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Unformatted text preview: AA311  Atmospheric Flight Mechanics Autumn 2010 University of Washington Homework 1 Solutions Problem 1 Not applicable. AA311  Atmospheric Flight Mechanics Christopher Lum Printed by Mathematica for Students Problem 2 Using the equation of state r = p R T = 1.2 atm 1.01325 ä 10 5 N m 2 1 atm K 287 J kg K O H 300 K L ρ = 1.2 ∗ 1.01325 ∗ 10 5 287 ∗ 300 1.4122 The specific volume is simply v = 1 ê r v = 1 ê ρ 0.708117 So we have r = 1.4122 kg m 3 v = 0.708117 m 3 kg Clear @ v, ρ D 2 hw01_solutions.nb Printed by Mathematica for Students Problem 3 Part a. We can first convert temperature to Rankine T R = T F + 459.67 H convert fromFarenheit to Rankine L T = 85 + 459.67 544.67 Using the equation of state r = p R T = J 2116 lbf ft 2 N K 1716 ft lbf slug R O H 544.67 R L ρ = 2116 1716 ∗ T 0.00226394 So the density is r = 0.00226394 slug ë ft 3 We can easily compute the volume. V = H 10 m L H 5 m L H 3 m L … 3.281 ft m » 3 V = 10 ∗ 5 ∗ 3 ∗ H 3.281 L 3 5297.98 So the volume is V = 5297.98 ft 3 So the total mass is m = r V m = ρ V 11.9943 The weight of this in lbf is given by W = m a = 11.9943 slugs * 32.2 ft s 2 hw01_solutions.nb 3 AA311  Atmospheric Flight Mechanics Christopher Lum Printed by Mathematica for Students W = m ∗ 32.2 386.216 So we have m = 11.9943 slugs W = 386.216 lbf Part b....
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 Fall '09
 Thermodynamics, Trigraph, Imperial units, Atmospheric Flight Mechanics

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