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Solution Manual For Fundamentals of Aerodynamics, 6th Edition - Document preview page 1

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

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Solution Manual For Fundamentals of Aerodynamics, 6th Edition

Boost your textbook learning with Solution Manual For Fundamentals of Aerodynamics, 6th Edition, providing the answers and solutions you need to succeed.

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Page 1 of 16
Solution Manual For Fundamentals of Aerodynamics, 6th Edition - Page 1 preview imageCHAPTER1"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
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Solution Manual For Fundamentals of Aerodynamics, 6th Edition - Page 2 preview imageDownloadedfromStudyXY.com®+StudyXYSdYe.o>\|iF’prE\3SStudyAnythingThisContentHasbeenPostedOnStudyXY.comassupplementarylearningmaterial.StudyXYdoesnotendroseanyuniversity,collegeorpublisher.Allmaterialspostedareundertheliabilityofthecontributors.wv8)www.studyxy.com
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Solution Manual For Fundamentals of Aerodynamics, 6th Edition - Page 3 preview imageTEA=[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
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Solution Manual For Fundamentals of Aerodynamics, 6th Edition - Page 4 preview image1g|[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
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Solution Manual For Fundamentals of Aerodynamics, 6th Edition - Page 5 preview imagec?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
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Solution Manual For Fundamentals of Aerodynamics, 6th Edition - Page 6 preview image1.5C=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
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Solution Manual For Fundamentals of Aerodynamics, 6th Edition - Page 7 preview imageZ2JXeoSAE=.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
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Solution Manual For Fundamentals of Aerodynamics, 6th Edition - Page 8 preview imageI=(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
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Solution Manual For Fundamentals of Aerodynamics, 6th Edition - Page 9 preview image£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
Page 10 of 16
Solution Manual For Fundamentals of Aerodynamics, 6th Edition - Page 10 preview image=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
Page 11 of 16
Solution Manual For Fundamentals of Aerodynamics, 6th Edition - Page 11 preview imageroMiVia,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
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Solution Manual For Fundamentals of Aerodynamics, 6th Edition - Page 12 preview imageor,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
Page 13 of 16
Solution Manual For Fundamentals of Aerodynamics, 6th Edition - Page 13 preview image_(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
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Solution Manual For Fundamentals of Aerodynamics, 6th Edition - Page 14 preview image11w=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
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Solution Manual For Fundamentals of Aerodynamics, 6th Edition - Page 15 preview imageThesecondintegralisevaluatedfromtheleadingedgetothetrailingedgealongthebottomsurface.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
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Solution Manual For Fundamentals of Aerodynamics, 6th Edition - Page 16 preview image19/6_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
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