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Apparent power, true power, reactive power and power factor

  1. Oct 16, 2015 #1
    I calculated the apparent power, true power, reactive power and power factor by using:

    S=VI
    S=17684.5 at an angle of 135

    In rectangular, that's = -1261.83+j1261.83


    This is probably a stupid question, but I have

    S= 17684.5
    P= - 1261.83
    Q=1261.83

    power factor = -0.707

    In the marking instructions, it has the real power (P) as 1261.83 instead of what I've got - NEGATIVE 1261.83

    Also, the power factor in the marking instructions is 0.707, not negative 0.707. When I calculate the power factor by simply doing cos(135, I get a negative.

    Do I just not include the negatives or what, I don't understand.
     
  2. jcsd
  3. Oct 16, 2015 #2

    anorlunda

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    Gold Member

    135 relative to what?

    You get to choose the direction of +I which determines the direction of +P. If you don't like the sign, choose the other direction.
     
  4. Oct 16, 2015 #3
    So my answers are correct, it can be - or +?
     
  5. Oct 16, 2015 #4

    donpacino

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    Gold Member

    it also depends on whether you are talking about power supplied or power used.

    If you are referring to a load, saying it has -700W used is wrong (in most cases). Make sure it is correct in the context of the question asked.
     
  6. Oct 16, 2015 #5
    In my opinion if 135 degrees is the angle between V and I then it is wrong because maximum possible it is only

    90 degrees-positive or negative. Conventionally, if the current lags the voltage -inductive load-then the angle

    positive and if it is capacitive it is negative. In any case cosines is positive but sinus could be positive or negative

    So P has to be positive always. Q may be negative.

    It seems to me it is not the right voltage [it is not the same phase as the current].
     
  7. Oct 16, 2015 #6
    The load could be:

    pure resistive cosfi=1 sinfi=0 fi=0 degrees

    pure inductive: cosfi=0 sinfi=1 fi=+90 degrees

    pure capacitive cosfi=0 sinfi=-1 fi=-90 degrees

    If it would be mixed load then the angle could be more than -90 and less than +90
     
  8. Oct 22, 2015 #7
    It depends on the reference direction you chose.If you considered a load,and the reference direction of current is entering towards positive reference terminal of load voltage you will not get an angle more than 90 degree, and the active power(absorbed) is always positive.However for a power source,angle more than 90 is possible and you will get negative active power in that case.This means source is supplying power.

    So if your example refers to a load, you chose the current reference direction away from positive reference direction of load voltage and hence negative active power.Because in this case VICos(phi) referes to active power supplied, which is negative for a load.
    If your example is of a power source, it possible for the source to absorb or provide active power.In that case it depends on how you chose current direction.If you chose current direction away from positive terminal and still you got negative active power, it means source is absorbing power.
    I
     
  9. Oct 22, 2015 #8
    If it is about a generator and V it is the output voltage, I agree with it. According to the sign convention the active power P is negative-if the generator supply the power, as usual. Of course, in a fault case, the generator turns into a motor and the power-maintaining the same sign rule-could be positive.
    However, in my opinion, the angle between I and V cannot be more than 90 degrees-or less than -90.
     
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