L-G fault in 3-ph system 1

[mab2008]

In a 3-ph system with neutral not solidly grounded, why the phase voltage of unfaulted lines gets increased 1.732 times its nominal voltage in the event of a L-G fault.
Mathematically, it is clear but what actually is happening in reality?

 

[AMBMI]

It's the "worst case". In "real world" you have always a fault resistance so you get something less than 1.732 Un.

 

[mab2008]

My actual question is the "reason" that leads the rise in voltage of the unfaulted lines.

 

[burnt2x]

Consider your voltage triangle with one side clipped to ground potential. If you are going to measure the voltage from ground to any phase (unfaulted), you will see the full line-line voltage. The line-line voltage equals 1.732 X phase voltage. That's it.

 

[tem1234]

Between phase and ground, there is VLL/1.73 in normal condition and VLL between 2 phase. If one phase is grounded, there is 0 V between this phase and the ground, and still VLL between this phase and the other 2, so the voltage between these 2 phase and the ground is VLL.

You can look at:

http://www.i-gard.com/appguides.htm

lots of good info, in the beginning of the application guide named "ground fault protection on ungrounded and HRG system", they explain that

 

[Marmite]

Any basic electrical text book will answer the question. The reasons are Ohms law and vector algebra. If you pass a current through an impedance you get a voltage drop across it. Imagine the normal 3 pointed star of three equal voltages displaced by 120 degrees, with tips labelled A,B & C. The neutral point is in the middle. If you get a fault to ground involving one of the phases, lets say A, the neutral point moves graphically nearer to the faulted phase, dependant on the relative impedances of the fault and the return path back to the system neutral earth point. For a solid bolted fault with no impedance the neutral point moves all the way to A, as they are now at the same voltage, being solidly bolted. Imagine picking up the neutral point and moving it nearer to A. What happens to the lengths of B and C relative to the new neutral point position? They get longer, upto a maximum of root three times the normal phase to earth voltage.
Regards
Marmite

 

[waross]

A very basic observation:

In a 3-ph system with neutral not solidly grounded, why the phase voltage of unfaulted lines gets increased 1.732 times its nominal voltage in the event of a L-G fault.

The phase voltage DOES NOT get increased.
The voltage to ground from the two unfaulted phases increases.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[Zogzog]

Look at this slide

 

[Runsor]

AMBMI said that with a non-zero fault resistance in an isolated network you get something less than 1.732 Un = sqrt(3) Un for the healthy phase-to-earth voltages. Yes, that is mostly true, but the relation of Uo and fault resistance is actually quite complex.

In fact with certain earth fault resistance one of the two healthy phase-to-earth voltages will rise up to sqrt(3) + 5.2 % => 1.82 Un.
This maximum happens, when the fault resistance is about 37 % of the total ground reactance Xcap of the network.

 

[slavag]

Hello.
"In fact with certain earth fault resistance one of the two healthy phase-to-earth voltages will rise up to sqrt(3) + 5.2 % => 1.82 Un.
This maximum happens, when the fault resistance is about 37 % of the total ground reactance Xcap of the network".
It's really new for me. Ransor, I assume is same for Petersen coil system. Tanks a lot for information, it's probably explained me several cases.
Best Regards.
Slava