Fault Current on Generators

[AppleJaxJap]

I have a generator for a project. in this example it is a 2000KW/2500KVA .8 PF and 480V so the full load amps @ 80% PF is 3007A

However, if this same generator was running at unity power factor then the most amps that I could get from it is 2406A.

The X”d for this unit is let’s say 10%

This unit has an HRG and there are only 3 wires no neutral. so only LLL and LL faults need to be considered That gives us a 3P fault current of 30,070A

My question:
A bolted fault is just a low resistive path why do I need to consider power factor on the faulted path?
In my way of thinking the bolted fault should happen at unity power factor and the available FC should be 24,060A and not 30,070A.
No literature shows this so I realize I must be missing something. What is it?

 

[ScottyUK]

You're neglecting the dominant impedance in the fault current loop, i.e. the generator reactances. Convert the X''d value into an ohmic value and evaluate the series combination of subtransient reactance and the resistance of a bolted fault.

 

[waross]

The Amps at unity power factor are limited by the power of the prime mover. Under short circuit conditions at the generator terminals the only impedance in the circuit is the generator impedance. The power factor of the load is immaterial as the load has been shorted out.
The rated current is 3007 Amps. That is what the % impedance is based on.
The power factor of the short circuit current is based on the X/R ratio of the generator. Short circuit currents are typically at low power factors. Definitely much below 80%. The prime mover, aided by inertia will have no trouble developing the initial fault current, followed quickly by voltage collapse and a corresponding reduction in current in the absence of independent excitation.

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

 

[AppleJaxJap]

Thanks for the response. I thought about that actually. The X/R ratio of this unit is about 9. Adding a resistance downstream (however small that may be) would only lower the X/R ratio on the circuit.

 

[waross]

The conductors will add some impedance (reactance plus resistance) but the load will be shorted by the fault.

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

 

[waross]

Hi dogzzz;
You may want to review your current calculations re 80% PF and unity PF.
This is a 2000 KW generator not a 2500 KW generator as per your example.
Tuny449 calculated the currents at 80% PF and unity PF correctly.

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

 

[ScottyUK]

Hi Bill,

The example does usefully show how the same alternator behaves at 0.8 and unity power factors though: the fault current is predominantly a machine characteristic, not a load characteristic.

 

[waross]

Hi Scotty.
For those of us familiar with alternator we can figure out what the poster means, sort of.
For those not so familiar, the post is confusing.
The OP had the currents figured correctly. The currents have been corrected with the wrong values.
Not the same alternator.
One issue is with the left hand information box. For a 2500 KVA the max current should be 3007 Amps not 2406 Amps.
Second issue is with the right hand information box.
As you know well, at unity power factor, the KW does not jump to 2500 KW to match the 2500 KVA, rather the KVA drops to 2000 KVA to match the 2000 KW.
He is not using the same alternator. He is using a 2500 KW alternator, not a 2500/2000 KVA/KW alternator.

There are two limits on a generator output. One is the maximum safe current rating of the windings, in this case 3007 Amps.
The impedance is determined based on this value of 3007 Amps.
This machine may safely deliver up to 3007 Amps at any power factor. Hovever at a power factor of 50 % the KW will drop to 1250 KW.
Diesel generators above about 10 KVA or 15 KVA are usually rated at 80% PF.
A safe current of 3007 Amps will yield an output of 2000 KW at 80% PF.
However the KW of a generator is limited by the power of the prime mover.
I have seen 600 KW sets lose RPM at just a few KW above 600 KW.
Other sets have an oversized engine and are able to safely produce more than the rated KW.


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

 

[ScottyUK]

Hi Bill,

You're right - there are some inconsistencies in that table. It looks like a different engine in the two cases, with the same alternator. I'd focused on a couple of other boxes and missed that one. Oops.

Assuming the same alternator in both cases: during the subtransient period of the fault the alternator itself essentially defines the fault behaviour. The AVR would just be starting to respond to the fault and the highly inductive field would have barely changed, so it's the machine and its magnetic state that determines the fault behaviour in that first few cycles. A more heavily excited machine will behave slightly differently to one with lower excitation at the moment of the fault, but that's about the only difference in the two cases for the subtransient period of the fault assuming the alternator is the same machine in both.

 

[dogzzz]

All the calculations were done keeping in mind that the Prime mover is not an obstruction , which means Prime mover can deliver above 2500 Kw.
Thus it was a same machine operation at .8 and 1 pf.

