Problem 3.3 - From table A.3 air 0.4292 SL = air 0.526 kg m...

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Problem 3.3 [Difficulty: 2] Given: Data on flight of airplane Find: Pressure change in mm Hg for ears to "pop"; descent distance from 8000 m to cause ears to "pop." Solution: Assume the air density is approximately constant constant from 3000 m to 2900 m. From table A.3 ρ SL 1.225 kg m 3 = ρ air 0.7423 ρ SL = ρ air 0.909 kg m 3 = We also have from the manometer equation, Eq. 3.7 Δ p ρ air g Δ z = and also Δ p ρ Hg g Δ h Hg = Combining Δ h Hg ρ air ρ Hg Δ z = ρ air SG Hg ρ H2O Δ z = SG Hg 13.55 = from Table A.2 Δ h Hg 0.909 13.55 999 × 100 × m = Δ h Hg 6.72 mm = For the ear popping descending from 8000 m, again assume the air density is approximately constant constant, this time at 8000 m.
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Unformatted text preview: From table A.3 air 0.4292 SL = air 0.526 kg m 3 = We also have from the manometer equation air8000 g z 8000 air3000 g z 3000 = where the numerical subscripts refer to conditions at 3000m and 8000m. Hence z 8000 air3000 g air8000 g z 3000 = air3000 air8000 z 3000 = z 8000 0.909 0.526 100 m = z 8000 173m =...
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This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.

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