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HW 8 - ASE 362K Assignment 8 Thursday.I March 6 2008 1 A...

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Unformatted text preview: ASE 362K, Assignment 8 Thursday.I March 6, 2008 1) A jet engine of an aircraft flying at M = 0.9 ingests an airflow of 40kg/s through the inlet of area Ai = 1.5m? The ambient temperature and pressure at this altitude are 230K and 10‘l Nm'2 respectively. Note the streamline pattern. The cross-sectional area of incoming flow (called the "capture area", A3) 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. ...,;r:.rma.gmms:z~zuis.w; N 1"'»'. my: 7 r e 2) A supersonic wind tunnel is sketched below. Po and T0 are 6 x 105 Nm'2 and 320K respectively. Calculate the test section Mach # and flow velocity. f “tr‘fim. T “ u h = _ Wamw C -D not-Lea is ‘73? A1 , 3) A tunnel operator, recently reprimanded for incompetence, is given a second _ ‘ chance and asked to determine the velocity in a new wind tunnel just acquired by his lab. He performs some measurements and claims that the test section velocity is 232 m/s. Has he done a better job this time? ' M56 (5' “I Had. anb‘N’ hi “it sfisnah'on cHaM‘K" ' ll- n_k_.. h In‘LI‘rh-n‘n_ . 1) . T; (3)4»? Hm; SEFIC Pressure w‘ er “(Let _...._.PL..> Wt weed H~e sBequmlfl Pressm a:ch we, - Nada 'H‘ - WE know M OJ? Am (= 0'?) so wave M-eok F: calwiafi. Am b 5° we com HR We. .._FO\\-'\'o . Am / AL E (5421“ ML- he SSS/«aha; __FrGSSu—re onk la‘ cud 'c‘ M“ iv: H‘e Sam: . t .\ ~ _..SUH‘CL We Camemsucw $0... 0“ FE ' L M‘H Lye 6"“ = fa Us» Ac. afid PW"‘0. '- K" - We (0M (5?)? ea ¢_U_o~ {\(M VOJFQMAQEZS‘ 77"“.5‘ 2‘)“ u; A“. Pa 7- ______——'O+ 7- O‘/S' Hal»; \D RT“ (7.8?) (230) u : Ha Imam?“ -— <0®[(t—w>(22%>(2zo>$uz Skefikec’x We {40.9 Cafhue area :3. ‘4an \“Mi enmu/Q wk).- So We {40.x H has PC ‘- P05. 3_ {'67 'K (0" IVM-Z .3 - [‘1‘ 3:2! [733 [H (_2>(.‘+52]3'3 'Z. I'SOS'X ’0‘? NM- 2 ,.PI-' = {-505‘ xm‘? - {.509 Po. {-0 x10“? pm ‘9 Co réaPC/v‘d) g H 1 Q‘g q-rg HTS QM H75 " X2 Tn \\ 3] MSWN.CHAHGe¢. 7‘ o'oq' A/AA' (H:O.Q-l> = “4- if? ISM-e5) AAA! (11-235) 2 (3-23;- ng2' (Foo) 6.4M -— (ALA "’ A? P‘Im' Ax (“3513' 13-275 -: [Lena-K (ABEG' A”; .. 0 WHO? 0v Atm' " ‘°°“‘+ (A3998! Ax (WI-P) ~43 E H z 0 '?~ “ll—11$ = MT;qu am ~. Jxram = M2 To 1 l‘V‘Yf.‘ “Ts 1 .—. @o-o‘f*°‘><l6'§l°\> ...
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