@ Scotty UK
You said " the fault current is predominantly a machine characteristic, not a load characteristic.
My Reply : I agree

@ waross
You said
For those of us familiar with alternator we can figure out what the poster means, sort of.
For those not so familiar, the post is confusing.
My Reply : No comments , but if its so confusing then I better take it out
------------
You said
The OP had the currents figured correctly. The currents have been corrected with the wrong values.
Not the same alternator.
My Reply : It is the same alternator , assuming that the Prime mover is not an obstruction , which means Prime mover can deliver above 2500 KW.
Thus it was a same machine operating at .8 and 1 pf.
-----------
You said
One issue is with the left hand information box. For a 2500 KVA the max current should be 3007 Amps not 2406 Amps.

My reply : Yeah you can say that .
------------
You said :
Second issue is with the right hand information box.
As you know well, at unity power factor, the KW does not jump to 2500 KW to match the 2500 KVA, rather the KVA drops to 2000 KVA to match the 2000 KW.
He is not using the same alternator. He is using a 2500 KW alternator, not a 2500/2000 KVA/KW alternator.

My Reply : It is the same alternator , assuming that the Prime mover is not an obstruction , which means Prime mover can deliver above 2500 KW.
Thus it was a same machine operating at .8 and 1 pf.
----------------------------
You said :
There are two limits on a generator output. One is the maximum safe current rating of the windings, in this case 3007 Amps.
The impedance is determined based on this value of 3007 Amps.
This machine may safely deliver up to 3007 Amps at any power factor. However at a power factor of 50 % the KW will drop to 1250 KW.

My reply : What you said is very generic , to be more precise , generator capability curve and generator damage curve should be referred to know as to in
what conditions the generator can work , and in which it cannot ?
--------------------------
You said :
Diesel generators above about 10 KVA or 15 KVA are usually rated at 80% PF.
A safe current of 3007 Amps will yield an output of 2000 KW at 80% PF.
However the KW of a generator is limited by the power of the prime mover.
I have seen 600 KW sets lose RPM at just a few KW above 600 KW.
Other sets have an oversized engine and are able to safely produce more than the rated KW.

My reply : Correct , but as I said better to look in generator capability curve as it will tell you the rating of Prime Mover and you will come to know
if your engine is over sized or not .
----------------

Example Capability Curve attached below

 

[waross]

See you in the Pub Scotty. We're not needed here.

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

 

[ScottyUK]

My round. ;-)

 

[waross]

Back to the Original Post which has to do with the available fault current.
The fault current is dependent on both the rated current, the rated voltage or EMF and the %Impedance.
The actual impedance is a machine characteristic and does not change.
The rated current is used by definition and does not change.
The rated voltage and the % Imp may change.
The basic operating limit on an alternator is the rated KVA.
KVA is derived from rated current times rated voltage times 1.73
The first limit on current is I[sup]2[/sup]R loss in the armature winding. There are other limits that are subject to change.
The fault current is determined by the alternator characteristics and the operating voltage.

This set is rated at 2500 KVA at 480 Volts. This gives a maximum current of 3007 Amps. Bolted fault current = 30,070. Amps
The rating is subject to change.
This may be a prime rated set. A prime rated set may be operated at 110% output for one hour out of 12 hours or 24 hours.
As a prime rated set, the maximum allowable current will be 3007 x 1.1 = 3308 Amps. However the %Imp is calculated based on the set rating and the bolted fault current is still 30.008 Amps.
If this is a diesel set it may be suitable for operation at 416 Volts. The KVA rating will drop to 416 x 3007 x √3 = 2700 KVA.
The rated maximum current will still be 3007 Amps but now there is less EMF to drive the current, the bolted fault current will drop and the %Imp will increase in the ratio of 480:416. The new %Imp will be 10% x (480/416) = 11.5%
Operated at 480 volts it takes 48 Volts to drive 3007 Amps through a bolted fault. 48V/480V = 0.1 or 10%
Operated at 416 volts it still takes 48 Volts to drive 3007 Amps through a bolted fault. 48V/416V = 0.115 or 11.5%

Considering the alternator rating as a factor of the prime mover power is interesting but not always valid.
I mentioned a 600 KW set that would not put out over 600 KW.
This was a tired old set, older than many of us here. As a prime power set it should have been capable of 660 KW easily.
It may have had a top end overhaul at 15,000 hours and was now long overdue for a 30,000 hour major overhaul. New or overhauled it would easily produce over 110% for one hour.
I suspect that the fuel may have been light as well but have no evidence of this.
A change in the specific gravity of the fuel will change the power of the prime mover. Available input HP is actually quite variable. In Canada, a 10% or more difference between summer fuel and winter fuel is common.
Note: the time limitation has more to do with the life expectancy of the prime mover than with the alternator.
The point: Yes, many sets are capable of over 100% of rated KW when new. The rating is intended to apply to end of life, not new sets.

