Transformer sizing conductors and lights at park

[Prototype1]

I have a transformer 150KVA Primary - 12470 GRDY /7200 V and Secondary 208Y/120 V. ( I understand that GRDY is grounded wye(Y) primary configuration with 12,470V between each phase and the neutral, and 7200 v between any two phases.) (208Y/120 V is three phase, four wire secondary configuration, 208 v between each phase and 120 v between any phase and the neutral.)

If it is supervised location for the transformer than you are going to select next lower standard size: if it is (primary voltage > 600V and secondary voltage > 600V). But if the (secondary voltage <= 600V), select next Higher standard size, fuse/circuit breaker.)

Conditions: Primary side of transformer
Overcurrent protection at primary side (Primary > 600V) : Rating of primary fuse at point A = 300 % of primary. FLC or next standard size or Rating of primary circuit breaker at point A = 600%. FLC or next standard size. - Unsupervised

• Rating of Primary Circuit Beaker, so next higher standard size. - Unsupervised

Calculation: Primary side
(Primary > 600V): Rating of primary circuit breaker at point A = 600%

I did Full load current: 150000 VA / 12470 V*1.732 = 6.945 A
Rating of primary Circuit Breaker = 6*6.945 = 41.67 A Circuit Breaker. from table 240.6(A)

conductor sizing: 6.945 A * 1.25 = 8.68 A??? Doesn't make sense
Table 310.16, 75°C: AWG ???? What I should do?

Conditions: Secondary side of transformer
Overcurrent protection at secondary side (Secondary Voltage <= 600V)

• Rating of Sec. Fuse / Circuit Breaker at Point B= 125% of Sec. Full Load Current or Next higher Standard size. -Unsupervised

Overcurrent Protection at Secondary Side (Secondary Voltage >600V):

• Rating of Secondary. Fuse at Point B= 250% of Sec. Full Load Current or Next higher Standard size. -Unsupervised
• Rating of Secondary. Circuit Breaker at Point B= 300% of Sec. Full Load Current. -Unsupervised

Calculation: Secondary side
Overcurrent protection at secondary side (Secondary Voltage <= 600V):

I did Full load current: 150000 VA / 208V*1.732 = 416.73 A
Rating of primary Fuse/Circuit Breaker: 416.37 A * 1.25 = 520.46 A
520.46 A, so next standard size of circuit breaker = 600A Circuit breaker from table 240.6(A)

conductor sizing: 416.37 A * 1.25 = 520.46 A
Table 310.16, 75°C: 520.46 A. So, I select (2) 300 kcmil conductors in parallel. (285A+285 = 570 A)
Table 250.122 for grounding: #1 G AWG.

Is this how you size your conductors for transformer?

I have 68 lights (tennis court lights, trail lights and other lights) and the lights total watts = 8400 Watts. ( I got total light watts like this: example: 20 trail lights poles * 45 watts = 900 watts + 10 bollards * 38 watts = 380 Watts - I add (900+380 = 1280 watts).

How do I go by to size conductors for lights?

 

[tortuga]

I have 68 lights (tennis court lights, trail lights and other lights) and the lights total watts = 8400 Watts. ( I got total light watts like this: example: 20 trail lights poles * 45 watts = 900 watts + 10 bollards * 38 watts = 380 Watts - I add (900+380 = 1280 watts).

How do I go by to size conductors for lights?

I'll start with the lights and what version of the NEC are you using 2023?
Those lights are likely a continuous load and that will affect all your calcs, See 220.11(B) and 210.19.

 

[Prototype1]

I'll start with the lights and what version of the NEC are you using 2023?
Those lights are likely a continuous load and that will affect all your calcs, See 220.11(B) and 210.19.

Yes, I am using 2023 NEC. The lights are continuous load. So, I need to size the conduit and conductors going to new electrical panel and going to 68 lights around the park. I attached a example drawing of it to visual it. When you say it will affect my calculations. Can you please tell or give example?

 

[tortuga]

​	​	​	​	​	​	​	​
Feeder​	​	Lamp Watts​	Ballast / Driver nameplate Amps​	Branch circuit voltage​	VA each​	QTY​	Line total (VA)​
​	Tennis court lights Fixture Sch. A​	1000​	5.5​	208​	1144​	12​	13728​
​	Pool area Lights Fixture Sch B​	400​	3.9​	120​	468​	24​	11232​
​	New Path Lights​	45​	0.5​	120​	60​	40​	2400​
​	Total lighting​	​	​	​	​	​	27360​
​	25% Of continuous load​	​	​	​	6840​	​	6840​
​	Total Feeder A2 panel A2 load​	​	​	​	​	​	34200​
​	​	​	​	​	​	​	​
​	​	​	​	​	​	​	​
Branch Ckts​	​	​	​	​	​	​	​
​	​	VA​	125% of VA​	CKT size (A)​	CKT size VA​	Lights per ckt​	Circuits req.​
​	Tennis court lights Fixture Sch. A​	1144​	1430​	30​	6240​	4​	3​
​	Pool area Lights Fixture Sch B​	468​	585​	20​	2400​	4​	6​
​	New Path Lights​	60​	75​	20​	2400​	32​	2​
​	​	​	​	​	​	​	​
​	​	​	​	​	​	​	​

 

[Prototype1]

I have this lights.
Pickleball court – Total Watts – 3,584 (8 lights)
Tennis – Total Watts – 2,406 (6 lights)
Trail Lights – 45 Watts * 26 Poles = 1170 Watts (26 Poles lights)
Bollards – 38 Watts * 20 Bollards = 760 Watts (20 Bollards lights)
Disk Lights – 60 Watts * 8 = 480 Watts (8 lights)

So, to get VA from watts. I have to do equation Watts / Power Factor = VA?

