<p><span style="font-size:14px">?We have seen that a power system is rich of inductive reactance. Resistive component is negligibly small. Refer to single phase equivalent circuit of a typical power system network as shown in figure 4.25.</span></p>

<p><span style="font-size:14px"><img alt="" src="https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16164831591616483159.jpg" style="height:491px; width:800px" /></span></p>

<p><span style="font-size:14px">When fault occurs, the load and system capacitance are shorted. Using Kirchoff&rsquo;s voltage law, we can write,</span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(L\frac{di}{dt}+Ri=e=E_m\sin(\omega t+\theta)\)</span>&nbsp; &nbsp; &nbsp;&nbsp;(Eq 4.18)</span></p>

<p style="text-align:justify"><span style="font-size:14px">The voltage waveshape is shown in figure 4.26.</span></p>

<p style="text-align:justify"><span style="font-size:14px"><img alt="" src="https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16161466691616146669.jpg" style="height:491px; width:800px" /></span></p>

<p style="text-align:justify"><span style="font-size:14px">Equation 4.18 is an equation written for fault current i when there is a sudden short circuit. This equation is a differential equation, and the solution is the sum of complementary function (transient component) and the particular integral (steady-state component). The transient solution can be obtained by writing,</span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(L\frac{di}{dt}+R\hspace{0.1cm}i=0\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(\dot{.\hspace{0.045in}.}\hspace{0.3cm}\frac{di}{i}=\frac{-R}{L}\hspace{0.1cm}dt\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px">Integrating,</span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(ln\hspace{0.1cm}i=\frac{-Rt}{L}+K\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;</span></p>

<p style="text-align:justify"><span style="font-size:14px">Where K = Constant of integration. </span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(\dot{.\hspace{.045cm}.}\hspace{0.1cm}ln\hspace{0.1cm}i=ln\hspace{0.1cm}e^{\frac{-Rt}{L}}+ln\hspace{0.1cm}A\)</span>&nbsp; &nbsp;&nbsp;</span></p>

<p style="text-align:justify"><span style="font-size:14px">Where A = Constant,&nbsp;<span class="math-tex">\(ln\hspace{0.1cm}A=K\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(\dot{.\hspace{.045cm}.}\hspace{0.1cm}\hspace{0.1cm}i=A\hspace{0.1cm}e^{\frac{-Rt}{L}}\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; (Eq 4.19)</span></p>

<p style="text-align:justify"><span style="font-size:14px">It is seen that the transient component is the exponentially decaying component or d. c. offset as it is called. It decays to a very small value after about 15 to 30 cycles depending upon time constant L/R. The transient component is not d. c., it decays after about 30 cycles as stated. But for the operation of an instantaneous overcurrent relay which operates within 1 to 3 cycles, this transient component can be assumed to remain constant. That is why the transient component is named d. c. offset.</span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px">For obtaining steady-state solution,</span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(i=C\cos(\omega t+\theta)+D\sin(\omega t+\theta)\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(\frac{\mathrm{d}i}{\mathrm{d}t}=-C\hspace{0.1cm}\omega \hspace{0.1cm}\sin(\omega t+\theta)+D\hspace{0.1cm}\omega \hspace{0.1cm}\cos(\omega t+\theta)\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px">Substituting these values of i and&nbsp;<span class="math-tex">\(\frac{\mathrm{d}i}{\mathrm{d}t}\)</span>&nbsp;in Equation 4.18,</span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(-C\hspace{0.1cm}\omega \hspace{0.1cm}L\sin(\omega t+\theta)+D\hspace{0.1cm}\omega \hspace{0.1cm}L\cos(\omega t+\theta)\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(+R \hspace{0.1cm}C\hspace{0.1cm} \cos(\omega t+\theta)+ R \hspace{0.1cm}D\hspace{0.1cm} \sin(\omega t+\theta)\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(=E_m \hspace{0.1cm} \sin(\omega t+\theta)\)</span></span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px">Equating coefficients of sin and cos terms,</span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(R\hspace{0.1cm}C+D\hspace{0.1cm}\omega\hspace{0.1cm}L=0,\)</span>&nbsp; &nbsp; &nbsp; &nbsp; </span></p>

