Help Me Understand - Single Phase Load on Three Phase Panel

[deke997]

Hi guys,

I am new to this forum, and I am DEFINITELY not an expert in this field. I'd like your insight on this.

I am trying to figure out how to do calculations when using a single phase load on a three phase panel.

If I have a 500kVA 208Y/120v transformer, how do I calculate the max single phase load? All loads are resistive and single phase. No motors or three phase loads anywhere.

I THINK that I should be able to use a full 500 kW in this case, but I'm not sure.

IF I wanted to use 120v, I think that I would be able to use 500k/120/3 = 1389 Amps per phase. In this case, since 120v is line to neutral, it would also be 1389 Amps per line, correct? That is, if I were take an ammeter and measure the current in any of the incoming lines at full load, would it also be 1389 Amps? If this is correct, then I would need an 1800 Amp main breaker on the box, right?

However, IF I wanted to use 208v for everything, then that would be 500k/208/3 = 801 Amps per phase. Is this correct? How would I calculate the current per line in this case (assuming everything was balanced)? That is, how would I calculate the expected ammeter reading on any of the incoming lines at full load? When I try to think about it, it gets confusing. If I'm using 208v, then all the breakers in the box will be 2-pole breakers on AB, BC, and CA. It seems like each phase should have a component of it's load on two different configurations, then. For instance, part of the load on line "A" comes from AB, and part comes from CA. If this is the case, how do you find the sum? I assume you need to add them vectorially somehow? (but HOW?)

If everything is balanced, my intuition says that it should still be 801 Amps per line. Would this mean that I would need a 1000 Amp main breaker?

Now, assuming the above is correct, in order to keep it up to code, I believe that I would need to go from the transformer into a distribution panel. From there, I could split the load into a few 200A or 400A panels and then go from there. Is this right? Even in this case, I think the math will be the same and simply proportional according to how I split it up.

 

[ron]

You will be able to load the 500kVA three phase transformer (which is really three 166.67kVA single phase transformers jammed together) to 500kVA.

Be sure to distribute your single phase loads evenly across phases as you describe.

 

[kwired]

Hi guys,

I am new to this forum, and I am DEFINITELY not an expert in this field. I'd like your insight on this.

I am trying to figure out how to do calculations when using a single phase load on a three phase panel.

If I have a 500kVA 208Y/120v transformer, how do I calculate the max single phase load? All loads are resistive and single phase. No motors or three phase loads anywhere.

I THINK that I should be able to use a full 500 kW in this case, but I'm not sure.

IF I wanted to use 120v, I think that I would be able to use 500k/120/3 = 1389 Amps per phase. In this case, since 120v is line to neutral, it would also be 1389 Amps per line, correct? That is, if I were take an ammeter and measure the current in any of the incoming lines at full load, would it also be 1389 Amps? If this is correct, then I would need an 1800 Amp main breaker on the box, right?

However, IF I wanted to use 208v for everything, then that would be 500k/208/3 = 801 Amps per phase. Is this correct? How would I calculate the current per line in this case (assuming everything was balanced)? That is, how would I calculate the expected ammeter reading on any of the incoming lines at full load? When I try to think about it, it gets confusing. If I'm using 208v, then all the breakers in the box will be 2-pole breakers on AB, BC, and CA. It seems like each phase should have a component of it's load on two different configurations, then. For instance, part of the load on line "A" comes from AB, and part comes from CA. If this is the case, how do you find the sum? I assume you need to add them vectorially somehow? (but HOW?)

If everything is balanced, my intuition says that it should still be 801 Amps per line. Would this mean that I would need a 1000 Amp main breaker?

Now, assuming the above is correct, in order to keep it up to code, I believe that I would need to go from the transformer into a distribution panel. From there, I could split the load into a few 200A or 400A panels and then go from there. Is this right? Even in this case, I think the math will be the same and simply proportional according to how I split it up.

No, it would be 500k/208/1.732 = ~1389, still same amps per leg.

 

[charlie b]

First of all, it would help if you forever hereafter cease all use of the phrase, "amps per phase." Current that leaves the transformer on Phase A is going to return to the transformer on Phases B and C. It is the same current. Secondly, 1800 amps is not a standard breaker setting. See 240.6. I would use 1600 amps as the main breaker downstream of a 500 KVA transformer.

 

[kwired]

First of all, it would help if you forever hereafter cease all use of the phrase, "amps per phase." Current that leaves the transformer on Phase A is going to return to the transformer on Phases B and C. It is the same current. Secondly, 1800 amps is not a standard breaker setting. See 240.6. I would use 1600 amps as the main breaker downstream of a 500 KVA transformer.

That is how one generally should approach it for service and feeder calculations but a wye system with only a single line to neutral load - current does return to the neutral point and not to either the other two "phase conductors".

Severely unbalance line to neutral loads and you have a majority of current flowing back to neutral as well.

 

[wwhitney]

However, IF I wanted to use 208v for everything, then that would be 500k/208/3 = 801 Amps per phase. Is this correct? How would I calculate the current per line in this case (assuming everything was balanced)? [. . .] I assume you need to add them vectorially somehow? (but HOW?)

