Correct Formula to Calculate Single Phase Current 2

[bxny]

I need some help to determine the correct way to calculate the current in a 480V, single phase circuit, which services ten roadway lighting fixtures. The circuit is fed from a 2-pole (phases a-b), 480V, circuit breaker in a 480Y/277V, 3-phase, 4-wire panelboard. There are 10 - 480V rated fixtures on the circuit, 304VA each = 3040VA total circuit load. There are two additional circuits (phases c-a and b-c)fed from this same panel, with the same number of fixtures on them. I contend that the correct way to calculate the current in this circuit, for the purposes of determining the cable size, is from the formula: I = 3040/480 = 6.34A. Am I not correct?

 

[buzzp]

Cable size is determined by your breaker size. Also, voltage drop issues might require a larger cable to reduce the voltage loss.

 

[bxny]

Agreed. However, what about the calculation? Is the formula correct?
Thanks.

 

[Skogsgurra]

The only problem is to know which cable you are referring to. If it is the cable feeding the armatures from a-b, b-c and c-a, then you are correct. If it is the cable feeding the distribution panel, then you should multiply by sqrt(3) or 1.732 to get the total current in each phase.

 

[bxny]

It is the two conductor cable between the two-pole circuit breaker and the load (which is, in this case, 10 lighting fixtures @ 304VA each).
Thanks.

 

[rbulsara]

Curren in each 2-pole circuit will be 6.34A.

If you have the same load between each 'pair' of the three phases, then the total load will be

3*3040 VA an is balanced

the line current in the Input feeder to the panel will be

(3*3040 VA)/(1.732*480V)=11A,which is same as what skogspurra said.

There are other ways to explain this, but then you should consult some basic text books on polyphase circuits.

 

[buzzp]

I agree with your calculation for the cables mentioned (and the other two posts).

 

[mc5w]

Another point is that if you are increasing the size of the hot wires for any reason you have to proportionately increase the equipment ground size. For your case 20 amp circuit breakers would be fine and will be less likely to nuisnace trip when the cable capacitance is first energized. That is, for a line of this length you will have additional inrush current because of cable capacitance which means that you need to use heftier power relays. The equipment ground in your case would need to be the same size as your hot wires unless one kind is aluminum and the other kind copper.

You should also install a current transformer and cheap (such as Simpson) ammeter on each branch circuit to measure how much current is leaking to ground. It is very easy for 40 feet (12 meters) of broken underground wiring to leak 7 amps at 120 volts into the soil during a drought. The meter will resgister some ground return current when the system is new because of cable capacitance and some of it will be harmonics and some line frequency - record this value. If your electric bill goes way up or something has a tingle voltage you can figure out which circuit is the culprit right away. You would then put a clamp-on ammeter around both main hot wires at each pole to find the segment that is leaking electricity.

Mike Cole, mc5w@earthlink.net