Floating Systems: Earth Faults & Detection

[fonz]

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.

 

[zoki85]

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?

< SNIP >

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.

 

[fonz]

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.

Thanks could you elaborate on capacity?

 

[zoki85]

Thanks could you elaborate on capacity?

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.

 

[fonz]

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.

Ok so once it reaches steady state though zero current will flow?

 

[zoki85]

Ok so once it reaches steady state though zero current will flow?

zero sequence current

 

[fonz]

zero sequence current

Ok thanks for the help but I think you are massively complicating what was actually a very simple question on electrical fundamentals.

 

[jim hardy]

@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

 

[jim hardy]

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...

 

[Aussielec]

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?

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.

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

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.

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.

 

[fonz]

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.

That is a massive help thank you.

 

[dlgoff]

If an entire system was floating wrt Earth ...

How about "wrt the surrounding air"?

 

[jim hardy]

I don't know which is more impressive

that demonstration of
"capacity"

or the skill of that pilot...

 

[zoki85]

Ok thanks for the help but I think you are massively complicating what was actually a very simple question on electrical fundamentals.

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⋅C_e​

where:
C_e...phase to Earth capacity (under assumption C_A=C_B=C_c=C_e)
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⋅V_p=V​

 

[zoki85]

I don't know which is more impressive

that demonstration of
"capacity"

or the skill of that pilot...

The skill of the pilot most definitely

 

[fonz]

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

 

[zoki85]

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?

-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.

 

[fonz]

-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.

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?

 

[zoki85]

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?

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.

 

[fonz]

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.

Ok I don't understand that at all could you please elaborate?

 

[sophiecentaur]

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

Absolutely NOT. The total potential difference is 17V (by any normal definition of the term '17V source'.
If you're having a problem with the +16 and -1V thing, imagine you have a 5m ladder. Its 5m from top to bottom, wherever the ladder is placed upright - up a mountain, on the ground., If its base is 1m below ground (-1m) then its top will be at +4m . The difference is still 5m and falling down to the bottom will feel the same wherever it is. How would you make a17V supply have -1V and +16V on its terminals (with respect to Earth, of course)? Just connect the negative terminal to a -1V source

It's worth noting that "Volts" usually means a voltage with respect to Earth. PD always refers to the 'voltage' between two terminals and is the more correct term - when you need to be picky.

 

[jim hardy]

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?

In theory world, the rigorously correct answer is " indeterminate " ; unknown, undefined..

Voltage is potential difference.
You have no idea what is the absolute potential of either Earth or of your floating voltage device.
So the difference between them is unknown.

In practical world
the act of connecting a voltmeter between them would bring them to same potential when charge flows from one to the other through the meter;
so by the time the meter settled and you got your reading, it'd indicate zero.

You've felt that phenomenon in wintertime - remember walking across carpet and getting zapped when you touch a doorknob?

It is important to remember the definition of electric potential: the work done on a unit charge in bringing it from infinity to wherever you are.
Nobody is going out to infinity to get that measurement for us, so we arbitrarily choose a more convenient point for our reference.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elepe.html#c3

 

[fonz]

In theory world, the rigorously correct answer is " indeterminate " ; unknown, undefined..

Voltage is potential difference.
You have no idea what is the absolute potential of either Earth or of your floating voltage device.
So the difference between them is unknown.

In practical world
the act of connecting a voltmeter between them would bring them to same potential when charge flows from one to the other through the meter;
so by the time the meter settled and you got your reading, it'd indicate zero.

You've felt that phenomenon in wintertime - remember walking across carpet and getting zapped when you touch a doorknob?

It is important to remember the definition of electric potential: the work done on a unit charge in bringing it from infinity to wherever you are.
Nobody is going out to infinity to get that measurement for us, so we arbitrarily choose a more convenient point for our reference.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elepe.html#c3

Thanks Jim this kind of makes sense. Although in your second statement you say that a voltmeter would measure zero as it would bring them to the same potential. Somebody else made the statement that measuring volts to Earth would be V/2 so I'm not sure how these two statements are related?

 

[zoki85]

Thanks Jim this kind of makes sense. Although in your second statement you say that a voltmeter would measure zero as it would bring them to the same potential. Somebody else made the statement that measuring volts to Earth would be V/2 so I'm not sure how these two statements are related?

