Calculate IDMT over Current Relay Setting (50/51)

  • 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>)

  • 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

(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:

  • 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 Current 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

 

About Jignesh.Parmar
Jignesh Parmar has completed M.Tech (Power System Control), B.E(Electrical) from Gujarat University. He has more than 13 years experience in Power Transmission-Power Distribution-Electrical energy theft detection-Electrical Maintenance-Electrical Projects(Planning-Designing-coordination-Execution). He is Presently associate with one of the leading business group as a Deputy Manager at Ahmedabad,India. He is Freelancer Programmer of Advance Excel and design useful Excel Sheets of Electrical Engineering as per IS,NEC,IEC,IEEE codes. He is technical Author for "Electrical Mirror" and "Electrical India" Magazines. He is Technical Blogger and Familiar with English, Hindi, Gujarati, French languages. He wants to Share his experience & knowledge and help technical enthusiasts to find suitable solutions and updating themselves on various Engineering Topics.

35 Responses to Calculate IDMT over Current Relay Setting (50/51)

  1. elango.k says:

    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

    • engelect says:

      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.

      • elango.K says:

        Hello sir,
        I am not clear with this explanation, can you explain with some example.
        thankyou

    • RAJASEKHAR says:

      WHEN THE RELAY IS SET ON Low set , when fault current is marginally more than the full load current it will take more time to trip the breaker like 750 msec

      WHEN THE FAULT CURRENT goes 10 times , Fault should get cleared in 50 m sec to protect the EQUIPMENT

  2. elango.k says:

    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

  3. vilas says:

    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

  4. Naresh Kumar says:

    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 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.

    • 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.

  5. nagendra mathur says:

    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

  6. Shakeel Ahmad says:

    Dear Jignesh,

    what is the best location of REF and SBEF CTs on Transformer side? i have 1000kVA Transformer and we are planning to mount the CTs either in LV cable box, REF on 1C630 Neutral Cable and SBEF on 1C240 Earth Cable OR Below LV cable box in Enclosure. Your suggestion please..

  7. elango.k says:

    Hello sir- what is the use of low over current(i>) setting and high over current setting(i>>i),
    sir i requesting you to post some notes on this topic
    thankyou

  8. kamlesh says:

    why we can not maintain of power factor one. if we mantaitain 1 and above what is problem in transformer and power.

    • EXP. ENG. Aqeel says:

      Dear K. 1- It is impossible to make P.F = 1 because ( P.F= COSᵠ ) and -1 ≤ cosᵠ ≤1 .
      2- To avoid over compensating and not to lead to advance P.F is better to make it less or equal 1 ,it means always lagging .

  9. P.Koteswar Rao says:

    Dear Sir,
    Please advice regarding df / dt relay minimum and maximum setting details.
    We are using ABB make, REF type relay for 11kv grid feeder.
    We are going to synchronize 30MW,11kv DG with this grid breaker.
    Power failure time, I wish to isolate grid feeder.
    Please advice.

  10. Suraj Ram says:

    sir can you please give an example how do understand it on breaker trip curve or Time current curve

  11. Prabakaran says:

    Sir, thank u for your notes on Normally inverse curve. By the same way plz give the formula for Extermly inverse curve.

  12. sartajalam says:

    sir,i need testing procedure of over current relay.

  13. jagdish ghogare says:

    Dear parmar Sir can i get your phone number so that i can take some help from you

  14. Murali krishnan says:

    Dear Jignesh,

    please advise where we can find the min fault level data.

  15. Rohaida says:

    Dear Sir,
    Your “Electrical Thumb Rules-(Part-8)” is referred.

    The details: Feeder Fault current Min11KA and Max 22KA.

    How can we calculate the Min Fault Current and Max Fault Current?

    I’m working on Overcurrent relay setting of Primary Transformer Protection (2000KVA with 6% impedance) with 200/5 CT ratio.

    Can you guide me on how to calculate min & max fault current.

  16. Bernard Farrugia says:

    Why in the standard IDMT td(I) calculation you haven’t divided by Beta i.e. by 2.97? as regulated by the IEC…

    Thanks

  17. Prageesh says:

    Dear Sir,

    In an excellent manner you have explain 50/51N relay setting. Can you please guide to set the 87 relay for 110/33KV, 50MVA, 50Hz transformer.

  18. mukund harne says:

    The details provided are really very useful.

  19. Aamir Javaid Malik says:

    Dear Sir,

    In your example for ‘calculating setting of IDMT over-current rely’ mentioned above, kindly guide me how would you prevent relay tripping from transformer inrush current when feeder breaker is switched on. You may assume any capacity of transformer installed in the outgoing feeder in above example.

  20. jegatheshwaran says:

    Sir,
    I would like to known how relay can be wiried and implemented and settings and how its configured in relay panel.
    And i had a doubt, this realy implementation and configuration and setting are different from one another??????
    If it’s different plz give the notes to get the idea about that please sir
    Thank you…….

  21. Muthukumaran says:

    operating time of low current operating time is higher than high current setting

  22. Pradnya says:

    Respected sir,

    For IDMT relay having CT ratio 300:1, load current 470A, maximum fault current is 2380A and minimum fault current is 720A, then how to select upper and lower boundaries on TMS & Plug setting of relay.

    Kindly guide me sir, looking forward for your positive reply.

  23. kader says:

    please clarify 0.14 / ( (7.33)0.02-1) = 3.44 Amp , how?

  24. Purvish shah says:

  25. lamho66@yahoo.com.hk says:

    Sir,
    I would like to know why there are some differences between the theoretical operating time and the experimental one? Why is that lower injection current with larger error?
    Thank you!

  26. muneeb says:

    Respected Sir,
    please guide, the following values are given as above example, are these also to be calculated?

    “””””””Time Delay (TMS) is 0.125 Sec, Relay Curve is selected as Normal Inverse Type.”””””

    because they are just mentioned and used in calculation, but not calculated.

  27. Solve the problem:
    The plug setting of a negative sequence relay is 0.2A. The CT ratio is 5: 1. The minimum
    value of line to line fault current for the operation of the relay is:

  28. Mohit krushna pradhan says:

    Sir please clear that,in Higher over current setting calculation,how you have considered relay plug setting is 2.5? and Please clear how you considered TMS is 0.125 for low over current setting?

    Thanks & Regards
    Mohit k Pradhan

  29. para says:

    Dear Sir,
    i have I> relay setting (IDMT)
    300A 0,15 NI
    What does it mean number 0,15 ?

    Thank you in advance

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