A 2-pole, 50 Hz, 11 kV, 100 MVA alternator has a moment of inertia of 10,000 kg.m². The value of inertia constant, H is

Concept:

Inertia constant, H = (Stored kinetic energy in MJ)/(Rating of the machine)

Stored K.E \(= \frac{1}{2}I{\omega ^2}\)

I is the moment of inertia

ω = 2πf

Calculation:

P = 2, f = 50 Hz, V = 11 kV, P = 100 MVA

Moment of inertia, I = 10,000 kg-m²

ω = 2πf = 2π (50) = 100 π

= 314.15 rad/sec

Stored K.E. \(= \frac{1}{2} \times 10,000 \times {\left( {314.15} \right)^2}\)

= 4.93 × 10⁸J = 493 MJ

\(H = \frac{{493\;MJ}}{{100\;MVA}}\)

⇒ H = 4.93 sec