HW 4 - ASE 362K, Assignment No 4 1) 2) 3) 4) 5) 6)...

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Unformatted text preview: ASE 362K, Assignment No 4 1) 2) 3) 4) 5) 6) Thursday, Februag 7, 2008 Review the section in Anderson on normal shock waves. A blunt nosed projectile is moving at Mach 4 at sea—level where the static pressure P = 1 atmosphere, and the static temperature T = 20°C. What is the projectile speed in m/s? What are the static pressure, static temperature and the velocity downstream of the normal shock wave? What is the Mach # downstream of the wave, M2? 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? 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 complicated 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. 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 (MK/M22) = M12 — M22. The cylinder on the right is flying at sea—level. (P1=latm=105 Nm"2). Given that the drag comes entirely from the pressure on face "A", what 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]. 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 #. P,+P,F_1_ il 2 p1 [Uzi 2 (y—l)+(y+1)-% p1_(y+1)+(y_1)%' Show that the change in internal energy can be written as 62 — 61 = and that the density ratio can be written as 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 J/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 Anstner ml& 6;) 'NS 1 ‘ial'm ileaézqgéfi [(9le T‘ 2 390K 'qlgzsafsglg u‘,:, Soomls . . H‘ =. 15 0.0 Ms “1:1 . “1'- 0‘53"? Pr-Mm a -. assign N5 Q) [1,41 Hi=o~zz$f (Di—‘44:. P; = to 33m. 1-] \5 \Su3\- VOA-TM? (don‘me . Wm could do (\r «(a \‘ke 7-0.5?» (Back ‘3 0v via ‘1“, ‘WWW N-‘na'e («Icahn-2- - of \SMaV ‘75 make WW9 W Max» uukak Sam M MM|‘/§\A‘q\’\~§6 8m (OM/(0‘ A0 (L ’—'.S£L ‘iibc\a-~ a3 Hos = Vw/ouu So VaQ = am H00- 1. ad, = E (\.+>(23?>(2°\1>3|" b HoQ = \nbna "' [27'1 ~l3- Kim: was r3= 1352M.“ 1 “(min (um; ‘iHS’W HY?“ = “a w- (“1‘5 ("900 H} g (,,eq)"l x o'H-‘ES‘ UL -— 011”; " ‘69-‘3- M" c [(c.w)(zn>(ug§-})1“z (0%“) __ ‘goo -2 ~15 9 ~\ C looo~l5 oLfc/z. .K'VJ'P‘ocH-s —— Scwg decMWako‘L. EB H PM «BM M‘ ‘- 14- r72 =. O'Lf‘Ss. m, z I6 Hz =o'3gz how Ma}. ‘ \‘V- (0'1) “‘1 thF - 0 ‘2 on H\ "> a“ i <“ ‘1"1~Z o-2. <<< \H-Hf‘ “ H I) E12 Wen‘ SWuaLT—ws) m cow. Cheri. wkeflv ‘oaw; (noun ‘LW Kn com ~ uxbkaFdJ W 0&0 CWQBMF.) v..o\o,FEd_ cox/~51» oaa LEV. Q.) M do (‘6‘. WM (3‘ W P\ So check E 5&2 H" onme km 196%. duv'aE-‘ol How about)!“ wuss CT‘H “T” ‘37 TBA 2. T4 \6:\ “‘1 ‘2. ’— 3°°E ‘+ ('2)(‘+)3 - s‘H-Okr T01. = T; {1+ \6':\ NJ) 1 . = 60%(\ + (-2>(-5??>1> -=- gmffi T04 w (Wredv- (5mm M) N, 5° 5M?- H-vwb WW.» B1 WW5 unuk. €‘hnu HA w T; C H'- (mm; cud- hab- G 5 we are (Md (>.l amo! L41 5.: we. 562. lé‘ CWFMah—s (LEM; :- C'2~O§Q>(28w.g> k3 [N15 _ 5%3’242... P‘ a?) 0‘ 5 " -= 1013;00 (Soc) (ZEDC 30°) 63)”) z ‘ ‘1:- “> I/ 6 HM 118/1000 .3 OWL? 4 \/6 N M (Md (QJMa/E 90‘ aflJk—xgaw (Wk dM;LIfO§M % ‘ —. Pox ‘— 10) (H- -L H‘>3.> \l J 3 p 9 H H“; \ L 5W9 ’6 p03- Ca‘vadr ? €°l. acre): ox nw-~a\ skew/L. WW _C__ Sum (21-. =SIIQ’LQ G, . be, 6 can ‘ '2. 'z. 0450‘ r. Ciel-f '- 0:, swfi uz: Hiaz a, [77-3 (puz z. 9H1}; = Ch" 6 - F ' Or (A = Cflz ._.——— 6 f, (A _ *.. 'L h 7. 5 w- ch‘rwwu G F’ ’1: 1 J u has @ \‘5 C} M. at .. Eda”! cc P01:— P.Ll+‘6;‘M23 L 3'8’ ,1 P. [er (-2>('64)3 = l-SzH- P. “WM-C ULR“ ‘31 a fiOv-rwn‘ At H :30 shod» NW6 akeé‘d «5 («ma Pr aw» fa“ A w\‘\ H! We (RanQWh—s pPPJDM 3 WC “M Amsmm 3 the skocL WM oi FOL. . EO} ak H13 W's) :— 0.3183 Pas 7’“ PUA=er11 '2. 3 ,- P01 -. PL (l+(-z>(o.>> b 9 We! ux w; mmeML—m eag- Miz ”’ PL-P‘ (71> ® ez~e\ ez ...
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HW 4 - ASE 362K, Assignment No 4 1) 2) 3) 4) 5) 6)...

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