
#1
Jan2313, 10:20 AM

P: 4

Hi All,
Good morning. we had a 20MW, 13.8 KV wye generator and it has a neutral grounding resistor. Some how we failed to connect the neutral grounding resistor and during a rain storm we burnt the generator. Even though the neutral is not connected to the neutral grounding resistor it was in direct contact with the enclosure which is metal. I just want to know how much current and voltage flows through the neutral during phase to ground fault. Please some one help me. Thanks in advance, Krishna 



#2
Jan2313, 12:14 PM

PF Gold
P: 1,528

The current can be upwards to the full supply current on ground faults, voltages can spike to several thousand in a short spike. Sort of like inductive kicks utilized by car starters.




#3
Jan2313, 12:24 PM

P: 4

Thank you very much for your response.




#4
Jan2313, 12:36 PM

PF Gold
P: 1,370

Phase to ground faultI'm not sure you will get a specific answer at this forum. When you purchase your next generator, you should have the manufacturer make sure everything is wired up correctly. Round trip airline tickets for a qualified electrician to fly anywhere in the world are probably a lot cheaper than $10,000,000.00 20MW/13.8 KV=1450 amps Short circuiting your generator > i >>>>>>>> 1450 amps which translates to "the current will have been much much much much greater than 1450 amps." ps. Current flows, voltage doesn't. 



#5
Jan2313, 12:53 PM

P: 532

Are you saying the Gen Neutral was connected directly to ground  or there was no (solid) connection between the Gen Neutral and Ground?
I am curious because if there was no connection  then you would need 2 fault locations to get a ground fault  although the insulation level between the neutral conductor may not have been rated for full line voltage. I have seen many faults get repaired and turned on  only to immediately fail again ( now with people standing around watching) because the original situation was not fully understood. The amount of fault current available would have been much more than the phase current  and a ground fault can be intermittent / sparking  that can cause the V spikes Mordred mentions . Nevertheless  the situation sounds to have been extremely hazardous  and I would say very lucky to have not caused a fatality. Please have a qualified person review the repair and final installation. It is my experience that these situations do not come about via a single error or lapse of judgment  but a series of mistakes in thought, judgment and physical mistakes. And arrogance leads to them all. 



#6
Jan2313, 08:03 PM

Sci Advisor
P: 3,139

So do i understand you had a phase to ground fault while the neutral was grounded by accident?
Were you connected to grid at the time? Was voltage regulator in manual or automatic? If in manual, the current that flows is about (rated amps X machine's short circuit ratio), short circuit ratio ratio being probably slightly less than 1. Short circuit ratio should be in your generator manual and given as either S C R (abbreviation for short circuit ratio), or synchronous impedance which is reciprocal of short circuit ratio. If in auto, multiply the above result by ratio of maximum field amps (which is set by a voltage regulator limit) to normal field amps for generator rated output. That's the quick and dirty way to get a ballpark answer. Your generator people will do a better calc using transient and subtransient impedances to calculate the short term transient currents and that'll will be higher than above method. Method above will give steady state current. Transient will be maybe 3X10X higher, but too brief to do much welding. But above gets you in ballpark for the machine's contribution. If you were connected to grid some more came in backward through transformer. I hope your metering was working okay  that kind of trouble can lead to overflux... http://www.ccjonline.com/archives/s...okgenerators/ good luck  old jim 



#7
Jan2413, 07:18 PM

P: 4

Thank you for the replies.
The neutral of the generator was supposed to be connected throught the neutral grounding resistor. Our electricians forgot to make this connection. The unit was running perfectly and is on grid. as the neutral is not connected to the neutral grounding resistor the 51N never got activated. what ever damage that can happen has already happened. the neutral cable was touching the frame of the grounding resistor. I just want to know how to calculate the amount of current that flows through the neutral in case of a phase to ground fault. Thanks 



