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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.
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    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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  • 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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