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]

Just to clarify and for simplicity of the exercise:
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?

 

[jghrist]

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.

You add Iab + Ibc + Ica to get In. You should add Ia + Ib + Ic. Since you have no Ø-n loads, you will always get zero for In.

 

[jghrist]

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?

The angle of the PP-2,4 current would be -150, not 30.
ØA current would be 18.25A.
The burn2x spreadsheet does not include any Ø-n loads.

 

[burnt2x]

I made changes to the worksheet I posted earlier to include Ø-n loading. Please see if I got it right.

The results are:
Ia = 18.25A @ -175.28 degrees (wrt to V_{AB{/sub])
Ib = 20.78A @ 0.00 degrees,
Ic = 0, and
In = 3.00A @ 120.00 degrees
(same as the values posted by jghrist)}

 

[bjenks]

I think we are now all on the same path now, thank you so much. Burn2x I see some things in your spreadsheet that don't make sense to me:
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.

 

[jghrist]

burn2x,
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.

 

[bjenks]

everyone, I think we got it now. Thank you for your patients on helping me go over stuff I learned in college 25 years ago (although I sometimes wondered if I ever really did)... These software program are making it so we don't know if they are giving us the corrent information. This program I use was incorrect and I assumed it was correct.

 

[bjenks]

This a side notice, but without the single phase load the line current was 20.8A on A & B. Then you add an additional 3A on phase A and now your line current went down to 18.22 and 20.77A respectively. From a simple electric meter standpoint it looks like you are using less power even though you just added a load...

 

[bjenks]

Found another error on the spreadsheet. Phase C had Ib instead of Ic in the formula. Please look this over and see if you see an error. Notice that adding a load to phase-N causes the current to go down.

 

[waross]

Some comments on the power factor phase angles of the various loads may explain the reduced currents.
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

 

[jghrist]

These software program are making it so we don't know if they are giving us the corrent information. This program I use was incorrect and I assumed it was correct.

If the program uses the simplified NEC kVA method, then it may give different answers than a rigorous analysis, but still not be wrong. As Rafiq stated in an earlier post, the results will be conservative which is not bad.

 

[bjenks]

I started from scratch on the spreadsheet as I found two other errors. 1) I believe you have to replace PF angle from the actual angle. The orginal was just the opposite. 2) The Current nodes were in the wrong direction compared to the Voltage.

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.

 

[burnt2x]

bjenks,
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.

 

[bjenks]

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.

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.