75KVA Padmount transformer feeding 800A service

topgone

Senior Member
Right, but often you need to know the exact number. Higher than actual numbers can increase the arc flash energy.
Given a higher number will ensure that you will secure electrical equipment that are rated above or at that given number! But the actual fault available is lower, is it not? Fault current withstand and arcflash protection do not sit well with each other, I was told!
 

mbrooke

Batteries Not Included
Location
United States
Occupation
Electricity
Given a higher number will ensure that you will secure electrical equipment that are rated above or at that given number! But the actual fault available is lower, is it not? Fault current withstand and arcflash protection do not sit well with each other, I was told!

Lower actual fault current can increase the amount of time a device takes to open increasing the incident energy.

Remember that services are already over sized by a factor or 2-3 times. POCOs on the other hand like to undersize their transformers for the actual load being served. %Z goes up as trafo kva goes up (none linear increase in fault current). Approximately a 1:4 kva ratio difference.

The result can be a main device where the instantaneous range or sub cycle range is above what the utility can provide.

Differential protection will solve weak source, but its not applied in all gear.
 

topgone

Senior Member
Lower actual fault current can increase the amount of time a device takes to open increasing the incident energy.

Remember that services are already over sized by a factor or 2-3 times. POCOs on the other hand like to undersize their transformers for the actual load being served. %Z goes up as trafo kva goes up (none linear increase in fault current). Approximately a 1:4 kva ratio difference.

The result can be a main device where the instantaneous range or sub cycle range is above what the utility can provide.

Differential protection will solve weak source, but its not applied in all gear.
Not sure i follow your thought process there! The thing with POCOs is that they will give you numbers that they have determined in advance from their tables (if they don't have an updated system model)! There is no mention of oversizing transformers there. The rest follows i.e. tripping time wrt to trip setting and actual fault current! Been there, done that thing!
 

mbrooke

Batteries Not Included
Location
United States
Occupation
Electricity
Not sure i follow your thought process there! The thing with POCOs is that they will give you numbers that they have determined in advance from their tables (if they don't have an updated system model)! There is no mention of oversizing transformers there. The rest follows i.e. tripping time wrt to trip setting and actual fault current! Been there, done that thing!

There is no such thing as a one size fits all table.
 

Hv&Lv

Senior Member
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Engineer/Technician
Here is one. This is the email from the POCO:

" The table I’m looking at shows a max short circuit current of 37,600 Amps at the transformer for a 3-phase 500kVA 480/277 V transformer. "

Calculated off dataplate using infinite buss: 11,558

Here is another one:

" The fault current at the secondary spades of the transformer is: 111,000a."

Calculated off date plate using infinite buss: 77,105.

Its just really annoying. Yeah I concur providing some extra for next size up or lower impedance down the road is prudent, but some of these values are out of control.
I hate to rehash this but I was looking at the numbers here..
People use cheat sheets a lot (charts).
Doesn’t know they know how to use them.

Some POCO charts show to divide the kva by three and use the l-g voltage.
So for a 500 you have 166.66kVA
Going on with the calc it has (assuming ~5.2 Z) ~ 11,548A
About in line with what you have..

It almost seems as though they are using ~4.x Z as default and multiplying by three incorrectly.
Either that or they assume about a 1.5-9%Z...

So your 11,558 becomes 34,674A
They may be looking at 12,533 and multiplying by three.
I know it’s wrong, but I’m trying to figure out where their number comes from.
 
I know it’s wrong, but I’m trying to figure out where their number comes from.
Yeah that's what drives me crazy, is where they are getting these values from. I admit this case is particularly egregious. Pole and submersible units seem to have much lower impedance than pads, maybe he was looking under the pole/UG banks chart? 35k is about right for a 3x166 pole/UG bank at 480/277
 

mbrooke

Batteries Not Included
Location
United States
Occupation
Electricity

topgone

Senior Member
Here is one. This is the email from the POCO:

" The table I’m looking at shows a max short circuit current of 37,600 Amps at the transformer for a 3-phase 500kVA 480/277 V transformer. "

Calculated off dataplate using infinite buss: 11,558

Here is another one:

" The fault current at the secondary spades of the transformer is: 111,000a."

Calculated off date plate using infinite buss: 77,105.

Its just really annoying. Yeah I concur providing some extra for next size up or lower impedance down the road is prudent, but some of these values are out of control.
That 37,600A fault amps available is correct with a 500 kVA 480/277-240 transformer! Say, the %Z of the transformer is 3.2%, the fault at the secondary will be 500,000/(1.732 x 240)/0.033 = 37,600A (at the 240V side)!
If the transformer in question is an MV-LV with secondary voltage at 480/277V, the fault current available would then be 500,000/(1.732 x 480)/0.032 = 18,800A for a transformer with %Z = 3.2%! Your 11,558A number is good with a 4% Z also.

Second, the fault on the primary side depends on the POCOs system impedance and the voltage at the point-of-common-coupling (POCC)! You cannot compute for the available fault at your connection point (primary side) using the your load impedance alone even if your load are purely motors (at 100% motor load). If it were, your motor load's contribution to the system fault would be 500,000/(1.732 x 480)/0.2 = 3,007A only (at the 480V side and assuming you neglect the impedance of the lines and the per unit impedance of the motors set at 20%)!
 

mbrooke

Batteries Not Included
Location
United States
Occupation
Electricity
That 37,600A fault amps available is correct with a 500 kVA 480/277-240 transformer! Say, the %Z of the transformer is 3.2%, the fault at the secondary will be 500,000/(1.732 x 240)/0.033 = 37,600A (at the 240V side)!
If the transformer in question is an MV-LV with secondary voltage at 480/277V, the fault current available would then be 500,000/(1.732 x 480)/0.032 = 18,800A for a transformer with %Z = 3.2%! Your 11,558A number is good with a 4% Z also.

Second, the fault on the primary side depends on the POCOs system impedance and the voltage at the point-of-common-coupling (POCC)! You cannot compute for the available fault at your connection point (primary side) using the your load impedance alone even if your load are purely motors (at 100% motor load). If it were, your motor load's contribution to the system fault would be 500,000/(1.732 x 480)/0.2 = 3,007A only (at the 480V side and assuming you neglect the impedance of the lines and the per unit impedance of the motors set at 20%)!
How many feet of POCO service conductor?
 
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