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[tex]\pi[/tex]Ns ) / f
= (2[tex]\pi[/tex]*3000 ) / 50 = 377.14 rad/sec
determine the kinetic energy of the machine
K.E = [tex]\frac{1}{2} * Jw^{2} _{s}[/tex]
= 1/2 * 11000 * ( 377.14 ) ^2
= 5500 * 142234.58 = 782.290 MJ