I need to do calculations to get Branch circuits too?

 

[tortuga]

So, to get VA from watts. I have to do equation Watts / Power Factor = VA?

It used to be a bigger issue back in the days of Metal halide/ HPS park lighting 220.10(B) says you need enough wire /feeder size to supply the VA of the ballast or LED driver as well as the lamp.

This info is typically from the data sheet or nameplate if existing equipment in amps.
I take the amps X the branch circuit voltage = VA.
So in my case I have existing 208V connected 1000W metal halide on the court, I have to go by the amps on the ballast tap for 208 which is in my case 5.5 amps. 5.5 X 208 = 1144VA.
The led path lights say "45 watt" but then led driver says 1/2 Amp @ 120V so I use the 1/2 amp value not the "lamp watts".

 

[david luchini]

Beaker, so next higher standard size. - Unsupervised
Calculation: Primary side
(Primary > 600V): Rating of primary circuit breaker at point A = 600%

I did Full load current: 150000 VA / 12470 V*1.732 = 6.945 A
Rating of primary Circuit Breaker = 6*6.945 = 41.67 A Circuit Breaker. from table 240.6(A)

conductor sizing: 6.945 A * 1.25 = 8.68 A??? Doesn't make sense
Table 310.16, 75°C: AWG ???? What I should do?

50A c/b would be the maximum allowable size.

The conductor needs only to have an ampacity of 6.945A.

The smallest 15kV conductor size is #2. See 311.60, not 310.16. A #2 conductor has an ampacity well above the required 6.945A.

Secondary side
Overcurrent protection at secondary side (Secondary Voltage <= 600V):

I did Full load current: 150000 VA / 208V*1.732 = 416.73 A
Rating of primary Fuse/Circuit Breaker: 416.37 A * 1.25 = 520.46 A
520.46 A, so next standard size of circuit breaker = 600A Circuit breaker from table 240.6(A)

conductor sizing: 416.37 A * 1.25 = 520.46 A
Table 310.16, 75°C: 520.46 A. So, I select (2) 300 kcmil conductors in parallel. (285A+285 = 570 A)
Table 250.122 for grounding: #1 G AWG.

600A c/b would be the maximum allowable size. (I'd be inclined to go with 500A or even 400A depending on the load.)

The secondary conductor ampacity must be at least 600 for the 600A secondary c/b, per 240.21(C). So, (2) 350kcmil conductors in parallel.

The "ground" (supply side bonding jumper) would be #2 per 250.102(C).

 

[Prototype1]

50A c/b would be the maximum allowable size.

The conductor needs only to have an ampacity of 6.945A.

The smallest 15kV conductor size is #2. See 311.60, not 310.16. A #2 conductor has an ampacity well above the required 6.945A.

600A c/b would be the maximum allowable size. (I'd be inclined to go with 500A or even 400A depending on the load.)

The secondary conductor ampacity must be at least 600 for the 600A secondary c/b, per 240.21(C). So, (2) 350kcmil conductors in parallel.

The "ground" (supply side bonding jumper) would be #2 per 250.102(C).

Calculation: Primary side
(Primary > 600V): Rating of primary circuit breaker at point A = 600%

I did Full load current: 150000 VA / 12470 V*1.732 = 6.945 A
Rating of primary Circuit Breaker = 6*6.945 = 41.67 A, and **45 A Circuit Breaker. from table 240.6(A)**

conductor sizing: 6.945 A * 1.25 = 8.68 A ( I thought you nee to time it by 125%) to get FLC and get the conductor size?
Table 310.16, 75°C: ?? ( I did not see 311.60 in NEC 2023 book, it directly goes from Article 310 to Article 312)

Secondary side
Overcurrent protection at secondary side (Secondary Voltage <= 600V):

I did Full load current: 150000 VA / 208V*1.732 = 416.73 A
Rating of primary Fuse/Circuit Breaker: 416.37 A * 1.25 = 520.46 A
520.46 A, so next standard size of circuit breaker = 600A Circuit breaker from table 240.6(A)

conductor sizing: 416.37 A * 1.25 = 520.46, **Okay I understand the conductors should have ampacity of at least 600 A because of circuit beaker is 600A.
Table 310.16, 75°C: **600A: so, I select (2) 350 kcmil conductors in parallel.**
Table **250.102** for grounding: #2 G AWG. **Question: So, if it is undergrounded conductor and in parallel. We will be using table 250.102(C)(1) and not Table 250.112??**

What about the lights? Can you help me on it?

