16/11/2013
C al culate IDM T over Cur r ent Relay Setti ng ( 50/51) | El El ectr i cal Notes & Ar ti cl es
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Calculate IDMT over over Curr Current ent Relay Setting Setting (50/51) (50/ 51) OCTOBER 11, 2013 9 COMMENTS COMMENTS (HTTP://ELECTRICALNOTES.WORDPRESS.COM/2013/10/11/CALCULATE-IDMT-OVERCURRENT-RELAY-SETTING-5051/#COMMENTS) Calculate Calcula te setting of IDMT over over Current Relay for following following Feeder Feeder and CT Detail Feeder Detail: Detail: Feeder Load Current 384 Amp, Feeder Fault current Min11KA and Max 22KA. installed on on feeder feeder is 600/1 Amp. Relay Error E rror 7.5%, 7.5 %, CT Error 10.0%, 10 .0%, CT over over shoot shoot CT Detail: Detail: CT installed 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 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 Relay High High Current Cur rent setting :Plug : Plug setting of Relay is 2.5 Amp and Time 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 Current Setting: Setting: (I>) Over Load Curr ent (In) = Feeder Feeder Load Current Cur rent X Relay Relay setting = 384 X 125% =480 =480 Amp Required Required Over Load Relay Plug Plug Setting= Sett ing= Over Over Load Cur rent (In) / CT Primary Pr imary Cur rent Require Required d Over Over Load Relay Relay Plug P lug Setting = 480 4 80 / 600 = 0.8 0. 8 Pick up Setting of Over Current Cur rent Relay Relay (PMS) (I>) (I>)= = CT Secondary Current Curr ent X Relay Relay Plug Setting Pick up Setting of Over Current Relay (PMS) (I>)= 1 X 0.8 = 0.8 Amp Plug Setting Multiplier Multiplier (PSM) = Min. Min. Feeder Fault Fault Curr C urrent ent / (PMS X (CT Pri. Pr i. Current / CT Sec. Current)) Plug Setting Multiplier (PSM) = 11000 / (0.8 X (600 / 1)) = 22.92 el ectr ical notes.wor dpr ess.com/2013/10/11/cal cul ate- i dmt- over - cur r ent- r elay- setti ng - 5051/
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16/11/2013
Calculate IDMT over Current Relay Setting (50/51) | Electrical Notes & Articles
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
(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 2.5 = 2.5 Amp Plug Setting Multiplier (PSM) = Min. Feeder Fault Current / (PMS X (CT Pri. Current / CT Sec. Current)) Plug Setting Multiplier (PSM) = 11000 / (2.5 X (600 / 1)) = 7.33 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 / ( (7.33)0.02-1) = 3.44 Amp Here Time Delay of Relay (TMS) is 0.100 Sec so Actual operating Time of Relay (t>) = Operating Time of Relay X TMS =3.44 X 0.100 =0.34 Sec Grading Time of Relay = [((2XRelay Error)+CT Error)XTMS]+ Over shoot+ CB Interrupting Time+ Safety Total Grading Time of Relay=[((2X7.5)+10)X0.100]+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 = 0.34 + 0.58 = 0.85 Sec
Conclusion of Calculation: electricalnotes.wordpress.com/2013/10/11/calculate-idmt-over-current-relay-setting-5051/
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16/11/2013
Calculate IDMT over Current Relay Setting (50/51) | Electrical Notes & Articles
Pickup Setting of over current Relay (PMS) (I>) should be satisfied following Two Condition. (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) 0.8 > =(480/600) = 0.8 >= 0.8, Which found OK For Condition (2) 0.125 <= 11000/600 = 0.125 <= 18.33, Which found OK Here Condition (1) and (2) are satisfied so Pickup Setting of Over Curr ent Relay = OK Low Over Current Relay Setting: (I>) = 0.8A X In Amp Actual operating Time of Relay (t>) = 0.271 Sec High Over Current Relay Setting: (I>>) = 2.5A X In Amp Actual operating Time of Relay (t>>) = 0.34 Sec
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9 Responses to Calculate IDMT over Current Relay Setting (50/51) electricalnotes.wordpress.com/2013/10/11/calculate-idmt-over-current-relay-setting-5051/
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16/11/2013
Calculate IDMT over Current Relay Setting (50/51) | Electrical Notes & Articles
elango.k says: October 11, 2013 at 4:32 pm Hello sir- what is the use of low over current(>i) setting and high over current setting(>>i), sir i requesting you to post some notes on this topic thankyou Reply engelect says: October 12, 2013 at 11:01 pm dear elango, Just sharing my knowledge, Always remember that we must limit ENERGY (thermal) of fault. Since it is energy it is always consist of Power (P) which equal with current (I) and consist of t (time). low over current is to limit thermal withstand of electrical equipment. Whereas high over current setting is to limit short circuit withstand. short circuit withstand is not only dealing with thermal but it is also dealing with electromagnetic force withstand of an electrical equipment, such as busbar, bus duct, etc. Faster clearing big fault (it is dealing high current setting) then we secure our structure of busbar/busduct to prevent system catastrophic. Setting high over current setting shall be wise to distinguish between real short circuit and starting current. Reply elango.K says: October 16, 2013 at 3:03 am Hello sir, I am not clear with this explanation, can you explain with some example. thankyou elango.k says: October 11, 2013 at 4:36 pm Hello sir- what is the use of low over current(>i) setting and high over current setting(>>i), sir i requesting you to post some notes on this topic thankyou Reply vilas says: October 13, 2013 at 12:41 pm DEAR SIR CAN U PL GIVE EXAMPLE OF IDMT RELAY SETTING CALCULATION FOR 132/33 KV 25 MVA TF WITH ONE I/C & 3 FEEDERS OF 200 A EACH WITH RELAY SELECTION CURVE Reply Naresh Kumar says: October 24, 2013 at 11:29 am Dear Sir, What is the size of 3-core,11KV cable required for 1000 KVA, 11/.415 KV transformer.By simple calculation for HT side current of 52 AMP, 3 core, 25 SQ mm cable will suffice. But i have noticed electricalnotes.wordpress.com/2013/10/11/calculate-idmt-over-current-relay-setting-5051/
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16/11/2013
Calculate IDMT over Current Relay Setting (50/51) | Electrical Notes & Articles
that at some places 180 SQ. mm cable being used.Does HT cable selection is related with fault level and HT cable at 11 kv should not be used less than 150 Sq. mm. Pl. provide your valuable comments on HT cable selection for transformers. Thanks. Naresh. Reply Jignesh.Parmar says: October 24, 2013 at 2:45 pm For HT Cable one important criteria which need to be consider is Short circuit level. Normally 11KV S.C level is 20 to 25KA at it varies according to Electricity provider company.. and normally 150 Sq.mm Cable S.C falls in 17 to 20KA. For HT cable you have to consider S.C level of utility first than Current carry capacity of Cable. Reply Naresh Kumar says: October 25, 2013 at 9:57 am Thanks. nagendra mathur says: November 5, 2013 at 11:03 am Please let me know the details of selection of Neutral grounding resistance. What should be nGR value for a 20 MVA, 132/6.6 KV Trnsformer, primary side fault 40 MVA for 1 sec Reply
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