Ra and EFLI relationship question.

[La Poste]

Greetings all.
I would like to ask a question about how Ra links to EFLI, what is the connection?

When you measure Ra properly with a decent earth resistance tester is the reading you get equivalent to R2?

If you had an EFLI tester and you got a value of 100 Ohms would that mean that if you measured the value of R1 and it came out at 1 ohm would that make your Ra value 99 ohms?

Does the Ra value measured using an earth resistance tester substitute the R2 value in your calculations for EFLI?

I am just trying to understand the connection between Ra and R2, Ra being the soil resistance and R2 being the resistance of CPC back to the star point of the transformer.

I'm really trying to get my head round earth rods and Ra values.

Thanks.

 

[SirKit Breaker]

I wont profess to be the expert on earth rod testing, but i struggle to understand fully your question. I would have a good look at GN3, this gives a very clear and concise instructions about testing earth rods. This will probably also answer your questions.

Cheers........Howard

 

[La Poste]

Apologies for not using the right terminology, I'm getting rusty.

I mean does Ze equal Ra so that when you are trying to find the earth fault current you use Ra instead of Ze.

For instance if you wanted to find the earth fault current at the main intake at the fuse board would you use the formula:

230Volts/(impedance of line conductor from transformer + Ra)

and would this give you your earth fault current?

Basically is Ra used instead of Ze in calculations of earth fault current.

 

[La Poste]

So let's say I turn up somewhere with my earth resistance tester and I have been given the figures for the impedance of the line conductor from the transformer to the point at which I am measuring.

I do my earth rod tests using the spikes in the ground method and I come up with a figure of 50 ohms.

The impedance of the line conductor is 0.5 ohms from the paperwork I have been given and I want to work out the earth fault current at this particular point.

I measure the voltage and it comes out at 240 volts.

Do I simply use the formula.

I = 240/(50+0.5) = 4.75 amps.

 

[La Poste]

Actually thinking about it a little more there is also an earth spike at the transformer and the resistance of the transformer windings which I will forget in this instance. So let's say the transformer rod is 20 ohms, our own rod is 50 ohms and the line conductor is 0.5 ohms, voltage 240 volts. Would the earth fault current be.

I = 240/(50+20+0.5) = 240/70.5 = 3.4 amps.