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Given the following three-phase system:

schematic

simulate this circuit – Schematic created using CircuitLab

Suppose to set 2 wattmeters as follows:

  1. \$P_{21}\$ which measures current \$I_2\$ and voltage \$V_{21}\$
  2. \$P_{23}\$ which measures current \$I_2\$ and voltage \$V_{23}\$

Show that $$P_{21} + P_{23} = 3 P_2 = 3V_{20}*I_2$$ holds true.

Here is the proof with the necessary assumptions:

$$V_{21}*I_2+V_{23}*I_2 = (V_{20}-V_{10})*I_2 + (V_{20} - V_{30})*I_2 = 2V_{20}*I_2 -(V_{10}+V_{30})*I_2$$

Now, assuming that $$V_{10}+V_{20}+V_{30} = 0$$ (from here named assumption 1)

The above equation yelds $$P_{21} + P_{23} = 3V_{20}*I_2 = 3P_2 $$

My question is: what assumptions on the three phase system should hold true in order for the above assumption 1 and, consequently, the proof, to be true? For instance, should it be symmetrical? Should it be balanced? Or both?

My reasoning is that since assumption 1 should hold true, the system should be symmetrical since the vector sum of phase voltages in a symmetrical system is always equal to zero (check the triangle of voltages below where in my case N=o and E1 = V10 and so on...)

enter image description here

I guess the real question boils down to this: since the o point is the centroid of the triangle of line voltage phasors, and the phase voltages is a half median, is there a property of a generic triangle that states that the sum of half of each median (the half closer to the median's angle) is always zero for each triangle?

I hope I made my question clear. If not please let me know in the comments, I'll try to explain better.

Note 1: * is used to express the dot product.

Note 2: All the voltages and currents are phasors.

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  • \$\begingroup\$ The proof does require the system to be balanced. The geometric proof for the phasor diagram is something worth looking into. \$\endgroup\$ – alphasierra May 3 '17 at 19:57
  • \$\begingroup\$ @alphasierra, to be clear, my understanding of "balanced system" means that each phase CURRENT should be equal in magnitude and shifted of 2pi/3 with respect to the other. In a "symmetrical system" each phase VOLTAGE should be equal in magnitude and shifted of 2pi/3 with respect to the other. A symmetrical system isn't necessarily balanced and viceversa. Assumption 1, which is key in this proof, does not rely on any phase current, therefore I can not understand how the system must be balanced. If it were symmetrical then assumption 1 is surely true, but is line voltage symmetry truly needed? \$\endgroup\$ – mickkk May 3 '17 at 20:19
  • \$\begingroup\$ As I use the terminology it would refer to the amplitudes being the same, and the angles all being 120 degrees (2pi/3 radians). I guess I used the term "balanced" incorrectly as that is supposed to also mean the loads are identical, but as you already pointed out the relationship is independent of current. \$\endgroup\$ – alphasierra May 3 '17 at 20:25
  • \$\begingroup\$ @alphasierra exactly! Same amplitude and angle shifted by 120 degrees. The only difference is that "balanced" (in the terminology of my books) refers to currents only being as such while "symmtrical" refers to line voltages only. Sorry I did not mean to be pedantic :) but I did not want to confuse the two either. \$\endgroup\$ – mickkk May 3 '17 at 20:40
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I found the answer! Since for a generic triangle it can be shown that the sum of E1, E2 and E3 is equal to zero (check here for the proof), assumption 1 is always true and therefore no assumptions are needed on the three phase system in order for the above proof to hold true.

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