62 amps short circuit current?

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mbrooke

mbrooke

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Alright, I'm getting 62 amps short circuit current on my furthest light pole. Did I do my math righthis number sounds awfully low.


For the POCO: 75 kva 3 phase transformer, 1% Z, 23,132 amps 3 phase fault.

Using ohms law I get a transformer value of: 0.00519 ohms


For the service:

Plastic PVC conduit, 600MCM copper- (0.039x0.039)+(0.023x0.023)=0.00205 root= 0.0452769256906871 /1000 x 25 feet = 0.0011319231422672 ohm

400MCM copper- (0.040x0.040) + (0.033x0.033)= 0.002689 root = 0.0518555686498567‬ /1000 x 25 feet= 0.0012963892162464‬ ohm

For the feeder:

# 3 copper in steal conduit- (0.059x0.059)+(0.25x0.25)= 0.065981 root= 0.2568676702117259 /1000 x 100 = 0.0256867670211726‬ ohm

# 8 copper EGC- (0.065x0.065)+(0.78x0.78)= 0.612625‬ root = 0.7827036476214992 ‬ /1000 x100= 0.0782703647621499 ohm

The branch circuit:

#12 copper THHN/THWN, PVC conduit-

live- (0.054x0.054)+(2.0x2.0)= 4.002916 root= 2.000728867188156‬ /1000 x 450 = 0.90032799023467ohm

ground- 0.90032799023467 ohm

Grand total= 1.912231424058514‬ ohms


Using ohms law at 120 volts I get 62.75396 amps. o_O

Here is a single line diagram:

https://imgur.com/iC6K9mj

And the table used:

https://imgur.com/ADFOJDS


 
Last edited:
Here is the time current curve for a bolt on
mikeames said:
I don't think your taking into account the let through on the utility transformer. I think this site lays it out simply https://iaeimagazine.org/magazine/features/calculations/calculating-short-circuit-current/

It appears you simply took 120v and the load side resistance and divided. That's not correct.
Click to expand...

I took the short circuit current of the transformer and then used ohms law (120 volts) to get 0.00519 ohms. Not sure if I did it right.
 
Yea I just edited my post. With that length on 12 it makes sense. You wont get much. In my earlier years I remember shorting out a circuit like that, and listening to it hum for seconds before the breaker tripped. After thinking about it I learned intuitively how distance and a small conductors can influence, AFC.
 
mikeames said:
Yea I just edited my post. With that length on 12 it makes sense. You wont get much. In my earlier years I remember shorting out a circuit like that, and listening to it hum for seconds before the breaker tripped. After thinking about it I learned intuitively how distance and a small conductors can influence, AFC.
Click to expand...

Yup, and that is my concern. A ground fault would produce 60 volts to remote earth, and its best the breaker trips in under a second.
 
mikeames said:
After thinking about it I learned intuitively how distance and a small conductors can influence, AFC.
Click to expand...
I would think that with the small gauge conductor at 450', the voltage has dropped so much that the amperage increases , trying to accommodate the load demand therefor over heating the breaker. Just curious what is the wattage of the pole light and what was the voltage at that distant pole.
 
Another C10 said:
I would think that with the small gauge conductor at 450', the voltage has dropped so much that the amperage increases , trying to accommodate the load demand therefor over heating the breaker. Just curious what is the wattage of the pole light and what was the voltage at that distant pole.
Click to expand...

Right, but you can still have 5% VD and not trip a breaker.
 
mbrooke said:
Right, but you can still have 5% VD and not trip a breaker.
Click to expand...
I would agree, but to have a 62 A spike before tripping the breaker, something odd going on there, not just a voltage drop unless those light are a pretty massive wattage, I'd almost suggest there could be a fault bleeding off through damaged insulation.
 
Another C10 said:
I would agree, but to have a 62 A spike before tripping the breaker, something odd going on there, not just a voltage drop unless those light are a pretty massive wattage, I'd almost suggest there could be a fault bleeding off through damaged insulation.
Click to expand...

Thats the scenario I'm planning for. A bolted fault at the post.
 
This is all fairly normal. It actually happens all the time. The math I usually use to demonstrate is think about a 100 A, 480 V motor on a circuit breaker. So depending on jurisdiction a 300 A breaker might be typical. If we have 1 ohm of impedance in the wire and bonding then we get 277 V divided by 1 ohm or 277 A, so the breaker will never trip on ground fault. If we used magnetic only, it definitely won’t trip. In the example the overload relay will respond, eventually. Same as your case.

The issue here is we have different faults (ground fault, short circuit, overload/overcurrent) but we are double dipping by trying to use one device for everything. The (monadjustable) short circuit protection is clearly just plain inadequate and doing nothing but checking off an inspectors box. The overcurrent is doing everything but it’s designed to protect wiring from melting which is not a fast process. The solution is to use ground fault protection in addition to the others.

At high currents NEC already recognizes the ground fault issue even before AFCIs and GFCIs became popular with the lawyers. But where they miss the mark is that ground faults not tripping anything is a much bigger issue. It probably would have become a rule a long time ago but for one issue that you are touching on. How do we turn the math into a simple calculation or a table? One of your obvious mistakes is you really need to use complex numbers first then convert to magnitude at the end but it works this time because all loads are resistive....in distribution we often do the opposite and ignore R since the impedances that drive everything are inductive. But the way engineers have done these calculations for years (ANSI or IEC) uses a lot of short cuts and estimation techniques. These are holdovers from the slide rule days and the computer models just copy the same stuff. It would be hard to jam it into NEC and we are bordering on engineering. It has already crept in the back door via the arc flash rule and SCCR labels. But doing breaker short circuit math takes it to a whole new level where engineers would be required for every job, I’m all for promoting the profession but at some point we need to be practical. I cannot imagine doing short circuit calculations on houses. That would be crazy.
 
