EV Charging Transformer Sizing Calculations

[wwhitney]

Is correct?

Conservative but not tight. For identical 2-wire loads of X amps, delta connected arranged as balanced as possible, then the magnitude of the line currents (A,B,C) will be:

(1) (X,X,0)
(2) (X, sqrt(3) X, X)
(3) (sqrt(3) X, sqrt(3) X, sqrt(3) X)
(4) (sqrt(4+sqrt(3)) X, sqrt(4 + sqrt(3))X, sqrt(3) X)
(5) (sqrt(4+sqrt(3)) X, 2 sqrt(3) X, sqrt(4+sqrt(3)) X)
(6) (2sqrt(3) X, 2 sqrt(3) X, 2 sqrt(3) X)
(7) (sqrt(12+sqrt(3))X, sqrt(12+sqrt(3))X, 2 sqrt(3) X)
. . .

So if you're using, say 40A EVSEs (which need 50A branch circuits), 4 of them would give a required transformer rating of 40 * sqrt(4 + sqrt(3)) * 208V * sqrt(3) = 34,500 VA. Versus 40 * 208 * 3 = 24,960 VA for 2 or 3 EVSEs, and twice that for 5 or 6 EVSEs.

Cheers, Wayne

 

[ggunn]

Conservative but not tight. For identical 2-wire loads of X amps, delta connected arranged as balanced as possible, then the magnitude of the line currents (A,B,C) will be:

(4) (sqrt(4+sqrt(3)) X, sqrt(4 + sqrt(3))X, sqrt(3) X)

. . .

So if you're using, say 40A EVSEs (which need 50A branch circuits), 4 of them would give a required transformer rating of 40 * sqrt(4 + sqrt(3)) * 208V * sqrt(3) = 34,500 VA. Versus 40 * 208 * 3 = 24,960 VA for 2 or 3 EVSEs, and twice that for 5 or 6 EVSEs.

Cheers, Wayne

I know how to calculate the unbalanced currents, but why wouldn't I want to size my transformer so that all three coils can accommodate the higher phase current since all three coils are identical?

 

[winnie]

In a related story...

I have a client who wants me to design an array of (4) charging stations which need to be fed by a 480V to 208V transformer. [...] I am assuming that for sizing the transformer I will need to size it as if there were six chargers. Is correct?

Essentially yes. If you do an exact vector analysis I think you might get a bit less than 6x single load kVA, but 6x is IMHO the correct approximation to use.

The Scott-T creating 2 phase would let you use 4x single load to size the transformer, but I'd bet that 4x custom, non-standard transformer would be more expensive than a 6x standard transformer.

Jon

 

[winnie]

I know how to calculate the unbalanced currents, but why wouldn't I want to size my transformer so that all three coils can accommodate the higher phase current since all three coils are identical?

That is what Wayne did.

He calculated the higher current seen on the more heavily loaded 2 legs, and applied that number to all three legs.

My assessment in post 43 was wrong.

Not doing the proper vector math, but approximating by simply assigning kVA by leg: 1 leg sees half kVA of each of 2 chargers. 2 legs see half of kVA of each of 3 chargers. In this approximation you assume the highest load on any one leg applies to all 3.

So to support 4 chargers you need a transformer rated 9/2 the kVA of a single charger, because each leg has to support 3/2 of a charger.

Jon

 

[wwhitney]

I know how to calculate the unbalanced currents, but why wouldn't I want to size my transformer so that all three coils can accommodate the higher phase current since all three coils are identical?

You do. That's how I came up with my example kVAs, using the highest of the 3 line currents (which equal the coil currents for a wye secondary; delta secondaries still confuse me).

And if you look at my example, you'll see that for (3, 4, 5, 6) EVSEs, the required transformer size works out to approximately (25, 35, 50, 50) kVA.

In other words 3n loads is most efficient, for small n the efficiency hit is not too bad at 3n+1, and then if you want to go to 3n+2 you get the same size required as 3n+3.

Cheers, Wayne

 

[Joethemechanic]

With all this EV stuff maybe we should bring back two phase.
Four phases 90 degrees apart.

That would make a good charging station or would it.
Sorry off topic

I always thought that would be a way to minimise some of the harmonics with all the single phase rectification going on. Although a delta-wye also shifts phases around.

