Fault Current Gone Wrong

I assume that the 140 KA is with the generators in parallel. Do they have to be run in parallel? Is the 140 KA at the main switch?

Yes they are in parrallel and hit a common bus at 140KA. And yes they want them in parallel even though our gear is now 200kaic and huge!

Dear Kellyshort,

To reduce S.C current we have to install reactor in series with generator line. You have to conduct short-circuit study in sizing reactor.

But, I am really confuse with your case, why the short-circuit current is to high for 2500 kW generator. Are these a special generator with very low subtransient/transient reactance?.. 140 kA total, mean contribution from each 2500 kW generator is 28 kA.

I have experience with steam turbine generator 13.8 kV, 12.5 MW, 0.8 PF with sub transient reactance around 13.5% and transient reactance 26%, the S.C contribution current from each generator is around 5 kA Symm. For 5 MW EDG we have subtransient reactance 22% and transient reactance is 34%.

Please check your generator subtransient and transient reactance to ensure both reactance values are correct?

Regards,

yes agree with rasam.. the fault is so high for a 2500MW generator..

Instead of opting for reactor you can go for generator transformer and parallel at the secondary side of the transformer then fault current can be brought to minimum.

Is the 5 number of generators to be connected in parallel (synchronised) to a common busbar.

Are the generators of the same make and specs

Is duty prime or standby.

Thanks

Dear Mr. Kellyshort,

I request you to give the detailed specification and application of the system. You have to specify Voltage, current, load application( whether using for infrastructure/ Industry(type)) what is the internal impedance of stator winding(individual generator)/ impedance of rotor, what is the excitation voltage and current.

you see if really want a correct answer you have to be very very specific and detailed. you may also need to give the electrical schematic for clear clarification.

Regards

My apologies for the lack of information. I wish I could post some pics of my skm and data sheet, but my reactances are .1194 x''d pu, .1804 x'd pu and my zero sequence is .0081. Each generator produces 26817A Isc 3p and 29486A Isc slg. The voltage for these units are 480V and 1800rpm @ 60hz, pf=.8 with a genset rating of 3125. All my equipment is within really close proximity. These are backup generators for a data center. My Short Circuit Ratio: 0.48 , Stator Resistance = 0.0012 Ohms, Field Resistance = 0.9703 Ohms. The generators are Diesel, Standby connected series star.
Excitation voltage: 12.98 Volts@ no load 52.73 Volts @ full load
Excitation current 1.19 Amps@ no load 3.99 Amps @ full load

Thanks for the advice after my post I realized I had a serious lack of information.

Kelly,

The kicker here is that you are generating at 480V and connecting the generator output directly to the distribution system. You don't have a transformer to choke the fault current, and your generators are right next to the load, so your impedance is limited to the stator. Your generator has a full load current of just over 3kA, so the windings are large diameter. Low voltage generation also means fewer winding turns, so less overall stator impedance. There's just no where to soak up that fault current.

I have a couple data centers on my system and dealt with similar issues during their construction. You're pretty much limited to 2 options:

• Add series reactors as rasam suggests (I normally put them in the neutral side of each generator phase before the star connection), or
• Design the distribution system for the calculated fault current (as it appears you already have).

The reliability requirements of data centers, particularly backup data centers, eliminates most other options. The owner's going to have to bite the bullet and pull out the checkbook one way or another.

Good luck.

Tony

Dear Kellytshort,

I agree with Mr. Pwr2thepeople comments, and now I give you very simple calculation to illustrate relation between reactor sizing and allowable voltage drop.

This simple calculation is to estimate the reactor impedance size and voltage drop at a certain current fault level.

For example we want to get 20 kA S.C fault level, with the following base: I base = 3125 A, V base = 480 V.

X"d = 0.1194 pu, X'd = 0.1804.

Isc 3 phase current = 3125/0.1194 = 26.17 kA (your calculation result is 26.817 kA, your calculation more precise than my calculation).

Y is reactor impedance that I am looking for.

20 kA = 3125/(0.1194 + Y)

Y = (3125 – 2388)/20000 = 0.03685 pu.

V drop in pu = 0.0365 x (480/480) = 0.0365 pu

V drop = 0.0365 x 480 V = 17.69 Volt.

