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  • I having trouble understanding PU system especially "briding" the gap from PU system -> to Actual system
  • To illustrate this here is an example where I am given the one-line diagram for the three phases system: enter image description here

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:

enter image description here

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

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  • \$\begingroup\$ \$\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 \$\endgroup\$
    – skvery
    Mar 29 '20 at 12:39
  • \$\begingroup\$ ops my bad, thank you for spotting the typo!:) \$\endgroup\$ Mar 29 '20 at 13:22
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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.

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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.

The image and link below are from my YouTube channel:

enter image description here https://www.youtube.com/watch?v=ydtmzQhv--Q

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  • \$\begingroup\$ Chad - Hi, (a) Link-only answers are not usually allowed here. Please see here where it says: "Links are not a substitute for including information in your answer itself". (b) You are linking to your own YouTube channel so you must disclose that affiliation, which you didn't do here. Unless there is a clear benefit for linking to (i.e. promoting) your own off-site resource, it's better not to do that as your answer might get treated as spam. For these reasons, I recommend you rewrite or delete this answer. Thanks! \$\endgroup\$
    – SamGibson
    Feb 16 at 13:50
  • \$\begingroup\$ 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. \$\endgroup\$
    – Chad
    Feb 16 at 13:54
  • \$\begingroup\$ I understand your submission and it is welcomed. \$\endgroup\$
    – Chad
    Feb 16 at 13:59
  • \$\begingroup\$ @SamGibson Will you then kindly remove the down vote since you have edited my post already? \$\endgroup\$
    – Chad
    Feb 16 at 14:09
  • \$\begingroup\$ @Chad - Hi, "I am not providing a link to promote a YouTube Channel" Whether you believe you are promoting your YouTube channel or not, you are linking to your own channel and that means you need to disclose the affiliation, in accordance with the site rule I linked above. I have added that disclosure into the answer for you now. (The downvote isn't from me.) \$\endgroup\$
    – SamGibson
    Feb 16 at 14:11
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First I would notice that you got difference in phase angles and this has no relation to the per unit system as angles are expressed in degrees/radians regardless of using per units or actual values.

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.

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