hw09_solutions - AA311 - Atmospheric Flight Mechanics...

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AA311 - Atmospheric Flight Mechanics Autumn 2010 University of Washington Homework 9 Solutions Problem 1 Part a. We assume that on first approximation, the horsepower generated is proportional to the altitude P altitude P sl = r altitude r sl where P altitude = power produced at altitude P sl = power produced at sea level r altitude = air density at altitude r sl = air density at sea level The service ceiling is defined as where the rate of climb is at 100 ft/min. The appropriate plot is shown below So we see that service ceiling º 24720 ft absolute ceiling º 26850 ft Part b. AA311 - Atmospheric Flight Mechanics Christopher Lum Printed by Mathematica for Students
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We can plot the curve of RoCe - 1 vs. h as shown below As can be seen, the area under the curve between 6000 ft and 20000 ft is approximately D T º 1.4708 ä 10 3 seconds º 24.5 minutes (time to climb) Part c. We can graph the range vs. V as shown below 2 hw09_solutions.nb Printed by Mathematica for Students
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part i: maximum range From previous results, we know that the maximum L ê D occurs when the aircraft is flying at V = 144.6721 ft ê s H L ê D L max = 13.6059 at V = 144.6721 ft ê s CLoverCDmax = 13.6059; Vinfty = 144.6721; We can calculate the weight of the fuel as W f = 65 gal À 5.64 lbf gal À Wf = 65 5.64 366.6 So the empty weight of the plane is W 1 = W 0 - W f where W 0 = gross weight of airplane = 2950 lbf hw09_solutions.nb 3 AA311 - Atmospheric Flight Mechanics Christopher Lum Printed by Mathematica for Students
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W0 = 2950; W1 = W0 Wf 2583.4 For this propeller driven aircraft, the range is given by Eq.6.67 R = h C L c C D ln J W 0 W 1 N (Eq.6.67) where h = 0.8 c = 0.45 H lbf L H hp L H hr L À 1 hp 550 ft lbf ê s À ÿ 1 hr 60 min ÿ 1 min 60 sec = 2.27273 ä 10 - 7 Å 1 ft W 0 = gross weight of airplane = 2950 lbf W 1 = empty weight of airplane
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hw09_solutions - AA311 - Atmospheric Flight Mechanics...

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