Short Circuit Current at Light Post Help 2

[Mbrooke]

I'm getting only 62 amps short circuit current on my furthest light pole. Did I do my math righ? This 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.

Here is a single line:

https://imgur.com/iC6K9mj

 

[waross]

Using the voltage drop tables and the fault current returning on a #12 AWG conductor, I get 36.5 Amps just on the 120 Volt section.
450 feet may be a little bit too far for a 120 Volt circuit.
I hope that these are LED lights.

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

 

[Mbrooke]

They are LED. What software are you using? 35 amps is really low.

 

[waross]

I'm too old for soft ware.
I used the voltage drop tables in the Canadian code for a 1% drop at 10 Amps.
From that I calculated the AC effective resistance and from that the voltage at the 120 Volt panel with a line to neutral short.
If you used close to zero impedance for the fault return path than your 62 Amps is reasonable.
I don't know present practice but at one time some cities were grounding the neutral at each standard to lessen the touch potentials that could result from a line to standard fault with a long clearing time.


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

 

[waross]

75 KVA is 208 Amps full load current at 120 Volts.
At full load current you should be using regulation rather than impedance.
I assumed full voltage at the last panel to get a ballpark figure.
For a solution, I would calculate the voltage drop to each panel at full connected load.
Under the CEC you are allowed 5% voltage drop total at the end of the line.
Of that 5% neither feeders nor branch circuits may exceed 3%.
So, by code the minimum voltage at the final panel must be 97% of 120 Volts or 116.4 Volts.
If this is at the design stage, I would consider a higher voltage, or other methods to reduce the voltage drops on the branch circuits.
Years ago, in one city, it was common for #2 AWG conductors to leave the panel.
As the branch circuit progressed down the line of light standards and the load lessened, the conductor size would be reduced until the last few spans would be #12 AWG.




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

 

[Mbrooke]

I know, but I'm interested in short circuit values. 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. Assumption is that L-N fault at the spades will be equal to a 3 phase L-L-L fault with an infinite impedance. Table 9 assumes 75*C (I think) so that would make up for any MV source impedance I think.

 

[magoo2]

Mbrooke,
If you look at the fault current at the transformer on a phase to neutral basis, 120 V/23132 A = 0.00518 ohms like you calculated.

I don't have a table of small conductor sizes and I can't follow your equations. My suggestion is to simplify the setup and just consider the 450 ft of #12 copper wire. Normally the impedance is given in ohms per 1000 ft. The reactance is pretty insignificant so you can go with just the resistance. Total for copper wire is approximately equal to R/1000*0.45*2. The factor of 2 considers the neutral and phase wires. This is the lion's share of the impedance so you can see if this result is close to the 1.8 ohms that you show.

 

[Mbrooke]

This is the table I used:

https://imgur.com/ADFOJDS

 

[magoo2]

R/1000 = 2
X/1000 = 0.054

I'll ignore X because R/X is almost 40(37) so X is not significant.

Using just R
R/1000 * 0.45 * 2 = 2 * 0.45 * 2 = 1.8 ohms

Using just the #12, you get the available fault current of 120/1.8 or 66.7 A

So it looks like your calculations are close enough to being correct.

 

[waross]

450 Feet to the standard = 900 feet of #12 AWG copper.
From an internet resistance calculator 1.429 Ohms.
Current at 120 Volts = 84 Amps
That is DC resistance, not AC resistance.
The actual AC Amps will be less.
The transformer is rated at 208 Amps per phase and feeds a 100 Amp panel.
Get a grip on reality.
This is an overload as far as the transformer is concerned.
It is far short of tripping the breaker instantaneously.
The transformer impedance is not used at this current level, the transformer regulation is used.
It is less PU than the PU impedance.
This is fed from a 20 Amp breaker.
Assume 80% loading.
That will be 16 Amps.
The circuit voltage drop will be about 45 Volts with a distributed load.
Time to forget transformer Ohms and get back to basics, including the electrical code.
Once you get your voltage drop under control you will not have a problem with fault currents.
And by the way, don't forget to check the catalogue for the price of breakers that will handle over 20,000 Amps of fault current.
By the way, 450 = 0.90032799023467ohm for #12 AWG copper is acceptable.

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

 

[Mbrooke]

The transformer is rated at 208 Amps per phase and feeds a 100 Amp panel.
Get a grip on reality.

Where are you getting this from? Its a 400 amp rated main panel, with a 400 amp main disconnect.

This is an overload as far as the transformer is concerned.

What utility sizes their transformers to the service disconnecting means? Typically utility transformers come out to be 1/3 to half of the service rating, with deliberate overloading anticipated between 110 to 200% based on duration and ratio difference between peaks and troughs. They could feed this service with 3 10kva pole pigs if deemed acceptable on their part.

