AIC fault changes

[mannyb]

our power decided at last minute to change the location of our meter loop and service closer to there utility pole. so instead of being 410' away from utlity pole. now the service will be 40' away from utility pole. which will be lot closer. our local inspector said to check the value in fault calculations if it changed we need to resubmit those affected pages to building dept. Everything else stays the same. but I noticed the values in caluculation dont match as far as fault #1 not being 115' its more like 410' +/-. #2 and # 3 alot longer also. My question besides the error in length, how do they get these values in first place?

 

[GoldDigger]

It looks like the second and third calculations each take the output number of the previous calculation as their input. Apparently these are subpanels one step down from each other but very close, like distribution panels or MCCs next to the main?
I would guess that the 115 feet was a guesstimate number that was known to be less than the actual length and yet large enough to produce acceptable fault current numbers.
Replacing 115 with 40 would correct Fault #1, and then using the successive output numbers as the new input numbers in the Fault #2 and #3 calculations without changing the F or M numbers will give you a conservative number.
If the results are going to be problematic, see whether putting in the right lengths for 2 and 3 helps.

 

[mannyb]

It looks like the second and third calculations each take the output number of the previous calculation as their input. Apparently these are subpanels one step down from each other but very close, like distribution panels or MCCs next to the main?
I would guess that the 115 feet was a guesstimate number that was known to be less than the actual length and yet large enough to produce acceptable fault current numbers.
Replacing 115 with 40 would correct Fault #1, and then using the successive output numbers as the new input numbers in the Fault #2 and #3 calculations without changing the F or M numbers will give you a conservative number.
If the results are going to be problematic, see whether putting in the right lengths for 2 and 3 helps.

thanks for replying. here is the Skinny

1.utility pole to meter loop/disconnect within 30' with PVC fault #1
2. from meter can to disconnect 3' metallic conduit fault #2
3. servicce disconnect to disconnect 370' fault #3
4/0 wire.

. we are going to resubmit to A and E firm but im just trying to figure out the calculations in general. Where do you get the 3624 amps in fault #1. do you get from utlity provider and what do you ask for??

 

[GoldDigger]

Where do you get the 3624 amps in fault #1. do you get from utlity provider and what do you ask for??

I think that is the only acceptable source for the information. Who you ask depends on your POCO, and you ask for available fault current at the meter (or service point?). If it is three phase, there may be more than one number.

 

[mannyb]

I think that is the only acceptable source for the information. Who you ask depends on your POCO, and you ask for available fault current at the meter (or service point?). If it is three phase, there may be more than one number.

Thanks for input. Is this information usually provided by A and E firm or is this something the EC can perform and or provide.

 

[kwired]

I think that is the only acceptable source for the information. Who you ask depends on your POCO, and you ask for available fault current at the meter (or service point?). If it is three phase, there may be more than one number.

Thanks for input. Is this information usually provided by A and E firm or is this something the EC can perform and or provide.

To be extremely accurate you need to know about things you don't have easy access to, but you can get a reasonable worst case value just by knowing transformer kVA rating, transformer impedance, conductor properties (primarily size, type length) and use of some easily available on line calculators.

 

[GoldDigger]

To be extremely accurate you need to know about things you don't have easy access to, but you can get a reasonable worst case value just by knowing transformer kVA rating, transformer impedance, conductor properties (primarily size, type length) and use of some easily available on line calculators.

Would you guess at the transformer properties by observation or would you get that information from POCO?

 

[kwired]

Would you guess at the transformer properties by observation or would you get that information from POCO?

I have done both, whatever works. If I am there at any time they are there with the doors open I like to verify the KVA and impedance for myself, it is right on the nameplate, but behind the doors when they are closed and locked.

Trying to find out in advance usually don't work, all you get is kVA they ordered, they won't know impedance until they see the unit (though they probably can find out if they would either contact manufacturer/supplier or look at data sheets - they usually won't).

 

[mannyb]

I have done both, whatever works. If I am there at any time they are there with the doors open I like to verify the KVA and impedance for myself, it is right on the nameplate, but behind the doors when they are closed and locked.

Trying to find out in advance usually don't work, all you get is kVA they ordered, they won't know impedance until they see the unit (though they probably can find out if they would either contact manufacturer/supplier or look at data sheets - they usually won't).

on a new service and power company doesnt provide AFC. whats worse case? where or how would you get AFC to start calculation ??

 

[kwired]

on a new service and power company doesnt provide AFC. whats worse case? where or how would you get AFC to start calculation ??

If you know they are providing (for example) a 100 kVA transformer, you can start with lowest impedance transformers of that rating and have a pretty good idea of what worst case may be, throw in your conductor sizes and distances and if you are well under - you are probably safe, if you are close - you may want to try to gain more information before making purchasing decisions on your gear.