 

[garbo]

With all this EV stuff maybe we should bring back two phase.
Four phases 90 degrees apart.

That would make a good charging station or would it.
Sorry off topic

Two phase was exactly what it says two phases180 degrees apart. Has been over 30 years since I connected a two phase motor up. In my area two phase service was 5 wires 240 volts. The center tap of each of the two phases were connected together and grounded to form a neutral ( now called grounded conductor ) to supply 120 volts. We had 2300 volt three wire two phase feed an old building that had all two phase equipment. Three phase is cheaper & more confident then any 4 wire two phase systems. When I purchased my last 30 amp 4 fuse ( 2 phase ) safety switch over 45 years ago it cost more then double the cost of a three phase 3 fuse safety switch and only one of the three supply houses that I dealt with had one in stock.

 

[ggunn]

You do. That's how I came up with my example kVAs, using the highest of the 3 line currents (which equal the coil currents for a wye secondary; delta secondaries still confuse me).

And if you look at my example, you'll see that for (3, 4, 5, 6) EVSEs, the required transformer size works out to approximately (25, 35, 50, 50) kVA.

In other words 3n loads is most efficient, for small n the efficiency hit is not too bad at 3n+1, and then if you want to go to 3n+2 you get the same size required as 3n+3.

Cheers, Wayne

OK; in my situation there are (4) 48A chargers; at 48A of current draw per charger, and sizing the transformer assuming (6) chargers for balance I calculate that I would need a minimum of 34.6kVA.

Ia = Ib = Ic = (96A)(sqrt3) = 166.3A @ 208V
(208V)(166.3A) = 34.6kVA

But a question: For continuous use considerations I need to use (48A)(1.25) = 60A OCPD and wiring for the chargers; do I need to do the same for the transformer?

 

[winnie]

OK; in my situation there are (4) 48A chargers; at 48A of current draw per charger, and sizing the transformer assuming (6) chargers for balance I calculate that I would need a minimum of 34.6kVA.

Ia = Ib = Ic = (96A)(sqrt3) = 166.3A @ 208V
(208V)(166.3A) = 34.6kVA

But a question: For continuous use considerations I need to use (48A)(1.25) = 60A OCPD and wiring for the chargers; do I need to do the same for the transformer?

My understanding is that transformers are rated for use at 100% of their nominal rating (presuming proper cooling, etc.)

But I think you should look back at Wayne's calculation, because I think you have 2 errors going on which will leave you with an undersized transformer.

Start with 3 of your 48A chargers in a delta configuration, leading to:
48A*sqrt(3) = 83.1A
Now you add a 4th charger, connected to 2 of the 3 legs. Ignoring the phase difference (which would _reduce_ the total), then on the heavily loaded legs you have:
83.1+48 = 131.1A

IMHO sizing the transformer for 131A per phase makes sense, not the 166A needed for 6 chargers.

But if you have a 131A 208V three phase transformer, that gives 131*208*sqrt(3) = 47kVA

Double check me! I've just started a new job and might be smoking something by accident.

-Jon

 

[wwhitney]

(1) (X,X,0)
(2) (X, sqrt(3) X, X)
(3) (sqrt(3) X, sqrt(3) X, sqrt(3) X)
(4) (sqrt(4+sqrt(3)) X, sqrt(4 + sqrt(3))X, sqrt(3) X)
(5) (sqrt(4+sqrt(3)) X, 2 sqrt(3) X, sqrt(4+sqrt(3)) X)
(6) (2sqrt(3) X, 2 sqrt(3) X, 2 sqrt(3) X)
(7) (sqrt(12+sqrt(3))X, sqrt(12+sqrt(3))X, 2 sqrt(3) X)
. . .

So if you're using, say 40A EVSEs (which need 50A branch circuits), 4 of them would give a required transformer rating of 40 * sqrt(4 + sqrt(3)) * 208V * sqrt(3) = 34,500 VA. Versus 40 * 208 * 3 = 24,960 VA for 2 or 3 EVSEs, and twice that for 5 or 6 EVSEs.

