Transformer Primary OCPD Sizing

[TheElectrician]

Hello,
I am doing a solar project and trying to size my primary OCPD.
I have a 480V step-down transformer to 208V. The 480V is the inverter side and 208V is the utility side.
I need a 340KVA transformer using my inverter FLA but the next size available is 500KVA. I am trying to understand what values should I use to size my OCPD and conductors on the 208 side. Should I be using the 340KVA or the transformer nameplate which is 500KVA.
Thank you for any assistance.

 

[jim dungar]

The transformer does not produce or consume current. For the most part, the conductors and OCPD simply need to be sized to the loads.
HOWEVER, you do need to consider the inrush current of a transformer when making your decisions.

Allow me to be a little loose with the math, just to illustrate my point.
Transformer full load amps = 500A
Transformer inrush = 10X FLA
Protective device Instantaneous pick up point = 15X continuous amps.

So your 'primary side' protective device and conductors would need to be a minimum of (500 x 10)/15 = 333.33A, which rounds up to 350A, in order to just turn the transformer on. But if your OCPD had a max Inst. rating of 10X, you would need a 500A device.

 

[jap]

The transformer does not produce or consume current.

I don't know about "consume" current, but, a transformer windings are a load in themselves regardless of whether or not a load is attached to the secondary.

JAP>

 

[jim dungar]

...a transformer windings are a load in themselves ...

Do you routinely account for the losses of transformer windings when selecting your overcurrent protection and your conductor sizes?

 

[jap]

No, just sayin.

JAP>

 

[Carultch]

The transformer does not produce or consume current. For the most part, the conductors and OCPD simply need to be sized to the loads.
HOWEVER, you do need to consider the inrush current of a transformer when making your decisions.

Allow me to be a little loose with the math, just to illustrate my point.
Transformer full load amps = 500A
Transformer inrush = 10X FLA
Protective device Instantaneous pick up point = 15X continuous amps.

So your 'primary side' protective device and conductors would need to be a minimum of (500 x 10)/15 = 333.33A, which rounds up to 350A, in order to just turn the transformer on. But if your OCPD had a max Inst. rating of 10X, you would need a 500A device.

How realistic are the 10X inrush and 15X instantaneous trip figures? Do these apply to most transformers, breakers, and fuses across the industry?

Because I've checked your calculation, maintaining the 10X and 15X factors, and it seems like just rounding up to the next standard transformer KVA rating (as you usually have to do) is generally not enough to require any larger OCPD on the grid-side, than you'd already be needing for 125% of the operating current that you'd be using anyway.

1.25/(10/15) = 1.875, and the ratio between adjacent standard KVA ratings is generally between 1.2 and 1.67. There are a couple of exceptions where the ratio is 2, and even then rounding up to the next standard OCPD is enough. So in general, as long as 10X and 15X are typical of the industry, and given the 125% continuous load factor applying by default, I wouldn't expect this calculation to impact a design.

It seems that where you would find this calculation affecting a design, is if you are using a transformer rating that is in excess of simply rounding up to the nearest standard kVA. E.g. you not only round up to 500kVA, you also skip that size, and use 750kVA. I could see specifying this for future expansion, but for that, you'd already be specifying a grid-side circuit for the full capacity you eventually expect. The only other reason I'd see you doing this, is if you were using a reconditioned transformer, and the one that is available is a size larger than you need.

 

[jim dungar]

Inrush currents for transformers can vary quite a bit although 12X is a common plot point on time current curves. Your vendor should be able to provide more accurate data. 15X may be achievable with electronic trip circuit breakers, while thermal magnetic breakers are more likely to be only 10X.

Start with your transformer then work backwards to your breaker.

 

[GoldDigger]

The transformer does not produce or consume current. For the most part, the conductors and OCPD simply need to be sized to the loads.
HOWEVER, you do need to consider the inrush current of a transformer when making your decisions.

Allow me to be a little loose with the math, just to illustrate my point.
Transformer full load amps = 500A
Transformer inrush = 10X FLA
Protective device Instantaneous pick up point = 15X continuous amps.

So your 'primary side' protective device and conductors would need to be a minimum of (500 x 10)/15 = 333.33A, which rounds up to 350A, in order to just turn the transformer on. But if your OCPD had a max Inst. rating of 10X, you would need a 500A device.

When using a transformer designed (and optimized) for step down use with the magnetizing current being supplied to the low voltage side, you have a greater problem with inrush. The inrush current can be up to four times the corresponding inrush of the designed primary side. This is separate from the change in inrush current between primary energization and secondary energization.
In the OP's case, for example, if the inrush on the higher voltage side would be 5x the FLA of that winding, the inrush on the lower voltage side may be 20x the FLA of the low voltage winding.
This is one of the performance details that led the Code to prohibit "backwards" use of a transformer unless the manufacturer specifically allows such use in their documentation and instructions. The problems related to energizing the secondary can be minimized by proper design, which is not usually the most efficient core and winding design for primary-only energization.