or1.248 atrnx10"N/m'(or0.659 atm3.2po_1.5x106_",.,p5x104FromTable4.1'Fod@nearinterpolation)Also fromTableA.1:Hence,T,T:2.643(linear interpolation)/r\ro=r[+l:2oo(2.64T=ltt6E\ t./

Loading page image...

3.3' a:JAT:rftr.4xr7r6x500) :1096ff/secM:yla=3000/1096=(2.74i,From Eq. (3.37)(M*;z:y+l2.4:3.63.4FromTableA.2,forMr:3;pzlpr:10.33,pzlpr:3.857,po,/po,=0.3283,andW4.4:.s2.rhus,pz:pr@zlp):(lxt.0txtO511tO.:3):1.043x106N/m2(or10.33atm)Thus,//f.\Pz=PrIgI:(1)(l.olxto)1t0.:3):\p,/./^\_Pz:pr |,,| :1.23(3.lsT=wqUffi\pr)12:o:=#:#:F6m-pzR(4.744)(287)u:Jtar,=,[(L4)Q87)Q66)=554.8 m/secw-a2M2=(554.9)(0.4752)=EOI.OrnAidFrom TableA.l,forMr:3,Po,/p1:36.73a1dT",lTr:2.8.Hence:(P"\tiu,n.nnurn.Ao^\po,:pltijfr:(1.01xrc\Q6.73X0.32s3):h218 "' 10fr-/*-1.i''(-Lr)'*'-'M-:Tq

Loading page image...

3.5,1.22x10'(a)pJp=ffii#:7.2r.Fromrablee.t:F<s3,7222(b)pJp: ffi:3.413.FromTableA.l;wefindthis wouldcorrespondtoM-=1.45.However,sincethisis supersonic,anormalshock sitsin front ofthePitottube.Hence,Poisnowthetotalpressurebehindanormalshockwave. Thus wehavetouseTableA.2.1)))p^-/p,=t-z-!-'-:3.412.FromTableA.2'M;= l.jl^"?2116'l3tg:n.85.FromTableA.2'M;=Til(c)Po,/Pr=.,OZO3.6For theshockcompression, useMriISisentropiccompression,usepzlpt:(v1lv1)-7:NORMALSHOCK COMPRESSION"(pzpr)-tMrpzlil=vzlYrPzlqraparameter,alongwithTableA.2.FortheISENTOPIC COMPRESSIONvzlvrwlPr1.0t.21.5t.72.02.53.0I1.341.862.202.67J.JJ3.8610.7450.5370.4550.3750.300.2591l.5l2.463.214.s07.1310.3I0.7s0.500.400.300.2011.502.64.3.615.49.51

Loading page image...

Notethat:t(a)Foragivendecreaseinvolume,a shockwaveyieldsahigherpressureandthereforemany beconstruedtobemore"effective"thananisentropiccompression.(b)Ontheotherhand,becarisethe entropyincreasesacrosstheshock,ashockwaveislessefficientthanan isentropiccompression.Toobtainagivenpzlptitrequiresmorework viaashock.(c) Note thatfor vzlvrnearunity,theshockandisentropic curvesareessentiallythesame.This isareflectionthat theentropyincreaseacrossweakshocks(Mr<1.3) isnegligible.3.7FromTableA.1foM-::g,fJT=289.9;To:(289.8)(270):78246"4Thistemperatureis almost8times thesurfacetemperatureofthesun.Longbeforethistemperature is evenreached, the airwilldissociateadiorize,andyis no longer constant,nor

Loading page image...

anywherecloseto1.4.Theabove answer istotallyinaccurate!Sincedissociationandionizationrequire energy, the actual temperature is considerablyless,morelike11,000"K.The propercalculation of thisanswer mustbe madewithtechniquesdiscussedinChapter14.3.8(a)FromTableA.3 forM1:2.0:fr:0.363,+:os2us,+:0.7s34FromTableA.1 forMr:2.0:ft:,.*-T.Hence:(r*t/m\ro*=I+-It+lr,=f+^.)(r.8X288)=653.4oK(Tq/\T,/\0.793ry'/Thisis thetotaltemperatureattheexit,sinceforchokedconditionsattheexit,M2=IandhenceTo,:To*,p2:p*, etc. Attheinlet,(1,:1.8Tr:(1.8X288):518.4"K.g:cp(t,-T",):1005(653.4-581.4):1.357xIpz=p*:p1/0.3636=#H=WdTz:T*=T110.5289:^^:F@0.5289(b)FromTableA.3,forM1:0.2:prlp*:2.273,TrlT* :0.2066,1,/To*:0.7736FrcimTableA.l,forM1:0.2:

Loading page image...

