EV Charger loads in 208Y/120V Panel Schedule

[wwhitney]

Not. It's still one phase (see your own first sentence) with a polarity difference relative to the center tap.

That first sentence was a quote of mine that he was disagreeing with, he just left out the quote tags.

Cheers, Wayne

 

[LarryFine]

That first sentence was a quote of mine that he was disagreeing with, he just left out the quote tags.

Gotcha, but it still works.

 

[Jpflex]

Correct.


No, the 120-vs-180-degree difference manifests itself by having 208v instead of 240v between

Not. It's still one phase (see your own first sentence) with a polarity difference relative to the center tap.

One leg to center common would give you one phase voltage if talking y system (as you say single phase) but I did not know you could call this single phase because the system is 3 phase And your utilizing 1/3 phase

When you say single phase system I think of split single phase transformer with one inductor secondary coil split at center for neutral.

 

[Jpflex]

Since we are on the subject what has been confusing to me is for example if a y system transformer rated 50kva, 460/240? Is mathematically calculated at about 120 amperes at secondary but is this output for ab bc ac line to line phases? Which would add to 120 i x 3 = 360 amperes total?

 

[wwhitney]

Since we are on the subject what has been confusing to me is for example if a y system transformer rated 50kva, 460/240? Is mathematically calculated at about 120 amperes at secondary but is this output for ab bc ac line to line phases? Which would add to 120 i x 3 = 360 amperes total?

I infer from your "about 120A" that you are talking about a transformer with a 240V 3 phase secondary. Let's say it's a 3 wire 240V delta secondary.

Looking at it from kVA, 50 kVA total gives 50/3 kVA for each of the 3 windings AB, BC, CA. And those are 240V windings, so the maximum coil current is 50,000/(3*240)= 69A. Then at that load if you look at one of the legs A, B, or C, the two currents from the two coils are 60 degrees out of phase and add to give a factor of sqrt(3), rather than 2. So the maximum leg current then is 120A (which is what you'd use to size your conductors for the fully loaded case).

However, it's not meaningful with a delta secondary to add up the 3 currents of 120A, that never happens. You could say we have 69A @ 240V times 3 windings, so we can connect up to 208A of 240V 2-wire loads (as long as they can be divided into 3 balanced sets of 69A each)."

But if you switch the secondary to a 240Y/138V system (a bit weird), now you have a secondary neutral. And you can connect 2-wire loads either L-L for 240V or L-N for 138V (again, weird, not likely to have 138V loads, but say you did). So in this case you could say that you can power up to 360A of 138V loads connected L-N (as long as they can be divided into 3 balanced sets of 120A each).

Hope this helps and hasn't make it more complicated.

On the "single phase" vs "3 phase", it comes down to what you mean by "system," I think. Sure if you're looking at the collection of wires at the transformer secondary, and you have a 3 wire delta or a 4 wire wye, that's a 3 phase system. But if you restrict your attention to a single 2 wire load and just consider those 2 wires, that subsystem is just single phase. So you'd call the load single phase, and it doesn't care what the configuration is upstream, just that it gets the voltage it needs.

Cheers, Wayne

 

[LarryFine]

To add, if you take two lines and the neutral from a 3ph 208Y/120 source, you would get 1ph 208/120. You would have two 1ph supplies at 120v to the neutral, but only one 1ph 208v supply between the lines.

A load that uses both lines as well as the neutral, like a range or dryer, can support 120v lights and timers as well as 240v heating elements, although the latter will only operate at 75% of design rating.

If you really wanted to, you could recreate a 3ph supply using two lines and the neutral, like the POCO does for an open-delta supply, something you can not do even with a 1ph 120/240v power source.

 

[Jpflex]

I infer from your "about 120A" that you are talking about a transformer with a 240V 3 phase secondary. Let's say it's a 3 wire 240V delta secondary.

Looking at it from kVA, 50 kVA total gives 50/3 kVA for each of the 3 windings AB, BC, CA. And those are 240V windings, so the maximum coil current is 50,000/(3*240)= 69A. Then at that load if you look at one of the legs A, B, or C, the two currents from the two coils are 60 degrees out of phase and add to give a factor of sqrt(3), rather than 2. So the maximum leg current then is 120A (which is what you'd use to size your conductors for the fully loaded case).

However, it's not meaningful with a delta secondary to add up the 3 currents of 120A, that never happens. You could say we have 69A @ 240V times 3 windings, so we can connect up to 208A of 240V 2-wire loads (as long as they can be divided into 3 balanced sets of 69A each)."

But if you switch the secondary to a 240Y/138V system (a bit weird), now you have a secondary neutral. And you can connect 2-wire loads either L-L for 240V or L-N for 138V (again, weird, not likely to have 138V loads, but say you did). So in this case you could say that you can power up to 360A of 138V loads connected L-N (as long as they can be divided into 3 balanced sets of 120A each).

Hope this helps and hasn't make it more complicated.

On the "single phase" vs "3 phase", it comes down to what you mean by "system," I think. Sure if you're looking at the collection of wires at the transformer secondary, and you have a 3 wire delta or a 4 wire wye, that's a 3 phase system. But if you restrict your attention to a single 2 wire load and just consider those 2 wires, that subsystem is just single phase. So you'd call the load single phase, and it doesn't care what the configuration is upstream, just that it gets the voltage it needs.