In practice, it is often simpler to find a solution path that avoids vectors, but if you choose to do the vector calculation you'll get the same answer. As an example, the vectorial computation for line B is as follows:

Say the A-B load current is 801 amps, and the B-C load current is also 801 amps. If the loads are identical (so that any phase shift between current and voltage is the same for both loads), then the two current vectors specified will be 120 degrees apart. However as written one of those current vectors is defined with the sense of current flowing to B (say the A-B vector), and the other current vector is defined with the sense of current flowing from B (the B-C vector). To add the currents, we need two vectors defined with the same sense of current flow, so one of the vectors gets negated before adding. [Negation meaning multiply by -1, or flip the vector around 180 degrees.]

That negation gives us two vectors to add that are only 60 degrees apart. There are a number of ways to do the vector addition: arithmetically by writing one vector as (801, 0) and the other vector as (801 * cos 60 deg, 801 * sin 60 deg), adding the components, and determining the resulting length. Or you could do it geometrically by drawing the two vectors tip to tail, drawing the resulting vector, and using a little trigonometry to compute the resultant length. Either way you do the vector computation, you'll end up with a current of 801 * sqrt(3), or 1387 amps. The discrepancy from your single phase calculation of 1389 amps is entirely due to rounding errors.

Cheers, Wayne

 

[hbendillo]

When you have a three-phase system with single phase loads your main wire and breaker sizes in panels are determined by the highest single phase load. 500 KVA is not going to be evenly distributed across the phases in a typical power system. Sometimes the unbalance can be significant. If so it is best to try to balance the load as much as possible by moving your single phase loads around in the various panels.

 

[deke997]

Thank you all for your help! I believe I understand now. I wasn't properly incorporating sqrt(3) into my calculations. I got it now.

500 KVA is not going to be evenly distributed across the phases in a typical power system. Sometimes the unbalance can be significant. If so it is best to try to balance the load as much as possible by moving your single phase loads around in the various panels.

My load will be highly balanced by design.

1800 amps is not a standard breaker setting. See 240.6. I would use 1600 amps as the main breaker downstream of a 500 KVA transformer.

Thanks for pointing this out. I will be constantly using the entire 1389 Amps, so a 1600A breaker wouldn't work. I would either have to use a 2000A breaker or wire multiple panels to the transformer.

Speaking of this, does anyone know if I'll run into any issues if I wire multiple panels? For example, if I wired a 1200A panel and a 600A panel directly into the transformer, is there anything I need to watch out for? I believe I could wire three 600A panels with no worries as well, assuming there is space in the transformer, correct? Does anyone know the limit to the number of panels which can be wired in directly to the transformer (2? 4?)?

 

[GoldDigger]

Speaking of this, does anyone know if I'll run into any issues if I wire multiple panels? For example, if I wired a 1200A panel and a 600A panel directly into the transformer, is there anything I need to watch out for? I believe I could wire three 600A panels with no worries as well, assuming there is space in the transformer, correct? Does anyone know the limit to the number of panels which can be wired in directly to the transformer (2? 4?)?

Just watch the size of the conductors from transformer to panels. Unless they each have adequate ampacity for full transformer secondary current you will need to conform to tap rules to the panel mains.

 

[kwired]

Thank you all for your help! I believe I understand now. I wasn't properly incorporating sqrt(3) into my calculations. I got it now.



My load will be highly balanced by design.



Thanks for pointing this out. I will be constantly using the entire 1389 Amps, so a 1600A breaker wouldn't work. I would either have to use a 2000A breaker or wire multiple panels to the transformer.

Speaking of this, does anyone know if I'll run into any issues if I wire multiple panels? For example, if I wired a 1200A panel and a 600A panel directly into the transformer, is there anything I need to watch out for? I believe I could wire three 600A panels with no worries as well, assuming there is space in the transformer, correct? Does anyone know the limit to the number of panels which can be wired in directly to the transformer (2? 4?)?

Is the transformer supplying the service or is it a separately derived system? makes a difference in which code sections apply.

 

[augie47]

If its a SDS as opposed to a service, you need to take your primary protection and Art 450.3 into account. The transformer itself needs to be protected. If you exceed 125% of the rated load on the secondary such as with multiple panels then you need to make sure that the primary is protected at 125%.

 

[Ultrafault]

The cliffnotes version of these posts is: Power (assuming unity power factor KVA = KW) in a three phase system, delta or wye, is equal to 1.73xV(line to line)xI(Line). Or in this system 500,000=1.73x208xIL.

Sent from my SM-N910P using Tapatalk

 

[GoldDigger]

The cliffnotes version of these posts is: Power (assuming unity power factor KVA = KW) in a three phase system, delta or wye, is equal to 1.73xV(line to line)xI(Line). Or in this system 500,000=1.73x208xIL.

Sent from my SM-N910P using Tapatalk

Which, for those who are more comfortable looking at line to neutral loads from the wye perspective, is mathematically identical to 3 x (V_{line to neutral} x I_line)

 

[Ultrafault]

Which, for those who are more comfortable looking at line to neutral loads from the wye perspective, is mathematically identical to 3 x (V_{line to neutral} x I_line)

Absolutely, however, I like the simplicity of the orignal formula becuase it works for all cases with no conversions necessary.

Sent from my SM-N910P using Tapatalk

 

[deke997]

Is the transformer supplying the service or is it a separately derived system? makes a difference in which code sections apply.

If its a SDS as opposed to a service, you need to take your primary protection and Art 450.3 into account. The transformer itself needs to be protected. If you exceed 125% of the rated load on the secondary such as with multiple panels then you need to make sure that the primary is protected at 125%.

It is an SDS

 

[kwired]

It is an SDS

Then secondary overcurrent protection likely comes into play as Augie mentioned.