MAE305_practice2s

MAE305_practice2s - M&AE 3050 Practice Exam 2 Fa 2009...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: M&AE 3050 Practice Exam 2 Fa 2009 This is for practice only. You do NOT need to submit. 1. We wish to consider the effect of altitude on the difference between true airspeed and calibrated airspeed, making the following comparisons. (a) Determine the i. true airspeed; the ii. equivalent airspeed; and the iii. calibrated airspeed for flight at an altitude of h = 12 , 000 m in the standard atmosphere when the flight Mach number is M ∞ = 0 . 80. (b) Repeat the calculation of true, equivalent, and calibrated airspeed for flight at the same Mach number, but at standard sea level conditions. Solution (a) At a height of h = 12 , 000 m in standard atmosphere, the density is ρ = 0 . 31194 kg/m 3 , the pressure is P = 1 . 9399 × 10 4 Pa and the temperature is T = 216 . 66 K. At standard sea level, air density is ρ s = 1 . 225 kg/m 3 , pressure is P s = 101 , 325 Pa and temperature is T s = 288 . 16 K. For flight at M ∞ = 0 . 8, i. The true airspeed can be found by relating the speed of sound to the Mach number V t = M ∞ a = M ∞ radicalbig γRT V t = M ∞ radicalbig (1 . 4)(287)(216 . 66) = 236 . 04 m/s ii. We calculate the equivalent airspeed as follows V e = V t radicalbigg ρ ρ s V e = (236 . 04) radicalBigg (0 . 31194) (1 . 225) = 119 . 11 m/s iii. The calibrated airspeed is given by V 2 cal = 2 a 2 s γ − 1 bracketleftBigg parenleftbigg P − P P s + 1 parenrightbigg γ- 1 γ − 1 bracketrightBigg Where a s is the speed of sound at standard sea level a s = radicalbig γRT s = radicalbig (1 . 4)(287)(288 . 16) = 340 . 27 m/s To find the pressure drop P − P we employ the isentropic relation for pressure P P = bracketleftbigg 1 + parenleftbigg γ − 1 2 parenrightbigg M 2 bracketrightbigg γ γ- 1 P P = bracketleftbigg 1 + parenleftbigg 1 . 4 − 1 2 parenrightbigg (0 . 8) 2 bracketrightbigg 1 . 4 1 . 4- 1 = 1 . 5243 P P − 1 = 0 . 5243 P − P = 0 . 5243 P = 0 . 5243(1 . 9399 × 10 4 ) = 10 , 172 Pa We then calculate the calibrated airspeed V 2 cal = 2(340 . 27) 2 1 . 4 − 1 bracketleftBigg parenleftbigg (10172) (101 , 325) + 1 parenrightbigg 1 . 4- 1 1 . 4 − 1 bracketrightBigg V cal = 126 . 66 m/s (b) Repeating the approach from 1a using standard sea level conditions we obtain i. V t = 272 . 21 m/s ii. V e = 272 . 21 m/s iii. V cal = 272 . 21 m/s 2. The North American Mustang P-51 was one of the first aircraft designed to incorporate a “laminar flow” wing. For the purpose of estimating skin-friction drag on this aircraft, we will approximate the wing as a flat plate having a rectangular planform of span b = 37 ft and mean chord ¯ c = 6 . 3 ft. We wish to estimate the skin friction drag on the wing at a speed of V = 440 mph at h = 25 , 000 ft in the standard atmosphere for several different scenarios, as described below....
View Full Document

This note was uploaded on 10/02/2010 for the course MAE 3050 taught by Professor Caughey during the Fall '08 term at Cornell.

Page1 / 8

MAE305_practice2s - M&AE 3050 Practice Exam 2 Fa 2009...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online