A couple fault current questions

[electrofelon]

I am doing a 277/480 service where the utility will be installing a 500KVA transformer. They gave me a fault current of 37K. After calculating the reduction from the service entrance conductors, I still was a little over my breaker ratings, so I asked for a photo of the name plate to run the calculation using actual transformer impedance (5.2%) to see if that would give me a better deal, and using that figure with 80% power factor in the MH calculator, I got 14.5K. Does that seem right? I just want to make sure I am not making a silly mistake somewhere - I wasnt expecting that much o a difference. Presumably the 37K figure is from a chart with the lowest possible impedance unit they could ever conceivably use?

The other question is, while double checking my formulas, I came across something in a cooper/bussman publication that I dont understand. They have a chart for "typical transformer impedance values" and it has two columns, one is "%Z" and the other is "suggested X/R ratio used for calculation." Could someone explain those two values?


Thanks!

 

[mivey]

I am doing a 277/480 service where the utility will be installing a 500KVA transformer. They gave me a fault current of 37K. After calculating the reduction from the service entrance conductors, I still was a little over my breaker ratings, so I asked for a photo of the name plate to run the calculation using actual transformer impedance (5.2%) to see if that would give me a better deal, and using that figure with 80% power factor in the MH calculator, I got 14.5K. Does that seem right? I just want to make sure I am not making a silly mistake somewhere - I wasnt expecting that much o a difference. Presumably the 37K figure is from a chart with the lowest possible impedance unit they could ever conceivably use?

The other question is, while double checking my formulas, I came across something in a cooper/bussman publication that I dont understand. They have a chart for "typical transformer impedance values" and it has two columns, one is "%Z" and the other is "suggested X/R ratio used for calculation." Could someone explain those two values?


Thanks!

The value the POCO gave is not always tied to the exact transformer you are looking at and is not always tied to the present system configuration.

The %Z = sqrt(%R^2 + %X^2)

 

[Ingenieur]

It is the internal Z of the xfmr that limits fault current
expressed in per unit (%) = actual value/base value
a math method/convention used to simplifiy calculation and interpretation of results

it is measured as follows
short the secondary
apply voltage to primary
raise until sec = rated i
eg
primary v is 1000
secondary v is 500
kva 1000
i sec rated = 1155
say it takes 50 v on primary to get 1155 A on shorted sec
pu Z = 50/1000 = 0.05 or 5%
it can be shown this equates to v/i or Z

now keep raising v to rated or 1000/50 = 20 times
that means sec i is 20 times larger, ie 20 times rated sec i
this is the sc current assuming infinite bus 20 x rated current 23094 A

or
kva = sqrt 3 x v x i / pu Z, ie at primary rated v with sec shorted

i sc = kva/(sqrt 3 x v x pu Z)
= 1000000/(1.732 x 500 x 0.05) = 23094
divide again by pf if desired

your numbers
500000/(1.732 x 480 x 0.052) = 11566
with power factor 11566/0.8 = 14457

the math can be shown to derive the above but only confuses the issue

as noted above Z is the magnitude of R and X
X/R is a unitless ratio and can be used to derive actual R and X for more precise calculations
it also gives the ph angle arctan(X/R), info lost with pu Z only
this can give info about rate of rise and decay
usually not important when looking for cb ratings

 

[augie47]

Locally when POCO provides a fault-current they base the number on the "worst case scenario" for replacement transformers they have available and not necessarily for the exact unit they are installing.
Getting the local AHJ to accept the lower number, if applicable, is another matter.

 

[Ingenieur]

What type of xfmr
3 pole mounted cans
pad mount
etc

 

[electrofelon]

thanks you all for the replies. So, in general, I would use X/R for more precise calculations, for example If I wanted to calculate asymmetrical fault current, motor contributions..is that fair?

Getting the local AHJ to accept the lower number, if applicable, is another matter.

Do you think so Augie? I would think an AHJ might even prefer to see a value right on the transformer nameplate and the calculations be based off that. To show an inspector the value from a utility person would generally be showing him a print out of an email - seems like a lot of honor system there.

I'll post the photo of the nameplate.

