Load Calc for 480V Single Phase Load

[jimma34]

I need to calculate electrical load for some (non-standard) electrical heating equipment, that will operate at 480V, single phase. I'm not an EE and I'm not sure of the correct method to calculate loads for a 480V, single phase application.

Here's the short summary: We are going to have 3 banks of resistance heaters. To get the watt output we need from the heaters, they are going to be fed at 480V, single phase. Based on the resistance, the heaters should draw ~160A per bank at 480V. We will feed each bank of heaters from 2 legs of a standard 3-pole breaker and balance the 3 banks of heater loads across phases (A-B, B-C, C-A) for the three breakers, so the load on the transformer is balanced.

My question is what is the proper method to calculate the total amperage and kVA with the 480V single-phase power to size the transformer as well as OCP devices?

I think the total amperage is simply: {phase amperage * (Sqrt 3)}, so Atotal= 160A*(Sqrt3) = 277A.

But, I am not sure how to calculate total power (kVA) for this case to account for the two pole feed.

Is it:

Ptotal= 480V*{160A*(Sqrt3)}/1000 = 133 kVA or,

Ptotal = 480V*{277A*(Sqrt3)}/1000 = 230 kVA

or, have I totally missed the boat on this?

I would appreciate any help anyone can offer. Thanks.

 

[stevenal]

230

 

[logbook]

You are over-complicating it. Even though it appears to be a three-phase system you can treat the phases independently.

Each load has a voltage of 480V, a current of 160A , and is resistive. That's easy= 76,800W. You have three, so that is 230,400W. All this business of adding the phasors of these 120degree phase shifted things is not actually relevant, except for sizing the phase conductors.

 

[sreid]

A caution. Resistance heaters draw much more current when they are cold, at least double the hot current.

 

[LionelHutz]

If the resistors are balanced across the phases a quick calc for total amperage is total power divided by voltage divided by sqrt(3).

In this case, each heater is equivalent to a leg of a delta connected load. The phase current is the delta current times sqrt(3).

In this case, you would have 230400W of heater and 277A per phase.

You need a transformer to supply your total wattage plus some additional safety margin. You have 230kW so I would expect 300kVA to be suitable. If you're doing some sort of phase control then you may have to upsize this to account for harmonics.

 

[tinfoil]

PLEASE, PLEASE, please take note of sreid's caution:

The 'cold' draw on many heaters is often much higher than the 'hot' draw.

If you not have a way to guarantee a 'staggered start' of such heaters (while simultaneously keeping decent phase balance), allow for this extra load in your cable sizings.

 

[jimma34]

Thanks for all of the responses. I will go with 230kVA in my calcs.

I am aware of the in-rush current loading on the resistance heaters. We will most likely ramp them up genetly to minimize in-rush load, but will size the conductors to handle the max load as well.

Thanks again to all!

Jim