1,:1.00gT,.n*_ft"-)fT",\( t\ro*=t-r"JtiJt':Iorrru1(l'008)(288):r672K1,:1.008T1:1.008(2SA;:290.3"K9:cp(f., -1,)=1005(1672-290.3)=1.389xl0oioulPz:P* =p110.273=Iatm2.273Tz=T*:T1/o.2o6u:#=h39aB..3.9Islugair+0.06 slugfuel:1.06slugof mixture= 424loR0.044.5xlos\n:#:2.547.x107ft-lbper slugof mixturec.=A-Q.4)(1716):6006ftlb'y-l0.4slugoR.n."q-2.547xl07'or-to,-%-6006FromTableA.1:AtM1:0.2,4,= Tr:1.0081,=1.008(1000):l008oR1,=4241+To,:4241+1008:5249"RFromTableA.3:ForM1:0.2,prlp*:2.273,TrlTo*:0.2066,T",flon=g,l

Loading page image...

To*:T",10.1736:1008/0.1736:5S06Rl,[o*:5249/5806= 0.904From TableA.3:For t",/To*:0.904,(by interpolation)N{r:0594,pz/p*:1.433,T21T*=0.9895n,=;?fpr:(1*r(#)(10):tu0;dt,:-ftT11=(o.e8er,(#J(rooo)=F7sed3.10 4,=.To*:5806'R(fromprob. 3.9)e:cp(T",-T",):6006(5306- 1003)=2.88xlOiftlb/slugLetF:fueVair ratiobymass.o:lA(4'5xlot):2.88xro7-(1+FA)FA=2.88x107i4.5x108-2.88x10?)=0.06840.03(4.5x108)3.11q::1.311x107ftlbper slugof mixture1.311x10?6006:2182oR1.03qco4,-1,=q,_T",T*q*However,_2182T*^o1,:.To*forchokedflow

Loading page image...

Also,H= l*r:l(sincetr4z:1)T22To*:l2Tz:1.2(4800):5760"R.,'L, -2182T"*5760:0.621T",q*FromTableA.3:For+:0.6212,3.12FromTableA.4:For M2:0.5,ry=\.069, pzlp* =2.l3gandT2lT*:1.143.4L,*=4L=,***=1.069+4(0.q0-t(40)!or.o,DDDFromTableA.4:For4L'*D=41.069,m=0.12iTrlT*:l.l96,prlp*:8.675n,=fr #,,:(8.67s)(#3J(rarn)=Eooad3.13FromTableA.4:=2.63V,AgTTFForMr:2.5.'iLtJrD:0.432,L=0.2921.T'-P*'T'F= 0.5333,po,/po*FromTableA.l:ForM1:2:5,!"'-:17.0gpr

Loading page image...

4Lrn-T*flo=0.432*4ff=0.132FromTableA.4:ForY:0.132,M;=l-4q,pzlp*:0.6125,T21T*:0.8314,po,/ po*-r.169*:;i#p,:(0.6t2r,(#)(os):m48d,,=hft,=(0.8314)(#)(5zo;=Fro.FEpo, =#Hfn'=rr.rur>ffi(r7.0e)(0.s):ltrs8;d3.14-L=10.Also,p1=100atm.Hence,p2:10ugn.Iftheexitischoked,the4p*=p2=Pz/10atm.Thus,ptlp*:100/10:10.FromTableA.4, forp1lp*=10,4L,*I:55.83DD=4inches:0.333ft.55.83Da.l-=-='4f,ss.83(0.333)=FgoE4(0.00t3.15StartingwithEq.(3.95),repeatedbelow:dp+pudu:-(ll2p*,Tandnoting that(pr'\t'pu2:(rp)|\YP)a-(3.es)

Loading page image...

Eq. (3.95)becomesdp*Mdu_-A,I'flgl)pu2(D/Fromp:pRTdp dodTppTFrompu:constdp:-dupuCoining (A2)and(A3)dp_dT-dupTuSubstitute(Aa)into(Al).rdr*(yMz_l)!* -tM'Tu2From the adiabatic energy equationfr+{=const2dh+udu:codT*udu:A-dT*udu=0y-l-l-g.4du Idr*14,$=oy-I T/RTuy-lTu(Al)(42)(A3)(As)t:.f:...iof,dr'')M'jgT:-(T-r'uFrom thedefinitionof Machnumber

Loading page image...

