<ul>
	<li style="text-align:justify">
	<p><span style="font-size:14px"><strong>Phase Relays:</strong></span></p>
	</li>
</ul>

<p style="margin-left:48px; text-align:justify"><span style="font-size:14px">The phase relays are meant for sensing triple line or double line faults (i.e., faults not involving ground).</span></p>

<p style="margin-left:48px; text-align:justify">&nbsp;</p>

<p style="margin-left:48px; text-align:justify"><span style="font-size:14px"><strong><u>Plug-Settings:</u></strong></span></p>

<p style="margin-left:48px; text-align:justify">&nbsp;</p>

<p style="margin-left:48px; text-align:justify"><span style="font-size:14px"><strong>Plug-settings are decided considering the following rules:</strong></span></p>

<p style="margin-left:48px; text-align:justify">&nbsp;</p>

<ol>
	<li style="text-align:justify">
	<p><span style="font-size:14px">The plug-setting must not be less than the maximum load current including permissible continuous overload unless monitored by an undervoltage relay or unless voltage-controlled overcurrent relay is used.</span></p>
	</li>
</ol>

<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/16252256391625225639.jpg" style="height:375px; width:800px" /></span></p>

<p>&nbsp;</p>

<ol start="2">
	<li style="text-align:justify">
	<p><span style="font-size:14px">While estimating plug-setting, an allowance must be made of the fact that the pick-up of an IDMT overcurrent relay varies from 1.05 to 1.3 times the plug-setting as per standards. Consider figure 7.6 to understand this rule. Current flowing in the radial feeder of figure 7.6 is 400 amp (maximum load current). Hence, plug-settings of relays R<sub>1</sub>, R<sub>2</sub> and R<sub>3</sub> are set at 100 % according to rule 1 above. Now, due to overload or fault somewhere in 415 V distributor emanating from bus D, 550 amp current is passing through radial feeder. With reference to allowable limits in pick-up, it is quite possible that pick-up of relay R<sub>3</sub> is 650 amp (1.3 times P. S. of R<sub>3</sub>) and pick-up current of relay R<sub>2</sub> is 525 amp (1.05 times P. S. of R<sub>2</sub>). This means that relay R<sub>2</sub> will pick-up and R<sub>3</sub> will not at all start. However, time of operation of R<sub>2</sub> is set to be more than that of relay R<sub>3</sub>, R<sub>2</sub> will operate and R<sub>3</sub> will not operate in this case. This is against the requirement of selectivity of protective relaying system. Hence, if selectivity is to be maintained,</span></p>
	</li>
</ol>

<p style="margin-left:48px; text-align:justify">&nbsp;</p>

<p style="margin-left:48px; text-align:justify"><span style="font-size:14px">Pick-up of R<sub>2</sub> &gt; Pick-up of R<sub>3</sub></span></p>

<p style="margin-left:48px; text-align:justify">&nbsp;</p>

<p style="margin-left:48px; text-align:justify"><span style="font-size:14px">or,&nbsp;<sub><span class="math-tex">\(1.05\times P.S.of\hspace{0.1cm}R_2&gt;1.3\times P.S.of\hspace{0.1cm}R_3\)</span></sub></span></p>

<p style="margin-left:48px; text-align:justify">&nbsp;</p>

<p style="margin-left:48px; text-align:justify"><span style="font-size:14px"><span class="math-tex">\(\dot{.\hspace{0.095in}.}\)</span>&nbsp;&nbsp;<span class="math-tex">\(P.S.of\hspace{0.1cm}R_2&gt;\frac{1.3}{1.05}\times P.S.of\hspace{0.1cm}R_3\)</span></span></p>

<p style="margin-left:48px; text-align:justify">&nbsp;</p>

<p style="margin-left:48px; text-align:justify"><span style="font-size:14px">In figure 7.6, P. S. of </span></p>

<p style="margin-left:48px; text-align:justify"><span style="font-size:14px">&nbsp;<span class="math-tex">\(R_2&gt;\frac{1.3}{1.05}\times 500\hspace{0.1cm}amp\)</span></span></p>

<p style="margin-left:48px; text-align:justify"><span style="font-size:14px">&nbsp; &nbsp; &nbsp;<span class="math-tex">\(&gt;619\hspace{0.1cm}amp\)</span></span></p>

<p style="margin-left:48px; text-align:justify"><span style="font-size:14px">&nbsp; &nbsp;<span class="math-tex">\(&gt;123.8%\)</span>&nbsp;&nbsp;of 500 amp (CT primary rating of relaying point R<sub>2</sub>).</span></p>

<p style="margin-left:48px; text-align:justify">&nbsp;</p>

<p style="margin-left:48px; text-align:justify"><span style="font-size:14px">Hence, 125 % shall be set. Similarly, P. S. of R<sub>1</sub> is set at 175 %, based on same consideration. </span></p>

<p style="margin-left:48px; text-align:justify">&nbsp;</p>

<p style="margin-left:48px; text-align:justify"><span style="font-size:14px">i.e., P. S. of &nbsp;<span class="math-tex">\(R_1&gt;\frac{1.3}{1.05}\times P.S. of \hspace{0.1cm}R_2.\)</span></span></p>

<p style="margin-left:48px; text-align:justify">&nbsp;</p>

<p style="margin-left:48px; text-align:justify"><span style="font-size:14px"><img alt="" src="https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16252256881625225688.jpg" style="height:489px; width:800px" /></span></p>

