Calculating wattage for 208VAC 1PH with multiple loads

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e5j

New member
Greetings,

I am trying to get as exact a measurement of power (W) as possible for a 280VAC-1Phase circuit feeding two power supplies in a load sharing configuration.

For the purposes of measurement I am using a pair of IEC320 power cables with outer insulation stripped away so I can use an inductive current meter on the individual phase wires. I wouldn't use such a cable in an everyday/production setting.

I have a single system with two load sharing power supplies (autoranging 120-240, 50-60Hz).
The two supplies are plugged to the same 280VAC-1Phase source (L1, L2, Ground).
I do not know how well tuned the phase shift between L1 and L2 are. IOW, I can only assume they are shifted 120 degrees.

Here are the readings I get when the system is under full load:
Power supply 1: L1=5.1A, L2=4.9A
Power supply 2: L1=5.0A, L2=4.9A

Since I can only measure one leg at a time and the load has minor fluctuations I am assuming that the L1 current draw is the same on both power supplies.

I have read up on this and I understand from Google searches and other topics on this forum that calculating 280V can be somewhat black magic.

So to get the closest to 280VAC-1Phase power consumption for this system would it be:

((120 * ((Ips1_l1 + Ips1_l2)/2) + (120 * ((Ips2_l1 + Ips2_l2)/2))

((208 * ((Ips1_l1 + Ips1_l2)/2) + (208 * ((Ips2_l1 + Ips2_l2)/2))

(208 * ((Ips1_l1 + Ips1_l2 + Ips2_l1 + Ips2_l2)/4))

or something entirely different?

If I look at each leg it appears I am pulling ~20A of 120V which would be 2400W. This is a much higher number than I was expecting and it is my impressions that running at 208VAC-1PH adds some efficiency and power savings. These numbers make me second guess everything which is why I am here. :)

Anyone willing to clear the air for me?
 

AdrianWint

Senior Member
Location
Midlands, UK
Firstly, from the info given, you can't determine the real power (W) only the apparent power (VA).

The formulae you are trying to apply is for three phase power. Although you are taking power from two legs of a three phase circuit, your LOAD is only single phase.

Your figures suggest that the line current is 5A per supply, so each unit is drawing 5 x 208 = 1040 VA. Without knowing the power factor, thats the best you can do.
 
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