Athletic Field Light question 480v

[gobblerhuntr]

Our city is building 4 soccer fields and putting four poles per field with 11-1500 watt lights per pole at 480 volts. The fields are shaped in a rectangular fashion with a pavillion in the middle of all fields where the transformer and controls will be located as well as a consession stand. In figuring the amps per pole I was doing

11 x 1500= 16500 watts
16500 watts / 480 / 1.732 = 19.85A
Throw in some ballast amps and get around 22A per pole

We are going to run about 400 feet to the closest pole from the transformer and then about 300 to the next, so I have to carry approx 50A for 400 feet then 25A 300 feet. I am going to stay with a single wire size for the whole run. My question is not the wire size but one of the square root of three calculation. The lights require 480 which will be derived from two wires and we will also carry a ground to the pole, so we will use a triplex cable. The poles come prewired for the lights to actually plug in, I was hoping to have a schematic of this so I could visually see the connection point in hopes it would accept all three legs of the 480 and balance the load at each pole.

Is this considered a three phase load even though we are only going to use two legs per pole? I would think so but we will have to balance the windings on the transformer between all the poles to keep our load even at the pot.

I had a drawing of the fields but it is at the office and I can't get to it right now. Can someone set me straight on the square root of three in this application.

 

[dpc]

If you have only two hot wires, that is a single-phase 480 V circuit, assuming your light fixtures are 480 V.

You don't have to worry about the root of three. Just add up the amps drawn by each fixture to get a total.

But I'm not sure if I've interpreted your description correctly or not, because you also talk about balancing the load.

If this is a three-phase system, you will want to have equal loading across each pair of hot wires.

 

[gobblerhuntr]

The lights are 480 volt and my impression is single phase from the supplier. We are looking at setting a 750 kVA 13200:277/480 volt pot to feed all these lights and consession stand.

The lighting manufacturer gave us a amp draw of 22 amps which figures as I have shown above. I just don't see the root three thing on this type of load. I first came up with 35 amps per pole by leaving the root three out. You can see where this would be a substantial difference in wire size by treating it for 35 amps instead of 22.

Balancing the load would be back at the pot on the 480 volt bushings. If it takes two wire for the 480 and I have a three phase pot I would need to utilize all three bushings in order to keep the windings balanced, correct??

 

[cranky108]

If the poles are wired for two wire, then the square root of 3 does not apply.
If the poles are wired 3 phase delta, then the square root of 3 applies.

 

[dpc]

750 kVA - must be a heck of nice concession stand. But this is almost certainly a three-phase transformer. The lights are single-phase (confirm 480 V vs 277 V), but the service is 3-phase, so you need to do three-phase load calcs.

I'd suggest splitting up the lighting load across all three phases to keep the phase loading as even as possible.

I did a similar system in the past and I took three-phase power to each light pole (it only takes one more wire) and split up the load on each pole.

You might want to bring in someone with some three-phase power experience to help you out.

 

[waross]

If your installation was single phase, then the current would be 11 x 1500 / 480 = 35 Amps. (Plus losses)
If the manufacturer has given you a figure of 22 Amps, you most probably have a group of single phase fixtures arranged for a three phase feed.
You will have 12 x 1500W / 480V / 3^2 = 21.7 Amps (plus losses) on the highest loaded phase. You must use 12 until you are able to distribute 11 fixtures evenly over three phases.
The load on A-B will be 4 x 1500 /480 = 12.5 amps.
The load on B-C will be 4 x 1500 /480 = 12.5 amps.
The load on "B" phase will be 12.5 x root 3 rather than 12.5 x 2 because of the phase difference between A-B and B-C.
The current on B-C will be 3 X 1500 / 480 = 9.4 Amps.
As a result, the current in "A" and "C" will be slightly less than the current in "B". The currents are not equal so root 3 may not be used. Vector addition must be used to determine the current in "A" and "C"
No matter. The current in "B" will be the highest and will determine the cable size and protection and voltage drop.
I would "roll" the connections so that the more heavily loaded phase from each lamp standard is connected to a different phase. Vector addition must be used to calculate the sum of unequal currents on three phase. Root 3 may only be used to sum equal currents that are displaced 120 degrees.
Don't worry about the unequal loads. Determine the greatest phase loading and use that for sizing.
For transformer lading, determine the phase with the heaviest load and use root three to determine the transformer loading. Don't try to add phase currents or use root 3 when the currents are unequal.
You may add the number of fixtures on each phase and apply root 3 to the highest number to find the maximum safe loading on the transformer.
If any one else can express this differently or more clearly, please do.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[EEJaime]