Arithmetic error applying the law of cosines for vector addition. That should be :

(1) (X,X,0)
(2) (X, sqrt(3) X, X)
(3) (sqrt(3) X, sqrt(3) X, sqrt(3) X)
(4) (sqrt(7) X, sqrt(7) X, sqrt(3) X)
(5) (sqrt(7) X, 2 sqrt(3) X, sqrt(7) X)
(6) (2 sqrt(3) X, 2 sqrt(3) X, 2 sqrt(3) X)
(7) (sqrt(19) X, sqrt(19) X, 2 sqrt(3) X)

And the example with 40A EVSEs is then 40 * sqrt(7) * 208 * sqrt(3) = 38,130 VA

Cheers, Wayne

 

[ggunn]

My understanding is that transformers are rated for use at 100% of their nominal rating (presuming proper cooling, etc.)

But I think you should look back at Wayne's calculation, because I think you have 2 errors going on which will leave you with an undersized transformer.

Start with 3 of your 48A chargers in a delta configuration, leading to:
48A*sqrt(3) = 83.1A
Now you add a 4th charger, connected to 2 of the 3 legs. Ignoring the phase difference (which would _reduce_ the total), then on the heavily loaded legs you have:
83.1+48 = 131.1A

IMHO sizing the transformer for 131A per phase makes sense, not the 166A needed for 6 chargers.

But if you have a 131A 208V three phase transformer, that gives 131*208*sqrt(3) = 47kVA

Double check me! I've just started a new job and might be smoking something by accident.

-Jon

I did find an error in my math. When I did the calcs correctly for the four chargers, using Iab = Ica = 48A and Ibc = 96A, I found that Ic = Ib = 130A and Ia = 83.1A. Sizing the transformer so that none of the coils is overloaded would be (130A)(208V) = 27.04kVA

 

[wwhitney]

OK; in my situation there are (4) 48A chargers; at 48A of current draw per charger, and sizing the transformer assuming (6) chargers for balance I calculate that I would need a minimum of 34.6kVA.

For 6 EVSEs, you can just compute the minimum transformer size as 48*208*6 = 59,904VA. Which is the load for either single phase or 3 phase, since you have a multiple of 3 and can make a balanced 3 phase load.

Ia = Ib = Ic = (96A)(sqrt3) = 166.3A @ 208V
(208V)(166.3A) = 34.6kVA

That last calculation needs a sqrt(3) because it's a 3 phase transformer; then you would get the same 59.9 kVA I got.

But for 4 EVSEs, the maximum line current is sqrt(7)*48A = 127A. So your transformer needs to be 127A*208V*sqrt(3) = 45,750 VA. Since the standard sizes are 45 kVA and 75 kVA, looks like you'll need a 75 kVA anyway, which would easily support 6. Unless you can justify the 2% overload on the basis that the probability is low and the transformer could handle it anyway.

But a question: For continuous use considerations I need to use (48A)(1.25) = 60A OCPD and wiring for the chargers; do I need to do the same for the transformer?

Yes, you'd use 125% OCPD and conductors on both the primary and secondary side, because the load is continuous. But you don't need a 125% factor on the transformer size itself, since I believe transformer ratings are already continuous. And I understand this discrepancy to be one of the reasons that you are allowed to oversize the transformer primary protection at 125%.

Cheers, Wayne

 

[wwhitney]

I did find an error in my math. When I did the calcs correctly for the four chargers, using Iab = Ica = 48A and Ibc = 96A, I found that Ic = Ib = 130A and Ia = 83.1A.

I got Ic = Ib = 127A after correcting my vector math error.

Sizing the transformer so that none of the coils is overloaded would be (130A)(208V) = 27.04kVA

Still missing the sqrt(3) factor. As a check, the transformer can't possibly be smaller than 4 * 208 * 48 = 39,936 kVA, as that's the power (for power factor 1) actually delivered to the loads.

Cheers, Wayne

 

[ggunn]

I did find an error in my math. When I did the calcs correctly for the four chargers, using Iab = Ica = 48A and Ibc = 96A, I found that Ic = Ib = 130A and Ia = 83.1A. Sizing the transformer so that none of the coils is overloaded would be (130A)(208V) = 27.04kVA

I did find an error in my math. When I did the calcs correctly for the four chargers, using Iab = Ica = 48A and Ibc = 96A, I found that Ic = Ib = 127A and Ia = 83.1A. Sizing the transformer so that none of the coils is overloaded would be (127A)(208V) = 26.42kVA

Ib = sqrt(Ibc^2 + Iab^2 + (Iab)(Ibc))
= sqrt((48A)^2 + (96A)^2 + (48A)(96A))
= sqrt(2304A^2 + 9216A^2 + 4608A^2))
= sqrt(16128A^2) = 127A

 

[ggunn]

I got Ic = Ib = 127A after correcting my vector math error.

