Single Phase Load Calculations on Three Phase system

[jbwhite]

This question is about sizing the branch circuit breaker for a parking lot lighting system.

I have simplified the numbers for ease of understanding so do not worry if the actual wattages are not correct.

The example is a group of parking lot lights each drawing 1200 watts at 240 volts. We will be connecting them to a Three Phase 240 volt breaker. Each single phase fixture will be connected to the Three Phase feeder in a round robin fasion.( first A,B: Second B,C: Third C,A ect)

I planned to find the maximum number connected to any two legs and use a single phase calculation to determine the circuit amps.

EG. 13 fixtures would give me 5 as the max between any two legs. So I multiply 5 X 1200 and get 6000 then multiply by 1.25 for continous duty and get 7500 then divide by 240 to get 31.25 amps total load.

So I would use a 40 amp breaker, and use the 31.25 amps for VD and other calculations.

Is this correct?

If not, what is the proper way?

 

[Dennis Alwon]

Hey JB -- long time

I would go about it differently. I would find the maximum on one phase-- this would be 9 (I think). Then I would multiply by 1200. That gives us 10800.

Divide by 2 to get the load on a single phase--- 5400. Then I would multiply by 3 and get 16200. Divide by 240*1.73= 39 amps.

Then I would multiply by 1.25

 

[iwire]

I would determine the VA of one fixture, multiply that by the total number of fixtures for the total VA required.

Divide that VA by the voltage and 1.73 that will be you VA per phase.

Now if you need to you can add the 25% for continuous loads.

15 - 1000 watt fixtures.

15 x 1000 = 15, 000 / 240 = 62.5 / 1.73 = 36 amps per phase.

This of course assumes you will keep it balanced, but even with the circuit not perfectly balanced (If you had 16 or 17 total fixtures) you will get 'close enough' for branch circuit determination.

 

[bob]

The example is a group of parking lot lights each drawing 1200 watts at 240 volts. We will be connecting them to a Three Phase 240 volt breaker. Each single phase fixture will be connected to the Three Phase feeder in a round robin fasion.( first A,B: Second B,C: Third C,A ect)

JB
You have 13 lamps, you would have 4 lamps x 1200 x 3 = 144000 watts
144000/240 x 1.73 = 34.7 amps line current + the 1 for a total of 13 lamps.
The load on the one lamp is a vector addition and not added directly to the
34.7 although it would not be a large error if you did so.

 

[Dennis Alwon]

And I thought this was going to be an easy one instead we get 3 different answers to the same question.

 

[augie47]

answers

answers

And I thought this was going to be an easy one instead we get 3 different answers to the same question.

why should this thread be any different ?

 

[cadpoint]

I had to remeber:

.... "I have simplified the numbers for ease of understanding so do not worry if the actual wattages are not correct." ...

I thought that second cup was never going to kick in ... LOL

 

[iwire]

And I thought this was going to be an easy one instead we get 3 different answers to the same question.

I bet we could get 9 more answers and each one them will work ....as always (at least in my mind) it comes down to speed vs accuracy.

We could figure it out very carefully and precisely and if my job was an Engineer I bet I would do it that way.

But my job is 'git er done' and many times a quick a dirty method works fine. :smile:

 

[Limey Pete]

JB
You have 13 lamps, you would have 4 lamps x 1200 x 3 = 144000 watts
144000/240 x 1.73 = 34.7 amps line current + the 1 for a total of 13 lamps.
The load on the one lamp is a vector addition and not added directly to the
34.7 although it would not be a large error if you did so.

I am confused as to why this last lamp would be a "vector addition"
The only vector I see that need to be used is if we were trying to find final neutral current, but as this circuit doesn't even require a neutral.

I believe the very first post had the right calculation. Think of it this way; If I had 5 x 1200 lamps across A &B phase how could the addition or subtraction of loads across either A&C or B&C possibly affect the current draw (and hence load) on A&B phase. and we would need to take the worst case scenario of 5 lamps max. per phase as I don't see getting 3 pole breakers or wire of different sizes/ratings per phase.

OOOPS! EXCEPT; I just read the first post again and saw he didn't allow for 1.73 in the equation.

Pete.

 

[donw]

I planned to find the maximum number connected to any two legs and use a single phase calculation to determine the circuit amps.

EG. 13 fixtures would give me 5 as the max between any two legs. So I multiply 5 X 1200 and get 6000 then multiply by 1.25 for continous duty and get 7500 then divide by 240 to get 31.25 amps total load.

So I would use a 40 amp breaker, and use the 31.25 amps for VD and other calculations.

I agree with this calculation for BRANCH calculations - breaker size, VD, etc, but not for feeder/panel calculations. That's when you need the 1.73 factor.
EDIT: Ooops! I just noticed that this will be a three-phase (tied) breaker, so you should calculate the max 3-phase current and set your breaker accordingly.

 

[bob]

I am confused as to why this last lamp would be a "vector addition"
The only vector I see that need to be used is if we were trying to find final neutral current, but as this circuit doesn't even require a neutral.Pete.

