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Motor/System Charging current

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raisinbran

Electrical
Joined
Sep 27, 2004
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67
Location
US
If I wanted to install a resistance grounded system (480 volt or 4160 volt), I know that there is a theoretical maximum resistance that is recommended. If I remember correctly, it is based on the system charging current. Can anyone help me with this issue?
Regards,
Raisinbran
 
For a high-resistance grounding system, the resistance should be roughly equal to the shunt capacitance of the system. So the ground fault current through the resistor would be equal to the system charging current. This helps reduce the high voltages that can occur during repetitive restrikes on arcing faults.

Normally this will be less than 5 amps.
 
This is for a mining application where the ground fault current is limited to 0.5 amperes. It trips at 200 ma. This type of a system is used quite often, but it appears to violate some of the equations used for sizing NGR that I have seen.

This is on a 480 volt system, but the same is true for the 4 kV systems used in many underground mines. Those cables are shielded, so I would think that the capacitance is considerably higher than the cables used at 480 volts. There have been no problems that I am aware of, but maybe that is because the cable is switched off with the motor, and the capacitance issues go away.

I am just curious if someone can shed some insight on this or if these high resistance grounded systems (0.5 ampere limit) are used in other industries.

Regards,
Raisinbran
 
0.5 amps may be OK. - it just depends on the size and configuration of the system.

In general, it is better to have a little too much current than too little current, in terms of stabilizing the neutral point and limiting overvoltages.

The calculation to determine the exact system capacitance is difficult because the parameters are hard to come up with.


 
Raisinbran,
Depends on the cables you are using. Please provide more info so we can give specific answers. If this is a mining underground, the capacitance of the cable used determines the charging current.
Code:
    C     =    0.0169 × e (  ln ( D÷d )) in µF/1000ft
            where:
               C = capacitance of cable
               e = 2.3 for XLPE; 2.8 for EPR
               D = diameter over the insulation
               d = diameter under the insulation
      Icharging = 2 x pi x f x C x Eo		
             where:
                Ic = mA / 1000 feet
                 f = frequency (Hz)
                 C = capacitance in F/1000ft
                Eo = line-to-ground voltage (kV)
 
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