3 Phase Feeder Capacitive Charging Current 2

[mrg397cs]

The normal line to ground capacitive charging current for one phase is the line to neutral voltage divided by the capacitive reactance in ohms. Let's assume that each phase has a current of one amp when the system is unfaulted ( Ia +Ib +Ic = 1). When one of the phases is grounded, the total system charging current is 3 times normal phase to ground current.

During a fault on one of the phases in a feeder, and the voltage on the other phases stepped up 1.73 times line to neutral, will a zero sequence current transformer/detector on the feeder see 1 amp or 3 amps charging current?

 

[gordonl]

I would say 1.73A if each phase normally has a 1A charging current.

I don't follow the Ia+Ib+Ic=1 during normal operation, I would think Ia+Ib+Ic=0 because it is a balanced three phase capacitive load.

Then when you get a fault your charging current on the high phases increases to 1.73 times the normal charging current. (1.73X more voltage same capacitance.) But this is unbalanced to ground, so I would think I would see 1.73A on the ground to balance the charging currents.

I had a false trip on a resistively grounded system with sensitive ground line differential protection. The ground differential was set lower than this unbalanced charging current, so I bumped it up. This was my best guess, but perhaps there is a large flaw in my reasoning, please feel free to point it out.

 

[rsherry]

I think mrg probably meant Ia=Ib=Ic=1 for the unfaulted case.
Are you considering here the currents in an unloaded feeder during a fault on another feeder connected to the same busbar?

 

[busbar]

One amp [for 480V] but all the experts say it needs to be empirically verified. Just don’t pop any of those Fluke 1kV fuses—they are pricey. Check for absence of neutral-shift voltage, then do the current measurement.

See:

http://www.neiengineering.com/papers/paper1JN.pdf

§VI. DESIGN CONSIDERATIONS …the resistor and equivalent capacitance are in parallel. Since the resistor is placed in the neutral of the transformer, the zero-sequence current will flow through it. As a result, that current makes the apparent resistance of the grounding resistor appear to be three times as large.

General references..

http://www.geindustrial.com/products/manuals/GEI-72116.pdf

http://www.sea.siemens.com/pde/product/Swbd_pdfs_3_29/LV_High_Resistance_Grounding.pdf

Chapter 6 of Beeman’s 1955 Industrial Power Systems Handbook

 

[jbartos]

Comment on the previous posting paper. The medium voltage high resistance system grounding usually uses a single phase distribution transformer with the resistor and relaying on its secondary to avoid the medium voltage to propagate to the relaying circuits.

 

[jbartos]

Suggestion: Draw a circuit diagram that would include for example:
1. Delta - Wye transformer
2. Downstream circuit breaker
3. Transmission line
4. Transmission line capacitances to ground
5. High-resistance distribution transformer in the transformer wye grounded neutral
6. Ground fault, e.g. in phase A downstream of the circuit breaker, or another circuit breaker, e.g. branch circuit circuit breaker.
Then, IFault=IR-jIC
Where
IC=IACN+IBCN+ICCN
IACN=0
IR will actually flow through the neutral and distribution transformer primary.

 

[jbartos]

Comment: My high-resistance system grounding methodology is similar to:

http://www.ipc-resistors.com/pdf/highres_grd.pdf

(up to my proprietary secrets)