 

[topgone]

I am doing a 277/480 service where the utility will be installing a 500KVA transformer. They gave me a fault current of 37K. After calculating the reduction from the service entrance conductors, I still was a little over my breaker ratings, so I asked for a photo of the name plate to run the calculation using actual transformer impedance (5.2%) to see if that would give me a better deal, and using that figure with 80% power factor in the MH calculator, I got 14.5K. Does that seem right? I just want to make sure I am not making a silly mistake somewhere - I wasnt expecting that much o a difference. Presumably the 37K figure is from a chart with the lowest possible impedance unit they could ever conceivably use?

The other question is, while double checking my formulas, I came across something in a cooper/bussman publication that I dont understand. They have a chart for "typical transformer impedance values" and it has two columns, one is "%Z" and the other is "suggested X/R ratio used for calculation." Could someone explain those two values?

Thanks!

What is the transformer primary voltage? How far is your distribution panel from the transformer?

• 11.6 kA is the rms short-circuit fault at the transformer terminals given the %Z and an infinite bus supply, FLA/Z = 601/0.052 = 11.6kA.
• The 37 kA available fault current means the system available MVA at your point of common coupling is 884 MVA if your primary distribution voltage is 13.8 kV or 266.6 MVA at 4160V!
  *MVAsc = Isc available from system X 1.732 X system voltage (37,000 x 1.732 x 13,800 = 884 MVA
• if your 480/277 panel is 200 feet from your transformer and you are going to use 3 - 250MC per phase, your available short-circuit fault at your distribution panel will be around 9.7 kA!
  *(total impedance = system Z + transformer Z + impedance per phase of line; Isc = secondary voltage per phase/ impedance per phase)
  
  
  system impedance (referred to 480v) = kV^2/MVA = 0.48^2/884 = 0.000261 ohms;
  
  transformer impedance = 0.052 x .480kV^2/0.5MVA = 0.0239616 ohms
  
  line impedance (3 by 250MCM per phase) = 0.066 ohms per 1000 ft. x 200 feet/1000 / 3 cond. = 0.0044 ohms
  
  total impedance at 480 panel = 0.000261+0.0239616+.0044 = 0.0286226 ohms, then
  
  Isc = (480/1.732)/0.0286226 = 9,666.7 amperes = 9.7 kA!​

The X/R figure tells you how much the assymetric fault current will be if the fault occurs when the voltage wave is at its peak. With an X/R data, you can compute for the peak assymetric fault factor using the formula:
factor = K x 1.414, where: K =1.02 + 0.98exp{-3/(X/R)}-->this formula is a simplified one.
Hope that helps.

 

[electrofelon]

Hope that helps.

Yes thank you!

So in general, in a situation like this, how would y'all spec the AIR of your equipment? I guess it comes down to balancing cost vs providing for future changes/expansion? I have 14.5K going off the actual transformer, and 37K going off the generic utility figure (not counting conductor reductions). I wouldn't want to cut it too close to the low figure, and think I should provide at leaast 10% each for impedance tolerance and higher voltage conditions.

 

[petersonra]

Yes thank you!

So in general, in a situation like this, how would y'all spec the AIR of your equipment? I guess it comes down to balancing cost vs providing for future changes/expansion? I have 14.5K going off the actual transformer, and 37K going off the generic utility figure (not counting conductor reductions). I wouldn't want to cut it too close to the low figure, and think I should provide at leaast 10% each for impedance tolerance and higher voltage conditions.

I think you are asking for trouble if you do not use the numbers the POCO gave you. They are based on the worst case for the size of service you have. Suppose the xfmr blows up and they replace it with a different one? That happens now and then.

 

[mivey]

I think you are asking for trouble if you do not use the numbers the POCO gave you. They are based on the worst case for the size of service you have. Suppose the xfmr blows up and they replace it with a different one? That happens now and then.

I agree. You must consider the POCO practices. Many include possibilities of the lowest impedance found in stock equipment as well as allowing for a size-up.

That said, electrofelon could ask them to verify the fault current value to be sure they gave the right one for his configuration. Just using local data, a 750 kVA UG pad would top out at around 17 kA but a 750 kVA transclosure or a 750 kVA overhead bank would top out at around 36 kA.

