# 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 re

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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_________

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Concept:

The angle variation with respect to time during fault with constant accelerating power(Pa) is given as

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

Where δo = Initial rotor angle

Pa = Accelerating power

M = Angular momentum

Mpu = 2GH/ωs = H/180f

G = Rating of the generator in pu

H = Inertia constant

Calculation:

Given δo = 5°, H = 2 MJ/MVA, t = 0.02 second, f = 50 Hz, Pa = 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°

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