Amp Calculations based on unbalanced loads
Amp Calculations based on unbalanced loads
(OP)
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?
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?






RE: Amp Calculations based on unbalanced loads
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.
RE: Amp Calculations based on unbalanced loads
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
RE: Amp Calculations based on unbalanced loads
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?
RE: Amp Calculations based on unbalanced loads
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Amp Calculations based on unbalanced loads
Then take the A column and divide by the line voltage.
RE: Amp Calculations based on unbalanced loads
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.
RE: Amp Calculations based on unbalanced loads
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
RE: Amp Calculations based on unbalanced loads
Loads in delta or wye does not matter much. See
thread238-261481: 3 Phase Power Monitor, if you want get some more exercise.
Rafiq Bulsara
http://www.srengineersct.com
RE: Amp Calculations based on unbalanced loads
Please see attached worksheet. Hope this helps.
RE: Amp Calculations based on unbalanced loads
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
RE: Amp Calculations based on unbalanced loads
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
RE: Amp Calculations based on unbalanced loads
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.
RE: Amp Calculations based on unbalanced loads
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Amp Calculations based on unbalanced loads
What is the expected current on the neutral?
I was surprised by the actual neutral current.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Amp Calculations based on unbalanced loads
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.
RE: Amp Calculations based on unbalanced loads
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
RE: Amp Calculations based on unbalanced loads
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?
RE: Amp Calculations based on unbalanced loads
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.
RE: Amp Calculations based on unbalanced loads
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.
RE: Amp Calculations based on unbalanced loads
This is a 208/120 V single phase panel (PP). Circuit PP-1 on phase A is a receptacle rated at 360VA. Circuit PP-2, PP-4 on phase A-B is two L-L loads that total 4320VA, but the panel schedule shows on Circuit PP-2 2160VA and Circuit PP-4 2160VA (4320*.5).
They then calculate the line current on phase A as:
The load on PP-1 is 3A phase-to-neutral (360/120). The load on PP-2,4 is 20.8A phase-to-phase (4320/208). These two vectors are 30 degrees apart. We'll assume phase-to-neutral is in the positive X direction.
The PP-1 vector is (3, 0).
The PP-2,4 vector is (20.8A cos 30, 20.8A sin 30), or (18.01, 10.4).
Add those together and you get (21.01, 10.4).
The current is the magnitude of the vector. sqrt(21.01^2+10.4^2) = 23.44.
I used burn2x spreadsheet and came up with 17.77A at 180deg.
Do you think they did it right for line A Current?
RE: Amp Calculations based on unbalanced loads
RE: Amp Calculations based on unbalanced loads
ØA current would be 18.25A.
The burn2x spreadsheet does not include any Ø-n loads.
RE: Amp Calculations based on unbalanced loads
The results are:
Ia = 18.25A @ -175.28 degrees (wrt to V[sub]AB{/sub])
Ib = 20.78A @ 0.00 degrees,
Ic = 0, and
In = 3.00A @ 120.00 degrees
(same as the values posted by jghrist)
RE: Amp Calculations based on unbalanced loads
1) I think you are referencing to Vab, what confuses me is that I think I normally see Van as the reference.
2) I was under the impression that Vab = Va*sqr(3)with a positive 30 degree phase shift. Ia=Iab*sqr(3) with a negative 30 degree phase shift. You seem to be saying just the opposite in your first column. So Van should be -30 and not 30 as you show. If we were putting Va at 0 then I think it would be right. I believe this needs to be corrected in your spreadsheet or reference Va at 0.
I think you keep getting the right answer as others, but your phase angles are off to me... I updated your spreadsheet to make what I believe the correct phase angles.
RE: Amp Calculations based on unbalanced loads
The phase relationship between Ø-Ø and Ø-n is incorrect. If Vab is 0°, then Van would be -30°.
bjenks,
Any of the voltages, Ø-Ø or Ø-n, can be the reference and have a 0° angle.
RE: Amp Calculations based on unbalanced loads
RE: Amp Calculations based on unbalanced loads
RE: Amp Calculations based on unbalanced loads
RE: Amp Calculations based on unbalanced loads
Generally, adding a phase to neutral load will not reduce the line current of line to line loads.
If you are mixing inductive and capacitive loads then addition of a load may cause the current to drop, regardless of the connection, line to line or line to neutral.
You may have calculated a "C" phase load as an addition to the "A" phase current or you may have made a sign error or an angle error.
Or, you may be adding a capacitor to an inductor.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Amp Calculations based on unbalanced loads
RE: Amp Calculations based on unbalanced loads
After fixing that, I was able to update the actual current and now have the same answer as the software program and it now makes sense.
real imag
Iab 18.00 10.39
Ica 0.00 0.00
Ian 3.00 0.00
Ia 21.00 10.39
Mag Angle(˚)
Ia 23.43 26.33
real imag
Ibc 0.00 0.00
Iab 18.00 10.39
Ibn 0.00 0.00
Ib -18.00 -10.39
Mag Angle(˚)
Ib 20.78 -150.00
I shouldn't need to do any updates from here as I tried all types of different situations and they all lined up with what logic would suggest. Here is the last spreadsheet update.
RE: Amp Calculations based on unbalanced loads
Please try punching-in PF's other than lagging loads (negative PF) and you won't get calculations done because you set PF data validation to "between" 0.000001 and 1. Suggest you reset PF data input validation to "less than" 1.000000009 or so to get the logic right.
I guess this is one reason why we didn't get the correct vector for currents since we missed to consider PF sign wrt lagging/leading PFs.
Please see attached file.
BTW, thanks for getting the exercise closure. It's been a long time since I get to do some calculations because we relied too much on softwares.
RE: Amp Calculations based on unbalanced loads
You are correct, I should have added that to make this complete. If I remember correctly, Leading PF would add to the original angle and Lagging would subtract. With that in mind I added this feature to the spreadsheet.