 

[Ingenieur]

If the xfmr will stay of the same the type, ie, pad mnt liquid cooled and rating v/kva using the Z for the sc calcs and up sizing 1 rating is not an unreasonable engineering judgement

good sc primer http://www.electricalmanuals.net/files/APP-MANUALS/GE/GET-1008L.pdf
page 17
has a nice curve for the max offset based on X/R
eg if X/R = 5 then max offset = 1.27 x symetrical (both rms)

 

[beanland]

Fault current power factor?

Fault current power factor?

I do not see how you can use power factor for fault current; there is no load. The power factor of the fault current is controlled entirely by the X/R ratio of the transformer and the medium voltage system.

And, the POCO will consider worst case. Is the 500kVA transformer sized based on the service rating or is it about 50% of the service? What if the load grows and they install a 750kVA transformer? What if the transformer fails and what they have handy is a 1000kVA? What if the impedance of the new transformer is lower? If you use the POCO value, then if the AFC exceeds that, you are not at fault for installing undersized equipment.

I also agree though that you can consider the length of the service conductors. The POCO will give the AFC at the transformer. Reduce it for the service conductors because those are under your control.

 

[Ingenieur]

I do not see how you can use power factor for fault current; there is no load. The power factor of the fault current is controlled entirely by the X/R ratio of the transformer and the medium voltage system.

And, the POCO will consider worst case. Is the 500kVA transformer sized based on the service rating or is it about 50% of the service? What if the load grows and they install a 750kVA transformer? What if the transformer fails and what they have handy is a 1000kVA? What if the impedance of the new transformer is lower? If you use the POCO value, then if the AFC exceeds that, you are not at fault for installing undersized equipment.

I also agree though that you can consider the length of the service conductors. The POCO will give the AFC at the transformer. Reduce it for the service conductors because those are under your control.

if they double the capacity the fault current is going to double for constant Z
anything could happen but what are the parameters now
there is a reason they are stamped on equipment
if you design for every contingency you'll waste the clients money


pf is a factor (lol)
the x/r is another way of expressing it
pf = cos(arctan x/r)

in most cases it can be ignored since pf is ~0.8 or better or x/r = 0.75 or 7.5/10
offset of only ~1
not until an x/r of 8 does it hit 1.4
and this is at 1/2 Hz
most molded case cb's will trip in a 3-6 cycle range with a bolted fault
by then the fault i will have settled to the symmetrical value

no matter what the utility gives you for a value as the design engineer you are responsible
it may mitigate your exposure somewhat in litigation but try suing a utility

I would have no issue spec'ing a cb in the 25k aic range
no risk
but each must make their own decision
I hesitate to squander my clients money to cover my butt
that is what we get paid for

 

[topgone]

Yes thank you!

So in general, in a situation like this, how would y'all spec the AIR of your equipment? I guess it comes down to balancing cost vs providing for future changes/expansion? I have 14.5K going off the actual transformer, and 37K going off the generic utility figure (not counting conductor reductions). I wouldn't want to cut it too close to the low figure, and think I should provide at leaast 10% each for impedance tolerance and higher voltage conditions.

Firstly, when you use the assumption of an "infinite bus" supplying your system, you are on the safe side. Since your PoCo gave you a figure (37 kA), it can be shown that the prospective fault current at your secondary terminals will be lower than 11.6 + 2.4 kA = 14 kA!
Second, doing it again (without the cable impedance and using the primary distribution voltage of 12.47 kV):
At the secondary terminals:

System impedance = kV^2/MVAsc = (0.48)^2/(37,000 x 1.732 x 12,470V) = 0.002957 ohms
Transformer impedance = 0.052 x (.48^2)/0.5MVA = 0.0239616 ohms
Total impedance = 0.024257 ohms
Then your 3-phase fault = (480/1.732)/0.024257 = 11.4 kA!
Motor contribution = 4 x FLA = 4 x 601 = 2,404 A ~ 2.4 kA
So your total fault at secondary is = 11.4 + 2.4 = 13.8 kA.​

Finally, you can choose a breaker with an ampere interrupting rating of at least 14 kA. But because of the way breakers are tested, (tester calibration is done with a shorting bar on the test cables), it is possible that a breaker stamped '14 kA' might actually have an interrrupting capacity of just 10kA (because of the breaker contact resistances different from that of the shorting bar)!
The next higher available breaker AIR I see is a 22 kA breaker.

