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(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
\(\begin{aligned}
I_{r m s}(0.05 S) & =\sqrt{I_{a c}^{2}(0.05)+i_{d c}^{2}(0.05)} \\
& =\sqrt{(2.851)^{2}+(5.238)^{2} e^{\frac{-2(0.05)}{0.10}}} \\
& =4.269 \text { Per Unit }=(4.269)(1.732)=\underline{\underline{7.394}} \mathrm{kA}
\end{aligned}\)
Correct Answer
verified
(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
\(\begin{aligned}
I_{r m s}(0.05 S) & =\sqrt{I_{a c}^{2}(0.05)+i_{d c}^{2}(0.05)} \\
& =\sqrt{(2.851)^{2}+(5.238)^{2} e^{\frac{-2(0.05)}{0.10}}} \\
& =4.269 \text { Per Unit }=(4.269)(1.732)=\underline{\underline{7.394}} \mathrm{kA}
\end{aligned}\)
Correct Answer
verified
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(a) 
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(c) Using Eq. (7.1.11) and (...View Answer
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(a)
(b)
(c) Using Eq. (7.1.11) and (...
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(c) Using Eq. (7.1.11) and (...
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Correct Answer
verified
(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,
\(\begin{aligned}
i_{d c}(t) & =\sqrt{2} I^{\prime \prime} e^{-t / T_{A}}=\sqrt{2}(3.704) e^{-t / 0.10} \\
& =5.238 e^{\frac{-t}{0.10}} \text { per unit }=\underline{ }=226.8 e^{\frac{-t}{0.10}} \mathrm{kA}
\end{aligned}\)
Correct Answer
verified
(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,
\(\begin{aligned}
i_{d c}(t) & =\sqrt{2} I^{\prime \prime} e^{-t / T_{A}}=\sqrt{2}(3.704) e^{-t / 0.10} \\
& =5.238 e^{\frac{-t}{0.10}} \text { per unit }=\underline{ }=226.8 e^{\frac{-t}{0.10}} \mathrm{kA}
\end{aligned}\)
Correct Answer
verified
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.
11eec41d_0e52_8a56_ae1f_d1423d85de1e_TB10648_00
Correct Answer
verified
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.
11eec41d_0e52_8a56_ae1f_d1423d85de1e_TB10648_00
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