<p style="text-align:justify"><span style="font-size:14px"><strong>Example 6.4:</strong> Figure 6.13 shows a single line diagram of power system network. The settings of relays are as follows:</span></p>

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<p style="text-align:justify"><span style="font-size:14px"><img alt="" src="https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16251274441625127444.jpg" style="height:390px; width:800px" /></span></p>

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<p style="text-align:justify"><span style="font-size:14px">Relays R<sub>2</sub> and R<sub>3</sub>: IDMT overcurrent relays giving normal inverse characteristic.</span></p>

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

<p><span style="font-size:14px">P. S. of relay R<sub>3</sub> = 100 %&nbsp;of relay rating.</span></p>

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<p><span style="font-size:14px">P. S. of relay R<sub>2</sub> = 75%&nbsp;of relay rating.</span></p>

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<p style="text-align:justify"><span style="font-size:14px">TMS of R<sub>3</sub> = 0.2</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">TMS of R<sub>2</sub> = 0.1</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">Relay R<sub>1</sub>: Voltage controlled IDMT overcurrent relay giving normal inverse characteristic.</span></p>

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

<p><span style="font-size:14px">P. S. at 100 % rated voltage = 75%&nbsp;of relay rating. P. S. of the relay reduces to 40%&nbsp;of set value if voltage collapses below 70 % of rated voltage. TMS of R<sub>1</sub> = 0.075.</span></p>

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<p style="text-align:justify"><span style="font-size:14px">Find out the time of operation of all the three relays for a three-phase fault immediately after relaying point R<sub>3</sub>.</span></p>

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<p style="text-align:justify"><span style="font-size:14px"><strong>Solution:</strong> The reactance diagram for a three-phase fault at F is as shown in figure 6.13(a).</span></p>

<p style="text-align:justify"><img alt="" src="https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16295208431629520843.jpg" style="float:left; height:330px; width:800px" /></p>

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<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(Fault\hspace{0.1cm}MVA=\frac{60}{0.406}\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px">&nbsp; &nbsp; &nbsp; &nbsp; =147.78 MVA.</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">Fault current,&nbsp;I<sub>f</sub> = 7756 amp.</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">PSM of R<sub>3&nbsp;</sub><span class="math-tex">\(=\frac{7756}{1000}\)</span>&nbsp;=7.756</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">Top of Relay R<sub>3&nbsp;</sub>=&nbsp;<span class="math-tex">\(\frac{3}{log_{10}7.756}\times0.2\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= 0.674 sec</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">PSM of R<sub>2&nbsp;</sub>=&nbsp;<span class="math-tex">\(\frac{7756}{0.75\times 5000}\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = 2.068</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">Top of Relay R<sub>2</sub></span></p>

<p style="text-align:justify"><span style="font-size:14px"><sub>=&nbsp;<span class="math-tex">\(\frac{3}{log_{10}2.068}\times0.1\)</span></sub></span></p>

<p style="text-align:justify"><span style="font-size:14px"><sub>=0.95 sec</sub></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">R<sub>2</sub> backs up R<sub>3</sub> and R<sub>1</sub> backs up R<sub>2</sub> for the fault at F. If R<sub>3</sub> and R<sub>2</sub> both fail to operate, R<sub>1</sub> has to operate. But by that time the generator reactance would have entered into steady-state region (X<sub>s</sub>) from transient region (X&#39;<sub>d</sub>&nbsp;).</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">&nbsp;<span class="math-tex">\(= \frac{60}{X_s\hspace{0.1cm}at\hspace{0.1cm}60\hspace{0.1cm}MVA\hspace{0.1cm}base\hspace{0.1cm}+\hspace{0.1cm}0.1}\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(=\frac{60}{1.53+0.1}\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px">=36.80 MVA.</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">I<sub>f</sub> on 6.6 kV side = 3220 amp</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">P. S. of Relay R<sub>1</sub> =0.75&nbsp;&times; 7500&nbsp;&times; 0.4 = 2250 amp, as voltage collapses due to fault.</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">PSM of R<sub>1&nbsp;</sub>=&nbsp;<span class="math-tex">\(\frac{3220}{2250}\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= 1.43</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">Top of R<sub>1</sub> = 1.45 sec</span></p>