IDMT: A power transformer feeds an 11 KV bus bar through a circuit breaker. The bus bar feeds two outgoing feeders through two circuit breakers. Current transformers housed in the transformer & the feeder’s circuit breakers are assumed to have turns ratios of 1000/1 & 500/1 respectively. The earth relays on all circuit breakers have plug settings of 20%, the time multiplier setting being 0.2 for the feeders. A single phase to earth fault of 1000A occurs on one of the outgoing feeders. Find the operating time of the feeder relay & suggest suitable time multiplier, which could be adopted for the transformer relay in order to ensure adequate discrimination. Use the standard IDMT time current characteristics, attached. Take a minimum grading margin of 0.3 seconds. Solution: Given that PMS (plug setting) = 20% = 0.2 TMS (for feeders) = 0.2 CM = Current multiplier =
fault current 1000 = =10 A CT ratio∗PMS 500∗0.2
From the graph when I (on x-axis) = 10 & TMS =0.2 Then operating time = 0.5 sec (for feeder) Operating time for transformer = operating time for feeder + margin = 0.5+0.3=0.8 sec fault current 1000 CM (for transformer)= CT ratio∗PMS = 1000∗0.2 =5 A With operating time = 0.8 & current 5 , from graph TMS for transformer = 0.2 10 MVA transformer, feeds an 11 KV bus bar through a circuit breaker. The bus bar feeds two outgoing feeders through other circuit breakers. Current transformers housed in the transformer & the feeders circuit breakers are assumed to have turns ratios of 525/1 & 325/1 respectively. The earth relays on all circuit breakers have plug settings of 20%, the time multiplier setting being 0.2 for the feeders. A single phase to earth fault of 404A occurs on one of the outgoing feeders. Find the operating time of the feeder relay & suggest suitable time
multiplier, which could be adopted for the transformer relay in order to ensure adequate discrimination. Use the standard IDMT time current characteristics (attached). Take a minimum grading margin of 0.4 seconds. Solution: Given that PMS (plug setting) = 20% = 0.2 TMS (for feeders) = 0.2 fault current 404 = =6.2 A CM = Current multiplier = CT ratio∗PMS 325∗0.2 From the graph when I (on x-axis) = 6.2 & TMS =0.2 Then operating time = 0.7 sec (for feeder) Operating time for transformer = operating time for feeder + margin = 0.7+0.4=1.1 sec fault current 404 = CM (for transformer)= CT ratio∗PMS 525∗0.2 =3.85 A With operating time = 1.1 & current 3.85 , from graph TMS for transformer = 0.2 In the system shown below relays on all circuit breakers have plug settings of 100%, the time multiplier setting being 0.1 for R3. Given the 3-phase fault levels shown, determine suitable time multiplier settings for relays R2 &R1.
Use the standard IDMT time current characteristics (attached). Take a minimum grading margin of 0.4 seconds.
Solution (not sure !!): PMS=100%=1 TMS = 0.1 for R3
CM=
fault current 1100 = =22 A CT ratio∗PMS 50∗1
Then from graph, operating time =0.22 sec R2 operating time = R3 operating time + margin = 0.22+0.4=0.62 sec CM for R2 =
2300 100∗1 =23 A
Then TMS of R2 from graph using time = 0.62 & current =23,, is 0.2 R1 operating time = 0.62 + 0.4 = 1.02 CM= 13000/400 = 32.5 TMS for R1= 0.4
