PSol4 - A: = 2%, gfilmfimz F fiffiu‘p Wag—M3)...

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Unformatted text preview: A: = 2%, gfilmfimz F fiffiu‘p Wag—M3) Subs-cafe surest-sonic: Norman! ShOCk' '2. __L_ Ma. M:+ 3"" - fl 5 "' 2r 9'_. a" %TM0 1 :1! iii-‘9’" aha-0 ._o- 15'" - ' "' As M3 M5 J3?!- _ ETfiJ _. 0 37795 3 13‘ = J [—7- J-urJ-awsas‘l __ At 0.31795 tar-(1+5 1.665 = W'ffiOfiMszp Mpg—.1335 D 6—7. A new supersonic passenger aircraft is being designed for flight Mach num— ber 2.5 at an altitude where the ambient pressure and temperature are 9 kPa and 220 K, respectively. The engine inlet configuration shown below allows for double oblique shock deceleration followed by a zone of subsonic decel- eration. The Mach number is 0.5 at the engine inlet plane Losses in the subsonic diffuser are neglected. Determine: a. The Mach numbers M. and M1 in the zones (9 and C2) shown on the drawing. the wave angles 9. and 61. also shown on the drawing, the overall stagnation pressure ratio I’m/p... the t‘NCrtl” static pressure ratio p,{p... the velocity ratio ('5sz for the subsonic diffuser. and the cross-sectional area {41ml} at the engine inlet plane if the engine mass flow rate is 500 ngS. 59.3.”? B An- fir Donald- L- . try-50 and (JD-12:3 ’ ' M.—I.75 5:17.5" 0’ =Gu-° lea—6447.5 Cc) E11" P01 :Ca,89)(0.90)1_ 20.80 _____..—- 3; .15 c! fi .. P i. '+ MM r” H" 2‘5 =. () . _ 75:; m =o.ao H—ffi, [1.5 e L . I; mt. P} (Pea—I M; ) fiLCH-L‘; M?) = 0.58 flea “5(9) = 103.5 kPa. — 2- 1;: 13 nearly/omgm) = 47: K ‘. ._. FE/g'r; = I03.5/(o:ze7x‘l-7f) = 0.765 kg/ms .0 ll ,3 til '11 :.l I .0 U’t f: :R G} on 23 3e :3 ll :18 m/s m ‘- a. 4’0.- '- 0.75512: 35 = 3.00 m e-tz. A ramjet burner is being designed for an overall stagnation temperature ratio of4 and outlet Mach number of 0.85. Two possibilities are being considered: a. The burner cross-sectional area is to be constant. b. The burner cross-sectional area is to be varied in the stream direction in such a way as to maintain the Mach number uniform within the burner. As a first approximation. assume the static presSure varies linearly with duct cross-sectional area. For both cases determine the ratio of outlet and inlet stagnation pressures. For the second case determine the ratio of outlet and inlet areas. Here ignore the effect of frictional losses and take 1; = 1.3. (a) d&:l K=O 1% :zl-l- M4=0-95‘ E=L3 Me in» 81—“ ME [1 “Man = 2' a \l HQJML H‘M‘t- \j H- 0.15 M: Solving ifera'tively M2 = 0.248 “3 511.6%! ; 1 1"... 3 85.) 03 + 3'" M 3’" |+t.3 0.2491,] l+o.i$(a. 1- %: = 0.836 93 __-———— P1A1’ Pit-A'l- 'f' EP‘JA —' D ‘4‘ 1: Wit-pun. —m3“1. [FAA = (sting-«mag 56 l 1A a: 1+2Qf43+mtfi (a W- 1 +g-(gs—1) + two—Ag) 2 Can'l'r'nuify an Arum: =6a¢fi+ %= P2 _. M— at H- fl M fill— m— M:J-n..!.—.—;’§—L 2 ® . - ° I z COMblfllfls CD and Q) .4) +KM:(l "IE-:4 M: “L “'"M‘DL-lf 7%): m m: (new. Marga—1 H + I: t ‘far M.‘._=Mq.$619.5"J 1721=+J K=o, 'r=t.3 05 #1” = 14-56 3..., _ it 1-,... A3. ___, JH- = 0.578 IO- 2. A sounding rocket is to be launched vertically from the earth‘s surface. It is to be designed for a 350-kg payload that must not suffer an acceleration greater than 53 during the burning period. The maximum propellant mass is 1080 kg, and the structural efficiency factor a is 0.1. The solid propellant rocket motor that will be designed for this vehicle is expected to have a spe- cific impulse of 250 s. Takingg a g, and neglecting aerodynamic drag. determine: a. The minimum allowable burning period, b. The maximum height attainable, c. The height allowable if the maximum payload acceleration were limited {045 E21 (IO—15') u: —Jl-:e cufkn—Jéyfi -aet a: t = LLe M— ._ a: — CI-a-i'fi, a is Maximum 4dr 'l:——— 111:. ' -- .1 Mr! :I lOOO ’ ML. 2:350 K3 and 6 0 Ms M: ms = “I K3 €=o.l = m “M 3 wu—l-tooo s to o+lll+350= 3J7 ms...th W 250( 3J7-13': so.“ For 03:“ 5 1': 1+ 5 (For 2.9% = 4— ‘tb = 108,55) 3: _ 8 z Eci.(loh:na1= ugh: R) _ act—béqRWLflR-iv 23¢ a. 7-__ 1 R4) MK "i = egg %‘..#;Ca-= ) =1 5:: I519]: —— ‘— 2. Egg-5+1 For M = 5 Jpn, 3_1_7——3.17-l-i c 2 3.127_§§——————-—~ i’lmax - 39l(250> 6 = 256 l-r-M For QNCS= ;+, ‘1qu = 2—29 KM ...
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PSol4 - A: = 2%, gfilmfimz F fiffiu‘p Wag—M3)...

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