<p style="text-align:justify"><span style="font-size:14px">&nbsp;<span class="math-tex">\(R\hspace{0.1cm}D-C\hspace{0.1cm}\omega\hspace{0.1cm}L=E_m\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(\dot{.\hspace{.045cm}.}\hspace{0.3cm}C= \frac{-D\hspace{0.1cm}\omega \hspace{0.1cm}L}{R}\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(\dot{.\hspace{.045cm}.}\hspace{0.3cm}R\hspace{0.1cm}D+ \frac{D\hspace{0.1cm}\omega^2 \hspace{0.1cm}L^2}{R}=E_m\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(\dot{.\hspace{.045cm}.}\hspace{0.3cm}D[\frac{R^2+\omega^2\hspace{0.1cm}L^2}{R}]=E_m\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(\dot{.\hspace{.045cm}.}\hspace{0.3cm}D=\frac{R\hspace{0.1cm}E_m}{R^2+\omega^2\hspace{0.1cm}L^2}\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;</span></p>

<p style="text-align:justify"><span style="font-size:14px">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<span class="math-tex">\(C=\frac{-\omega\hspace{0.1cm}L\hspace{0.1cm}E_m}{R^2+\omega^2\hspace{0.1cm}L^2}\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px">Referring to impedance triangle shown in figure 4.27,</span></p>

<p style="text-align:justify"><span style="font-size:14px"><img alt="" src="https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16161467031616146703.jpg" style="height:491px; width:800px" /></span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(\cos\phi=\)</span>&nbsp;Power factor,&nbsp;&nbsp; </span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(Z=\sqrt{(R^2+\omega^2\hspace{0.1cm}L^2)}\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<span class="math-tex">\(\sin\phi=\frac{\omega\hspace{0.1cm}L}{Z}\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;</span></p>

<p style="text-align:justify"><span style="font-size:14px">&nbsp;<span class="math-tex">\(\cos\phi=\frac{R}{Z}\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px">These relations give the final steady-state solution as follows:</span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(i=\frac{E_m}{Z}[\sin(\omega\hspace{0.1cm}t+\theta)\cos\phi\)</span></span></p>

<p style="text-align:justify">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<span class="math-tex">\(-\cos(\omega\hspace{0.1cm}t+\theta)\sin\phi]\)</span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(=\frac{E_m}{Z}\hspace{0.1cm}\sin(\omega\hspace{0.1cm}t+\theta-\phi)\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(=I_m\hspace{0.1cm}\sin(\omega\hspace{0.1cm}t+\theta-\phi)\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;</span></p>

<p style="text-align:justify"><span style="font-size:14px">Where&nbsp;<span class="math-tex">\(\frac{E_m}{Z}=I_m\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px">Total solution is </span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(i=A\hspace{0.1cm}e^{-\frac{RT}{L}}+I_m\hspace{0.1cm}\sin(\omega\hspace{0.1cm}t+\theta-\phi)\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px">Now at t = 0 (just before fault inception), i = 0&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (ignoring pre-fault current)</span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(0=A\hspace{0.1cm}+I_m\hspace{0.1cm}\sin(\theta-\phi)\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(\dot{.\hspace{0.045in}.}\hspace{0.1cm}A\hspace{0.1cm}=I_m\hspace{0.1cm}\sin(\phi-\theta)\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px">Therefore,</span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(i\hspace{0.1cm}=I_m\hspace{0.1cm}[\sin(\omega t+\theta-\phi)+sin(\phi -\theta)e^{\frac{-Rt}{L}}]\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px">It is seen that the fault current is a sinusoidal wave superimposed on exponentially decaying d. c. offset.</span></p>

<p style="text-align:justify"><span style="font-size:14px">Now, the fault can occur at any instant of voltage wave. In figure 4.26, fault is shown to have occurred at <span class="math-tex">\(\theta=45^0\)</span> approximately. If fault occurs at e = 0, <span class="math-tex">\(\theta=0^0\)</span>and if we assume &omega;L &gt;&gt; R (the assumption which is not far from reality), <span class="math-tex">\(\phi=90^0\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(i=I_m\hspace{0.1cm}[\sin(\omega t-90)+e^{\frac{-RT}{L}}]\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px">The waveshape will be as shown in figure 4.28. The waveshape shows that there will be maximum asymmetry. If fault occurs at e = E<sub>m</sub>, &theta; = 90<sup>0</sup>.</span></p>

<p style="text-align:justify"><span style="font-size:14px"><img alt="" src="https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16161467471616146747.jpg" style="height:491px; width:800px" /></span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(i=I_m\hspace{0.1cm}\sin\hspace{0.1cm}\omega t\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px">The waveshape of fault current will now be symmetrical w. r. t. time axis. But one never knows the instant at which fault will occur. Hence, a switchgear engineer (or a protection engineer) has to design the power system for worst asymmetry as shown in figure 4.28 wherein the peak of the current will be 2 I<sub>m</sub> (or 2.8 &times; I<sub>rms</sub>). This is known as dynamic current.</span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><strong>The significance of this asymmetry in waveshape is many-fold: </strong></span></p>

