120/208V single phase panel amperes

[slice]

Hello all I have calculated the load on a single phase 3 wire 208v panel to be ~21000VA. In order to determine the current rating of the panel I believe all I have to do is divide 21000 by 208. This comes up with a current of ~100A, which would lead me to specify a 125A panelboard. This is assuming the loads are balanced between the two phases. I think this is a fairly simple question however I have been looking at past threads and have uncovered several similar situations and they refer to the Oregon fudge factor. I don't believe I need to take this into account here. Any input is appreciated.

Thanks

 

[augie47]

I can only speak for myself but if you gave me a load calculation of ≈21000 at 208v I would not accept a 100 amp panel. 100 amp would not suffice for 21000 or any higher number and ( ≈ ) certainly could be higher.

 

[don_resqcapt19]

Assuming that the load calculations have been per Article 220, and understanding that they are very conservative, I would probably look to see of I could cut 200va out of the calculation.

 

[winnie]

If you have a calculated load of 21000 VA, and a 208V supply, then taking 21000/208 = 101A is a _conservative_ result.

It is only correct if you are supplying 208V line-line loads.

If you are supplying 120V line-neutral loads, then the current would only be 87.5A.

The difference between the two values has to do with the phase angle difference between the supply and the loads. This phase angle difference is the source of the 'Oregon Fudge Factor'.

If you have a mix of 120V and 208V loads then the actual current will be between these two values.

-Jon

 

[Smart $]

Agree with Jon.

Additionally, not mentioned are continuous and noncontinuous loads. But that may not enter the determination if this is a dwelling load. Not that it is impossible, just unlikely for most.

Anyway, to get a true load calculation, you'll have to go through the calculation process tracking the 120V L-N and 208V L-L loads, so at the conversion to ampere stage you can divide them separately, then add the ampere values together.

 

[Dennis Alwon]

Did the OP change the panel size from 100 amps to 125 amps? I have not seen him mention 100 amp panel

 

[Strathead]

Agree with Dennis and want to add. Just be like the typical design Engineer today. Need a 100 amp panel stick in a 200, you aren't paying the bill. Makes for less paperwork. I have seen more panels sized for connected load than not in recent years. My current job as a 4000 amp main and the largest utility transformer available is rated for around 2900 amps

 

[Smart $]

Did the OP change the panel size from 100 amps to 125 amps? I have not seen him mention 100 amp panel

No.

But his post leads one to believe he is questioning whether a 100A panel would be sufficient and compliant. What other purpose would there be for mentioning the Oregon Fudge Factor?

 

[petersonra]

I think a 100 amp panel is likely OK.

I would still probably put in a 125A panel. Not for code reasons, but as a matter of practicality.

 

[kwired]

As has been mentioned if this is all 120 volt loads being supplied and is equally balanced between the two phases current on each phase is only 87.5 amps. Also don't forget current on the neutral is also 87.5 amps and it is considered a current carrying conductor for conductor ampacity calculations and adjustments.

Throw in some 208 volt loads and you will need to calculate those separately and add them to the 120 volt calculations for a total load, and take any unbalance on the 120 volts into consideration as well.

 

[winnie]

Ahh, the different reads when reading between the lines. slice: come back and tell us what you are concerned about!

As I read the OP's question, the calculation suggested that 100A would be just barely acceptable, and that a 125A panel was going to be used...but that the point of the 'oregon fudge factor' was confusing. So I read the question as 'I have calculated that I need about 100A and will use a 125A panel, would the fact that this is 2 legs of a 3 phase supply make this wrong?'

My answer was that the calculation in the OP was conservative and that the proposed installation would be fine.

-Jon

 

[marmorcam]

where those 87.5A come from?

If you have a calculated load of 21000 VA, and a 208V supply, then taking 21000/208 = 101A is a _conservative_ result.

It is only correct if you are supplying 208V line-line loads.

If you are supplying 120V line-neutral loads, then the current would only be 87.5A.

The difference between the two values has to do with the phase angle difference between the supply and the loads. This phase angle difference is the source of the 'Oregon Fudge Factor'.

If you have a mix of 120V and 208V loads then the actual current will be between these two values.

-Jon

 

[winnie]

If you have 21000VA of unity power factor 120V loads, evenly distributed on two legs of a 208/120V supply, then the current on each leg is simply:
21000/2/120 = 87.5

If you have 21000VA of unity power factor 208V loads, placed on two legs of a 208/120V supply, then the current on each leg is:
21000/208 = 101A

The difference between these two answers is because you are looking at two _different_ load arrangements, with different phase angles placed on the supply.

-Jon

 

[petersonra]

I think I would go with a 125A panel as well if they are all 208V loads, just because it gives you some flexibility and future expansion potential. But as another poster suggested, you are well within the 100A limit if you have all 120V loads.

Or you could find some way to get rid of 200VA of calculated load and use the 100A panel as another poster suggested.

Is there any real price difference?

 

[marmorcam]

But you are supposed to have three legs if it is a 120/208v panel?

I would have said that each leg would carry 7000VA (21000/3) and then divide 7000/120=58.33 A or that assumption is just in case that 120 and 208 are mixed? I do not have it very clear!

And I have read before, that it would be the same (if the panel is totally equilibrated) that 21000/(208*sqrt3)=58.3 A

I know that I am missing somo concepts, but I do not know which ones they are!!

Thank you very much for your answer!

