<p style="text-align:justify"><span style="font-size:14px"><strong>Example 6.3:</strong> Refer figure 6.12. The data for the numerical is given below:</span></p>

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

<p style="text-align:justify">&nbsp;</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/16251273671625127367.jpg" style="height:374px; width:800px" /></span></p>

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

<ol>
	<li style="text-align:justify">
	<p><span style="font-size:14px">Relay R is normal inverse IDMT relay provided with high-set instantaneous unit.</span></p>
	</li>
</ol>

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

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

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

<p style="text-align:justify"><span style="font-size:14px">&nbsp;IDMT unit: 50-200 % of relay rating in 7 equal steps.</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">H. S. Inst. Unit: 400-2000 % of relay rating in steps of 50 %.</span></p>

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

<ol start="2">
	<li style="text-align:justify">
	<p><span style="font-size:14px">C. T. ratio: 500/1 amp.</span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px">Full load current of the feeder (transmission line) = 300 amp.</span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px">Overload withstand: 10 % continuous.</span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px">Operating time required for a fault current of 5000 amp is 0.7 second.</span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px">H. S. unit should operate for a minimum fault current of 6125 amp.</span></p>
	</li>
</ol>

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

<p style="text-align:justify"><span style="font-size:14px">Calculate relay settings (P. S., TMS and setting of H. S. unit).</span></p>

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

<p style="text-align:justify"><span style="font-size:14px"><strong>Solution:</strong> Full load current = 300 amp.</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">Considering allowable overload, I = 330 amp.</span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(P.S.=\frac{330}{500}\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;=0.66 or 66%</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">The next step available (from steps 50-75-100-125-150-175-200 %) w. r. t. this calculation is&nbsp;75 %.</span></p>

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

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

<p style="text-align:justify"><span style="font-size:14px">P. S. = 75 %</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">Top (Time of operation) of the relay</span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(=\frac{3}{log_{10}PSM}\times TMS\)</span></span></p>

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

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(PSM=\frac{I_f}{P.S.}\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<span class="math-tex">\(=\frac{5000}{0.75\times500}\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;=13.33</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">Therefor, Top of the relay</span></p>

<p style="text-align:justify"><span style="font-size:14px">&nbsp;<span class="math-tex">\(=\frac{3}{log_{10}PSM}\times TMS\hspace{0.2cm}of\hspace{0.2cm}the\hspace{0.2cm}relay\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px">= 0.7</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">TMS of the relay = 0.262.</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;&nbsp;&nbsp; </span></p>

<p style="text-align:justify"><span style="font-size:14px">TMS is available from 0 to 1.0 in steps of 0.05.</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;&nbsp;&nbsp;&nbsp; </span></p>

<p style="text-align:justify"><span style="font-size:14px">Therefore, TMS selected = 0.3.</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">High short-circuit current = 6125 amp = 12.25&nbsp;&times; C.T. rating.......(<span class="math-tex">\(\frac{6125}{500}\)</span>&nbsp;=12.25)</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">H. S. setting has to be selected as 1250 %.</span></p>