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

Answer:

782.290 MJ

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 = $\frac{1}{2} * Jw^{2} _{s}$

     = 1/2 * 11000 * ( 377.14 ) ^2

     = 5500 * 142234.58 = 782.290 MJ