Load VA

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ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Consulting Electrical Engineer - Photovoltaic Systems
I am assisting a colleague with the design of an EV charger array, and he has a tool that calculates the loading of the panel supplying the chargers. It asks for the VA for each pole of each breaker and I want to make sure that we are feeding it correctly. The supply for the panel is 208V three phase and each charger is a 48A load on two of the phases on a two pole 60A breaker. In his panel app we are entering (208V)(48A) = 9984VA for each breaker pole, but is that correct, or should each pole get half that number?
 
Each pole would get half that value. Each would have the same current value (48 amps), since the current leaving one pole returns via the other pole (i.e., it's the same current).
 
Each pole would get half that value. Each would have the same current value (48 amps), since the current leaving one pole returns via the other pole (i.e., it's the same current).
Thanks, Charlie!
 
Unless your panelboard is protected by a 100% rated breaker, you need to put in a total of 208V * 60A for each EVSE (half of that to each breaker pole the EVSE is connected to, as Charlie stated). Because while the actual current will only be 48A, as a continuous load the OCPD for the panel needs the 125% factor (unless it is 100% rated). And since the OCPD for the panel has to protect the busbars, the busbars need the 125%.

How many EVSEs? If it's a multiple of 3, you can use the fact that (3) 60A 2-wire loads in a delta configuration require line currents of 60A * sqrt(3) = 103.9A. Skipping VA calculations altogether (if the panel supplies only EVSEs).

Cheers, Wayne
 
Unless your panelboard is protected by a 100% rated breaker, you need to put in a total of 208V * 60A for each EVSE (half of that to each breaker pole the EVSE is connected to, as Charlie stated). Because while the actual current will only be 48A, as a continuous load the OCPD for the panel needs the 125% factor (unless it is 100% rated). And since the OCPD for the panel has to protect the busbars, the busbars need the 125%.

How many EVSEs? If it's a multiple of 3, you can use the fact that (3) 60A 2-wire loads in a delta configuration require line currents of 60A * sqrt(3) = 103.9A. Skipping VA calculations altogether (if the panel supplies only EVSEs).

Cheers, Wayne
He tried to get the customer to get the customer to settle for three chargers, but he was adamant about having four.
 
He tried to get the customer to get the customer to settle for three chargers, but he was adamant about having four.
Is this the same project as we discussed here: https://forums.mikeholt.com/threads...zing-calculations.2576420/page-3#post-2865853 ?

For 4 EVSEs as in the OP, and nothing else, the minimum panel/OCPD/conductor size is 60A * sqrt(7) = 159A.

Apparently the VA method is non-conservative here, as for it I get 60A * (120V * sqrt(3) / 120V) * (1/2) * 3 EVSEs on worst case line = 156A.

Cheers, Wayne
 
Each pole would get half that value. Each would have the same current value (48 amps), since the current leaving one pole returns via the other pole (i.e., it's the same current).
If its the same voltage and same current why half?
 
Unless your panelboard is protected by a 100% rated breaker, you need to put in a total of 208V * 60A for each EVSE (half of that to each breaker pole the EVSE is connected to, as Charlie stated). Because while the actual current will only be 48A, as a continuous load the OCPD for the panel needs the 125% factor (unless it is 100% rated). And since the OCPD for the panel has to protect the busbars, the busbars need the 125%.

How many EVSEs? If it's a multiple of 3, you can use the fact that (3) 60A 2-wire loads in a delta configuration require line currents of 60A * sqrt(3) = 103.9A. Skipping VA calculations altogether (if the panel supplies only EVSEs).

Cheers, Wayne
The OP commented that the it needed the VA for each pole of each breaker, so they should be taking 48A * 208V * (1/2). That would give them the VA for each pole.

Why are you using 60A for your calculations? The 60A breaker is covering the 125% of the continuous load. to size the main breaker, it would be 125% of the continuous load which would be (48A*208V*4chargers) / (208V*sqrt(3)) *125%... unless someone sees an issue with that.
 
Why are you using 60A for your calculations?
The final answer will be proportional both to the 48A individual load figure and the 125% continuous load factor. I just chose to multiply them together at the beginning, then the 125% factor is done with. If you prefer to do your calculations with 48A as the input, then you'll need to multiply by 125% at the end. Same difference.

