<p style="text-align:justify"><span style="font-size:14px">In this case all relays R<sub>1</sub>, R<sub>2</sub> and R<sub>3</sub> are instantaneous overcurrent relays. Fault current is inversely proportional to impedance up to fault point.</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">i.e.,&nbsp;<span class="math-tex">\(I_f =\frac{E}{Z}\)</span>&nbsp;&nbsp;Where, E = Induced emf of the generator.</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">This is the formula for finding fault current for a three-phase fault. As the impedance of transmission line is proportional to line length, fault current is inversely proportional to distance between a relaying point and a fault point. Hence relay R<sub>1</sub> is set to reach up to bus B, R<sub>2</sub> to reach up to bus C and so on.</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">i.e., Pick-up setting of relay R<sub>1</sub></span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(= \frac{I_f\hspace{0.2cm}at\hspace{0.2cm}B}{CT_1\hspace{0.2cm}ratio}\)</span></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">Where I<sub>f</sub> at B = Fault current for three-phase fault at bus B And CT<sub>1 </sub>ratio = CT ratio of CT<sub>1</sub>.</span></p>

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

<p style="text-align:justify"><span style="font-size:14px">Similar is the formula for pick-up setting of relays R<sub>2</sub> and R<sub>3</sub>. Thus, R<sub>1</sub> will not operate for fault in section II of radial feeder. Similarly, R<sub>2</sub> will protect for faults only in line segment BC (section II). In this way, the discrimination for faults in different zones (or sections) is achieved. This method of discrimination suffers from many disadvantages as follows:</span></p>

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

<ol>
	<li style="text-align:justify">
	<p><span style="font-size:14px">Unrealistic setting accuracy requirement: Protective relays are not supposed to be very accurate. The tolerance in pick-up value of <u><span class="math-tex">\(\pm\)</span></u>&nbsp;5%&nbsp;(or as bad as&nbsp;<u><span class="math-tex">\(\pm\)</span></u>10%&nbsp;so often) is allowed by standards. This means that relay R<sub>1</sub> (and hence R<sub>2</sub> and R<sub>3</sub> also) will either not reach up to bus B or will reach beyond bus B. None of these can be permitted in protection practice. As, if R<sub>1</sub> underreaches, some portion of line will be unprotected and if it overreaches, it is unwanted operation for faults in another zone (or section).</span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px">This method does not provide remote back-up protection, which is a major disadvantage.</span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px">Let us first define new terms.</span></p>
	</li>
</ol>

<p style="margin-left:24px; text-align:justify">&nbsp;</p>

<p style="margin-left:24px; text-align:justify"><img alt="" src="https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16251397341625139734.jpg" style="height:485px; width:800px" /></p>

<p style="margin-left:24px; text-align:justify">&nbsp;</p>

<p style="margin-left:24px; text-align:justify">&nbsp;</p>

<p style="margin-left:24px; text-align:justify">&nbsp;</p>

<p style="margin-left:24px; text-align:justify"><span style="font-size:14px">Z<sub>s</sub> = Source Impedance = Impedance from source to relaying point. (Refer figure 7.3).</span></p>

<p style="margin-left:24px; text-align:justify">&nbsp;</p>

<p style="margin-left:24px; text-align:justify"><span style="font-size:14px">Z<sub>l</sub> = Line Impedance = Impedance from relaying point to fault.</span></p>

<p style="margin-left:24px; text-align:justify">&nbsp;</p>

<p style="margin-left:24px; text-align:justify"><span style="font-size:14px">For R<sub>1</sub>, the reach is up to bus B, therefore, Z<sub>l</sub> = Z<sub>AB</sub>.</span></p>

<p style="margin-left:24px; text-align:justify">&nbsp;</p>

<p style="margin-left:24px; text-align:justify"><span style="font-size:14px">It can be appreciated that Z<sub>s</sub> for relays R<sub>2</sub> and R<sub>3</sub> are different and not equal to that for R<sub>1.</sub></span></p>

