Question

A 50 Hz generating unit has H – constant of 2 MJ/MVA. The machine is initially operating in steady-state at synchronous speed and producing 1 pu of real power. The initial value of the rotor angle δ is 5° when a bolted three-phase to ground short circuit fault occurs at the terminal of the generator. Assuming the input mechanical power to remain at 1 pu, the value of δ in degrees, 0.02 second after the fault is_________

Answer 1

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 × 0.02²

δ  = 5.9°