Consider the airplane shown in Figure The airfoil used in it

Consider the airplane shown in Figure. The airfoil used in its wing is very similar to a double wedge as shown in figure. The pressure and shear distribution are shown in Figure. The mean chord of the airfoil is 1.8m. Based on the information provided, estimate per unit of span: the Normal Force The Axial Force The Lift Force The Drag Force The Leading Edge Moment The C/4 Moment The center of Pressure

Solution

Assuming that the Normal, Axial, Lift and Drag forces (for unit span) respectively are : N, A, L, D. And P, T being Pressure and Shear stress:

L = N cos Asin
D = N sin + Acos

Differential equations for upper and lower surfaces of the air foil:

dN_u = (P_u cos T_u sin ) ds_u
dA_u = (P_u sin + T_u cos ) ds_u
dN_l = (P_l cos T_l sin ) ds_l
dA_l = (P_l sin + T_l cos ) ds_l

ds sin = -dy ds cos = dx

The upper surface could be logically divided into two parts for easy integration (and length of each being approximated to 1, hence S=1, based on the chord length of 1.8)

dN_u = (-84 cos(5) - 300S^_0.5 sin(5)) ds_u + (-84 cos(5) - 280S^0.5 sin(5)) ds_u

= (-83.68 - 26.14 S^0.5) ds + (-83.68 - 24.4 S^0.5) ds = (-167.36 - 30.5S^0.5) ds = -

N_u = -167.36 - 30.5 x S X S^0.5 x 0.5 = -182.61

dN_l = (+84 cos(3) - 280S^_0.5 sin(3)) ds_u + (+84 cos(3) - 290S^0.5 sin(3)) ds_u

N_l = 167.76 - 7.457 = 160.31

dA_u = (-84 sin(5) + 300S^_0.5 cos(5)) ds_u + (-84 sin(5) + 280S^0.5 cos(5)) ds_u

A_u = -14.642 + 288.88 = 274.24

dA_l = (+84 sin(3) + 280S^_0.5 cos(3)) ds_u + (+84 sin(3) + 290S^0.5 cos(3)) ds_u

A_l = 8.792 + 284.60 = 293.40

N = -22.31 kN

A = 567.64 kN

Lift and Drag could be calculated based on :

L = N cos Asin
D = N sin + Acos