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:

Always resolve forces carefully: lift balances both kite’s weight and vertical component of thread tension. Then apply lift coefficient formula.

Answer 1

Correct Answer: 0.5

Step 1: Convert data to SI units.
- Weight of kite: $W = 100\,g = 0.1\,kg \times 9.81 = 0.981 \,\text{N}$. - Tension: $T = 450\,g = 0.45\,kg \times 9.81 = 4.4145 \,\text{N}$. - Wind speed: $V = 60\,\text{km/h} = \frac{60 \times 1000}{3600} = 16.67 \,\text{m/s}$. - Area: $A = 0.045 \,\text{m}^2$. - Air density: $\rho = 1.2 \,\text{kg/m}^3$.

Step 2: Force balance on kite.
At steady flight, vertical force balance: \[ L + T \sin 45^\circ = W. \] But careful: The thread tension supports the kite partly upward and horizontally. Actually, the lift force generated by wind must balance vertical forces: \[ L = W + T \sin 45^\circ. \]

Step 3: Compute vertical requirement.
\[ L = 0.981 + 4.4145 \times \sin 45^\circ. \] \[ = 0.981 + 4.4145 \times 0.707 = 0.981 + 3.122 = 4.103 \,\text{N}. \]

Step 4: Lift coefficient formula.
Lift force: \[ L = \tfrac{1}{2}\rho V^2 A C_L. \] So, \[ C_L = \frac{2L}{\rho V^2 A}. \]



Step 5: Substitute values.
\[ C_L = \frac{2(4.103)}{1.2 (16.67^2)(0.045)}. \] First compute denominator: \[ 1.2 \times (278) \times 0.045 = 15.012. \] Numerator: \[ = 8.206. \] So, \[ C_L = \frac{8.206}{15.012} = 0.546. \] Rounded to 2 decimals: \[ C_L = 0.54. \] Final Answer:
\[ \boxed{0.54} \]