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ASE 362K - HW 3 Solutions

ASE 362K - HW 3 Solutions - Problem 4 You need to consider...

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Problem 4. You need to consider two different regimes: subsonic and supersonic. For the subsonic case you use the compressible Bernoulli’s equation. For the supersonic case there is a bow shock upstream of the Pitot tube and so you use the Rayleigh Pitot formula.
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Blue: From Isentropic Flow Relations Red: Normal Shock Relations Black: Calculated a=sqrt(rR*(T1/T0)*T0) v=M1*a M1 T1/T0 M2 P1/P02 T0 a v (m/s) 0.1 0.998004 0.1 0.993031 300.6 347.1887 34.71887 0.2 0.992063 0.2 0.972497 302.4 347.1887 69.43774 0.3 0.982318 0.3 0.93947 305.4 347.1887 104.1566 0.4 0.968992 0.4 0.895614 309.6 347.1887 138.8755 0.5 0.952381 0.5 0.843019 315 347.1887 173.5944 0.6 0.932836 0.6 0.784004 321.6 347.1887 208.3132 0.7 0.910747 0.7 0.720928 329.4 347.1887 243.0321 0.8 0.886525 0.8 0.656022 338.4 347.1887 277.751 0.9 0.860585 0.9 0.59126 348.6 347.1887 312.4698 1 0.833333 1 0.528282 360 347.1887 347.1887 1.1 0.805153 0.91177 0.468857 372.6 347.1887 381.9076 1.2 0.776398 0.84217 0.415368 386.4 347.1887 416.6265 1.3 0.747384 0.785957 0.368515 401.4 347.1887 451.3453 1.4 0.718391 0.739709 0.327951
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  • Spring '07
  • DOLLING,D
  • UCI race classifications, Tour de Georgia, Shock wave, Pitot tube, Normal Shock Relations, isentropic flow relations

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