Safely Maximize Power from a 100amp 3-Phase Delta Panel

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gar

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Ann Arbor, Michigan
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EE
170319-1327 EDT

EricJ:

I think you are still confused. Do you really have a degree in electrical engineering? Or is it civil engineering? If electrical, then what kind of basic AC circuit theory courses did you have?

I'll take a stab at it. I think this is because each leg is 120 degrees out of phase from the other. If you have a load LL between leg A and leg B, leg A will supply some current and leg b will also supply some current. So even though the total current across the load is 92.4 amps, the individual legs will only see 80 max.
With a single load connected between legs A and B you really only have a single phase circuit.

All alone your first sentence is a correct answer to Smart $'s question. But with your following description it is clear you did not understand his question.

The current is not across the load. Current is thru a load.

Your load between leg A and leg B has the same current flowing out of the leg B end of the load as it has flowing into the leg A end of the load. Smart $ calculated this for you at 46.2 A. That is the current flowing thru your line to line load.

If there is only this one line to line load on the three phase source, whether wye or delta, then the current in the supply legs A and B are equal and equal to that single load current.

Add another identical load between leg B and leg C as is connected between legs A and B. Now what happens to the current in leg B. That is basically what Smart $'s question is. And that is why he is asking about 2*46.2 not equaling 80.

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Smart $

Esteemed Member
Location
Ohio
I'll take a stab at it. I think this is because each leg is 120 degrees out of phase from the other. If you have a load LL between leg A and leg B, leg A will supply some current and leg b will also supply some current. So even though the total current across the load is 92.4 amps, the individual legs will only see 80 max.

Am I in the ballpark Smart$?
Yes, in the ballpark. LL load currents are 120 degrees out of phase but the currents as seen at the connections are only 60 degrees out of phase because one of the two currents, more importantly its angle, is measured backwards. Two equal magnitude currents at 60 degrees out of phase combined will yield a current that is √3 times either component magnitude and at an angle halfway in between. To prove this all you need do is draw the waveforms of the component currents then mathematically add the magnitude at each instance. To simplify, we use vector math to calculate more-complex combined currents.
 
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Smart $

Esteemed Member
Location
Ohio
...


I'll take a stab at it. I think this is because each leg is 120 degrees out of phase from the other. If you have a load LL between leg A and leg B, leg A will supply some current and leg b will also supply some current. So even though the total current across the load is 92.4 amps, the individual legs will only see 80 max.


Am I in the ballpark Smart$?
Upon review, Gar has caught the error of your circuit description...
...Add another identical load between leg B and leg C as is connected between legs A and B. Now what happens to the current in leg B. That is basically what Smart $'s question is. And that is why he is asking about 2*46.2 not equaling 80.
 
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