<p style="margin-left:24px; text-align:justify">&nbsp;</p>

<p style="margin-left:24px; text-align:justify"><span style="font-size:14px">It is obvious that, Fault current,&nbsp;<span class="math-tex">\(I_f=\frac{E}{Z_s+Z_l}\)</span>&nbsp;.&nbsp;Now if&nbsp;<span class="math-tex">\(Z_s&gt;&gt;Z_l\)</span>,&nbsp;<span class="math-tex">\(I_f=\frac{E}{Z_s}\)</span>,&nbsp;and fault current is not proportional to line length. It is approximately constant for fault at any location of line. In this case, discrimination between faults in different sections (or zones) is not possible. Instantaneous overcurrent relays are affected the most by Z<sub>s</sub>/Z<sub>l</sub> ratio and are of no use for high Z<sub>s</sub>/Z<sub>l</sub> ratio.</span></p>

<p style="margin-left:24px; text-align:justify">&nbsp;</p>

<ol start="4">
	<li style="text-align:justify">
	<p><span style="font-size:14px">The relaying system using instantaneous overcurrent relays does not exploit short-time fault current withstand capability of the equipment to be protected as instantaneous overcurrent relays operate within 1-3 cycles.</span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px">It also does not take into account the transient stability limit of the machines. As such, the generators can feed the fault current for some time (known as critical clearing time which may be few cycles or a fraction of a second) without loosing synchronism as decided by equal area criterion. The transient fault may vanish within this time.</span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px">Instantaneous overcurrent relays suffer from the problem of transient overreach, because of which there can be an unwanted operation of the relay for faults external to its assigned zone.</span></p>
	</li>
</ol>

<p style="margin-left:24px; text-align:justify">&nbsp;</p>

<p style="margin-left:24px; text-align:justify"><span style="font-size:14px">We have discussed earlier that the fault current is asymmetric with reference to time axis following the instant of fault inception. The fault current may be asymmetric or may not be so depending on instant of voltage waveshape at which fault occurs, as already discussed. Hence, the relays are always set for symmetrical rms value of fault current. When asymmetry is present in fault current, current measured by relay is higher (2.8 times the rms value if the relay operates within time equal to zero) than the symmetrical rms fault current. The d. c. offset component of fault current decays exponentially but this decay takes 15-30 cycles depending upon L/R ratio of the power system being protected. The instantaneous overcurrent relay operates within 1-3 cycles and therefore it measures a current which is higher than symmetric rms value. In other words, the relay sees the fault nearer than the location where it is located. Thus, the relay overreaches. This overreach is transient as there will be no overreach after 15-30 cycles. Hence time-delay relays do not suffer from this disadvantage.</span></p>

<p style="margin-left:24px; text-align:justify">&nbsp;</p>

<p style="margin-left:24px; text-align:justify"><span style="font-size:14px">For this reason, the instantaneous overcurrent relay has to be set for the three-phase &nbsp;fault current with maximum generation (all generators are in operation) at 80 % of line section to be protected, assuming transient overreach of the relay as 20 %. However, in this case, remnant 20 % will be unprotected. Hence, instantaneous overcurrent relays alone cannot be used for radial feeder protection, but they are used in conjunction with time-delayed relays so that 80 % of line section can be protected instantaneously and remnant 20 % section protected by time-delayed unit of a protective relay.</span></p>

<p style="margin-left:24px; text-align:justify">&nbsp;</p>

<p style="margin-left:24px; text-align:justify"><span style="font-size:14px">The formula for deciding setting of instantaneous high-set unit is as follows:</span></p>

<p style="margin-left:24px; text-align:justify">&nbsp;</p>

<p style="margin-left:24px; text-align:justify"><span style="font-size:14px">Setting of H. S. Inst.&nbsp;<span class="math-tex">\(Unit=\frac{100B}{100-A}\)</span>&nbsp;</span></p>