Capability curves are interesting and in some cases colourful. But what what are they based on?
Well, a prime rated set will actually be a larger set, derated.
A 2500 KVA prime rated set will actually be a 2750 KVA set. Get the basic capability curve for the 2750 KVA set and look at the capability of a 2500 KVA set inside that curve.
Using the same set as a prime rated, 416 Volt set? Look at the capability of that 2170 KVA set inside the capability curve of a 2750 KVA set.
But what does it matter? A bolted fault is so far outside any capability curve that discussion of capabilities becomes moot.

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

 

[dogzzz]

Hello All ,
My idea is not to let anybody down , but just want to learn from you guys .
So , if you think I am wrong then please correct me , but please dont make fun , because
like generator I also have my " DAMAGE CURVE ".
Please find my views in red and there are some questions asked , if you can reply , I will be greatful , but if you are not interested then also its ok .

======================================================
Back to the Original Post which has to do with the available fault current.
The fault current is dependent on both the rated current, the rated voltage or EMF and the %Impedance.
The actual impedance is a machine characteristic and does not change.
[highlight #EF2929]-------I am not saying you are wrong , but there is more to it , as explained below

------[/highlight]
The rated current is used by definition and does not change.
The rated voltage and the % Imp may change.
[highlight #EF2929]---------
I dont know what do you mean when you say rated voltage may change . If the generator is 11KV , it is supposed to produce L-L 11KV ,
please correct me if I am not understanding .
Yeah the % impedance change from STran to Tran to Syn Imp.
---------[/highlight]
The basic operating limit on an alternator is the rated KVA.
KVA is derived from rated current times rated voltage times 1.73
The first limit on current is I2R loss in the armature winding.
[highlight #EF2929]-----
Thats true I2R , and this info is provided by the generator vendor in the form of generator damage curve .
------[/highlight]
There are other limits that are subject to change.
The fault current is determined by the alternator characteristics and the operating voltage.
[highlight #EF2929]--------
Ok , but why dont you say to check the capabilty curve for all this limitations .
--------[/highlight]
This set is rated at 2500 KVA at 480 Volts. This gives a maximum current of 3007 Amps. Bolted fault current = 30,070. Amps
The rating is subject to change.
This may be a prime rated set. A prime rated set may be operated at 110% output for one hour out of 12 hours or 24 hours.
As a prime rated set, the maximum allowable current will be 3007 x 1.1 = 3308 Amps.
However the %Imp is calculated based on the set rating and the bolted fault current is still 30.008 Amps.
[highlight #EF2929]------
Ok
-----[/highlight]
If this is a diesel set it may be suitable for operation at 416 Volts. The KVA rating will drop to 416 x 3007 x √3 = 2700 KVA.
The rated maximum current will still be 3007 Amps but now there is less EMF to drive the current, the bolted fault current will drop and the %Imp will increase in the ratio of 480:416. The new %Imp will be 10% x (480/416) = 11.5%
Operated at 480 volts it takes 48 Volts to drive 3007 Amps through a bolted fault. 48V/480V = 0.1 or 10%
Operated at 416 volts it still takes 48 Volts to drive 3007 Amps through a bolted fault. 48V/416V = 0.115 or 11.5%
[highlight #EF2929]------
Can you please explain , how you reached on this conclusion """ Operated at 416 volts it still takes 48 Volts to drive 3007 Amps through a bolted fault""""
------[/highlight]
Considering the alternator rating as a factor of the prime mover power is interesting but not always valid.
I mentioned a 600 KW set that would not put out over 600 KW.
[highlight #EF2929]----
OK , but we always make sure that our PM works for 110% for 30 minutes , when we design our offshore power plants.
----[/highlight]
This was a tired old set, older than many of us here. As a prime power set it should have been capable of 660 KW easily.
It may have had a top end overhaul at 15,000 hours and was now long overdue for a 30,000 hour major overhaul. New or overhauled it would easily produce over 110% for one hour.
[highlight #EF2929]--------
I see , thanks for sharing your exp.
--------[/highlight]
I suspect that the fuel may have been light as well but have no evidence of this.
A change in the specific gravity of the fuel will change the power of the prime mover. Available input HP is actually quite variable. In Canada, a 10% or more difference between summer fuel and winter fuel is common.
[highlight #EF2929]---------
I see , again good to know for me
--------[/highlight]

Note: the time limitation has more to do with the life expectancy of the prime mover than with the alternator.
The point: Yes, many sets are capable of over 100% of rated KW when new. The rating is intended to apply to end of life, not new sets.

