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Gustavo Narez MAE 3306-‐001 Dr. Chudoba Jan. 27, 2012 Problem 1 An airfoil has section lift, drag, and quarter-‐chord moment coefficients
given by the following equations:
𝐶! = 5.0𝛼 + 0.3 𝐶! = 0.2𝛼! + 0.004 𝐶!!/! = −0.04− 0.01𝛼
where 𝛼 is the angle of attack in the radians. Find the center of pressure and the aerodynamic center of the airfoil for angles of attack of -‐5, 0, 5 and 10 degrees.
Solution.
!!"!= 0.25−
!!!!
!!= 0.25−
!!!!
!! !"#!!!! !"#!
= 0.25−−0.04− 0.01𝛼
5.0𝛼 + 0.3 cos𝛼 + (0.2𝛼! + 0.004) sin𝛼
𝛼(deg) 𝛼 𝑟𝑎𝑑 𝑥!"
𝑐
-‐5 -‐0.0873 -‐0.034261 0 0 0.3833333 5 0.0873 0.301885 10 0.1745 0.278507
𝑥!"𝑐 = 0.25−
𝐶!!!,!
𝐶!,!= 0.25−
𝐶!!!,!
(𝐶! ,𝛼 + 𝐶!)cos𝛼 − (𝐶! − 𝐶!,!) sin𝛼
𝐶!!
!,!= −0.01 𝐶!,! = 5.0 𝐶!,! = 0.4𝛼
= 0.25−−0.01
(5+ 0.2𝛼! + 0.004)cos𝛼 − (5.0𝛼 + 0.3− 0.4𝛼) sin𝛼
𝛼(deg) 𝛼 𝑟𝑎𝑑 𝑥!!
𝑐
-‐5 -‐0.0873 0.2520 0 0 0.2520 5 0.0873 0.2520 10 0.1745 0.2521
Gustavo Narez MAE 3306-‐001 Dr. Chudoba Jan. 27, 2012 Problem 2 Compute the absolute temperature, pressure, density, and speed of sound for the standard atmosphere defined in table 1.2.1 at a geometric altitude of 35,000 meters. Solution
𝑍 =𝑅!𝐻𝑅! + 𝐻
=6,356,766𝑚 35,000𝑚6,356,766𝑚 + 35,000𝑚
= 34,808𝑚
𝑍! = 11,000𝑚 𝑇! = 288.15 𝐾 𝑇!! = −6.5 !
!"
𝑇! = 𝑇! + 𝑇!! 𝑍! − 𝑍! = 288.15 𝐾 − 6.5 11 − 0 = 216.65𝐾
𝑝! = 𝑝!𝑇!𝑇!
!!!!!!!
= 101,325 𝑁 𝑚!216.650𝐾288.150𝐾
−9.806645 𝑚 𝑠2287.05 𝑚2 𝑠2•𝐾(−0.00065 𝐾 𝑚)
= 22,632 𝑁 𝑚!
𝑍! = 20,000 𝑚 𝑇! = 216.650 𝐾 𝑇!! = 0.0 𝐾/𝑘𝑚 𝑇! = 𝑇! + 𝑇!! 𝑍! − 𝑍! = 216.650 𝐾
𝑝! = 𝑝! 𝑒!!! !!!!!
!!! =22,632 𝑁 𝑚! 𝑒𝑥𝑝 −9.806645 𝑚 𝑠2 20,000−11,000 𝑚287.05 𝑚2 𝑠2•𝐾 216.650𝐾) = 5,474.9 𝑁/𝑚2
𝑍! = 32,000𝑚 𝑇! = 216.650 𝐾 𝑇!! = 1.0 𝐾/𝑚
𝑇! = 𝑇! + 𝑇!! 𝑍! − 𝑍! = 216.650 𝐾 + 1.0𝐾𝑘𝑚
32.000 − 20.000 𝑘𝑚 = 228.65 𝐾
𝑝! = 𝑝!𝑇!𝑇!
!!!!!!!
= 868.02 𝑁/𝑚!
𝑍! = 34,808𝑚
𝑇! = 228.65 𝑇!! = 2.8𝐾𝑘𝑚
Gustavo Narez MAE 3306-‐001 Dr. Chudoba Jan. 27, 2012 𝑇 = 𝑇! + 𝑇!! 𝑍 − 𝑍! = 228.65 + 2.8
𝐾𝑘𝑚
34.808 − 32.000 𝑘𝑚 = 236.51 𝐾
𝑝 = 𝑝!𝑇𝑇!
