A 100-MV A, 22-kV, 50-Hz synchronous generator is operating open circuited and is excited to give rated terminal voltage. A 3-phase (symmetrical) short circuit develops at its terminals. Neglecting dc and double frequency components of current.
(a) find the initial current, and
(b) find the current at the end of two cycles and at the end of 10s.
Given:
Base 100 MV
X_{d}=1.0 pu , X_{d}^{\prime}=0.3 pu , X_{d}^{\prime \prime}=0.2 pu\tau_{d w}=0.03 s , \tau_{f}=1 s
(a) Initial current, I^{\prime \prime}=\frac{E_{f}}{X_{d}^{\prime \prime}}=\frac{1}{0.2}=5 pu
I_{\text {Base }}=\frac{100 \times 1000}{\sqrt{3} \times 22}=2624 A
Then I^{\prime \prime}=5 \times 2624=13120 A
(b) I^{\prime}=\frac{E_{f}}{X_{d}^{\prime}}=\frac{1}{0.3}=3.33 pu
I=\frac{E_{f}}{X_{d}}=\frac{1}{1}=1 pu
I_{S C}(t)=(5-3,33) e^{-t / 0.03}+(3.33-1)^{e-t / 1}+1
=1.07 e^{-t / 0.03}+2.33 e^{-t}+1
t(2 \text { cycles })=1.67 e^{-0.04 / 0.03}+2.33 e^{-0.04}+1
t(2 \text { cycles })=1.67 e^{-0.04 / 0.03}+2.33 e^{-0.04}+1
= 3.68 pu or 9656 A
I_{S C}(10 s )=1.67 e^{-10 / 0.03}+2.32 e^{-10}+1
= 1.0001 pu or 2624 A (steady-state is practically reached)