Solution Manual For Fundamentals of Aerodynamics, 6th Edition

Preview (16 of 178 Pages)

100%

Solution Manual For Fundamentals of Aerodynamics, 6th Edition - Page 1 preview image

Loading page ...

CHAPTER1"11@p=Ro12X0GeRT(287)(203)op1058rarHRI®PR(123x10°)(1716)SOIRTE.1.2N=-[I(pucosO+1,sin0)ds,+["0-7,sin0)d17I.(p,cos@—1,sinb)ds,(1.7)dscos6=dxdssin6=-dyHence,N=-[Fa+[7rr)i=-f,,Guepoder[1@rr)dyTETEN=[1[Pu-p)=pldt[|(tT,dyDividebygeS=gs¢(1)N_Lge(Bee)(zee)I(vz)arCPcCIEqe9.cE\q,dq,Cp=EN(c5,-¢,)dxdt[I(es,+e)dyccThisisEq.(1.15).1I~StudyXY

Solution Manual For Fundamentals of Aerodynamics, 6th Edition - Page 2 preview image

Loading page ...

DownloadedfromStudyXY.com®+StudyXYSdYe.o>\|iF’prE\3SStudyAnythingThisContentHasbeenPostedOnStudyXY.comassupplementarylearningmaterial.StudyXYdoesnotendroseanyuniversity,collegeorpublisher.Allmaterialspostedareundertheliabilityofthecontributors.wv8)www.studyxy.com

Solution Manual For Fundamentals of Aerodynamics, 6th Edition - Page 3 preview image

Loading page ...

TEA=[I(-puSin®+1,cosO)ds,™.+[I(p,sin®+1,cosd)ds,(1.8)TETEA=Gepodyt[1Grn,TE<A=(pup)=-pldy+[more,)dxAgm--.|A_Lpm(2Po).(2be)ydJ(Ev2)oxquecEL®q.ced.gq.1em1reca=3[I(ep,ep)dy+=I.(cq,—cq)dxThisisEq.(1.16).Mg=[I[(pucosB+1,SINB)X—(pySind-7,cosO)y]dsyTE..:+|[-p,cosO+1,sinB)x+(p,sind+1,cosB)y]ds,TETEMies[|[pu-pdxdc-[|(tr)xdyTETE+fespdydy+|Gute)yax”TETEMie=[©[(pu-p)=p,Po)Ixdx-[©(ntt)xdyTETE+fPep)=,op]yay+[mtydxDividebyguc’:My1m=(Rezee)E53Lope(==|—=-xdx-—+Ixdqc’[ie|9.q.c?Jiea)2—StudyXY

Solution Manual For Fundamentals of Aerodynamics, 6th Edition - Page 4 preview image

Loading page ...

1g|[py=Pu)[Pr=Pa1pm(7,7%vals(=e)[2|r1(frye13TEen,=[fC=Coxax-[(C+C,)xdyTE©+f(C=Cydy+[7(Cy4C,)vax)ThisisEq.(1.17).13MossTe—_—SooRy~~;&/%//TLVE4~~~~~/EdyC~FTMie=-[*60-p)@)(x-G,-pa)|]xdxc?Mie==Po)>N=I;®-pu)dX=~pu)©3

Solution Manual For Fundamentals of Aerodynamics, 6th Edition - Page 5 preview image

Loading page ...

c?ve[6-05]X=0TTTm2a)eN(P,P)©13Foraflatplate,6=0inEgs.(1.7)—(1.11).Hence,N=[py-padx=[1[2x10°e1)?+119%10°)dioev3x211N'=-2x10[5x+x]!+[1.19x10%]!=[.12x10A=[waxes[17317024288x0)dx..AT=[1274x")=[1274NL'=N'cosa-A’sina.=1.12x10°cos10°—1274sin10°=[.105x10D'=N'sino+A’cose=1.12x10°sin10°+1274cosa=2.07x10°N]Mie=[°fpu-pdxax=[|2x10°Ge?-119%10%xdx..LoxxPf.+2x10C55ho[0595x1038;=[578x10°NM'ys=M'g+L’(c/4)=-5.78x10°+1.105x10°(0.25)=[3.02x10"N/gq_4xpMeC378x10)_geN112x104Study

Solution Manual For Fundamentals of Aerodynamics, 6th Edition - Page 6 preview image

Loading page ...

1.5C=CpCOSUL-CySINOL=(1.2)cos12°~(0.3)sina.=C4=CpSina+¢,COST=(1.2)sin12°+(0.3)cosa=1.6cy=c,cosa+cqsino|Also,usingthemoreaccurateN'ratherthanL’inEq.(1.22),wehave.SoMaofon)®4ON4Ug,Hence:o®)CnXep/C-2.00.04981.0900.250.412.00.440.3364.00.6390.3066.00.8460.2938.01.070.28410.01.2430.27712.01.4020.27114.01.520.2665Study

Solution Manual For Fundamentals of Aerodynamics, 6th Edition - Page 7 preview image

Loading page ...