The transformer impedance is not used at this current level, the transformer regulation is used.

It is still part of the ground fault loop path.

That will be 16 Amps.
The circuit voltage drop will be about 45 Volts with a distributed load.

A circuit can be loaded below that. 4 amps would have me at 5% VD, which is ok for a LED sign or a few lights.

Time to forget transformer Ohms and get back to basics, including the electrical code.

Check out the NESC, different rules when heading in reverse from the service disconnecting means.

Once you get your voltage drop under control you will not have a problem with fault currents.
By the way, 450 = 0.90032799023467ohm for #12 AWG copper is acceptable.

4 amps will keep my voltage drop under control (FWIW remember the NEC does not mandate voltage drop be kept to 5%) however breaker clearing time remains an issue. There is no 5 amp breaker available and tapping off a loaded circuit still gives the same results.

And by the way, don't forget to check the catalogue for the price of breakers that will handle over 20,000 Amps of fault current.

Not needed- 25 feet of URD will drop SCC below 22kaic, and series combination ratings can even let me get away with 10kaic past the main breaker. RK low peak fuses aren't bad either- but not needed here.

 

[waross]

A 75 KVA three phase transformer is 25 KVA per phase.
25 KVA at 120 Volts = 208.333 Amps. You can't change the transformer rating by adding a 400 Amp panel.
The utility may choose to install whatever size transformer that they deem the most cost effective.

Did I do my math righ?

Your arithmetic may be right but your procedure is wrong.
You cannot add impedances with different X:R ratios.
That is also the reason that the transformer impedance is not the correct value.
The impedance is only valid for a short circuit at the transformer terminals.
Take the root of the sum of the resistances squared and the sum of the inductive reactances squared.
Your fault current will still be in the same ballpark.
Consider a GFI breaker as the most economical solution.

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

 

[Mbrooke]

A 75 KVA three phase transformer is 25 KVA per phase.
25 KVA at 120 Volts = 208.333 Amps. You can't change the transformer rating by adding a 400 Amp panel.
The utility may choose to install whatever size transformer that they deem the most cost effective.

I don't understand how (why) the size of the transformer is being called into question. The code requires that the service be sized per article 220. 99% of the time the NEC's load calcs result in values much higher than what the POCO sets down or hangs on their poles.

Your arithmetic may be right but your procedure is wrong.
You cannot add impedances with different X:R ratios.

This is why I ask here! Do you know the right way to go about doing calculating these values?

Consider a GFI breaker as the most economical solution.

That is until it the GFCI fails...

 

[Mbrooke]

I don't know present practice but at one time some cities were grounding the neutral at each standard to lessen the touch potentials that could result from a line to standard fault with a long clearing time.

Not to side track, but how does this work? Wouldn't it make things worse from step potential?

 

[waross]

Sorry we got off on the wrong foot.
I made some unsupported assumptions with the inevitable result.
I plead exhaustion.
I spent the weekend branding cattle.
Usually there are many hands to share the work and it's a lot of fun.
This year it was still fun, but with the COVID restrictions it was less than half as many workers as normal.
Everyone was exhausted.
Re: % voltage drop.
The CEC and the NEC are fairly well harmonized but it slipped my mind that one of the differences is that % voltage drop is a rule under the CEC but only a suggestion in the NEC.
If the fault current returns on the equipment grounding conductor the supply conductor and the equipment bonding conductor form a voltage divider.
When the equipment grounding conductor is the same resistance as the supply conductor, this can create a touch potential of 50% of the circuit voltage.
The circuit grounding conductor is often two gauge sizes smaller than the supply conductor creating an even higher touch potential.
Connecting each standard to a local grounding electrode will reduce the impedance of the ground return path and thus reduce the touch voltage.
I would normally consider the fault current in the #12 conductors first.
In this case where the fault current is so low I would not bother calculating the feeders.
On the other hand, for ASCC, I first calculate the transformer ASCC. If that is within the capability of my switchgear I would go no further.
If it is over the switchgear rating, I then consider the impedance of the feeders. That will often drop the ASCC to the switchgear capability.


The size of the transformer is to illustrate that the fault current is very small in relation to the transformer rated current.
While the branch circuit is overloaded, the effect on the transformer is to add some resistive load.
You must resolve the transformer impedance into resistive and inductive reactive components and add the resistance of the load to the transformer resistance.
Most of the impedance of your transformer is inductive reactance.
Adding a high resistance load and the circuit impedance becomes mostly resistive.



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

 

[Mbrooke]

Its ok and I fully understand, trust me on this

I agree that it does form a resitive divider, and with that its a good idea the breaker disconnect in less than 1 second to prevent bodily injury should someone be in contact with the pole.

How does the high inductance of the trafo alter the short circuit value?

