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What is the theoretical maximum symmetrical fault current?

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    What is the theoretical maximum symmetrical fault current?

    What is the theoretical maximum symmetrical fault current that would be available at the secondary terminals of a 3-phase 12,470-480Y/277V, 1000kva, 6.25% impedance transformer?

    I am having problems solving this question. Please advise.

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  3. SecondGen's Avatar
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    Fault Current = Full Load Amps (FLA) / % Impedance (Z)
    FLA = VA / L-L Voltage x 1.732

    Here is a thread that explains the equations step by step -> http://testguy.net/threads/1341-Easy...-fault-current

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    Short-circuit

    Here the information:
    12,470-480Y/277V
    1000kVA, 6.25% impedance

    another way:

    For a impedance of 6.25%, your necessary voltage will be 6.25% of 480 Volts = 30 Volts
    What that means:
    With secondary shorted, if you apply 30 Volts, you will have full transformer current, which is:
    Click image for larger version. 

Name:	Current.jpg 
Views:	623 
Size:	16.9 KB 
ID:	35 (remeber to multiply by 1000 (KVA to VA conversion)
    I = 1202.85 A
    from here you calculate your transformer Impedance
    Z = V/I......Z = 30/1202.85.......Z = 0.025 Ohms.
    Your max short-circuit current:
    I = V/Z.......I = 480/0.025 = 19200 A
    Attached Thumbnails Attached Thumbnails Current.jpg  


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    Thanks for the help.

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  9. rofo42 is offline Junior Member
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    Quote Originally Posted by joaogemal View Post
    Here the information:
    12,470-480Y/277V
    1000kVA, 6.25% impedance

    another way:

    For a impedance of 6.25%, your necessary voltage will be 6.25% of 480 Volts = 30 Volts
    What that means:
    With secondary shorted, if you apply 30 Volts, you will have full transformer current, which is:
    Click image for larger version. 

Name:	Current.jpg 
Views:	623 
Size:	16.9 KB 
ID:	35 (remeber to multiply by 1000 (KVA to VA conversion)
    I = 1202.85 A
    from here you calculate your transformer Impedance
    Z = V/I......Z = 30/1202.85.......Z = 0.025 Ohms.
    Your max short-circuit current:
    I = V/Z.......I = 480/0.025 = 19200 A
    The link that Second Gen posted doesn't state anything about the additional steps you took dealing with the impedance.

    Can anyone confirm?

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  11. joelcallow is offline Junior Member Pro Subscriber
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    Motor winding Results

    Winding test results on a motor indicate possible degradation when the output waveforms show:
    Winding waveform symmetry
    Noticeable difference in winding waveforms
    Chopped waveforms on grounded windings
    Chopped waveforms on phase ungrounded windings

    Having trouble getting an answer for this.

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