Another C10 said:
I would think that with the small gauge conductor at 450', the voltage has dropped so much that the amperage increases , trying to accommodate the load demand therefor over heating the breaker. Just curious what is the wattage of the pole light and what was the voltage at that distant pole.
Click to expand...
Current will only go up with an inductive load when the voltage drops. With a resistive load if the voltage goes down so does the current. Additionally even with inductive the impedance of the small wire and and long run will limit the current. In the same context you could plug a motor load such as a compressor into the end of the long run and it may fail to start under load, but not trip the breaker.
 
mikeames said:
Current will only go up with an inductive load when the voltage drops. With a resistive load if the voltage goes down so does the current. Additionally even with inductive the impedance of the small wire and and long run will limit the current. In the same context you could plug a motor load such as a compressor into the end of the long run and it may fail to start under load, but not trip the breaker.
Click to expand...

Inductors by definition are impedances, too. If it was just an inductor current would go down, too. I=V/Z. However in a motor the torque curve drops. So the motor slips more which increases the current draw.

Similarly if the light has an electronic ballast or an LED driver these will increase current draw to compensate even though these are resistive loads.
 
paulengr said:
This is all fairly normal. It actually happens all the time. The math I usually use to demonstrate is think about a 100 A, 480 V motor on a circuit breaker. So depending on jurisdiction a 300 A breaker might be typical. If we have 1 ohm of impedance in the wire and bonding then we get 277 V divided by 1 ohm or 277 A, so the breaker will never trip on ground fault. If we used magnetic only, it definitely won’t trip. In the example the overload relay will respond, eventually. Same as your case.

The issue here is we have different faults (ground fault, short circuit, overload/overcurrent) but we are double dipping by trying to use one device for everything. The (monadjustable) short circuit protection is clearly just plain inadequate and doing nothing but checking off an inspectors box. The overcurrent is doing everything but it’s designed to protect wiring from melting which is not a fast process.
Click to expand...

Sadly, I have to agree. This is rather common in long runs...

The solution is to use ground fault protection in addition to the others.
Click to expand...

I'm prepared to debate that.😁


At high currents NEC already recognizes the ground fault issue even before AFCIs and GFCIs became popular with the lawyers. But where they miss the mark is that ground faults not tripping anything is a much bigger issue. It probably would have become a rule a long time ago but for one issue that you are touching on.
Click to expand...

The thing is they already know, hence why GFCIs and special-purpose GFCIs are gradually being mandated even in applications where historically missing EGCs have not been common.


How do we turn the math into a simple calculation or a table? One of your obvious mistakes is you really need to use complex numbers first then convert to magnitude at the end but it works this time because all loads are resistive....in distribution we often do the opposite and ignore R since the impedances that drive everything are inductive. But the way engineers have done these calculations for years (ANSI or IEC) uses a lot of short cuts and estimation techniques. These are holdovers from the slide rule days and the computer models just copy the same stuff. It would be hard to jam it into NEC and we are bordering on engineering. It has already crept in the back door via the arc flash rule and SCCR labels. But doing breaker short circuit math takes it to a whole new level where engineers would be required for every job, I’m all for promoting the profession but at some point we need to be practical. I cannot imagine doing short circuit calculations on houses. That would be crazy.
Click to expand...

Not unless you condense everything to a simple table, with a simple equation like Zs=Ze+(R1+R2). Which is what I'm trying to do for electricians, but, well, I need help.

The bigger issue is the NFPA mandating AFCIs and GFCIs to fix an issue that either doesn't exist or can be solved with simple math.
 
The fault current is 62A. The branch circuit breaker rating is 20A ie the fault current is more than 3 times the CB rating. So no issue. The 20A CB would trip almost instantaneously.
 
Sahib said:
The fault current is 62A. The branch circuit breaker rating is 20A ie the fault current is more than 3 times the CB rating. So no issue. The 20A CB would trip almost instantaneously.
Click to expand...


The breaker would take between 5 and 30 seconds to trip. Such long duration violates the IEC body graph:

https://en.wikipedia.org/wiki/Electrical_injury#/media/File:IEC_TS_60479-1_electric_shock_graph.svg

In the Indian Electrical Code this would also violate Table 41.1 I'd think.
 
Sahib said:
The fault current is 62A. The branch circuit breaker rating is 20A ie the fault current is more than 3 times the CB rating. So no issue. The 20A CB would trip almost instantaneously.
Click to expand...
That is not necessarily true. The available short circuit currents calcs are way conservative. So 60 amps might really be 40 amps and that might take minutes or hours to trip.

No matter what the code explicitly requires an effective ground fault. This is not an effective gf path so is not code compliant.
 
petersonra said:
That is not necessarily true. The available short circuit currents calcs are way conservative. So 60 amps might really be 40 amps and that might take minutes or hours to trip.

No matter what the code explicitly requires an effective ground fault. This is not an effective gf path so is not code compliant.
Click to expand...

Correct, the code wants a breaker to trip based on 250.4 (A) (5) and 250.4 (B) (4). However, it doesn't actually specify a trip time.

https://imgur.com/t5RbR6p

So in theory 40 amps would not violate the NEC, even though it is a real danger.
 
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