 

[electrofelon]

Great replies, thanks a lot.

I hesitate to squander my clients money to cover my butt
that is what we get paid for.....anything could happen but what are the parameters now
there is a reason they are stamped on equipment
if you design for every contingency you'll waste the clients money

I agree with you a lot on this. How far do we go to provide for every possible contingency? Ok sure the transformer could fail and the utility could replace it with one with less impedance or one twice as large because its "lying around". In my 18 years in the trade I dont recall ever hearing of that happening BUT I have HAVE seen the utility reroute their distribution so now the transformer is not half a block away, or upsize a pole bank that is serving multiple customers. Maybe we should start putting in fused discos universally since fuses will series rate with many more combinations and brands of CB's. Many times prominent well respected forum members have made statements such as, "Ill be happy to come back in 18 months and make additions or changes to their electrical system - and charge them for it." Just being a bit of a devils advocate here, and not just latching on to what I want to hear. I think there is a balance in providing for "reasonable" changes in fault current, but someone has to pay for that insurance.

I do not see how you can use power factor for fault current; there is no load.

The mike holt calculator has a power factor field, and it makes a substantial difference in the fault current value. Perhaps others can explain the theory behind it.

If you use the POCO value, then if the AFC exceeds that, you are not at fault for installing undersized equipment.

I dont see any NEC requirement to use values other than that on the equipment at the time of installation.



In this specific case, The MDP will have 35K AIR so I am about at the chart value. The issue is a sub. I got a price $800 lower for a different brand, so no series rating, and the AIR is 14K. Its a decent 125 foot run, but will still be under using the utility chart value but well over using the nameplate value. I am feeling quite comfortable using this panelboard. Of course it is likely easier for someone not paying the bills to think otherwise.

 

[mivey]

At the secondary terminals:

System impedance = kV^2/MVAsc = (0.48)^2/(37,000 x 1.732 x 12,470V) = 0.002957 ohms
Transformer impedance = 0.052 x (.48^2)/0.5MVA = 0.0239616 ohms
Total impedance = 0.024257 ohms
Then your 3-phase fault = (480/1.732)/0.024257 = 11.4 kA!
Motor contribution = 4 x FLA = 4 x 601 = 2,404 A ~ 2.4 kA
So your total fault at secondary is = 11.4 + 2.4 = 13.8 kA.​

The 37 kA was at 480 volts.

 

[topgone]

The 37 kA was at 480 volts.

You got it wrong there. He said he has a 35kA rated protection on the high side and a 14kA off the 500kVA transformer. Methinks he wants to justify his choices and see if his hunch has some backing using calculations.
If you have observed, I mentioned about how breakers are tested and stamped with AIR (amperes interrupting rating). He wanted to choose the lower end of the interrupting rating range. So I cautioned against it knowing the interrupting capacities of breakers might be very much lower that what interrupting rating they are stamped with.

 

[mivey]

You got it wrong there. He said he has a 35kA rated protection on the high side and a 14kA off the 500kVA transformer.

No sir. He got the 37 kA figure from the utility at their 480 volt secondary. He did not agree and made his own 480 volt fault calc and got 14 kA.

37 kA at 12.47 kV would be abnormal. We generally try to hold the fault at the station to 10 kA or less but I have seen some at the station in the 20 kA level. 37 kA at 12.47 kV at a downline customer site would just be wierd.

 

[topgone]

37 kA at 12.47 kV would be abnormal. We generally try to hold the fault at the station to 10 kA or less but I have seen some at the station in the 20 kA level. 37 kA at 12.47 kV at a downline customer site would just be wierd.

Precisely. And the only explanation how 37 kA would exist is that it is actually the available short-circuit amperes on the 12.47 kV side.

 

[Bugman1400]

Precisely. And the only explanation how 37 kA would exist is that it is actually the available short-circuit amperes on the 12.47 kV side.

37kA @ 12.47kV sounds high unless you're near large generation or a large substation.

I also have never used PF as part of my fault calcs. Impedance Z% is the total of R%+jX% not P and Q.