Step and Touch Potentials

[Dobber1978]

All,

I am trying to do some step and touch potential calculations following the methods outlined in IEEE std 80, 2000.

The values I am getting are not very reasonable, I think the main problem I am having is calculating my maximum grid current, and right now I am using the available fault current of 50000Amps.

Any suggestions???
Thanks
Jeff

 

[cuky2000]

There are two-step and touch potentials to be calculated:

1) Allowable values: Are mainly function of the native soil and surfacing resistivity, the average weight of the person (typ 50 kg) and the duration of the SC.

2) Actual values: Depend on the SC fault, ground grid configuration, resistivity, etc.

For safe conditions, the allowable values should be higher than the actual step and touch potentials.

_{NOTE: If you provide more information about the values of the parameter there is a possibility that some one will be able to help.}

 

[Dobber1978]

I don't have a problem with the allowable values, I found the formulas for those and they seem to make sence to me.

It's the actual values I am getting, i will try to provide a few of the system parameters that i am using.

Transformer 13.8kV to 600V, 1MVA,

Avalable Fault Current, 50000Amps, this was given to me and I believe that it's the upstream fault current with a little bit of safety built into it

My grid calculations are 0.7202 ohms, 6 x 7 meters, 500MCM, 1 meter spacing,

Estep70 = 2633V
Etouch70 = 824V, did these both for 70kg person

GPR works out to 36010V, using the 50kamps, much higher then the touch, went on to do my mesh voltage = 131012V, again way too high, and step voltage = 407770V again a little high.

 

[jghrist]

For a small station such as yours, most of the fault current will usually return through the 13.8 kV neutral instead of through the earth. You have to consider the split of the fault current to get reasonable results.

If there is no neutral, then the fault current would be restricted by the grid resistance. You can't get a GPR higher than the phase-to-ground voltage. If the grid were insulated from the earth (grid resistance infinite), then the voltage of the grid with respect to remote earth would equal the phase-to-ground voltage. Conversely, you cannot get 50000A fault current with a 0.7202 ohm grid resistance if the only return path is through the grid. The highest it could be, even with zero line impedance is 13800/sqrt(3)/0.7202 = 11063A.

There is also something wrong with your mesh and step voltages. They cannot be higher than the GPR. The mesh voltage is the voltage between the grid and the earth surface in the middle of the mesh. The earth surface voltage cannot be lower than zero (remote earth). Similarly, step voltage is the voltage difference between two points on the surface one meter apart. The highest earth surface voltage has to be lower than the grid voltage and the lowest has to be higher than zero.

 

[Cooda76]

Hi
I have studied IEEE std for this stuff but never happend 2 get chance 2 work but a curosity after jghrist reponse:
As you said:
If no neutral, GPR would not go beyond Ph-g voltage, the normal acceptable value of GPR according to Canadian Electrical Code is 5000 v that means if we consider the fault is on primary 13.8 Kv the fault current would go through neutral to grid and probably to downstream transformer or generating station who is feeding it but if u said there is no neutral than there is no neutral connection then GPR cannot be more than Ph-G vol so that means if fault occurs on 600 V then GPR will be less than 600/sqrt(3) and if Single line 2 ground fault occurs on 13.8 Kv then max GPR will be 13.8KV/Sqrt(3). Am I correct in assuming what I just wrote as per you arguments.if yes why?
Sorry for interfering in ur question Dobber.

 

[cuky2000]

The calculated step and touch potentials using the total fault current will be higher than the actual values resulting in some cases excessively conservative voltage level to meet in the grounding design.

As mentioned above, a more realistic approach is obtained using the injected current into the earth, which is a fraction of total current based in the number of incoming and outgoing feeders.

See if the enclose graph could assist to estimate the current split factor to determine the fraction of the current injected into the earth.

 

[jghrist]

Cooda76,

The GPR is the voltage of the grid during a Ø-grd fault with respect to remote earth. Remote earth is at the ground point of the source. The voltage of the grid during a fault is the same as the voltage of the phase wire that is faulted to it. The voltage of the phase wire is somewhere between the Ø-grd voltage at the source and zero (voltage of the ground point of the source). There is no way for the GPR to be greater than the Ø-grd voltage at the source.

 

[cflatters]

Dobber

Just a quick observation

If you have a 13.8/0.6kV transformer substation, then surely the highest theoretical GPR you can get is assuming a SLG fault on the 13.8 side, ie 13.8/1.732 = 7,967V, on the 600V side, the highest GPR is 600/1.732 = 346V, neither the step or touch potentials can exceed this.

The GPR cannot possibly exceed the driving point voltage at the point of the fault.

Therefore the value of 36,010V clearly has a problem.

Also, assuming a pessimistically low transformer impedance of 4.5% and infinite bars on the HV, the fault level would be 21.38 kA symmetrical, approximately the same for SLG, your 50kA quoted value is well over the top.

 

[Dobber1978]

cflatters,
I was aware that I was having a little problem and that is why I was here asking questions, I think the biggest problem I have is calculating the fault level, I will have to continue to look into this calculation when I get a little more time,
Thanks All

 

[fwipl]

One of your reply indicates yu are using,Transformer 13.8kV to 600V, 1MVA,
The general requirement is to check the touch & step potential if the equipment is in the open switch yard.
For low voltage system there is no need to check for touch & step pots.
You have to define the duration of 50kA of fault current.Accordingly the grid conductor size can be selected.
For MV systems you can play with many things to 1. Restrict the fault current with the NGR's 2. reduce the withstand duration of fault current by the ground grid conductor & suitably apply the protection system.
3.increasing the copper in the grid.