Current Division Factor

[lume7006]

Dear friends,

We are involved in another grounding grid analysis for an industrial system.
Now, we are trying to calculate the current division factor, which for a transmission network it seems very easy to determine following some equations.
However, for an industrial network in medium voltage, how do you calculate it?
what factors do you use to do it?

Any help or suggestion will be welcome

Best Regards,

 

[rcw retired EE]

"Industrial network in medium voltage" ?

What is the voltage, type of grounding, short circuit level an dtype of distribution - i.e. insulated shielded cables, overhead lines, bus duct? What type of locations you are concerned about for step and touch voltages? Are they outdoor, open air subs in a gravel surfaced switchyard, or industrial metal clad switchgear in electrical rooms?

The division factor depends on the type of distribution and system grounding method (high impedance, solidly grounded, low impednace, resonant, etc.). You would have to calculate the current split between the earth, cable shields and any sysem grounding wires run with the cables.

But my other questions are trying to find out what sort of ssytem installation you have. For most industrial systems, step & touch is not a concern due to the lower voltages and the installation location.

 

[jghrist]

Calculating the current division accurately in a transmission system is not easy either. It depends on the impedance and mutual coupling among phase conductors, ground wires, earth return, and on the resistance of tower grounds and remote substation grounds. We use SES SPLITS or FCDIST software to calculate it.

http://www.sestech.com/

 

[7anoter4]

I think a fair appreciation could be done following IEEE Std 80/2000 ch.15.9 "Computation of current division". But as rcwillson said
in an industrial zone touch and step potential it is not a concern. DIN/VDE 0141 recommends some "reducing factors"
for different elements connecting the area with remote earth as cable shields ,static wires, pipes and rails, but for an industrial zone
it is not necessary to calculate at Ground Potential Rise at all. EPRI EL 904-1 contains some indications in ch. 5 in a subchapter
GROUND CURRENT COUPLING

 

[lume7006]

Hello colleagues,

I know that related to step and touch voltages there is no problem, because the fault current flows thorugh the "four wire" according to NEC and as it does not flow to earth there are no step and touch voltages.

However, this customer wants us to determine the size of the conductor for the whole grid mesh based on the maximum fault current magnitude which is in low voltage (52 kA), then, he is requesting us to determine:
Current Division Factor
Decrement Factor

We have determine Current Division Factor for substations at transmission or distribution level, but not for industry ate low-medium voltage.

Another aspect is related to the point 11.3.3 that IEEE Std allows to size the down leads conductors with a larger size than the ones buried because once the current is on the grounding grid it is diverted through several paths. Have you ever used this criteria to size the grid conductors?

Best Regards

 

[jghrist]

If all of the fault return current flows through the ground wire, then the current division factor is 0% (none flowing through earth). In reality, if there is a remote MV source, some of the return current for a MV ground fault will flow through the earth. It doesn't make any difference to the current whether the system is "industrial" or "utility".

 

[7anoter4]

I agree with jghrist, of course.
But if we return to the last sentence of Lume7006 post from 7:32 in IEEE 80/2000 ch. 11.3.3 " Additional conductor sizing factors" states:
"The down leads from the equipment to the grid may be subjected to the total fault current into the grid,
while the grid divides this current so that each conductor segment in the grid is only subjected to some
fraction of the total fault current. Thus, the down leads may have to be larger than the grid conductors or
may have to be multiples from the equipment to the grid to have sufficient ampacity for the total fault
current."
That means you may reduce the local area current in some way.
In a power station following the IEEE Std 665/1995 Annex C[informative]:
"Division of current for small interior grids"
in order to appreciate an interior grid current the following criteria was checked:
1)The current is divided proportional with the area ratio
2)The current is divided proportional with conductor length.
3) Supposing that the GPR for this area will be the same as this calculated for entire station
Using a simplified example the conclusion was:
the 3-rd criterion is better. The small area current was calculated by dividing GPR by local area ground resistance and then
verifying the touch and step voltage compared with entire station area results.
I never use any criterion in order to reduce the local current and so to reduce the copper cross-section but I considered for the main
area grounding conductor connecting the grounded parts of transformer , switchgear and so on, the entire short-circuit current.
As explained in IEEE-80/2000 the mechanical strength is the minimum limit.

 

[lume7006]

I thank your ideas, are always helpful

7another4, I would like to give you more data in order to understand our situation. This specific project is a bit complicated and I will explain you why:

1) The largest fault current magnitude is on a low voltage transformer (at 460 V)
2) We are using this magnitude to size the conductor for the whole ground grid, however, according to our customer "idea" this current may flow only from the earthing point of the switchgear to the neutral of the transformer to "close the circuit".
As you can see, there are two grounding points, one at the switchgear and the seond at the transformer neutral (X0 terminal).

3) Then, this current is not causing the step and touch voltages because it is not flowing through earth.
The current magnitude that will flow through the grounding grid causing the step and touch voltages is at 4.16 kV and it is limited to 800 A by a resistor installed at the neutral of this transformer.

4) As you can see, the step and voltage potentials are not a problem, our only problem is the size of the conductor which may not tolerate a large current for a while without exceeding the 250 celsius degrees and all the side effects on connectors and so on.

5) Because of that, we have thought that determining the current distribution on the grounding grid could be an alternative to prove that the whole grounding grid will not be exposed to the large fault current, in fact using a commercial software (previously mentioned by one of the colleagues), we found that by injecting 53 kA into the down lead, this current is divided and the surrounding segments of the grounding grid to the injection point are taking 27 kA as a maximum, this magnitude will not shown a temperature over 250 C and of course connector neither.
But, we are not sure if this can be done despite being mentioned on IEEE Std 80-2000

I thank your feedback

 

[7anoter4]

In a Phase-to-Ground short-circuit the reactance between the cable supplying the switchgear and the Ground return path will be such high
that the short-circuit current will be limited but the touch potential at the Switchgear will be elevated.
A good practice is to run a grounding cable of a cross-section not more than live supply conductors and close to these conductors.
The short-circuit current has to be the same as for live conductors.
As jghrist explained, only 4.16 KV remote system may supply a short-circuit current through the Ground and to produce a GPR.

 

[jghrist]

At 480 V, you would normally have a ground conductor carried with the phase conductors, and/or a metallic raceway, that would carry most of the ground return current. This would make your calculated 53 kA for the downlead and 27 kA for grid conductors conservative. In any case, either by calculation or by use of IEEE-80, using 27 kA for sizing the grid conductors is proper.