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Tripping Curves and M.C.B.s.

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A Schneider catalogue that I have has a technical section on tripping curves. It discusses a C20 M.C.B. It states that it will interrupt a current of 100 Amps (20 times the rated current) in:


0.45 seconds at least


6 seconds at most.


For tripping currents exceeding 20 times the rated current the time-current curves do not give a sufficiently precise representation. The breaking of high short circuit currents is characterized by the current limiting curves, in peak current and energy.


The total breaking time can be estimated at 5 X the value of the ratio (I2t)/(I)2. Is that I squared t divided by I squared? Just what is that?


Z.
  • 0
    Link


    If you look carefully, the second I is wearing a little hat, thus: Î. It is not defined in the text nor in section 2 of BS 7671.
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    In this case, (I2t) is the let-through energy of the device quoted by the manufacturer, or obtained from BS EN 60898 or BS EN 61009, and I2 in the denominator (divisor) is the square of the prospective fault current.


    Example


    For a B32, the BS EN 60898 maximum let-through energy (I2t) is 45,000 A2s. The lowest prospective fault current in the circuit is (measured or calculated to be) 1000 A.


    Therefore, I2=1,000,000, and t ≈ 45,000/1,000,000, so t ≈ 0.045 s.


    Usually, the manufacturer's quoted let-through energy (I2t) is much lower than BS EN 60898, and so the approximate tripping time will be less than the value calculated from the BS EN 60898 data.
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    Chris Pearson:
    Link


    If you look carefully, the second I is wearing a little hat, thus: Î. It is not defined in the text nor in section 2 of BS 7671.


    Well yes Chris, you are right, I had to use a magnifying glass to see that. I thought that it was just ( and ). So, what does the hat mean?


    Z.


  • 0
    gkenyon:

    In this case, (I2t) is the let-through energy of the device quoted by the manufacturer, or obtained from BS EN 60898 or BS EN 61009, and I2 in the denominator (divisor) is the square of the prospective fault current.


    Example


    For a B32, the BS EN 60898 maximum let-through energy (I2t) is 45,000 A2s. The lowest prospective fault current in the circuit is (measured or calculated to be) 1000 A.


    Therefore, I2=1,000,000, and t ≈ 45,000/1,000,000, so t ≈ 0.045 s.


    Usually, the manufacturer's quoted let-through energy (I2t) is much lower than BS EN 60898, and so the approximate tripping time will be less than the value calculated from the BS EN 60898 data.


    Thanks G.K. I am trying to get my head around energy let through. I am partially there. The numbers are difficult to visualize as they are either so BIG or so small. Or the figures have no units.


    Z.


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    I like the tripping curves on page 11/5 of the Schneider catalogue. The graphs show time (S) as the vertical line and I/In and the horizontal line. Curves for types B, C and D devices are shown with coloured areas for the operation of their 3 to 5, 5 to 10 and 10 to14 times In ranges. Applicable  AREAS are shown which helps clarity. I was wondering what the horizontal "I/In" meant. "In" would make sense. I presume that it means I or In, rather than I divided by In.


    Z.
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    I was wondering what the horizontal "I/In" meant. "In" would make sense. I presume that it means I or In, rather than I divided by In.

    Without looking at the context I suspect they mean the fault current divided by the rating of the device - i.e. how many "times" the rating we're talking about. Often that approach is useful for protective devices as many currents are defined as multiples of the nominal rating - 1.13x or 1.45x or 5x or 10x. Often that means you can produce one graph or table for the whole range of devices rather than having to produce one for each rating.

       - Andy.

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