Question

Consider the expression for $\mathbf{i}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$given by

localid="1655289475813" $\mathbf{i}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\sqrt{\mathbf{2}}{\mathbf{l}}_{\mathbf{rms}}\mathbf{[}\mathbf{sin}\mathbf{(}\mathbf{\omega t}\mathbf{-}{\mathbf{\theta }}_{\mathbf{z}}\mathbf{)}\mathbf{-}\mathbf{sin}\mathbf{\left(}{\mathbf{\theta }}_{\mathbf{z}}\mathbf{\right)}{\mathbf{e}}^{\mathbf{-}\frac{\mathbf{\omega R}}{\mathbf{X}}\mathbf{t}}\mathbf{]}$

Where localid="1655289479517" ${\mathbf{\theta }}_{\mathbf{z}}\mathbf{=}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\begin{array}{c}\mathbf{X}\\ \overline{\mathbf{R}}\end{array}\mathbf{\right)}$

(a) Forlocalid="1655289482614" $\left(\begin{array}{c}X\\ \overline{R}\end{array}\right)$equal to zero and infinity, plot localid="1655289486080" $\mathbf{i}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$as a function oflocalid="1655289491779" $\mathbf{\left(}\mathbf{\omega t}\mathbf{\right)}$.

(b) Comment on the dc offset of the fault current waveforms.

(c) Find the asymmetrical current factor and the time of peak ,localid="1655289494958" ${\mathbf{t}}_{\mathbf{p}}$in milliseconds, for localid="1655289498467" $\left(\begin{array}{c}X\\ \overline{R}\end{array}\right)$ratios of zero and infinity.

Short answer

(a) Therefore, the plot for the current $i\left(t\right)$is shown below.

(b) The waveform is purely sin wave for $\frac{X}{R}=0$and the dc offset achieves the maximum value for$\frac{X}{R}=\infty$.

(c) The asymmetrical current factor for$\frac{X}{R}=0$is $14.14\mathrm{A}$and peak time is $4.2ms$.

The symmetrical current factor for$\frac{X}{R}=\infty$is $2.8283\mathrm{A}$and peak time is $8.33ms$.

Step-by-step solution

Write the given data from the question.

The expression for current $i\left(t\right)$is given as follows.$\mathbf{i}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\sqrt{\mathbf{2}}{\mathbf{l}}_{\mathbf{rms}}\mathbf{[}\mathbf{sin}\mathbf{(}\mathbf{\omega t}\mathbf{-}{\mathbf{\theta }}_{\mathbf{z}}\mathbf{)}\mathbf{+}\mathbf{sin}\mathbf{\left(}{\mathbf{\theta }}_{\mathbf{z}}\mathbf{\right)}{\mathbf{e}}^{\frac{\mathbf{-}\mathbf{\omega R}}{\mathbf{X}}\mathbf{t}}\mathbf{]}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}$

Here $\mathbf{X}$is the inductive reactance,$\mathbf{R}$is the resistance, $\mathbf{\omega }$is the supply angular frequency and$\mathbf{t}$is the time.

The equation to calculate the angle ${\theta }_z$is given as follows.

${\mathbf{\theta }}_{\mathbf{z}}\mathbf{=}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\begin{array}{c}\mathbf{X}\\ \overline{\mathbf{R}}\end{array}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{(2}\mathbf{)}$

Let frequency of the system is $60\mathrm{Hz}$.

plot i(t) as a function of (ωt).

(a)

Calculate the angle${\theta }_z$.

Substitute $0$for $\frac{X}{R}$into equation (2).

$\begin{array}{l}{\theta }_z={\tan }^{-1}\left(0\right)\\ {\theta }_z=0\end{array}$

Calculate the peak timewhen $\frac{X}{R}=0$.

