ASE_362K_Assignment_7_Solutions

ASE_362K_Assignment_7_Solutions - ASE 362K, Assignment No 7...

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Unformatted text preview: ASE 362K, Assignment No 7 Thursday= March 8, 2007 1) A jet engine of an aircraft flying at M = 0.9 ingests an airflow of 40kg/s through the inlet of area A,- = 1.5m2. The ambient temperature and pressure at this altitude are 230K and 10" Nm‘2 respectively. Note the streamline pattern. The cross-sectional area of incoming flow (called the "capture A“) is not the same as the physical area of the inlet entry A. This is because the engine does not require such a large mass flow and "spillage" occurs (i.e. flow around nacelle exterior). Since the incoming flow is subsonic and the stream-tube area is diverging the flow is compressed even before it reaches the inlet (called "external compression"). Calculate the ratio of the pressure at the entry Pi to the ambient pressure Pa. 2) A supersonic wind tunnel is sketched below. Po and T0 are 6 x 105 N111'2 and 320K respectively. Calculate the test section Mach # and flow velocity. lat * 9 at. T “r - l II N " . 4 AF WY C -D ISOZLLE. 5 7 9 H ' 3) A converging-diverging nozzle (see below) has Ala/A: = 2.0. The nozzle exhausts to atmospheric pressure. What value of settling chamber pressure PD will generate a Mach number of 0.7 at the nozzle throat? [10 points] 4) A gas (y = 1.4) is flowing along -- - - a 2-D parallel duct at a Mach # of - .7 ' I 0.04. The passage-way then converges (Ae/Ai = 0.0767). At Ae the gas exhausts to vacuum. Calculate Me. Suppose Ae was reduced further so that Ae/Ai = 0.060. Explain what will happen to the flow. 5) Pitot pressure measurements are being made along the centerline of a nozzle operating at the design condition (in this case Me = 3). On the axes below show how the pitot pressure would way along the centerline, given that Po (settling chamber) is fixed. 1‘ :l I. fix...- —. —. “é.- .— lb .- _. — _____ r». w Hae til-€1- __._,P;.r_> we weed H~e SESAQLWA PrfSSWE and H“; _ Elma 'H‘ We know M or Am C: 0-9 5a — LP"— Weed E CGIMGE Am ‘5 So 04% Conn HR. ‘14: __ ra‘n‘o . Ao- /A'L E ML- “'9- Sfisnot'lcv“ -Ffessw'e 01" '03 and 'a' will Ive He SQM'C. Sudan. We Ccmfresaxcvé {AN-n... {DU Fo— lc‘.‘ “in [3-2 _\S;€»nFr'-olouc and lake; 1 “k c. WQ (ON-v: SE)? 6.: bvo. {\fow. .S\_u-Q..,J $050m9min Thus. Lug“ (3)“ U5 G“ —. ff. ._ no“F - 0 (SI HSIJ RSV“ (?B?>(23€> < 3k eFdwcl 14—0 —— <0-:§|>(aH-6> f—‘ga. AR " m1' We Cnva k.) chC Morani ofl-e Q 90 H“! ‘Hva “MM HM: MW: our-9o. (La/“(Dresses OU‘”S\ol€ 0+ “we 5‘96. _.N0u3 :- 51‘ ——w— AG- fi’y Ac. .M‘ O'q J ._._ A“ = (.000! AA: Ex _ So Ax '-'- 1/( oat-‘1 7- = "5 a I 5km A0». 363 .TM ((6%) = A; o-cm .91: ‘7" = [-56% _ AX Had- I}: “Ye M E x.) Hamel. Rue p-OfiCk (HM! :2 Poe Than; ' r»;- Po. P; "— Po; 3 - .2 '3 [H- ‘o'; no} _ {.gos xm‘F {-0 (10‘1" H ('6q X {04" [H 2>(.Lr52]3'3; A r? UK [WM-To A A _. ———~ F; -= 1%» = i} f 3-3” '5 g '2:- U (0 1-9 3.5 TS : 0W4 Oi a ) F; T5 T$ Ti 3 Hm H 1 .5 ; XQT 01M 528 '_ E M lo / 15 C H- % 2.8 2. HZ. 1-5 : 310/6 H—( - ex 2 «>3 1 C ( 1-i)63% 8 (2. 9) OZ QCZZ’S MIME) U: r 3 a) A“; U T5 ...— \\ \1‘7 1'3- : 1 I: H (1)0133 f G Spy-«1 HM Mow .5 at.» where subscw—tc 5 l» ;. maybe/(r. ) amok Po ‘5 Hm: Jam/«c. wa‘hem ‘ @ Gave”- MI 0W“)! AZ 0/ 2‘. MI. W M. Ma FR prov, _a|91r:> Hat H Cwnp;/.ovo‘(—u:3 LE A_/l’-'\* -.: 2.426 1;: ‘5. (lea-r 0v“- {Dawn-rib Lav . We (on—3 t2— w-a'Er/selofl: I-> r Ac {’9‘ P" 4? 1L l—b wlll 3w: vb Me sun-m CM 01°F.“ SMQ H; I) Luau-as— ‘\ .—_ “Jr-1+? A“ 01H? Ac New Ac D- wma ML“, r; 670 :5 A»- Nm our H= O‘OLf-l (0+ ugh-y) A; /fi:K : MM»? 02‘ OK (OWE-caer 3 I/LHW it). n‘f’f'ah’oi FE 81+ a.) he“, [‘12-— o-oq— L; 4__ :u' we mwv wk Ae =- 6-fi7o 0+ A; Se? He "'- 1. gul- w~e km 591(- fxe «\r 670 lo‘fiou: 6-‘10/0. F‘M um'u J [Amway LSM_>\'eM\, M04 N434» Ha. g/Vow Aém'mnb Ad/Ae = [/06 -— (5'66 Tame) Saxb w{\\ Lr Nomad km. 0.0% 1; 7. 1H: o-oq— Ac/A" : [LP-H"? M: 0-0; A/ficfi :AK‘C’IH' :ngcw|:> Wigwam-:5 [6‘66 —m—-crsr —_, 2-)96 AMI—vW-teY lI—r-t-ré 3—. 013'. I"); -- 0er -— (0135602.) “:— o'ol-[v —" 0-003 ‘- 0'03?~ W te Me. Suppose was reduced W1 appen to the flow. so that Ae/Ai = . 5) pitot pressure measnrements are being made along the centerline of a nozzle Operating at the deSLgn conchtlon (in this case Me = 3). On the axes below show how the pitot pressure would vary along the centerline, given that Po (settling chamber) is fixed. -.: _- € In I 7 --\--Iu _ —n q —- ————— (“-0 w A.) as 91¢» 15 SHOE/Omit) {filék probe reach, Po O‘Q‘ BOW) CT... COHJLCZ‘A“ SHA'CQ. {JEWWOpl'CD Once. ("W0 1—; 5“p‘€.“¢°w'6 0\ shock. Ewm w: WV oJf Pray? Wok "NM PraLc MEGDW‘? Hfie— sBSMtFo». FPOb)M‘Q OJY Hm? have) O‘OWHJWQQW 5+ We Pin-)9: he? 538nm???“ flyffijgm 0+ UPSFED‘W‘ ...
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This note was uploaded on 02/07/2011 for the course ASE 362K taught by Professor Dolling,d during the Spring '07 term at University of Texas at Austin.

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ASE_362K_Assignment_7_Solutions - ASE 362K, Assignment No 7...

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