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What are the major reasons for failures of first generation numeric relay?

Short Answer

Expert verified

The failure reason of the first generation numeric relay is slower processing speed of the protection circuit in comparison to what is demanded by the software.

Step by step solution

01

Define the numeric relay

The numerical relays are the important protection devices in the power system that are provided with the software based electrical fault detection method. The numerical relay are configure for the different input and the output to detect the fault at various levels which give them the other name of microprocessor relay.

02

Determine the answer

The first generation numerical relay had several disadvantages as mentioned below.

  1. The relay has old software technology that made them open to hackers.
  2. Due to number of features offered by the numerical relay like the speed, sensitivity and error free output. The numerical relay was not working properly and the software that were in conjunction were not operating fast as demanded by the software.

Therefore, the failure reason of the first generation numeric relay is slower processing speed of the protection circuit in comparison to what is demanded by the software.

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Most popular questions from this chapter

(a) Draw the protective zones for the power system shown in Figure 10.45. Which circuit breakers should open for a fault at (a) P1, (b) P2, and (c) P3?

Evaluate relay coordination for the minimum fault currents in Example 10.4. For the selected current tap settings and time dial settings, (a) determine the operating time of relays at B2 and B3 for the 700 A fault current. (b) Determine the operating time of relays at Bl and B2 for the 1500 A fault current. Are the fault-to-pickup current ratios โ‰ฅ2.0(a requirement for reliable relay operation) in all cases? Are the coordination time intervals โ‰ฅ0.3seconds in all cases?

A simple system with circuit breaker-relay locations is shown in Figure 10.49. The six transmission-line circuit breakers are controlled by zone distance and directional relays, as shown in Figure 10.50. The three transmission lines have the same positive-sequence impedance of j0.1 per unit. The reaches for zones 1, 2 and 3 are 80, 120 and 250% , respectively. Consider only three-phase faults. (a) Find the settings Zrin per unit for all distance relays. (b) Convert the settings in V if the VTs are rated 133 kV : 115 V and the CTs are rated 400 : 5 A. (c) For a fault at location X, which is 10% down line TL31 from bus 3, discuss relay operations.

Figure 10.49

Figure 10.50

Line impedances for the power system shown in Figure 10.47 areZ12=Z23=3.0+j40.0, and Z24=6.0+j80.0. Reach for the zone 3B12 impedance a relay is set for 100% of line 1-2 plus 120% of line 2-4. (a) For a bolted three-phase fault at bus 4, show that the apparent primary impedance โ€œseenโ€ by the B12 relays is

Zapparent=Z12+Z24+(I32/I12)Z24

Where (I32/I12)is the line 2-3 to line 1-2 fault current ratio. (b) If |I32||I12|, does the B12 relay see the fault at bus 4? Note: This problem illustrates the โ€œinfeed effect.โ€ Fault currents from line 2-3 can cause the zone 3 B12relay to under reach. As such, remote backup of line 2-4 at B12is ineffective.

Figure 10.47

The CT of Problem 10.5 is utilized in conjunction with a current sensitive device that will operate at current levels of 8 A or above. Check whether the device will detect the 1300 A fault current for cases (b) and (d) in Problem 10.5.

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