Is the voltage drop accepted in your system?

You mentioned the SKM, in my opinion SKM is the software that can be used for S.C study, so you can add reactor in your calculation. How reactor impedance value you are looking for will depend on the allowable voltage drop in your system.

Regards,

Dear Mokamel,

Sorry I cannot see and read your reply. Can you tell me why?

Regards

Dear Guest,

I appreciate for your correction for my simple calculation even both data coming from the same source. Please do not hesitate to correct me.

1. "The rated full load current for each generator=2500/(1.732 x 0.48 x 0.8 )=3759 Amps"

In my simple calculation the generator rating is 3125A.

Each generator produces 26817A Isc 3p, simple calculation 26.17 kA and your calculation is 31.48 kA. I realize simple calculation is not precise, others?

2. I draw the SLD as shown above. Is that correct SLD?

3. The short-circuit value in section 1, is 2 x 31.48 kA. I am not with you in this case, because contribution current from other generators are eliminated, except if the tie rector between section 1 and section 2 opened. See SLD above.

Bus sizing shall consider the contribution short-circuit from other short-circuit sources, it is including load if there is any motor higher than 50 hp or several 25 HP motors. I think, in this case may not be existed.

4. No voltage drop problems on sections, excitation system will compensate terminal voltage drop. If a voltage drop problem still present study to improve power factor.

I do not know how much current to be reduced to meet Mr Kellytshort requirement.

In selecting reactor sizing a voltage drop is limitation as one criteria. If the voltage drop will not have any impact to the system we can select reactor easily.

If it is handling by excitation yes I agree, but also it is has limitation, the generator manufacturer will inform how high voltage can be increased. If the voltage increase, the Per Unit voltage also increase and it is proportionally with increasing short-circuit current.

If G1 & G2 off, it will be supplied from other buses, the worst case is taken from section 3, in my opinion we have also to consider a voltage drop.

Best regards,

Dear Rasam,

Thanks for your passing over my answer and thanks for your carefully revision.

Kindly note the username mokamel or guest are the same person.

1- I consider the value 3125 Amps. which, indicated in Kellytshort mail regarding to the electrical data for generator is the setting value-as a generator loading factor. but in case of short circuit generator will act without regarding to thermal protection systems so, I expect the worst case in study-full load Amps based on data avaliable. for more precise calculation i should to consider the resistance effect but studying still under worst case.

2- Yes, you read thoughts in my mind! Good.

3- Good for this note, but i consider when we select short circuit rating of bus and given that calculated short circuit which is 63kA we select the next standardizied value which is 70kA, any way your are correct, and for clarifications the total equivalent subtransient reactances seen by the equivalent source is 0.0597 pu and 0.05348 pu with respecting to your note, the first value lead to 63kA and the second value lead to 70.3kA. as you now reverse feding effect of motor during SC can be neglected with respect to the value comming from generators, also you can set reverse power relay on motor feeder to avoid this action.

4- I think this problem of voltage drop present when a reactor is connected in series in the path to load, in my proposal a reactor connected between loading sections and in this case manoeuvres are performed to get optimum loading and optimum sharing between sections in the presence of excitation system constraints which give a reasonable terminal voltage. the percentage of short circuit increasing as the percentage of terminal voltage increasing.

In case of G1&G2 off Yes, we should consider voltage drop in this case, but when he design and select 5 generator to fed the loads, no way when two generators are off the all loads are still fed. you can study to select a vital load on section#1 to fed from section#3 this can be ocuured by interlocking between the incoming circuit breaker to section#1 and outgoing circuit breakers to vital load.

Regards,

Mohammed K.

Dar AL-Handasah^Cairo,Egypt International Consultancy Group

Can't get my comments to show up and my last one was a Monday comment for sure, I know how we got the .073728

Thanks again you guys are great!

Wow thank you so much!

A few questions? Why is the value of the series reactor 3.5x's X''d

How does one convert to find Zbase at .073728 ohm and then to get .082mH ?

I completely understand why we have to not worry about voltage drop because of the excitation.

Dear Kellytshort,

Greetings,

when you need to get the short circuit on generator terminal bus, only divide the the rated full load current by X''d (in perunit value).I_sc=I_{full load generator current}/X''d .

