Using complex numbers to model 3 phase AC

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Discussion Overview

The discussion centers on the use of complex numbers and phasor diagrams to model three-phase AC voltages, specifically focusing on the relationship between line-to-neutral (L-N) and line-to-line (L-L) voltages in a Y-connected transformer. Participants explore theoretical calculations and practical considerations regarding voltage magnitudes and phase angles.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a calculation involving phasors to derive the L-L voltage from L-N voltages, suggesting that the result should be approximately 400V and questioning the role of √3 in this context.
  • Another participant clarifies that the specified 230V is RMS and discusses the amplitude of the signal, indicating that the signal can be expressed in terms of sine functions.
  • There is a discussion about the exactness of using √3 in calculations, with one participant noting that in ideal balanced conditions, √3 is exact, while in practice, phases may not be perfectly balanced.
  • A participant provides a calculation for L-L voltage based on phasor diagrams, arriving at a magnitude of approximately 398.4V and relating it to the cosine of the angles involved.
  • Another participant describes the phase relationships among the three lines, indicating that when one line is at zero volts, the others are at ±60°, leading to a specific calculation for the voltage between two lines.
  • There is a challenge regarding the addition of phasors, with one participant questioning the treatment of negative signs in vector addition and whether they should be subtracted or not.
  • A later reply emphasizes that phasors represent voltages from the origin and that the line-to-line voltage should be considered as a voltage difference rather than a simple sum.

Areas of Agreement / Disagreement

Participants express differing views on the treatment of phasor addition and the implications of phase angles in calculations. There is no consensus on the exactness of using √3 in practical scenarios, as some acknowledge real-world imbalances while others focus on ideal conditions.

Contextual Notes

Limitations include the assumption of balanced phases in some calculations, the dependence on specific definitions of voltage (RMS vs. peak), and unresolved mathematical steps in the derivation of L-L voltage from L-N voltage.

Guineafowl
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TL;DR
How to derive the voltages seen in a three phase, Y connected transformer.
Assume a transformer as above, with 230V L-N, and I want to work out the L-L voltage. A phasor diagram will show me that the voltages are 120° out of phase.

(230∠0°) + (230∠120°) = (230cos0 + j230sin0) + (230cos120 + j230sin120) = 230 + (-115 + j199.2)

115 + j199.2 = 230∠60

What I’m looking for is, of course, 400V. I know √3 is involved, but how, and is that an exact calculation or a useful rule of thumb?

This is not homework.
 
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The specified 230 Vac is RMS, so the signal amplitude is 230 * √2 * Sin( ).
 
Guineafowl said:
Summary:: How to derive the voltages seen in a three phase, Y connected transformer.

is that an exact calculation or a useful rule of thumb?
How exact do you need to be? In reality, the three phases are never exactly balanced.

Edit: But in the ideal balanced case, ##\sqrt{3}## is exact.
 
Last edited:
anorlunda said:
How exact do you need to be? In reality, the three phases are never exactly balanced.
It’s not important, just a theoretical question really.I think I have it worked out. Because, on the phasor diagram, the L-N voltages oscillate either side of the origin (0V), the angle between each phasor and the nearest is actually 60°.

So:
VL-L = (230∠0°) + (230∠60°) = (230) + (230cos60 + j230sin60)

= (230) + (115 + j199.2) = 345 + j199.2

|VL-L| = 398.4 V (∠30°)

EDIT: And because the magnitude of L-L voltage depends on cos 30°, and that of L-N depends on cos 60°, the ratio of voltages (400:230) is similar to the ratio of cos 30:cos 60, which is 1.73, or ≈√3.
 
Last edited:
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When L1 is at zero volts, L2 will be at +60°, and L3 will be at –60°.
The L2 to L3 voltage will then be 2 * Sin( 60° ) = 2 * √(3/4) = √3.
 
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According to vectorial diagram:
 

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Babadag said:
According to vectorial diagram:
Thanks - I see you’ve ignored the (-) sign when adding the vectors. I assumed that if the voltage magnitudes are of opposite sign they subtract. Is that not correct?
 
Guineafowl said:
I see you’ve ignored the (-) sign when adding the vectors.
Phasors point from the origin, neutral or ground.
The Line to Line voltage is therefore not the sum, it is the voltage difference.
 
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