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Open Book Midterm

Open Book Midterm - ASE 362K Mid-Semester Exam Tuesday...

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Unformatted text preview: ASE 362K Mid-Semester Exam Tuesday, March 18th , 2008 Open Book — Answer all questions [40 points] 1) A convergent nozzle, shown below, is being run such that the Mach number in the exit is equal to one. The air is supplied by a large tank which is at a stagnation pressure of P0. The stagnation pressure P0 is then doubled. By how much will the mass flow through the nozzle change? [Note: The Mach number at the exit remains fixed at unity] (10 points) lexii- Flame l l l———D Mex}? =- 1'00 2) A thin flat plate is at angle of attack of 8 degrees in a Mach 3 airstream. Using “shock—expansion” theory calculate the drag coefficient of the plate. Shock-expansion theory just assumes that there are leading and trailing edge shocks and expansions as shown in the sketch below. (10 points) 3) Air enters a ramjet combustion chamber with a velocity of 100m/s and a static temperature of 400K. Given that: a) for air, y = 1.4, R = 287 J/kgK, and CD = 1005 J/kgK, b) that air behaves as a perfect gas with a constant specific heat, c) assuming that there is no friction and d) the heating value of the fuel is 40 MJ/kg. (i) What is the maximum amount of heat that can be added in the combustion chamber without reducing the mass flow rate? (ii) For the qmax in part (i) above calculate the fuel-air ratio. (iii) If the fuel-air ratio was to be increased by 10% calculate the reduction in the mass flow rate for the same inlet stagnation pressure and temperature. [Note: when you consider the flow through the combustion chamber you can neglect the fuel flow rate in comparison with the air flow rate.] (20 points) 4:3 [“ 0m? 3 WC 0L5>i nw‘evJ-d; 30v; we. 0|)‘T804 ‘3 Prm Haml- mop) ﬂow.) n2) wok-UL ca“. kt exfare>ycad \$».~LP)3 0.) w.» ‘— (MA/”t , Cam also (at ex (:ﬁcho‘! 9V; LEV-M5 S SE naFTo... o e :. P where P’— Po — K/g—t \ZT‘ (11-33sz a. (la-‘53 H" L 0 q r. QM :- JKRT’ H where T“? T—o (bi-5’s r1" 7’ vi —_ 30.13,: (0.5;?) Pr JTO 12 H- T°>Xo 52 awe CWJLg-My‘ \"Lrev. M = CWJSMV x R, A V; Po «5 chukka) n3 ”mix afoukk- —— Pf a3 Ik .5 ccc‘TaéJ3 Fae, va- Hyena yum a shod... Md 9Xpowuvov; 05‘" Ha: Wmhé& @0130 (pvdr “1‘26 do no? antrk Hue chat/>- T7»? ﬁnale; (M04 full-i JQWA cm Wt PN).>W\$ Lekwd hm: ‘emwz 90432 5koch. amok 3(PQMJ\W4 9] H H V‘ 0; 4 F a. 11;): ‘h‘e H- 01 1,... '2~ O [00 V1 2: oo 1'11 )CZWX‘ > [m . ‘> .> > g ‘ Lr-ag’rr (1-0 mm .M We ‘8 :4 Wm 6: DD» We A W ‘36 be? haw- C Air-W \zu CW We m 938‘ 1;. CE (To; ‘TO|> n)» m; (Mrsmm 3 fue\> .1 (Ioogzgl‘SS’Z-l— 1+0l+~q€r> #03000) ooo u 0 0 p .9 0\ the am mm ”WM L: Q~9(o-ozow> =— 0-0225 Few Hue emu-\$3 eel/n we would kw ‘17»).—T'5\ = (o-oszs}(%o)ooo;ooo> —.. |1°13L-5K loos 0v \$w§cz 7:2. =To’k H53“ ”3* would be \1q5.5+ Lu-O‘P'o‘g 6’- |?Oi~H-K T 0 .1 7.0‘ H . g .. H-Olvr q \$ 6’)“ 0‘ _. T ’“ 9-02.44- O oHl-l- 23q 13‘ -; t -:. 0 — P:\- H, T‘ Pol ‘ 1.0%05- F; " ”#0044” K “918 _. So T\‘-’- {H‘o‘l‘ ) Cl~ohl+> 0.2mm ' \ Haw hvw‘nbth db AN" O’TSIWQ l - Lon—hr 0‘ -' TX PO| _. l'olflf P) ”z A rah/121‘) EMT/1 gw 4))284 Xandfz (1) r33 94 T' \$0 ak “’3‘0% "bk/Peon: mi ‘7:- 0'230‘ ‘1‘: To.) 1 —_ 04\$;ng = [Too-MK (1-0119 P == (POI) ‘ O ‘16! P04. (1 cums) and 0* Drama” (We‘d-w“ M‘ .-.. 0 liq-01”- Tz (T2) ‘ —_ (o 01%?)(1;,>= Lrool’f (roslvr) P7 03°) ‘ = 0957' 9M. (1 own-9 —D')\A-> {KAI-norm): =2 m 10% mot-{3‘50“ M or. ; RO‘X, Po F N Mjo%> T.“ 3 1 Mar Tug/o =(o qstZPoX (013°!) (La-0° > = 0.0167. (0 (‘16?)l3/I ('0 2m») H-OO DH w; 2-22. may; ...
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Open Book Midterm - ASE 362K Mid-Semester Exam Tuesday...

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