Its a vector sum because the lamp added is a phase load. What you need is the line load. Thats what the breaker sees.

 

[Limey Pete]

Its a vector sum because the lamp added is a phase load. What you need is the line load. Thats what the breaker sees.

It makes me feel just marginally better that I managed to edit my own mistake before you corrected me.
Very bad day today!, I shall be content to read and not reply to further threads until I have had at least 8 hours sleep. :-?

 

[hardworkingstiff]

......

Is this correct?

I don't believe it is.

If you have three columns, A B & C and 13 rows, put your loads down under the columns as you would connect them. You would have 9 connections on each A & B and 8 connections on C. Each connection is 5-amps. A & B would be loaded at 45-amps and C would be loaded at 40-amps.

EDIT to add:

If you take 13 fixtures at 1200 watts = 13*1200 = 15,600 total watts.

15,600 watts / (240*1.732) = 37.5 amps. But this is not correct because the load is not balanced on the 3 phases.

15 fixtures would balance, 15 * 1200 = 18,000 / (240*1.732) = 43.3-amps.

Someone smarter than I will have to explain the difference between 43.3-amps and 45-amps.

 

[bob]

It makes me feel just marginally better that I managed to edit my own mistake before you corrected me.
Very bad day today!, I shall be content to read and not reply to further threads until I have had at least 8 hours sleep. :-?

Pete
You have 13 lamps, you would have 4 lamps x 1200 x 3 = 144000 watts
144000/240 x 1.73 = 34.7 amps line current + the 1 for a total of 13 lamps.
The load on the one lamp is a vector addition and not added directly to the
34.7 although it would not be a large error if you did so.
If I did it correct the vector sum = 37.64 and the addition as you stated is
20 + 5 = 25 x 1.73 = 43.25. Not a big difference.

 

[jbwhite]

Thanks for the insight so far.

Is there an easy way to explain how much to add if I do install 13 or 14 fixtures leaving one or two that are not balanced.

Assuming the first 12 are connected in a complete round robin and the last two are connected AB and BC.

Also what would happen if the first 12 were balanced, and then or some of the loads on one phase were open due to lamp failure or something. Could my load current on the other two phases rise enough to trip the breaker?

 

[Smart $]

Is there an easy way to explain how much to add if I do install 13 or 14 fixtures leaving one or two that are not balanced.

Assuming the first 12 are connected in a complete round robin and the last two are connected AB and BC.

No easy way, as the line current is affected by the total number of fixtures connected to each line. For instance, how much one would add would be different for 16 and 17 fixtures even though only one or two again cause the imbalance.

The following depicts how I would figure the scenario in the OP, and assumes a power factor of 0.9. Different fixtures' power factors can vary substantially, so do not use the 0.9 value as a given.


Click on image if you want to download the Excel file.

To determine line current from load current I use the following formulas:

Also what would happen if the first 12 were balanced, and then or some of the loads on one phase were open due to lamp failure or something. Could my load current on the other two phases rise enough to trip the breaker?

No... that is, assuming no fault condition is present. Open shorts and lamp failures will decrease the load current on connected phases.

 

[Dennis Alwon]

Yikes---------

All of this is very interesting but the reality is that most of us electricians don't understand nor do they use those formulas. Surely the NEC or master exams don't expect those type of calculations with vectors and whatnot.

I don't, in any way, intend to demean or belittle anyones efforts or abilities. In fact I am quite impressed by what is here even if I don't fully comprehend it.

It just seems there must be a simplier way to caluculate the loads than figuring most of the factors that are being brought into the equations.

We have had a number of different responses and different enough that the calculated wire size would be different in some of the scenarios.

I never considered myself good with calculations---sure give me the formulas and the values and I can do the math but how is the average Joe electrician suppose to be able to calculate an answer when none of us here at the forum has gotten the same answer.

Is there no right answer? What is the NEC answer to this question?

Do we really need to put vectors in to solve the problem?

Do we need to use the formula that Smart used above? It is another language to me.

I'd be lost-- mind as well give up my license. Maybe this is why I should stick with residential. I rarely have to do any calculations.

 

[Mr. Bill]

There's always a right answer. But people will always disagree over that answer. I don't think the NEC explains many calculation methods. An engineering text book would be of better use in finding the answer. If you want an exact number then I do think vector calculations are needed. But approximations work well most of the time.

Also note 210.23(C). If you're going to use a 40A circuit for lighting then the luminaires need to have heavy duty lamp holders.

 

[Jomaul]

13 fixtures = max of 5 connected to any 2 legs
1200w * 5 = 6000 watts
6000watts / 2 = 3000 watts per phase
3000 * 3 = 9000 watts total load
9000 / 240 * 1.73 =21.65
21.65 * 1.25 = 27.06 amps

Could this be correct?

 

[boater bill]

Verify with the lamp manufacturer that a 40 or 50 amp breaker is acceptable. This would require individual fusing at each pole in my experience. I know you are adding in the ballast loads for the real calculations. HPS and MH ballasts add a significant load to the lighting load.