3-Phase utility tied inverter feeding single phase load

[philly]

Trying to wrap my head around the following concept and am not quite seeing it. Hoping others can shed some light.

Hypothetical scenario....I have a 480/277 utility transformer feeding facility (lets say 100kVA for simplicity) and on secondary of transformer I have a 480V bus. On 480V bus I have (2) bi-directional AC/DC power supplies. The DC side of each power supply is connected through common DC bus.

The first power supply (#1) is connected L-N at 277V and draws current as a load from the 480V bus. The second power supply (#2) is a 3-phase connection to the 480V bus and circulates power back onto the 480V bus from supply #1. So power supply is 480V Bus >>supply #1 >> DC Bus ->>supply #2>>back to 480 bus and then circulating back over to input of supply #1.

My question is what happens to the 3-phase AC current that is being supplied from supply #2 and circulates back on to 480V bus with only a single phase load at input of supply #1. I know the neutral of the utility transformer will allow 3-phase inverter to connect to common 480V bus with L-N load at supply #1 being supported by transformer neutral but i'm having a hard time envisioning current flow?

 

[synchro]

... My question is what happens to the 3-phase AC current that is being supplied from supply #2 and circulates back on to 480V bus with only a single phase load at input of supply #1. I know the neutral of the utility transformer will allow 3-phase inverter to connect to common 480V bus with L-N load at supply #1 being supported by transformer neutral but i'm having a hard time envisioning current flow?

Lets assume the supplies are 100% efficient and all of the DC current out of supply #1 is drawn by the DC side of supply #2 (i.e., no net current onto the DC bus). Also the PF of supplies is unity. It seems in your example that there'll be no net circulation, but instead a power transfer from the output of the 480 line conductor and neutral that's loaded by supply #1 over to the other two 480 lines which drive the utility.

Say that supply #1 is drawing an input current I from the phase A and neutral of the 480 bus. Then supply #2 will drive each of the phases A, B, and C with a current 1/3 I . Therefore the net current drawn from the utility on phase A will be 2/3 I, and supply #2 will drive the utility with 1/3 I on phases B and C. The current on the neutral will be I .
Overall there will be very unbalanced "loading" between the phases as seen by the utility because the current will be in-phase with the voltage on phase A like with a resistive load, but at 180° on phases B and C because they are being driven with currents from supply #2 (although with half the current on phase A).