Phase kva

[hhsting]

I have 50A load 208V single phase. Turn that to kva is 10.4kva. The 10.4kva if I do panel schedule I get 5.2kva each phase. The system voltage is 208/120V three phase.

The phase kVA is 5.2kVA. If I turn that phase 5.2kva to amps I should get 50A each phase but I don’t. I get 5.2/.12= 43.33A.

Is their something I am doing incorrect?

 

[retirede]

A single phase load cannot be thought of as 1/2 of that load “on each phase.”

The load is the load, period.

 

[hhsting]

A single phase load cannot be thought of as 1/2 of that load “on each phase.”

The load is the load, period.

Then how would one do each phase load cal in panel schedule to see if panel is balanced or not? I dont follow if it is not split in half then what is going on?

 

[charlie b]

Yes, the panel schedule should show 5.2 KVA on each of the two phases that feed this load. That is not the same as saying the 50 amps is essentially 25 amps on each phase. The 50 amps that leaves the panel on Phase A will return to the panel on Phase B. So it's the same amps. The problem with your math is due to the fact that this load does not have a neutral wire. Therefore, the system's 120 volts phase-to-neutral does not come into the equation. So you don't divide the 5.2 KVA that each phase will carry by 120 volts. Instead, you divide the entire 10.4 KVA that the two phases will carry by the 208 volts between the two phases. Thus, 10.4 KVA / 208 V = 50 amps. That should surprise nobody. since this is how you calculated the 10.4 KVA in the first place.

 

[Besoeker3]

Yes, the panel schedule should show 5.2 KVA on each of the two phases that feed this load.

What if the two phases are different loads?

 

[hhsting]

Yes, the panel schedule should show 5.2 KVA on each of the two phases that feed this load. That is not the same as saying the 50 amps is essentially 25 amps on each phase. The 50 amps that leaves the panel on Phase A will return to the panel on Phase B. So it's the same amps. The problem with your math is due to the fact that this load does not have a neutral wire. Therefore, the system's 120 volts phase-to-neutral does not come into the equation. So you don't divide the 5.2 KVA that each phase will carry by 120 volts. Instead, you divide the entire 10.4 KVA that the two phases will carry by the 208 volts between the two phases. Thus, 10.4 KVA / 208 V = 50 amps. That should surprise nobody. since this is how you calculated the 10.4 KVA in the first place.

I still dont follow. If each phase see 5.2kva and you cant use 120V then what can you use to divide? What voltage is each phase if not 120V?

 

[synchro]

The currents from L-L loads are at +- 30 degrees from the L-N voltages. If you have balanced L-L loads then each L-L load will contribute cos(30°) = √3/2 ≈ 0.866 of its current to the line current. So if you had three 50A 208V loads distributed across all phases then the line current on each phase will be 2 x (0.866 x 50A) = 86.6A. And 3 x 120V x 86.6A = 31.2 kVA which is the same result as 3 x 208V x 50A = 31.2 kVA.

But consider the simple case where you have just one 208V 50A L-L load on the three phase system, and no other loads of any kind. Then the line currents on two of the phases would be 50A because another L-L current which is at 60 degrees from it has not been added to it.

 

[winnie]

The issue here is that you are trying to do straight arithmetic when for an _exact_ result you need to do vector math. You've simple stumbled into one of the corner cases where the difference between the simple arithmetic and the vector math result is quite large.

When you have a large panel with lots of loads, the vector math issues tend to average out. So when you do a panel schedule a perfectly reasonable approximation is to assign a simple arithmetic VA value to each load connected to a phase, add up all the VA values for each phase, and call that the VA placed on that phase. This is generally a good enough approximation, but it is _not_ exact.

In the OP's case, to do an exact calculation not only would you need to know the amp rating of the load, you would also need to know the phase angle of the load. But for the sake of discussion, I will assume that the load is a perfect resistor. I will further assume that this is the _only_ load placed on the transformer supplying the 120/208V. In this case the load is 50*208 = 10.4 kVA just as originally calculated.

As Charlie says, 50A flows from phase A through the load to phase B. Since the phase A coil is providing 120V and is seeing 50A, the load on that coil is 6kVA. Similarly the load on coil B is 6kVA. This is a total of 12kVA transformer loading! Sounds like a paradox. The load is a pure resistor with a power factor of 1, consuming 10.4 kVA (and thus 10.4 kW) but the transformer loading is 12 kVA.