<p style="margin-left:48px; text-align:justify">&nbsp;</p>

<ol start="3">
	<li style="text-align:justify">
	<p><span style="font-size:14px">The relay shall reach at least up to the end of next protected zone. In figure 7.5, relay R<sub>1</sub> shall reach for an L-L fault at bus C with minimum generation (and relay R<sub>2</sub> shall reach up to bus D with same conditions). This is required to ensure back-up protection.</span></p>
	</li>
</ol>

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

<p style="text-align:justify"><span style="font-size:14px">It is obvious that the higher of the two plug-settings offered by rule 1 and rule 2 shall be considered for setting. This value of plug-setting must satisfy rule 3. If this is not satisfied, one has to use voltage-controlled overcurrent relay to satisfy rule 3.</span></p>

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

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

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

<p style="text-align:justify"><span style="font-size:14px"><strong><u><img alt="" src="https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16252257151625225715.jpg" style="height:727px; width:800px" /></u></strong></span></p>

<p style="text-align:justify"><span style="font-size:14px">Relay R<sub>3</sub> of figure 7.5 has to be coordinated with fuse (or MCCB) characteristic. The TMS of R<sub>3</sub> will be decided based on this grading. TMS of R<sub>2</sub> has to be chosen for the highest fault current that concerns relay R<sub>2</sub> and R<sub>3</sub>. This fault current is a triple line fault immediately after relaying point R<sub>3</sub> with maximum generation (this fault location will be at bus C as circuit breaker of relaying point R<sub>3</sub> is usually just 10 to 20 meters far from bus C). The value of this fault current is I<sub>f</sub> as shown in figure 7.7. The time-current characteristics of relays R<sub>2</sub> and R<sub>3</sub> are also shown in the figure 7.7. Obviously, R<sub>2</sub> should operate later than relay R<sub>3</sub> but discrimination time margin &ldquo;af&rdquo; shall be as small as possible as speed of operation is also a requirement of protective relaying system.</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">Referring to figure 7.7,</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">oa = Standard time of operation of relay R<sub>3</sub> = 0.23 sec.</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">This assumes fault current equal to 20&nbsp;&times; P.S.&nbsp;of relay and TMS = 0.1.</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">The relay may not operate at the end of this time exactly as certain time errors are allowed by standards. As per standards, following time tolerances are allowed:</span></p>

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

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

<table cellspacing="0" class="MsoTableGrid" style="border-collapse:collapse; border:none; margin-left:123px; width:900px">
	<tbody>
		<tr>
			<td style="border-bottom:1px solid black; border-left:1px solid black; border-right:1px solid black; border-top:1px solid black; vertical-align:top; width:178px">
			<div style="text-align:center"><strong><span style="font-size:28px">PSM</span></strong></div>
			</td>
			<td style="border-bottom:1px solid black; border-left:none; border-right:1px solid black; border-top:1px solid black; vertical-align:top; width:172px">
			<div style="text-align:center"><strong><span style="font-size:28px">Time Tolerance</span></strong></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:178px">
			<div style="text-align:center"><span style="font-size:28px">Up to 2</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:172px">
			<div style="text-align:center"><span style="font-size:28px">Not Defined.</span></div>

			<div style="text-align:center"><span style="font-size:28px">Relay shall operate</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:178px">
			<div style="text-align:center"><span style="font-size:28px">2-4</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:172px">
			<div style="text-align:center"><span style="font-size:28px"><span class="math-tex">\(\pm12.5\)</span>&nbsp;%</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:178px">
			<div style="text-align:center"><span style="font-size:28px">4-20</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:172px">
			<div style="text-align:center"><span style="font-size:28px"><span class="math-tex">\(\pm7.5\)</span>&nbsp;%</span></div>
			</td>
		</tr>
	</tbody>
</table>

<p>&nbsp;</p>

<p>&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px">Therefore, ab = - 7.5 % error in operating time of relay R<sub>3</sub> = 0.017 sec.</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">The relay R<sub>3</sub> has, hence, operated at the expiry of time ob. Relay, then, gives signal to circuit breaker. CB takes certain time to clear the fault. Good SF<sub>6</sub> circuit breakers take 2 cycles to clear the fault (contact separation, arcing and final arc quenching).</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">Therefore, bc = Breaker time = 2 cycles = 0.04 sec.</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">At the end of time oc, the fault is cleared and current flowing through the relay is zero. But the relay R<sub>2</sub> also had sensed the fault and had started. It should reset, but due to mechanical inertia in electromagnetic relays and electrical inertia (due to inductance of relay circuitry, current reduces to zero exponentially) in static/numerical relays, it keeps on counting time for a little time (after clearance of the fault) known as overshoot or overtravel. This is de.</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">de = 5 cycles in electromagnetic relays</span></p>

<p style="text-align:justify"><span style="font-size:14px">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 1 cycle in static/numerical relays</span></p>

<p style="text-align:justify"><span style="font-size:14px">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 0.02 sec.</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">ef = + 7.5 % &nbsp;time error of the relay R<sub>2</sub>.</span></p>

<p style="text-align:justify"><span style="font-size:14px">&nbsp;&nbsp;&nbsp;&nbsp; = 0.036 sec.</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">It is very obvious that points c and d shall not coincide or overlap.</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">cd = safety margin = 0.1 sec.</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">Hence for all practical purposes, we will use a fixed discrimination time margin of 0.25 sec between time of operations of two successive relays.</span></p>

<p style="margin-left:48px; text-align:justify">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;</p>

<p style="margin-left:48px; text-align:justify">&nbsp;</p>