hw9 sol

hw9 sol - n. 23 swimwear: {mean Jflemh‘ ‘ 11.23 At a...

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Unformatted text preview: n. 23 swimwear: {mean Jflemh‘ ‘ 11.23 At a given instant of time. two of the span-e pressure waves, each moving at the speed of sound, emitted by a point source moving with constant / — velocity in a fluid at rest are shown in Fig. 1311.23 @ Determine the Mach number involved and indi- — 9"“ cate with a sketch the instantaneous location Of the point source. I '—-—0.15 m— . . FIGUR 1111.2 7/): Mad: Imméer anacmkd wrft, flue mg 'm of flu: parhf' source xiv W/Vea’ 1:} Me 51:9.ch above is add/y 4529an wiflu 53.1,”? as Shawn Maw, Md 2 5m or From five fiefr‘: (3&ch we mic MM 5?}; 4', = 0.0, m ____ 0. I’M 1 our». +1 77m; fi.d/m)/§J5'mr£) = (a; m) 1 W I = (a,w.9(o./€m) z 0.0/67”, (0.09m) —“ 0 5734’ -= '0’”, = 0.9-9? 0 0M7»: M 11.2.“! At the seashore, you observe a high— speed aircraft moving overhead at an elevation of 10,000 ft. You hear the plane 8 s after it passes directly overhead. Using a nominal air temper- ature of 40 °F, estimate the Mach number and speed of the aircraft. x=vé fire Mach numéer is ire/«And #‘o 2%: any/c K by Eg- “3‘7. 73w; MR=-—/— : V fink C 24/50 hm? '-'- g. V'f (ambz‘méj £75- Mimi 2 We aér‘m}: {MW _ Ema” 054’ (5‘00 £){/.4z) z mag 12 L 5 Fur/{her H 0 ’0 v : (Ma) c =(2.a:){ro% 5.) = 227 E. 11.39 The stagnation pressure and tempera- ture of air flowing past a probe are 120 kPa (abs) and 100 °C, respectively. The air pressure is 80 kPa (abs). Determine the air speed and Mach number considering the flow to be (a) incom- pressible; (h) compressible. (A) Assume? Ihcampessié/c f/ow we use Kerm/lz‘f ail/(aim (£57.17) 7'13 confide)" 194:: 56091: and sfifnafibn flak: and gel“ _ Z — Ipa Wifh fine idea! g4; 29544? 9);; 915 5131:, (5594) we oéfim’ _ f; _. __ 2 g m: K J and wméilnfiy 4%: - f and 2 We 05750.2 y = zffi-F) R72 F5 or V a )(373k) .— I” 5720 mam] (x g; ) " =7} 1 .m an Mad: new», we need 9-5 Ma = E = V (g) C. [1.972% 7} defermrée 7‘ we we ma. quafibn of Mean (fijfifljfiafim 7- : 7* __ V1064) = 373g _, [zr7fffiq-;) (1511”) '3; Z 15!? 2(I-#)(285.9 m a}: ‘9" T =33?-SK #34 (0)14) Wr'flx 59.3 We ova/m}: _. 257 3 Ma '5‘ 0.7.25 1| (286-7 42 )(332590-4) I fish/K 7 N if. m ) Si {5) For comp/ESSFb/c 7%“; P 5’0 fiP¢(a.és} --- = ——-— = 0. 6 7 F: /20 M; (44:) and 2%»: Fig.9.} Wt. wad A50 15w Fig. DJ MEL read I z: (7.39 75 and Ma! 7- :02? )(373k)= 332 K flux, V: and 11.45 An ideal gas is to flow menu-apically from a large tank where the air is maintained at a temperature and pressure of 59 “F and 80 pain to standard aunuspheric discharge conditions. Describe in general terms the kind of duct involved and determine the duct exit Mach number and velocity in ft/s if the gas is air. To dew‘ermxhc. Hm dud‘ exif Mach nuMéer Mam} )weuge 53.05“? 0r- fiw m}, Fifi-DJ. Thus, 0) M4 3 Var/ac, a; 4 flack”. 0K Exit (2) '9fo a 7'2 daffrmzée exg‘f vcfpch‘y, DEX”) w: axe. Vex); = (Ma-915'!- ) Ca”: M499;- RQHL 4e (‘3) sullen; s 7; “if fi-r 2- / __ ('4‘ _ +( 2 )ley,‘;. ‘2 or ‘Fb’alr 7;!” = 76/;92': yam 14m Fig.0.) 1% Man”) (’5) P .7 ’ Jae—xii- : ——Ff Fin" n 0./5’3¢? r gaps; find fibers 34m Fig 04} m Corrapana’fnj We?! are, :- /- M‘fmf é. and 7:015?- 3 0 62 77 (Can't!) (con’é) Haem wififi 6737.5 we 06152er - e , ‘56! gm — (mama: 2 32?