What is happening is that even though the current through the load is flowing in phase with the A-B voltage, the current through the transformer coils is _not_ flowing in phase with their respective N-A and B-N voltages. In the transformer coils there is a power factor less than 1.

To do a more accurate panel schedule, instead of assigning a VA value to each load/phase you would assign a current vector (current magnitude in amps and phase angle in degrees). Then for each phase you would do the vector sum of all of these current values, to get the total current flowing on each phase.

Even this would not be perfectly accurate, but it would be a closer result. However in general the simple arithmetic sum of load VA values is accepted as good enough, unless you look at the corner cases of extremely unbalanced loading.

-Jon

 

[winnie]

What if the two phases are different loads?

The two phases cannot have different loads if the only thing connected is a L-L load.

50A connected from phase A to phase B means 50A flowing on each.

-Jon

 

[xptpcrewx]

I have 50A load 208V single phase. Turn that to kva is 10.4kva. The 10.4kva if I do panel schedule I get 5.2kva each phase. The system voltage is 208/120V three phase.

The phase kVA is 5.2kVA. If I turn that phase 5.2kva to amps I should get 50A each phase but I don’t. I get 5.2/.12= 43.33A.

Is their something I am doing incorrect?

Think of it like this. With 10.4-kVA split in half, so too would the voltage be split in half with the voltage being 104-V (not 120-V).

 

[Besoeker3]

The two phases cannot have different loads if the only thing connected is a L-L load.

50A connected from phase A to phase B means 50A flowing on each.

-Jon

"I have 50A load 208V single phase"per the OP?
So it isn't single phase?

 

[winnie]

"I have 50A load 208V single phase"per the OP?
So it isn't single phase?

I'm sure you've asked me this and I've answered this before, so I presume you are prompting me to explain it to others.

In the US, it is not uncommon in apartment buildings for the entire building to have 3 phase 208/120V wye service, but for two 'legs' of this three phase service to be brought to the individual apartments.

Each apartment is described as having 120/208V single phase service.

However this pseudo single phase service has different phase angles present, and will lead to exactly the sort of calculation issues that the OP is describing.

With 50A flowing on the two legs of such a system, you could either be supplying 10.4 kVA of L-L load, or 12 kVA of L-N load.

-Jon

 

[Besoeker3]

However this pseudo single phase service has different phase angles present, and will lead to exactly the sort of calculation issues that the OP is describing.

With 50A flowing on the two legs of such a system, you could either be supplying 10.4 kVA of L-L load, or 12 kVA of L-N load.

-Jon

So not really single phase. That was my point.

 

[wwhitney]

So not really single phase. That was my point.

But not typical 3 phase, either, as the 3 voltage pairs are not all the same. So in US terminology it is called single phase, as it is only used for single phase loads.

Cheers, Wayne

 

[jim dungar]

So not really single phase. That was my point.

IMO it is single phase because there is only a single Line-Line voltage. This methodology has not failed me in the 45 years I have been using it.

There have been many threads on this topic.

 

[Besoeker3]

But not typical 3 phase, either, as the 3 voltage pairs are not all the same. So in US terminology it is called single phase, as it is only used for single phase loads.

Cheers, Wayne

Not typical 3-phase, I agree. UK and many other countries are just simply single phase, no typical about it.

 

[gar]

210713-1406 EDT

My take on this is ---

If you have two wires as your source, then you have single phase, or DC. And DC is really the limit of a very low frequency.

As soon as there are more than two wires, then it is a multiphase system. One or more of the phases could be in phase with other phrases.

If two wires originated from exactly the same source, then it might be reasonable to call both of those a single phase, if there was no phase shift between them.

If I have a single phase source, and run two sets of wires from that source, but one set of wires is 10 ft long, and the other is 5 miles long, and I compare the outputs of those two transmission lines, then I have two different phases.

If you have three wires from a 3 phase wye source, and one is the neutral, then any two wires from that source are a single phase. But more than two wires and it becomes a multiphase source.

If I have two sine waves from AC sources of identically the same frequency, then I have two phases, and these could be displaced in phase anywhere from 0 to +/- infinity.

.

 

[wwhitney]

As soon as there are more than two wires, then it is a multiphase system. One or more of the phases could be in phase with other phrases.

The usual terminology is based not on the number of individual phases, but the number after conflating phases that are scalar multiples of each other (including negative multiples). That explains the naming convention of single phase (including L-N-L), 2 phase, and 3 phase.

Cheers, Wayne