Calculating 3 Phase Loads

[Hire]

I am working on a panel schedule for 120/208 volt panel. I found an old thread posted by Charlie B. #post-2054753

Charlie stated For a single phase, 208 volt load, you take 208 time the current then divide by two and put this value in for each of the two phases.
I don't understand how this works. If I put and ammeter on each leg of the 208 load I will get the same amperage on each leg.
So if I have a 30-Amp load I would think the VA for each leg should be 30 x 208 = 6,240 VA per phase. What am I not understanding?

 

[Strathead]

Ok, let me give this a try. Eliminate the word phase from this equation. Replace it with the word pole or leg. Each leg in reference to the common point of the supplying transformer is 120 volts. Since the two legs are 120 degrees apart instead of opposite or 180 degrees, the voltage from leg to leg is 120*1.73 instead of 120*2. So a 120 volt va for one leg to neutral would be 120v * amps. The va for 208 volts would be 208v * amps. But since none of that work (VA) is being shared with the third leg the va is split between the 2 legs it is drawing from, with no cancellation from the third leg.

 

[wwhitney]

I am working on a panel schedule for 120/208 volt panel. I found an old thread posted by Charlie B. #post-2054753

Charlie stated For a single phase, 208 volt load, you take 208 time the current then divide by two and put this value in for each of the two phases.
I don't understand how this works. If I put and ammeter on each leg of the 208 load I will get the same amperage on each leg.
So if I have a 30-Amp load I would think the VA for each leg should be 30 x 208 = 6,240 VA per phase. What am I not understanding?

At the end of our accounting process, we want the sum of the VA per leg to match the sum of the VAs of all our loads. If you assigned the total load VA to each of the two legs, that would be double counting.

Try working through this example. Here are 3 different loading scenarios that all draw 10A on each leg:

- (3) 10A 120V L-N loads, one on each leg
- (3) 5.77A 208V L-L loads, one on each pair of legs
- (1) 10A 3P3W load connected L-L-L

When you do the VA accounting for each of these cases, they should all give the same answer of 1200 VA per leg.

Cheers, Wayne

 

[ggunn]

Ok, let me give this a try. Eliminate the word phase from this equation. Replace it with the word pole or leg. Each leg in reference to the common point of the supplying transformer is 120 volts. Since the two legs are 120 degrees apart instead of opposite or 180 degrees, the voltage from leg to leg is 120*1.73 instead of 120*2. So a 120 volt va for one leg to neutral would be 120v * amps. The va for 208 volts would be 208v * amps. But since none of that work (VA) is being shared with the third leg the va is split between the 2 legs it is drawing from, with no cancellation from the third leg.

What he said. Another way to look at it is that if the neutral is not involved the load is at the midpoint of a symmetrical voltage divider network, and the current through it is multiplied by half the phase to phase voltage to get the VA per breaker pole.

 

[wwhitney]

This VA accounting method is used ultimately to determine the load current per leg. And as such, it is a non-conservative approximation to the actual answer.

The simplest way to see it is non-conservative is to consider a single L-L load of say 5.77A. That makes 1200VA, which we split to 600VA on each of the two legs. If that's the only load, we now divide 600VA per leg by 120V per leg to get 5A per leg. Yet we know that the load current will actually be 5.77A on each leg.

The reason to do things this way is that the accounting does capture the way the L-L currents from two different identical loads cancel in the common line conductor.

Cheers, Wayne

 

[wwhitney]

For a more complicated but more conservative VA accounting method, keep 6 separate VA totals, one each for A, B, and C, and one each for AB, BC, and CA. Then

1) Each L-N load VA gets assigned to A, B, or C accordingly
2) Each L-L load VA gets assigned to AB, BC, or CA accordingly
3) For balanced 3 phase loads L-L-L, assign 1/3 of the VA to each of A, B, and C.

Now at the end you have 3 VA totals for AB, BC, and CA. Take the smallest of those, and subtract that value from each of the AB, BC, and CA totals and add it to each of the A, B, and C totals. This is just moving VA around in the accounting, and it represents the possibly balanced portion of the L-L loads.

That leaves you with one of AB, BC, and CA equal to 0 VA. For ease of presentation I'll assume it's CA; the other cases are comparable. The line currents are now approximated by:

I_A = VA_A / 120V + VA_AB / 208V
I_B = VA_B / 120V + min(VA_AB,VA_BC) / 120V + |VA_AB-VA_BC| / 208V
I_C = VA_C / 120V + VA_BC / 208V

This will be conservative (any error is on the high side) if the L-L loads all have the same power factor. In which case min(VA_AB,VA_BC) represents the load on leg B where current from AB and BC partially cancel, and |VA_AB-VA_BC| represents the rest of the load from the higher of the two values.

Cheers, Wayne

 

[Hire]

I want to thank you all for your responses and the all help me understand.
I also want to thank you all for the quick response much appreciated.

 

[Hire]

Now would it be possible to print this thread?

 

[Dennis Alwon]

The easiest way is to highlight all the threads and choose copy and then special paste. Paste without format.

 

[Hire]

Never mind I just printed it!

 

[Dennis Alwon]

Never mind I just printed it!

I sent you a pm with the text from the thread---oh well

 

[Hire]

Just though I would share my panel schedule

 

[charlie b]

I don't understand how this works. If I put an ammeter on each leg of the 208 load I will get the same amperage on each leg.

Yes you will, and that is a key point. The current from phase A to the load and the current returning from the load along phase B are, in fact, the same current. The (regrettably popular) notion of adding currents on two or three phases to get a "total current" is utter nonsense. As Wwhitney pointed out in post #3, that is double counting.

It sounds odd, perhaps, but adding amps to amps, hoping for an answer in amps, will often give you an incorrect result. That is especially true when you have a mix of 1, 2, and 3-pole breakers. On the other hand, adding power (in KVA) to power (again in KVA) will give you the right results.

 

[Oneshop]

Yes you will, and that is a key point. The current from phase A to the load and the current returning from the load along phase B are, in fact, the same current. The (regrettably popular) notion of adding currents on two or three phases to get a "total current" is utter nonsense. As Wwhitney pointed out in post #3, that is double counting.

It sounds odd, perhaps, but adding amps to amps, hoping for an answer in amps, will often give you an incorrect result. That is especially true when you have a mix of 1, 2, and 3-pole breakers. On the other hand, adding power (in KVA) to power (again in KVA) will give you the right results.

Charlie, I just joined so I can attach these two links for anyone that comes across this thread. I have referred back to these two links over the years, numerous times, trying to wrap my head around this very issue. In both, you clearly articulate how to approach this question. I'm just an architect so by nature I know a little about a lot and the little I know about this always wants me to assume you can add amps, that you can take a load on a 120/240 2 pole and divide by 2 and call it a day or on a 120/208 3 pole take a load divide by 3 and call it a day. Maybe there's something intuitive about that or maybe it is just nonsense but obviously questions centered around this continue to pop up. Anyway thanks!

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Wwhitney, I am glad you posted the simple way above. It's a clear way to visualize the current is the same.