<ol>
	<li style="text-align:justify">
	<p><span style="font-size:14px">When fault occurs, all the equipment of the power system (generators, transformers, isolators, circuit breakers, busbars etc.) have to withstand this dynamic current as this current can be a reason for electrodynamic failure of the equipment and, in turn, bursting of the equipment.</span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px">In the era of very fast operating relays (one cycle) and breakers having very low breaker time (one or two cycles), the circuit breaker will be interrupting the asymmetric fault current. Hence, while specifying, a switchgear engineer shall have to ask for higher breaking capacity than the symmetrical breaking capacity calculated in foregoing section. The multiplying factor for specifying asymmetric breaking capacity for breakers is tabulated in Table 4.3 below:</span></p>
	</li>
</ol>

<table align="center" border="1" dir="ltr" style="width:800px">
	<caption>
	<div style="text-align:center"><span style="font-size:28px">Table 4.3: Multiplying Factor for Asymmetric Breaking Current</span></div>
	</caption>
	<thead>
		<tr>
			<th scope="col" style="border-color:black; vertical-align:top; width:197px">
			<div><span style="font-size:28px">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Circuit Breaker Time</span></div>
			</th>
			<th scope="col" style="vertical-align:top; width:209px">
			<div><span style="font-size:28px">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Multiplying Factor</span></div>
			</th>
		</tr>
	</thead>
	<tbody>
		<tr>
			<td style="border-bottom:1px solid black; border-left:1px solid black; border-right:1px solid black; border-top:none; vertical-align:top; width:197px">
			<div style="text-align:center"><span style="font-size:28px">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; More than 8 cycles</span></div>
			</td>
			<td style="border-bottom:1px solid black; border-left:none; border-right:1px solid black; border-top:none; vertical-align:top; width:209px">
			<div style="text-align:center"><span style="font-size:28px">1.0</span></div>
			</td>
		</tr>
		<tr>
			<td style="border-bottom:1px solid black; border-left:1px solid black; border-right:1px solid black; border-top:none; text-align:center; vertical-align:middle; width:197px">
			<div style="text-align:center"><span style="font-size:28px">5 cycles</span></div>
			</td>
			<td style="border-bottom:1px solid black; border-left:none; border-right:1px solid black; border-top:none; vertical-align:top; width:209px">
			<div style="text-align:center"><span style="font-size:28px">1.1</span></div>
			</td>
		</tr>
		<tr>
			<td style="border-bottom:1px solid black; border-left:1px solid black; border-right:1px solid black; border-top:none; vertical-align:top; width:197px">
			<div style="text-align:center"><span style="font-size:28px">3 cycles</span></div>
			</td>
			<td style="border-bottom:1px solid black; border-left:none; border-right:1px solid black; border-top:none; vertical-align:top; width:209px">
			<div style="text-align:center"><span style="font-size:28px">1.2</span></div>
			</td>
		</tr>
		<tr>
			<td style="border-bottom:1px solid black; border-left:1px solid black; border-right:1px solid black; border-top:none; vertical-align:top; width:197px">
			<div style="text-align:center"><span style="font-size:28px">&nbsp;2 cycles</span></div>
			</td>
			<td style="border-bottom:1px solid black; border-left:none; border-right:1px solid black; border-top:none; vertical-align:top; width:209px">
			<div style="text-align:center"><span style="font-size:28px">1.4</span></div>
			</td>
		</tr>
		<tr>
			<td style="border-bottom:1px solid black; border-left:1px solid black; border-right:1px solid black; border-top:none; vertical-align:top; width:197px">
			<div style="text-align:center"><span style="font-size:28px">1 cycle</span></div>
			</td>
			<td style="border-bottom:1px solid black; border-left:none; border-right:1px solid black; border-top:none; vertical-align:top; width:209px">
			<div style="text-align:center"><span style="font-size:28px">1.6</span></div>
			</td>
		</tr>
	</tbody>
</table>

<ol start="3">
	<li style="text-align:justify">
	<p><span style="font-size:14px">The instantaneous overcurrent relays and distance relays (in first zone) will overreach due to this d. c. offset. This will be discussed later.?</span></p>
	</li>
</ol>

<p><span style="font-size:14px">?</span></p>