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Primary Fault Current Question

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  1. anderssok09 is offline Junior Member Pro Subscriber
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    Primary Fault Current Question

    Question:
    1000kva 34.5kv to 480v secondary transformer 5% imp. Apply 480v to primary side and ground secondary. What is primary fault current?

    Answer: 4.65A

    I don't know what I am doing wrong, I am using the full load amps and fault current equations and I am not arriving at this answer.

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  3. jrm5116 is offline Junior Member
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    Quote Originally Posted by anderssok09 View Post
    Question:
    1000kva 34.5kv to 480v secondary transformer 5% imp. Apply 480v to primary side and ground secondary. What is primary fault current?

    Answer: 4.65A

    I don't know what I am doing wrong, I am using the full load amps and fault current equations and I am not arriving at this answer.
    One way to do it is to convert the impedance percentage back to the actual impedance value to solve with the new voltage.

    So from the nameplate SBase = 1000kVA and VBase = 34.5kV, then ZBase = VBase2/SBase
    ZBase = 34.5kV2/1000kVA = 1190
    Since the impedance is 5%, the actual impedance seen by the high side is
    ZBase*5% or 1190*.05 = 59.5 Ohms

    Now we just have an impedance and a voltage we are putting across it to find the current V=IR.

    So VLN/R or I = (480/1.732)/59.5 Ohms = 4.65A

    Another way to do this would be to find the primary side fault current at given levels, and multiply by the ratio of voltage you are applying.

    IPrimary Fault @ 34.5 = 1000kVA/(34.5kV*1.732*.05) = 334.71A

    334.71A * (480/34500) = 4.65A

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  5. anderssok09 is offline Junior Member Pro Subscriber
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    Quote Originally Posted by jrm5116 View Post
    One way to do it is to convert the impedance percentage back to the actual impedance value to solve with the new voltage.

    So from the nameplate SBase = 1000kVA and VBase = 34.5kV, then ZBase = VBase2/SBase
    ZBase = 34.5kV2/1000kVA = 1190
    Since the impedance is 5%, the actual impedance seen by the high side is
    ZBase*5% or 1190*.05 = 59.5 Ohms

    Now we just have an impedance and a voltage we are putting across it to find the current V=IR.

    So VLN/R or I = (480/1.732)/59.5 Ohms = 4.65A

    Another way to do this would be to find the primary side fault current at given levels, and multiply by the ratio of voltage you are applying.

    IPrimary Fault @ 34.5 = 1000kVA/(34.5kV*1.732*.05) = 334.71A

    334.71A * (480/34500) = 4.65A
    Thanks!

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