Amp Calculations based on unbalanced loads

[bjenks]

Lets say you have a three phase 208/120V supply and you calculate the kVA on phase A, B, C as following: 7.42, 4.29, 2.62. Now you need to determine the actual current per phase so you can get your wire sizes (don't worry about 1.25% of cont load).

How would you calculate the current per phase?

I get VA/(120*cos 30) = 71.39A for A, 41.28A for B, 25.21A for C.

I am being told I am wrong as the answer is 62.7A, 37.2A, 21.8A. They say it is because for 1 phase loads on a three phase system you have to multiply by cos 30 degree phase shift. What am I missing?

 

[bjenks]

I see that my formula was wrong... I was copying from an e-mail of mine and messed up. Still got the wrong answer with my formula.

Another example is on a 208/120 single phase panel with A phase at 2.52kVA and B phase at 2.16kVA. I am told it is 23.8A and 20.8A respectively, but I get 21A and 18A respectively.

 

[waross]

2520VA / 120V = 21 A
2160VA / 120 V = 18 A
If you are trying to combine the currents from 120V single phase loads with the currents from 208V single phase loads, it gets real messy real quick.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[bjenks]

I am just trying to fill out a panel schedule that has kVA on each circuit. Then at the bottom, I want to have a total kVA for each phase and Current for each phase. So just assuming all the loads are kw per NEC. The current calculation would change based on the panel being single phase or three phase and being delta/wye.

I am using a third party software that does the panel schedule and it doesn't match what I think it should be. They just say it changes because of 30 deg phase shift... What you would put for the first option on the three phase panel?

 

[waross]

Single phase loads may be line to neutral connected or line to line connected. Line to neutral loads are straightforward, line to line loads are not.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[bjenks]

For NEC there isn't any power factors and kva = kw. So I don't think we need to be doing symetrical components. I keep thinking all I need to do is add all the A phase kva together for one column, then B phase for another column.
Then take the A column and divide by the line voltage.

 

[burnt2x]

What Bill said regarding line-to-neutral loading is very true - straightforward!

But, assuming your initial data were "phase loads" (connected line-to-line), Load AB =7.42kVA; Load BC =4.29kVA; and Load CA =2.62kVA:
phase AB amps= 7420/208 = 35.67A,
phase BC amps= 4290/208 = 20.63A, and
phase CA amps= 2620/208 = 12.6A.

The resulting line amps will be:
Line A = 41.80A;
Line B = 48.76A, and
Line C = 28.77A
And that is, if I have read your problem right.

 

[bjenks]

burnt2x, can you explain how you calculated the Line A at 41.80A. I must be having a brain freeze as this should be simple to understand and I am missing something.

Based on the article below:
on a Delta system Ia=VA*SQR(3)/Vab (VA is L-L)
on a Wye system Ia=VA/Va (VA is L-N)

I am only interested in line current since that is the conductor I am sizing for per NEC. These are just NEC 220 loads on a panel schedule being totalled for each lug then the line current is being calculated for each lug to properly size the conductor.

http://eece.ksu.edu/~starret/581/3phase.html#11

 

[rbulsara]

Follow what waross (Bill) said and NEC. Simple arithmetic sum per NEC will result in slightly conservative results, which is always good. You also need to size all three conductors the same, so highest of line current applies.

Loads in delta or wye does not matter much. See
thread238-261481, if you want get some more exercise.

Rafiq Bulsara

http://www.srengineersct.com

 

[burnt2x]

Bjenks,
Please see attached worksheet. Hope this helps.

 

[rockman7892]

burnt 2X

When I divide the L-L KVA's amoungst phases I get the same answers that you did in your first post. But when I add the individual phase currents I get the same answers you come up with in your worksheet which are different from your fist post. Why are the answers when you add these currents vectorally slightly different then when you divide the kVA amoungst phases and then calculate currents?

Obviousy if you just divide currents amoung phases first and then add you will come up with a higher number then the two methods above.

