Another 208 3 Phase Question

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bf111d

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I have an application where a 208v 3ph circuit is being brought to a server cabinet power distribution unit where it breaks down the 3ph circuit into 3 separate 208v single ph circuits through 3 2pole breakers. Circuits connections are of course phases AB, BC, & CA. I am measuring the individual currents at the original 3ph source ph A,B,C.

Can I use the following calculations to get the current draw on the single phase connections?

Iab=(Ia+Ib-Ic)/2
Ibc=(Ib+Ic-Ia)/2
Ica=(Ic+Ia-Ib)/2

Thanks
 
bf,The way i see the connections at the power supply is in a delta config.The formula for phase current in delta is IP= IWx1.73.If you had 16 amps @ 208 in each winding,you would have 27.68 amps in each phase.Assuming they are all balanced.If there unbalanced then i dont know how to figure it.
Rick
 
additional info

additional info

the circuits are unbalanced for sure. The end single phase circuits have different loads connected. This wiring scheme is becomming a standard application for PDU strips serving high density blade server cabinets. It reduces wiring costs while increasing KW capacity from that of traditional 120v feeds.
 
bf111d said:
I have an application where a 208v 3ph circuit is being brought to a server cabinet power distribution unit where it breaks down the 3ph circuit into 3 separate 208v single ph circuits through 3 2pole breakers. Circuits connections are of course phases AB, BC, & CA. I am measuring the individual currents at the original 3ph source ph A,B,C.

Can I use the following calculations to get the current draw on the single phase connections?

Iab=(Ia+Ib-Ic)/2
Ibc=(Ib+Ic-Ia)/2
Ica=(Ic+Ia-Ib)/2

Thanks
Click to expand...

I don't believe you can add the magnitudes measured at the source to calculate the current in the loads. You also have to take the current vectors into account and the currents entering and leaving at the source.
 
bf111d said:
I have an application where a 208v 3ph circuit is being brought to a server cabinet power distribution unit where it breaks down the 3ph circuit into 3 separate 208v single ph circuits through 3 2pole breakers. Circuits connections are of course phases AB, BC, & CA. I am measuring the individual currents at the original 3ph source ph A,B,C.

Can I use the following calculations to get the current draw on the single phase connections?

Iab=(Ia+Ib-Ic)/2
Ibc=(Ib+Ic-Ia)/2
Ica=(Ic+Ia-Ib)/2

Thanks
Click to expand...

You can do it this way:

Measure the current at the output of each breaker, or switch the breakers on one at a time and measure the phase currents.
 
bf111d said:
... I am measuring the individual currents at the original 3ph source ph A,B,C.

...
Click to expand...
Why not just measure the single phase currents directly, say at the 2P CB's?
 
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