Question 15.6: A kite weighing 3.2 N has a planform area of 0.6 m² and is f......

Given data:

Weight of kite                                                      W = 3.2 N

Planform area of kite                                        A = 0.6 m²

Velocity of wind                                                  U_{\infty}=20 \mathrm{~km} / \mathrm{h}=\frac{20 \times 1000}{60 \times 60}=5.556 \mathrm{~m} / \mathrm{s}

Tension in the string                                           T = 5.6 N

Density of air                                                        ρ = 1.22 kg/m³

The arrangement is schematically shown in Fig. 15.5.

Resolving the tension in the string into horizontal and vertical components, we have

T \sin 30^{\circ}=F_D

or                                            F_D=5.6 \sin 30^{\circ}=2.8 \mathrm{~N}

T \cos 30^{\circ}+W=F_L

or                                              F_L=5.6 \cos 30^{\circ}+3.2=8.0496 \mathrm{~N}

Using Eq. (15.14), the coefficient of drag is computed as

F_{D}=C_{D}A{\frac{1}{2}}\rho U_{∞}^{2}        (15.14)

C_D=\frac{F_D}{\frac{1}{2} \rho U_{\infty}^2 A}

=\frac{2.8}{\frac{1}{2} \times 1.22 \times 5.556^2 \times 0.6}=0.2478

Using Eq. (15.13), the coefficient of lift is computed as

F_{L}=C_{L}A{\frac{1}{2}}\rho U_{∞}^{2}    (15.13)

C_L=\frac{F_L}{\frac{1}{2} \rho U_{\infty}^2 A}

=\frac{8.0496}{\frac{1}{2} \times 1.22 \times 5.556^2 \times 0.6}=0.7125