Stator Ground-Fault Protection

Because they manual said that 11 minutes was too long.

Jeez! Read the manual or call the manufacturer. You don't even tell us the model of the generator.

Here, Let the GREAT FUNGILI figure it out.

OpEn YoUr MiNd..... OpEn YoUr MiNd..... OpEn YoUr MiNd.....

The risk falls to the system protection engineer that specified the 10 minute time delay. With a setting this high it is probable that the generator has a high impedance grounding scheme involving either a resistor or Neutral Grounding Transformer (NGT)/resistor combination to limit the ground fault current to a relatively low value of 5-10 Amps. This limits the amount of damage due to the ground fault. For more information get a copy of the GE Protection Guide.

dear sir

the generator grounded through transformer 18kv/240v

Just as I suspected, so now take the turns ratio and calculate the effective resistance, and when you divide the L-N voltage of the generator by the effective resistance you'll know approximately how much current flows in a winding during a fault. How much damage will it cause?

dear sir

thank you very much for your response.please can you help me to calculate how much current flows in a winding during the fault.

Please review posts #2 & 5, I have already provided all the information on how to perform the calculations, you now need to do some work to gather the required machine parameters and plug them in to the formulas.

You will learn more and become a better student/designer by doing it yourself rather than having someone here do it for you. "Give a man a fish and he eats for a day, teach a man to fish and he eats for life." Start fishing.

Dear sir

thank you very much for your advice

Because someone arbitrarily decided to adjust it to 10 minutes either by mistake or to circumvent machine protection tripping of the generator?

Yes, it could be very detrimental to the generator.

Does the generator protection settings comply with OEM specifications?

If yes;

Contact the OEM for answers and justification documentations.

If no;

Return the settings to factory OEM specifications and attempt a generator start.

If the generator trips on GF;

Identify if a fault or faults exist then locate and correct/repair as needed.

If there are not any system ground fault(s) identified;

Review and verify generator protection equipment is fully functional and properly programmed/set up for the application.

If the protection equipment is properly programmed/set up for the application and you suffer false GF trips;

The protection equipment is suspect and should be replaced or sent to the OEM factory for repair and calibration.

Hope this helps you. Good luck and stay safe.

It is not unknown for a basic power system with a number of outgoing open-wire lines to operate on the basis that an earth fault on one line is tolerable for a short time.

An earth fault may be due to a wind-blown broken tree branch etc.

Usually, there is an alarm relay, sensing unbalanced voltage to earth, to wake up the operator.

There is no way to know which line has the fault, except to manually open each feeder breaker in turn & reclose - if it is not the one. On finding the faulty line, its breaker is left open and the lineman called to inspect the line. Except for the faulty line, no service gets cut for more than seconds.

If the generator tripped instantly in response to earth fault, all services would be cut - the ten minutes allows time to try the outgoing breakers - it is also likely the earthing transfo and resistor are only rated for 10 minutes.

A generator for such a system would be insulated to withstand one pole earthed.

There is a risk to the generator running with one pole earthed - a sparking fault causes spike overvoltages detrimental to transformers etc also.

There is also a risk that an animal or person may come close to the earth fault location and be electrocuted - a fault cannot be left for long.

67model

It is hard to imagine the GF protection trip delay time would be set at 10 minutes as most protection schemes require 15 cycles or less in order to minimize damage to equipment.

At 10 minutes trip delay time the generator current would definitely be at or close to system "bolted-fault" magnitude minus the drop across the resistor for way too long.

I wonder if the OP has possibly mistaken a generator startup parameter with a protection parameter but who knows?

While it may be "...hard to imagine...", it used to be common practice where continuity of service is more important than possible equipment damage. As the above referenced GE manual states, "... In spite of the foregoing (problems), many power companies are willing to risk the possibility of additional damage until they can conveniently remove the faulty generator from service..." With the advent of microprocessor relay systems coupled with fault signature analysis, this particular arrangement will be relegated to a back-up function "just in case" the primary electronic relay fails, utility engineers are obsessed with "protection in depth".

My response illustrates the problem with answering any question without considering/knowing the context of the OP's or reader's background. In this case, the specific relay is a very small part of a multi-layer in-depth generator protective scheme that includes many other independent systems that would act very quickly if major damage was occurring. It's all spelled out in the GE manual, sorry for the confusion.

I think you are assuming that the generator earth fault protection is being used for that purpose, rather than as part of a simple, low cost, system. I think the "high resistance earthing" & earth fault protection scheme hardware is just adapted to the system requirements.

I work out the earth fault current as follows [ it will answer one of ASSER's questions anyhow!] :-

Assume generator is 11 kV line to line, that is 11/√3 = 6350V line - star point.

With one phase shorted to earth, 6350V will be applied to the primary of the 18000:240V transfo.

The secondary voltage will be 6350 x 240/18000 = 85V

The secondary load resistance is marked as 0.336 ohm, so the secondary current will be 85/0.336 = 252 amps and the resistor loss 85 x 252 = 21400 watts.

That is not very much for an oil immersed or air cooled resistor.

The primary current will be 252 x 240/18000 = 3.36 amps

This will certainly limit the damage if there is an earth fault in the generator, the I²t will be 3.36² x 600 = 6800 amp²second with 10 minute delay.

Compare the situation with a 200 amp earth fault limiting resistor and an 0.1 second trip time (do not forget the excitation has to be killed, there is still remanent field flux & the generator run-down time will be minutes) -- 200² x 0.1 = 4000 A²s. I note that the I²t of a 32A, 415V fuse is about 4000 A²s.

In any case, in a simple system like this, the risk is taken that the generator might have an internal earth fault before it has paid for itself.

Regards,

67model