If the OP's kVA's were actual phase L-N kVA's then he'd have:

Ia= 61.83
Ib= 35.75
Ic= 21.83

 

[waross]

I think that we are between a rock and a soft place.
The soft place is an everyday calculation for a panel feeder selection.
First, all currents are assumed to be equal to the highest current. (The conductors will be the same size and must be adequate for the maximum phase load.)
The Phase current is multiplied by 1.73 to find line current. The resulting figure will be safely conservative. Line to neutral currents may be added to this figure directly.
The Rock. If a rigorous solution is required, phase angles of the loads must be considered. The power factor and phase angle of each load current must be determined. The currents must then be added vectorially.
From this result you may select feeder sizes, based on the greatest phase current.



Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[burnt2x]

Rockman,

The first post was done by hand using a calculator with limited digits especially when calculating sine/cosine values. The worksheet(computer calculation) is set up to a higher precision causing a bit of discrepancies when arriving at the final values (round-off errors).

FWIW, yes, the final values differ by about 1 amp or so;

41.8 amps against 43.4 amps in the worksheet results,
48.76 amps against 49.37 and
28.77 versus 29.07 amps in the worksheet.

 

[waross]

As an exercise in vector math, great. In the real world, it is unusual to find all he loads on such a panel to be running at unity power factor. You have ignored power factor and the resulting errors may be much greater than the rounding errors.


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[waross]

As a real world real example, I measured the load on a 120:240 Volt single phase panel. I found 7 amps on one leg and 9 amps on the other leg.
What is the expected current on the neutral?
I was surprised by the actual neutral current.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[burnt2x]

Bill,
Please look at the worksheet that was attached. Just punch-in the correct PF in the cell provided and the phase/line currents will be correct. Since no PF was mentioned, I put UPF in each VA load.

 

[waross]

Sorry, I stand corrected. For the sake of browsers of this thread, it may be well to have mentioned the effect of power factor.
In my single phase example, instead of the textbook answer of 2 amps, I got a neutral current of about 6 amps.
One load was a loaded motor with a good power factor, the other load was an unloaded motor with a very bad power factor.
It was a dramatic illustration of the importance of considering the power factor in current additions.
Respectfully
Bill

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[rockman7892]

Burnt2x

Is it just conincidence then that I came up with the same answers as you did in your first post by just dividing the total L-L kVA between phases and calculating line current with L-G voltages?

For instance for AB load of 7.42 I didvied this in half and put 3.71kVA on phase A and for the CA load of 2.62kVA I divided this in half and put 1.31kVA on phase A. Phase A then has a total load of 5.02kVA and dividing this by 120V I come up with 41.83A. Is this just a coincidence that this comes out to the same answer you got by vectorally adding with not that many signifigant digits in your first calc? Obviously when adding vectorally I get different answers that match those in your worksheet.

If you had a three phase load, would you divide the kVA across each of the three phases?

Is dividing the kVA amoungst phases the simple way for performing calcs necessary for the NEC, as opposed to just adding currents from loads on each phase?

 

[bjenks]

Burnx2, thank you for the spreadsheet and now I understand. I am in no position to say this based on my education level, but I think you had an error on "c" and it should be changed to what is below. You got the correct mag, but the phase angle was wrong.

At "c":
Ic = Ibc - Ica real imag
Ibc -10.320 -17.875
Ica -6.303 10.917
Ic -4.017 -28.792
Ic 29.07

I liked your spreadsheet so much that I combined it with another spreadsheet I created for neutral calculations. See if you see a mistake on it as I added a vector graph.

 

[burnt2x]

rockman,

If it were physically feasible to spread-out the loads, i.e. 7.42kVA is a group of single phase loads, why not? I just took the data as it was presented. Other than that, I don't have a definite answer.

By the looks of it, the OP should have tried balancing the loads to avoid the complexity and that's how it's normally done. I believe that the loads were group loads being supplied by a 3-panel panel remote from each group loads, as in 3 buildings supplied single-phase from a 3-phase panel.