# Problem understandin per unit system

The first objective is to find the current going from 2 to 3 in pu. When I am given the line voltage at bus 3 and the complex power of the load: $$V_3 = 400\ kV\require{enclose}%\enclose{phasorangle}{a+b}$$ $$S_3 = (77+j14) MV\!A$$ in the following calculations I will use: $$S_{base}=100~~MV\!A$$ $$V_{base}=400~kV$$ using that the formula for pu power in a three-phase per unit system is: $$S_3=V_3\cdot I_{23}^*$$ $$I_{23}= (\frac{S_3}{V_3})^*$$

$$I_{23} =(0.77-j0.14) pu = 0.78\enclose{phasorangle}{-10.3^\circ}~pu$$ In the above calculation I let $$\V_3\$$ be the reference angle. If I wanted to find the actual current I would need to time the per unit value by base value: $$I_{base} =\frac{S_{base}}{\sqrt{3} \cdot V_{base}}$$ $$I_{base} =144.3\ A$$ Hence: $$\{I_{23}\}_A = \{I_{23}\}_{pu}* \{I_{base}\}_A = 112.9\enclose{phasorangle}{-10.3^\circ}~A$$ Question 1: Is it correct that $$\\{I_{23}\}_A\$$ is the current flowing form bus 2 to bus 3 with $$\V_3\$$ (Line-Line) as reference angle set to 0? (of course the angle could be +120 and -120 also)

Now let's say I would like to first find the actual current then convert to per unit value. Using the line voltage for the red phase as reference (same as before) angle, assume a star load: Line current equals phase current in star connection, for the red phase: $$\{I_{23}\}_A =I_{RN}= (\frac{S_{3\phi}}{3\cdot V_{RN}})^*$$ $$V_3 = V_{RN} \cdot\sqrt{3} \enclose{phasorangle}{30^\circ}\ V$$ $$\{I_{23}\}_A = (\frac{S_{3\phi}}{\sqrt{3}*V_{3}\enclose{phasorangle}{30^\circ}\ V})^*$$ $$\{I_{23}\}_A = 112.96\enclose{phasorangle}{-40.3^\circ}\ A$$ Getting per unit value: $$\{I_{23}\}_{pu} = \frac{\{I_{23}\}_A}{I_{base}}$$ $$\{I_{23}\}_{pu} = 0.78 \enclose{phasorangle}{-40.3^\circ}\ pu$$ Question 2 Why the difference in result (angle) from method 1 and method 2. I get the same magnitude but I get a difference in 30 degrees in the angle, why do I get that difference? What have I done wrong? My thoughts on question 2: I can see that in the second case if I used $$\V_{RN}\$$ as reference angle I would get the same angle, but on the other side in both cases, I used the line voltage for the red phase as the reference angle.

Sorry for my poor English, but thanks for the help in advance:) Literature suggestions to get a good understanding of per unit systems would also be greatly appreciated:D

• $\require{enclose}%\enclose{phasorangle}{-30^\circ}$Just looking quickly, I suspect your current angle should be $\enclose{phasorangle}{-40.3^\circ}$ in the second part, or is it $\enclose{phasorangle}{-30^\circ}$ in the voltage Mar 29, 2020 at 12:39
• ops my bad, thank you for spotting the typo!:) Mar 29, 2020 at 13:22

You must note that the per unit system is not consistent with phase shifts through transformers, or star or delta connected loads.

The actual phase shift is only calculated in the end if it matters.

Simple clear explanation on how to solve per unit problem in power system.

It is important to note that phase angles have no effect when calculating your per unit impedances, they are however important for illustration, as you could be dealing with inductive or capacitive impedances.

• Hi, I hear you, I am not providing a link to promote a YouTube Channel but rather a link to a Tutorial on per unit problem, I am sure anyone interested in how these per unit are solved will want to explore all possible avenues.
Feb 16, 2021 at 13:54
• I understand your submission and it is welcomed.
Then, when you draw a one-phase diagram you draw it for one single phase. So voltages are phase-to-ground voltages; you just multiply them by $$\\sqrt 3\$$ for convenience. In your second example you put line-to-line voltage as reference and introduce 30 degree shift in phase-to-ground voltage. This phase shift just propagated to the current.
So, actually $$V_3=\sqrt 3 V_{RN}$$ is correct transform from voltages in real 3-phase system to voltages in one-phase equivalent.