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Unformatted text preview: AA311  Atmospheric Flight Mechanics Autumn 2010 University of Washington Homework 2 Solutions Problem 1 Part a. From the problem statement we know that g = 24.9 m ë s 2 R = 4157 J kg K T = 150 K g0 = 24.9; R = 4157; T = 150; Since the atmosphere is assumed to be isothermal, from Eq.3.9, we have p p 1 = exp I g R T H h h 1 LM (Eq.3.9) We would like to find the distance, h , where the pressure, p , is half of the sea level pressure, p 1 . So we would like to find 0.5 p 1 p 1 = exp I g R T H h h 1 LM note: h 1 = 0 (sea level) Å 1 2 = exp J g h R T N ln I Å 1 2 M =  g h R T h =  ln I Å 1 2 M R T g AA311  Atmospheric Flight Mechanics Christopher Lum Printed by Mathematica for Students h = − Log B 1 2 F R T g0 17 357.9 Part b. Once again, since the atmosphere is isothermal, we can compute the pressure, p , as a function of altitude, h , using Eq.3.9 and assuming that p 1 = 10 atm = 10 * 1.01325 ä 10 5 N ë m 2 p = p 1 exp I g R T H h h 1 LM where h 1 = p 1 = 10 atm = 10 * 1.01325 ä 10 5 N ë m 2 h = desired height Once we know the pressure and temperature at the given altitude, we can use the equation of state to find the density r = p R T Part c. With our newly constructed table, we can find the pressure at an altitude of h = 17357.9 m. We first find the two points closest to this, we see that these correspond to x 1 = 17000 m ﬂ y 1 = 5.1392 ä 10 5 N ë m 2 x 2 = 18000 m ﬂ y 2 = 4.9380 ä 10 5 N ë m 2 We can use our interpolation equation y d = y 2 y 1 x 2 x 1 H x d x 1 L + y 1 where x d = h part, a = 17 357.9 m So we obtain y d = 5.0672 ä 10 5 N ë m 2 This is not exactly half the sea level pressure, this is due to the fact that the interpolation is a linear approximation of a nonlinear...
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 Fall '09
 Imperial units, sea level, Atmospheric Flight Mechanics, Altimeter

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