{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw04_solutions

# hw04_solutions - AA311 Atmospheric Flight Mechanics Autumn...

This preview shows pages 1–4. Sign up to view the full content.

AA311 - Atmospheric Flight Mechanics Autumn 2010 University of Washington Homework 4 Solutions Problem 1 We can find pressure at free stream as using Appendix A with h G = 3000 m p = 7.0121 ä 10 4 N ë m 2 pinfty = 7.0121 10 4 ; Because the flow is assumed isentropic, the relationship between freestream pressure and total pressure is found using Eq.4.74 p o = p J 1 + g- 1 2 M 2 N g g- 1 Minfty = 0.5; γ = 1.4; po = pinfty 1 + γ − 1 2 Minfty 2 γ γ− 1 83 178.4 We can apply Eq.4.74 to the point on the wing p o p w = J 1 + g- 1 2 M w 2 N g g- 1 p w = p o J 1 + g- 1 2 M w 2 N g g- 1 where M w = 0.9 Mw = 0.9; pw = po I 1 + γ− 1 2 Mw 2 M γ γ− 1 49 180.1 So we have p w = 49 180.1 N ë m 2 Clear @ pw, Mw, po, γ , Minfty, pinfty D AA311 - Atmospheric Flight Mechanics Christopher Lum Printed by Mathematica for Students

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Problem 2 We can obtain conditions at 25,000 ft from Appendix B r = 1.0663 ä 10 - 3 slug ë ft 3 ρ infty = 1.0663 10 3 ; The problem states that the aircraft is flying with a velocity of 800 ft/s, this is assumed to be the true velocity. So the dynamic pressure experienced by the aircraft is q = Å 1 2 r V true 2 Vtrue = 800; qinfty = 1 2 ρ infty Vtrue 2 341.216 At sea level, the density is r s = 2.3769 ä 10 - 3 slug ë ft 3 . So to experience the same dynamic pressure, the equivalent airspeed is obtained by solving q = Å 1 2 r s V e 2 V e = J 2 q r s N 1 ê 2 ρ s = 2.3769 10 3 ; Ve = 2 qinfty ρ s 1 ê 2 535.827 So we see that V e = 535.8 ft ê s Clear @ Ve, ρ s, qinfty, Vtrue, ρ infty D 2 hw04_solutions.nb Printed by Mathematica for Students
Problem 3 Part a. Assuming laminar flow, the drag force over one side of the plate is given by D è f , l = C f q S recall: C f = 1.328 Re L = 1.328 Re L q S recall: Re L = r V c m = 1.328 J r V c m N 1 ë 2 q S recall: q = Å 1 2 r V 2 = 1.328 J r V c m N 1 ë 2 I Å 1 2 r V 2 M S = 1.328 J r c m N 1 ë 2 V 1 ë 2 I Å 1 2 r V 2 M S D è f , l = 1.328 2 r S J r c m N 1 ë 2 V 3 ê 2 Multiplying by two to obtain the drag over the entire plate we obtain D f , l = 1.328 r S J r c m N 1 ë 2 V 3 ê 2 So we see that laminar skin friction drag is proportional to V 3 ê 2 Part b.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern