3 Phase Full load current / Phase current and Line current

[ssirobotics]

We have an industrial machinery that uses 208VAC 3 phase power (3 wire + Earth ground).
I need to figure out the Full load current and power consumption in KWatts to specify a UPS unit.

Each phase is fairly well balanced and draws about 7 Amps peak each during normal operation.
For extra margin I figure about 11 Amps Full load each phase.

To get the total Line current I did following:
Phase current * Sqrt(3) = Line current
11A * 1.732 = 19A

To calculate power consumption in kW I did following:
(Line current * VAC)/1000 = kVA
(19A * 208VAC)/1000 = 3.952 kVA
kVA * .7 = kW
3.952 kVA * .7 = 2.766 kW

I assume Full Load Current of a 3 Phase machinery is Line current and not phase current.
Can someone verify that this is correct?

I'm confused, since some UPS manufactures use following formula to figure out power consumption in kW:
(Full load current * Line VAC * 1.732 * .7)/1000 = kW
19A * 208VAC * 1.732 * .7)/1000 = 4.785 KW

 

[AFab]

your formula calculates watts for single phase, when calculating 3 phase
1.732 x Volts x Amps x P.F.
in 3 phase power each cycle is 120deg apart from each other therefor you have to use the 1.732 multiplier.

 

[Skogsgurra]

ssirobotics,

Your calculation does not take the three phases into account. The formula with sqrt(3) is correct.

Think of it like this: There are actually three single phase systems. Each one has a line current and a line-neutral voltage. The line-neutral voltage is line-line (or system) voltage/sqrt(3). So, the apparent power in each phase is line current times system voltage/sqrt(3). If you sum up the three phases you get three times that value. And factoring out sqrt(3) leaves the correct result Apparent power = line current times system voltage times sqrt(3).

You should be careful with loads that contain motors because motors usually have (or can have) high starting currents. If the machinery needs to do actual work while fed from the UPS, you should do a more thorough analysis with a power analyser to find out the actual needs. Or overrate the UPS substantially.

 

[ssirobotics]

Thanks for your resonses guys.
The system does not use single or 3 phase motors.
I should be safe this way.

I think I understand that I need to use sqrt(3) when using the 3 phase power calculation.

However, I thought that I already did this when I figured 11 Amps for each phase.
phase Amps * sqrt(3) = Line Amps
11Amps * sqrt(3) = 19Amps

Then when I calculate VA I need use the sqrt(3) again?
19Amps * 208VA * sqrt(3)

I just want to make sure that it's OK to use sqrt(3) twice.

 

[Skogsgurra]

No, just once. Apparent power = line current times system voltage times sqrt(3). System voltage = line-line voltage.

 

[rbulsara]

You error is in multiplying 11A with sqrt(3) to get 19A. Although your final answer is of 3.9 kVA correct, it was not arrived at by correct understanding.

Your UPS mfr has the right formula, but they are using 19A that you provided them which is not correct.

The line current is the current you will measure in each line with a clamp-on meter. According your original post it appears that you 'measured' 7 amps in each line.

Also the rated amperes for a 3 phase machine is for each line.

The formula with sqrt(3) is the corrent one where the V is the line to line voltage and current is the line current, which is 7 amps (or 11 if you are allowing for margin) in your case.

If I were you I would size the UPS for 1.25% of the rated (nameplate) current of the machine. This is ,of course, assuming it does not have motors.

How did you arrive at 0.7 pf? It sounds low.

 

[stevenal]

This thing you are calling "total line current" doesn't exist. "Line current" and "phase current" are used interchangeably (11 A). The sqrt(3) in the formula comes from the fact there are three phases and the ratio of the line to line voltage to line to neutral. Looks like you've got the formula right anyway, getting 3.95kVA. There is something called line to line current, but it doesn't really come into play here.

 

[peebee]

Here's another way of looking at it: like 3 single-phase 120v circuits.

Each single phase circuit draws 120v x 11a = 1320va = 1.3kVA

The total motor draws 3x that: 120v x 11a x 3 = 3960va = 3.96kVA.

The nice thing about thinking about it this way is you no longer have to worry about the sqrt(3).

The reason you don't need to think about sqrt(3) anymore is because 120 x sqrt(3) = 208. By thinking of this as (3) 120v 1-phase circuits, you've done the same thing as multiplying 208 x sqrt (3).

 

[ssirobotics]

Thanks to everyone for helping me with this.
I understand it much better now.

To clarify one more thing.
We need to make up a Nameplate for this machinery that states the "Full Load Current".

I'm now assuming it will need to state 11Amps?

bulsara,
the 0.7 is not for pf. I used it to convert VA into Watts

 

[rbulsara]

ssir:

Definition of power factor is kW/kVA.

0.7 is not a conversion factor. You need to determine pf by measuring both kVA (amps and V) and kW with a wattmeter.

If your machine does not have motors or transformer, I would assume a high power factor such as 0.95 or even 1.0.

Power factor depends on nature of the load, inductive load had lagging pf, resistive loads have unity pf (1.0) and capactitive load will have leading pf.

 

[peebee]

Re: "I'm now assuming it will need to state 11Amps?"

Well -- it would probably want to state a value in the ballpark of 11 amps. . . .

What is this machinery? Did you buy it or build it? What's in it? Is there no existing vendor nameplate with this value?

You shouldn't guess at this number, or apply a factor to measured load to come up with it. You should be able to determine it from the ratings of the motors, transformers, lights, or whatever you have in there. . . .

 

[ssirobotics]

rbulsara, peebee,
It's a custom machinery. We did not buy it.
Inside is a robot and several servo drives, plc & pneumatics etc. So it did not come with a vendor nameplate.
I assume that Power Factor is most likely close to 95 or 1.

Makes sense.
I'll gather the ratings of each of the individual components and add them up to establish total full load current.

 

[peebee]

Is there a DC power supply? Is there anything else in there that is NOT powered from the DC power supply?

 

[ssirobotics]

peebee,
there are 3 DC power supplies (208VAC single phase to 24VDC).
These power all the (low current) solenoids for the pneumatics, sensors, DeviceNet components and some of the smaller servo drives.

The rest are all 208 VAC single phase devices.
For instance, robot, other servo drives, PLC, PC and monitor.

 

[advidana]

just an extra tip:

Ok you, now that you have the kva and kw size of the UPS.

What happens after the utility fails then the UPS pick the load. the ups run under batteries. Now the utility starts up and now you have the load of the UPS plus the charging current of the Batteries for the UPS. My point is don't forget to size the ckt feeding the UPS to include both loads.

Also the ups is a new source of power for the system. UPS neutral needs be grounded at the load side to the building system grounding system.