,]:a2M2:M2yRTThus,dr:M*1dTuM2T(A7)Substitute(.4.6)into (A7)du-dM -1rn-ttM2iuuM2"u^duJOIVeIOr_.u+:#u*+(y-r)rvr't-'Substitute(,4.6)and(A8) into (A5)-(y-r)M'#U*+(y-r)att-,+(TM,-t,gtl*:(y-r)rvr,t-,tM'z-|-Tr,/P+Mrl#tt*)<r-rlrl-'=ry(atr.)M--t)#u*+(y-r)M,r-'=-ry(9or,finally+#(1-M'?)#tt*jtr-r)rur't''whichis Eq. (3.96).(A8)g=cp (To,z-To,r)

Loading page image...

AtM=2.5,fromTablea.3,lU-=0.7101.'Tt€SincecpTo,tisthetotalenthalpyofthegasenteringthe duct,and qisgivenas30%oofthis totalenthalpy, we haveq=0.3toI,,Hencek:(l.3xo.7ror):0.e23T*FlomTableA.3,for+:O.g23,wehavetot,,0.3:l''T*;

Loading page image...

vlz4.1Ia1:1ffi:{14rc(520t&r:1118fl/secM1=v1la1:335511118:3.0Mo,:M1 sinP:3.0sin(35)"=1.72FromTableA.l.Pz/pr:3.285Tzttr=1.473:Mn, :0.6355?P2:3.285 (2000):@OtAlfrT2:1.473(5201:@M,,:M1cosp:3cos(35o1:2.457w1:2.457(1I 18)=2747fVsec:w2a2:JIRT2=JQW=1357.3fflsec.u2:Mo,a2=0.6355(1357.3):862.6ff/secTan(F -0):wlwz:862.612747= 0.314p-0=17.4,,,..e=p- 17.4:35-17.4=

Loading page image...

vr=rli]**r'=@:Es79m;d4.2Mn=2Fromthe O-B-Mdiagram:F:39.3oMn,:MrsinB=2sin39.3"= I.2668FromTableA.2:po, lpo,:FW4.3F1o-the O-B-M diagram,at M1:3,0r*:34.1oandp= 66o.Hence,the wedgecan benolarger thanahalf-angleof34.1e.:iMn,=MrsinF= (3.0)sin66-o=2.74FromTableA.2:pzlpr:8.592pr=1.01x10sN/mz (standifdsealevelpressure)pz=bp:8.592(1.01x10s1=8.678xlgs+prm'Thisis themaximum pressure.Anyhigherpreslure wouldconespondtoawedgehalf-angle>0,or.,theshockwouldbe detached.4.4P'-5.FromTableA.2,PrMn,2.2sinP=:j'': =,p=38.94o'Mr3.5Yn,=2.2,Mor:M1sinp,Thus,From the O:B-M diagram,fo{M1'.-3,5

Loading page image...

4.5Fromthe O-B-M diagram,lB:3S.1Mn,:M1 sinB:(3.0)sin3go=l.g5FromTableA.2,forMn,=1.85;pz/pr =3.g26,T2/Tr:l.56gand M".:0.6057P2:3.826p1:3.826efi6)=18096h/f,Tz=1.569Tr:1.569(5te;=@v,:ffi=ffi:lr-td4.60-___:l5z=2ooM,=9.6Cr;,QFromthe0-p-Mdiagram,gt:34Mo,:Mrsin34o=(3.6X0.559)=2.01FromTableA.2:M,.:0.5752M2=.y:"'::= '0'5752:2.378x2.38-sn(p-0)sin(34-20)F.q-* e-B-Mdiagram,forIvIz=2.38and02:20",p1-=,j'\,.

Loading page image...

4.7From theO-p-Mdiagram:ForMr= 2.g and 0:30o,0r=Ilo.Mn,Mrsin30o:2.9 (0.5)_1.4FromTableA.2: Mn,:1.4:pz/pr=2.12,T2/Tr=1.255,po,/po,:0.95g2,Mn,:0.7297M-n.ra^11'Itdz::;:"':. --0"739-7:2.272sin(p _0)sin(30_Il)Frome-p-Mdiagram:Fortr42=2.272and02:Ilo,pz= 35.5o.Forthereflected shock.designatethe upstreamnormalMach number asMn,,.Mn,':I\42sinp2:2.272sin(35.5")=1.319FromTableA'2'forMn-'=r.3r9;ptlpz:1.g66,Tzfiz= r.204,po,rpo,=0.97Sg,Mo.:0.776vr,:ffi=*ffi:lr.8,o,:33pr:(1.866)e.r2)(r)=F.e56;dPzPtTTt,=ifrr:(t.2o4xl.25s)(3oo):F53JE'2rl

Loading page image...

