McMac
Member
- Location
- St. Louis, MO USA
A 480Y inverter states 40A per phase. In a wye config the phase current is equal to line current (what I size my conductors to).
In a Delta config the line current is phase current x √3 (correct?).
My question is for calculating line current on the delta side (240v).
Is it
(a) the 40A line current (which is the same and stated as phase current on the spec sheets) from the 480Y is a simple and direct calculation. At 240D my line current is now 80A and I size my conductor to 125% of that (100A). Leaving the phase current on the delta is then less by √3. But phase current is more applicable to the transformer and not where my concern lies.
Or
(B) the 40A phase current from the 480Y is the simple and direct calculation. At 240D the phase current is 80A. Leaving that my line current is then greater by √3. Which means I would need to size conductors even larger. In this example 80x√3=138.6Ax1.25=173.3A. Wow I hope this a silly concern based on incorrect application of the knowledge.
Full disclosure, I hope (a) is right because this is how I have been designing, and I haven't had any issues. But when I was forced to put three single phase inverters on a 240 delta, I did my research and learned that at least for that application I did needed to factor the line current to be the inverter max x √3 to properly size my conductors. But should I have been applying this same consideration for my 3P 480wye inverters when xfrming to 240delta? I hope not.
I appreciate your help in adressing this concern of mine. Thanks in advance for your time.
Sent from my SAMSUNG-SM-G930A using Tapatalk
In a Delta config the line current is phase current x √3 (correct?).
My question is for calculating line current on the delta side (240v).
Is it
(a) the 40A line current (which is the same and stated as phase current on the spec sheets) from the 480Y is a simple and direct calculation. At 240D my line current is now 80A and I size my conductor to 125% of that (100A). Leaving the phase current on the delta is then less by √3. But phase current is more applicable to the transformer and not where my concern lies.
Or
(B) the 40A phase current from the 480Y is the simple and direct calculation. At 240D the phase current is 80A. Leaving that my line current is then greater by √3. Which means I would need to size conductors even larger. In this example 80x√3=138.6Ax1.25=173.3A. Wow I hope this a silly concern based on incorrect application of the knowledge.
Full disclosure, I hope (a) is right because this is how I have been designing, and I haven't had any issues. But when I was forced to put three single phase inverters on a 240 delta, I did my research and learned that at least for that application I did needed to factor the line current to be the inverter max x √3 to properly size my conductors. But should I have been applying this same consideration for my 3P 480wye inverters when xfrming to 240delta? I hope not.
I appreciate your help in adressing this concern of mine. Thanks in advance for your time.
Sent from my SAMSUNG-SM-G930A using Tapatalk