partial_solution3

partial_solution3 - W 07%” 3 Wm 5 EL 5 Gavan : M 2 2L}...

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Unformatted text preview: W 07%” 3 Wm 5 EL 5 Gavan : M 2 2L} ks/éml To 1 To = Q‘ToSF KP” d?“ 2533 Mac At 1 Soxmfiml EQF’POQ = ODSOWPQ Y: 113 w) U: :W flu with; : ‘Oééwmam {swarm t WM , 1‘3 (1% )m l /S (athmct) (mm? ’ W “ ‘Rw‘aMW—v-“h wwwwmmmumm - F : :831.s~r : MW CF“ '{N (2~§33X106)CSXISLU or cw f%<%;r>‘?‘ [cl—(5;) J] W figure 38 M3 ,3, -13.} leg/5w b\ Ce: mm + (PM—Tie, )Aawih/M ; ()5; x103 + <?M—Io)*3.?DXL¢>1/Yh m /P%] a, [Pam Tait? (DR) UM (gs/g3 ) Ce, (Vatéee) 1’13 (‘Lméc 300 30 25.5 1?? o 65% 7. u X193 5‘. 60 {DD 10 85 73% éé’lo 6310003 310 F I. 4%); -CQ FPO (Pia ) 300 lIQ/U 2—00 13 so ’ Q lgmm .' 493:0.002 smfa, .. ' ‘ .2.._L%oa_.,azo.5:Q.“,.s¢5.o.~(01d8{¥2.0;..(“0439main, . .. . 1/1215 my 2 2. Wm: +0.. M (crow—mzfskto‘“. ' : 3,9614x105N $anqu ., I ~ H m I I 155% j ‘r $3 I W»? f: Maximum Exhaust Velocity Given: totem mam: Em: a) Maximum possLbLe ue/c" {or a QLVCVL proseLLtz wt b) range of ue/c* {or tgPLc-m VRLLAES ofy Assumgtions: wotl/I'ng Ly» addition to Law Anal! sis: a) "Maximum ue/c" From the notes (or text) for an mm mm, the exit veLothg is gavel/L b5: So the mt'Lb is ue : 2 1/2 2 A?” 71; .1294” c* y—l . 7+1 ., '. ‘po‘ This 0M5 deprast on 'y (a PropeLLawt ProPerta) and the pressure {ratio (a wozzLe geometry/5 issue). This w’LLL be maximized whem the‘cxwwsim ratio is essemt’LaLLU Lwflmi‘ce (pg—>0), or. f \1/2 x 7 \ y+1ZI _\ _ 2 ’ 2 /' 47H} 5) Range A reason/Lame YMVLQE {om/Ls 1.2-1.67. 50' Lie/c”:flax = 2.25 ——1.62 Emgfiicatfions: ’I) ’m oLass motes, showed c*~1 .1-1.4 a0, so ue,max~2—3 a0 2) me does not go to Lwflmitg vame BVEVL {or Lmflmte axPamsLom, lain/Leth max/95 that can be produced is Limited to (Amount o{thsmaL EMI’QH avaiLa‘oLe (1.5., mm TED) ...
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This note was uploaded on 12/12/2010 for the course MECH 351 taught by Professor Chekhov during the Fall '10 term at Concordia Canada.

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partial_solution3 - W 07%” 3 Wm 5 EL 5 Gavan : M 2 2L}...

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