 

[mannyb]

If you know they are providing (for example) a 100 kVA transformer, you can start with lowest impedance transformers of that rating and have a pretty good idea of what worst case may be, throw in your conductor sizes and distances and if you are well under - you are probably safe, if you are close - you may want to try to gain more information before making purchasing decisions on your gear.

the service will be a 200a 240v single phase service with pvc conduit and 4/0 thhn wire ( per drawing ) what would i need from POCO to start my calculation. the transfomer will be pole mounted. what question do ask the service coordinator to get that info. I am trying to work out problem just for personnel use.

 

[pkelectrical]

They figured that fault current at transformer's secondary terminals given the kVA rating and impedance and they got 3623 AMPS RMS.
Not knowing the KVA of transformer or impedance I can give you hypothetical example:
50kVA /240V = 208 amps (full load secondary current)

Now to figure it out short-circuit current at the transformer’s secondary terminals:
Assume hypothetical impedance of 5.75% for the 50kVA transformer
208 amps /0.0575 = 3623 AMPS RMS (it is what they have)

They used f and m values to calculate the available short circuit current at the meter based on how many conductors per phase, their resistance length and the voltage supplied. The f equation is different for single phase and 3 phase.

F= (2x 410’(length of conductors)x 3623(short circuit rating))/
(12844 (C value of service conductors in this case)x240(voltage)) = 0.964

Notice that their f falue is 0.27 because it is based on 115’ of conductors, I did based on 410’ based on what you said
M= 1/((1+0.964 (f value))=0.509

Available short circuit at the meter = 3623 x 0.509=1845 AMPS RMS (the short circuit based on 410’ conductors)
Note they have 2852 AMPS RMS because they used 115’ instead of 410’

Now if you have the conductors at 40’, you can plug the 40 into the f equation and you will get f value of 0.094
The M value will be 0.914
The available short circuit at the meter will be =3311 AMPS RMS

 

[kwired]

They figured that fault current at transformer's secondary terminals given the kVA rating and impedance and they got 3623 AMPS RMS.
Not knowing the KVA of transformer or impedance I can give you hypothetical example:
50kVA /240V = 208 amps (full load secondary current)

Now to figure it out short-circuit current at the transformer’s secondary terminals:
Assume hypothetical impedance of 5.75% for the 50kVA transformer
208 amps /0.0575 = 3623 AMPS RMS (it is what they have)

They used f and m values to calculate the available short circuit current at the meter based on how many conductors per phase, their resistance length and the voltage supplied. The f equation is different for single phase and 3 phase.

F= (2x 410’(length of conductors)x 3623(short circuit rating))/
(12844 (C value of service conductors in this case)x240(voltage)) = 0.964

Notice that their f falue is 0.27 because it is based on 115’ of conductors, I did based on 410’ based on what you said
M= 1/((1+0.964 (f value))=0.509

Available short circuit at the meter = 3623 x 0.509=1845 AMPS RMS (the short circuit based on 410’ conductors)
Note they have 2852 AMPS RMS because they used 115’ instead of 410’

Now if you have the conductors at 40’, you can plug the 40 into the f equation and you will get f value of 0.094
The M value will be 0.914
The available short circuit at the meter will be =3311 AMPS RMS

Only thing I disagree with is your assumed impedance of transformer of 5.75%. 1.5 - 2% is probably more realistic value that you would end up with on that size of unit. If you don't know what it will be start with low impedance for estimating purposes - if your numbers at low impedance allow the equipment you wish to use the fault current will only get lower if a higher impedance transformer gets installed.

 

[pkelectrical]

Only thing I disagree with is your assumed impedance of transformer of 5.75%. 1.5 - 2% is probably more realistic value that you would end up with on that size of unit. If you don't know what it will be start with low impedance for estimating purposes - if your numbers at low impedance allow the equipment you wish to use the fault current will only get lower if a higher impedance transformer gets installed.

Yeah I just did this as "hypothetical" example. I wanted to get the 3326 right so I had to work backwards and got 5.75% so I used that in the example to show him how the calculation works so he can figure it out later when he has the right info.

 

[kwired]

Yeah I just did this as "hypothetical" example. I wanted to get the 3326 right so I had to work backwards and got 5.75% so I used that in the example to show him how the calculation works so he can figure it out later when he has the right info.

You said they got 3326( actually you said 3623), is they the POCO?

One thing the fault current calculators or formulas commonly used don't consider is limitations on the POCO distribution - is possible your 3623 figure had some of those factors built into it. Motor contribution also isn't always in a calculator or formula either.