277/480 Calculation

[dsmith411]

I'm figure transformer size for some parking lot lighting on a 277/480v. A-has 52 amps, B-has 50 amps, C- 48 amps. Do I us 277 in calculating what Kva transformer I need or use 480v ? I'm going to heat the lights with 277v. If you would show me the best way to calculate this.
Thanks

 

[iwire]

Fixture wattage x number of fixtures / 1000 = load KW Use a transformer larger than that.i am skipping power factor just because I am a rebel.


If I had 480 available I would use it in place of 277, you should be able to use smaller conductors at 480.

 

[Pharon]

I'm not sure where you're getting those amperage readings from, but roughly the kVA of load you have is 52A/(.480 x sqrt(3)) = 62.6kVA

Next transformer size up from that would be a 75kVA, but I would put in a 112.5kVA just to be safe, and to allow for future load growth.

 

[dsmith411]

277/480 calculation

277/480 calculation

The amp readings are from adding the wattage of each light per circuit. I came up with 43.1 Kva. Is that correct ?
480 x 52A X 1.73=43.18 kw

 

[Pharon]

If you have the fixture wattage then go with iwire's suggestion.

Also, my math was wrong -- yours is right. Sorry about that.

 

[lielec11]

The amp readings are from adding the wattage of each light per circuit. I came up with 43.1 Kva. Is that correct ?
480 x 52A X 1.73=43.18 kw

Your math is correct. 45kva is the next size up but go with 75.

 

[iwire]

The amp readings are from adding the wattage of each light per circuit. I came up with 43.1 Kva. Is that correct ?
480 x 52A X 1.73=43.18 kw

How many fixtures and how many watts each?

 

[texie]

The amp readings are from adding the wattage of each light per circuit. I came up with 43.1 Kva. Is that correct ?
480 x 52A X 1.73=43.18 kw

Assuming the current of 52 amps, I agree with the math. And as you did, you have to figure on the highest current on any one phase. You mention that you derived the current from the wattage. If these lights are anything other than purely resistive, you need that account for the PF as the transformer and all components of the circuit will see the line current, not just the watts. For many typical HID lights this could be a substantial difference of current VS wattage. For example, if the PF of the fixture is .8 you will need 54 KVA. The power drawn may be 43 KW but the load on the transformer and other circuit components in current will be 54 KVA.

 

[mbrooke]

Use the highest current from the 3 phases. Use 52 amps in this case. However, other loads and ballast type play a role. Some ballasts like auto lag ballasts draw more current on start up. If I was you Id do 480 volts since its cheaper in terms of wiring but in the end thats your call.


Id go like this: 52amps x 480= 24960 x 1.73= 43.21 A 45kw transformer, assuming this is the max load your legs will see. Out of curiosity is this a dry type unit or an oil filled pad mount? If you have a pad mount you have some overload capability for intermittent loads.