ASE_362K_Assignment_3_Solutions.PDF

# ASE_362K_Assignment_3_Solutions.PDF - ASE 362K, Assignment...

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Unformatted text preview: ASE 362K, Assignment No 3 Thursday, February 8, 2007 1) Review the section in Anderson on normal shock waves. 2) A blunt nosed projectile is moving at Mach 4 at sea-level where the static pressure P = 1 atmosphere, and the static temperature T = ZOEC. a) What is the projectile speed in m/s? b) What are the static pressure, static temperature and the velocity dewnstream of the normal shock wave? c) What is the Mach # downstream of the wave, M2? d) How much does the downstream Mach number change if we double, then quadruple the ﬂight Mach #? e) What is M2 if we accelerate to M1 = 00? 3) This question, on the attached sheet, is from Quiz #1, given when I taught this class in spring 2006. It was in the closed book part. Try it without using your notes. Note that to answer these questions you don’t have to remember normal shock equations and be able to do c0mplicated arithmetic in your head. . ..you just need to bear in mind the conservation equations and other basic things that are “fundamental” to normal shock waves such as the stagnation temperature is constant across the wave, the stagnation pressure goes down etc ...... ..and which should be second nature to you. 4) A normal shock occurs in a ﬂuid which is not a perfect gas. In fact, the pressure and density are related by the expression p [dp/dp] = c, where c = a constant. Show that in this case the upstream and down stream Mach numbers are related by the equation log (Mlz/Mgz) = M[2 —- M22. £9.52 A 5) The Cylinder on the right is ﬂying at sea-level. (1‘31=latm=105 Nm"2). Given that the drag comes entirely M C from the pressure on face "A", what ' 6 is the ratio of the drag at Mach 3 to that at Mach 0.8? [You can assume the pressure on face A is uniform]. 6) A normal shock wave is, in essence, a thermodynamic process (namely a compression) and the change in properties across it can be expressed solely in terms of thermodynamic variables, rather than introducing a "ﬂow" quantity, the Mach #. a+af1 1i 2 laugﬂ P (y — l)+ (2/ + 1):; and that the density ratio can be written as & = p, (y+1)+(y—1)%' Show that the change in internal energy can be written as (22 — el 2 In a student laboratory measurements are being made downstream of a normal shock wave. The conditions upstream of the shock wave were measured by the instructor and are all correct. The wave forms in air for which R = 287J/kg, Cp = 1005 I/kg and Y = 1.4. Some of the measurements are shown on the attached sheet. If you think an error has been made say so and indicate the error, or errors. If you think the measurements are okay, say so (NN points). Note: no answer = no points) N5 Answer iii in i] \$5 Suﬁ- (-0qu? (G'W‘O‘WW ' Lia“ (mid al° l-y v.1 h“ R5115 (Back ’3 QAoLuaox.) Ov vu'k V1,, 'www (aw-Me calwlala. 0r duct- ‘7: may“. 5w baa human “ﬂank aux M'- MM.\P“‘q\-M 3m cowld Ac ('1. (Watson-Ml “3a.. .Llpﬂaw- a) Man ~.—. Von/om 3o VaQ = aloQHoo- I: ad, -_-_ L (l-'+>(28’?>CZ°CS>3 "' b Hm "'=‘+ khan = ‘3}; "*1" 5) Pa. ._ War (“‘2’”) = “1M” ’6’”) 72‘ M ‘S'H Pilp‘ ; PL: lg'SaLM : [Hg-f. 153): 1+ («£20621 . (Zia->06) ._ WW“ 7;: (Lr-o46‘i>CZC\3> : [1%S’}W (warn: 3:] ‘h Wear Shunt-Tom) we COM Cheri. wine-Hav- COVE-M; CW5v¢aLR {om ‘noUV-r Leah ulbkaﬁdb W OHM:- CamouaAJL-Fa vsolo‘ﬁd, (1') M do ncL MM (3‘ W P‘ so cannch c‘nc‘ck E 5&2 N- comnguﬂ-Qw anme haw-1: bet—w U;FIO‘FCA. How Oi‘pw)!‘ amass CTo. =“n-.)? T54 = T4 (1+ \6-3 mg) 7' 1:” EMEHC‘ZDC‘Q} *— 5H—OH' T01. = TLC-1+- \"1\ H17") 1 _ -—- mo +(-z)(-s-w)‘) » mm- Tcm .3 cwwd— (maﬁa. TL; H, M 5° SMtHw-ib was» Ira-r Mara—Z) u-a'hA ‘81th H,“ W TL (ﬂ- m1.) owl- Hal- “w ‘Fn-w 1i LM‘ T15 H2 :5 she’s) 2'. . H 1.”) we could (OkiwAaJI Pc\ “(a—5%» W améwsnm 0E3 shocL. P0; =— 19; Cl+ -L H‘>3'§ 1‘ 7- WW at; P03. -_ P,_(H-2.H’;>3.; t 7+ 33( 1-0653} 5" s a ma. an. I002. cm...» I? ‘99:. “((13); 5ko0L vow: . at. nW---m\ ‘35 Maﬁa?» M; h” .13 (2660.“ HI 9 a“ (oi/{wax at) O'HI "" qul "_"" o PI+ 01‘1‘?‘ ‘- Fiﬁ" (Dzth "—'——- g em- 69 Pl-n = e -——— C9 n. afaeral all: :quz —- H15— Smaz :32». A? ._._ 3...; [LU ‘ C’ b? e w-W __' Mat magma-MS. £0119 .,. Cjolf I (’ km “(9 C “49739 ‘* C (050.2) —_ c le\$(hL1/hf -———® MR6) “‘1 6 " [‘35 till) ﬁzz..le 0v- (obvlrlar— HF—hz? ___ B’qu Q PGI 1 F’.]_I+~:_:H~12 7. 3-" = P. [1+(-2)('5‘*)3 3 '-= [53-H- P1 ' haw-c um“ ‘31- a “Wmn‘ shoal. wane akﬁéol 6 Qua Pf “and F0“; A an.“ be? “ac CEOHQWW prPath-ﬂ' 3 WC how-3 clawan 3 HM- 5kocL mm d. [’01. . Pea. aw M=1 W“) “ 0.3183 Fm ‘0” - me1 Pan: Ps (Ii-3:} H1”) It P61 -.». PL ([+c-z)(°a))3’° -~ gm P, \ﬁﬁna Q: A -.-. (0 357.3%) 6 ‘7‘) P\ '7— F,\ 1' Hal-es G‘Do Ill-3" P01/ P; '0; Wm; SulaSF\IVuE MOA ream-Mob}; (3‘ 1 XI :— XJ—J —- \_/_1 X'l) V7. e2. (9) P; {y-12+( {14) IF. (V+D*(Y") PUP: ...
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## ASE_362K_Assignment_3_Solutions.PDF - ASE 362K, Assignment...

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