Transcript

Calculate IDMT over Current Relay

Setting (50/51)
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Calculate setting of IDMT over Current Relay for following Feeder and CT Detail
Feeder Detail: Feeder Load Current 384 Amp, Feeder Fault current Min11KA and
Max 22KA.
CT Detail: CT installed on feeder is 600/1 Amp. Relay Error 7.5%, CT Error 10.0%,
CT over shoot 0.05 Sec, CT interrupting Time is 0.17 Sec and Safety is 0.33 Sec.
IDMT Relay Detail:
IDMT Relay Low Current setting: Over Load Current setting is 125%, Plug setting
of Relay is 0.8 Amp and Time Delay (TMS) is 0.125 Sec, Relay Curve is selected as
Normal Inverse Type.
IDMT Relay High Current setting :Plug setting of Relay is 2.5 Amp and Time
Delay (TMS) is 0.100 Sec, Relay Curve is selected as Normal Inverse Type

Calculation of Over Current Relay Setting:
(1) Low over Current Setting: (I>)
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Over Load Current (In) = Feeder Load Current X Relay setting = 384 X 125%
=480 Amp
Required Over Load Relay Plug Setting= Over Load Current (In) / CT Primary
Current
Required Over Load Relay Plug Setting = 480 / 600 = 0.8
Pick up Setting of Over Current Relay (PMS) (I>)= CT Secondary Current X
Relay Plug Setting
Pick up Setting of Over Current Relay (PMS) (I>)= 1 X 0.8 = 0.8 Amp
Plug Setting Multiplier (PSM) = Min. Feeder Fault Current / (PMS X (CT Pri.
Current / CT Sec. Current))
Plug Setting Multiplier (PSM) = 11000 / (0.8 X (600 / 1)) = 22.92
Operation Time of Relay as per it’s Curve
Operating Time of Relay for Very Inverse Curve (t) =13.5 / ((PSM)-1).
Operating Time of Relay for Extreme Inverse Curve (t) =80/ ((PSM)2 -1).
Operating Time of Relay for Long Time Inverse Curve (t) =120 / ((PSM) -1).
Operating Time of Relay for Normal Inverse Curve (t) =0.14 / ((PSM) 0.02 -1).
Operating Time of Relay for Normal Inverse Curve (t)=0.14 / ( (22.92)0.02-1) = 2.17
Amp
Here Time Delay of Relay (TMS) is 0.125 Sec so
Actual operating Time of Relay (t>) = Operating Time of Relay X TMS =2.17 X
0.125 =0.271 Sec
Grading Time of Relay = [((2XRelay Error)+CT Error)XTMS]+ Over shoot+
CB Interrupting Time+ Safety
Total Grading Time of Relay=[((2X7.5)+10)X0.125]+0.05+0.17+0.33 = 0.58 Sec
Operating Time of Previous upstream Relay = Actual operating Time of Relay+
Total Grading Time Operating Time of Previous up Stream Relay = 0.271 + 0.58 =
0.85 Sec

85 Sec Conclusion of Calculation:            Pickup Setting of over current Relay (PMS) (I>) should be satisfied following Two Condition.02-1) = 3. Operating Time of Relay for Long Time Inverse Curve (t) =120 / ((PSM) -1). Which found OK For Condition (2) 0.5A X In Amp Actual operating Time of Relay (t>>) = 0.5 Amp Plug Setting Multiplier (PSM) = Min.33)0. Feeder Fault Current / (PMS X (CT Pri. Operating Time of Previous up Stream Relay = 0.5)+10)X0.(2) High over Current Setting: (I>>)                 Pick up Setting of Over Current Relay (PMS) (I>>)= CT Secondary Current X Relay Plug Setting Pick up Setting of Over Current Relay (PMS) (I>)= 1 X 2.100 =0.58 Sec Operating Time of Previous upstream Relay = Actual operating Time of Relay+ Total Grading Time.33.58 = 0.8 >= 0.5 X (600 / 1)) = 7.14 / ( (7.44 Amp Here Time Delay of Relay (TMS) is 0. Operating Time of Relay for Extreme Inverse Curve (t) =80/ ((PSM)2 -1).125 <= 18. Current)) Plug Setting Multiplier (PSM) = 11000 / (2. Operating Time of Relay for Normal Inverse Curve (t) =0.100 Sec so Actual operating Time of Relay (t>) = Operating Time of Relay X TMS =3.8A X In Amp Actual operating Time of Relay (t>) = 0.271 Sec High Over Current Relay Setting: (I>>) = 2. (1) Pickup Setting of over current Relay (PMS)(I>) >= Over Load Current (In) / CT Primary Current (2) TMS <= Minimum Fault Current / CT Primary Current For Condition (1) 0.100]+0.44 X 0.34 + 0.05+0.5 = 2.34 Sec . Current / CT Sec.02 -1).14 / ((PSM) 0.5 / ((PSM)-1).17+0. Operating Time of Relay for Normal Inverse Curve (t)=0.34 Sec Grading Time of Relay = [((2XRelay Error)+CT Error)XTMS]+ Over shoot+ CB Interrupting Time+ Safety Total Grading Time of Relay=[((2X7.33 = 0.8.125 <= 11000/600 = 0.8 > =(480/600) = 0.33 Operation Time of Relay as per it’s Curve Operating Time of Relay for Very Inverse Curve (t) =13. Which found OK Here Condition (1) and (2) are satisfied so Pickup Setting of Over Current Relay = OK Low Over Current Relay Setting: (I>) = 0.