Can someone explain how to size a transformer neutral resistor for limiting fault current? For example: What size neutral resistor would be required to limit the fault current of a 13,800-480Y/277V 1500kVA Transformer to 200A?
Can someone explain how to size a transformer neutral resistor for limiting fault current? For example: What size neutral resistor would be required to limit the fault current of a 13,800-480Y/277V 1500kVA Transformer to 200A?
This question comes from the level 2 practice exam..!?
im curious my self
Dan
277 V/200 A=1.385 OHMS to limit the ground fault current to 200 A on the 480Y/277 V system.
Some related information in case you are designing a resistance grounded system as opposed to answering a test question:
1. A 200 A max ground current limit would be more appropriate for a MV system (e.g., 4.16Y/2.4 kV).
2. In a LV system, a high resistance grounding scheme is employed with a resistor typically sized to limit ground fault current to < 10 A (the current must be > the system capacitive charging current e.g., cables, motors, transformers,.. capacitances).
3. The minimum size of the conductor from the supply transformer neutral to the neutral grounding resistor is stated in the NEC.
4. When investigating HRG system viability there are some application considerations such as:
- line to neutral loads like 277 V lighting must first be supplied via an isolation transformer's
- overvoltage protection of adjustable speed drives can be problematic
Several manufacturers have NGR whitepapers on the internet.
this Question came on the NETA 2 practice test and I am wondering if there is a different approach to what i have. here is the question
11. What size neutral resistor would limit the fault current of a 4.16Y/2.4 kV 3000kVA Transformer to 200A?
My reasoning.
the Y on 4.16Y tells me it is a Y configuration, therefore between the common point of my three windings and the ground, i will put a resistor and a current of 200A going through it from the Y winding common point to the ground.
- Now the voltage in a Y configuration is as follow VLine = 1.73 Vphase. A this point I am interested in the Vph which is; Vphase= Vline/1.73 => Vphase = 4160 /1.73 = 2402v. This voltage is the one i will have from the uncommon leg of one winding to the ground. At this point have ohm law of one unknown resistor in series with one winding with a voltage of 2402v across them. Since, the winding infos are not given i will neglect that and just divide my phase voltage by the limiting current doing so, R=2402/200 =12.01 ohm as suggested by Testguy correct or no? thanks
I think you have an error when choosing the voltage you selected in your calculation. You stated the transf is a delta wye, 13.8kv to 480y/277. So seeing how there is not a neutral on the hv side of the transformer, I am supposing the resistor is supposed to get connected to the low voltage side. And seeing how the device is called a neutral to ground resistor I think you would need to use the hot to neutral voltage of 277 for your calculation. And this makes more sense by the fact that, with a NGR you are trying to limit your fault current to ground, and the only continuity from phase to ground is via the neutral.
I haven’t dealt performed a calculation to size a NGR before but I thought I might be right and wanted to help. Someone let me know if I’m right!