As Steve wrote, "reasonable" depends on what you are calculating it for.sbyrne said:
I come up with:
Phase A @ 66.4A
Phase B @ 49.6A
Phase C @ 54.3A
View attachment 228
panelboard calculations...
- Thread starter sbyrne
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I come up with:
Phase A @ 66.4A
Phase B @ 49.6A
Phase C @ 54.3A
View attachment 228
I used TurboCAD for the ones posted earlier in this thread. They can be drawn in practically most any vector graphics or CAD application. For instance, the following image depicts a quick and incomplete example using Canvas, a graphics application. However, the better applications for such would feature numerical entry of lengths and angles, and also provide snapping and measurment or dimension functions. CAD applications offer better precision.Zifkwong said:
[edit to add] If you thought it was a program where one just plugs in the numbers and presto, out comes the diagram... well, that's simply not so.
View attachment 230
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wasasparky
Senior Member
A - 49.6
B - 66.4
C - 54.3
B - 66.4
C - 54.3
winnie
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sbyrne said:
Smart$'s approach is the 'correct' way to do this calculation.
However when doing these sort of calculations on a _real_ system, you have to deal with the fact that real measurements always have errors.
The example says a 480V system. But 480V nominal could reasonably be anything from 440V to 520V, possibly more. Depending upon the load characteristics, the actual current to the load will change with the changing input voltage.
The supply voltage need not be perfectly balanced; so you can't assume that the three phase load is drawing perfectly balanced current.
Finally the loads themselves have a certain amount of manufacturing tolerance and variability. We are told that we have a single phase 20A load, but hardware with a 480V 20A nameplate probably draws _about_ 20A, with some variation.
I would say that both Smart$ vector approach, and Steve's 'lumped VA per phase' approach both give numbers that fall well in the center of the range of values that actual measurement would give. The 'lumped VA for a presumed balanced panel' is probably even close enough.
-Jon
P.S. Smart$'s phasor angles for the single phase loads simply presume that the load is unity power factor, but connected _line to line_ rather than _line to neutral_. Draw a balanced 'wye', with lines that start at the origin at 0, 120, and 240 degrees. Now draw the outer triangle, with the tips of the wye forming the vertices of the triangle. What is the angle of the lines that form the edges of the triangle?
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kingpb
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sbyrne said:
Steve has to be correct, because Smart has phase A and B flipped.
According to the OP:
A = 49.3 A
B = 66.1 A
C = 54.0 A
Although if you want to be conservative, then simply add the currents per phase:
A = 51 A
B = 71 A
C = 56 A
winnie said:
You confused me, this is a 1 phase delta, there is no neutral.
The way I understand things, on a 3 phase delta connection, the line currents will be at 0 (A), -120 (B), and -240 (C) degrees (assuming unity PF). The phase currents will be at 30 (A), -90 (B), and -210 (C) degrees. again, assume unity.
So if I wanted to show the LINE currents of a 15 amp single phase load (across A-B) in phasor notation I would draw a phasor with magnitude 15 at angle 0 and a second phasor with magnitude 15 at angle -120. Am I wrong?
Got any other hints?
wasasparky
Senior Member
"So if I wanted to show the LINE currents of a 15 amp single phase load (across A-B) in phasor notation I would draw a phasor with magnitude 15 at angle 0 and a second phasor with magnitude 15 at angle -120. Am I wrong?"
Yes
There is only one current, one phase angle.
Yes
There is only one current, one phase angle.
sbyrne said:
Not quite! The line currents will be the vector sums of the balanced loads and the unbalanced loads, therefore the separation will not be 120 degrees.
The 36A balanced currents constitute a large portion of the line currents, but the unbalanced currents affect the phase angles as well as the magnitudes.
The dashed lines in the phasor diagram represent the line currents.
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wasasparky
Senior Member
sbyrne said:
1. It's not my diagram
2. There were multiple loads in that problem, and calculating current differences..., I thought you were talking if there was just the one load...
Because:
Because:
It is the same current, but its direction has been redefined (reversed), therefore the phase angle has been shifted by 180 deg. Same argument for the 20A single phase current.
Each of these currents flows in two of the three line, therefore they must be added into two line currents.
Because:
sbyrne said:
It is the same current, but its direction has been redefined (reversed), therefore the phase angle has been shifted by 180 deg. Same argument for the 20A single phase current.
Each of these currents flows in two of the three line, therefore they must be added into two line currents.
rattus said:
rattus said:
I understand what you are saying (i think). Where I'm having trouble is, How do I determine the angle on the single phase loads? I understand that my phasors will be 15 amp at "X" degrees and 15 amp at "X + 180" degrees. How do I find "X"?
Thanks for all your help everyone!
sbyrne said:
For resistive loads, all load currents will be in phase with their respective phase voltages. The balanced load currents will split into two components 60 degrees apart. For example,
Ia = 36A@0 = 20.8A@30 + 20.8A@-30
Vac = Vrms@+30
Vab = Vrms@-30
This does not mean that the phase voltages are 60 degrees apart. The sense of one of the phase voltages is reversed for ease of calculation. That is,
Vba = Vrms@-150 = -Vab
One must be extremely careful to correctly define the sense (direction) of the various voltages and currents; otherwise your results will be grossly wrong.
Actually worse than that... I put the 20A load on A-C, rather than B-C, so it was essentially flipped and rotated... but thanks all the same for catching my mistake, kingpb. (HEY, IWIRE BOB... PLEASE NOTE THE PRECEDING STATEMENT; apologies for the jaunt to others) Nevertheless, I still come up with the same amps but in different order and angle:kingpb said:
Phase A = 49.6 @ -351.3?
Phase B = 66.4 @ -117.8?
Phase C = 54.3 @ -250.6?
View attachment 232Phase B = 66.4 @ -117.8?
Phase C = 54.3 @ -250.6?
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Smart $ said:
OK I marked it on the spread sheet I keep on you.
Perhaps the following will help:sbyrne said:
View attachment 233
In pdf format:
View attachment 234
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