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A2 Power System : Short Circuit Calculations

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  • เผยแพร่เมื่อ 5 มี.ค. 2019
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ความคิดเห็น • 23

  • @itanongmokayengineer
    @itanongmokayengineer 3 ปีที่แล้ว +3

    super clear explanation this helps me a lot thank you so much for sharing your brilliant mind

  • @jaykesapalaran5434
    @jaykesapalaran5434 2 ปีที่แล้ว +2

    Thank you for sharing your knowledge ❤️ very much informative and easy to understand 💌 THANK YOU...

  • @sambathbunkh
    @sambathbunkh 10 หลายเดือนก่อน +2

    Good explanation

  • @sandipkumarprajapati3189
    @sandipkumarprajapati3189 ปีที่แล้ว +1

    Thank you Professor!

  • @lythous
    @lythous 4 ปีที่แล้ว +11

    Hi. I think you made a mistake in your calculations.
    I_FB = (20000 / sqrt(3)) / 33.47
    = 345 A @ 20kV :: 17.25 kA @ 400V

    • @tomaszpiwowar7657
      @tomaszpiwowar7657 3 ปีที่แล้ว +1

      You right!

    • @atmsc4nd4l
      @atmsc4nd4l 3 ปีที่แล้ว +1

      agree.

    • @Brandlead
      @Brandlead 3 หลายเดือนก่อน

      605A at 110kV right not the primary of the 20kV,0.8MVA transformer?

  • @shabeerahamed1501
    @shabeerahamed1501 4 ปีที่แล้ว +3

    If we use cable which is connected to the transformers, we need to add those impedences in the SCC.

  • @DebjitDas93
    @DebjitDas93 4 ปีที่แล้ว +1

    In grid impedence calculation and Numerator in KV & denominator in MVA? How? If KV, then bring MVA into KVA

  • @madhuvimal88
    @madhuvimal88 3 ปีที่แล้ว +1

    Hai..
    Can you please validate with etap value..
    What inputs have taken in etap and that particular input how to use in manual calculation..

  • @jfsimon1981
    @jfsimon1981 3 ปีที่แล้ว +3

    Good day,
    Is ths is mistake at 4'04" Isc,3P = (20000/1.732)/33.47 = 345 A (not 605A) ?
    After calculating I found out different results:
    Short circuit level at 20 kV: 605 A -> 345 A
    Short circuit level at 400 V: 30 kA -> 17.25 kA
    Thank you,
    Jean-François

    • @willgould4417
      @willgould4417 ปีที่แล้ว +2

      I got the same result as you, think its a mistake

  • @aymane9489
    @aymane9489 3 ปีที่แล้ว +3

    Very good explanation, but I have a question. Since both of the grid's and the 110KV line's impedences are behind the transformator, why did you only multiply the line impedence in square of "the transformation ratio" but not the grid impedence. I meant by transformation ratio U2/U1. Thank you.

    • @jfsimon1981
      @jfsimon1981 3 ปีที่แล้ว

      Hi,
      He calculated the grid impedence seen from 20 kV (20kV^2/4000MVA). Yes the units match in this case since kV^2=MVA.
      Jean-François

  • @arijitmahanta235
    @arijitmahanta235 ปีที่แล้ว +1

    I think there is calculation mistake during calculation of fault current at 20 kV side for fault at B.
    It is coming 345 amps not 605 ams

  • @mahmoud7142
    @mahmoud7142 4 ปีที่แล้ว

    Thankfully for highclass explain

  • @surendrakverma555
    @surendrakverma555 3 ปีที่แล้ว

    Very good

  • @atmsc4nd4l
    @atmsc4nd4l 3 ปีที่แล้ว

    Fault current at bus B must be 17.25cis(-90)kA.

  • @noobzgamer4358
    @noobzgamer4358 2 ปีที่แล้ว

    How if fault on bus 110kv before Transformator?

  • @roufriaz1662
    @roufriaz1662 4 ปีที่แล้ว

    Bts133 transistor replacement no

  • @aymane9489
    @aymane9489 3 ปีที่แล้ว

    You still didn't answer my question sir.