We agree on that after I corrected mine.

Still missing the sqrt(3) factor. As a check, the transformer can't possibly be smaller than 4 * 208 * 48 = 39,936 kVA, as that's the power (for power factor 1) actually delivered to the loads.

Hmmm, cogitating on that... I am obviously missing something.

 

[ggunn]

We agree on that after I corrected mine.

Hmmm, cogitating on that... I am obviously missing something.

OK; you are correct, of course; sorry it took me so long to get it through my thick skull.

P = (V)(I)(sqrt3) = (127A)(208V)(sqrt3) = 45.75kVA

 

[petersonra]

That's not correct, 625.41 requires that branch circuit and feeder OCPD be sized with a 125% factor. And I would say an EVSE is a continuous load, so the usual 125% factors required by 210.19/210.20, 215.2/215.3, and 230.42 would also apply.

Cheers, Wayne

I do not think these factors apply to transformer sizing though.

 

[ggunn]

OK; you are correct, of course; sorry it took me so long to get it through my thick skull.

P = (V)(I)(sqrt3) = (127A)(208V)(sqrt3) = 45.75kVA

To recap, there is an existing 225A 480/277V panel that has plenty of unused capacity which feeds a 15kVA transformer via a 3P-20A breaker which in turn feeds a 50A 208/120V panel that has only (4) 20A single pole breakers in it. I do not know how heavily loaded the 1P-20A breakers are. If all four of the EV chargers are running at maximum capacity, which is possible, they will draw (see above) 45.75kVA, which is about 127A on two of the three phases.

If the chargers were all the loading on a 480V to 208V transformer powering the EV chargers, a 50kVA transformer would do it (full load current = 138.79A), but they want to combine the existing 208/120V loads with the car chargers with an upsized transformer and 208/120V panel. The existing four 1P-20A breakers, if fully loaded, would add 9.6kVA to the 45.75kVA EV charger loading for a total of 55.35kVA. Seeing as this sort of design is a bit out of my wheelhouse and I don't know the loading on the existing breakers, in order to err on the side of caution I have designed it with a 75kVA transformer - the next standard size up from 50kVA.

My client is asking me if, in order to save money and space in the tiny electrical room, a 50kVA transformer will suffice. Table 450.3(B) says the maximum OCPD is 125% of full load current; does that mean that I can safely pull 125% of full load current (138.79A X 1.25 = 173.49A) through a 50kVA transformer?

 

[wwhitney]

Table 450.3(B) says the maximum OCPD is 125% of full load current; does that mean that I can safely pull 125% of full load current (138.79A X 1.25 = 173.49A) through a 50kVA transformer?

Code-wise, the calculated load still needs to be no more than 50 kVA. The 125% factor on the primary OCPD is for (a) inrush current and (b) supplying continuous loads via a non-100% rated breaker, since the transformer rating is continuous.

I do not know how heavily loaded the 1P-20A breakers are.

That's going to be critical to determining whether you can use a 50 kVA transformer, so you'll need to do a load calc for the (4) 20A 120V branch circuits. Because treating them each as 9600 VA turns out to just barely exceed the 50 kVA capacity.

If all four of the EV chargers are running at maximum capacity, which is possible, they will draw (see above) 45.75kVA, which is about 127A on two of the three phases.

They will draw 127A on two of the 3 lines, but the 3rd line will only see 48A * sqrt(3) = 83A, so the load won't be a full 45.75 kVA.

A 50 kVA transformer has an allowable line current of 139A. That means you have 56A of capacity on line 3, but only 12A of capacity on lines 1 and 2 (actually we can sharpen that a bit if we assume the 120V load is power factor 1, see post to follow). So once you have load calcs for the (4) 120V branch circuits, you can see if you can allocate them among the lines not to exceed the remaining capacity of (12, 12, 56).

Cheers, Wayne

 

[winnie]

Is a 50 kVA 480:208/120V transformer a stock item with your supplier? Standard sizes are 45 and 75 kVA.

-Jon