Deck 6: Symmetrical Faults

Question 1

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.

Answer 1

(a) $\bar{Ƶ}=R+j \omega L=0.5+j 2 \pi(60) 3 \times 10^{-3}=0.5+j 1.131$
$=1.2366 \angle 66.15^{\circ} \Omega ; Ƶ=1.2366 \Omega, \theta=66.15^{\circ}$
$I_{a c}=V / Ƶ=220 / 1.2366=177.9 \mathrm{~A}$
(b) $I_{r m s}(0)=I_{a c} k(0)=177.9 \sqrt{3}=308 \mathrm{~A}$
(c) Using Eq. (7.1.11) and (7.1.12)
$\begin{gathered}X / R=1.131 / 0.5=2.262 ; k(\tau=5 \text { cycles })=\sqrt{1+2 e \frac{-4 \pi(5)}{2.262}} \\k(5 \text { cycles }) × 1.0 \\I_{r m s}(5 \text { cycles })=I_{a c} k(5 \text { cycles })=177.9 \mathrm{~A}\end{gathered}$
(d) Using Eq. (7.1.1)
$\begin{aligned}V(0) & =\sqrt{2} V \sin \alpha=244 \text { Volts } \\\alpha & =\sin ^{-1}\left(\frac{244}{\sqrt{2} 277}\right)=38.53^{\circ}\end{aligned}$
Using Eq. (7.1.4)
$\begin{aligned}i_{d c}(t) & =-\frac{\sqrt{2} V}{Ƶ} \sin (\alpha-\theta) e^{-t / T} \\i_{d c}(t) & =-\frac{\sqrt{2}(220)}{1.2366} \sin \left(38.53^{\circ}-66.15^{\circ}\right) e^{-t / T} \\& =116.62 e^{-t / T} \\\mathbf{T} & =L / R=3 \times 10^{-3} / 0.5=6 \times 10^{-3} \text { seconds } \\i_{d c}(t) & =116.62 e^{-t /\left(6 \times 10^{-3}\right)} \mathrm{A}\end{aligned}$

Question 2

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

Answer 2

(a) $\bar{Ƶ}=1+j 3=3.1623 \angle 71.57^{\circ} \Omega$
$I_{a c}=\frac{V}{Ƶ}=\frac{4000}{3.1623}=\underline{\underline{1265 . \mathrm{A}}}$
(b) $I_{r m s}(0)=1265 \sqrt{3}=\underline{2191.A }$
(c) $\frac{X}{R}=3 \quad k(5)=\sqrt{1+2 e^{-4 \pi(5) / 3}}=1.0$
$I_{r m s}(5)=k(5) I_{a c}=1.0(1265)=1265 . \mathrm{A}$
(d) $\alpha=\sin ^{-1}\left(\frac{300}{4000 \sqrt{2}}\right)=3.04^{\circ}$
$\begin{aligned}T & =\frac{L}{R}=\frac{X}{W R}=\frac{3}{(2 \pi 60)(1)}=7.958 \times 10^{-3} \mathrm{~S} \\i_{d c}(t) & =\frac{-\sqrt{2}(4000)}{3.1623} \sin \left(3.04^{\circ}-71.57^{\circ}\right) e^{-t / T} \\i_{d c}(t) & =\underline{\underline{1665 . e^{-t /\left(7.958 \times 10^{-3}\right)}} \mathrm{A}}\end{aligned}$

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.

Answer 3

(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

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.

Question 5

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