— - and wIM £3. 3‘ we CWMG 754w“ um, = 0.3 J (17/5 {we )(?22 “aw/w) = {580 H 5.4;}. “,e 7 1 _:£,_ I}.— 51'u,. __ J]. I? cant/ergmj-a’fu/wfmj nay/e, Jr refuu’m' éccaam ML “if f/aw z': rapusam'o. . 11.47 Upstream of the throat of an isentropic converging— diverging nozzle a! section [1), V1 = 150 m/s, p, = 100 Wang). and T, = 20 °C. Ifthe discharge flow is supersonic and the throat area is 0.1 m3, determine the mass flowratc in kg/a for the flow of air. we, defermxhe. fin: Mach numéer m‘ SeCfi'on(!] wr‘fln {MdJI '-'- .5 = _V’ (I) Q E: 1"; 1a For me ya: rave/wed H- is hie/y fia!‘ Ma, 2': £555 mm {.0 be cause. 14 is low. mug, as; flow m‘ Hm firrmf :3 deea’ aims. M: {Mania} flaw is suésom’c and fire, {raw}? flow 5: Superwmk. EV mass f/awrak M: us: 59.1% 6‘0 {p 061517;}, I ‘ m = ,0 A” V’ (2,1 For flormf lie/chy, 1/”,r we use V": W (3) To 0613,}, 7* we. we git/A63- ms; T‘r: 7; (‘1’) 0r ‘Ffl'rdrr’ r“ I F‘a I = . 7- -]:(va/ug 019% 5m 190/ film: 10) ) 75 defamxht 2; we. we Eng-ié. Ms) Z = w (*2:ij a _ ____——__ {Va/ac of fig». Fig.9.: 29-.— MQ‘) K?) To dc+erma€p¢ (a! we use, 7%: r‘a’ex/ 9:5 ejuafibn m": sin/c (Eff/J) 73:05, f — P 533: (8) w; (45:. 55.12.62. 71.“; i K )h (if) (A) For m}- we as: £3" +1., 467%»; M4 z (/57: g!) I (2519 Mfm 1% /_N_ 5i. a) Me)! 7» flow ,3 51mm a; la: mmf. an E7. 7 M; on», 79., (airmaner 1,31“; (.5; (Win! for M4, := 0. 5’?‘ 292;; 7‘; .: = 305 K (0.?6 ) WE”: £j-5’ wc. 56191:}: fl” = (205 g){a.£3;33} (can?) Fray» Fri). we obi"; WF‘H‘ fie he)!” “C H?- p" 1000!“: 5 = ———d ) {/5 fifi/xés) 4.97 and wifl'u £7. {0 F! .7: [H5 Mfdéa)?(0.5292&’) = 60.3 fife/£65) 711811 w)”: 3 J; = [we :00? 1‘1) /’ ————”" — = (235.? m Final! I Wm; g? 11.49 (See Fluids in the News article titled “Rocket nozzles," Sec« tion 11.4.2.) Comment on the practical limits of area ratio for the diverging portion of a converging—diverging nozzle designed lo achieve supersonic exit flow. Ham A}. DJ we see. flu} flu; A/fii"c v3“. Ma curve éuame: Way Site/J wf'fll than”??? Ira/ac: 025M; (m7 {uric 25cm {:11 fl/A’V needed 7‘» achieve oven find/l 905:: 9'. Ma km!) 'SRjjeflfi'nj fflbflca/ bluff)" 1‘0 areal. dfwryence P456 3'; we“: devices“ Fir Cardin/11: , an}? 53. (A?!) fine. A/fli‘ dime-act Hugo headed 74v- Ma = 5‘ is 3749/ 11.65 The Mach number and stagnation pressure of air are 2.0 and 200 just upstream of a normal shock. Estimate the stagnation pressure loss across the shock. We wanf‘ ‘fD deft/rmzhe 7%: thnazfi'oh presure fair acmfl a mom-ml rhock, W _ P Px}; .. {.— 121 {I} y Pa’x( 5x I '3 7E) defermrinc fin n’afindfibn pram/e mfib w: Me xx. HE. 57712:, 1}) FWm}fé=H)M/a have 76%“ F”? 9"“ 76" M”! = 2'0 J 11.68 A total pressme probe like the (me shown in Video V3.8 is inserted into a supersonic air flow. A shock wave forms just up- stream of the impact hole. The probe measures a total pressme of 500 Watts]. The stagnation temperamIe at the probe head is 500 K The static pressure upstream of the shack is measured with a wall tap to be 100 Watts). From these data, estimate the Mach num- ber and velocity of the flow. 77151 is like _Edep_/¢. rib/9. We "my, F‘Fg.P-‘I Wm“ KW 500 fififl") —- 7f M4 Mafia) and read 11.70 An aircraft cruises at a Mach number of 2.0 at an altitude of 15 km. Inlet air is decel— erated to a Mach number of 0.4 at the engine compressor inlet. A normai shock occurs in the inlet difiuser upstream ofthe compressor inlet at a section where the Mach number is 1.2. For isentropic diffusion. except across the shock, and ' for standard atmosphere determine the stagna- tion temperature and pressure of the air entering the engine compressor. 