Question

A horizontal cylinder of 1.0 m diameter is placed transversely at the aft of a ship and is completely immersed in water. The cylinder rotates at 100 RPM and inflow velocity is 10 m/s. Water density is 1000 kg/m\(^3\). Assuming an ideal planar flow, the lift force per unit length acting on the cylinder is ____________ kN/m (rounded off to one decimal place).

Hint:

Rotating cylinders generate strong lift due to Magnus effect; circulation increases with angular speed.

Answer 1

Correct Answer: 163.3

Using Kutta–Joukowski lift theorem: \[ L' = \rho\, V_\infty\, \Gamma \] The circulation for a rotating cylinder is: \[ \Gamma = 2\pi R^2 \omega \] Given:
Diameter = 1 m → \(R = 0.5\ \text{m}\)
RPM = 100 →
\[ \omega = \frac{100 \times 2\pi}{60} \approx 10.47\ \text{rad/s} \] Thus: \[ \Gamma = 2\pi (0.5)^2 (10.47) \approx 16.45\ \text{m}^2/\text{s} \] Lift per unit length: \[ L' = 1000 \times 10 \times 16.45 \approx 164500\ \text{N/m} \] Convert to kN/m: \[ L' \approx 164.5\ \text{kN/m} \] Thus, the value lies within: \[ \boxed{163.3\ \text{to}\ 165.5\ \text{kN/m}} \]
Final Answer: 163.3–165.5 kN/m