Cheers, Wayne

Im going to have to take some time to take this all In as it’s a lot to review

 

[Tulsa Electrician]

Hi all. I recently received this panel schedule to review. I'm having trouble understanding how/why the total connected load came out to 72.7kVA.
For referene, the 20A/1P breaker is a 6A 120V load, and the 60A/2P breaker is a 48A 208V load.



According to my math:
(6A *120V) = 720VA
(48A * 208V) = 9,984 VA

(720VA + 9,984VA) * (6 chargers) = 64.2 kVA

This was an explanation I got:


I would greatly appreciate it if someone can tell me what i'm missing here.

Thank you!

Should be 5 amp *120 volt= 600va
The original should 5 amp

 

[Jpflex]

“ Looking at it from kVA, 50 kVA total gives 50/3 kVA for each of the 3 windings AB, BC, CA. And those are 240V windings, so the maximum coil current is 50,000/(3*240)= 69A”

Here on your quote, the windings you’re mention are 240 volt phase leg to phase leg, therefore would not voltage have to be multiplied by 1.732?

50,000 KVA / 3 = 16,666 VA per phase

Then 16,000 va / 240v x square root of 3 or 1.732 = 40 i each phase?

 

[Tulsa Electrician]

Should be 5 amp *120 volt= 600va
The original should 5 amp

Showed

 

[wwhitney]

“ Looking at it from kVA, 50 kVA total gives 50/3 kVA for each of the 3 windings AB, BC, CA. And those are 240V windings, so the maximum coil current is 50,000/(3*240)= 69A”

Here on your quote, the windings you’re mention are 240 volt phase leg to phase leg, therefore would not voltage have to be multiplied by 1.732?

Knowing when to use sqrt(3) in 3 phase can certainly be hard to remember, and some of my approach is based on minimizing when I have to use it to just a couple cases that I can remember. The most familiar place it pops up is that in a wye system, the L-L voltage is sqrt(3) times the L-N voltage.

In the sentence you quoted above, we already have the L-L voltage as 240V, and we're calculating the current through the coil, not through the legs. So once I divide the 50 kVA rating in 1/3 for the 3 coils, it's now a single phase, 2-wire problem. 50/3 kVA, 240V across the coil, divide to get the current in the coil, no sqrt(3).

In the next step I took two coil currents (say AB and CA) and added them up to get the leg current (on A). Those legs currents are out of phase, so that is a 3-phase type computation, and the sum of the two identical magnitude currents isn't twice the current, but sqrt(3) times the current. That's the other place I have to remember to use the sqrt(3).

The diagram below might help with visualizing things. In particular at that node labeled "A", you know that the sum of the currents (as vectors) has to be 0. And the fact that I_BA and I_AC aren't pointed in the same direction is the reason that their sum (when equal in magnitude) isn't twice their value.

Cheers, Wayne

 

[Jpflex]

Thanks. I recall in one of my books that delta has line voltage equal to phase voltage but in y line voltage is 1.732 x phase voltage.

I’m going to review all this and go back to the book. It is hard to retain all this even after finishing several books on the subject

 

[LarryFine]

Thanks. I recall in one of my books that delta has line voltage equal to phase voltage but in y line voltage is 1.732 x phase voltage. .

And the opposite for the current. That's why Kva works out for either.

 

[Jpflex]

Knowing when to use sqrt(3) in 3 phase can certainly be hard to remember, and some of my approach is based on minimizing when I have to use it to just a couple cases that I can remember. The most familiar place it pops up is that in a wye system, the L-L voltage is sqrt(3) times the L-N voltage.

In the sentence you quoted above, we already have the L-L voltage as 240V, and we're calculating the current through the coil, not through the legs. So once I divide the 50 kVA rating in 1/3 for the 3 coils, it's now a single phase, 2-wire problem. 50/3 kVA, 240V across the coil, divide to get the current in the coil, no sqrt(3).

In the next step I took two coil currents (say AB and CA) and added them up to get the leg current (on A). Those legs currents are out of phase, so that is a 3-phase type computation, and the sum of the two identical magnitude currents isn't twice the current, but sqrt(3) times the current. That's the other place I have to remember to use the sqrt(3).

The diagram below might help with visualizing things. In particular at that node labeled "A", you know that the sum of the currents (as vectors) has to be 0. And the fact that I_BA and I_AC aren't pointed in the same direction is the reason that their sum (when equal in magnitude) isn't twice their value.

Cheers, Wayne

View attachment 2564483

300,000va transformer 480v Secondary 3 phase Assume Y generator

Does this figure look correct for secondary output and would secondary be sized to 120.28 amperes for each phase pairs a to b And b to c and C to a?

Output total 360.85 i

3 phases x 120.28 i ?

 

[wwhitney]

300,000va transformer 480v Secondary 3 phase Assume Y generator

Does this figure look correct for secondary output

If by figure you mean drawing, there's no drawing.

would secondary be sized to 120.28 amperes for each phase pairs a to b And b to c and C to a?

Output total 360.85 i

3 phases x 120.28 i ?

300 kVA / 3 coils = 100 kVA per coil. If each coil is supplying 2-wire 480V loads, that's up to 208A amps per phase pair AB, BC, CA. Each leg A, B, or C would carry up to sqrt(3) of that, or 361A.

So the 361A figure is correct, but it's per leg. There's no 120A figure associated with the fully loaded transformer.

Cheers, Wayne