Capability curves are interesting and in some cases colourful. But what what are they based on?
[highlight #EF2929]----------
They are based on
(1)Stator Copper Loss (stator heating):
The maximum allowable heating of the stator sets a maximum
phase current
IA for the machine. It’s equivalent to set a maximum
apparent power for the machine. (Power factor is irrelevant.)
(2) Rotor Copper Loss (rotor heating):
The maximum allowable heating of the rotor sets a maximum
field current
IF for the machine. It’s equivalent to set a maximum
EA for the machine.
(3) Prime-mover’s Power Limit.
------------------------------[/highlight]
Well, a prime rated set will actually be a larger set, derated.
A 2500 KVA prime rated set will actually be a 2750 KVA set. Get the basic capability curve for the 2750 KVA set and look at the capability of a 2500 KVA set inside that curve.
Using the same set as a prime rated, 416 Volt set? Look at the capability of that 2170 KVA set inside the capability curve of a 2750 KVA set.
But what does it matter? A bolted fault is so far outside any capability curve that discussion of capabilities becomes moot.

[highlight #EF2929]---------------------
You are right , generator capability was mentioned , because since limitations of alternator were getting discussed , so capability curve became essential.
Offcourse , gen capabilty curve is not helpful in knowing the machine behaviour while a short circuit bolted fault occurs
---------------------[/highlight]

 

[ScottyUK]

Long-term readers, it's like the second coming of jbartos...

 

[waross]

The original questions:
Question:
A bolted fault is just a low resistive path why do I need to consider power factor on the faulted path?
Answer: No,the load is shorted out by the bolted fault.
In my way of thinking the bolted fault should happen at unity power factor and the available FC should be 24,060A and not 30,070A.
Answer: KVA is limited by the set characteristics. The PF of the fault current is determined by the X/R ratio of the set. KW may be limited by the prime mover power. Use the current derived from the KVA as that reflects the set characteristics and the set Ohmic impedance.
End of answer to the original question.

Hi dog.
Let me try to walk through your comments and questions.
The actual impedance is a machine characteristic and does not change.
-------I am not saying you are wrong , but there is more to it , as explained below
"available FC should be 24,060A and not 30,070A."
By definition, the Available Short Circuit Current is a defined term and is the symmetrical current. It is the voltage required to drive rated current through a bolted fault. It may be expressed as a percentage of rated voltage or as a PU value.
The ASCC is used to select equipment such as switchgear, fuses, breakers, bus gutters and anyting else that is required to have a short circuit withstand rating.

Yes, the Asymmetrical short circuit current may approach 2.8 times the symmetrical current, but that is considered when the equipment is rated. In the case of a generator or transformer with an unusually high X/R ratio you should enquire further.
Possibly someone with more familiarity with testing and rating equipment for Available Short Circuit Current will give us the limiting X/R ratio that equipment is tested at.
The actual fault current is the comprised of a symmetrical component and an asymmetrical component. in the event of a fault symmetrical component is always there. The asymmetrical component depends on the point on wave that the fault occurred.

"I don't know what do you mean when you say rated voltage may change . If the generator is 11KV , it is supposed to produce L-L 11KV ,
please correct me if I am not understanding .
Yeah the % impedance change from STran to Tran to Syn Imp."
My comments on this are based in part on many hours spent studying generator specs and comparing actual model numbers of alternator ends and diesel engines.
Lets take a generator end and look at the possible applications.
I will choose a Three phase, 600 KVA, 480 Volt, Generator end and base the calculations on an impedance of 10% Note: these are the maximum ratings off the generator end, not the ratings as applied to an assembled gen-set. I'll skip many of the formulas. You know most of them. Feel free to ask about any that you do not recognise.

The maximum current will be 722 Amps
The maximum voltage of a nominal 120 Volt winding will be 139 Volts.
The impedance is 10% at 480 Volts.
By definition, it will take 10% of rated voltage to drive a symmetrical current of 722 Amps through a bolted fault. That means that 48 Volts will drive 722 Amps through a bolted fault.

Now back to our assumed alternator and the new ratings for some typical applications.

