Question

Consider a circular cylinder of diameter 0.5 m and length 2 m, rotating in clockwise direction at a speed of 100 rpm in a flow of velocity 2 m/s. Assume the density of the fluid as 1.225 kg/m\(^3\) and \( \pi = 3.14 \). By Kutta-Joukowski theorem, the lift force on the cylinder is \(\_\_\_\_\_\) N (rounded off to the nearest integer). 
 

Hint:

For problems involving the Kutta-Joukowski theorem: 1. Use the formula \( F_L = \rho \cdot V \cdot \Gamma \cdot L \) for lift force calculations.
2. Ensure the angular velocity (\( \omega \)) is converted from rpm to rad/s correctly.
3. Include the length of the cylinder in the final lift force calculation.

Answer 1

Step 1: Recall the Kutta-Joukowski lift formula.
The lift force (\( F_L \)) on a rotating cylinder is given by: \[ F_L = \rho \cdot V \cdot \Gamma \cdot L, \] where: - \( \rho = 1.225 \, \text{kg/m}^3 \) is the fluid density, - \( V = 2 \, \text{m/s} \) is the free-stream velocity, - \( \Gamma \) is the circulation around the cylinder, - \( L = 2 \, \text{m} \) is the length of the cylinder. The circulation (\( \Gamma \)) is given by: \[ \Gamma = 2 \pi r^2 \cdot \omega, \] where: - \( r = \frac{\text{diameter}}{2} = \frac{0.5}{2} = 0.25 \, \text{m} \) is the radius of the cylinder, - \( \omega \) is the angular velocity in radians per second. Step 2: Calculate the angular velocity (\( \omega \)).
The angular velocity (\( \omega \)) in radians per second is calculated from the rotational speed (\( N \)) in rpm: \[ \omega = \frac{2 \pi N}{60}. \] Substitute \( N = 100 \, \text{rpm} \): \[ \omega = \frac{2 \cdot 3.14 \cdot 100}{60} = \frac{628}{60} \approx 10.47 \, \text{rad/s}. \] Step 3: Calculate the circulation (\( \Gamma \)).
Substitute \( r = 0.25 \, \text{m} \) and \( \omega = 10.47 \, \text{rad/s} \) into the circulation formula: \[ \Gamma = 2 \cdot 3.14 \cdot (0.25)^2 \cdot 10.47. \] Simplify: \[ \Gamma = 2 \cdot 3.14 \cdot 0.0625 \cdot 10.47 = 4.11 \, \text{m}^2/\text{s}. \] Step 4: Calculate the lift force (\( F_L \)).
Substitute \( \rho = 1.225 \, \text{kg/m}^3 \), \( V = 2 \, \text{m/s} \), \( \Gamma = 4.11 \, \text{m}^2/\text{s} \), and \( L = 2 \, \text{m} \) into the lift formula: \[ F_L = 1.225 \cdot 2 \cdot 4.11 \cdot 2. \] Simplify: \[ F_L = 1.225 \cdot 16.44 \approx 20.11 \, \text{N}. \] Round to the nearest integer: \[ F_L = 20 \, \text{N}. \] Conclusion: The lift force on the cylinder is \( 20 \, \text{N} \).