Question 1

For Test Bank Problem 7.3, determine (a) the instantaneous symmetrical fault current in kA in phase $a$ of the generator as a function of time, assuming maximum dc offset occurs in this generator phase; and (b) the maximum dc offset current in kA as a function of time that can occur in any one generator phase.

Accepted Answer

(a) Using Eq. (7.2.1) with the transformer reactance included, and with $\alpha=0^{\circ}$ for maximum dc offset
$\begin{aligned}
i_{a c}(t) & =\sqrt{2}(1.0)\left[\left(\frac{1}{0.27}-\frac{1}{0.40}\right) e^{-t / 0.05}+\left(\frac{1}{.40}-\frac{1}{1.6}\right) e^{-t / 1.0}+\frac{1}{1.6}\right] \sin \left(\omega t-\frac{\pi}{2}\right) \
& =\sqrt{2}\left[1.204 e^{-t / 0.05}+1.875 e^{-t / 1.0}+0.625\right] \sin \left(\omega t-\frac{\pi}{2}\right) \text { Per Unit }
\end{aligned}$
The generator base current is:
$I_{\text {baseL }}=\frac{S_{\text {rated }}}{\sqrt{3} V_{\text {rated } L}}=\frac{1500}{\sqrt{3}(20)}=43.3 \mathrm{kA}$
Therefore:
$i_{a c}(t)=61.23\left[1.204 e^{\frac{-t}{0.05}}+1.875 e^{\frac{-t}{1.0}}+0.625\right] \sin \left(\omega t-\frac{\pi}{2}\right) \mathrm{kA}$
where the effect of the transformer on the time constants has been neglected.
(b) From Eq. (7.2.5) and the results of book Problem 7.4,
$\begin{aligned}
i_{d c}(t) & =\sqrt{2} I^{\prime \prime} e^{-t / T_{A}}=\sqrt{2}(3.704) e^{-t / 0.10} \
& =5.238 e^{\frac{-t}{0.10}} \text { per unit }=\underline{ }=226.8 e^{\frac{-t}{0.10}} \mathrm{kA}
\end{aligned}$

Question 2

A $500-\mathrm{kV}$ three-phase transmission line has a $2.2-\mathrm{kA}$ continuous current rating and a $2.5-\mathrm{kA}$ maximum short-time overload rating, with a $525-\mathrm{kV}$ maximum operating voltage. Maximum symmetrical fault current on the line is $30 \mathrm{kA}$. Select a circuit breaker for this line from Table 7.10.in the text

Accepted Answer

From Table 7.10, select the $500 \mathrm{kV}$ (nominal voltage class) breaker with a $40 \mathrm{kA}$ rated short circuit current. This breaker has a $3 \mathrm{kA}$ rated continuous current.

Question 3

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.

Accepted Answer

(a) Neglecting the transformer winding resistance,
$I^{\prime \prime}=\frac{E q}{X_{d}^{\prime \prime}+X_{T R}}=\frac{1.0}{0.17+0.10}=\underline{\underline{3.704}} \text { Per unit }$
The base current on the high-voltage side of the transformer is:
$\begin{aligned}
& I_{\text {base } H}=\frac{S_{\text {rated }}}{\sqrt{3} V_{\text {H rated }}}=\frac{1500}{\sqrt{3}(500)}=1.732 \mathrm{kA} \
& I^{\prime \prime}=(3.704)(1.732)=\underline{\underline{6.415}} \mathrm{kA}
\end{aligned}$
(b) Using Eq (7.2.1) at $t \quad 3$ cycles $0.05 \mathrm{~S}$ with the transformer reactance included:
$\begin{aligned}
I_{a c}(0.05) & =1.0\left[\left(\frac{1}{0.27}-\frac{1}{0.40}\right) e^{\frac{-0.05}{0.05}}+\left(\frac{1}{0.40}-\frac{1}{1.6}\right) e^{\frac{-0.05}{1.0}}+\frac{1}{1.6}\right] \
& =2.851 \text { Per Unit }
\end{aligned}$
Using Eq (7.2.5),
$i_{d c}(t)=\sqrt{2}(3.704) e^{-t / 0.10}=5.238 e^{-t / 0.10} \text { Per unit }$
The rms asymmetrical current that the breaker interrupts is
$\begin{aligned}
I_{r m s}(0.05 S) & =\sqrt{I_{a c}^{2}(0.05)+i_{d c}^{2}(0.05)} \
& =\sqrt{(2.851)^{2}+(5.238)^{2} e^{\frac{-2(0.05)}{0.10}}} \
& =4.269 \text { Per Unit }=(4.269)(1.732)=\underline{\underline{7.394}} \mathrm{kA}
\end{aligned}$

Question 4

Repeat Example 7.1 with $\mathrm{V}=4 \mathrm{kV}, \mathrm{X}=3 \Omega$, and $\mathrm{R}=1 \Omega$.

Accepted Answer

The answer of Repeat Example 7.1 with \(\mathrm{V}=4 \mathrm{kV}, \mathrm{X}=3...

Question 5

In the circuit of Figure $7.1, V=220$ volts, $L=3 \mathrm{mH}, \mathrm{R}=0.5 \Omega$, and $\omega=2 \pi 60$ $\mathrm{rad} / \mathrm{s}$. Determine (a) the rms symmetrical fault current; (b) the rms asymmetrical fault current at the instant the switch closes, assuming maximum dc offset; (c) the rms asymmetrical fault current 5 cycles after the switch closes, assuming maximum dc offset; (d) the dc offset as a function of time if the switch closes when the instantaneous source voltage is 244 volts.

Accepted Answer

The answer of In the circuit of Figure $7.1, V=220$...

Exam 6: Symmetrical Faults

Repeat Example 7.1 with $\mathrm{V}=4 \mathrm{kV}, \mathrm{X}=3 \Omega$ , and $\mathrm{R}=1 \Omega$ .

Exam 6: Symmetrical Faults

Repeat Example 7.1 with V=4kV,X=3Ω\mathrm{V}=4 \mathrm{kV}, \mathrm{X}=3 \OmegaV=4kV,X=3Ω , and R=1Ω\mathrm{R}=1 \OmegaR=1Ω .

Repeat Example 7.1 with $\mathrm{V}=4 \mathrm{kV}, \mathrm{X}=3 \Omega$ , and $\mathrm{R}=1 \Omega$ .