A diesel engine is directly coupled to an AC generator as sh

A diesel engine is directly coupled to an AC generator, as shown in figure (Q2) and the moment of inertia of the rotating pans is 80 kgm^2 The full load output of the engine is 185KW at 720RPM. The torque required by the generator in overcoming friction and windage has been estimated as (400 + 0.376 omega^2) Nm where \"omega\" is the angular speed (rad s) of the engine. If the output torque of the engine is assumed constant and equal to the full load. Calculate the time taken for the engine to reach 95% of its rated full load speed after start-up. the engine speed in RPM after 3s the number of revolutions completed after 3s Comment on the lubrication process of the engine bearings and the importance of rapidly attaining full speed

Solution

Power = 2*pi*N*T/60

185*10³ = 2 * 3.14 * 720 * T / 60

Torque T = 2454.88 Nm

Angular speed w = 2*pi*N / 60

= 2 * 3.14 * 720 / 60

= 75.36 rad/s

a)

95% of rated full load speed = 0.95*75.36 = 71.592 rad/s

At this speed, friction loss in generator = 400 + 0.376* 71.592² = 2327.16 Nm

Net torque = 2454.88 - 2327.16 = 127.72 Nm

Torque = I*alpha

127.72 = 80*alpha

Angular acceleration, alpha = 1.6 rad/s²

alpha = (w - w₀) / t

1.6 = (71.592 - 0) / t

Time t = 44.84 s

b)

w = w₀ + alpha*t

= 0 + 1.6*3

= 4.8 rad/s

w = 2*pi*N / 60

4.8 = 2 * 3.14 * N / 60

N = 45.86 RPM

c)

w² = w₀² + 2*alpha*theta

4.8² = 0 + 2 * 1.6 * theta

theta = 7.2 rad

= 7.2 / (2*pi) revs

= 1.146 revs

d)

Since the acceleration is pretty slow, at low speeds the oil pressure does not build up fast enough. Hence, we need to use an external system for lubrication at low speeds.