Per Unit Analysis Confusion

Discussion in 'Homework Help' started by Xtremeth, Feb 9, 2014.

  1. Xtremeth

    Thread Starter New Member

    Feb 9, 2014
    I have this problem I'm trying to figure out, doing the problem normally versus using per unit gives me different answers!

    So I have a three-phase motor that is rated for 20MVA at 13.8kV and it is connected to a bus at 1<0 degrees pu. The system base Sb=100MVA and Vb=13.8kV.

    To find the PU current:

    I = S / (sqrt(3)*V) = (20MVA/100MVA) / (sqrt(3)*(13.8kV/13.8kV)) = 0.1155 pu

    Another way of doing it is to find the actual current:

    I = S / (sqrt(3)*V) = 20MVA /(sqrt(3)*13.8kV) = 836.739 A
    The base current = S / (sqrt(3)*V) = 100MVA / (sqrt(3)*13.8kV) = 4183.7 A
    The PU current using this approach is 0.2 pu

    Comparing the two answers (0.1155 pu) and (0.2 pu, there is a factor of sqrt(3) somewhere...

    Can anyone tell me which is right and why the two different approaches yield different answers? Was my assumptions wrong?

    Thank you
  2. WBahn


    Mar 31, 2012
    I'm not familiar with "per unit analysis", having never taken a power systems course, so I'm not following everything you are saying. What do you mean by 1<0 degrees pu?

    Are the currents in the two methods both line currents? Or is one phase current?

    If you could spend a bit of time briefly describing what "system base" means and what "per unit analysis" is, I can probably take a decent shot at helping you. Plus, there's a good chance you'll spot your problem as you try to explain it to me.
  3. Hector Berlioz

    New Member

    Feb 9, 2014
    The per-unit system is a method of expressing quantities in an electrical system (e.g. voltage, current, impedance, etc) as a proportion of pre-defined base quantities. By definition, the per-unit value of a quantity is the ratio of the original quantity to its base value.
  4. Xtremeth

    Thread Starter New Member

    Feb 9, 2014
    Per unit is a lot to explain with my words, I'm not very good at explaining. But the wikipedia page on Per Unit will explain it.

    Anyways, it's a method of analyzing currents and voltages in a power system with transformers. It simplifies the hassle of carrying the turns ratio around through each step of solving the problem.
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    The first answer [0.115pu] is wrong.

    While it's reasonable to use either the 3-phase or per phase values I think it's simpler to use the former. Then you don't have to introduce the √3 factor at all. Not that this is the issue with your solution. The first equation as you apply the method, is incorrect.

    If you think about this particular problem it should be clear that for the same voltage for the supply system and the load, the per-unit load current will equate directly to the ratio of the load MVA and the system base MVA.
  6. badgerbun

    New Member

    Mar 30, 2015
    TNK is correct!
    Even though this is over a year old , more people will have the same confusion, yes the first answer in wrong, when you do conversions with only power MVA and voltage KV , the {sqrt(3)} cancels each other out so is ignored
    the correct way to do the first answer would be
    (20MVA/100MVA) = 0.2 pu
    since the base voltage is the same.