Standby power, 480 volts: 600 KVA, Rated Current= 722 Amps; imp%= 10%
Prime power, 480 Volts: 545 KVA , Rated Current= 702 Amps; imp%= 9.1%
Standby power, 416 Volts: 520 KVA, Rated Current= 722 Amps; imp%= 11.5%
Prime power, 416 Volts: 472 KVA, Rated Current= 702 Amps; imp%= 10.45%
Standby power, 400 Volts: 500 KVA, Rated Current= 722 Amps; imp%= 12%
Prime power, 400 Volts: 455 KVA, Rated Current= 702 Amps; imp%= 10.91%

Now the same set may also be used for 50 Hz service.
At 50 Hz, the inductive reactance will be lower and the Ohmic impedance will be lower.
That will yield another table of ratings, ALL FOR THE SAME BASIC alternator.


On the matter of KVA, KW and rated voltage.
I had an application for a standby set for service at 120:208 Volts.
I don'r remember the exact rating but let's say 100 KVA.
The load was fairly constant and easily calculated.
Rather than a 120:208 Volt set, we were supplied with a 100 KVA, 120:240 Volt four wire delta set.
When the set was reconnected for 120:208 the KVA rating dropped to 208V/240V x 100 KVA = 87 KVA. Not enough for our connected loads.
Our options were:
1> Send the set back and have it replaced with the proper set. That would entail the cost of ocean shipping and probably unrecoverable import duties.
2> operate the set at 240 Volts delta and buy a transformer bank to derive 100 KVA at 120:208 Volts.
3> Curtail some of the load when on standby power. This was our final choice.
The capability limit of rotor heating would be the same BUT the PERCENTAGE of the rated KVA would change to reflect the change in rated KVA.

A comment on capability curves: For large alternators, particularly paralleled generators, capability curves are a must and protection to prevent operation outside of the curves is a must.

For smaller diesel generators operated islanded it is unusual to see a capability curve. I admit that if you are in the habit of demanding capability curves your regular suppliers will readily supply them, but I have spec'ed a lot of diesel generators, installed a lot more and checked the published specs for even more without seeing an easily accessible capability curve. We use nameplate data and basic published data but we are aware that unusual loads may need a second look.
It would take something like a combination of an unusually low power factor load demanding the full KVA capability of the alternator to put the alternator outside the PU capability curve.
In the real world;
Such loads are rare.
The PU capability curve is valid for only one or two applications. For operation at reduced KVA rating or reduced voltage, the capability in the lagging PF region is extended somewhat proportionally.
I admit that if you are designing parallel generator applications, some extra protection may be prudent.
Again, in practice, protecting against over-fluxing and under-fluxing by the AVR may be all that is required. In some cases the basic protection afforded by the AVR may be adequate.

If you are using a generic capability curve it may not be applicable to all possible ratings of the set without adjustment.
Even with smaller diesel sets, in parallel operation, a malfunctioning AVR and or an inappropriate throttle setting may result in damage due to operation outside the capability curve.

OK , but we always make sure that our PM works for 110% for 30 minutes , when we design our offshore power plants.
Good idea, but the industry standard for prime power and the alternator is 110% for 60 minutes.
But if you need 600 KVA, buy a 600 KVA Prime power rated set. It will allow 110% for one hour as part of the spec and without overloading either the prime mover or the alternator.
The prime mover will be sized for the greater possible load.
Standby rated sets will often be sized very close to the load requirements.
Notwithstanding, some engineers will order a set with the prime mover sized at 125% to allow full output when the engine is old and tired, and more probably full output on light fuel.

A good time to mention fuel. The heat value of diesel and related fuels is closely related to the specific gravity.
At one time I was looking at specs for a prime power set in a remote location where fuel was quite expensive. Fuel economy was a factor.
I looked at fuel consumption specs for 8 or 18 brands. Among 7 of the sets there was not enough difference to influence a decision.
The other set showed a 10% saving across the board. Almost too good to be true. It was too good to be true. Reading the fine print I discovered that while the majority of manufacturers used the same standard grade of fuel for testing, one conducted their tests with a fuel that was 10% heavier.
If you do some research on grades and specific gravities of diesel fuel you will find a lot of good information.
However I have not seen a text on generators that mentions specifically that a gen-set may not produce full output on light fuel.
You may need that extra 10% more often to accommodate light fuel than for overloading.
I have a tendency to make silly typo's so don't hesitate to question something that may be a silly mistake.

On the subject of rotor damage, we did have heat related rotor damage on a 1500KVA set. This was the result of uncorrectable system load unbalances. Uncorrectable? Interesting case. Ask if you dare.
And by the way, If my friend ScottyUK posts any corrections, believe him. Scotty works on very large machines while I work on mostly diesel stuff below about 1500 KVA. Scotty is more conversant than I with capability curves.

