240v-delta / 480v-delta
- Thread starter Mike01
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Mike01 said:In a 240∆/120V-3?, 4W system the voltage ?B-N (high leg) measurers 208V. How is this calculated? And what would happen if you connected a 480∆-3? and provided a center tap ground? What would your voltage be on that system?
It can be calculated by using triginometry. You have a right triangle whose hypotenuese "C" is 240V and the short leg "A" is 120V. One forumla for the long leg "B" is SQRT(C^2 - A^2) = SQRT((240*240)-(120*120)) = 208V.
There are other methods, but the end result is a factor of 1.73 x L-N (yes the same value we use for WYE systems). So a 480V high leg would yield B-N = 416V.
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Mike01 said:
The formula is for a right triangle so it always works. But in a 240/120 or the 480/240 center tapped delta system, from the OP, the short leg "A" is equal to 1/2 the long leg "C", which is not the case in your question.
For a center tapped 208V delta: C=208V, A=104V, B=179.92
When voltages are represented as vectors (or phasors as some prefer to call 'em), i.e. where length equates to magnitude and direction correlates with phase angle (the point in the reference cycle where [positive] peak voltage of secondary waveforms occurs), one can use geometry and trigonometry to determine voltage and phase angle from any two connection points.Mike01 said:
For the answer to your question regarding a wye configuration, refer to the following diagram...
View attachment 1097
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How so? ...given VAN as the reference voltage.rattus said:
Also see "Positive and negative angles". Using negative angles is actually more correct in phase angle description than is positive angles, since B and C phases do lag A phase.
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On positive and negative angles I emphasize the latter part of the following passage:rattus said:
The given line in my diagrams are the ones noted to be at 0?.
The corner designation on traditional delta diagram are clockwise, left to right A, B, then C... exactly how I have it. Adherence to the convention of 0? always to the "east" while keeping traditional delta corner orientation is impossible.
Why, I ask?
Why, I ask?
Smart, this phasor stuff is difficult enough without redefining the reference axis, and I see no reason to ever do so. I can't see any justification in Wikipedia to do so either. Just rotate your diagrams such that east is east and west is west; then they would be perfect.
My argument is that you should orient the arrows correctly, connect the arrows properly, and your diagram is complete.
Edited to add a reference:
Smart here is an excerpt from my AC Circuits text.
…….is shown a phasor V that makes an angle theta with the reference axis (directed horizontally to the right)……………
Nuf sed.
Why, I ask?
Smart, this phasor stuff is difficult enough without redefining the reference axis, and I see no reason to ever do so. I can't see any justification in Wikipedia to do so either. Just rotate your diagrams such that east is east and west is west; then they would be perfect.
My argument is that you should orient the arrows correctly, connect the arrows properly, and your diagram is complete.
Edited to add a reference:
Smart here is an excerpt from my AC Circuits text.
…….is shown a phasor V that makes an angle theta with the reference axis (directed horizontally to the right)……………
Nuf sed.
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