HW 10 - ASE 362K, Assignment N0 10 Thursday, March 27th...

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Unformatted text preview: ASE 362K, Assignment N0 10 Thursday, March 27th 2008 1) Review the section in Anderson on “Diffusers” before you try the two problems below. In my old version of Anderson it is Chapter 5, Section 5. It is a more detailed discussion of what we covered in class today, namely the “starting problem” for supersonic wind tunnels. 2) A moderately hypersonic wind-tunnel has a nozzle area ratio of Ae/At1 = 104.1 The throat area At; = 1 cm2. What is the minimum area ofthe diffuser throat, At;, which will permit the tunnel to start? 3) A continuous flow supersonic wind-tunnel is designed to simulate the following conditions: M=2.5, P=0.3 x 105 Nm‘2 T=230K. The cross sectional area of the test- section is 180 cm2. The supersonic diffuser is built with a [11911 geometry. Calculate the areas of the nozzle and diffuser throats. How much power is needed to M the tunnel? How much power is needed to operate it in a steady state mode? First, mad mic below Note on wind tunnel operation: In practice we know that there are frictioerhock losses between the nozzle and the diffuser, so that the entropy of the air increases and the stagnation pressure decreases. Compensation for this loss in stagnation pressure is accomplished by power input to the system. The air from the receiver is therefore allowed to enter the intake of a compressor, where the stagnation pressure of the air is raised to its original value. To minimize the amount of power needed by the compressor, and to maintain a constant stagnation temperature the air is cooled in a heat exchanger before entering the compressor (the compressor is assumed here to compress the gas via an M) ijSW-u Ae/Ae. —_ (0+! 5 He -— 10 ’1‘; 3m m cam (seem Ham» MN be \ouobz eMcMSL (2 pa.” (1“ we“; {4M chA. Haw ué NS m: “at (E‘- R¢L1> m' Atz/Aér. : Pm/Pcz Vs (M:‘9—> POL/‘9°\ : 0-013'1 Z Cay-1 REL 1 ._.__.l K Ael 'OIY‘Z (Jo/P: or of“ .2 TO " SW's'f _> Po -— €11-gg’xto3/vm At“ H=7~-§) A/Ps" = 2-636} Anozzle ‘ _I_8’__O = 6‘3'267- ch}- 1-6“?- . h — at z r101»an m v. RA‘X [L5kq] = 6‘11H3I5. Wraq¥ Max 5m LEA-P‘> ‘X r 0\ Jr A A‘K‘WW ’ %z A Huron»- ., ( __L_ >(m-m9 > amaze»? ‘>WM5 LT-~¢\ sm'wf- > Hac- 5LOCL ch‘w; LY; “ac {-579- 5ech‘cv. aw H I; 2.5‘ {49v “Add. {2,219M ~_ o-q—cmoz PM t; (Owflreaaov 1'— P =— r\;\ (hex-“06> l c Mam 0.; = 953mqu «Abut- DC ‘ o ; sfiank’a, ' ouLpuV “wwws Pa- ) P =— WKCP(T,,x—T:L> = w: C\QT:;LE,C —.] To; H” h'm Cowpm>31ow .é [>ngp'é flak Y'VY Toxin = (Rm/Pa.) 6 "EL —- §|?<( Lem)” -'— Lr-‘UrZH' \9 = (61\>(l003">(q.2ur-2>[$_19_§ _ 4'] Zu--L wkx'd. CWrP>fl0vola t? H -: 2.1 A!— M: 2-1 Pal/Pm(/\/.s> = -62&( Po; ~. (-628t>(§I2-59X103)Nn:‘ ’— 3Z\-°\S' x \O.5 Nov-2 1—04; 7- 5|?“S’ ('62666‘18’6 If-ggw . t (6-1\>((oo§>(H-§‘S> Eli-{‘41 Lf'S‘l ...
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HW 10 - ASE 362K, Assignment N0 10 Thursday, March 27th...

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