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Gustavo Narez MAE 3306-Ββ001 Dr. Chudoba Jan. 27, 2012 Problem 1 An airfoil has section lift, drag, and quarter-Ββchord moment coefficients
given by the following equations:
πΆ! = 5.0πΌ + 0.3 πΆ! = 0.2πΌ! + 0.004 πΆ!!/! = β0.04β 0.01πΌ
where πΌ 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 -Ββ5, 0, 5 and 10 degrees.
Solution.
!!"!= 0.25β
!!!!
!!= 0.25β
!!!!
!! !"#!!!! !"#!
= 0.25ββ0.04β 0.01πΌ
5.0πΌ + 0.3 cosπΌ + (0.2πΌ! + 0.004) sinπΌ
πΌ(deg) πΌ πππ π₯!"
π
-Ββ5 -Ββ0.0873 -Ββ0.034261 0 0 0.3833333 5 0.0873 0.301885 10 0.1745 0.278507
π₯!"π = 0.25β
πΆ!!!,!
πΆ!,!= 0.25β
πΆ!!!,!
(πΆ! ,πΌ + πΆ!)cosπΌ β (πΆ! β πΆ!,!) sinπΌ
πΆ!!
!,!= β0.01 πΆ!,! = 5.0 πΆ!,! = 0.4πΌ
= 0.25ββ0.01
(5+ 0.2πΌ! + 0.004)cosπΌ β (5.0πΌ + 0.3β 0.4πΌ) sinπΌ
πΌ(deg) πΌ πππ π₯!!
π
-Ββ5 -Ββ0.0873 0.2520 0 0 0.2520 5 0.0873 0.2520 10 0.1745 0.2521
Gustavo Narez MAE 3306-Ββ001 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
π =π !π»π ! + π»
=6,356,766π 35,000π6,356,766π + 35,000π
= 34,808π
π! = 11,000π π! = 288.15 πΎ π!! = β6.5 !
!"
π! = π! + π!! π! β π! = 288.15 πΎ β 6.5 11 β 0 = 216.65πΎ
π! = π!π!π!
!!!!!!!
= 101,325 π π!216.650πΎ288.150πΎ
β9.806645 π π 2287.05 π2 π 2β’πΎ(β0.00065 πΎ π)
= 22,632 π π!
π! = 20,000 π π! = 216.650 πΎ π!! = 0.0 πΎ/ππ π! = π! + π!! π! β π! = 216.650 πΎ
π! = π! π!!! !!!!!
!!! =22,632 π π! ππ₯π β9.806645 π π 2 20,000β11,000 π287.05 π2 π 2β’πΎ 216.650πΎ) = 5,474.9 π/π2
π! = 32,000π π! = 216.650 πΎ π!! = 1.0 πΎ/π
π! = π! + π!! π! β π! = 216.650 πΎ + 1.0πΎππ
32.000 β 20.000 ππ = 228.65 πΎ
π! = π!π!π!
!!!!!!!
= 868.02 π/π!
π! = 34,808π
π! = 228.65 π!! = 2.8πΎππ
Gustavo Narez MAE 3306-Ββ001 Dr. Chudoba Jan. 27, 2012 π = π! + π!! π β π! = 228.65 + 2.8
πΎππ
34.808 β 32.000 ππ = 236.51 πΎ
π = π!ππ!
!!!!!!!
= 868.02ππ!
235.51 πΎ 228.65 πΎ
!!!!!!!
= 605.19 π/π!
π =ππ π
=605.19 π/π!
287.05π!/π ! β’ πΎ( 236.51 πΎ)= 0.00891 ππ/π!
π = πΎπ π = 1.4 β’ 287.0528 π!/π ! β’ πΎ(236.51) = 308.297 π/π
Gustavo Narez MAE 3306-Ββ001 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 π» = 95,000 ππ‘ . 3048
πππ‘
= 28,956 π
π! = 11,000 π
π =π !π»π ! + π»
=6,356,766π 28,956 π6,356,766π + 28,956 π
= 34,808π
π! = 11,000π π! = 288.15 πΎ π!! = β6.5 !
!"
π! = π! + π!! π! β π! = 288.15 πΎ β 6.5 11 β 0 = 216.65πΎ
π! = π!π!π!
!!!!!!!
= 101,325 π π!216.650πΎ288.150πΎ
!!.!"##$% ! !!!"#.!" !! !!β’!(!!.!!!"# ! !)
= 22,632 π π!
π! = 20,000 π π! = 216.650 πΎ π!! = 0.0 πΎ/ππ π! = π! + π!! π! β π! = 216.650 πΎ
π! = π! π!!! !!!!!
!!! =22,632 π π! ππ₯π !!.!"##$% ! !! !",!!!!!!,!!! !!"#.!" !! !!β’! !"#.!"#!)
= 5,474.9 π/π! π! = 28,956π π! = 216.650 πΎ π!! = 1.0 πΎ/π
π = π! + π!! π! β π! = 216.650 πΎ + 1.0πΎππ
28.956 β 20.000 ππ = 224.65 πΎ
π = π!π!π!
!!!!!!!
= 1,586.27 π/π!
π = 224.65 πΎ β95Β°π πΎ
= 404.37 Β°π
π = 1,586.27 π/π! 0.02088543 πππππ‘!
ππ! = 33.13
πππππ‘!
= 0.230 ππ π
Gustavo Narez MAE 3306-Ββ001 Dr. Chudoba Jan. 27, 2012
π =ππ π
=1,586.27 π
π!
287.05 π! π ! β’ πΎ 224.65 πΎ= 0.0246
πππ! 0.001940320
π ππ’πππ‘!πππ!
= 0.000047729π ππ’πππ‘!
π = πΎπ π = 1.4 β’ 287.0528 π!/π ! β’ πΎ(224.65 πΎ) = 300.468ππ
10.3048
ππ‘π
= 985.79ππ‘π