Question

Es Ef #3 A hinged flap and slat is used to change the lift and moment characteristics of a symmetric thin airfoil. For the cases given below, find al- and Cmera: Case os Of 0 0.15 O 5° 2 0 0.15 0 10° 3 0.15 0.15 5° 5° 4 0.15 0.15 10° | 10° 1

Answer 1

ㅠ T 0 Given hinged flap slat given Es, Et, ds, os to find &L-of Cmcle Noul from thin air foil theory for ageneral airfoil. CL= 2T (0- A0+ 4112) for slap only to=-(1-1) or Ap=2 singl xdis & A2= sihol di is slate are there Ao = -(1-04 106 + (-42) ds A1 = (2 sin 6-86)-(2 sinda.rs) ㅠ 4 A2 = sin201.04 - Sin202.05 to find & 1402 as (-Et) x6 = & Ci- cos,) => cos4, = 281 ficospe - 1-2Es 2 IT (PA0+ 4) 27ra + 217 [86 €1 - 04)- 05 (147)]+2 [df sind, Y2=0 = AO AL fire-o) = (66 (1-432) - 04(1- ) +(88.singh – og.sinó, Es=0, E5= 0,15, os= 0,05 = 5° D Es=0, E5= 0,15, 05-0,08 = 59 cos, 2 (015)-1 013-) = -0.7 2013 COS$2 = 1-2(0) = 1 + $2=0'corad [0 (1-02) 0.087 (1- 2034 6 346 + [ o(sin 82) 0 ds T 5 01087267 rad 010872 67 rad 13404273 = 243 46rad XL=0 01087 ( sin 134,42+ "] de-0.022)) + (-0.01972) -0.04787 = d2=0 fa mala = II (A2-41) ma4 IT 0.1663 - mely 4 L [oso sin2º1 2 sin 01) ds (sin 262 - 2 sin42)] = [(0.0672) Esin (2x134.427) -2 sf n (24.427)7 - 007 $ 0.0842E-0199979) – 1.4282]} → ® Es = 0, E8=0:15 &s= 0, df = 10° = 001745 rad Again. LL- 07 egno cosdi 2661 2 (0.15)-1 = -067 = 134.427° = 2.346 vad coso2= 1- 2 Es = 1- 210) = 1 +$20 rad. 0.1745 sin134.427 22=0 for ) - 011445 (1 - 26346 sc0.04419).+ (- 0.039667) 01083857 ans. 4 Cmcly - [6.18451[sin kk 134.424) – 2 sin ( 134.421)-001) - 4 [ 601745)(-0.99979) – 27014414)] =-2587 dimillany we can calculate for 34 y steps: of calculate di4d2 from Est Es using equo eq" f IT the formula. 2) cale. &L=0 - using 3. cale ore 14 using fermula - eq" ®. Scanned with CamScanner

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