<p style="margin-left:24px; text-align:justify">&nbsp;</p>

<p style="margin-left:24px; text-align:justify"><span style="font-size:14px">Where A = % overreach of instantaneous overcurrent relay.</span></p>

<p style="margin-left:24px; text-align:justify"><span style="font-size:14px">&nbsp;B = Magnitude of three-phase symmetric RMS fault&nbsp;Current at the end of feeder section (in CT secondary&nbsp;Amp)</span></p>

<p style="margin-left:24px; text-align:justify">&nbsp;</p>

<p style="margin-left:24px; text-align:justify">&nbsp;</p>

<p style="margin-left:24px; text-align:justify"><span style="font-size:14px"><img alt="" src="https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16260061041626006104.jpg" style="height:386px; width:800px" /></span></p>

<p style="margin-left:24px; text-align:justify"><span style="font-size:14px">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</span></p>

<p style="margin-left:24px; text-align:justify"><span style="font-size:14px">For example, refer to figure 7.2 and data as follows:</span></p>

<p style="margin-left:24px; text-align:justify">&nbsp;</p>

<p style="margin-left:24px; text-align:justify"><span style="font-size:14px">CT ratio of CT<sub>1</sub> = 500/1 amp, Fault current for three-phase fault at bus B with maximum generation is 5000 amp, % overreach of instantaneous overcurrent relay R<sub>1</sub> = 10 %.</span></p>

<p style="margin-left:24px; text-align:justify">&nbsp;</p>

<p style="margin-left:24px; text-align:justify"><span style="font-size:14px">Hence, <span class="math-tex">\(B=\frac{5000}{500}\)</span></span></p>

<p style="margin-left:24px; text-align:justify"><span style="font-size:14px">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;=&nbsp; 10 amp.</span></p>

<p style="margin-left:24px; text-align:justify">&nbsp;</p>

<p style="margin-left:24px; text-align:justify"><span style="font-size:14px">Therefore, setting of H. S. Inst. <span class="math-tex">\(Unit=\frac{100\times 10}{100-10}\)</span>&nbsp;=11.11 amp or 1111%&nbsp;of 1 amp. With setting range 400-2000 % in steps of 50 %, setting selected is 1150 % (or 1200 % if steps are of 100 %).</span></p>

<p style="margin-left:24px; text-align:justify">&nbsp;</p>

<p style="margin-left:24px; text-align:justify"><span style="font-size:14px">With the advent of static and numerical relays, this problem of transient overreach is solved to certain extent. We have seen that,</span></p>

<p style="margin-left:24px; text-align:justify">&nbsp;</p>

<p style="margin-left:24px; text-align:justify"><span style="font-size:14px"><span class="math-tex">\(i=I_m[sin(\omega t+\theta-\phi)\)</span></span></p>

<p style="margin-left:24px; text-align:justify"><span style="font-size:14px">&nbsp; &nbsp; &nbsp;&nbsp;<span class="math-tex">\(+sin(\phi-\theta)e^{\frac{-Rt}{L}}]\)</span></span></p>

<p style="margin-left:24px; text-align:justify">&nbsp;</p>

<p style="margin-left:24px; text-align:justify"><span style="font-size:14px">This asymmetry is present in fault current waveshape, but if this fault current is differentiated before feeding to a protective relay, the d. c. offset will be attenuated.</span></p>

<p style="margin-left:24px; text-align:justify">&nbsp;</p>

<p style="margin-left:24px; text-align:justify"><span style="font-size:14px"><span class="math-tex">\(\frac{di}{dt}=I_m[\omega\hspace{0.1cm}cos(\omega t+\theta-\phi))\)</span></span></p>

<p style="margin-left:24px; text-align:justify"><span style="font-size:14px">&nbsp; &nbsp; &nbsp; &nbsp;<span class="math-tex">\(-\frac{R}{L}sin(\phi-\theta)e^{\frac{-Rt}{L}}]\)</span></span></p>

