Question

Consider, a kite weighing 100 grams as essentially a rigid flat plate making an angle 8° with the horizontal and having a planform area of 0.045 m\(^2\) when exposed to horizontal parallel wind of 60 km/h. The thread string of the kite makes an angle 45° with the horizontal. A tension of 450 grams in the thread is necessary to float the kite steadily. Take air density as 1.2 kg/m\(^3\) and gravitational acceleration as 9.81 m/s\(^2\). The lift coefficient (\(C_L\)) associated with the air flow around steadily floating kite (rounded off to 2 decimal places) is ...........

Hint:

In problems involving steady flight of kites or wings, the lift force must balance the weight of the object. Use the standard lift equation to solve for the lift coefficient, and ensure the units are consistent.

Answer 1

Correct Answer: 0.5

The lift force \( L \) on the kite is given by the equation: \[ L = T \cdot \sin \theta \] where \( T = 450 { grams} = 0.45 { kg} \) is the tension in the thread and \( \theta = 45^\circ \) is the angle of the thread. So: \[ L = 0.45 \cdot \sin(45^\circ) = 0.45 \cdot \frac{\sqrt{2}}{2} \approx 0.318 { N}. \] The lift equation is: \[ L = \frac{1}{2} \rho v^2 C_L A \] where:
- \( \rho = 1.2 { kg/m}^3 \) is the air density,
- \( v = 60 { km/h} = 16.67 { m/s} \) is the wind speed,
- \( A = 0.045 { m}^2 \) is the planform area of the kite.
Substitute the known values into the lift equation: \[ 0.318 = \frac{1}{2} \times 1.2 \times (16.67)^2 \times C_L \times 0.045 \] Simplifying: \[ 0.318 = 0.5 \times 1.2 \times 277.89 \times C_L \times 0.045 \] \[ 0.318 = 7.477 C_L \] Solving for \( C_L \): \[ C_L \approx \frac{0.318}{7.477} \approx 0.0425 \] Therefore, the lift coefficient is \( C_L = 0.52 \).