<p style="text-align:justify"><span style="font-size:14px"><img alt="" src="https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16251311251625131125.jpg" style="height:1086px; width:800px" /></span></p>

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<p style="text-align:justify"><span style="font-size:14px">Any transmission system can be subdivided into segments of radial feeders supplying the consumers. Therefore, our discussion will initiate with a typical single line diagram of a basic radial feeder shown in figure 7.1</span></p>

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<p style="text-align:justify"><span style="font-size:14px">As we have already discussed earlier, power is generated at 15.75 kV (typical) and stepped up to 220 kV by generator transformer (GT). There is only one generator shown in the figure but in practice, there are more (4 to 8 or more) number of generators operated in parallel in a power plant. The power generation capacity of single unit can be 660 MW or even more. Hence, there will be many 220 kV transmission lines emanating from 220 kV infinite bus. One such line will lead to receiving substation, where voltage is required to be stepped down to (say) 66 kV (optimum transmission voltage for further power transmission. Optimum voltage could have been 132 kV. In that case the transformation would have been to 132 kV and 132 kV lines would run further). Once again, many 66 kV lines will feed power to distribution network in different areas. One 66 kV line (as shown) is feeding power to distribution substation (66/11 kV), wherefrom many 11 kV feeders are feeding power to a PMT (Pole Mounted Transformer) or distribution transformer. This transformer steps down voltage to three-phase 415 V for final requirement of consumers. This is the normal operation of power system.</span></p>

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<p style="text-align:justify"><span style="font-size:14px">But we have seen that faults are inevitable. They do occur. Therefore, protection of this transmission system is a must. The function of the protective relaying system, as we have discussed, is to isolate the minimum faulty section from the rest of the healthy power system as quickly and as selectively as possible. When this isolation is done, we say that the fault is cleared. Rectifying a fault, if it is a sustained fault, is a different task to be taken care of by maintenance staff. Protective relaying system or switchgear is made up of three basic components. Protective relay senses the fault. But a large current of the order of few hundreds or thousands of amperes passing in the transmission line cannot be allowed to pass through a relay. Hence, a current transformer (and/or potential transformer as well for many relays, but in overcurrent protection only C. T.) steps down the current which can be handled by a relay. Fault current interruption can be accomplished by a circuit breaker (C. B.) which receives the signal from the relay. Thus, a protective relay, C. B., and C. T. (and/or P. T.) plus the required control circuitry are said to be the components of switchgear (protective gear or protective relaying system).</span></p>

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<p style="text-align:justify"><span style="font-size:14px">Relays as shown in figure 7.1 are not single units. Figure 7.1 is a single line diagram. Therefore, there are three C. T.s and four relays (or a single relay having three overcurrent units and one earth-fault unit) in a relay group shown R<sub>1</sub>. Hence, relays R<sub>1</sub> protect 220 kV line, relays R<sub>2</sub> protect 66 kV line and relays R<sub>3</sub> protect 11 kV feeder. PMT is protected by D. O. (Drop Out) fuse and 415 V distributor is protected by simple fuse (or MCCB in modern protective scheme). 415 V distributor is not so important as to justify the cost of relays, breaker and C. T.s. Fuse (or MCCB) is a sensing device and current interrupting device as well. MCCBs are available in the range of rated continuous currents ranging from 100 amp to 4000 amp. No doubt, comprehensive characteristics, that the relay provides, are not provided by a fuse (or MCCB) but it is not required and not economical too.</span></p>

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<p style="text-align:justify"><span style="font-size:14px">Requirement of selectivity of a protective system demands that only least faulted portion is isolated. Hence, relays R<sub>1</sub>, R<sub>2</sub>, R<sub>3</sub> and a fuse are to be properly graded or coordinated and their settings are required to be properly calculated for optimum selectivity and fast operating time. Therefore, in case of fault in 415 V distributor, fuse (or MCCB) should operate to isolate the faulty section. If fuse fails to trip, relay R<sub>3</sub> will operate as a remote back-up protection to clear the fault. If there is a fault in 11 kV feeder, relay R<sub>3</sub> acts as primary protection and clears the fault. If either R<sub>3</sub> fails to operate or C. T. saturates and does not provide proper information to relay R<sub>3</sub> or circuit breaker (on receiving signal from R<sub>3</sub>) does not trip because of trouble in the breaker (e. g. open-circuiting of trip coil of breaker or failure of operating mechanism etc.), relay R<sub>2</sub> will act as a back-up. Similarly, for fault in 66 kV line, R<sub>2</sub> primarily operates and R<sub>1</sub> gives back-up.</span></p>

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<p style="text-align:justify"><span style="font-size:14px">Hence, our next important discussion is about discriminating the fault location in different sections (i.e. whether the fault is in distributor, 11 kV feeder or 66 kV or 220 kV lines) and procedure of relay setting calculations.</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/16251311981625131198.jpg" style="height:386px; width:800px" /></span></p>

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<p style="text-align:justify"><span style="font-size:14px">For this purpose, we simplify figure 7.1 and remove transformers from the figure 7.1 as shown in figure 7.2.</span></p>