<p style="margin-left:24px; text-align:justify">&nbsp;</p>

<p style="margin-left:24px; text-align:justify"><span style="font-size:14px">This means that d. c. offset is attenuated to R/L times (as L&gt;&gt;R in power systems) the original value of d. c. offset. Cosine wave is sinusoidal only, so it does not change steady-state component but for phase-shift which overcurrent relay does not measure (in other words, overcurrent relay is not affected by this phase-shift). If R/L is 0.1, d. c. offset is attenuated to 10 %. This means that the relay which has % overreach of 10 % will now overreach by only 1 %. Double differentiation will make the relay practically immune to the problem of transient overreach.</span></p>

<p style="margin-left:24px; text-align:justify">&nbsp;</p>

<p style="margin-left:24px; text-align:justify"><span style="font-size:14px">Differentiation of current can be realized in static relays by using op-amp as a differentiator. In numerical relays it is yet faster because it takes samples every millisecond (a typical sampling interval is one millisecond).</span></p>

<p style="margin-left:24px; text-align:justify">&nbsp;</p>

<p style="margin-left:24px; text-align:justify"><span style="font-size:14px"><span class="math-tex">\(\frac{di}{dt}=\frac{i_2-i_1}{\Delta t}\)</span>&nbsp;,</span></p>

<p style="margin-left:24px; text-align:justify">&nbsp;</p>

<p style="margin-left:24px; text-align:justify"><span style="font-size:14px">Where i<sub>1</sub> = First Sample,</span></p>

<p style="margin-left:24px; text-align:justify"><span style="font-size:14px">i</span><sub>2</sub><span style="font-size:14px"> = Second Sample,</span></p>

<p style="margin-left:24px; text-align:justify"><span style="font-size:14px"><span class="math-tex">\(\Delta t\)</span>&nbsp;=&nbsp;Sampling Interval</span></p>

<p style="margin-left:24px; text-align:justify">&nbsp;</p>

<ol start="7">
	<li style="text-align:justify">
	<p><span style="font-size:14px">Type of Fault: It is known that the magnitude of fault current for three-phase fault (L-L-L) is more than that for double line fault (L-L). Different relays (Earth-fault or Ground relays) will operate for L-g and L-L-g faults and these relays require a separate treatment which we will be discussing later in this module.</span></p>
	</li>
</ol>

<p style="text-align:justify"><span style="font-size:14px">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<span class="math-tex">\(I_{LLL}=\frac{E}{Z_1}\)</span>&nbsp;and&nbsp;</span></p>

<p style="text-align:justify"><span style="font-size:14px">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<span class="math-tex">\(I_{LL}=\frac{\sqrt3 E}{Z_1+Z_2}\)</span>&nbsp;</span></p>

<p style="text-align:justify"><span style="font-size:14px">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;&nbsp;<span class="math-tex">\(=\frac{\sqrt3E}{2Z_1}\)</span></span></p>

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

<p style="text-align:justify">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<span style="font-size:14px"><span class="math-tex">\(I_{LL}=0.866I_{LLL}\)</span></span></p>

<p>&nbsp;</p>

<p style="margin-left:24px; text-align:justify">&nbsp;</p>

<p style="margin-left:24px; text-align:justify"><span style="font-size:14px">Now, if relay R<sub>1</sub> is made to reach up to bus B for double line fault, it will overreach for triple line fault and if it is set to reach up to bus B for triple line fault, it will underreach for double line fault. This statement is true for relays R<sub>2</sub> and R<sub>3</sub> also for similar faults at bus C and D respectively.</span></p>

<p style="margin-left:24px; text-align:justify">&nbsp;</p>

<p style="margin-left:24px; text-align:justify"><span style="font-size:14px">Because of these problems involved with the use of instantaneous overcurrent relays, such relays alone are not used for transmission line protection. However, they are used in conjunction with time-delayed relays having in-built high-set inst. unit to trip the line instantaneously for faults in the vicinity of relay location (usually 80 % of line section).</span></p>

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