Transformer Sized for Load?

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fifty60

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USA
I am having to add a transformer to a piece of equipment. The transformer is 3KVA, and will take 575V down to 230V. I will be powering three 1HP motors on the secondary. When I size the disconnect for the equipment, which also includes heater loads and compressors, should I use the FLA of the transformer primary (3000/575) or should I use the actual FLA's of the motors?

By the way, the motors fed by this transformer are all 1HP, and are not the largest motor inside the equipment.

If I used the Amp load of the motors on the transformer secondary, would I have to use the same ratio of 575/230 to calculate the amps being drawn on the primary? The amps would be reduced by that ratio going from secondary to primary.
 

Jraef

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Electrical Engineer
Yes, your current ratio follows the voltage ratio.

A 3kVA transformer is however not going to be big enough to start 3 x 1 HP motors in my opinion, especially if there are other loads on it. The quick and dirty rule of thumb is that if you are starting across the line, 3kVA / HP is necessary if you want to keep the VD during starting less than 5% on the secondary side. You can go smaller if you can stagger the starting (or they don't all run at the same time anyway), you can tolerate LV during starting or if you are using soft starters or VFDs, but I would never go below 1.5 kVA / HP.
 

fifty60

Senior Member
Location
USA
Jreaf, thanks for the help. I was mistaken however, there are only two 1HP 3-phase motors. I do not have the motor specs in front of me, but the amps * voltage of each motor came out to approximately 1000VA each. With 2 motors, i have 2000VA, so using a 3KVA transformer should be plenty right?

The transformer and motor are also 3 phase. I would think that 3 phase motors would initially create less inrush current than a single phase motor. Is that correct? For 3 phase there is a naturally occuring rotating magnetic and electric field, whereas with the single phase there has to be a shaded pole or start capacitor to get the rotation started. Is that a correct analysis?

Would taking the full FLA of the transformer, and not the transformer load, be correct for sizing the equipment disconnect? Or should i take the transformer load, and reduce the amps by the primary to secondary ratio and add that to the other branch circuits in the equipment?
 
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fifty60

Senior Member
Location
USA
My question can be summed up as follows: I have a branch circuit inside of my equipment that consists of a 3KVA transformer. The 3KVA transformer has 2 motors across its secondary. Is it acceptable to use the 3KVA/Line voltage= FLA of Transformer as the FLA for the branch FLA contribution from the transformer branch to the overall equipment FLA?

Or, would it be better to take the FLA of the loads across the secondary, and step that down with the primary/secondary ratio and use that as the branch full load contribution?
 

david luchini

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Location
Connecticut
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Engineer
My question can be summed up as follows: I have a branch circuit inside of my equipment that consists of a 3KVA transformer. The 3KVA transformer has 2 motors across its secondary. Is it acceptable to use the 3KVA/Line voltage= FLA of Transformer as the FLA for the branch FLA contribution from the transformer branch to the overall equipment FLA?

Or, would it be better to take the FLA of the loads across the secondary, and step that down with the primary/secondary ratio and use that as the branch full load contribution?

The motors are the load, not the transformer. Use the motors as your basis for the equipment load.
 

fifty60

Senior Member
Location
USA
So my primary to secondary ratio is 575/230= 2.5. So if my load is drawing 6A, would I then make my contribution to the equipment fla 6/2.5 = 2.4A?
 

GoldDigger

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Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
Yes. If you have a large transformer which is lightly loaded you may need to consider the magnetizing current of the transformer too. (It will decrease the apparent power factor seen by the primary feed and therefore increase the current some.)
 
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