tryinghard
Senior Member
- Location
- California
The high school I graduated from (my old alma mater) has asked me to help in any way to upgrade their football field lighting. I am glad to help at least design the circuitry and installation; seems like all schools are looking for donations to save money. The lighting levels have already been engineered to achieve a 50fc average by the manufacture but I would like to verify my calculations.
There are a total of 4 light standards and 46 luminaires; I will be feeding the controller with 480V, 125A, and 3W circuit. One side of the field will have two light standards each with 11 luminaires and I am going to circuit these through a common existing underground conduit. The circuit is 480V 3W and the lamps are 1500W each and I am assuming a 90% power factor, I have calculated the farthest light standard as follows: I=P/(Ex1.73x.90) or 22=16,500/(480x1.73x.90) so my circuit is 30A=22x1.25. Distance is 550? therefore: VD=1.73xKxIxD/CM or 16.41 (2.2%)=1.73x12.9x22.08x550/16,510 as #8, this calculation reveals a 30A circuit with 3-#8?s are okay. I do have a few questions though.
Q1: If PF=Watts/VA do I have enough information above to arrive at this without simply assuming 90%?
Q2: Should I calculate the ?K? factor rather than plugging 12.9 for the conductor?
Q3: I will be combining two 3-phase circuits in an existing 1-1/2? underground conduit, with this I understand table 310-15(B)(2)(a) as I can now have an allowable ampacity for #8 at 80% of 50A which is 40A; isn?t this only applicable to my actual load 22A and not my circuit breaker 30A?
There are a total of 4 light standards and 46 luminaires; I will be feeding the controller with 480V, 125A, and 3W circuit. One side of the field will have two light standards each with 11 luminaires and I am going to circuit these through a common existing underground conduit. The circuit is 480V 3W and the lamps are 1500W each and I am assuming a 90% power factor, I have calculated the farthest light standard as follows: I=P/(Ex1.73x.90) or 22=16,500/(480x1.73x.90) so my circuit is 30A=22x1.25. Distance is 550? therefore: VD=1.73xKxIxD/CM or 16.41 (2.2%)=1.73x12.9x22.08x550/16,510 as #8, this calculation reveals a 30A circuit with 3-#8?s are okay. I do have a few questions though.
Q1: If PF=Watts/VA do I have enough information above to arrive at this without simply assuming 90%?
Q2: Should I calculate the ?K? factor rather than plugging 12.9 for the conductor?
Q3: I will be combining two 3-phase circuits in an existing 1-1/2? underground conduit, with this I understand table 310-15(B)(2)(a) as I can now have an allowable ampacity for #8 at 80% of 50A which is 40A; isn?t this only applicable to my actual load 22A and not my circuit breaker 30A?
