Load using only two wire of a three phase system

[cbr600]

Hi Everyone. I have a question on the correct equation to calculate how much current runs through a phase to phase configuration.

a 4000W heater is connected to phase A and B on a 120/208V 3 phase panelboard. To size the breaker, should I be using 4000/(SQRT3*208) or 4000/(208) to determine the current through each phase?

Single phase and three phase calculations I understand, but this two wire three phase configuration is a bit confusing..:blink:

 

[Dennis Alwon]

I would do 4000/208

 

[GoldDigger]

I would do 4000/208

Assuming the heater is rated at 4000W at 208V, that will certainly do it.

If the heater is rated 4000W at 240V, then the actual current will be 4000*(208/240)*(1/208) = 4000/240.

For the OP: The SQRT(3) comes in when you have to add the current vectors when more than phase of line-to-line loads terminate on the same line.
So if you do have several single phase loads on different phases you cannot just numerically add the currents.

 

[david luchini]

If the heater is rated 4000W at 240V, then the actual current will be 4000*(208/240)*(1/208) = 4000/240.

I should think if the heater is rated 4000W at 240V, and it is connected to 208V, then the actual current will be 4000*(208/240)^2*(1/208) = 4000/277

 

[GoldDigger]

I should think if the heater is rated 4000W at 240V, and it is connected to 208V, then the actual current will be 4000*(208/240)^2*(1/208) = 4000/277

You are absolutely right. I had a feeling that I was leaving something out, and you found it!
Another way of looking at it would be that the current at 240 would be 4000/240 and the current at 208 will be (4000/240)*(208/240).

 

[Dennis Alwon]

The fact is the OP never gave us that info. but certainly that would change things. From what the op was asking I think he was confused by the 3 phase and only using single phase.

 

[cbr600]

I should think if the heater is rated 4000W at 240V, and it is connected to 208V, then the actual current will be 4000*(208/240)^2*(1/208) = 4000/277

I actually never thought about that. The schematic only shows two 2000 watt heater in parallel connected to Phase A and B. Should i check the nameplate of the heater to find the rated voltage?

Also, in general would a load across only two phases in a 3 phase system would be considered single phase load? thanks

 

[david luchini]

Also, in general would a load across only two phases in a 3 phase system would be considered single phase load? thanks

Yes, it would.

 

[cbr600]

thanks everyone. great forum BTW :thumbsup:

 

[GoldDigger]

I actually never thought about that. The schematic only shows two 2000 watt heater in parallel connected to Phase A and B. Should i check the nameplate of the heater to find the rated voltage?

Since the schematic shows the heating elements from phase to phase (and does not explicitly show a 120/240 system), it could either be designed for 208Y/120 or for 240 delta. But it is more likely, in the absence of any other information, that it is 240. Depends on what other components besides the heaters is in the system.