Insulated Neutral Voltage Stree

[Arej85]

This question is to help my understand of an insulated neutral system and what happens during an earth fault.

I understand that the system is earthed through capacitance, and that voltage will charge and discharge through the fault. This will cause voltage stress in the functioning phases due to constantly changing voltages.

I was looking at this website - An Overview Of Grounding System (Ungrounded) | EEP

The equation they derive is : If = IC2 + IC3 = IC = j3CwV1

It is stated that due to this equation the healthy phases undergo severe voltage stress.

I think i am correct in thinking that voltage cant rise above the line to line voltage of a system. So i was wondering why the equation would cause voltage stress?
Would it cause stress due to an increase of power due to the current increase?

Any help is appreciated as my knowledge is limited.

 

[Darkwood]

Welcome to the forum Arej85, what area are you training in, this is not part of the standard course work format for an Electrician.

 

[Arej85]

Hi Darkwood,

I am training to be an ETO, essentially a ships electrician.

 

[marconi]

This question is to help my understand of an insulated neutral system and what happens during an earth fault.

I understand that the system is earthed through capacitance, and that voltage will charge and discharge through the fault. This will cause voltage stress in the functioning phases due to constantly changing voltages.

I was looking at this website - An Overview Of Grounding System (Ungrounded) | EEP

The equation they derive is : If = IC2 + IC3 = IC = j3CwV1

It is stated that due to this equation the healthy phases undergo severe voltage stress.T

I think i am correct in thinking that voltage cant rise above the line to line voltage of a system. So i was wondering why the equation would cause voltage stress?
Would it cause stress due to an increase of power due to the current increase?

Any help is appreciated as my knowledge is limited.

The equation you refer to does not explain why the two remaining healthy phases undergo voltage stress. What causes the voltage stress (ie increased electric field strength surrounding each of these two phases) is in two parts.

The first is the rise in potential of each of these conductors with respect to earth caused by connecting the faulty phase to earth through a the low impedance conduction path. You can see from Fig 6 how the potential differences of the two healthy phases to earth change from V2 and V3 when faultfree to v2 and v3 when Phase 1 has an earth fault; simply the voltage changes from the phase voltage to the line voltage. Thus the electric field strength in any insulation surrounding the healthy phases, and thus the electric stress in the dielectric is increased by the square root of three. This is the case for a constant impedance phase-earth fault. Cables are rated by two voltage figures eg 600V/1000V referring to inter conductor and conductor earth potential differences.

The second is if the phase-earth fault is an arcing type fault, - ie not constant but regularly striking and extinguishing as the emf in the fault loop undulates, - will mean that the earth fault loop current is a chopped or interrupted current waveform, rich in harmonics. The earth fault loop circuit has resistance, loop inductive and capacitive reactances. It is possible for series resonance to be excited and thus large currents to flow limited only by the conductor and fault resistances. Even larger emfs than the phase voltage can thus appear across the winding of phase 1 which will further raise the potential of the healthy phases V2 and V3 with respect to earth with consequent risk of overstressing insulation to the point of failure at weak points.

 

[Arej85]

Thanks marconi,

This has helped me to understand what is going on and has provided me with further topics of reading!

 

[marconi]

Thanks marconi,

This has helped me to understand what is going on and has provided me with further topics of reading!

Good. Some suggestions for further reading:

a. Virtual neutral in polyphase systems.
b. Neutral point displacement voltage.
c. Impedance grounded polyphase systems.
d. Neutral grounding transformer.
e. Millman's theorem to study star connected impedances with common point earthed.
f. How to detect earth faults in isolated polyphase systems.
g. I think you will be intrigued by the Ferranti effect in long cables.
h. Never a bad thing to refresh one's understanding of LCR circuits and how they respond in steady state and also to impulse and step changes in voltage/current.

 

[Arej85]

Thank for the suggested topics marconi!

Can ask a question relating to what happens to healthy phases...

During an earth fault, the loads between phases will still see the correct voltages, but if a motor was connected to one of the healthy phases, would its phase winding's experience this increased 1.73 voltage?

Also what happens to motors that are connected to the phase with the fault?

thanks

 

[marconi]

Thank for the suggested topics marconi!

Can ask a question relating to what happens to healthy phases...

During an earth fault, the loads between phases will still see the correct voltages, but if a motor was connected to one of the healthy phases, would its phase winding's experience this increased 1.73 voltage?

Also what happens to motors that are connected to the phase with the fault?

thanks

Dear Arej85,

Regret delay in replying - I have been resettling my wife at home after a long stay in hospital.

You are asking some good questions but I would like to see you attempt to answer them also. That way you will develop a deeper understanding about applying theory to practice.

Taking you last questions, here is how to proceed. First state what 3 phase system you are studying: is it a 3 wire or 4 wire system for example? In what follows I am assuming a 3 wire balanced supply system. What type of motor and how many phases? I am assuming a single phase motor.

Draw the 3 phasors representing the 3 line emfs. Now consider how you can connect a single phase motors windings to these 3 lines. You will soon see two of the three ways are equivalent (but not the same) for the earth fault situation.

Now consider the potential difference(pd) between the ends of the windings to study the potential difference between turns of the windings.

Then consider the pd between turns of the windings and the former they are wound on. What is the nature of this impedance path - resistive, inductive or capacitative or a combination? - between the windings and their former, the former being earthed (-to what?). Is it localised or distributed along the windings? Is it possible to consider this impedance path by an equivalent circuit which localises and lumps the distributed impedance? (Answer is yes - but what form does it take? Clue Pi or T?) How do the nodes of this equivalent circuit connect to the 3 phase 3 wire network? How do you connect in the earth for fault free and earthed line? Now draw some phasors to represent the emfs and pds similar to Fig 6 in the reference in #1. Now consider the pd between earth and any point along the conductor of the windings. How do you draw a vector to represent this pd? Do this for fault free and then for an earthed phase situations. Draw a graph of this pd as a function of point along the winding from one end to the other.

A bit rushed but have a stab and get back to me when you have with your thoughts.

 

[marconi]

Amend Figure 6 to Figure 2 in what I wrote earlier.

 

[Arej85]

Hi marconi,

Thanks for the information and glad things are back on track!

I will look over what you wrote in more detail as soon as i can. I have a few reports and a project on going at the moment.

Looking back at my question it was very vague. I was referring to a balanced 3 wire insulated neutral system, with a 3 phase induction motor connected.

I think i was getting mixed up with what happens when an insulated neutral system goes to earth. Say the generator was star configured. I made a star(generator) to star drawing with the second star representing the capacitive earthing of the system.

Say the system is 400V line to line under normal conditions (no earth fault) I was unable to get my head around why the line conductors carrying 400V were not stressing their insulation at 400V, but if my thinking is right the distributed capacitance creates a neutral point (like a star point) through the system therefore only 400 / 1.73 = 231V is 'seen' by the insulation.

I was thinking that because the two healthy phases under an earth fault increase by 1.73 x phase voltage that this would increase the stress on equipment insulation.

So i guess that in an insulated neutral system the insulation is rated at full line to line voltage, but operates at line voltage / 1.73 due to the distributed capacitance ?