 

[david luchini]

( I did not see 311.60 in NEC 2023 book, it directly goes from Article 310 to Article 312)

Sorry...315.60 in NEC 2023

So, if it is undergrounded conductor and in parallel. We will be using table 250.102(C)(1) and not Table 250.112??

Even above ground and not in parallel, you would still you 250.102(C).

 

[Prototype1]

50A c/b would be the maximum allowable size.

The conductor needs only to have an ampacity of 6.945A.

The smallest 15kV conductor size is #2. See 311.60, not 310.16. A #2 conductor has an ampacity well above the required 6.945A.

600A c/b would be the maximum allowable size. (I'd be inclined to go with 500A or even 400A depending on the load.)

The secondary conductor ampacity must be at least 600 for the 600A secondary c/b, per 240.21(C). So, (2) 350kcmil conductors in parallel.

The "ground" (supply side bonding jumper) would be #2 per 250.102(C).

Calculation: Primary side
(Primary > 600V): Rating of primary circuit breaker at point A = 600%

I did Full load current: 150000 VA / 12470 V*1.732 = 6.945 A
Rating of primary Circuit Breaker = 6*6.945 = 41.67 A, and **45 A Circuit Breaker. from table 240.6(A)**

conductor sizing: 6.945 A * 1.25 = 8.68 A ( I thought you nee to time it by 125%) to get FLC and get the conductor size?
Table 310.16, 75°C: ??

I did not see 6.945 A in 315.60 tables. I check from pages 70-207 to 70-215. I am confuse as 6.945 A is very low ampacity. Now, I am confuse.

 

[david luchini]

conductor sizing: 6.945 A * 1.25 = 8.68 A ( I thought you nee to time it by 125%) to get FLC and get the conductor size?
Table 310.16, 75°C: ??

No reason to multiply the transformer primary rated current by 125%. See 235.202(A) in the 2023 NEC.

I did not see 6.945 A in 315.60 tables. I check from pages 70-207 to 70-215. I am confuse as 6.945 A is very low ampacity. Now, I am confuse.

6.945A is the minimum required ampacity for the transformer feeder. You need a conductor that has an ampacity that is 6.945 or greater.

#2 awg is the minimum size conductor for 15kV, and #2 has an ampacity much higher than 6.945.

 

[Prototype1]

I am probably wrong so please correct me. When you say 15KV, do you mean the transformer? It is 150 KVA Primary though? I think I am lost where the 15 KV is coming from.

 

[david luchini]

I am probably wrong so please correct me. When you say 15KV, do you mean the transformer? It is 150 KVA Primary though? I think I am lost where the 15 KV is coming from.

You said the transformer primary is 12,470V. You will need a conductor that has 15,000V (15kV) insulation.

 

[Prototype1]

Sorry, I did not think about that. I need look into it. I want keep working on this but I have to go to meeting. Are you sir going to available after 5 PM? It would be great if you there to answer my dumb questions.

Thank you for you help till now. I was really annoyed by this.

 

[Prototype1]

You said the transformer primary is 12,470V. You will need a conductor that has 15,000V (15kV) insulation.

Are you looking at Table 315.60(C)(1)? 5001-35000 Volts ampacity?

 

[Prototype1]

You said the transformer primary is 12,470V. You will need a conductor that has 15,000V (15kV) insulation.

Okay, got you. So, if my transformer primary is 15,001 V and will I need a conductor that is 28,000 (28kV) insulation? Looking at tables 315.10

No reason to multiply the transformer primary rated current by 125%. See 235.202(A) in the 2023 NEC.

6.945A is the minimum required ampacity for the transformer feeder. You need a conductor that has an ampacity that is 6.945 or greater.

#2 awg is the minimum size conductor for 15kV, and #2 has an ampacity much higher than 6.945.

Okay, I read the 235.202(A) and it says "ampacity of feeder conductors shall not be less than the sum of the nameplate rating of the transformers supplied when only transformers are supplied." So, the ampacity of 6.945 A is required ampacity. Still confuse on the wording but get grasp of it.

For #2 AWG at 15kV, I found that in table 315.12(A) Minimum size of conductors. 8001-15,000 volts is AWG #2. [not in Table 315.60(C)(1)]

 

[augie47]

I missed your calculated load. Note you do not have to size your secondary at the transformer max output but can size it to the load. Something to consider cost wise

 

[Prototype1]

I missed your calculated load. Note you do not have to size your secondary at the transformer max output but can size it to the load. Something to consider cost wise

I don't have to size conductors for secondary side of transformer at max output? So, not 208 but use 120 V to size it?

 

[augie47]

Calculate the load in kw....
add in a "growth" factor if desired
On you 208/120 3 phase system Total kw/360= amps load
Select the main breaker panel you wish to use based on that load + growth
Size transformer secondary conductors to match that breaker.

That said, unless there is some significant load on the transformer (existing or proposed) the number make no sense as you are looking at a load of less than 10kva on a 150 kva transformer."????

 

[Prototype1]

Hmm, I need make sense on calculate the Total kw/360 = amps load and selecting the main breaker panel. I will message with my work (what I did).

Yes, the transformer is already existing on park and there is existing load on that transformer. It is feeding the building near it.