**Concept:**

The angle variation with respect to time during fault with constant accelerating power(P_{a}) is given as

\(δ (t) = {δ _o} + \frac{1}{2}\left( {\frac{{{P_a}}}{M}} \right){t^2}\) ....(1)

Where δ_{o} = Initial rotor angle

P_{a} = Accelerating power

M = Angular momentum

M_{pu} = 2GH/ω_{s} = H/180f

G = Rating of the generator in pu

H = Inertia constant

**Calculation:**

Given δ_{o} = 5°, H = 2 MJ/MVA, t = 0.02 second, f = 50 Hz, P_{a} = 1pu

Rating of the generator(G) is always asumed as 1 pu, then

\(M = \frac{{2 × 1}}{{180 × 50}} = \frac{1}{{4500}}\;pu\)

Substitute all the values in equation(1), we get

δ = 5 + 0.5 × 4500 × 0.02^{2}

**δ = 5.9° **