Practice Exam - Michael Naper 2010.04.21 AOE 3104 Exam #2...

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Unformatted text preview: Michael Naper 2010.04.21 AOE 3104 Exam #2 Problem 1: Given: = 80 = 44,000 = 2,500 = 150 2 = 2.0 Find: a) the minimum and maximum speed at sea level b) the parameters 0 and (assuming a parabolic drag polar). c) the maximum climb rate at sea level d) the absolute ceiling altitude Michael Naper 2010.04.21 AOE 3104 Exam #2 Problem 1: (continued) Solution: Governing Equations: = 2 , 2 3 = 3/2 , = 2 3 4 0 30 3/2 = 2 = 2 4 30 - = = , ( ) = , ( )- 1 2 Michael Naper 2010.04.21 AOE 3104 Exam #2 Problem 1: (continued) Calculations: a) = 2.377 10-3 2 2 500 3 150 2 2.0 = 83.7 = 230 = 83.7 b) = 40 = 22,000 = 2 2,500 2.377 10-3 3 4 0 150 2 30 30 3/2 3 = 125 = 2 2,500 2.377 10-3 3 150 2 4 0 = 0.01580 = 0.0588 Michael Naper 2010.04.21 AOE 3104 Exam #2 Problem 1: (continued) c) 44,000 = - 22,000 = 8.8 2,500 d) - 1 = 22,000 2 = 44,000 - 1 2 - 22,000 2,500 44,000 =0= = 0.630 15,000 Michael Naper 2010.04.21 AOE 3104 Exam #2 Problem 2: Given: = 60 = 600 2 0 = 0.025 = 0.8 0 = 40,000 = 5,000 = 6,000 1 = 1.0 = 3,600 Find: a) the maximum range (in statutory miles) and corresponding endurance (in hours) b) the maximum endurance (in hours) and corresponding range (in statutory miles) Solution: Governing Equations: 2 1 2 1 1 =2 0 2 - 1 2 1 = 1 0 ln 1 = = 0 3 0 Max Range: Max Endurance: = 0 3 4 = 2 0 Michael Naper 2010.04.21 AOE 3104 Exam #2 Problem 2: (continued) Calculations: = 60 2 =6 600 2 = 1 = 0.0663 0.8 6 a) = 0.025 = 0.354 3 0.0663 4 0.025 = 0.0333 3 = =2 2.377 10-3 2 3 600 2 1 1 -1 3,600 0.354 2 0.0333 1 40,000 1 2 - 35,000 1 2 = 1,967,000 = 373 = 1 1 -1 3,600 0.354 40,000 ln 0.0333 35,000 = 5,110 = 1.420 b) = 0.025 = 0.614 0.0663 = 2 0.025 = 0.05 = 1 1 -1 3,600 =2 2.377 10-3 0.614 40,000 ln 0.05 35,000 2 3 600 2 = 5,900 = 1.640 1 1 -1 3,600 0.614 0.05 1 2 40,000 1 2 - 35,000 1 2 = 1,726,000 = 327 Michael Naper 2010.04.21 AOE 3104 Exam #2 Problem 3: Given: = 90 000 = 35 2 0, = 20 000 2 = 0.02 + 0.05 , = 2.0 = 0.02 = 5 000 Find: a) the ISA sea-level takeoff distance (in meters) using the lift coefficient for maximum acceleration where = 0 = 1.2 b) explicit expressions for the modified parameters and Solution: Governing Equations: = 2 2 , 2 2 - 2 1 - 1 ln + ln 2 2 2 - 2 - = = = 0 - 1 = - + 2 0 + = - Michael Naper 2010.04.21 AOE 3104 Exam #2 Problem 3: (continued) Calculations: a) 2 90 000 = 45.8 1.225 35 2 2.0 3 = = 1.2 45.8 = 55.0 = 0.02 = 0.2 2 0.05 2 = 0.02 + 0.05 0.2 = 9.81 2 = 9.81 2 90 000 = 0.022 = 1.984 20 000 - 0.02 90 000 1 1.225 3 2 35 2 0.022 - 0.02 0.2 + 0 = 2.53 = 4.21 10-5 = 9.81 2 20 000 + 5 000 - 0.02 90 000 = 1 2 4.21 10-5 = 624 ln 1.984 1.984 - 4.21 10-5 0.25 55.0 2 + ln 2.53 - 4.21 10-5 0.25 55.0 2.53 - 4.21 10-5 55.0 2 2 Michael Naper 2010.04.21 AOE 3104 Exam #2 Problem 3: (continued) Calculations: b) = : - ( + ) + cos = = 0: - - + sin = 0 = - - sin - - - - sin + cos = + cos + sin 2 1 - - - = 2 0 + cos + sin 1 = - - - + 2 2 0 + cos + sin - 1 = - + 2 = ...
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