Ok, so I had a problem about a three-phase system with the following data



$$\cos \phi=0.85$$

where V is the phase-phase voltage, S is apparent power, and \$\phi \$ is positive (inductive load).

Ok I was asked to find out the phase-neutral voltages, the phase-phase voltages and the currents.

I had no trouble doing it so I just post the outline of my work.

Phase-neutral voltages

$$V_a = \frac{220}{\sqrt{3}} e^{j0}$$ $$V_b = \frac{220}{\sqrt{3}} e^{-j120^o}$$ $$V_c = \frac{220}{\sqrt{3}} e^{j120^o}$$

Phase-phase voltages

$$V_{ab} = V_a-V_b= 220 e^{j30^o}$$ $$V_{bc} = V_b-V_c= 220 e^{-j90^o}$$ $$V_{ca} = V_c-V_a= 220 e^{j150^o}$$


$$I_{a} = \frac{S}{\sqrt{3}V} e^{-j31.7^o}= 393.6 e^{-j31.7^o}$$ $$I_{b} = 393.6 e^{-j151.7^o}$$ $$I_{c} = 393.6 e^{j88.3^o}$$

And the impedance for each load (a,b and c) is $$Z= \frac{V_a}{I_a}= 322.7 e^{j31.7^o}$$

Now they say for me to imagine that the system becomes unbalanced with b becoming 1.1Z and c becoming 0.9Z

$$I_n= I_a+ I_b + I_c = \frac{V_a}{Z} + \frac{V_b}{1.1Z} + \frac{V_c}{0.9Z} = 68.5 e^{j61.8^o}$$

Now I was asked to redo this using the pu system with given base values


$$S_{base}=100 MVA $$

This lead me to

$$V_{phase-neutral-base}=220/\sqrt{3} kV$$ $$I_{base}= \frac{S_{base}}{\sqrt{3}V_{phase-phase-base}}=262.4A$$ $$Z_{base}= \frac{V_{phase-phase-base}}{\sqrt{3}I_{base}}=484 \Omega$$

For phase-neutral voltages I have to divide my previous value by \$ V_{phase-neutral-base}\$ leading to

$$V_a = 1 e^{j0}$$ $$V_b = 1 e^{-j120^o}$$ $$V_c = 1 e^{j120^o}$$

As for phase-phase voltages, I divide by \$V_{phase-phase-base}\$ leading me to

$$V_{ab} = 1 e^{j30^o}$$ $$V_{bc} = 1 e^{-j90^o}$$ $$V_{ca} = 1 e^{j150^o}$$

Now my question is: why if I apply the formulas $$V_{ab} = V_a-V_b$$ $$V_{bc} = V_b-V_c$$ $$V_{ca} = V_c-V_a$$ using pu values the equations don't verify. Is it because I use different base values for each and this is just a question of proportionality?

However, then I continued my problem and obtained the rest of the values:


$$I_{a} = \frac{S}{V} e^{-j31.7^o}= 1,5 e^{-j31.7^o}$$ $$I_{b} = 1.5 e^{-j151.7^o}$$ $$I_{c} = 1.5 e^{j88.3^o}$$

Everything looks fine here, because if I multiply the currents by \$I_{base}\$ I obtain the values I did before.

$$Z= \frac{V_a}{I_a}= \frac{1 e^{j0}}{1,5 e^{-j31.7^o}} = 0.667 e^{j31.7^o}$$

Then again, multiplying by \$Z_{base}\$ we obtain the value before.


$$I_n= I_a+ I_b + I_c = \frac{V_a}{Z} + \frac{V_b}{1.1Z} + \frac{V_c}{0.9Z} = \frac{1 e^{j0}}{0.667 e^{j31.7^o}} + \frac{1 e^{-j120^o}}{1.1 \times 0.667 e^{j31.7^o}} + \frac{1 e^{j120^o}}{0.9 \times 0.667 e^{j31.7^o}} = 0.261 e^{j61.8^o}$$

And multiplying by \$I_{base}\$ I obtain the same value as before.

So in this last calculations using the pu values works pretty well and I obtain values equivalent to the calculations I did before. The only calculation that's failing is the one about the voltages phase-phase. Why does this happen? What is the mathematical subtlety I am missing here?



Your work is good. You are using 2 different voltage bases to get your results (pasted in below). That is why they are different in per unit.

enter image description here

If you take the results of your calculation using the phase-neutral voltages (Va-Vb) and convert them to the ph-ph voltage base you used, then they will agree. In other words, the results of Va - Vb is 1.73 per unit in magnitude. That is on a voltage base of 220kV/1.732. To convert that per unit result to the ph-ph base you used above, multiply by (220kV/1.732 divided by 220kV).

The ph-ph voltage base is root 3 larger than the phase-neutral. When working problems - do not mix. Just use one voltage base and carry on. That way, when you convert back to actual volts you won't make a mistake.

UPDATED: Showing base calculations.

enter image description here

  • 1
    \$\begingroup\$ Hello! Thank you for your answer. Ok, I think I get it: because, even though some are phase-phase and others are phase-neutral, they are still voltages, and therefore, the same base must be used; however in my other calculations, I'm computing currents and impedances, and therefore, adding components in phase-phase base will produce me coherent results, since the current base and the impedance base were derived from the postulated voltage phase-phase base. \$\endgroup\$ – Granger Obliviate Mar 1 at 1:53
  • 1
    \$\begingroup\$ Granger - i updated my answer to show base calculations. When you pick a ph-ph voltage base you automatically set the ph-neutral base because they are related by root 3. So, if you calculate the current and impedance base quantities with either approach you will get the same results. \$\endgroup\$ – relayman357 Mar 1 at 15:59

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