A $1500-\mathrm{MVA} 20-\mathrm{kV}, 60-\mathrm{Hz}$ three-phase generator is connected through a $1500-\mathrm{MVA}$ $20-\mathrm{kV} \Delta / 500-\mathrm{kV}$ Y transformer to a $500-\mathrm{kV}$ circuit breaker and a $500-\mathrm{kV}$ transmission line. The generator reactances are $\mathrm{X}_{d}^{\prime \prime}=0.17, \mathrm{X}_{d}^{\prime}=0.30$, and $\mathrm{X}_{d}=1.5$ per unit, and its time constants are $\mathrm{T}_{d}^{\prime \prime}=0.05, \mathrm{~T}_{d}^{\prime}=1.0$, and $\mathrm{T}_{\mathrm{A}}=0.10 \mathrm{~s}$. The transformer series reactance is 0.10 per unit; transformer losses and exciting current are neglected. A three-phase short-circuit occurs on the line side of the circuit breaker when the generator is operated at rated terminal voltage and at no-load. The breaker interrupts the fault 3 cycles after fault inception. Determine (a) the subtransient current through the breaker in per-unit and in kA rms; and (b) the rms asymmetrical fault current the breaker interrupts, assuming maximum dc offset. Neglect the effect of the transformer on the time constants.

Question

Quizplus · Accepted Answer

The Answer is (a) Neglecting the transformer winding resistance,

I^{\prime \prime}=\frac{E q}{X_{d}^{\prime \prime}+X_{T R}}=\frac{1.0}{0.17+0.10}=\underline{\underline{3.704}} ext { Per unit }

The base current on the high-voltage side of the transformer is:

\begin{aligned}& I_{ ext {base } H}=\frac{S_{ ext {rated }}}{\sqrt{3} V_{ ext {H rated }}}=\frac{1500}{\sqrt{3}(500)}=1.732 \mathrm{kA} \& I^{\prime \prime}=(3.704)(1.732)=\underline{\underline{6.415}} \mathrm{kA}\end{aligned}

(b) Using Eq (7.2.1) at

t \quad 3

cycles

0.05 \mathrm{~S}

with the transformer reactance included:

\begin{aligned}I_{a c}(0.05) & =1.0\left[\left(\frac{1}{0.27}-\frac{1}{0.40} ight) e^{\frac{-0.05}{0.05}}+\left(\frac{1}{0.40}-\frac{1}{1.6} ight) e^{\frac{-0.05}{1.0}}+\frac{1}{1.6} ight] \& =2.851 ext { Per Unit }\end{aligned}

Using Eq (7.2.5),

i_{d c}(t)=\sqrt{2}(3.704) e^{-t / 0.10}=5.238 e^{-t / 0.10} ext { Per unit }

The rms asymmetrical current that the breaker interrupts is

\begin{aligned}I_{r m s}(0.05 S) & =\sqrt{I_{a c}^{2}(0.05)+i_{d c}^{2}(0.05)} \& =\sqrt{(2.851)^{2}+(5.238)^{2} e^{\frac{-2(0.05)}{0.10}}} \& =4.269 ext { Per Unit }=(4.269)(1.732)=\underline{\underline{7.394}} \mathrm{kA}\end{aligned}

A 1500−MVA20−kV,60−Hz1500-\mathrm{MVA} 20-\mathrm{kV}, 60-\mathrm{Hz}1500−MVA20−kV,60−Hz Three-Phase Generator Is Connected Through A 1500−MVA1500-\mathrm{MVA}1500−MVA 20−kVΔ/500−kV20-\mathrm{kV} \Delta / 500-\mathrm{kV}20−kVΔ/500−kV

A $1500-\mathrm{MVA} 20-\mathrm{kV}, 60-\mathrm{Hz}$ Three-Phase Generator Is Connected Through A $1500-\mathrm{MVA}$ $20-\mathrm{kV} \Delta / 500-\mathrm{kV}$