AerE 344: Undergraduate Aerodynamics and Propulsion Laboratory. Lab Instructions

Transcription

1 ArE 344: Undrgraduat Arodynamics and ropulsion Laboratory Lab Instructions Lab #08: Visualization of th Shock Wavs in a Suprsonic Jt by using Schlirn tchniqu Instructor: Dr. Hui Hu Dpartmnt of Arospac Enginring Iowa Stat Univrsity Offic: Room 51, How Hall Tl: huhui@iastat.du 1

2 ArE344 Lab09: Visualization of Shock Wavs in a Suprsonic Jt by using Schlirn tchniqu Why look at flow through a nozzl? Th nozzl is on of th most important physical systms on a rockt. Whatvr tchnology drivs th propllant, th nozzl is whr that combustion nrgy is turnd into thrust, and th rockt is ultimatly dpndnt on th nozzl for its prformanc. A givn nozzl will produc th most thrust at a spcific altitud: only whn th xit prssur of th nozzl and th ambint prssur ar matchd dos th nozzl rach its pak prformanc. At this point, calld third critical, flow xits smoothly from th nozzl with no shock wavs. At altituds abov or blow th dsign altitud, shock wavs will dvlop outsid, or insid th nozzl, rspctivly. This would prsnt no difficulty if rockts wr mainly usd at on altitud, but nozzl fficincy will drop off svrly abov or blow th altitud for which it was dsignd. This problm of matching th xit prssur to th ambint prssur is why th ara ratios of th Shuttl Main Engin and th Solid Rockt Boostrs ar so diffrnt. Th SRBs, which run through th lowr portion of th ascnt hav an ara ratio of about 7:1, whil th main ngins, which fir th ntir ascnt, hav a much highr dsign altitud and thus a highr ara ratio of ~77:1. This also th rason for th translatabl skirt som rockts mploy. Ths allow th nozzl to hav mor than on dsign altitud, and thus, travl furthr and fastr with lss propllant. hoto courtsy of NASA GSFC

3 Quasi 1D Nozzl Rviw Exampl: Want to find,t,m, tc. givn o,, and nozzl shap. Quasi Ara is allowd to vary along x coordinat, but flow variabls ar functions of x only. Start out with govrning consrvation quations: Mass: V ρ dv + ρu nds = 0 t (1.1) S Momntum: t V S S V Enrgy: U U q ρ + dv + ρ + U nds = pu nds + ρ dv + ρ( f U ) dv t t V S S V V (1.3) Assuming: 1. stady. inviscid 3. no body forcs 4. D flow ρudv + ρ( U n ) UdS = pds + ρ fdv + Fviscous (1.) Quasi 1D: Ara is allowd to vary but flow variabls ar a function of x only Mass: A, u, ρ A+dA,u+du, ρ +dρ 3

4 ( )( )( ) 0 ρua + u + du ρ + d ρ A + da = (1.4) ρua + ρua + ρuda + ρdua + d ρua + highr ordr trms = 0 (1.5) Dividd by through ρ ua da du d ρ + + = 0 A u ρ or (1.6) d( ρua) = 0 Momntum: da ρu A + ( ρ + d ρ)( u + du)( u + du)( A + da) = A ( + d)( A + da) + (1.7) ρu A+ ρu A+ ρu da + u Ad ρ + ρuadu + ρuadu = A A da Ad + da (1.8) ( ) uadu Ad u ρuda+ uadρ + ρadu + ρ = (1.9) d = ρudu (1.10) d d d ρ d udu a ρ dρ ρ dρ = = = (1.11) dρ u = du ρ a (1.1) Substituting quation (1.1) into (1.6) to gt: da du u + 0 du = A u a (1.13) which can b rarrangd to gt: da du u da du + 1 = + ( 1 M ) = 0 A u a A u (1.14) da du = ( M 1) A u (1.15) s which can b usd to dtrmin gnral flow bhavior in a convrging divrging nozzl, as blow: 4

5 M da du < 1 < 0 > 0 da <0 > convrging subsonic < 1 > 0 < 0 da >0 > divrging du <0 > suprsonic> 1 < 0 < 0 dclrating > 1 > 0 > 0 du >0 > acclrating Enrgy: W will not go through th drivation for th nrgy quation but, applying analysis as bfor will giv: dh + udu = 0 or c dt = udu (1.16) Total Static Mach Rlations Isntropic Rlations: ρ T = = ρ T γ γ (1.17) Static Total From Enrgy Equation with u o =0 (total) c T Rarranging To T γ R using c = 1, γ u1 = c T + (1.18) o 1 u = 1+ (1.19) c T 5

6 Insrting a To = γ RT and M = uagivs T u = 1+ (1.0) γ RT o T o = + M (1.1) T 1 = 1+ M γ 1 o 1 M γ ρ = 1+ ρ Givn Eq. s (1.1),(1.), and (1.3) w can now: (1.) (1.3) Find any static proprty in an isntropic flow givn Mach #, o,t o, o. Us/control known total conditions to find mach # through nozzl Ara Mach Rlations From mass u at A M = = 1 u = a, giving a ρ ua = ρua (1.4) A ρ a = A ρ u or A ρ ρo a = A ρo ρ u using isntropic rlations for th dnsity trms A 1 = 1+ M A M γ + 1 Mach # is a function of this ara ratio only. Must find A*. γ + 1 (1.5) (1.6) (1.7) 6

7 Compltly Subsonic Flow = atm (1.8) From isntropic rlation γ o M = 1 (1.9) Can now dtrmin A* and ntir Mach # distribution o If: 1 = Thn: A M = 0,, A = 0 NO FLOW A o A As incrass, M incrass, dcrass, A* incrass A Not that A/A* < 1 is not physically possibl. That is, aftr 1 st critical is rachd, must hav Amin = A* Suprsonic Flow Subsonic Flow ahad of throat. Follow suprsonic A/A* branch aftr throat. A = A t (1.30) A M = f A (1.31) γ 1 γ = o 1+ M (1.3) 7

8 Suppos w pick o so that = atm If w dcras o, thn < atm bcaus M is unchangd. Nd wak obliqu shocks to gt a small prssur jump. As o dcrass, nd strongr obliqu shocks until normal shock at xit, nd critical. As o dcrass, shock movs up th nozzl. Evntually gt to 1st critical. Incrasing o from 3 rd critical, > atm. Gt randtl Myr xpansion fan to gt prssur dcras Summary: For th nozzl usd in th lab: o,3rd o,nd o,1st 60 psig 0 psig 8psig 8

9 Visualization of Shock Wavs in a Suprsonic Jt by using Schlirn tchniqu (a). Bfor turning on th Suprsonic jt (b). Aftr turning on th Suprsonic jt Schmatic of th Z typ Schlirn systm usd in th prsnt xprimnt 9

10 Schlirn imags of th shock wavs in th suprsonic jt flow ArE344 Lab 09: Visualization of Shock Wavs in a Suprsonic Jt by using Schlirn tchniqu Writup Guidlins This is a dmonstration xprimnt. Thr is no lab rport rquird for this lab! 10