Experiments were conducted in a wind tunnel with wind velocity 50 km/hr on a flat plate of size 2m long and 1m wide. Take density of air as $1.15kg/m^3$.

Take coefficient of dag as 0.15, and coefficient of lift as 0.75. Determine -

i) Lift Force

ii) Drag Force

iii) Resultant Force

iv)Direction of the resultant force

v) Power exerted by the air on the plate

Solution :

Given :

1. Area of plate, $\quad A=2 \times 1=2 \mathrm{~m}^{2}$

2. Velocity of air, $V =50 \mathrm{~km} / \mathrm{hr} =13.89 \mathrm{~m} / \mathrm{s} $

3. Density of air, $\rho =1.15 \mathrm{~kg} / \mathrm{m}^{3}$

4. Value of $C_{D} =0.15$ and $C_{L}=0.75$

(i) Lift Force $(F_L)$

$$\begin{aligned} F_{L} &=C_{L} \times A \times \rho \times V^{2} / 2 \\ &=0.75 \times 2 \times 1.15 \times 13.89^{2} / 2\\ &=166.404 \mathrm{~N} \end{aligned}$$

(ii) Drag force $(F_D)$

$$\begin{aligned}F_{D} &=C_{D} \times A \times \rho \times U^{2} / 2 \\ &=0.15 \times 2 \times 1.15 \times 13.89^{2} / 2\\ &=33.28 \mathrm{~N} \end{aligned}$$

(iii) Resultant force $(F_{R})$

$$\begin {aligned} F_{R} &=\sqrt{{F_{D}^{2}}+{F_L}^2} \\ &=\sqrt{{{33.28}^{2}}+{166.404}^2} \\ &=169.67 \mathrm{~N} \end {aligned}$$

(iv) The direction of resultant force ( $\theta)$

The direction of resultant force is given by.

$$ \begin{aligned} \tan \theta &=\frac{F_{L}}{F_{D}}=\frac{166.38}{33.275}=5 \\ \theta &=\tan ^{-1} 5\\ &=78.69^{\circ} \end{aligned} $$

(v) Power exerted by air on the plate

Power = Force in the direction of motion × Velocity

$$ \begin{aligned} &=F_{D} \times U \mathrm{~N} \mathrm{~m} / \mathrm{s} \\ &=33.280 \times 13.89 \mathrm{~W} \\ &=462.26 \mathrm{~W} \end{aligned} $$