Z2JXeoSAE=.LACHR12.order.|.E.0F-:ROCTRDNRTESREDEAEeaSalle-.edio.4+~-\ienErmanEdmeestontsMEAPoleedenBRIi!i!|ZoZo:&oO.Bo.Aooll::.|NotethatX.,movesforwardasaisincreased,andthatitcloselyapproachesthequarter-chordpointintherangeofoof10°to14°.Athigherangles-of-attack,beyondthestall(au>16°),opwillreverseitsmovementandmoverearwardasacontinuestoincrease.Comparetheabovevariationwiththecenter-of-pressuremeasurementsoftheWrightBrothersononeoftheirairfoils,showninFig.1.28.1.7K=3(mass,length,andtime)fi(D,PusVeo,€,2)=0HenceN=35WecanwritethisexpressionintermsofN—K=5—3=2dimensionlessPiproducts:£2(IT,Ty)whereTT;=£5(PesVos©,D)Ih={4(pw,Veo,€,8)LetIT=psV'c!D:6Study

Solution Manual For Fundamentals of Aerodynamics, 6th Edition - Page 8 preview image

Loading page ...

I=(m£3tY5m2H=0mass:a+1=0a=-1length:-3a+b+c+1=0b=-2time:-b-2=0c=-2Hence:PE.EppV.cpV2DheeLetIL=pw’Vac?go!1=(m£30th2°t)'=0mass:a=0a=0length:-3a+1+b+d=0d=-1/2time:-1-2d=0b=-1/2Hence:\Ih=—=JegThus:DV,£2(TT,IT)=f;(2%)=0a.¢*egor:1.8Dw=1fi(PesVo,C,aur,Cp,©)K=4(mass,length,time,degrees)7Study

Solution Manual For Fundamentals of Aerodynamics, 6th Edition - Page 9 preview image

Loading page ...

£5(Dw,PasVoor€,8,Cy€4)=0Hence,N=7.ThiscanbewrittenasafunctionofN—K=7—4=3piproducts:f3=(I1;,IT,TT3)=0where:Th=£4(PeesVeo,€,Cp,D)IT=£5(PeesVeo,©,Cp,8);113=£5(Peo;Veos€,Cp,Cv)|Thedimensionsof¢,andc,arefey]=energy_(force)(distance)_(met?)mass(°)mass(°®)m(°)[ep]=£242(°)"where(°)degrees.ForIT:oopalVocp"D=1I1,(m£3CY(OF(LE)Om£H=1mass:i+1=0i=-1length:-3i+j+k+2n+1=0n=0"time:-j-2n-2=0ji=-2degrees:-n=0k=-2Hence:m=aorfly==ForIp:ThpeVio©cas"8

Solution Manual For Fundamentals of Aerodynamics, 6th Edition - Page 10 preview image

Loading page ...

=m£2)(2eh(ey(2OF(eelmass:i=0i=0length:-3i+1+j+2k+n=0k=0time:-1-2k-n=0n=-1degrees:-k=0j=0Hence:meea,ForIs:CoI;=PorVocpCyT=(m£2)(LY£5Or"(LHOymass:i=0i=0length:-3i+j+k+2n+2=0n=-1time:-j-2n-2=0j=0degrees:-n—1=0k=0Hence:I;==Wecantakethereciprocal,andstillhaveadimensionlessproduct.Hence,’.Ih===Y»Thus,or,9

Solution Manual For Fundamentals of Aerodynamics, 6th Edition - Page 11 preview image

Loading page ...

roMiVia,VT_100800,M,V,a,V,yT,200V200Hence,theMachnumbersofthetwoflowsarethesame.Rei_pve,(2)-ode[b_(12)100)1)[B00_4354Re,pVie,gy,pVe,VT1.739/\200/\2/V200TheReynold’snumbersaredifferent.Hence,thetwoflowsarenotdynamicallysimilar.1.10Denotefreeflightbysubscript1,andthewindtunnelbysubscript2.Fortheliftanddragcoefficientstobethesameinbothcases,theflowsmustbedynamicallysimilar.HenceM,=M,andRe;=Re;ForMachnumber:V,_V,a,a,SinceaowVT,wehaveVv,\4250JRESRB.A15[©]NoV,V,cForReynoldsnumber:oYPTEHiHyAssume,asbefore,thatpoJT.Hence£2Ye,_piVieJnoJT10Study

Solution Manual For Fundamentals of Aerodynamics, 6th Edition - Page 12 preview image

Loading page ...

or,pV:pV,(&J_(0414)250)©JITLe,2231or,LY:C34650)NesFinally,fromtheequationofstate:p,101x10°|T=2=—"—"++—=35193p=3573)|Egs.(1)—(3)representthreeequationsforthethreeunknowns,p,,Va,andT>.Theyaresummarizedbelow:v,—==16713[©]£22Ys3465@VTpT2=351.93)FromEq.3):02=351.9/T,“)Subst.(4)into(2):9)V.wa)=34.65(5)TTSubst.(1)into(5):22(16.7)=34.652Hence,11Study

Solution Manual For Fundamentals of Aerodynamics, 6th Edition - Page 13 preview image

Loading page ...