FromTableA.lforMr=2.g,po,/p1=27.14Po'-Po'Po'Po'-Po,p;p,:(o'9758)(o'ss82)Q7.l4xl)wherefotM:= l.g7t0",=30,=(6.398)(3.956)=25.ta,,rr-p:-Clean enoughwithin roundofferrors.Po,Es3-sudAsacheck,po,Po,/Pscan becalculated moredirectlyfromTableA.l=6.3984.8(a)Irrsr.s.PrFromTableA.2for]Wn,:2.57:Po,=0.4715*OMo, =0.5065.porFrom theO-p-Mdiagram,forMr=4and,p=40:0=26.3oyr:-r,.--0.5065-Sin(F-e)Si{A-26,,=2.139FromTableA.2,forMr=4.0,*,=0.13gg.FromTableA.l,forMr:4.0,Hence,Po,=L fo,=(o.l38go(rsl.8Xl)Mn,=MrsinF:4Sin40o:2.57Po,(b)

Loading page image...

FromTableA.2,forNh:2.139,no':0.6557po.p",:+I9nt:(0.6557x0.47r5xr5r.sxr)-Po,Po,PrNote thatthetotalpressureincase(b)ishigher thancase(a),indicatingamoreefficientshockcompressiontosubsonicflow forcase(b).Theupstreamtotalpressureis po,:(l5l.8xl)=151.8 atrn.Forcase (a)the loss is totalpressure=151.8-2r.07atrn:130.7 atut.Forcase(b)the lossin totalpresstre:151.8-46.93=104.9atm.Hence,case(b)experiences asmallerlos!inpo, ?ndtherefore is moreefficient.Thisis,fgenerallytrue.Obliqueshockinletsonjetenginesare moreeffectivedevices thantheoldernormalshock inlets.4.9From the O-B-M diagram,andFigure4.22;For0z=20?andMr:3,g=37.8o.Mn,=M1:sinP=(3) sin(37.8)=1.839;P'-3.783PrMo, :0.607g;Mr:,,Y,"', :o='90.t*-== l.ggS:u;r(6-qSin(37.s-20)For0g:l5oandMr:3,F:32.2Mn,:MrsinB:(3) sin(32.2):1.60;b:2.82PrMo,=0.6684;M3:Mo'slun//-e)_0.6684=2_26Sin(322-It.:..',l'..,'j',.,..i..:'.-,:;,';;:--':-:..:.':.'

Loading page image...

Fortheupstreamflowrepresentedfollowingcalculations:byregion2,plota press're deflectiondiagramfromthe0t'pM""=M2sinpPt'Pz9=Pa'Pzo:02-oa,PrPzPt30JJ.J37.24t.646.753.5FortheupstrOamflowfollowingcalculations:I1.091.201.32t.451.60M."=lrzl3sinp3.7834.6r5.727.068.6510.67&-=&-&-o:oa_03PrPrPr20t6l2840048t2t620I1.2r91.5131.8662.2862.820represefltedbyregion3,plotapresswedeflectiondiagramfromthe0+pP+Pz2629JJ36.84t.546.853.7I1.096r.231.351.501.651.82Ir.231.598t.962.4583.013.6982.823.474.515.536.938.4910.43-15-l I-7a-JI59048t2t62024

Preview Mode

This document has 362 pages. Sign in to access the full document!

Report

Study Now!

Document Details

Related Documents

A system is a group of parts that work together to

Solution Manual For Thermodynamics: An Interactive Approach, 1st Edition

Solution Manual for Introduction to Robotics: Mechanics and Control, 4th Edition

Solution Manual For Thermal Radiation Heat Transfer, 6th Edition

Mechanical Vibrations: Theory and Applications 1st Edition Solution Manual

Solution Manual For Theory Of Vibrations With Applications, 5th Edition

Solution Manual For System Dynamics, 4th Edition

Solution Manual for Machine Elements in Mechanical Design, 6th Edition

Solution Manual for Fluid Mechanics for Engineers, 2017 Edition

Solution Manual For Applied Strength Of Materials, 5th Edition

Company

Explore

Study Tools