Gobblerhuntr,
Waross has pretty much hit the nail on head. Most of the installations we've done from community fields through professionsl sporting venues use a very similar solution. You have a 480y/277V,3ph,4w system. Take advantage of it's efficiencies. Run a 3-phase feed to each lighting standard and terminate it on a 3-pole fused disconnect or breaker. Take fused, two wire leads from a series of terminal blocks in an enclosure on the pole up to the ballast-lamp assembly at the top of the pole, (or in the case of a particular commodity grade manufacturer, to the noisy ballast inside the same enclosure). Rotate the phase connections to each lighting head, (AB,BC,CA,AB,BC,...), so that the unbalanced load at each structure is minimized and virtually eliminated for the installation as a whole.

Your load of 1500W x 1.2(ballast loss)x 11=19800W
Assuming a power factor of 0.8, this is 24.75kVA,
or 29.8 Amps @480V,3ph. Again as Waross said one phase will in fact have one more connection than the other two, so this is slightly off, but not by much. Just wanted to note that the Amperage is a little higher than the figure you quoted, but that may be due to the conservative ballast factor of 20% loss that I used. This is not unusual for a 1500W Metal Halide fixture, but manufacturers differ.

However, you can run a single 480v, single phase feed to each pole and balance the load on a "per pole" basis. This results in a somewhat higher imbalance in that two phases will have an additional connection for the 16th pole.

Many ways to skin this particular cat, good luck.

 

[gobblerhuntr]

EEJamie,
It appears that the poles to be used are prewired with a terminating box 10' off the ground. This box has a two pole breaker in it in which we are going to run a two wire 480 plus a ground into. This will lead us to have to hook it up as you mentioned in the last part of your post. The lights are fused in this box before they go up the pole. I have included a diagram with the fields labeled as we will hook them up (phases). One field stands out but this is a first run at it and we are going to take a closer look. It appears that we will have 6 poles feed by AB and 5 each on BC and AC. We will have to do some studying on the load to make sure this will work. There is a 125 amp fused disconnect in the pavillion to control and protect each field separately and running a 1/0 cu to each side of the field. Then there is a 200A 480/277 panel going in for the pavillion itself, so total capacity is 700 amps. I am starting to think the 750 kVA padmount is overkill if all calcs are correct. I believe this will work fine but thought I would throw it out and see what others thought.

 

[gobblerhuntr]

Here is a .pdf of the file that was attached in the previous post.

 

[EEJaime]

I think you've got it pretty well covered. I would double check with the supplier however regarding the poles along the center aisle with the poles illuminating the fields in two directions. Are you certain there are two poles at each location? Side by side? I wouldn't put it past this particular manufacturer as they don't customize anything and they are not particularly concerned with what the finished installation looks like. I was just concerned that you didn't have these poles along the center with twice the number of fixtures on one pole with stacked Electrical Component Enclosures. I have a stadium where some poles ended up with three or four or even six enclusures stacked one over the other, (which within a year were producing quite a concerto in HUMMMM major, I'd say a good 110dB at the base of the pole).

If your maximum service capacity is 700 Amps, then you could drop the transformer down to a 500 kVA and size your service at 800 Amps. If the transformer is already purchased it will just end up underutilized. As this is a city project, you may want to verify that they do not require the added capacity for future expansion, etc.... Is this fed from a Utility Company? Or is this a City owned distribution system?

 

[gobblerhuntr]

EEJaime,
Your last two lines is a local joke to all of us, it is a city owned electric utility. Long story. Anywho I have dropped the transformer down to a 500 kVA and the poles in between fields are single poles, there will be a total of four between the fields. Thanks for your input.