# 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 – skvery Mar 29 at 12:39
• ops my bad, thank you for spotting the typo!:) – FluidProblem Mar 29 at 13:22