Question

A 2-pole, 50Hz, 20 kV turbo generator is rated 120 MVA and .95 power factor lagging. The machine rotor has a moment of inertia of 11000 kg*m^2. What is the kinetic energy of the machine

Answer 1

782.290 MJ

Step-by-step explanation:

Given Data:

number of poles ( p ) = 2-pole

Frequency ( f ) = 50 Hz

voltage ( v ) = 20 kV

power = 120 MVA

power factor (cos ∅ ) = 0.95

moment of inertia ( J ) = 11000 kg*m^2

First we calculate the value of

Ns = ( 120 * f ) / p

= ( 120 * 50 ) / 2 = 3000 rpm

Ws = ( 2
`\pi`Ns ) / f

= (2
`\pi`*3000 ) / 50 = 377.14 rad/sec

determine the kinetic energy of the machine

K.E =
`(1)/(2) * Jw^(2) _(s)`

= 1/2 * 11000 * ( 377.14 ) ^2

= 5500 * 142234.58 = 782.290 MJ