Derivation of this equation as below:

I_sc=V_ph/X''d = kV_L/1.732X''_d (kA) where, V_L line voltage in volt, V_ph phase voltage in volt, X''_d subtransient reactance in Ω.

Z_base = (kV_L)²/MVA, I_{full load generator current}(kA)= MVA/1.732kV_L

X''_d(Ω) = X^''_d(pu)xZ_base ,I_sc = kV_L/(1.732X^''_d(pu)xZ_base) = MVA x kV_L/(1.732X''_d(pu)xkV_L²),

hence Isc=1.732kV_Lx I_{full load generator current} x kV_L/(1.732X''_d(pu) x kV_L²)

Isc = I _{full load generator current}/X''_d(pu) (kA)

in our case study

P= 2.5MW, Powerfactor = 0.8, MVA(apparent power)= P(MW)/Powerfactor

S(MVA)=2.5/0.8 = 3.125

Z_base= (0.48)²/3.125 = 0.073728 Ω.

X_{series reactor}(Ω) = 3.5x0.1194x0.073728 = 0.030811(Ω)

0.030811= 2∏f x L where, f frequency(Hz), L inductance(Henry)

L=0.030811/(2x3.14x60)= 81.73 μH.

If you anlayze the hint in my reply#13 you will get X_{series reactor} = 3.5 X^''d.

Regards,

Mohammed K.

Dar AL-Hanadasah^Cairo,Egypt International Consultancy Group

I have come across similar problem while designing the DG sets for Software Building in India.Here it was 5 nos 2000 kVA,433 Volts and fault current was working out to be 104 kA.

What I have done is Designed to operate only sets 1 & 2 in parallel as group 1 and sets 4 and 5 to operated in parallel as group.Set no 3 will be stand by for group 1 or group2.I have made the control circuit in such a way only two sets in group and group will operate in parallel and feed to the load.Either 1&2,or 1&3,or 2&3 in group1 will operate.Similarly in group 2 it will be 4 &5 or 3&4 or 3&5 will operate in parallel.

In this method short circuit current will get drastically reduced to 42 kA and the devices like Busduct,switchgear,switchboards etc can be designed for 50 kA only.

Economically there will be good saving on capital cost.

This is only suggestion.

Comments are welcome.

I couldn't agree more that is the answer, the only problem is I am the only who thinks so. I am going to fight the good fight on this one because I have some 215kA on my single line to ground fault and they refuse to use a high impedance system because it sounds so so scary when it is actually safer in my opinion. I need only +-.0084 ohms to get my Isc 3p = Isc SLG because how can it be more dangerous when the Short Circuit currents are the same?

Dear Kellytshort,

Your client refused using high impedance system for reducing ground fault current (am I correct?), have you talked with them on low resistance grounding? If possible you can share with us about the scary on HRG or LRG (low resistance grounding).

It is not a suggestion, but for information only. The following are my consideration when selecting low resistance grounding:

1. Most of Short circuit fault are initiated by ground fault. The possibility of ground fault become high due to level voltage on the system are the same.

2. The ground fault current is limited to 400 A or 1000 A. Those values are still lower than generator rating.

3. CT box, neutral cable and ZCT might be bigger if we use resistance/impedance for getting 3 phase fault is equal to ground fault current.

I am as a viewer appreciated to all of you that already share the best information on the above subject, especially for Mr. Mokamel.

With best regards,

Yes I also pitched LRG as well and pretty much there thought process which in some situations is correct they want the lowest impedance to ground possible. Some consultant told them that and they are sticking to it no matter what. But get this I was able to convince them to allow me to run the neutral conductor longer and coil it up inside the generator and brace it like there is no tomorrow. So now all I have to do is calculate how many times I need to wrap the coil and what distances and etc. You are probably thinking well that is an impedance grounded system, but I guess it boils down to perception. Any thoughts on coiling up (9) sets of 500kcmil conductors?!? I pitched the fact that I could run a #1 around 100ft to get .0084 ohms which would limit my Isc SLG. But since I don't have my PE yet and it isn't my stamp the one who is stamping it refuses.