!!!!!!!
= 868.02𝑁𝑚!
235.51 𝐾 228.65 𝐾
!!!!!!!
= 605.19 𝑁/𝑚!
𝜌 =𝑃𝑅𝑇
=605.19 𝑁/𝑚!
287.05𝑚!/𝑠! • 𝐾( 236.51 𝐾)= 0.00891 𝑘𝑔/𝑚!
𝑎 = 𝛾𝑅𝑇 = 1.4 • 287.0528 𝑚!/𝑠! • 𝐾(236.51) = 308.297 𝑚/𝑠
Gustavo Narez MAE 3306-‐001 Dr. Chudoba Jan. 27, 2012 Problem 3 Compute the absolute temperature, pressure, density, and speed of sound, in English units, for the standard atmosphere that is defined in table 1.2.1 at a geometric altitude of 95,000 ft. Solution 𝐻 = 95,000 𝑓𝑡 . 3048
𝑚𝑓𝑡
= 28,956 𝑚
𝑍! = 11,000 𝑚
𝑍 =𝑅!𝐻𝑅! + 𝐻
=6,356,766𝑚 28,956 𝑚6,356,766𝑚 + 28,956 𝑚
= 34,808𝑚
𝑍! = 11,000𝑚 𝑇! = 288.15 𝐾 𝑇!! = −6.5 !
!"
𝑇! = 𝑇! + 𝑇!! 𝑍! − 𝑍! = 288.15 𝐾 − 6.5 11 − 0 = 216.65𝐾
𝑝! = 𝑝!𝑇!𝑇!
!!!!!!!
= 101,325 𝑁 𝑚!216.650𝐾288.150𝐾
!!.!"##$% ! !!!"#.!" !! !!•!(!!.!!!"# ! !)
= 22,632 𝑁 𝑚!
𝑍! = 20,000 𝑚 𝑇! = 216.650 𝐾 𝑇!! = 0.0 𝐾/𝑘𝑚 𝑇! = 𝑇! + 𝑇!! 𝑍! − 𝑍! = 216.650 𝐾
𝑝! = 𝑝! 𝑒!!! !!!!!
!!! =22,632 𝑁 𝑚! 𝑒𝑥𝑝 !!.!"##$% ! !! !",!!!!!!,!!! !!"#.!" !! !!•! !"#.!"#!)
= 5,474.9 𝑁/𝑚! 𝑍! = 28,956𝑚 𝑇! = 216.650 𝐾 𝑇!! = 1.0 𝐾/𝑚
𝑇 = 𝑇! + 𝑇!! 𝑍! − 𝑍! = 216.650 𝐾 + 1.0𝐾𝑘𝑚
28.956 − 20.000 𝑘𝑚 = 224.65 𝐾
𝑝 = 𝑝!𝑇!𝑇!
!!!!!!!
= 1,586.27 𝑁/𝑚!
𝑇 = 224.65 𝐾 ∗95°𝑅𝐾
= 404.37 °𝑅
𝑝 = 1,586.27 𝑁/𝑚! 0.02088543 𝑙𝑏𝑓𝑓𝑡!
𝑁𝑚! = 33.13
𝑙𝑏𝑓𝑓𝑡!
= 0.230 𝑝𝑠𝑖
Gustavo Narez MAE 3306-‐001 Dr. Chudoba Jan. 27, 2012
𝜌 =𝑃𝑅𝑇
=1,586.27 𝑁
𝑚!
287.05 𝑚! 𝑠! • 𝐾 224.65 𝐾= 0.0246
𝑘𝑔𝑚! 0.001940320
𝑠𝑙𝑢𝑔𝑓𝑡!𝑘𝑔𝑚!
= 0.000047729𝑠𝑙𝑢𝑔𝑓𝑡!
𝑎 = 𝛾𝑅𝑇 = 1.4 • 287.0528 𝑚!/𝑠! • 𝐾(224.65 𝐾) = 300.468𝑚𝑠
10.3048
𝑓𝑡𝑚
= 985.79𝑓𝑡𝑠
