3 phase amperage calculation

[kent grimsley]

If I have a 208v 3phase 3 watt circuit with 28.8 amps on each leg with 6 watts on each leg and a lose phase L1 to L3, is there a formula that I can calculate what the amoerage would be on L2?

 

[GoldDigger]

If I have a 208v 3phase 3 watt circuit with 28.8 amps on each leg with 6 watts on each leg and a lose phase L1 to L3, is there a formula that I can calculate what the amoerage would be on L2?

When you say a lose (loose? lost?) phase from L1 to L3, are you saying that it is an open delta? If so, that will have no direct effect on the line currents in each of the secondary phase lines.

I am not at all sure what your 3watt and 6 watt circuit figures refer to, especially in conjunction with 28.2A on each leg of 208V. Do you mean kilowatts instead? Even then the figures do not make sense together as you state them.
Once you provide a more complete and correct set of figures to work from, we can probably point you to a formula.
If this is an exercise or homework problem, we would very much prefer to see how you would try to approach the problem on your own and then coach you from there.

PS: Welcome to the Code Forum, Kent!

 

[kent grimsley]

thanks for the Welcoming, this is a extra credit school problem. Forgetting about the watts because I don't think it applies, I have a 208 v 3 phase circuit with 28.8 amps per leg under normal conditions.The question is what would be the amperage on L2 if i lost the L1 to L3 leg. I can not find a formula to help me calculate this

 

[GoldDigger]

thanks for the Welcoming, this is a extra credit school problem. Forgetting about the watts because I don't think it applies, I have a 208 v 3 phase circuit with 28.8 amps per leg under normal conditions.The question is what would be the amperage on L2 if i lost the L1 to L3 leg. I can not find a formula to help me calculate this

OK.
The first thing you need to ask yourself is what the voltages would be (on the load side) with no load applied.
Then calculate whether those voltages would be any different if the L1-L3 winding of the transformer secondary was open or closed.
Let us know what you get and we can go on from there.

 

[kwired]

IMO you need more detail of the connected load to determine the answer. A single three phase motor - can depend on if motor was running when the phase was lost, if motor was stopped when the phase was lost it probably will not accelerate at all and will draw somewhere near locked rotor current on remaining two phases. All resistive loads - you are going to lose the portion of current from the lost phase on the other two phases, but yet will not lose any watts on loads that are connected between the two good phases - sounds a little tricky doesn't it? Things change some if you have line to neutral loads but the way the question is worded I'm assuming there is no line to neutral load to consider.

 

[GoldDigger]

kwired:
The problem (as relayed) is not the loss of a phase line (like L3) but rather that one of the source windings (L1-L3) is open.
This converts the closed delta to an open delta. The effect of that on the currents in the other two windings is what is asked for.
To determine that we do not need to know the nature of the load, just that it is balanced, including phase angles, on all three legs.
At least that is the way I see it.

Sent from my XT1080 using Tapatalk

 

[kwired]

kwired:
The problem (as relayed) is not the loss of a phase line (like L3) but rather that one of the source windings (L1-L3) is open.
This converts the closed delta to an open delta. The effect of that on the currents in the other two windings is what is asked for.
To determine that we do not need to know the nature of the load, just that it is balanced, including phase angles, on all three legs.
At least that is the way I see it.

Sent from my XT1080 using Tapatalk

Could be.

When I see 208 volts in there, a delta system is not the first thing that comes to mind either though.

Am also confused with what is meant with the 3 watt and 6 watt figures in the OP.

If it is what I think you are saying and the question to be answered is what is the current on L2 ....the answer should be pretty simple, like possibly simple enough to make him go:slaphead:

 

[GoldDigger]

That is what I will try to point him toward when he comes back with the answers to my first question.
It sounds just right for an extra credit problem, since there is no "formula" exactly.

Sent from my XT1080 using Tapatalk

 

[Smart $]

If I have a 208v 3phase 3 watt circuit with 28.8 amps on each leg with 6 watts on each leg and a lose phase L1 to L3, is there a formula that I can calculate what the amoerage would be on L2?

What GD and K are eluding to is that a 208V 3? 3W (W meaning "wire" in this case) is typically and conventionally supplied from a 208/120V 3? 4W wye system supply source. I do not believe this has any bearing on your problem.

In your case they may be saying the 3? load is connected in a 6-wire configuration. In other words, three single phase loads, one connected L1-L2, another connected L2-L3, and the third connected L3-L1. If you draw this out, there will be two wires connected to each "leg".

I think for this problem you should assume loading to be resistive, non-reactive in nature. If I continue to explain more, I will essentially supply you with the answer... which we try to avoid here on the forum as it gives you an unfair advantage over others and unearned credit.

 

[kent grimsley]

I asked for more info on this and was told that it is a 208v 3 phase 3wire feeding a hanging unit heater. I believe that if I lost the L1 and L3 element I would still be feeding 2 elements with L2 and my amperage would not change. Am I wrong to think this way?

 

[Smart $]

I asked for more info on this and was told that it is a 208v 3 phase 3wire feeding a hanging unit heater. I believe that if I lost the L1 and [to] L3 element I would still be feeding 2 elements with L2 and my amperage would not change. Am I wrong to think this way?

No, you are not. IOW, you are thinking correctly.

*FWIW, a hanging 3? unit heater typically has a fan, often a driven with a 1? motor. The problem is stated in simple terms and does not account for which phase the motor may be connected to, nor its affect on the load current.

 

[kent grimsley]

Thanks, as far as the fan goes. based on the information given that it was a balanced load with 28.8 amps per phase wouldn't that mean that it was a 3 phase blower motor? If it was single phase motor wouldn't one phase be a little higher?

 

[Smart $]

Thanks, as far as the fan goes. based on the information given that it was a balanced load with 28.8 amps per phase wouldn't that mean that it was a 3 phase blower motor? If it was single phase motor wouldn't one phase be a little higher?

If it had a 3? blower motor, that would complicate the problem to the degree you could not provide a definitive answer, or have to cover multiple scenarios in your answer. Did you lose just the L1-L3 heating element or the element and the corresponding winding in the motor? For the former, L2 would remain the same. For the latter, L2 would change, but by how much would depend on how the motor reacts to losing a winding, so no concise answer without assumptions.

If it had a 1? blower motor, you are correct, the load would likely not be balanced, but even if it were, you would not know to which phase it was connected or whether it was lost in the process. If the blower motor was affected, that changes the air flow, which in turn will change the current through the heating elements.

Problems as such typically ask for the simple answer, ignoring the plethora of potential variables.

 

[kent grimsley]

Thanks Smart$, for the lack of any additional info given I will keep it simple and just say the amperage stays the same

 

[kwired]

Thanks Smart$, for the lack of any additional info given I will keep it simple and just say the amperage stays the same

Keep it simple - maybe a radiant heater with no blower - that would be realistic and simple.