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Machine Riddle No.5 - Power Swing Problem
The 20 KV side of power transformer is earthed by a low impedance neutral grounding resistor. The rated current of resistor is 250 A for 10 seconds.
According to operator's electrical reports, some times badly power swing occurs in electrical system especially after one sever earth fault.

What is your opinion? How can we  solve this problem?
Author : Hamid - From: IRAN
 
#1
Sun, August 24, 2008 - 06:38
Apparently ,the neutral point of 20 KV system has been grounded via 46 ohm resistance and in principle , if the neutral point to be grounded through resistance , in case of earth fault current it will fluctuate over earth and will take voltage. But on the other hand 46 ohm is considered as low impedance and doesn't seem to present very severe fluctuation.  
Author : AMIR KHASHAYAR - From: IRAN
 
#2
Sat, September 13, 2008 - 15:53
An important consideration in resistance-grounded systems is the power loss in the resistor during line-to-ground faults. The power loss in the grounding resistor is Ir.Er or Ir2.R. Ir and Er are both in terms of normal values per phase. The power loss obtained by their product is therefore in terms of normal value per phase.
Conseqcntly, if the power loss in the resistor is to be expressed in terms of total three-phase system kva rating, it must be divided by three. Thus resistor power loss in percent is R.Ir2/3, when Ir is in per unit and R in percent. The maximum power loss for this case is 89.3 percent of the system rated capacity if three times the resistance in the neutral has the same ohmic magnitude as the reactants determining the ground-fault current. If the generator reactance are lower, still higher power mill occur during grounds and may cause violent swinging of generator phase position or instability.
Resistances in this region are to be avoided, and since there is always some additional resistance in the fault, it is preferable that the grounding resistors alone be sufficient to carry well beyond the peak. Effect of system size and operating voltage on size of neutral resistor to limit ground fault current to one-quarter full-load system current.
The actual ohmic value of the grounding resistor required will vary widely depending upon the circuit voltage and system capacity. By “full load system current” is meant the summation of the rated currents of all generating capacity, converted to the voltage base of the line. In terms of the rated line currents, the fault currents may still be several times full load, depending upon the number of lines.

Therfore we can write:

I < 1/4 x 12000/(20 x 0.8 x 1.73)   , If = 100 A is correct
 
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