Section 6.5 1. A wing with an elliptical planform is flying through sea-level ai

Question

Section 6.5 1. A wing with an elliptical planform is flying through sea-level air at a speed of 45 m/s. The wing loading WIS 1000 N/m2. The wing is untwisted and has the same section from root to tip. The lift curve slope of the section mo is 5.7. The span of the wing is 10 m, and the aspect ratio is 5. Find the sectional-lift and induced-drag coefficients. Find also the effective, induced, and absolute angles of attack. What is the power that is required to overcome the induced drag of the wing?

Explanation / Answer

Solution :- To find:

                  Rho = 1.225kg/m3   ( density of air at sea level)

Since the wing is an elliptical platform the sectional lift coefficient is equal to wing lift coefficient.

         = (0.8062)2/(3.14*5)

   CDi = 0.041

      = 0.8062/5.7

    = - 0.041/0.8062

   = - 0.05 rad

     = 0.1414 + 0.05

     = 0.1922 rad

                                    Di = CDi * 0.5*rho*v2*S

= [(CL)2 * 0.5*rho*v2*S]/ (pi*AR)

Di = [{W/(0.5*rho*v2)*S}2* 0.5*rho*v2*S]/ (pi*AR)

                                          AR = b2/S       

                                          5 = 102/S,

                                          S = 20 m2

    W/S = 1000,     W = 1000*20 = 20000N

Equating and we get

                              Di = [1/(pi*0.5*rho*v2)]*[W/b]2

                                    = [1/(3.14*0.5*1.225*452)]*[20000/10]2

                                                      = 1026.54 N

Power Required           PR        = Di * v0

                                                 = 1026 * 45

                                                  = 46194 N

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