Single-Phase Loads on 3-Phase Supply

[gummo]

I am working on designing a power distribution panel that will supply power to UV lamp ballasts. The power coming into the distribution panel is 240VAC 3-phase. Suppose I have three single-phase loads (where the load is an arrangement of ballasts), each 48 Amps. If the three loads are balanced (one load across phase a-b, one load across a-c, and one load across b-c), will the full-load amps of the power distribution panel be 48 Amps? In other words, will I need to size the main 3-pole circuit breaker for a 48 Amp load (of course I have to divide by 0.8 per NEC/UL, which would be 60 Amp)?

Upstream of the panel, there is a 480/240VAC 3-phase transformer. Not sure if it is delta or wye connected, or if it even makes a difference. Someone was telling me that I need to multiply the 48 Amps by the square root of 3 to get the FLA.

Does anyone have any answers or insight on the above? Any help would be very much appreciated. Thank you.

 

[Smart $]

Someone was telling me that I need to multiply the 48 Amps by the square root of 3 to get the FLA.

That person advised you correctly.

 

[gummo]

Thanks Smart $. So it doesn't make a difference whether the upstream transformer is delta or wye connected?

I typically work on 3-phase motor panels where the minimum size of the circuit breaker would be the motor FLA divided by 0.8 - no square root of three factor. I never really thought of how much current is going through each individual line. With what you are saying, can you think of a motor as three separate single-phase loads with the individual line current being the motor FLA divided by square root of 3? Of course, I realize that motors are not resistive loads.

 

[jim dungar]

Thanks Smart $. So it doesn't make a difference whether the upstream transformer is delta or wye connected?

The only thing that matters,for sizing your branch, is how your loads are connected.

 

[Smart $]

Thanks Smart $. So it doesn't make a difference whether the upstream transformer is delta or wye connected?

Correct. 3? and 1? line-to-line loads operate the same on wye and delta secondaries, provided the secondaries have the same line-to-line voltage.

I typically work on 3-phase motor panels where the minimum size of the circuit breaker would be the motor FLA divided by 0.8 - no square root of three factor. I never really thought of how much current is going through each individual line. With what you are saying, can you think of a motor as three separate single-phase loads with the individual line current being the motor FLA divided by square root of 3? Of course, I realize that motors are not resistive loads.

The FLA of a 3? motor is the line current because the winding currents have already been summed for you, being considered a single load. However, if you consider say a six-lead delta or wye connected motor, it's FLA is dependent on which way you hook it up. In the delta configuration, you would connect each of three windings (two-leads each) in balanced fashion (A-B, B-C, and C-A). The current across each windings would be the FLA (line) current divided by √3. Connecting these three windings and their currents is comparable to connecting three 1? line-to-line loads in a balanced scheme. The current through the 1? loads, windings, or components if you will, gets multiplied by √3 to get the line current.

However, I should note this is only an accurate calculation for balanced loads.

 

[gummo]

Thanks Smart $ and Jim. I understand it now for a balanced load.