Ground fault for 50/0.025 CT

[tizi1963]

How to determine the ground fault pickup (50G) of GE369 Multilin relay for MV motor fused starter unit (300HP) at 4.16kV. The ground CT used for sensitive earth fault detection is 50/0.025 for high resistive grounded systems. What is the method to calculate or determine a threshold for ground current 50G for a MV motor with a 50:0.025 CT?
Thanks

 

[waross]

Look up the minimum pickup current of the relay and multiply by 50/0.025 or 2000.
With a ground fault current of 50 Amps you will have a maximum of 0025 Amps to operate the relay.
Are you sure that you have the current rating correct??

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[oldfieldguy]

The 50/0.025 GFCT is a common device with many modern microprocessor-based relays. Many of them can be programmed for operational parameters in primary units. Others use multipliers of the rated primary current. You can see what maximum ground fault current will be by comparing your phase to neutral voltage and the ohmic resistance of your transformer's neutral resistor.

The pickup for the GF element in your relay MUST be less than the maximum ground fault value, otherwise it will never see enough current to operate.

It is not a particularly critical setting. We typically start with ten amps, and may adjust either time or current values in the case of nuisance trips.

old field guy

 

[LionelHutz]

Since you have a 50:0.025 CT for high resistance grounded systems, the first step is to determine that you are connecting this starter to a high resistance grounded power system. I believe Multilin specifies that a high resistance grounded system is limited to 25A or less. Then, set the relay below this maximum ground fault current level. Typical I've seen on 5kV is using a resistor that allows 5A and setting the relay 0.8-1A below but it can vary depending on the power system.