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hw03_solutions

# hw03_solutions - AA311 Atmospheric Flight Mechanics Autumn...

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AA311 - Atmospheric Flight Mechanics Autumn 2010 University of Washington Homework 3 Solutions Problem 1 Part a. From Appendix B, at h = 50 000 ft, the temperature is T = 389.99 R Tinfty = 389.99; The free stream speed of sound is a = g R T where g = 1.4 R = 1716 ft lbf slug R T = 389.99 R a = H 1.4 L J 1716 ft lbf slug R N H 389.99 R L = 967.942 ft lbf slug We can convert these units, recall that F = m a so 32.2 lbf = 1 slug * 32.2 ft s 2 1 lbf = 1 slug ft s 2 So we have a = 967.942 ft slug slug ft s 2 a = 967.942 Å ft s a = 967.942; We can convert the velocity to ft/s AA311 - Atmospheric Flight Mechanics Christopher Lum Printed by Mathematica for Students

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V = 1500 mi hr 5280 ft mi ÿ 1 hr 60 min ÿ 1 min 60 s Vinfty = 1500 5280 1 60 1 60 2200 Then divide by the speed of sound to obtain the Mach number M = V a M = Vinfty a 2.27286 So we have M = 2.27 Part b. From the energy equation, we have (where denotes free stream values and 's' denotes values at the sensor) c p T s + Å 1 2 V s 2 = c p T + Å 1 2 V 2 Å 1 2 V s 2 = c p T - c p T s + Å 1 2 V 2 V s 2 = 2 c p H T - T s L + V 2 V s = I 2 c p H T - T s L + V 2 M 1 ê 2 = J 2 J 6006 ft lbf slug R N H 389.99 R - 793.32 R L + I 2200 Å ft s M 2 N 1 ê 2 = J 2 J 6006 ft lbf slug R N H 389.99 R - 793.32 R L + I 2200 Å ft s M 2 N 1 ê 2 2 hw03_solutions.nb Printed by Mathematica for Students
cp = 6006; Ts = 792.82; Vs = I 2 cp H Tinfty Ts L + Vinfty 2 M 1 ê 2 34.7281 So we have V s = 34.72 Å ft s Part c. So we see that flow velocity is very slow, so this sensor is probably near a stagnation point. Part d. From the energy equation, we have (where denotes free stream values and 's' denotes values at the sensor) c p T s + Å 1 2 V s 2 = c p T + Å 1 2 V 2 note: V s = 0 c p T s = c p T + Å 1 2 V 2 T s = T + 1 2 c p V 2 Ts2 = Tinfty + 1 2 cp Vinfty 2 792.92 Converting to Farenheit Ts2farenheit = Ts2 459.67 333.25 So we would expect the stagnation point to reach a temperature of T s = 333.25 o F = 792.92 o R Chicken should be cooked to at least 170 o F , so if we could attach the chick to the outside of the aircraft at a stagnation point, it should cook properly. Clear @ Ts2farenheit, Ts2, Vs, Ts, cp, M, Vinfty, a, Tinfty D Problem 2 The speed is slow enough to assume incompressible flow. We can calculate the density of the flow r = p R T hw03_solutions.nb 3 AA311 - Atmospheric Flight Mechanics Christopher Lum Printed by Mathematica for Students

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p = 2116.2; R = 1716; T = 70 + 459.67; ρ = p R T 0.00232827 So we see that the density is r = 0.00232827 slug ë ft 3 We can convert the velocity to ft/s V = 150 5280 1 60 1 60 220 The pitot tube will measure the total pressure, p o . From Bernoulli's equation p o = p + Å 1 2 r V 2 po = p + 1 2 ρ V 2 2172.54 So the pitot tube will measure p o = 2172.54 lbf ft 2 Clear @ po, p, ρ , V, R, T D Problem 3 Part a. The altimeter measures the pressure altitude. So from Appendix B with h = 8000, we see that the static pressure is p = 1.5721 ä 10 3 lbf ë ft 2 p = 1.5721 10 3 ; Since we know the outside temperature, we can calculate the density using r = p R T 4 hw03_solutions.nb Printed by Mathematica for Students
R = 1716; T = 500; ρ = p

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