#8
Jan2413, 11:00 PM

Sci Advisor
P: 3,139

http://www.geindustrial.com/publibra...3550Fgeneric page 5 gives approach, page 15 starts detailed procedure for that calculation. You'll have to find some of your machine constants. This link has some tables of "typical" transient and subtransient reactances. http://www.magergy.com/documents/Ebo...s/96316_03.pdf see figs 3.5 thru 3.12 lessee 20 mw at 13.8kv = 836 amps? If that's at .8pf, it's a 1045 amp machine? Let's call it a kiloamp. Subtransient reactance gives the initial spike, see fig 3.9 If we guess at 12% that'd be 1/.12 = 8.33 X rated , or 8.33 kiloamps Initial spike decays and transient reactance takes over in a very few cycles. See subtransient time constant in fig 3.10. Maybe two cycles, 33 msec for initial spike? Transient reactance is higher than subtransient, see fig 3.7 . Maybe 25% ? So after subtransient , current would become 1/.25 X rated, maybe 4 kiloamps. That'd go on maybe a half second  see transient time constant in fig 3.8. Transient reactance dies out leaving synchronous reactance in charge, fig 3.5. Maybe 150%? So steady state current would settle, after subtransient and transient time constants, at 1/1.5 X rated = 666 amps. .........EXCEPT THAT the voltage regulator will be raising excitation trying to restore voltage, and the transformer may be backfeeding the fault. So current will be higher  so the analyst will have to figure out how your particular voltage regulator responds. AND generator protection should have tripped it by now. Those calculations are tedious , and honestly i did not do them in my work. I only knew and worked with people who did. This "back of the envelope" will give you ballpark answers but for anything approaching accurate , you need details about the machine and transformer impedances and somebody who is well versed in that type calculation. ~8300 amps initially , plus some from the transformer  That's the best i can do for you. How did the damage look around that accidental neutral ground? You said "burnt the generator"  we melted the end out of one I hope yours was less spectacular. If all you had was alittle sparking there from neutral displacement, you might be okay. old jim 



#9
Jan2413, 11:44 PM

P: 4

Thank you very much for your response jim.
neutral is not grounded accidentally. the neutral was never connected to the grounding resistor, from which the 51N gets activated when ever there is flow of current through the neutral. as the neutral is not connected the 51N relay never got activated. what i was thinking was the neutral grounding resistor will dissipate the energy in case of a ground fault. if there is no neutral grounding resistor then there might be a huge explosion if the neutral is touching the ground during a ground fault in the generator. Correct me if i am wrong. Thanks Krishna 



#10
Jan2513, 04:46 PM

Sci Advisor
P: 3,139

Now  ten amps at 13.8kv is 138kw, so the resistor is physically quite large. The voltage across an arc is probably only forty volts or so, so at twenty amps you're putting less than a kilowatt into damaging the machine. Vaporized copper expands by the same gas laws as dynamite vapor. You've heard the "BOOM" from electrical explosions? Now  let's think about what situation i think you are saying you had, the neutral wire grounded instead of tied to resistor. That's just bypassing the neutral resistor, eh? So this question becomes '...what if neutral resistor is bypassed, ie your neutral wire touching touching ground in absence of a phaseground fault? What current flows then ?' Well, let us think about it  aloud. First thought step will be Kirchoff's Current Law: The current that flows into the neutral from ground is the sum of the currents that flows out of the individual phases into ground. What path do those currents take? Those currents flow mostly through the distributed capacitance of your generator windings, transformer windings and interconnecting wiring. That's the only connection... So the dominant impedance is that capacitance. Next thought step  Tesla's gift to EE's remember the three phases are 120 degrees out, so those capacitive currents add to zero. So neutral current will be quite small so long as there's not much imbalance. I'd wager it's less than an amp. Well, maybe two... carry on... Next thought step: Fourier's gift to EE's  remember your generator does not make a perfect sine wave, it has harmonic content like any other real genertor. Probably two or three percent in a machine that small. And much if not all of that will be third harmonic. Aha ! Third harmonic  that's 180 hz, only 5.555 milliseconds for a whole cycle , 360 degrees at 180 hz. Aha^2 ! 5.555 milliseconds is also 120 degrees (1/3 cycle)at 60 hz. What that means is this  the third harmonic of each generator phase is in phase with the third harmonic of the other two phases. Aha^3 ! So the third harmonic currents flowing into the neutral do not cancel out as did the fundamental, they directly add. SO  in normal operation, the current returning in the neutral is largely third harmonic. I have measured this with a spectrum analyzer and find neutral curent to be maybe thirty percent third harmonic, maybe the majority like 85%. It doesn't look a lot like the 60 hz sinewave produced by the machine. Using the amplitudes and ratios of fundamental to third harmonic in both phase voltage and neutral current, you can get a pretty good guess at what is the distributed capacitance of your generatortransformer system. >>> Don't Give Up On Me Yet  I haven't forgot your question...<<< I'm going to guess that in normal operation you'll find an amp or two of neutral current , at least 50% of it third harmonic. Harmonic content goes up with excitation. Now  what happens if you bypass the neutral resistor with no phase fault present? Basically, nothing. The dominant impedance in the circuit is still the distributed capacitance. Neutral current will change very little. Incidentally that's how you size a grounding resistor, make it a few times smaller than the impedance of your distributed capacitance. This is all described very clearly in IEEE's "Green Book", standard 142 i think it is. Every power plant guy needs to have his own personal copy. Ask your company to get you one. I'm sorry the answer turned out so long  it's just that i was trained by a stern mentor to always "answer the question that was asked" and i hope this seemingly endless post helped. old jim 


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