Substitute $0$for$\frac{X}{R}$into equation (1).

$$\begin{gathered} i\left(t\right)=\sqrt{2}{l}_{rms}[sin(\omega t-{\theta }_z)+sin\left({\theta }_z\right)e^{\frac{\frac{-\omega t}{X}}{R}}] \\ i\left(t\right)=\sqrt{2}{l}_{rms}[sin(\omega t-{\theta }_z)+sin\left({\theta }_z\right)e^{\frac{-\omega t}{0}}] \\ i\left(t\right)=\sqrt{2}{l}_{rms}[sin(\omega t-{\theta }_z)+sin\left({\theta }_z\right)e^{-\infty }] \\ i\left(t\right)=\sqrt{2}{l}_{rms}sin(\omega t-{\theta }_z) \end{gathered}$$

Differentiate the above equation with respect to$t$,

$\frac{di\left(t\right)}{dt}=\sqrt{2}{l}_{rms}\cos \left(\omega t-{\theta }_z\right)\omega$

Substitute $0$for $\frac{di\left(t\right)}{dt}$and$t_p$for localid="1655290345800" $t$into above equation equal to zero.

$$\begin{gathered} \sqrt{2}{l}_{rms}\cos \left(\omega t_p-{\theta }_z\right)\omega =0 \\ \cos \left(\omega t_p-{\theta }_z\right)=0 \\ \cos \left(\omega t_p-{\theta }_z\right)=\cos 90° \\ \omega t_p-{\theta }_z=90° \end{gathered}$$

Solve further as,

$$\begin{gathered} \omega t_p-{\theta }_z=90°\times \frac{\mathrm{\pi }}{180} \\ \omega t_p-{\theta }_z=\frac{\mathrm{\pi }}{2} \\ {t}_p=\frac{\frac{\mathrm{\pi }}{2}+{\theta }_z}{\omega } \\ {t}_p=\frac{\frac{\mathrm{\pi }}{2}+{\theta }_z}{2\pi f} \end{gathered}$$

Substitute $60\mathrm{Hz}$for$f$, and $0$for ${\theta }_z$into above equation.

$$\begin{gathered} t_p=\frac{{\displaystyle \frac{\mathrm{\pi }}{2}+0}}{2\mathrm{\pi }\times 60} \\ {t}_p=\frac{\mathrm{\pi }}{4\mathrm{\pi }\times 60} \\ {\mathrm{t}}_{\mathrm{p}}=\frac{1}{240} \\ {\mathrm{t}}_{\mathrm{p}}=4.2\mathrm{ms} \end{gathered}$$

Consider the current equation,

$i\left(t\right)=\sqrt{2}{l}_{rms}\sin \left(2{\mathrm{\pi ft}}_{\mathrm{p}}-{\mathrm{\theta }}_{\mathrm{z}}\right)$

Substitute $60\mathrm{Hz}$for$f$,$4.2ms$for $t_p$and $0°$for ${\theta }_z$into above equation.

$$\begin{gathered} i\left(t\right)=\sqrt{2}{l}_{rms}\sin \left(\left(2\mathrm{\pi }\times 60\times 4.2\times {10}^{-3}\right)\frac{180}{\mathrm{\pi }}-0\right) \\ i\left(t\right)=\sqrt{2}{l}_{rms}\sin \left(90°\right) \\ i\left(t\right)=\sqrt{2}{l}_{rms} \\ i\left(t\right)=1.414l_{rms} \end{gathered}$$

Calculate the angle${\theta }_z$.

Substitute $\infty$for $\frac{X}{R}$into equation (2).

$\begin{array}{l}{\theta }_z={\tan }^{-1}\left(\infty \right)\\ {\theta }_z={90}^{°}\end{array}$

Calculate the peak time when $\frac{X}{R}=\infty$.