The 60A breaker is covering the 125% of the continuous load. to size the main breaker, it would be 125% of the continuous load which would be (48A*208V*4chargers) / (208V*sqrt(3)) *125%... unless someone sees an issue with that.
Yes, the issue is that the load is not balanced, but the formula you used only works for balanced loads. These are 2-wire 208V loads, not 3-wire 3 phase loads.

Cheers, Wayne
 
Yes, the issue is that the load is not balanced, but the formula you used only works for balanced loads. These are 2-wire 208V loads, not 3-wire 3 phase loads.

Cheers, Wayne
Correct. My calculation took the 208V load first. Then compared that load to 3ph 208V and got the overall amps. The load would be balance out with 4 equal 2pole breakers stacked.
 
with 4 chargers you would have an unbalanced load. One phase would have 20kVA and the other 2 would have 10kVA each.

You also could balance it out a little better with 15kVA on 2 of the phases and 1 phase would have 10kVA.
 
with 4 chargers you would have an unbalanced load. One phase would have 20kVA and the other 2 would have 10kVA each.

You also could balance it out a little better with 15kVA on 2 of the phases and 1 phase would have 10kVA.

Iab = (2)(48.0A) = 96.0A
Ibc = (1)(48.0A) = 48.0A
Ica = (1)(48.0A) = 48.0A

Ia = sqrt(Iab^2 + Ica^2 + (Iab)(Ica)) = sqrt(9216A^2 + 2304A^2 + 4608A^2) = 127A = 26.4 kVA @208V
Ib = sqrt(Ibc^2 + Iab^2 + (Ibc)(Iab)) = sqrt(2304A^2 + 9216A^2 + 4608A^2) = 127A = 26.4 kVA @208V
Ic = sqrt(Ica^2 + Ibc^2 + (Ica)(Ibc)) = sqrt(2304A^2 + 2304A^2 + 2304A^2) = 83.1A = 17.3 kVA @208V
 
Ibc = (1)(48.0A) = 48.0A
. . .
Ia = sqrt(Iab^2 + Ica^2 + (Iab)(Ica)) = sqrt(9216A^2 + 2304A^2 + 4608A^2) = 127A
Note that 127 / 48 = sqrt(7) as per post #6. And for OCPD/wire sizing you'll need a 125% factor on your results, as EVSEs are continuous loads.

127A = 26.4 kVA @208V
Since these are the line currents, you'd multiply by 120V to get the VA. I.e. (127, 127, 83) A * 120V = (15.24, 15.24, 9.98) kVA, or 40.46 kVA total. Which makes sense as you have (4) EVSEs, each at 48A * 208V = 9.98 kVA.

Cheers, Wayne
 
... for OCPD/wire sizing you'll need a 125% factor on your results, as EVSEs are continuous loads.


Since these are the line currents, you'd multiply by 120V to get the VA. I.e. (127, 127, 83) A * 120V = (15.24, 15.24, 9.98) kVA, or 40.46 kVA total. Which makes sense as you have (4) EVSEs, each at 48A * 208V = 9.98 kVA.

Cheers, Wayne
Yes, I see that now but it's a non issue since the customer decided to size the feeders for the FLA of the 480V to 208V transformer. The individual conductors are indeed sized for 60A for the 48A chargers.
 
Iab = (2)(48.0A) = 96.0A
Ibc = (1)(48.0A) = 48.0A
Ica = (1)(48.0A) = 48.0A

Ia = sqrt(Iab^2 + Ica^2 + (Iab)(Ica)) = sqrt(9216A^2 + 2304A^2 + 4608A^2) = 127A = 26.4 kVA @208V
Ib = sqrt(Ibc^2 + Iab^2 + (Ibc)(Iab)) = sqrt(2304A^2 + 9216A^2 + 4608A^2) = 127A = 26.4 kVA @208V
Ic = sqrt(Ica^2 + Ibc^2 + (Ica)(Ibc)) = sqrt(2304A^2 + 2304A^2 + 2304A^2) = 83.1A = 17.3 kVA @208V
As wwhitney pointed out, the currents are correct but the kVAs are not.
 
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