"Somebody else" would be me? To refresh your memory:

-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.

If you connect just ONE classical voltmeter between one voltage source terminal and ground the voltmeter will quickly show reading 0 in most of the cases.
If you connect TWO voltmeters, first one between one terminal and ground, second one between other terminal and ground, and the voltage source is ideally balanced wrt ground, you'll have ±V/2 readings.

 

[jim hardy]

There do exist devices called "Electrostatic Voltmeters" that measure the electric field surrounding a charged object.
One of those could be used to measure the unknown potential difference posed in fonz's original post...

[QUOTE =www.google.com/patents/US4147981 ]
Description
BACKGROUND OF THE INVENTION
This invention relates to the electrical measurement art, and more particularly to a new and improved sensor for electrostatic voltmeters.

An instrument which measures electrostatic fields and electrostatic potentials on surfaces without current flow through the instrument is known as an electrostatic voltmeter which makes measurements in a non-contacting manner. These instruments comprise a probe or sensor assembly and an associated voltmeter wherein the probe converts the electrostatic field or surface potential to an a.c. voltage, the magnitude of which is proportioned to the field or potential being measured. This conversion is accomplished by a capacitance modulation process wherein the capacitive coupling between a surface associated with the probe and a surface associated with the field or potential to be measured is modulated or varied at a fixed periodic rate. A voltage difference existing between the two surfaces will induce an a.c. voltage on the probe surface.
[/QUOTE]

 

[dlgoff]

There do exist devices called "Electrostatic Voltmeters" that measure the electric field surrounding a charged object.
One of those could be used to measure the unknown potential difference posed in fonz's original post...

How neat and simple. I had no idea this existed.

 

[fonz]

"Somebody else" would be me? To refresh your memory:

If you connect just ONE classical voltmeter between one voltage source terminal and ground the voltmeter will quickly show reading 0 in most of the cases.
If you connect TWO voltmeters, first one between one terminal and ground, second one between other terminal and ground, and the voltage source is ideally balanced wrt ground, you'll have ±V/2 readings.

Thanks again zoki. I'm struggling with this though could you briefly explain why this is the case? If not at least refer to some text that might explain.

 

[zoki85]

Thanks again zoki. I'm struggling with this though could you briefly explain why this is the case? If not at least refer to some text that might explain.

I think your questions are answered and exemplified fine. As for the references, a typical textebook that teaches fundamentals of electrical engineering would suffice.

 

[jim hardy]

I'm struggling with this though

A traditional voltmeter has resistance.
Two voltmeters in series make a resistive voltage divider.
If they're of equal resistance, half the voltage will appear across each.
It's really that simple.
If the junction of those two meters happens to be grounded; well, that's what zoki is telling you.

Most ground detectors work on that principle.

Real world example: (Oh No, here comes another boring anecdote !)

In the power plant we had 125 volt battery power system for stuff that's really important like the emergency lube oil pump for Mr Turbine.
That battery itself was un-earthed so that a ground fault wouldn't trip something important.

To tell us when an accidental ground appeared we connected two 120 volt (25 watt) light bulbs in series across the battery, so each glowed a kinda dim orange. The junction of those lights we earthed.
Draw that on a piece of paper for yourself. It's just a voltage divider with middle grounded.
Now, should a ground fault come along on either the + or - side of battery , it'll put low resistance in parallel with one of the light bulbs.
Add that to your sketch.
So, one light goes out and the other goes to full brightness.
Those "DC Ground Lamps" were located prominently on the board so the operators knew right away we had a DC ground and could dispatch electricians to find it.
In the nuke plant we put similar lamps on our 120 VAC instrument power system which was also floating(and quadruple redundant).
Clamp-on ammeters were the basic instrument for chasing ground faults. Since lamp current returns through the faulted wire a sensitive meter around both wires of a circuit with a fault on it will report that lamp current.

I wanted to invent GFCI-type circuit breakers for such applications that'd illuminate a LED instead of tripping. I think somebody has done that but they aren't widely used. They'd be a godsend for maintenance men.

old jim

 

[Averagesupernova]

I always like your boring anecdotes Jim.

 

[dlgoff]

I always like your boring anecdotes Jim.

Ditto

 

[Windadct]

You can make the same with 3 Ph AC -- resistors and light bulbs. Still used today in older factories with 3 Ph ungrounded or high impedance grounded systems.