Substitute$\infty$for$\frac{X}{R}$into equation (1).

$$\begin{gathered} i\left(t\right)=\sqrt{2}{l}_{rms}[sin(\omega t-{\theta }_z)+sin\left({\theta }_z\right)e^{\frac{\frac{-\omega t}{X}}{R}}] \\ i\left(t\right)=\sqrt{2}{l}_{rms}[sin(\omega t-{\theta }_z)+sin\left({\theta }_z\right)e^{\frac{-\omega t}{\infty }}] \\ i\left(t\right)=\sqrt{2}{l}_{rms}[sin(\omega t-{\theta }_z)+sin\left({\theta }_z\right)e^0] \\ i\left(t\right)=\sqrt{2}{l}_{rms}sin(\omega t-{\theta }_z)+sin\left({\theta }_z\right) \end{gathered}$$

Differentiate the above equation with respect tolocalid="1655292943562" $t$,

$\frac{di\left(t\right)}{dt}=\sqrt{2}{l}_{rms}\left[\cos \left(\omega t-{\theta }_z\right)\omega +\cos \left({\theta }_z\right)\right]$

Substitute $0$for$\frac{di\left(t\right)}{dt}$,$90°$for ${\theta }_z$and$t_p$for $t$into above equation equal to zero.

$$\begin{gathered} \sqrt{2}{l}_{rms}\left[\cos \left(\omega t-90°\right)\omega +\cos \left(90°\right)\right]=0 \\ \cos \left(\omega t_p-90°\right)=0 \\ \cos \left(\omega t_p-90°\right)=\cos 90° \\ \omega t_p-90°=90° \end{gathered}$$

Solve further as,

$$\begin{gathered} \omega t_p=\left(90°+90°\right)\times \frac{\mathrm{\pi }}{180} \\ \omega t_p=\mathrm{\pi } \\ {t}_p=\frac{\mathrm{\pi }}{\omega } \\ {t}_p=\frac{\mathrm{\pi }}{2\mathrm{\pi f}} \end{gathered}$$

Substitute localid="1655294698921" $60\mathrm{Hz}$forlocalid="1655294716762" $f$into above equation.

$$\begin{gathered} t_p=\frac{1}{2\times 60} \\ {t}_p=\frac{1}{120} \\ {t}_p=8.33ms \end{gathered}$$

Consider the current equation,

$i\left(t\right)=\sqrt{2}{l}_{rms}\left[\sin \left(\omega t-{\theta }_z\right)+\sin \left({\theta }_z\right)\right]$

Substitute localid="1655295258789" $60\mathrm{Hz}$for$f$,$8.33\mathrm{ms}$for $t_p$and $90°$for ${\theta }_z$into above equation.

localid="1655296030298" $$\begin{gathered} i\left(t\right)=\sqrt{2}{l}_{rms}\left[\sin \left(\left(2\mathrm{\pi }\times 60\times 8.33\times {10}^{-3}\right)\frac{180}{\mathrm{\pi }}-90°\right)+\sin \left(90°\right)\right] \\ i\left(t\right)=\sqrt{2}{l}_{rms}\left[\sin \left(90°\right)+\sin \left(90°\right)\right] \\ i\left(t\right)=\sqrt{2}{l}_{rms}\times 2 \\ i\left(t\right)=2.8283l_{rms} \end{gathered}$$

The plot for the currentlocalid="1655296112442" $i\left(t\right)$is shown below.

Comment on the dc offset of the fault current waveforms.

(b)

In the fault current waveform for$\frac{X}{R}=0$,the dc offset is absent. Therefore the waveform is purely sin wave.

In the fault current waveform for$\frac{X}{R}=\infty$, the dc offset achieves the maximum value.

Calculate the asymmetrical fault current factor and peak time.

(c)

From the figure shown in the step 1, the asymmetrical current factor for $\frac{X}{R}=0$is$1.414\mathrm{A}$and peak time is$4.2\mathrm{ms}$.

The symmetrical current factor for$\frac{X}{R}=\infty$is$2.8283\mathrm{A}$and peak time is $8.33\mathrm{ms}$.