Meeting Touch Potential Limits on Long Cable Runs

[Kayaker11]

I have a medium-voltage cable run of almost 50,000 ft and because it is in an underground mine, the ground potential rise ("ground fault voltage") due to a ground fault at the remote end must be limited to 100V. The system is high-resistance grounded but with those kind of cable lengths, the impedance of the ground return path results in a touch potential far exceeding regulations. These kinds of cable lengths are typical for soft rock mining applications and feeding the underground with a HRG circuit is the norm. Has anyone ever had to do a touch potential test with these kinds of cable lengths? If so, what sort of touch voltage did you measure and where was your reference electrode.

Obviously reducing my NGR let-thru current isn't an option with this amount of cable, the cable charging currents are huge as well.

Any sort of anecdotes, especially involving testing and a mines inspector would be greatly appreciated!

 

[jghrist]

Are you including the EGC and/or cable shield in parallel with the ground return path?

 

[Kayaker11]

Yes. When you have distances like I'm talking, even intentional ground return paths, like those in cable grounds, shields, armour etc, don't have nearly low enough impedance to bring the touch voltage down below 100V. The inductance alone would bring Z up over the limit, let alone the resistance.

 

[7anoter4]

In my opinion, you could reduce the touch voltage if you would reduce the mine side Ground Grid resistance.
The HRG and the medium voltage cable impedance will reduce the homopolar component of the
short-circuit phase-to-ground current.
If the GPR [=I"K*Rgrid] is less than 100V then the touch voltage will be lesser.
Also you can test the touch voltage injecting a low-voltage current through the medium voltage cable connecting the other end directly to the Ground Grid. You can measure the low-voltage current and then the touch voltage between the different grounded parts and an auxiliary vertical electrode. Then multiplying the measured voltage by ratio of calculated I"k to low-voltage current you'll get the actual touch voltage. As an anecdote I did myself something like this using a cast iron water pipe of 30000 ft long well insulated and I get 10 A at the grid connection and I measured about 20 mV on the farthest corner.

 

[Kayaker11]

Thanks 7anoter4,

The problem with mining apps is that they drag the substations around on skids. Nothing is permanent so there aren't any ground grids established.

BTW, what's "homopolar"? I see you are using I"k for fault current. Something to do with IEC standards?

 

[7anoter4]

In a case of ground fault the currents are unbalanced. We use symmetrical components in order to solve the problem.
The symmetrical components are zerosequence, positivesequence and negativesequence.
One of these components is zerosequence, consisting of three phasors with equal magnitudes and with zero phase displacement.Sometime it is named "homopolar" that means "in the same phase". You are right, may be this term is used more in Europe. The name"homopolar" is also connected with electric machines, indeed.
For symmetrical components informations see -for instance:

http://www.intranet.eee.manchester....Power/Symmetrical comp introduction_PAC07.pdf

I"k may be I"k3- three-phase short-circuit or I"k1 -single phase short-circuit and I"k1 includes all three symmetrical component.
I"K3 does not contain zerosequence component and usually not Negativesequence too.
But HRG and cable impedance to ground will increase the Zo-zerosequence impedance [and in turn will reduce the total I"k1, of course.]

 

[jghrist]

You wouldn't normally take capacitance into account when calculating GPR, but in this instance the charging current is of the same order of magnitude as the fault current. How would the distributed capacitance affect the GPR?

If all grounded equipment is bonded together, then when would you experience the full GPR? Only if a remote ground were brought to the fault point and not connected to the cable shield or EGC. This might occur for communications cables, but the insulation of the communications cables should withstand more than 100 V of GPR.

 

[7anoter4]

It is a very interesting issue, jghrist, indeed. But since I don't like the distributed impedance I'll try something else. What if the three-core voltages will be -approximately- the same, from one end to another of the cable as the drop voltage may be neglected. So we are against an equilibrated system of voltages and the total capacitive current will enter the mine grounding will be close to zero.
As the cable has to be shielded the shield will collect the capacitive current and most of this will return to the source.
You are right, jghrist; one cannot experience the GPR voltage or measure it. What I said it is that Emash is always less than GPR. If you could measure the Ground Grid resistance you could calculate the GPR and if it is less than 100 V the touch potential will be lesser.
The Ground Grid resistance may be measured following different proceedings as for instance:

http://www.electro-specialties.com/technical/downloads/Grounding Measurements.pdf

I think, if it is possible, 4 vertical electrodes of 3/4 inch dia driven in the soil 10-12 ft depth will be
suitable[in a soil of 100 ohm.m one could get 35 ohm resistance].
As an example let's take a 15 kV XLPE insulated 3*500 MCM copper conductor cable.For 50000 ft the Z1=Z2=1.4+j*1.6 ohm and Zo=18.8+j*14 ohm .The capacitive reactance will be 430 ohm.
If the source will supply 15 kV the ground fault current I"k1 will be 203 A[HRG=0].
If all three-conductors will be connected in parallel Z=21.6+j*17.2 ohm.
If a source of 240 V 60Hz will be connected at remote end and the other end will be connected to
the above "4 electrodes grounding grid" the total impedance will be 54 ohm and the current Itest=4.4 A measured at grounding grid connection.
The touch potential could be appreciated using an auxiliary electrode and then multiply the result with the ratio I"k/Imeasured.
If GPR has to be less than 100 V then I"k1= 100/35=2.85 A.
If HRG=850 ohm I"k1=2.82 A.