SGP309 Numerical on Operating Time of IDMT relays

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  • Опубліковано 9 гру 2024
  • Lectures on Switchgear & Protection By Dr. Tirupathiraju Kanumuri, Assistant Professor, NIT Delhi
    Link for Material
    drive.google.c...
    Lectures on Power Systems: • POWER SYSTEM
    Lectures on EMFT: • Electro Magnetic Field...
    Lectures on Synchronous Machines:
    • Synchronous Machines
    Lectures on Special Electrical Machines: • Special Electrical Mac...
    Lectures on Transformers: • Transformers
    Lectures on DC Machines: • DC Machines
    Lectures on Induction Motors:
    • Induction Motors
    Lectures on Basic Electrical engineering:
    • Basic Electrical Engin...

КОМЕНТАРІ • 29

  • @sunilkatigari783
    @sunilkatigari783 4 роки тому +3

    Ur making these period of time very useful by uploading the videos thanking you sir.. From heartful

  • @mohankandukuri8293
    @mohankandukuri8293 3 роки тому +1

    Ur videos r very useful so many people sir 💐 💐

  • @madanbohara8476
    @madanbohara8476 2 місяці тому

    in the second qns, fault is originated form transformer side, so it must be
    actual operation time of transformer relay = actual operation time of feeder relay minus 0.5 sec

  • @surendrakverma555
    @surendrakverma555 2 роки тому

    Very good. Thanks 🙏🙏🙏🙏🙏

  • @ANURAGKUMAR-yd3wy
    @ANURAGKUMAR-yd3wy 3 роки тому +2

    Sir in third question, the fault originated from transformer. But we calculated TSM for feeder first i.e. feeder relay should operate first which is contradictory. Shouldn't we calculate transformer relay operating time first? (Question at 10:55)

  • @aryanzee3072
    @aryanzee3072 4 роки тому +2

    Sir I request you pls make a short video ,why we need different types of relays as we've Inverse relay then why we construct IDMT.

    • @TheManzico
      @TheManzico 3 роки тому

      @@lecturesinelectricaleng This short answer clarifies it. i was having same question. Thank you.

  • @moulimouli591
    @moulimouli591 4 роки тому

    thank you sir.

  • @i-techengineering2076
    @i-techengineering2076 3 роки тому

    How to check binary O/P during calibration?

  • @AmitkumarEVB
    @AmitkumarEVB 2 роки тому

    how we calculate the fault level for given capacity sub station like 2 x 40 MVA capacity sub station 132 kV/33 kV
    ??

  • @yipmanho3843
    @yipmanho3843 2 роки тому

    sir how to check the chart If PSM is 3.33 equal to 5.6 sec

  • @amnaimam7894
    @amnaimam7894 3 роки тому

    Thanks sir
    In Q2 : what the difference between transformer relay current & relay fault current?? Both are in the secondary of the same relay

  • @kris9811
    @kris9811 2 роки тому

    Hi can you make a video about protection coordination?

  • @fluxML
    @fluxML 4 роки тому

    Thank u so much sir for providing free access to this course.. It is helping me alot to clear my concepts.
    I want to clear my doubt regarding 2nd question tranformer section where we are selecting upper seeting that is 150%..this allows transformer to get overloaded upto 143%.so is it ok to do so?? Or its better to select the lower setting i.e 125%?? 17:18

    • @fluxML
      @fluxML 4 роки тому

      17:18

    • @fluxML
      @fluxML 4 роки тому

      Thank you sir.

    • @gkptechnicalclasses3518
      @gkptechnicalclasses3518 4 роки тому

      In that case transformer relay can't operate but feeder relay will operate and circuit will be tripped...

  • @nagamantriudaykumar4200
    @nagamantriudaykumar4200 4 роки тому

    Sir again relay ck t 5000/(1000:5) divide and another time 500/(400:5) divide some thing confuse please tell

  • @kousikmullapudi5126
    @kousikmullapudi5126 4 роки тому

    Sir where will be this PSM present in the relay circuit

  • @futureaptitude2350
    @futureaptitude2350 3 роки тому

    In table no given 8 value I mean 3.15

  • @gurpreetsinghee
    @gurpreetsinghee 4 роки тому

    Sir, how can we find fault current in practical.
    Suppose everything working fine

    • @er.sachidanandsah428
      @er.sachidanandsah428 Рік тому

      For this, all types of fault is done at particular point and fault with maximum values will be the actual fault current of that point. Usually LLL fault gives maximum current value and Lg fault gives maximum voltage value.