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

 

[ScottyUK]

The best document on capability curves that I've seen was published by the CEGB when I was still at school. It dates from the days when the industry in the UK was run by engineers and not accountants, so they took time to do it right even though some of the parameters are of limited practical consequence.

 

[waross]

Thanks Scotty. I haven't had time to study it in depth, but from a quick skim through it looks really good.
I noticed one point that I mentioned above. Use the actual machine parameters, not the nameplate information, as the machine may have been re-rated. In almost all diesel generators below about 1.5 or 2 MVA the machines are basically 480 Volt, standby power rated machines. That is generally the most heavily loaded and the highest voltage in that class. Prime power, continuous power and at all other lower voltages than 480 Volts the alternator is most likely re-rated.

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

 

[dogzzz]

Hello Scotty thanks for the wonderful book , I am planning to read it over the weekend and then may contact you .
Hello Bill ...please see my replies in RED ....


Let me try to walk through your comments and questions.
The actual impedance is a machine characteristic and does not change.
-------I am not saying you are wrong , but there is more to it , as explained below
"available FC should be 24,060A and not 30,070A."
By definition, the Available Short Circuit Current is a defined term and is the symmetrical current. It is the voltage required to drive rated current through a bolted fault. It may be expressed as a percentage of rated voltage or as a PU value.
The ASCC is used to select equipment such as switchgear, fuses, breakers, bus gutters and anyting else that is required to have a short circuit withstand rating.
==========
[highlight #EF2929]I agree[/highlight]
====================
Yes, the Asymmetrical short circuit current may approach 2.8 times the symmetrical current, but that is considered when the equipment is rated. In the case of a generator or transformer with an unusually high X/R ratio you should enquire further.
Possibly someone with more familiarity with testing and rating equipment for Available Short Circuit Current will give us the limiting X/R ratio that equipment is tested at.
The actual fault current is the comprised of a symmetrical component and an asymmetrical component. in the event of a fault symmetrical component is always there. The asymmetrical component depends on the point on wave that the fault occurred.
==========
[highlight #EF2929]May be a more better terminology would be to use the term called " Making Current " the asymmetrical current + symmetrical current = Peak Current (@ first half cycle) tells you that how much Amps the breaker should be able to withstand for first 1/2 cycle.[/highlight]
=========
"I don't know what do you mean when you say rated voltage may change . If the generator is 11KV , it is supposed to produce L-L 11KV ,
please correct me if I am not understanding .
Yeah the % impedance change from STran to Tran to Syn Imp."
My comments on this are based in part on many hours spent studying generator specs and comparing actual model numbers of alternator ends and diesel engines.
==========
[highlight #EF2929]I am glad to know that , and will sure take advantage of your experience[/highlight].
=========
"I don't know what do you mean when you say rated voltage may change . If the generator is 11KV , it is supposed to produce L-L 11KV ,
please correct me if I am not understanding .
Yeah the % impedance change from STran to Tran to Syn Imp."
My comments on this are based in part on many hours spent studying generator specs and comparing actual model numbers of alternator ends and diesel engines.
Lets take a generator end and look at the possible applications.
I will choose a Three phase, 600 KVA, 480 Volt, Generator end and base the calculations on an impedance of 10% Note: these are the maximum ratings off the generator end, not the ratings as applied to an assembled gen-set. I'll skip many of the formulas. You know most of them. Feel free to ask about any that you do not recognise.

The maximum current will be 722 Amps
The maximum voltage of a nominal 120 Volt winding will be 139 Volts.
The impedance is 10% at 480 Volts.
By definition, it will take 10% of rated voltage to drive a symmetrical current of 722 Amps through a bolted fault. That means that 48 Volts will drive 722 Amps through a bolted fault.

Now back to our assumed alternator and the new ratings for some typical applications.

Standby power, 480 volts: 600 KVA, Rated Current= 722 Amps; imp%= 10%
Prime power, 480 Volts: 545 KVA , Rated Current= 702 Amps; imp%= 9.1%
Standby power, 416 Volts: 520 KVA, Rated Current= 722 Amps; imp%= 11.5%
Prime power, 416 Volts: 472 KVA, Rated Current= 702 Amps; imp%= 10.45%
Standby power, 400 Volts: 500 KVA, Rated Current= 722 Amps; imp%= 12%
Prime power, 400 Volts: 455 KVA, Rated Current= 702 Amps; imp%= 10.91%

Now the same set may also be used for 50 Hz service.
At 50 Hz, the inductive reactance will be lower and the Ohmic impedance will be lower.
That will yield another table of ratings, ALL FOR THE SAME BASIC alternator.


