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hw05_solutions

# hw05_solutions - AA311 Atmospheric Flight Mechanics Autumn...

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AA311 - Atmospheric Flight Mechanics Autumn 2010 University of Washington Homework 5 Solutions Problem 1 Part a. For standard sea level conditions T = 288.16 K r = 1.225 kg ë m 3 p = 1.01325 ä 10 5 N ë m 2 Tinfty = 288.16; ρ infty = 1.225; pinfty = 1.01325 10 5 ; We can first calculate the Mach number M = V a where a = g R T H speed of sound L γ = 1.4; R = 287; Vinfty = 50; a = γ R Tinfty Minfty = Vinfty a 340.269 0.146943 So we see that the speed of sound is 340 m/s and M = 0.14 which is slow enough so we do not need to worry about compressib- lity. In this case, we can calculate the low mach number coefficient of pressure as C p ,0 = p - p Å 1 2 r V 2 where p = 9.5 ä 10 4 N ë m 2 AA311 - Atmospheric Flight Mechanics Christopher Lum Printed by Mathematica for Students

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p = 9.5 10 4 ; Cp0 = p pinfty fractionbar 1 2 ρ infty Vinfty 2 4.13061 So we have C p ,0 = - 4.13 Part b. If the velocity is increased to V = 150 m ê s , we can calculate the Mach number Vinfty = 150; Minfty = Vinfty a 0.440828 So we see that the Mach number is high enough that we need to apply compressiblity effects. The low speed coefficient of pressure is given as C p ,0 = p - p Å 1 2 r V 2 Cp0 = p pinfty fractionbar 1 2 ρ infty Vinfty 2 0.458957 We can now apply the Prandtl-Glauert rule to obtain the actual coefficient of pressure C p = C p ,0 1 - M 2 Cp = Cp0 1 Minfty 2 0.511321 So we have C p = - 0.511 Clear @ Cp, Cp0, Minfty, Vinfty, Cp0, p, Minfty, a, Vinfty, R, γ , pinfty, ρ infty, Tinfty D 2 hw05_solutions.nb Printed by Mathematica for Students
Problem 2 Part a. We are interested in the pressure coefficient of C p = p - p Å 1 2 r V 2 (Eq.1.1) If we assume that the flow is incompressible, we can apply Bernoulli's equation. p + Å 1 2 r V 2 = p + Å 1 2 r V 2 p - p = Å 1 2 I r V 2 - r V 2 M (Eq.1.2) Substituting Eq.1.2 into Eq.1.1 yields C p = Å 1 2 I r V 2 - r V 2 M Å 1 2 r V 2 = r V 2 - r V 2 r V 2 = 1 - r V 2 r V 2 recall: incompressible so r = r C p = 1 - V 2 V 2 V = 195; Vinfty = 160; Cp = 1 V 2 Vinfty 2 êê N 0.485352 So we have C p = - 0.485 This is not a function of density since it only depends on the ratio of velocities, so it would not matter where the wind tunnel is located Clear @ Cp, Vinfty, V D hw05_solutions.nb 3 AA311 - Atmospheric Flight Mechanics Christopher Lum Printed by Mathematica for Students

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Problem 3 From Appendix B, we find some free stream conditions of T = 483.04 o R p = 1.4556 ä 10 3 lbf ë ft 2 r = 1.7556 ä 10 - 3 slug ë ft 3 Tinfty = 483.04; pinfty = 1.4556 10 3 ; ρ infty = 1.7556 10 3 ; We can calculate the speed of sound a = g R T where g = 1.4 R = 1716 lbf ft slug R γ = 1.4; R = 1716; a = γ R Tinfty 1077.24 So we can find the free stream velocity V = M a Minfty = 0.205; Vinfty = Minfty a 220.835 So we have V = 220.8 ft ê s We can calculate the Reynold's number. To do this, we need to first find m . From Figure 4.34, we see that for T = 483.04 o R = 268.35 o K m = 1.65 ä 10 - 5 kg m s = 1.65 ä 10 - 5 kg m s ÿ À 2.20462262 lbm 1 kg À ÿ À 1 slug 32.2 lbm À ÿ 1 m 3.2808399 ft 4 hw05_solutions.nb Printed by Mathematica for Students
μ infty = 1.65 10 5 2.20462262 1 1 32.2 1 3.2808399 3.44332 × 10 7 So we have m = 3.44 ä 10 - 7 slug ft s

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hw05_solutions - AA311 Atmospheric Flight Mechanics Autumn...

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