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Floating Earth Faults

  1. Mar 2, 2015 #1
    If an entire system was floating wrt earth via infinite impedance and a person (grounded) touches a conductor am I right in saying nothing would happen due to the fact there is no return path through earth?

    Presumably the only way to detect an earth fault in this case is to measure the impedance of the supply wrt to earth and alarm if it goes below a certain value?

    In this case why aren't all distributions completely insulated from earth? Normally the negative is earthed and earth faults are cleared either by the supply fuse/breaker or by residual current detection.
     

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  2. jcsd
  3. Mar 2, 2015 #2
    No, you're not. There's capacity , and depending on voltage of the system and size of it, more or less current will flow at the point of fault.
     
  4. Mar 2, 2015 #3
    Thanks could you elaborate on capacity?
     
  5. Mar 2, 2015 #4
    Every conductor has electrical capacity (per definition C=ΔQ/ΔV). Line-to-earth capacities of other two healty phases are fed with current flowing throu the earthed point of a third phase.
     
  6. Mar 2, 2015 #5
    Ok so once it reaches steady state though zero current will flow?
     
  7. Mar 2, 2015 #6
    zero sequence current
     
  8. Mar 2, 2015 #7
    Ok thanks for the help but I think you are massively complicating what was actually a very simple question on electrical fundamentals.
     
  9. Mar 2, 2015 #8

    jim hardy

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    @fonz
    you already know the answer.

    Your automobile electrical system is connected to earth through only the tires
    they're not infinite impedance but pretty high.
    In wintertime you get a shock when exiting the car as its "capacity" discharges through your fingers and foot.
    That flow of current is a brief one.

    old jim
     
  10. Mar 2, 2015 #9

    jim hardy

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    ps there are good reasons to earth a system.

    The IEEE "Green Book", standard 142, is a down-to-earth (pardon the expression) of both the practical and theoretical sides of the subject.

    Try a search....
     
  11. Mar 3, 2015 #10
    If you leave the neutral point of the system floating your line to earth voltage will vary depending on phase to earth capacitance. Generally speaking the XC to earth will be far to large to create a solid voltage reference to earth. In which case yes, if you were touching one phase and earth at the same time you would not receive a shock. Though as above the capacity of the system also plays a large role in this.

    In systems operating with an unearthed neutral insulation monitoring devices are usually used to detect an earth fault.

    Unearthed systems are far too unstable to be used for public distribution. If you don't create a solid reference point for your system there's no way you can keep line to earth voltages within acceptable limits. Have think about what would happen if there was inadvertent contact between an MV/HV and and LV conductor. Or, a short circuit between primary and secondary in the transformer. Another one is arcing or interment LV earth faults that because of system capacitance, can build up large line to earth overvoltages which will very quickly overstress our conductor insulation.

    Hope this helps.
     
    Last edited: Mar 3, 2015
  12. Mar 3, 2015 #11
    That is a massive help thank you.
     
  13. Mar 3, 2015 #12

    dlgoff

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    How about "wrt the surrounding air"?

     
  14. Mar 3, 2015 #13

    jim hardy

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    I dont know which is more impressive

    that demonstration of
    "capacity"

    or the skill of that pilot....
     
  15. Mar 3, 2015 #14
    I agree. It is a simple question on electrical fundamentals. Giving formula for the fault current replaces thousand words of description:

    I = 2⋅√3⋅π⋅f⋅V⋅Ce

    where:
    Ce...phase to earth capacity (under assumption CA=CB=Cc=Ce)
    f... nominal frequency of 3 phase system
    V...nominal (line) voltage of 3 phase system

    More importantly, phase to earth voltages of other two phases rise by aprox. 73% in this case. In steady state their magnitude is:
    V'= √3⋅Vp=V​
     
  16. Mar 3, 2015 #15
    The skill of the pilot most definitely
     
  17. Mar 5, 2015 #16
    Another question; If I had a battery or any ideal voltage source that is open circuit and insulated from earth and I measure the voltage on each terminal wrt earth what would it be?

    For example if I had a 17V voltage source and measure the voltage to earth I would expect to get 17V on the positive terminal and -17V on the negative terminal is this correct?

    If I then connected a load to the two source terminals and measured the voltage again would it be the same?

    Thanks again
     
  18. Mar 5, 2015 #17
    -17/2 V and +17/2 V you're expected to measure IF you have two ideal voltmeters and the voltage source is ideally balanced.
     
  19. Mar 5, 2015 #18
    Thanks zoki i'm not sure I fully understand why though? Is it the same in both cases i.e. open circuit and with load?
     
  20. Mar 5, 2015 #19
    Becouse 17 V could mean -1 V and +16 V wrt earthing as well.
    Load (except short circuit) doesn't matter if you have ideal voltage source.
     
  21. Mar 5, 2015 #20
    Ok I don't understand that at all could you please elaborate?
     
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