On the matter of KVA, KW and rated voltage.
I had an application for a standby set for service at 120:208 Volts.
I don'r remember the exact rating but let's say 100 KVA.
The load was fairly constant and easily calculated.
Rather than a 120:208 Volt set, we were supplied with a 100 KVA, 120:240 Volt four wire delta set.
When the set was reconnected for 120:208 the KVA rating dropped to 208V/240V x 100 KVA = 87 KVA. Not enough for our connected loads.
Our options were:
1> Send the set back and have it replaced with the proper set. That would entail the cost of ocean shipping and probably unrecoverable import duties.
2> operate the set at 240 Volts delta and buy a transformer bank to derive 100 KVA at 120:208 Volts.
3> Curtail some of the load when on standby power. This was our final choice.
The capability limit of rotor heating would be the same BUT the PERCENTAGE of the rated KVA would change to reflect the change in rated KVA.

A comment on capability curves: For large alternators, particularly paralleled generators, capability curves are a must and protection to prevent operation outside of the curves is a must.

For smaller diesel generators operated islanded it is unusual to see a capability curve. I admit that if you are in the habit of demanding capability curves your regular suppliers will readily supply them, but I have spec'ed a lot of diesel generators, installed a lot more and checked the published specs for even more without seeing an easily accessible capability curve. We use nameplate data and basic published data but we are aware that unusual loads may need a second look.
It would take something like a combination of an unusually low power factor load demanding the full KVA capability of the alternator to put the alternator outside the PU capability curve.
In the real world;
Such loads are rare.
The PU capability curve is valid for only one or two applications. For operation at reduced KVA rating or reduced voltage, the capability in the lagging PF region is extended somewhat proportionally.
I admit that if you are designing parallel generator applications, some extra protection may be prudent.
Again, in practice, protecting against over-fluxing and under-fluxing by the AVR may be all that is required. In some cases the basic protection afforded by the AVR may be adequate.

If you are using a generic capability curve it may not be applicable to all possible ratings of the set without adjustment.
Even with smaller diesel sets, in parallel operation, a malfunctioning AVR and or an inappropriate throttle setting may result in damage due to operation outside the capability curve.
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[highlight #EF2929]That is a very good explanation , thanks a lot[/highlight]
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OK , but we always make sure that our PM works for 110% for 30 minutes , when we design our offshore power plants.
Good idea, but the industry standard for prime power and the alternator is 110% for 60 minutes.
But if you need 600 KVA, buy a 600 KVA Prime power rated set. It will allow 110% for one hour as part of the spec and without overloading either the prime mover or the alternator.
The prime mover will be sized for the greater possible load.
Standby rated sets will often be sized very close to the load requirements.
Notwithstanding, some engineers will order a set with the prime mover sized at 125% to allow full output when the engine is old and tired, and more probably full output on light fuel.

A good time to mention fuel. The heat value of diesel and related fuels is closely related to the specific gravity.
At one time I was looking at specs for a prime power set in a remote location where fuel was quite expensive. Fuel economy was a factor.
I looked at fuel consumption specs for 8 or 18 brands. Among 7 of the sets there was not enough difference to influence a decision.
The other set showed a 10% saving across the board. Almost too good to be true. It was too good to be true. Reading the fine print I discovered that while the majority of manufacturers used the same standard grade of fuel for testing, one conducted their tests with a fuel that was 10% heavier.
If you do some research on grades and specific gravities of diesel fuel you will find a lot of good information.
However I have not seen a text on generators that mentions specifically that a gen-set may not produce full output on light fuel.
You may need that extra 10% more often to accommodate light fuel than for overloading.
I have a tendency to make silly typo's so don't hesitate to question something that may be a silly mistake.
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[highlight #EF2929]Again that was a good info on oil , normally we use HFO(rare) , MDO( almost always ) , and in some projects duel fue[/highlight]l.
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On the subject of rotor damage, we did have heat related rotor damage on a 1500KVA set. This was the result of uncorrectable system load unbalances. Uncorrectable? Interesting case. Ask if you dare.
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[highlight #EF2929]Normally we have a Generator individual phase voltage Unbalanced Setting: 20% with Time Delay: 0.8 Sec , so in your case I am wondetring why that kind of functionlaity was missing and may I ask
what kind of protection relay you were using ?[/highlight]
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And by the way, If my friend ScottyUK posts any corrections, believe him. Scotty works on very large machines while I work on mostly diesel stuff below about 1500 KVA. Scotty is more conversant than I with capability curves.
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[highlight #EF2929]I have no issues with Sotty and you , I am here to learn and learn , the only problem I have with you and scotty is that you guys went to pub alone ....could have invited me......lol
Anyway thankyou very much[/highlight]
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[waross]