If you have 21000VA of unity power factor 120V loads, evenly distributed on two legs of a 208/120V supply, then the current on each leg is simply:
21000/2/120 = 87.5

If you have 21000VA of unity power factor 208V loads, placed on two legs of a 208/120V supply, then the current on each leg is:
21000/208 = 101A

The difference between these two answers is because you are looking at two _different_ load arrangements, with different phase angles placed on the supply.

-Jon

 

[Smart $]

But you are supposed to have three legs if it is a 120/208v panel?

I would have said that each leg would carry 7000VA (21000/3) and then divide 7000/120=58.33 A or that assumption is just in case that 120 and 208 are mixed? I do not have it very clear!

And I have read before, that it would be the same (if the panel is totally equilibrated) that 21000/(208*sqrt3)=58.3 A

I know that I am missing somo concepts, but I do not know which ones they are!!

Thank you very much for your answer!

As stated in the OP, it's a single phase panel. In IEEE parlance, that would be a 120/208V 1Ø 3W system. It is a subsystem of 3Ø system somewhere upstream, and quite likely a 208Y/120V 3Ø 4W system.

The 21kVA load is single phase, connected to two lines, not all three. Upstream where it is a true 3Ø system, it may be balanced out as three (3) equal loads of 21kVA for a total of 63kVA on the system, so 63kVA ÷ 360V = 175A, where 360V=120V×3=208V×sqrt(3). From there, we know the Line current of this balanced 3Ø load is 1.732 (sqrt of 3) times the Phase current (the 1Ø load current)... or 101A per Phase.

As an exercise for you, say on this 120/208V 1Ø 3W system, 10kVA is 120V loads distributed equally to the two lines and 11kVA is 208V load. If all loads are unity or equivalent power factor, what is the line current? This is where you will see that 120V and 208V loads have to be tracked separately to calculate approximate line current.

 

[marmorcam]

...I do not know if I am getting it right. if it is 120/208v 1 phase, 3 wire system I would say that when you distribute 10kVA on the 2 lines equally that means that you have 5kVA at each line. So 5kVA/120=41.67 + 5kVA/120=41.67<br> <br>I think that the 11kVA, 208V will be I=11000/208=52.9 A (single phase, no sqrt!)<br><br>So the current will be 41.67+41.67+52.9=136.24A<br><br>Probably is wrong, but still do not see why I would have to calculate 120v single phase divided by 120*2!! <br><br>

As stated in the OP, it's a single phase panel. In IEEE parlance, that would be a 120/208V 1Ø 3W system. It is a subsystem of 3Ø system somewhere upstream, and quite likely a 208Y/120V 3Ø 4W system.<br>
<br>
The 21kVA load is single phase, connected to two lines, not all three. Upstream where it is a true 3Ø system, it may be balanced out as three (3) equal loads of 21kVA for a total of 63kVA on the system, so 63kVA ÷ 360V = 175A, where 360V=120V×3=208V×sqrt(3). From there, we know the Line current of this balanced 3Ø load is 1.732 (sqrt of 3) times the Phase current (the 1Ø load current)... or 101A per Phase.<br>
<br>
As an exercise for you, say on this 120/208V 1Ø 3W system, 10kVA is 120V loads distributed equally to the two lines and 11kVA is 208V load. If all loads are unity or equivalent power factor, what is the line current? This is where you will see that 120V and 208V loads have to be tracked separately to calculate approximate line current.

<br>
<br>

 

[kwired]

...I do not know if I am getting it right. if it is 120/208v 1 phase, 3 wire system I would say that when you distribute 10kVA on the 2 lines equally that means that you have 5kVA at each line. So 5kVA/120=41.67 + 5kVA/120=41.67<br> <br>I think that the 11kVA, 208V will be I=11000/208=52.9 A (single phase, no sqrt!)<br><br>So the current will be 41.67+41.67+52.9=136.24A<br><br>Probably is wrong, but still do not see why I would have to calculate 120v single phase divided by 120*2!! <br><br><br>
<br>

The 120 degree phase angle is what is confusing you. A true 120/240 single phase system has 180 degree phase angle.

I don't know exactly how to explain it but consider a couple things that are easy to observe.

Take 2- 10 amp 120 volt loads and balance them on a 120/240 single phase system. If you measure current you will find 10 amps on each ungrounded conductor and no current in the neutral. If you disconnect the neutral there is no impact on operation of the circuit, each load still sees 120 volts drop across it.

Apply same loads to two lines and the neutral of a 208/120 Y system. You will measure 10 amps on all three conductors in this case. If you disconnect the neutral in this case you now only have 208 volts across the entire series instead of 240. The voltage across each load will drop to 104. This obviously will change the amount of power being drawn for resistance loads, for inductive loads power may remain same or about same but current will go up in order to maintain power being drawn.

 

[Smart $]

...I do not know if I am getting it right. if it is 120/208v 1 phase, 3 wire system I would say that when you distribute 10kVA on the 2 lines equally that means that you have 5kVA at each line. So 5kVA/120=41.67 + 5kVA/120=41.67<br> <br>I think that the 11kVA, 208V will be I=11000/208=52.9 A (single phase, no sqrt!)<br><br>So the current will be 41.67+41.67+52.9=136.24A<br><br>Probably is wrong, but still do not see why I would have to calculate 120v single phase divided by 120*2!! <br><br><br>
<br>

Using vector addition, which accounts for phase angle relationships, the magnitude calculates at 91.41A on each line... assuming unity or equivalent power factor.

The closest arithmetic method would be 41.67 + 52.92 for an approximated line current of 94.59A

PS: Please do not use html coding.

 

[petersonra]

I can't say I have ever seen a 120/208V single phase panel. Seen a lot of 120/240 Volt single phase panels wired up to 120/208 though.