73):, dcas/tra/r’ah process {A fin: hit/cf dram?- is Affirmed 7‘» £5 awake 2’72. since Me an; wnrikferrej ésenr'ngw'c. dim/Jim exceff across five flack. Thus; 7; = (washers/— and 1: T 7;, camp :h/ef v, cfiffliw 1i- lcl- {I} 7:9 deft/flatlth #16 diffiuey wef- fipfnafi’am fgmmfiflc m we? Fig. DJ WI”: “4: 2.0 and read T -- r..- O, 5' 7a. 5 (z ) ,4; :5- fi». elemfibn in Sézndara' ahaspémz we read 16% TdJ’oZe C-7_ 7“=_5{.5°c =2/6-5'K 77mg wiffl 53:. {and 2. we. obfim'a 7 = (inane) __ WK 3dime We} m - ___= 72; dcfermzhe flat: {figmfibn Pfcffarc at (Z: cempyeflav ride? Za— cong. vhf?! MUSCo _ P F - 7 - P , K " (“fly 5W”? (3) race ‘f-I- 0.1099” Id —-— _ . I r“, If!!! g 3 H flue}! [03x F‘” Quasar» :5an “e “5‘ g d'i'fiécnr thief f; 63%» rialcf‘ (4) (ilmuwr (am 5‘ were .é a =l-2nxra N 11 figmm’bm ml” at {5‘ m r QMMWM r5040 52/». Rue C2- (CGWHL) go’in thief We “9’95? Lfimmf * 79m 537.0.) 1% Mjm 52.0. {d flit/9.} ’ .m- t5/9f‘ Btu: 19:»... Fig, P,J we have, éz'Ffiucy {a} It} ID ediffluev kid I 60mm”? 53; Gland 5 1M: :26qu q AzrtX/o $694!) = 732000 L/{qg} h2- é’y . 774m, «rim 5?, 5’ w: a£fizm 5; Com, Fog/cf “[92 “agmgjaaMMH-Uflo) -.: 92,6ao’fi/flgéy : éwfl) To dekyrmine fire 515261: plenty: 42f flue; Wiener xiyzisr‘ we [EV/Mir F217. 12 I wile: qup and = at; and read fit»? rhizf = 0:5"? =. (0.8? )[41 Mmsflrs’z mam—j 11.72 A normal shock is positioned in the diverging portion of a frictionless. adiabatic, con- verging—diverging air flow duct where the cross section area is 0.1 ft‘ and the local Mach number is 2.0. Upstream of the shock. pa = 200 psia and In = 1200 “R. If the duct exit area is 0.15 ff. determine the exit area temperature and pressure and the duct mass flowrate. 7'0 defermxéc Me dun ex)? 7’c‘mfler4fi4re, 7i and FICJIUIE I J 5 I we need 5 and f3 , We can carry}, fries: rafib: 753w»: 7:2. r“; J' gz‘ Fig. DJ finawakj Me rig/q; of" M42. 73:: var/m: a/Mapwe 0615703 799»: Fifi-i:th Awfifi or. known Vfia/HC 07‘ if 5011de 961‘ 6?»: A" i1 = a A M“ (7, f1) 0} are Value of (I: obfizfied 74w». FJIf-D-X WI”. Mew/ac Of Me} “Vanna fray». Fig. 0.7 Wrm q A'wwn Van/we 07° Mex = 2.0. 77w: yam... [37.0.92 29» Macao May=0.53= andfmn Fifi-DJ we raga! fer/14c: =05? j 4, ,7; 52»: Me pné/en. Sfi/cmcnf fl: _ (2/591 "3+ m m and #m: mm EJJ We W A I; = 0.95.2 ):I-J’ :l.2 - ["5" (can'f) can?) Wifh -&'=/'3 we 9a? m Fij- £27.! A!‘ M42: 0-35! (2) .13 = 0-97 (3) 4:2 and E =o.92 (‘0 83:. 77’s Va/ue of 72:1 I: aéhrfied 14am 70: = 2. 59*! a = Q = z: : move (a) 77m“. vafuc of a: 1'; oéi‘auéed 7g)»: ; _ _ P. ’21-ng Lil/Jere. 5»? = 0.72 4:” 73m All Max = 2'0. 73m: 132 : (Zoo/“Wk 2(0'72 ) 5’ I'D.“ ) Mit- 535 Band; we 0675»; a: 7" 2'1 ._- (1200'12 (5.27 )E/zfa‘k 0,2 C51] ) _."'--_-——II WU": 5f: 4444’6 we him. and flea = ((32,054, )(N? 151-" [agfi’L/flw; «M a ' Ezm 3:2)0163 a) I}; : 0.57 5&1: 3“; Hun — J =: :— ...
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hw9 sol - n. 23 swimwear: {mean Jflemh‘ ‘ 11.23 At a...

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