Hi Dog;
"May be a more better terminology would be to use the term called " Making Current " the asymmetrical current + symmetrical current = Peak Current (@ first half cycle) tells you that how much Amps the breaker should be able to withstand for first 1/2 cycle.
========="
The ASCC rating on a breaker tells you that the breaker is capable of interrupting the worst case Asymmetrical current that may be encountered when interrupting the rated symmetrical current.
ASCC is easy to calculate. The simple calculation may be done by electricians and even advanced apprentices. A lot of smaller commercial installations may be designed by electricians. An even greater percentage of changes and additions may be designed by electricians.
All the information needed is on the transformer nameplate.
Few electricians have the training and math skills to calculate the worst case Asymmetrical current, or to combine the vector sum of the Symmetrical current component and the Asymmetrical current component.
Note: We have referred to Symmetrical currents plus Asymmetrical currents. This is a simplification and is not really accurate as there is a 90 degree displacement between the currents. Determining the vector sum requires the assistance of Pythagoras.
Engineers designing switching for very low PF loads (unusually high X/R ratios) may encounter instances when the fault current exceeds the value that the breaker is tested to and need to move up to the next higher rating. This is a special case that may never be encountered by the typical electrician.
In short, ASCC ratings are quick simple and safe. The needed information is readily available. The more involved calculations have been done in the background by the people testing and rating the breakers.

ROTOR DAMAGE.
I was looking after a small utility serving about 5000 customers on an island.
Our diesel plant was cobbled together from used equipment from two different offshore rigs.
We didn't have much protection. Breakers and fuses and reverse power.
We had about 1000 customers on an offshore reef.
They were served by about 1000 feet of undersea cable. I hesitate to call it "Submarine cable" as that implies a special construction.
We had direct burial cable consisting of three 15000 Volt rated, lead covered, paper and oil insulated single conductor cables twisted together in trefoil configuration.
One conductor failed and we served the reef with two phases. The load was residential with the addition of one seafood processing plant with a lot of three phase motors. During the outage, the plant ran on their own generators.
This gave us a current unbalance that we lived with rather than try to re-distribute the load on the other part of the system. Yes, at this time the unbalance may have been correctable but the story continues.
This was in the poorest country in Latin America and cost was an overriding concern.
Eventually a surplus cable was located and purchased. This was a single conductor submarine cable.
The new cable was laid paralle to and a few feet away from the existing cables.
Now we had two phases twisted together and the third phase not twisted and separated by a few feet.
This gave us a noticeable voltage unbalance on the reef. This may still have been correctable to some extent, but remember the seafood plant.
When the plant started up their refrigeration, The large motors acted as induction generators.
All induction motors do this. That is the source of the back EMF that limits the current. If the voltage on one phase is low, the induction generator attempts to correct the voltage by transferring energy from the two good phases to the low phase.
That is why a small voltage unbalance to an induction motor will result in a large current unbalance.
The path for this energy is the rotor and rotor heating results.
So, when the refrigeration equipment was running, the motors provided some correction of the voltage unbalance. This resulted in the load back at the generators being reduced on the odd phase with a corresponding increase in the load on the good phases.
The correction was dependant on the number and size of motors running.
During the lobster season the number of motors running was maximum. When the season ended ice making stopped and some large motors stopped.
Following the season, processing tapered off and as product was shipped, cold storage rooms were shut down.
Now comes the best part. We had an arrangement with the seafood plant. When our load was peaking and we were reaching the limit of our generating capacity our operator would phone the seafood plant operator and the plant would transfer over to their own generators for a few hours.
At that point we had high loading coupled with the loss of the correction provided by the seafood plant and peak unbalances.
After 5 or 6 years of operation, a PM check showed heat damage and a safe but dropping megger reading of the rotor. The rotor was repaired and put back in service for probably another 6 or 10 years.
The reason I say that the unbalance was uncorrectable is that even if we were able to achieve a good balance, as soon as the seafood plant started or stopped a motor our load balance would be lost.

Bill
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"Why not the best?"
Jimmy Carter