Inertia constant on 500 MVA base

Assumption: the paper does not give the rotor inertia. The standard version of this problem uses a moment of inertia of $J = 10{,}000$ kg·m², and that value is assumed here.

Step 1: Rating and speed

$$\begin{aligned} S &= \frac{P}{\text{pf}} = \frac{100}{0.85} = 117.65\ \text{MVA} \\ N_s &= \frac{120f}{P} = \frac{120\times50}{2} = 3000\ \text{rpm} \\ \omega_{sm} &= \frac{2\pi\times3000}{60} = 314.16\ \text{rad/s} \end{aligned}$$

Step 2: Stored kinetic energy

$$KE = \tfrac{1}{2}J\omega_{sm}^2 = \tfrac{1}{2}\times10^4\times(314.16)^2 = 493.48\times10^6\ \text{J} = 493.48\ \text{MJ}$$

Step 3: Inertia constant

On the machine's own rating: $H = 493.48/117.65 = 4.19$ MJ/MVA.

On a 500 MVA base:

$$H_{500} = \frac{493.48}{500} = 0.987\ \text{MJ/MVA}$$

Step 4: Angular momentum

$$M = \frac{GH}{180f} = \frac{KE}{180f} = \frac{493.48}{180\times50} = 0.0548\ \text{MJ·s/elec deg}$$

In per unit on the 500 MVA base, this is $0.0548/500 = 1.097\times10^{-4}$ pu·s/elec deg.

Answer: $H = 0.987$ MJ/MVA on 500 MVA (4.19 MJ/MVA on its own 117.65 MVA rating); $M = 0.0548$ MJ·s/elec deg.