 

[waross]

With a short circuit on the transformer terminals, the current is limited mostly by the inductive reactance.
The short circuit current is at a very low power factor.
At normal loading the resistance of the load limits the current.
If the impedance of the circuit is now calculated, the contribution of the transformer inductive reactance to the total circuit impedance is quite low.
The impedance of the transformer calculated from the transformer voltage drop with a resistive load is less than the short circuit impedance.
Link
The voltage regulation of the transformer is the percentage change in the output voltage from no-load to full-load. And since power factor is a determining factor in the secondary voltage, power factor influences voltage regulation. This means the voltage regulation of a transformer is a dynamic, load-dependent number.

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

 

[Mbrooke]

Alright, but what does all that lagging current do during a fault on large cables like 4/0? I take my simple math is not enough?

 

[waross]

Current is limited by the impedance of the entire circuit including the transformer internal impedance.
The transformer PU impedance or impedance voltage si/are used to calculate the steady state short circuit current in the event of a fault at the transformer terminals.
This is the worst case.
In the case of a fault downstream, the conductors add impedance which limits the current to below the worst case.
But you can't add impedances directly. The impedances of individual sections of the conductors represent directed values.
You may sum all the resistances and separately sum all the reactances and then apply Pythagoras' theorem to determine the impedance of the complete circuit.
Your math should be enough.
You may use your method to calculate the impedance of any section of conductors.
You may also use your method to calculate the voltage drop of any section of the circuit.
What you cannot do is add impedances with differing X:R ratios nor the resulting voltage drops.
The voltage drops will be at different phase angles and the directed sum will not equal the numerical sum.

A real world example of what phase angles can do to you:
I measured 11 Amps on one leg of a three wire, 120/240 Volt single phase circuit.
I measured 9 Amps on the other leg of the same circuit.
We all know from high school electricity that the current on the neutral must be 11 Amps minus 9 Amps or 2 Amps.
In high school they never confused us with power factor.
I did confuse a science teacher with power factor and a paper on the difference between KVA and kW as it related to an obsolete distribution transformer that the local electrical department had given me.
I doubt that he ever again gave an assignment with such a wide range of student choice of topic. grin

Later, in college or trade school we learned about power factor, but I don't remember it ever being specifically related to neutral current in single phase circuits, three phase circuits, yes.
Back to the real world example and the expected 2 Amps on the neutral.
You may draw a triangle with sides of 7 units, 9 units and 11 units.
This will be a representation of a head to tail vector sketch of the currents.
One side of the triangle will be close to horizontal.
I forget which.
On a purely resistive circuit the 9 Amp vector would return on the 11 Amp vector at 180 degrees and the result would be the 2 Amp difference that we are expecting.
All related tp phase angles, X:R ratios and power factor.
If you can get your head around this example it may help to explain the issues with your calculations.
I measured 7 Amps on the neutral.
What Total Foolishness. (WTF)
On closer examination I discovered that the load on one leg was a motor running at a very good power factor and the load on the other leg was a motor running at a very bad power factor.
Again, at different phase angles, currents and/or voltage drops may not be added arithmetically.
The arithmetic sum will be greater than the actual directed sum.
Likewise, at different phase angles, currents and/or voltage drops may not be subtracted arithmetically.
The arithmetic difference will be less than the actual directed difference.


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

 

[Mbrooke]

You may sum all the resistances and separately sum all the reactances and then apply Pythagoras' theorem to determine the impedance of the complete circuit.
Your math should be enough.

I may try this. How much does it vary from my method? I like your idea.

You may use your method to calculate the impedance of any section of conductors.
You may also use your method to calculate the voltage drop of any section of the circuit.
What you cannot do is add impedances with differing X:R ratios nor the resulting voltage drops.
The voltage drops will be at different phase angles and the directed sum will not equal the numerical sum.

Understood. This applies to short circuits as well, correct? In that the reactance in the circuit effects the entire circuit as apposed to looking at it in a "point to point" way?

If my method does not differ much from the mathematically correct method, would a "rule of thumb" multipler or divider cover these errors?

I'll come clean- I probably should have from the beginning. I want to make a table that electricians can use on site...

An electrician can, for example: a single L-G fault choose 0.016 ohms for the POCO transformer, 0.002 ohms for the service, 0.035 ohms for the feeder (phase + EGC) and 0.25 ohms for the branch circuit (phase + EGC) to its furthest point.

Simple ohms law would take you to the required destination.


This would allow an Electrician to calculate short circuit values in the field with simple math instead of complex engieering software even if it results in slight error, but still under the AIC equipment rating in most cases.

I notice there is a lack of short circuit and disconnection times missing from most prints and no way to get those values in the field easily.

Over dutied equipment and insufficient fault current to promptly open a breaker are two very real issues encountered in an electrician's career.

Regarding you last example- why doesn't the neutral ever overload? Sounds like it would based on where the math is headed.