Sizing UPS from 3-Phase 208V Panelboard using 120V Single Phase Loads Calculation

[EElmarko]

A client EE sent me an Excel spreadsheet where he has columns for each phase. The panelboard is 3-phase 208V but all loads are 120 single phase (single pole) loads. At the bottom of the spreadsheet he adds up the Amps for each phase which in the above cells of each phase column he was calculating by the consumer kW divided by the product of (voltage x (SQRT of # of Phases) x PF x Eff). For a one pole example that would be 0.6kW / (120V x SQRT(1) x 0.8 x 1.0). Now to clarify, he is applying a 0.8 Power Factor because this panelboard (PB) is deriving its power from an engine-generator set onboard a ship.

So now at the bottom of each phase column he then applies a "loading factor" as he calls it (most on this forum refers to it as Demand Factor). In the next cell below each he than coverts to W, (although in the spreadsheet the label reads "kW". He then adds up three phases for "Total kW". He then applies what he calls a "60% maximum usage" which I would define as design margin or contingency by dividing the Total kW by 0.6 to come up with what he calls "Minimum UPS Rating (kW)". I will in a separate future post the issue of UPS rating in KVA versus Load Analysis such as this in KW rating this equipment's PF & Eff considerations which comes up in design discussions often.

My question is whether his calculating method correct? From prior posts I have seen from fellows, the general consensus is when take this approach you still need to total based on KW rather than Amps because you have PB's that are not balanced per phase circuits. If you do follow this equation would you then apply the three phase factor and divide his total by 1.732 to get the minimum rating on a 3-phase power-rated UPS?

 

[charlie b]

Your client’s calculation process is invalid.

It is very risky to add amps to amps and expect to get a correct answer in amps. It works if all the loads are single phase and are fed by single pole breakers. That is the case here; that part of your client’s spreadsheet might have been OK. But it’s not OK in this instance, because of what was done with the total amps.

Applying the same power factor to each load (or in this instance to the total amps in each column) is inappropriate. A load is either purely resistive or it has some reactance; the load is what the load is. If there are motor loads or other loads that have inductive properties (a likely thing indeed, especially on a ship), then adding the amps on each phase to get a total amps for that phase, then converting those three values to KW, and finally adding the KW from the three phases to get a total KW, is flat-out wrong. It would be adding the KW of loads that have both resistance and reactance properties to the KW of loads that only have resistive properties but treating the latter group as though they also had reactive properties. It will give an answer that is larger than the right answer, and that may or may not be what is needed.

The only way to get the answers right each time is to do all math in units of KVA, and only convert to amps at the very end. That works even if you have a mix of single-phase single-pole loads, single-phase two-pole loads, and three-phase loads.

Let me offer what has come to be known as one of the “Famous Sayings by Some Famous Guy”:

“If you make an error in your reasoning or your math, and you obtain a result that is close to what correct reasoning or math would have given you, you don’t get to call that a success."

 

[JoeStillman]

The assumption that a generator always supplies kW at 0.8 power factor is wrong. The generator has two forms of rating, engine kW and alternator kVA. The power factor rating of the generator is the ratio of rated kW : rated kVA. It represents the worst the genset can tolerate. The operating power factor is determined by the load.