Question

The turbine rotor of a ship has a mass of 20 tonnes and a radius of gyration of 0.75 m. Its speed is 2000 r.p.m. the ship pitches 6° above and 6° below the horizontal position. One complete oscillation takes 30 seconds and the motion is simple harmonic.

Calculate :

1. The maximum couple tending to shear the holding down bolts of the turbine,

2. The maximum angular acceleration of the ship during pitching, and

3. The direction in which the bow will tend to turn while rising, if the rotation of the rotor is clockwise when looking from rear.

Answer 1

mass m = 20 tonnes = 20000 kg, k = 0.6 m, N = 2000 rpm, ω = 2πx 2000/60 = 209.5 rad /sec, φ = 6° = 6 x π/180 = 0.105 rad/sec, tp = 30 sec.

(i) Maximum gyroscopic couple:

w.k.t Mass moment of Inertia of the rotor,

I = m. k2 = 20000 x (0.6)2

I = 7200 kg-m2

Angular velocity of SHM,

ω1 = 2π/tp = 2π/30 = 0.21 rad/sec

Maximum angular velocity of precession,

ωPmax =ϕω1 = 0.105 x 0.21 = 0.022 rad/sec

w.k.t the maximum gyroscopic couple, Cmax = IωωPmax

Cmax = 7200 x 209.5 x 0.022

Cmax = 33185 N-m

(ii) Maximum angular acceleration during pitching:

w.k.t. Maximum angular acceleration during pitching

= ϕω2 = 0.105 x (0.21)2

= 0.0046 rad/sec

(iii) Direction in which the bow will tend to turn when rising:

We know that, when the rotation of the rotor is clockwise when looking from the left and when the bow is rising, then the reactive gyroscopic couple acts in the clockwise direction which tends to turn the bow towards right.