_(3519)(167)51,-COND)_from)PTTT(3465)FromEq.(1):V2=16.7JT,=1671696=[217.5=sec35193519kgFromEq.(3):==P07—=Orp=mp20pr]111py=pa-pgAh=1.01x10°-(1.36x10*)(9.8)(0.2)po=[7.43x10°N/m1.12Weight=Buoyancyforce+liftw=B+LB=(15000)(1.1117)(98)=1.634x10°N———~——volumeairdensityacceleration(m®)at1000mofgravity(kg/m®)(m/sec)Go=:PuVil=5(1.1117)(30)*=500N/m?S=nd¥4=n(14)%/4=153.9m®L=qaSCL=(500)(153.9)(0.05)=3487NHence:W=1.634x10°+3847=[1.67x101.13LetususetheformalismsurroundingEq.(1.16)inthetext.Inthiscase,cq=c,,andfromEq.(1.16),neglectingskinfriction12Study

Solution Manual For Fundamentals of Aerodynamics, 6th Edition - Page 14 preview image

Loading page ...

11w=I.(c,,-c,)ay1FromEq.(1.13)inthetext,Eq.(1)abovecanbewrittenas1pm.w=[,(Cc,-C,)Csinods)©)Drawapicture:i-6Followingoursign?iconvention,note1that0isdrawnViocounterclockwise-_—[47inthissketch,henceANitisanegativeangle,-6.Fromthegeometry:-B=m-¢Hence,sin(-8)=-sin0=sin(1-8)=cos¢SubstitutethisintoEq.(2),notingalsothatds=rd¢andthechordcistwicetheradius,¢=2r.FromEq.(2),1TEc=>I.(c,,-C,)cosdprdg1TEcq=5[I(c,,-c,)cospd_1TE1TEa=IC,cosdd>I.C,,cosdpdo3Considerthelimitsofintegrationfortheaboveintegrals.Thefirstintegralisevaluatedfromtheleadingedgetothetrailingedgealongtheuppersurface.Hence,=0atLEand7atTE.13Study

Solution Manual For Fundamentals of Aerodynamics, 6th Edition - Page 15 preview image

Loading page ...

Thesecondintegralisevaluatedfromtheleadingedgetothetrailingedgealongthebottomsurface.Hence,¢=27atLEand=attheTE.Thus,Eq.(3)becomesc=4|”c,cosodoLeC,cos¢d@)d2JoPuPRETPeInEq.(4),C,=2cos’§for0<¢<m/2TC,,=0for><g=n23C,,=2cos"¢for=<¢=<2m3zCc,=0forns¢s—Thus,Eq.(4)becomescq=[cos’do-[rrcos’od¢Sincecos’¢pd§=asing)(cos”¢+2),Eq.(5)becomes3a=[(5sing)eos”6+2127(5sind)(cos'p+21%11C=5H)-(3)D@.1.1414Study

Solution Manual For Fundamentals of Aerodynamics, 6th Edition - Page 16 preview image

Loading page ...

19/6_FLuib1—TLfp—_oA,—A,—7[IE---—(BoyaE—4]-—\SuBNERGED1B--——|mathy=—=xFru|Iho--——iIEEooa4,—h__od_————_ra\FfConsiderthearbitrarybodysketchedabove.ConsideralsotheverticalcylinderelementinsidethebodywhichinterceptsthesurfaceareadA,nearthetopofthebody,anddA,nearthebottomofthebody.ThepressuresondA;anddA,arep;andp;respectively,andmakesangles0;and0,respectivelywithrespecttotheverticallinethroughthemiddleofdA;anddA,.Thenetpressureforceinthey-directiononthiscylinderis:dF,=-p)cos6;dA;+p;cos0;dA;[€))]LetdA,betheprojectionofdA;anddA;onaplaneperpendiculartotheyaxis.dA,=cos0;dA;=cos0;dA,Thus,Eq.(1)becomesdF,=(py—pi)dA,2)Fromthehydrostaticequationh,p-pi=[7pedy®CombiningEgs.(2)and(3),hydF,=|Lopedyda,“However,dydA,=dV=elementofvolumeofthebody.Thus,thetotalforceintheydirection,Fy,isgivenbyEq.(4)integratedoverthevolumeofthebody15Study

Preview Mode

This document has 178 pages. Sign in to access the full document!

Report

Study Now!

Document Details

Related Documents

Investigating the Impact of Apparel Creation on En

The Sauer Condiment NYT

Understanding the Marine Corps Performance Evaluat

SERE 100

Harley Watson Livestock Terminology Student Handou

Navigating Blackboard-How to Access and Use the On

Understanding the Importance of Putouts and PO in

Alcoholic Beverage Trend

Human Nature - A Threat to Security

Humanities Final Exams

Company

Explore

Study Tools